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http://www.pagalguy.com/discussions/official-quant-thread-for-cat-2011-part-8-25073031?page=52 | 1,469,660,333,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827080.38/warc/CC-MAIN-20160723071027-00038-ip-10-185-27-174.ec2.internal.warc.gz | 632,170,326 | 18,906 | Last updated
Author
Pankaj Kamani @pkaman
Tags
Official Quant Thread for CAT 2011 [Part 8]
Order by:
Page 52 of 915
If we add the square of the digit in the tens place of the positive two digit number to the product of the digits of that number we get 52,and if we add the square of the digit in the unit's place to the same product of the digits we get 117.Find the two digit number??
i got till this step
x^2+xy=52
x(x+y)=52--(1)
y^2+xy=117
y(y+x)=117--(2)
after this hw to proceed the given answer is 30..pls explain in details with steps.
10x+y is the number
x*y+x^2 =52
x*y+y^2 =117
y^2-x^2 = 65
y+x = 13
y-x= 5
aading 2y = 18 y= 9
x= 4
so value is 49
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Ani1308 Says
hi..a little help in remainder part...find the rem of (3^5^7^9)/41
5^7^9 = 5k
so it is 4k+1
3*3*3*3*3 /41 remainder 38
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jain4444 Says
The sum of digits of a natural number (10^n 1) is 4707, where n is a natural number. The value of n is ?
The sum of digits of a natural number (10^n 1) is 4707, where n is a natural number. The value of n is ?
10^n-1 is obviously multiple of 9 and 11
so 4707/9 =523 (sum will be 9*n where n is no. by which 9 is repeated)
so value is 9999999999999999............523 times
P.S.:- jain sahab itna easy kaise de sakte hai
Commenting on this post has been disabled.
hi sry man...i dont have the answers bt i have one more question on the seme pattern...if u liketo practice dn..
5^25^125^3125 /11..
thanks all...
The remainder is 1.
E(11) = 10
25^anything has unit digit 5
5^(10K+5) % 11 = 5^5 % 11 = 25*25*5 = 45 = 1
Smoothest seas do not make tough sailors
Commenting on this post has been disabled.
hi sry man...i dont have the answers bt i have one more question on the seme pattern...if u liketo practice dn..
5^25^125^3125 /11..
thanks all...
Have fun in whatever you do, otherwise whats the point !!
Commenting on this post has been disabled.
E(41) = 40
We need to find 5^7^9 % 40
E(40) = 16
7^9%16 = 7
5^(16K+7) % 40 = 5 * 5^(16K+6)%8
Now, E(8 ) = 4
Therefore => (5^16K + 6) % 8 = 1
=> 5^(16K+7) % 40 = 5
3^(40K + 5) % 41 = (3*81) % 41 = -3 = 38
The remainder is 38
Ani1308 Says
but dude u cant apply euler in bold part...5 and 40 are nt co prime
I have not applied Euler in that part...I just calculated the remainder of 7^9 with 16....Then I represented 7^9 as (16K + 7), then I segregated common multiple from numerator and denominator. After that I applied Euler of 8....
What is the O.A.??
Smoothest seas do not make tough sailors
Commenting on this post has been disabled.
Ani1308 Says
hi..a little help in remainder part...find the rem of (3^5^7^9)/41
(3^5^7^9)/41
3 and 41 are coprimes.
=> E(41) = 40
5^7^9/40 => 40 = 5*8
=> 5^7^9/8 => E(8 ) = 4
=> 7^9/4 = 3^9/4 => -1 or 3 remainder
=> 5^3/8 = 125/8 => 5 remainder
5^7^9 => 5x = 8y+5 => 5 remainder
3^5/41 = 3*81/41 = 120/41 = 38 remainder
Sorry for x blunder.:-(
No problem.
• 1 Like
Commenting on this post has been disabled.
E(41) = 40
We need to find 5^7^9 % 40
E(40) = 16
7^9%16 = 7
5^(16K+7) % 40 = 5 * 5^(16K+6)%8
Now, E(8 ) = 4
Therefore => (5^16K + 6) % 8 = 1
=> 5^(16K+7) % 40 = 5
3^(40K + 5) % 41 = (3*81) % 41 = -3 = 38
The remainder is 38
but dude u cant apply euler in bold part...5 and 40 are nt co prime
Have fun in whatever you do, otherwise whats the point !!
Commenting on this post has been disabled.
Ani1308 Says
hi..a little help in remainder part...find the rem of (3^5^7^9)/41
E(41) = 40
We need to find 5^7^9 % 40
E(40) = 16
7^9%16 = 7
5^(16K+7) % 40 = 5 * 5^(16K+6)%8
Now, E(8 ) = 4
Therefore => (5^16K + 6) % 8 = 1
=> 5^(16K+7) % 40 = 5
3^(40K + 5) % 41 = (3*81) % 41 = -3 = 38
The remainder is 38
Smoothest seas do not make tough sailors
Commenting on this post has been disabled.
Ani1308 Says
hi..a little help in remainder part...find the rem of (3^5^7^9)/41
i am getting 38 ... oa?
C(41)=40
checking for 5^7^9 mod 40
C(40) = 4
7^9 mod 4= 3
so... 5^3 mod 40=5
and 3^5 mod 41 = 38
2012 :Joined NMIMS. Left after an year due to unfortunate circumstances. Target MBA 2016-18 :D
Commenting on this post has been disabled. | 1,595 | 4,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2016-30 | latest | en | 0.83433 |
https://gateoverflow.in/85624/arrives-at-office-at-10am-regularly-arrives-at-11-am-every-day | 1,511,245,512,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806317.75/warc/CC-MAIN-20171121055145-20171121075145-00078.warc.gz | 607,927,249 | 17,924 | +1 vote
134 views
A arrives at office at 8-10am regularly; B arrives at 9-11 am every day. Probability that one day B arrives before A? [Assume arrival time of both A and B are uniformly distributed]
A can come between 8 to 10 , which is 120 min duration . A can come on any instant from this 120 min window .
B can come between 9 to 11 , which is 120 min duration . B also can choose any instant for coming from this 120 min window
so tota possible cases of coming of A nd B =120*120
Now consider the case where B is coming early to A . here common window of 60 min which is from 9 to 10 .
we can choose two time instant from this window by 60C2 ways .
so probability is = 60c2/120*120 = 60*59
--------------= 59/480
2*120*120
selected
why are you choosing two time instants from 9-10??
and in this question,though the given answer is 1/8 which is very close to your answer but as it is given that data is uniformly distributed,does'nt we have to solve like that?
I choose two instant from 9 -10 because it is the common time window and two instant because one for A's arriving time and one for B's arriving time. and which value is less we take that for B's arriving time and other one for A's arriving . And it is satisfying the criteria .
And dont find any error in this method and in ans as well .
PS: I dont know the method of uniform distribution
thanks a lot..i got your point
+1 vote | 369 | 1,405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-47 | latest | en | 0.95542 |
https://nathanpeelphoto.com/how-many-days-in-a-year-2024/ | 1,685,245,236,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643462.13/warc/CC-MAIN-20230528015553-20230528045553-00004.warc.gz | 488,585,420 | 10,907 | # How many days in a year 2024?
The year 2024 is a leap year, with 366 days in total.
## How many days are there in the year 2025?
The year 2025 is a common year, with 365 days in total.
## Will there be 366 days in 2024?
To be a leap year, the year number must be divisible by four – except for end-of-century years, which must be divisible by 400. This means that the year 2000 was a leap year, although 1900 was not. 2020, 2024 and 2028 are all leap years.
## Is there 29 days in February in 2024?
The next leap day is February 29, 2024. 2020 was also a leap year, so the last leap day was February 29, 2020.
## How many days does the year 2023 have?
There are a total of 260 working days in the 2023 calendar year. To compute partial month salary amounts, take monthly budgeted salary amount X # of days worked that month / total working days for that month.
## Is there ever 367 days in a year?
By adding a second leap day (Friday, February 30) Sweden reverted to the Julian calendar and the rest of the year (from Saturday, March 1) was in sync with the Julian calendar. Sweden finally made the switch from the Julian to the Gregorian calendar in 1753. This year has 367 days.
## How far is 2050 years from now?
2050 is only about 30 years from now, which means it’s close enough that we can imagine it happening, but far enough away that we can’t confidently say what it will look like.
## Will 2028 will be a leap year?
2020, 2024 and 2028 are all leap years.
## Why 1900 is not a leap year?
There is a leap year every year whose number is perfectly divisible by four – except for years which are both divisible by 100 and not divisible by 400. The second part of the rule effects century years. For example; the century years 1600 and 2000 are leap years, but the century years 1700, 1800, and 1900 are not.
## Is 2028 a leap year or not?
To be a leap year, the year number must be divisible by four – except for end-of-century years, which must be divisible by 400. This means that the year 2000 was a leap year, although 1900 was not. 2020, 2024 and 2028 are all leap years.
## Did 2000 have a leap year?
The year 2000, like the years 1996 and 2004, is a leap year – with 29 days in February; but the years 1900, 1999, 2001, 2002, 2003, 2005 and 2100 are not leap years – and have only 28 days in February.
## Is there a year Zero?
There is no year 0. Jesus was born before 4 B.C.E. The concept of a year “zero” is a modern myth (but a very popular one). In our calendar, C.E. 1 follows immediately after 1 B.C.E. with no intervening year zero.
## Do we have year zero?
Historians have never included a year zero. This means that between, for example, 1 January 500 BC and 1 January AD 500, there are 999 years: 500 years BC, and 499 years AD preceding 500. In common usage anno Domini 1 is preceded by the year 1 BC, without an intervening year zero.
## Does February 30th exist?
February 30. February 30 or 30 February is a date that does not occur on the Gregorian calendar, where the month of February contains only 28 days, or 29 days in a leap year. February 30 is usually used as a sarcastic date for referring to something that will never happen or will never be done.
## Why is there no February 30th?
The Julian Calendar added a little more than 10 days to each year, making each month either 30 or 31 days long, except for February. To account for the entire 365.25 day-long year, one day was added to February every four years, now known as a “leap year.” During most years, this left February with just 28 days.
## How hot will the earth be in 2050?
Since 1880, average global temperatures have increased by about 1 degrees Celsius (1.7° degrees Fahrenheit). Global temperature is projected to warm by about 1.5 degrees Celsius (2.7° degrees Fahrenheit) by 2050 and 2-4 degrees Celsius (3.6-7.2 degrees Fahrenheit) by 2100.
## What will Earth be in 100 years?
In 100 years, oceans will most likely rise, displacing many people, and it will continue to become warm and acidic. Natural disasters like wildfires and hurricanes will continue to be very common and water resources could be scarce. NASA is researching earth to make observations that will benefit everyone.
## What happens every 400 years for leap year?
So, 29th February will come 97 times in 400 years.
## Will the year 2300 be a leap year?
Also, if the year can be evenly divided by 100, it is NOT a leap year, unless the year is also evenly divisible by 400. This means that 2000 and 2400 are leap years, but the years 1800, 1900, 2100, 2200, 2300 and 2500 are not considered leap years.
## What would happen if you were born on February 29?
In non-leap years, that day is March 1. So for someone born on February 29, the first day they can legally drive, vote, join the Army, buy alcohol or start collecting Social Security is presumably March 1 in non-leap years.
## How many leap years in 400 years?
So, 29th February will come 97 times in 400 years.
## How do leap year babies age legally?
His legal thinking is that February 29 is the day after February 28, so a person born on February 29 is legally considered to have aged one year on the day after February 28. In non-leap years, that day is March 1.
## What happens every 400 years?
29) pops up on the calendar only on leap years, once almost every four years. It has taken millennia for our calendar, called the Gregorian calendar after the pope who modified it in 1582, to evolve to include this tweak — 97 leap years every 400 years.
## Was there a year 666?
Year 666 (DCLXVI) was a common year starting on Thursday (link will display the full calendar) of the Julian calendar. The denomination 666 for this year has been used since the early medieval period, when the Anno Domini calendar era became the prevalent method in Europe for naming years.
## Who was born in the year 1?
Birth of Jesus, as assigned by Dionysius Exiguus in his anno Domini era according to at least one scholar. | 1,553 | 5,987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-23 | latest | en | 0.969753 |
https://www.physicsforums.com/threads/temperature-of-body-in-space.97579/ | 1,548,241,680,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584331733.89/warc/CC-MAIN-20190123105843-20190123131843-00072.warc.gz | 887,540,355 | 14,088 | # Temperature of body in space
1. Oct 30, 2005
### sid_galt
I want to determine the temperature of a body in space exposed to the Sun as it varies with time.
I tried this.
Let A be the area exposed to the Sun and 2A the total area of the body. Let 1380 W/m2 be the power of the sunlight falling on the body. Let $$\sigma$$ be the boltzmann constant, $$\epsilon$$ the emissivity of the body and T the temperature at a a particular instant of time, m the mass of the body and C the specific heat constant. Then
$$\displaystyle\frac{dT}{dt} = \displaystyle\frac{1380A - 2\sigma\epsilon A T^4}{mC}$$
$$\displaystyle\frac{mC}{A}\int\displaystyle\frac{1}{1380 - 2\sigma\epsilon T^4}dT = \int dt$$
I tried to integrate it on integrals.wolfram.com taking boltzmann constant as 5.6E-8 and emissivity as 0.7. The result was
$$\displaystyle\frac{0.156942mC}{A}(\arctan[0.00230862T]+arctanh[0.00230862T]) + C' = t$$
C' is here the integration constant
I dont know how to proceed further. Can anyone help please?
Thank you
Last edited: Oct 31, 2005
2. Oct 31, 2005
### sid_galt
3. Oct 31, 2005
### µ³
Are you sure it's T^4? My thermodynamics knowledge is limited but I thought the rate of heat transmition was proportional to deltaT, I might be wrong though. Otherwise, there is no analytic solution to the lower equation for T.
Edit - nevermind, it's blackbody radiation. Yeah, sorry, can't figure out what's wrong with either your physics or math. The last equation is not solvable for T.
Last edited: Oct 31, 2005
4. Oct 31, 2005
### sid_galt
But there must be someway to find temperature as a function of time for a body in space exposed to the Sun.
5. Oct 31, 2005
### Tide
Yes, there is. You can evaluate your integral numerically.
6. Oct 31, 2005
### sid_galt
But how do I integrate high temperatures into the equation, say if I want to evaluate it for a body with an initial temperature of 1000 K.
My Arctanh would give an unreal value for all temperatures higher than 434.78 K.
7. Oct 31, 2005
### µ³
if you notice that ArcTan[x]+ArcTanh[x] for x > 1 always give some value a - 1.5708i , well , make it so your integration constant takes out the imaginary part.
Edit- more specifically
for x>1
ArcTanh[x] = ArcTanh[1/x] -1/2pi*i
Edit: Another approximation:
for small x
ArcTanh[x] = ArcTan[x]
another nice identity:
(you might be able to solve for this actually)
$$tan^{-1}(x) +tan^{-1}(y) = tan^{-1}(\frac{x+y}{1-xy})$$
Last edit - I tried all the above, it doesn't work (not even with shoddy approximations).
Last edited: Oct 31, 2005
8. Oct 31, 2005
### sid_galt
I didn't notice that before. Thanks for the help. | 808 | 2,639 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-04 | latest | en | 0.879689 |
https://easetolearn.com/smart-learning/web/physics/mechanics/gravitation/gravitational-potential-energy/gravitational-potential/5032 | 1,708,946,866,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00359.warc.gz | 221,330,964 | 28,838 | # Gravitational Potential
Gravitational field:
In the case of non-contact force, the source mass and the test mass interact via a gravitational field. You can think of the gravitational force as a "command" and the gravitational field as a dialogue or speech used to give an order.
Gravitational field intensity
The strength of the gravitational field is called the gravitational field strength. This is the gravitational force acting on the unit mass.
where r cap represents the position vector of the test mass from the source mass. The intensity of the gravitational field depends only on the mass of the source and the distance of a unit of the test mass from the mass of the source.
The unit of gravitational field strength is N/kg.
The scale gives [M0L¹T–²]. The scale of gravitational field strength is identical to that of acceleration (from a gravitational perspective, we prefer to call it acceleration due to gravity).
The principle of superposition extends to the intensity of the gravitational field, for example
E = E1 + E2 + E3 + . . . . . . . + En
Thus, E1, E2, E3, . . . . . In are the gravitational field intensities at the point due to the nth particle of the system.
In the system, mass is always distributed in two different ways:
• Discrete mass distribution
• Continuous mass distribution
dE is the intensity of the gravitational field due to the main mass dm. The gravitational field formula is expressed as:
g = F/m
where, F = gravitational force and m = mass of the object.
Gravitational Field Intensity of a Point Mass:
Consider a point mass M, the gravitational force at a distance 'r' from it is given by the formula
Gravitational Intensity of the Gravitational Field created by the Ring:
Consider a ring of mass M with radius 'a'; the gravitational field at a distance x along its axis is found as:
Gravitational field intensity
Consider a small element along the circumference of the ring with mass 'dm'; the field strength resulting from that length element is given by the formula,
dE = Gdm/r²
The vertical components of the fields cancel each other due to ring symmetry, and only the horizontal components are preserved and added.
Gravitational field due to Uniform Spherical Shell:
Consider a thin uniform spherical shell in space of radius 'R' and mass 'M'. A 3D object divides space into three parts:
• Inside the spherical shell.
• On the surface of the spherical shell.
• Outside the spherical shell.
Our problem is to find out the value of gravitational field intensity in all these 3 regions
Outside the spherical shell
Consider a unit test mass at a point "P" at a distance "r" from the center of the spherical shell. Draw an imaginary spherical shell with point 'P' on its surface.
As we know, the gravitational field intensity at a point depends only on the mass of the source and the distance of the point from the mass of the source. We can say that the initial mass in the imaginary sphere is M and the separation distance is 'r'. That's what we get
E = -GM/r²
E -1/r²
On the surface of the spherical shell
Consider a unit test mass at a point "P" on the surface of the spherical shell at a distance "r" from the center of the spherical shell, where r = R. As mentioned above, the gravitational field intensity at the surface of the spherical shell is given by spherical shell
E = -GM/R²
Inside a spherical shell
If you consider a point inside a spherical shell, the entire mass of the shell is above the point. Draw an imaginary spherical shell around point 'P'; to leave this imaginary realm is zero.
We know that if the mass of the source is zero, the intensity of the gravitational field is also zero.
E = 0
Gravitational field due to a Uniform Solid Sphere:
Consider a uniform solid sphere of radius "R" and mass "M". Let's find the value of the gravitational field strength in all three regions:
• Within a solid sphere.
• On the surface of a solid sphere.
• Off a certain ball.
To find the gravitational field intensity at a point "P" at a distance "r" from the center outside the solid sphere, consider an imaginary sphere around P enclosing the entire mass "M".
E = – GM/r²
E -1/r²
On the surface of a solid sphere
To find the gravitational field intensity at point "P" on the surface of a solid sphere,
The distance from the surface point is r = R.
Then E = -GM/R² E = Constant.
Within a solid sphere
Find the gravitational effect at a point 'P' inside a uniform solid sphere at a distance 'r' from the center of the sphere. If we draw an imaginary sphere around this point, the mass of that imaginary sphere is given by 'm'.
For space (4/3) πR³, the existing mass is M; for volume (4/3) πr³ the existing mass is "m".
Since the density of a solid sphere remains the same at all times,
m = M × (r³/R³)
Then the gravitational field intensity at point "P" inside the solid sphere at a distance "r" from the center of the sphere is given by:
E = -Gm/r²
Where m is the initial mass on an imaginary sphere drawn around point "P". Substituting the value of m in the above equation, we get
E = -GMr/R³
E -r
Gravitational potential energy - formulas, derivatives, solved problems
Gravitational potential energy is the energy that an object possesses or gains from its change in position when it is in a gravitational field. Simply put, gravitational potential energy is the gravitational force or energy associated with gravity.
The most common example to help you understand the concept of gravitational potential energy is that you take two pencils, one of which is placed on a table and the other is held above the table. Now we can argue that a pencil in the air has more gravitational potential than a pencil on the table
Gravitational Potential Energy:
If a body of mass (m) is moved from infinity to a point under the gravitational influence of the initial mass (M) without accelerating it, the amount of work done in moving it to the initial field is stored in the form of potential energy. . . . This is called gravitational potential energy. It is denoted by the symbol Ug.
Explanation:
We know that the potential energy of a body at a particular location is defined as the energy stored in the body at that location. When the position of the body changes under the influence of external forces, the change in potential energy is equal to the amount of work done by the forces acting on the body. The work done by gravity does not depend on the path of change of position, so the force is a conservative force. All such forces also have a certain potential.
The effect of gravity on a body at infinity is zero; therefore, the potential energy is zero, called the reference point.
Gravitational Potential Energy Formula
The equation for gravitational potential energy is:
GPE = mgh
Where,
• m is the mass in kilograms
• g is the acceleration due to gravity (9.8 on earth)
• h is the height above the ground in metres
Derivation of Gravitational Potential Energy Equation:
Consider a source mass ‘M’ is placed at a point along the x-axis; initially, a test mass ‘m’ is at infinity. A small amount of work done in bringing it without acceleration through a very small distance (dx) is given by
dw = Fdx
Here, F is an attractive force, and the displacement is towards the negative x-axis direction, so F and dx are in the same direction. Then,
dw = (GMm/x2)dx
Integrating on both sides
If ri> rf then ΔU is negative.
Since the work done is stored as its potential energy U, the gravitational potential energy at a point which is at a distance ‘r’ from the source mass is given by;
U = -GMm/r
If a test mass moves from a point inside the gravitational field to another point inside the same gravitational field of source mass, then the change in potential energy of the test mass is given by;
ΔU = GMm (1/ri – 1/rf)
If ri> rf then ΔU is negative.
Expression for Gravitational Potential Energy at Height (h) –
Derive ΔU = mgh.
If a body is taken from the surface of the earth to a point at a height ‘h’ above the surface of the earth, then ri = R and rf= R + h, then,
ΔU = GMm [1/R – 1/(R+h)]
ΔU = GMmh/R(R + h)
When h<<R, then R + h = R and g = GM/R2.
On substituting this in the above equation, we get,
Gravitational Potential Energy ΔU = mgh
Note:
The weight of a body in the center of the earth is zero because the value of g in the center of the earth is zero.
At a point in the gravitational field where the gravitational potential energy is zero, the gravitational field is zero.
What Is Gravitational Potential?
The amount of work done in moving a unit test mass from infinity into the gravitational influence of source mass is known as gravitational potential.
Simply, it is the gravitational potential energy possessed by a unit test mass.
V = U/m
V = -GM/r
Important Points:
The gravitational potential at a point is always negative, and V is maximum at infinity.The SI unit of gravitational potential is J/s
Relation between Gravitational Field Intensity and Gravitational Potential
Integral Form:
V =
( If E is given and V has to be found using this formula)
Differential Form:
E = -dV/dr (If V is given and E has to be found using this formula)
(components along x, y, and z directions).
Gravitational Potential of a Point Mass
Consider a point mass M, the gravitational potential at a distance ‘r’ from it is given by;
V = – GM/r.
Gravitational Potential of a Spherical Shell
Consider a thin uniform spherical shell of the radius (R) and mass (M) situated in space. Now,
• Case 1: If point ‘P’ lies inside the spherical shell (r<R):
As E = 0, V is a constant.
The value of gravitational potential is given by, V = -GM/R.
Case 2: If point ‘P’ lies on the surface of the spherical shell (r=R):
On the surface of the earth, E = -GM/R2.
Using the relation
• Case 1: If point ‘P’ lies inside the spherical shell (r<R):
As E = 0, V is a constant.
The value of gravitational potential is given by, V = -GM/R.
• Case 2: If point ‘P’ lies on the surface of the spherical shell (r=R):
On the surface of the earth, E = -GM/R2.
Using the relation
over a limit of (0 to R), we get,
Gravitational Potential (V) = -GM/R.
• Case 3: If point ‘P’ lies outside the spherical shell (r>R):
Outside the spherical shell, E = -GM/r2.
Using the relation
V = -
over a limit of (0 to R), we get,
Gravitational Potential (V) = -GM/R.
Case 3: If point ‘P’ lies outside the spherical shell (r>R):
Outside the spherical shell, E = -GM/r2.
Using the relation
V = -
Gravitational Potential of a Uniform Solid Sphere:
Consider a thin, uniform solid sphere of radius (R) and mass (M) situated in space. Now,
• Case 1: If point ‘P’ lies inside the uniform solid sphere (r < R):
Inside the uniform solid sphere, E = -GMr/R3.
Using the relation
V = -
over a limit of (0 to r).
The value of gravitational potential is given by,
V = -GM [(3R2 – r2)/2R2]
• Case 2: If point ‘P’ lies on the surface of the uniform solid sphere ( r = R ):
On the surface of a uniform solid sphere, E = -GM/R2. Using the relation
V = -
over a limit of (0 to R) we get,
V = -GM/R.
• Case 3: If point ‘P’ lies outside the uniform solid sphere ( r> R):
Using the relation over a limit of (0 to r), we get, V = -GM/R.
• Case 4: Gravitational potential at the centre of the solid sphere is given by
V =(-3/2) × (GM/R).
Gravitational Self Energy
The gravitational self-energy of a body is defined as the work done by an external agent in assembling the body from the infinitesimal elements that are initially at an infinite distance apart.
Gravitational self-energy of a system of ‘n’ particles:
Let us consider n particle system in which particles interact with each other at an average distance ‘r’ due to their mutual gravitational attraction; there are n(n – 1)/2 such interactions, and the potential energy of the system is equal to the sum of the potential energy of all pairs of particles, i.e.,
Solved Problems
Example 1. Calculate the gravitational potential energy of a body of mass 10 kg and is 25 m above the ground.
Solution:
Given, Mass m = 10 Kg and Height h = 25 m
G.P.E is given as,
U = m × g × h = 10 Kg 9.8 m/s2 × 25 m = 2450 J. | 2,937 | 12,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-10 | latest | en | 0.870954 |
https://www.physicsforums.com/threads/finding-inverse-matrices-using-guass-approach.358005/ | 1,723,274,273,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00371.warc.gz | 739,593,492 | 19,269 | Finding inverse matrices using Guass approach
• thomas49th
In summary: The only thing that might have saved you some time would have been to recognize that the bottom row of I was not correct, because the bottom row of A was not all 0's.
thomas49th
Homework Statement
Use the determinate method and also the Guass elimination method to find the inverse of the following matrix. Check your results by direct multiplication
$$A =\left | \begin{array}{ccc} 2&1&0\\ 1&0&0\\ 4&1&2 \end{array}\right | =$$
Let's do Guass first
Homework Equations
Place A by I and attemp to get A into I. Everything I perform on A must be performed on I and when A is in I, the original I is the inverse?
$$A =\left | \begin{array}{ccc} 2&1&0\\ 1&0&0\\ 4&1&2 \end{array}\right |$$
$$I=\left | \begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}\right |$$
The Attempt at a Solution
Interchange rows 1 and R
R1 <-> R2
$$A =\left | \begin{array}{ccc} 1&0&0\\ 2&1&0\\ 4&1&2 \end{array}\right | =$$
$$I =\left | \begin{array}{ccc} 0&1&0\\ 1&0&0\\ 0&0&1 \end{array}\right |$$
Now R2 - 2R1
$$A =\left | \begin{array}{ccc} 1&0&0\\ 0&1&0\\ 4&1&2 \end{array}\right | =$$
$$I =\left | \begin{array}{ccc} 0&1&0\\ 1&-2&0\\ 0&0&1 \end{array}\right |$$
R3 - 4R1
$$A =\left | \begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&1&2 \end{array}\right | =$$
$$I =\left | \begin{array}{ccc} 0&1&0\\ 1&-2&0\\ 0&-4&1 \end{array}\right |$$
R3 - R2
$$A =\left | \begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&0&2 \end{array}\right | =$$
$$I =\left | \begin{array}{ccc} 0&1&0\\ 1&-2&0\\ -1&-4&1 \end{array}\right |$$
R3 / 2
$$A =\left | \begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}\right | =$$
$$I =\left | \begin{array}{ccc} 0&1&0\\ 1&-2&0\\ -1/2&-2&1/2 \end{array}\right |$$
but multiplying my A and new I together does not give me I? What have I done wrong?
Check the R3-R2 step. (-4)-(-2)=(-2) not (-4).
You made a mistake in the calculation right after what you show as R3 - R2.
The -4 in the last row of I should be -2.
ahh got it working by starting again and doing another combination... made a silly mistake in this jumble!
Thanks for the pointers guys. My other route still took just as long as this one. Do any of you any tricks to improve Guass Elimination efficiency or is it just experience.
Thanks
Tom
I don't know how you are doing the calculations, but your work was difficult to check with two separate matrices. The usual practice is to create an augmented matrix, starting with A in the left half and I in the right half. After you get to I in the left half, your inverse of A will be in the right half. Each augmented matrix will look something like this.
$$\left[ \begin{array}{ccccccc} 2&1&0&|&1&0&0\\ 1&0&0&|&0&1&0\\ 4&1&2&|&0&0&1 \end{array}\right ]$$
An "accounting" tip to make it easier to understand what you did is to use a notation that indicates which row changes when you add a multiple of one row to another. One way to do this is fairly verbose R1 <-- R1 - 3R2. You can abbreviate this to R1 - 3R2 if it's understood that it's always the first row listed that is added to. Obviously if you just switch two rows, it doesn't matter which one you list first, and if you replace a row by a multiple of itself, there's only one row involved, so there shouldn't be any confusion about which rows are involved.
Other than that, you row-reduced your matrices the way I would have, so I don't see anything that you could have done that would have economized your efforts.
What is the Gauss approach for finding inverse matrices?
The Gauss approach, also known as Gaussian elimination, is a method for finding the inverse of a square matrix by performing a series of row operations to transform the matrix into reduced row echelon form.
Why is finding inverse matrices important in scientific research?
Finding inverse matrices is important in scientific research because it allows for the efficient solving of systems of linear equations and can be applied to various mathematical models and simulations.
What are the steps involved in using the Gauss approach to find inverse matrices?
The steps involved in using the Gauss approach to find inverse matrices are:1. Augment the given matrix with the identity matrix2. Use row operations to transform the given matrix into reduced row echelon form3. If the resulting matrix is the identity matrix, then the original matrix is invertible and the transformed matrix is its inverse4. If the resulting matrix is not the identity matrix, then the original matrix is not invertible
What are the limitations of using the Gauss approach to find inverse matrices?
The Gauss approach may not work for matrices that are not square or for matrices that are singular (have a determinant of 0). It also involves a significant amount of computation, especially for larger matrices, which can make it time-consuming.
Are there any alternative methods for finding inverse matrices?
Yes, there are other methods for finding inverse matrices such as using the adjugate matrix, the Cramer's rule, or the LU decomposition method. Each method may have its own advantages and limitations, and the most suitable method may depend on the specific characteristics of the given matrix.
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2K | 1,695 | 5,753 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-33 | latest | en | 0.74703 |
http://mathhelpforum.com/advanced-algebra/165335-please-help-clarify-galois-proof.html | 1,516,227,337,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886979.9/warc/CC-MAIN-20180117212700-20180117232700-00540.warc.gz | 208,220,385 | 11,311 | I more or less copied the following notes from my professor:
Find the Galois group of $x^4-2$ over $\mathbb{Q}$.
Solution: $x^4-2$ has roots $\pm\sqrt[4]{2},\pm i\sqrt[4]{2}$. Note that since $G$ acts on the four roots, $G$ is isomorphic to a subgroup of $S_4$. The splitting field is $\mathbb{Q}(\sqrt[4]{2},i)$, so $|G|=8$. By Sylow, any two subgroups of $S_4$ of order $8$ must be isomorphic. So $G\cong D_8$.
Everything about this proof makes perfect sense to me except the part in bold: Why is $G$ a subgroup (isomorphic) of $S_4$? It cannot be simply because $G$ acts on the four roots (because for instance any group of any order acts on any set trivially). All I can tell from that is that there is a homomorphism from $G$ into $S_4$. So, how do I conclude that $G$ is a subgroup (isomorphic) of $S_4$?
Also, I'm a bit confused as to how it can be that $|G|=8$. For each $\alpha\in G$ is completely determined by $\alpha(i)$ and $\alpha(\sqrt[4]{2})$. Now, $1=\alpha(1)=\alpha(-i^2)=-\alpha(i)^2$, which seems to imply $\alpha(i)=i$. But then $\alpha(\sqrt[4]{2})$ must be one of the four roots. So $|G|\leq 4$, a contradiction. Where did I go wrong, I wonder?
Any help would be much appreciated!
2. Originally Posted by hatsoff
I more or less copied the following notes from my professor:
Everything about this proof makes perfect sense to me except the part in bold: Why is $G$ a subgroup (isomorphic) of $S_4$? It cannot be simply because $G$ acts on the four roots (because for instance any group of any order acts on any set trivially). All I can tell from that is that there is a homomorphism from $G$ into $S_4$. So, how do I conclude that $G$ is a subgroup (isomorphic) of $S_4$?
Any help would be much appreciated!
As you remark the part in bold is not completely correct. For it to be it should be that the Galois group acts transitively on
the four roots, and then it'd follow that the group is isomorphic to a subgroup of $S_4$.
There's another way to show that $G\cong D_8$: first, it's easy to see that $G$ cannot be abelian and that
its order is divisible by four and divides 8, so it is either $D_8\,\,or\,\,Q_8$. Nevertheless, this extension has a non-normal
subextension, $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}$ , from where it follows that there is a non-normal subgroup of $G$
of index 4, and since all the subgroups $Q_8$ are normal then we're done.
Tonio
3. Originally Posted by hatsoff
I more or less copied the following notes from my professor:
Everything about this proof makes perfect sense to me except the part in bold: Why is $G$ a subgroup (isomorphic) of $S_4$? It cannot be simply because $G$ acts on the four roots (because for instance any group of any order acts on any set trivially). All I can tell from that is that there is a homomorphism from $G$ into $S_4$. So, how do I conclude that $G$ is a subgroup (isomorphic) of $S_4$?
Also, I'm a bit confused as to how it can be that $|G|=8$. For each $\alpha\in G$ is completely determined by $\alpha(i)$ and $\alpha(\sqrt[4]{2})$. Now, $1=\alpha(1)=\alpha(-i^2)=-\alpha(i)^2$, which seems to imply $\alpha(i)=i$. But then $\alpha(\sqrt[4]{2})$ must be one of the four roots. So $|G|\leq 4$, a contradiction. Where did I go wrong, I wonder?
Any help would be much appreciated!
$\alpha(i)^2=-1\Longrightarrow \alpha(i)=\pm i$ , since $i^2=(-i)^2=-1$
Tonio
4. Okay, I see now what is going on. Thanks a bunch for the help! | 1,035 | 3,431 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 58, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-05 | longest | en | 0.927491 |
https://ru-facts.com/how-does-a-simple-dc-circuit-work/ | 1,719,274,452,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865490.6/warc/CC-MAIN-20240624214047-20240625004047-00775.warc.gz | 437,798,290 | 12,171 | # How does a simple DC circuit work?
## How does a simple DC circuit work?
A simple DC circuit consists of a current source (e.g. battery) and one or more “loads” (circuit elements). Each “load” absorbs electrical energy, converting it to some other form of energy, e.g. a light bulb emits heat and light energy, an electric motor performs mechanical work and emits heat.
### What are the three common components of a DC circuit?
The resistor is the main component of the DC circuit. A simple DC circuit is shown in the figure below which contains a DC source (battery), a load lamp, a switch, connecting leads, and measuring instruments like ammeter and voltmeter.
#### Can a parallel circuit be DC?
When two or more electrical components are connected in a way that one end of each component is connected to a common point and the other end is connected to another common point, then the electrical components are said to be connected in parallel, and such an electrical DC circuit is referred as a parallel DC circuit.
What is the formula of DC circuit?
Equations
Equation Symbols
P = I Δ V P = I \Delta V P=IΔV P P P is power, I is current, and Δ V \Delta V ΔV is electric potential difference
What are the rules for drawing circuit diagrams?
Circuit diagrams Always try to make the wires straight lines. Do not be tempted to make them wiggly because the whole point is to make it easier to see what is connected to what. If you have to draw a circuit diagram: draw the circuit symbols first, then.
## What is common in a DC circuit?
“Common” is a point of reference shared throughout a circuit, typically the node in reference to which every other node is described. In the case of a single-pole double-throw (SPDT) switch, common (also called the pole) is the terminal which the “normally open” and “normally closed” terminals switch to.
### How does a DC current flow?
Direct current (DC) is an electric current that is uni-directional, so the flow of charge is always in the same direction. As opposed to alternating current, the direction and amperage of direct currents do not change.
#### What is the common wire in a DC circuit?
The black wire in DC is called NEGATIVE or COMMON. In the british system theBLK. is called neutral.
What is the effective resistance of the DC circuit?
The effective resistance is basically the resistance of that single edge. In the study of electrical circuits, one is interested in the behaviour of idealized electrical com- ponents in a circuit in terms of voltage (or potential difference) and current [13, p. 11].
How does a DC Circuit work?
A direct current (DC) electric circuit consists of a source of DC electricity—such as a battery—with a conducting wire going from one of the source terminals to a set of electric devices and then back to the other terminal, in a complete circuit. A DC circuit is necessary for DC electricity to exist.
## How to read a circuit diagram?
In order to learn how to read a circuit diagram, it is necessary to learn what the schematic symbol of a component looks like. It is also necessary to understand how the components are connected together in the circuit.
### What is a direct current (DC) Circuit?
A direct current (DC) electrical circuit consists of a source of DC electricity with a conducting wire going from one of the source terminals to a set of electrical devices and then back to the other terminal, in a complete circuit.
#### What is a circlecircuit diagram?
Circuit diagrams or schematic diagrams show electrical connections of wires or conductors by using a node as shown in the image below. A node is simply a filled circle or dot.
Begin typing your search term above and press enter to search. Press ESC to cancel. | 800 | 3,745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-26 | latest | en | 0.939576 |
https://cracku.in/48-the-present-age-of-bob-is-equal-to-abbys-age-8-yea-x-ibps-po-2015-mock-1 | 1,721,290,069,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00468.warc.gz | 155,898,500 | 23,067 | Question 48
# The present age of Bob is equal to Abby’s age 8 years ago. Four years hence, the respective ratio between Bob’s age and Abby’s age will be 4 : 5 at that time. What is Bob’s present age?
Solution
Let the present ages of Abby be 'a' and Bob be 'b'
b = a - 8 ---> (1)
$$\frac{b+4}{a+4} = \frac{4}{5}$$
5b + 4 = 4a
5 (a - 8) + 4 = 4a
a = 36
b = 28
Hence, the present age of Bob = 28 years | 152 | 401 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-30 | latest | en | 0.902052 |
http://docplayer.net/10453557-Cost-benefit-analysis-a-methodology-for-sound-decision-making-student-lesson-document.html | 1,545,018,629,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828056.99/warc/CC-MAIN-20181217020710-20181217042710-00080.warc.gz | 65,761,147 | 31,134 | # Cost/Benefit Analysis. A Methodology for Sound Decision Making. Student Lesson Document
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## Transcription
2 Prior to checkout, you remember the sign that appeared in the drugstore s window: Soda pop 12- packs - \$2.50 each. Again, you use math to calculate that this would be a savings of \$1.00 per 12-pack; this would add up to a savings of \$4.00 if you purchase four 12-packs! Ask yourself: Is it worth it to make an extra trip into the drugstore in order to save \$4.00 instead of purchasing the soda at the grocery store? Presumably, \$4.00 is a significant enough savings for exactly the same item. In fact, many people would decide to get their soda pop at the drugstore. Heck, at this price, you could buy a fifth 12-pack and still save \$1.50 over the grocery store purchase! Discussion #2: Cost/Benefit Analysis is a practice that most of us use all the time when making decisions, even when we do not realize we are utilizing the tactic. So, let s define what it is. Cost/Benefit Analysis (CBA) - a comparative analysis of estimates of the costs and the benefits of undertaking a particular action in order to determine if taking the action is worthwhile. The procedure to complete an effective Cost/Benefit Analysis is a four-step process: 1. Try to determine all of the costs associated with taking a particular action. Try to determine all of the benefits associated with taking that action. 2. Try to estimate the monetary value of each cost and total them. 3. Try to estimate the monetary value of each benefit and total them. 4. Finally, compare the total costs versus the total benefits to determine if the action is worth taking. Note: These are seemingly simple steps; however, things are rarely as simple as they seem. The more accurate the monetary estimates of the costs and benefits are, the easier it will be to make a decision. Also, the greater the difference between the totals of the costs and the benefits, the easier the decision will be. Yet, when the costs and benefits totals are closer together, the decision becomes more subjective. Also, while the monetary value of costs may be a relatively easy thing to estimate, many times there are benefits that cannot be ignored and are difficult, if not impossible, to value in monetary terms. Again, this can make a decision subjective. Explanation: In the example of purchasing soda pop 12-packs, the difference in prices between the grocery store and the drugstore was substantial enough to make it a seemingly easy decision regarding where to purchase the soda. However, what if the difference in price per 12-pack was only a nickel instead of a dollar? The savings on purchasing four 12-packs would be merely 2
3 \$0.20. Depending on the individual shopper, this might not be a significant enough savings to make the trip into the drugstore. The decision on where to purchase the soda pop might be different because the CBA results changed the shopper s mind. Would a \$0.20 savings difference, instead of a \$4.00 savings, change your mind? Discussion #3: As previously stated, assigning a monetary value to some benefits may be difficult, if not impossible, to do. Let s take a look at an example that still produces a reasonably clear-cut result. Example: A small community, located in the vicinity of your neighborhood, is host to a troublesome intersection. Most of the time, it seems to be quite busy. Not only does heavy traffic move through the intersection, but it is also frequented by many people, especially small children. Perhaps it is even a school bus drop-off site? In any case, the intersection is clogged with vehicular and pedestrian traffic. It appears to be an accident waiting to happen. Residents continually petition the village council to install a traffic control device. A traffic light is preferred to control the flow of traffic and to help keep pedestrians safe. The village council has agreed and does a Cost/Benefit Analysis on installing the traffic light at this intersection. The costs are straightforward. \$8,500 for the traffic light and equipment needed to control it. \$2,500 for installation. \$1,000 per year to maintain the light and pay for the electricity to run it. Total costs = \$11,000 + \$1,000 annually The benefits are also straightforward, but assigning a monetary value to them is more difficult. A traffic light would almost certainly prevent injury or worse to pedestrians. How does one put a monetary value on a human life? Note: The village council would be hard pressed to find anyone who would disagree with the fact that if the traffic light saved even one life, the \$11,000 cost would be well worth the expense. Placing a monetary value on a human life would be difficult, if not impossible, to do; yet, almost everyone would agree that whatever the monetary value of that life is, it is much greater than \$11,000. Therefore, the project should be completed. Do you agree? 3
4 Discussion #4: Remember, step #1 of conducting a CBA is to: 1. Try to determine all of the costs associated with taking a particular action. Try to determine all of the benefits associated with taking that action. It can be difficult to assign monetary values to benefits; yet, those benefits cannot and should not be ignored. In addition, there may be some obscure costs involved that should also be taken into account. Opportunity Costs Opportunity costs are real and are usually a part of the decision-making process; yet, they are easily missed when completing a CBA. What are Opportunity Costs? Opportunity cost - the cost associated with forgoing the benefit of making a different decision. If there are two choices, Option A and Option B, but only one option can be selected, then the opportunity cost of choosing one option is the benefit of choosing the other option. Example: It s Friday night and you are looking forward to having a nice dinner at your favorite restaurant. Then, some friends contact you to invite to a movie tonight. Sounds like a blast! However, you only have enough money for dinner OR a movie, not both. So, you must make a choice: dinner OR a movie with your friends? Option A: If you choose to go to dinner, you will miss out on going to the movies. Option B: If you choose to go to the movies, you will miss out on dinner at your favorite restaurant. Explanation: The opportunity cost of choosing dinner is the movie with friends. The opportunity cost of choosing the movie is dinner at your favorite restaurant. Opportunity costs can be thought of as the consequences of selecting one action over another. In most CBAs, these opportunity costs are real and should always be taken into account. What would you decide to do? Discussion #5: Reflect back to the CBA completed based on the traffic light scenario. What if there is an opportunity cost wrinkle? As the village council works on the CBA to decide if installing the traffic light project should move forward, the council determines that the funds would need to come from a previously 4
5 approved project. The community park s equipment is in disrepair and needs to be fixed and/or replaced. The jungle gyms, swing sets, etc. have all seen better days and \$11,000 has already been approved and allocated to repair and upgrade the park. The current state of the equipment also presents a potential safety hazard to the people who frequent the park. The park upgrade project and the traffic light project are two choices. The council, however, is aware that there is only enough money to fund one. The opportunity cost of installing the traffic light is that the park upgrade would have to wait. The opportunity cost of upgrading the park is that the traffic light would have to wait. Both costs deal with the value of a human life or injury. The decision to install the traffic light is no longer so clear. Exercise: In order to solve this dilemma, use the four-step CBA process and include the opportunity costs of each choice. Total cost of installing a traffic light Opportunity cost of installing a traffic light Total cost of park repair Opportunity cost of park repair Benefits of installing a traffic light Benefits of repairing the park Respond: How would you rationalize your decision? Discussion #6: Future Value Not all opportunity costs of a decision are subjective and/or difficult to value in monetary terms. Some can have hard and fast monetary values that can be easily calculated if we remember to include them. Example - Mr. and Mrs. Jones have been planning necessary renovations/upgrades to their home for quite a while and have been saving to make them happen. They did their homework and determined the cost for the materials and the manual labor to be \$2,700. They have also determined that the upgrades would have a useful life of 10 years. The benefits they perceive are entirely subjective, but the pride in their home and the increased comfort they will enjoy for the 10-year period far outweigh the \$2,700 price tag. Fair enough. There is an opportunity cost the Jones family is missing. This should be calculated and taken into account. This is what we call the future value of the \$2,700 if they decided not to renovate. The \$2,700 they have set aside for the upgrade could alternatively be placed in a safe investment over 5
6 the same 10-year period. When the 10 years is up, the original \$2,700 investment will have grown. By choosing the renovation, Mr. and Mrs. Jones will be giving up the \$2,700 and the value of their \$2,700 after 10 years of growth. So, the true cost of their renovations is not only the \$2,700 for materials and labor, but also the opportunity cost of the investment growth of the \$2,700. Their upgrade will cost them substantially more than \$2,700 if they include this cost. But, the question remains, how much more? To accurately determine the future value of their \$2,700, we turn to a formula called the Future Value Formula, also known as the Compound Interest Formula. It looks like this: F = P*(1 + r) n Where: F = the future value of an investment P = the present value r = the interest rate n = the time frame (in years) Example: What is the future value of \$1,000 invested at 6% for 8 years? F= Future Value P= \$1,000 r = 6% or.06 n = 8 F = P*(1 + r) n F = 1000*(1 +.06) 8 F = 1000*( ) F = 1, Hands-on Activity #1 Now you try. Calculate the Future Value of the Jones family s \$2,700 if Mr. and Mrs. Jones decide not to renovate their home and instead invest in a fixed, 10-year safe investment paying 2.5%. If Mr. and Mrs. Jones renovate, they will pay \$2,700 and enjoy the upgrade for 10 years but will have \$0.00 in 10 years. Question: If they choose to invest, instead of renovate, how much will their investment be worth 10 years from now? This amount should be added to the \$2,700 as an opportunity cost of finishing the renovations. 6
7 Hands-on Activity #2 CBA Activity Tom and Patty Conley purchased their dream home 10 years ago. They both work, so with two incomes and excellent credit, they qualified for and received a \$200,000 mortgage to purchase their dream home. Now, 10 years later, economic times have changed. The U.S. economy has suffered through tough economic challenges, including a severe housing market collapse. As a result, interest rates on today s mortgages are lower than when the Conleys purchased their house. The Conleys are considering refinancing their mortgage to take advantage of these low interest rates. They believe refinancing will give them a lower monthly payment, which will give them more of their money to use for other things. The refinancing will save lots of interest over the long term. CBA can be used to help the Conleys analyze their opportunity. First, it would be a good idea to define a couple of terms. Refinancing a mortgage means to pay off an existing mortgage with a new one, usually because the terms of the new mortgage are more favorable than the old one. Monthly Mortgage Payment is the amount of money paid each month to pay back a mortgage loan. The pure monthly mortgage payment consists of principal (the money borrowed) and interest (the price paid for borrowing the money), or P & I. Some monthly mortgage payments consist of P & I, plus another amount called escrow, which is an estimate of the property taxes and property insurance that will be needed to be paid on the home. Some lenders require the annual insurance and property tax estimates to be totaled and divided by 12 (monthly amount), then added to the monthly mortgage payment as a payment into an escrow account that the lender will use to pay the insurance and taxes when they become due. In the case of Mr. and Mrs. Conley, the lender did not require a monthly escrow payment and the Conleys are responsible for paying their insurance and taxes when they are due. Therefore, the Conleys monthly mortgage payment consists only of P & I. Amortization Table is a table showing the amount of interest and principal being applied to a mortgage loan each month. There are 360 monthly payments due in a 30-year mortgage (12 x 30 = 360 months). Each payment is the same amount but how the funds are used changes each month. For each payment, the lender calculates the amount of interest the borrower owes on the money he/she has borrowed. The lender subtracts that interest amount from the mortgage payment. Whatever 7
8 money is left over, the lender applies to the outstanding balance, which reduces the amount the borrower now owes. Next month then, when the lender calculates the amount of interest due, it will be a bit less because the borrower owes a bit less. So, the lender takes a smaller amount from the mortgage payment to pay for interest and applies a larger amount to pay back the amount owed. Each and every month then, the borrower owes a bit less than he/she did the previous month; therefore, more of the mortgage payment is being used to pay back the loan, and less of the payment is being used to pay interest. The exact breakdown of the amounts of interest and principal being used out of each payment is presented in a table called an Amortization Table. Discussion #7: To complete the calculations necessary for this activity, we will need the help of a mortgage loan calculator. There are plenty of fine calculators available on the web (any one of which would probably be fine to use). One calculator to use is found on BankRate.com. Exercise: Let s begin by identifying a couple of important numbers. The Conleys original mortgage loan was for \$200,000. It is a 30-year mortgage at 8.625% interest, and their monthly payment consists of only principal and interest. Using the mortgage loan calculator, answer these questions: What is the Conleys current monthly mortgage payment? If the Conleys keep their existing mortgage, how much total interest will they end up paying after 30 years? It has been exactly 10 years since the mortgage was taken out (120 months). How much do the Conleys still owe on their loan? How much interest have they paid so far over the 10 years? If they keep their existing mortgage, how much interest will they pay from now until the end? Cost/Benefit Analysis(CBA) is a comparative analysis of estimates of the costs and the benefits of undertaking a particular action in order to determine if taking the action is worthwhile. 8
9 The procedure to do an effective Cost/Benefit analysis is a four-step process: 1. Try to determine all of the costs associated with taking a particular action. Try to determine all of the benefits associated with taking that action. 2. Try to estimate the monetary value of each cost and total them. 3. Try to estimate the monetary value of each benefit and total them. 4. Finally, compare the total costs versus the total benefits to determine if the action is worth taking. Step 1: 1. Try to determine all of the costs associated with taking a particular action. Try to determine all of the benefits associated with taking that action. Refinancing a mortgage means to pay off an existing mortgage with a new one, usually because the terms of the new mortgage are more favorable than the old one. There are costs associated with refinancing a mortgage. The Conleys are considering refinancing their mortgage to take advantage of lower interest rates. They believe refinancing will give them a lower monthly payment, which will give them more of their money to use for other things. Refinancing will save lots of interest over the long term. Step 2: 2. Try to estimate the monetary value of each cost and total them. Refinancing costs vary from state to state and lender to lender. For our purposes, we will list costs typically found in any refinancing. Also, some of the costs are based on the amount borrowed and some are fixed amounts. Cost Explanation Amount Application fee Initial cost of determining if a borrower qualifies for a loan. Can \$125 include fee for checking credit report. Loan origination fee 1.5% of loan amount Lender s fee for documenting and preparing the loan. Points 2 points 1 point = 1% of loan amount Bank s profit for establishing the loan Appraisal fee A current appraisal of the home s \$350 value done by a licensed appraiser. 9
10 Inspection fee Home inspection by a licensed inspector to determine the home s condition \$200 In the table that follows, we will fill in fixed costs and ask you to calculate those based on the loan amount. The new loan amount is the amount the Conleys still owe on their existing mortgage. Costs of a refinance: Bank attorney fee Conleys attorney fee Title work Survey TOTAL COSTS Bank s attorney reviews and signs off on the loan agreement and may conduct the closing. Borrower s attorney reviews the documents to protect the borrower. Searching the property s history to determine if the ownership is clear. Also, insures against future claims against ownership. An updated survey of the property is required. Total cost of the refinance \$500 \$500 \$800 \$125 Step 3: 3. Try to estimate the monetary value of each benefit and total them. Look at the new loan. How much do the Conleys owe on their existing loan? (You answered this question earlier. This will be the new loan amount.) The new loan will be a 30-year fixed rate mortgage at 4.75% interest. What will the Conleys new monthly mortgage payment be? If the Conleys keep their new mortgage, how much total interest will they end up paying after 30 years? 10
11 Step 4: 4. Finally, compare the total costs versus the total benefits to determine if the action is worth taking. We have information about the existing 30-year mortgage loan. We have information about the cost to refinance this loan into a new loan. We have information about the new loan. Now, analyze the information and determine what the benefit of refinancing the loan will be for Mr. and Mrs. Conley. One benefit the Conleys anticipate from refinancing is a lower monthly mortgage payment. Will the refinanced mortgage free up money in their monthly budget for other things? What is the Conleys current monthly mortgage payment? What will be the Conleys new monthly mortgage payment? Subtract (if positive, this will be the Conleys monthly savings): Were Mr. and Mrs. Conley right? Is the monthly savings a substantial benefit? Break-Even Period The Conleys will be saving money on their mortgage each and every month. However, it cost them a substantial sum to refinance and get a new mortgage. How many months of savings will it take before the Conleys recover the cost of refinancing the loan? Total Costs Monthly Savings equals months Another benefit the Conleys expect is that refinancing will save them a substantial amount of interest over the long term. Will it? If the Conleys keep their existing mortgage, how much interest do they still have left to pay? How much interest will they pay with the new mortgage? - Subtract 11
12 The difference here is how much interest they will save by refinancing. Do you consider it substantial? Answer: It appears the Conleys expectations were correct. They will save a substantial amount of money each and every month on their mortgage payment. They will reimburse themselves for the cost of the refinance in a reasonable amount of time and will ultimately save a substantial amount of interest over the long term. Note: You may have noticed that there is an opportunity cost missing from the analysis: the opportunity cost of investing the money they paid for the refinance, instead of spending it on refinancing. Opportunity costs are the consequences of choosing one course of action over another. The Conleys needed to reduce monthly spending. Saving big on interest over the long term was not a choice, but a necessity. This calculation was not included in the calculation of costs. But, they really should know what it would be, so Please calculate the Future Value of investing the Total Costs figure from the table above in a safe investment yielding 2.5% interest for 30 years. Conclusion: Cost/Benefit Analysis is a practice most of us use all the time when making decisions, even if we do not realize we are utilizing the tactic. Cost/Benefit Analysis (CBA) is a comparative analysis of estimates of the costs and the benefits of undertaking a particular action in order to determine if taking the action is worthwhile. Just because you can do something, does not mean you should do it. Not every idea is a good one. Look before you leap. This is good advice, wouldn t you say? Following a particular course of action can sometimes simply feel right. Subjective benefits can be difficult to quantify; therefore, these make it difficult to prove that feeling like the right thing to do is the right thing to do. 12
13 Extension Activities: Can you think of a situation where you just know what the right course of action is, but cannot prove that it is merely by reducing the action to dollars and cents? Can you think of a situation where what seems to be the right thing to do turns out to be the wrong thing to do? Thank you for your attention. 13
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### The following is an article from a Marlboro, Massachusetts newspaper.
319 CHAPTER 4 Personal Finance The following is an article from a Marlboro, Massachusetts newspaper. NEWSPAPER ARTICLE 4.1: LET S TEACH FINANCIAL LITERACY STEPHEN LEDUC WED JAN 16, 2008 Boston - Last week | 10,389 | 47,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2018-51 | latest | en | 0.922249 |
https://convert-dates.com/days-from/450/2024/06/12 | 1,718,840,269,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00895.warc.gz | 156,504,598 | 4,324 | ## 450 Days From June 12, 2024
Want to figure out the date that is exactly four hundred fifty days from Jun 12, 2024 without counting?
Your starting date is June 12, 2024 so that means that 450 days later would be September 5, 2025.
You can check this by using the date difference calculator to measure the number of days from Jun 12, 2024 to Sep 5, 2025.
September 2025
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September 5, 2025 is a Friday. It is the 248th day of the year, and in the 36th week of the year (assuming each week starts on a Sunday), or the 3rd quarter of the year. There are 30 days in this month. 2025 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 09/05/2025, and almost everywhere else in the world it's 05/09/2025.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 450 weekdays from Jun 12, 2024, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Jun 12, 2024, which falls on a Wednesday. Counting forward, the next day would be a Thursday.
To get exactly four hundred fifty weekdays from Jun 12, 2024, you actually need to count 630 total days (including weekend days). That means that 450 weekdays from Jun 12, 2024 would be March 4, 2026.
If you're counting business days, don't forget to adjust this date for any holidays.
March 2026
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March 4, 2026 is a Wednesday. It is the 63rd day of the year, and in the 63rd week of the year (assuming each week starts on a Sunday), or the 1st quarter of the year. There are 31 days in this month. 2026 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 03/04/2026, and almost everywhere else in the world it's 04/03/2026.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date. | 925 | 2,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-26 | latest | en | 0.915864 |
http://mathbitsnotebook.com/Algebra1/StatisticsReg/ST2ScatterPlot.html | 1,542,364,494,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743007.0/warc/CC-MAIN-20181116091028-20181116113028-00201.warc.gz | 207,813,002 | 4,417 | Scatter Plots & Line of Best Fit MathBitsNotebook.com Terms of Use Contact Person: Donna Roberts
A scatter plot is a graph of plotted points that shows a relationship between two sets of quantitative data (referred to as bivariate data). Scatter plots are composed of "dots" (points) on a set of coordinate axes. Do NOT connect the dots!
Statisticians and quality control technicians spend a good deal of time gathering sets of data to determine if relationships exist between the sets. Scatter plots are a popular and effective way of graphing data to display patterns, trends, relationships and an occasional extraordinary value located apart from the other values. Let's see an example.
Does studying for that Final Exam really help your score? Does one event really affect the other?
The scatter plot at the right appears to show that the longer students studied, the higher their examination scores.
According to this survey of 15 students studying for the same examination, it appears that the answer to our initial question is "yes", studying does affect your score. At least, the answer is "yes", for this particular group of students.
NOTE: A scatter plot is not necessarily a function. It is often the case where the same x-value may have more than one corresponding y-value, such as (6,70) and (6,80).
Notice how the data in the graph resembles a straight line rising from left to right. When working with scatter plots, if is often useful to represent the data with the equation of a straight line, called a "line of best fit", or a "trend" line. Such a line may pass through some of the points, none of the points, or all of the points on the scatter plot.
To see how to prepare a line of best fit by hand and a line of best fit with a graphing calculator, click the link below.
What is the line of best fit for our "studying affects scoring" problem?
When finding the line of best fit "by hand", different students may arrive at different answers. So who's answer is the best? You will need a graphing calculator to get the "best" answer.
The graphing calculator computed the line of best fit shown at the right. The equation is:
y = 4.722392638x + 50.99539877
Based upon this equation, we can predict scores given any number of hours spent studying.
Interpolate:
If you are making predictions for values that fall within the plotted values, you are said to be interpolating. For this problem, our plotted values range from x = 1 to x = 9.
Example: Predict the final examination score of a student studying for 5½ hours.
(Substitute the number of hours into the equation for x.)
Score: approximately 77
Extrapolate: If you are making predictions for values that fall outside the plotted values, you are said to be extrapolating. Be careful when extrapolating. The further away from the plotted values you go, the less reliable is your prediction. For this problem, outside of the plotted values would be x greater than 9 or x less than 1.
Example: Predict the final examination score of a student studying for 12 hours.
(Substitute the number of hours into the equation for x.)
Score: approximately 108
WOW!!! Great score!!
But is it realistic? It is very likely that the top score is 100.
So, in addition to yielding less reliable predictions, extrapolating
may also give completely unrealistic predictions.
For calculator help with scatter plots click here.
NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". | 775 | 3,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-47 | longest | en | 0.947547 |
https://www.coursehero.com/file/73103/Astro-1HW-5-Solutions-Winter-08/ | 1,529,902,802,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867424.77/warc/CC-MAIN-20180625033646-20180625053646-00399.warc.gz | 795,092,305 | 64,274 | Astro 1:HW 5 Solutions (Winter 08)
# Astro 1:HW 5 Solutions (Winter 08) - UCSB Winter 2008 Astro...
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UCSB Winter 2008 Astro 1 - Homework #5 Solutions Kevin Moore 2/9/08 1 How can we use the Moon’s orbital motion, plus the distance an apple falls in one second, to determine the dependence of the gravitational force formula on the distance between two bodies? The basic idea here is to realize that the distance an object falls in one second is directly proportional to acceleration, which is in turn proportional to the force of gravity. You didn’t need to do the actual calculation to find this, but I’ll go through it and show you can deduce that the force of gravity goes as 1 /r 2 . You may know from physics that for constant acceleration, distance is given by d = d o + v o t + 1 2 at 2 Thus the distance an object falls in one second is d Earth = 1 2 a (1 s 2 ), indeed directly proportional to acceleration. We also know that F = ma , so d F for an object falling in one second. Since the force of gravity is directly proportional to mass (see prob. 2), it doesn’t matter whether it’s the Moon or an apple up there - they all fall at the same rate. First we’ll get the distance an object falls on Earth’s surface: here a = 9 . 8 m/s 2 so d = 4 . 9 m Getting the distance the Moon ’falls’ in 1 sec is more involved: First we need to find how far the Moon moves in its orbit in one second: It’s angular speed is (in radians/sec) ω = 2 π 27 . 3 days 1 day 24 hr 1 hr 3600 s = 2 . 66 × 10 - 6 rad sec The Earth-Moon distance is D moon = 384 , 400 km so the Moon travels about (384 , 400)(2 . 66 × 10 - 6 ) = 1 . 02 km in 1 sec.
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# If a and b are two-digit numbers that share the same digits,
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If a and b are two-digit numbers that share the same digits, [#permalink]
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If a and b are two-digit numbers that share the same digits, except in reverse order, then what is the sum of a and b?
(1) a-b=45
(2) The difference between the two digits in each number is 5.
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Re: Number Properties - DS [#permalink]
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25 May 2010, 07:04
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srichaks wrote:
Q:If a and b are two-digit numbers that share the same digits, except in reverse order, then what is the sum of a and b?
(1) a-b=45
(2) The difference between the two digits in each number is 5.
OA:E
Any two digit integer $$n$$ can be expressed as $$n=10x+y$$, where $$x$$ and $$y$$ are digits, $$x>0$$.
Given: $$a=10x+y$$ and $$b=10y+x$$. Q: $$a+b=11(x+y)=?$$
(1) $$a-b=10x+y-10y-x=45$$ --> $$x-y=5$$. Multiple choices for x and y: (9,4), (8, 3), (7, 2), (6, 1). Not sufficient.
(2) $$x-y=5$$. Same info as above. Not sufficient.
(1)+(2) No new info. Not sufficient.
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Re: Number Properties - DS [#permalink]
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25 May 2010, 18:42
Thanks Bunuel!. Great explanation.
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Re: Number Properties - DS [#permalink]
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29 May 2010, 07:09
I dont understand the "except in reverse order" part???????
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Re: Number Properties - DS [#permalink]
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29 May 2010, 07:20
hardnstrong wrote:
I dont understand the "except in reverse order" part???????
It means that if digits in $$a$$ are $$xy$$, then in $$b$$ the digits are $$yx$$ (in reverse order).
According to the solution above there are 5 pairs of $$a$$ and $$b$$ possible satisfying both statements (94 and 49, 83 and 38, 72 and 27, 61 and 16). So multiple answer to $$a+b$$. Not sufficient.
Hope it's clear.
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Re: Number Properties - DS [#permalink]
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30 May 2010, 06:27
Bunuel wrote:
hardnstrong wrote:
I dont understand the "except in reverse order" part???????
It means that if digits in $$a$$ are $$xy$$, then in $$b$$ the digits are $$yx$$ (in reverse order).
According to the solution above there are 5 pairs of $$a$$ and $$b$$ possible satisfying both statements (94 and 49, 83 and 38, 72 and 27, 61 and 16). So multiple answer to $$a+b$$. Not sufficient.
Hope it's clear.
This is the confusing part in this question............ except in reverse order means they cannot be in reverse order (so if a = xy then b cannot be yx)
Dont you think "except in reverse order" takes question is completely different direction, or should it be "in reverse order"
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Re: Number Properties - DS [#permalink]
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30 May 2010, 06:40
hardnstrong wrote:
Bunuel wrote:
hardnstrong wrote:
I dont understand the "except in reverse order" part???????
It means that if digits in $$a$$ are $$xy$$, then in $$b$$ the digits are $$yx$$ (in reverse order).
According to the solution above there are 5 pairs of $$a$$ and $$b$$ possible satisfying both statements (94 and 49, 83 and 38, 72 and 27, 61 and 16). So multiple answer to $$a+b$$. Not sufficient.
Hope it's clear.
This is the confusing part in this question............ except in reverse order means they cannot be in reverse order (so if a = xy then b cannot be yx)
Dont you think "except in reverse order" takes question is completely different direction, or should it be "in reverse order"
Not a perfect wording - agree. "except in reverse order" here means "but in reverse order".
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Re: If a and b are two-digit numbers that share the same digits, [#permalink] 15 Sep 2018, 10:32
Display posts from previous: Sort by | 1,508 | 5,352 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-22 | latest | en | 0.806459 |
https://www.jiskha.com/display.cgi?id=1251913186 | 1,532,270,944,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593302.74/warc/CC-MAIN-20180722135607-20180722155607-00328.warc.gz | 918,525,014 | 5,283 | # Very easy math
posted by Very easy Math
my teacher told me that the inverse of addition was subtraction and that the inverse of subtraction was addition...
could you prove it to me
-x = (-1)x
((-1)x)^-1
I don't see how I'm suppose to get + x by taking the inverse of -x i've always been told in math to just to the opposite of subbtraction which is addition but my teacher is telling me that is a lie and that it's really the inverse of subtraction is addition but I don't see the reasoning behind it
basically can you prove to me that the opposite of subractiion is addition and vise versa??? by taking the inverses????
I don't get it...
like I can prove that the opposite of multiplication is division by taking the inverse and can prove it just by defintion
(5x = 2)5^-1 = x = 5^-1 (2)
that's how you prove that relationship is really just inverses but what about addition and subtraction how are the inverse relationships...???
Thansk
1. Writeacher
Use numbers.
6 + 2 = 8
Therefore, 8 - 6 = 2 or 8 - 2 = 6
How can you state those relationships in abstract terms?
2. Count Iblis
If x is some number and:
x + y = 0
then y is called an inverse (w.r.t. addition) of x
Then it follows from the same definition that x is an inverse of y. Now, what you need to prove is that inverses are unique. I.e. if for some given x
x + y = 0
and also
x + z = 0
you necessarily have y = z.
So, it then follows that the inverse of the inverse of x is x and it can't be anything else than x.
Then, if we denote the inverse of x by
-x, we can prove that:
-x = (-1)*x
THis is because:
x + (-1)*x =
1*x + (-1)*x =
(1 + (-1))*x =
0*x = 0
Here we have used that -1 is the inverse of 1.
So, (-1)*x satisfies the criterium the inverse of x which we always denote as
-x must satisfy and therefore
-x = (-1)*x
Then the fact that taking twice the inverse yields the same number implies that:
(-1)*(-1) = 1
3. Very easy Math
i agree with all of it but still don't see how
X + B = C
we can simply solve for B by simply multiplying the whole equation by B^-1 which we note as -B because????
(X + B = C)B^-1
B cancels out
X = B^-1 C
what allows us to say that B^-1 is equal to -B
4. bobpursley
You are confusing terms:
Inverse is not the reciprocal. You are using reciprocal (B^-1) is reciprocal.
Now it is confusing, because the inverse operation to multiplication is division, and the inverse to division is multiplication
http://www.mathsisfun.com/definitions/inverse-operation.html
Watch the usage to "inverse", a lot of folks really mean reciprocal when they use it.
## Similar Questions
1. ### Another Problem...PLz Help COunt Iblis
Statements Reasons 1. 3x - 7 = -4 2. 3x - 7 + 7 = -4 + 7 3. 3x + 0 = -4 + 7 4. 3x + 0 = 3 5. 3x = 3 6. (1/3) 3x = 3 (1/3) 7. (1/3) 3x = 1 8. 1x = 1 9. x = 1 Is this problem about explaining each step?
2. ### Math
Using the definition of subtraction to determine the subtraction equations that are related to each of the following addition equation: a.8+7=n b.14+x=25 c.r+s=t my answers are this: a. n-7=8 b.x=25-14 c.t-s=r did I do these right?
3. ### MATH 156
What models can be used to help explain the concepts of addition and subtraction of rational numbers?
4. ### math
What inverse operation is needed for the first step in solving the equation 12+5x=22?
5. ### math
An eighth-grade student claims she can prove that subtraction of integers is commutative. She points out that if a and b are integers, then a-b = a+ -b. Since addition is commutative, so is subtraction. What is your response?
6. ### Math--1 question
Name the inverse operations for each of the following operations. a. addition b. division c. multiplication d. subtraction | 1,057 | 3,713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-30 | latest | en | 0.969874 |
http://www.us-lotteries.com/Kentucky/Pick_3/Pick_3-Triples.asp | 1,386,827,975,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164554256/warc/CC-MAIN-20131204134234-00020-ip-10-33-133-15.ec2.internal.warc.gz | 609,294,417 | 6,513 | Frequency and draw dates of all triples (three identical digits) for Kentucky Lottery Pick 3 game. Is time to play triples?
# Kentucky Lottery Pick 3 Triples Data and Statistics
Updated on Wednesday, December 11, 2013 Triples are more appealing to play since the player does not have to worry about in what order to play them. The down side is that they are not drawn that often and, except when they are really late, it is hard to predict which one to play. Many Pick 3 players play all the ten triple numbers when they think that triples are due to come. On this page...
## How often are Kentucky Lottery Pick 3 triples drawn?
The chances of a triple-digit Pick 3 number being drawn are the same as any other straight number, that is, one in one thousand. However, the chances of any one of the ten triples of Pick 3 being drawn are one in 100. This means that we should expect a triple number once every 100 draws. All the winning triple-digit numbers of the Kentucky Lottery Pick 3 game drawn, as well as the draw dates, are listed in the table below. The last column of the list designated by NDAPT (Number of Draws After Previous Triple) represents the number of draws elapsed between triples and is intended to give you an idea of how often the triples are showing up in reality.
## Kentucky Lottery Pick 3 triples never drawn
There are no Kentucky Lottery Pick 3 triples never-drawn. ALL Pick 3 triples are drawn at least once.
## Analysis Results of Kentucky Lottery Pick 3 triples
Here is a summary of the analysis results of Kentucky Lottery Pick 3 triples. Theoretically, since the 10 triples make 1% of the total number of Pick 3 numbers, the percentage of triples drawn is expected to be close to 1%. Also, the average number of draws between triples is expected to be close to 100. Moreover, check if the number of draws since most recent triple number is well over 100. These three results could be an indication of whether or not to expect triples soon.
Kentucky Lottery Pick 3 Triples summary
( As of Wednesday, December 11, 2013)
• Total number of draws = 10238
• Number of triples drawn = 108
• Percent of total draws that triples are drawn = 1.055%
• Average number of draws between triples = 94
• Minimum number of draws between triples = 1
• Maximum number of draws between triples = 526
• Long-time-no-see triple number = 111(Sun Aug 16, 2009)
• Most recent triple number = 999(Wed Nov 13, 2013)
• Draws since most recent triple number = 53
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## Summary of individual Kentucky Lottery Pick 3 triples
The following table summarizes the data for each triple number. It lists, for each Kentucky Lottery Pick 3 triple number, the number of times it is drawn, the date when it was last drawn, and how many draws ago is it since its last appearance.
Triple Number Number of Times Drawn Date Last Drawn Number of draws since last seen 000 13 Sun Jul 1, 2012 981 111 10 Sun Aug 16, 2009 2931 222 9 Sun Jul 22, 2012 942 333 14 Tue Sep 17, 2013 158 444 13 Sat Jun 9, 2012 1022 555 10 Wed Mar 20, 2013 495 666 10 Wed Jul 24, 2013 261 777 11 Wed Sep 8, 2010 2211 888 10 Fri May 20, 2011 1739 999 8 Wed Nov 13, 2013 53
## List of all Kentucky Lottery Pick 3 triples drawn
Number Draw Date NDAPT* 111 Fri Nov 27, 1998 145 333 Thu Jan 7, 1999 41 000 Thu Jan 28, 1999 21 111 Thu Feb 11, 1999 14 444 Mon Mar 29, 1999 51 555 Fri Apr 2, 1999 8 333 Thu Jun 10, 1999 128 999 Wed Sep 22, 1999 184 999 Sat Oct 30, 1999 71 888 Tue Nov 23, 1999 44 222 Wed Aug 30, 2000 522 777 Sat Sep 16, 2000 32 333 Thu Nov 2, 2000 87 999 Fri Jan 5, 2001 118 222 Sun Jan 7, 2001 4 888 Mon Jan 15, 2001 15 888 Wed Feb 7, 2001 42 888 Thu Apr 5, 2001 107 444 Wed Jun 6, 2001 114 444 Tue Jun 19, 2001 24 777 Wed Aug 15, 2001 107 888 Wed Aug 29, 2001 26 000 Thu Sep 6, 2001 14 444 Thu Jan 24, 2002 261 111 Sat Jan 26, 2002 4 333 Tue Feb 26, 2002 56 000 Tue Mar 26, 2002 53 666 Tue Apr 9, 2002 25 111 Fri Jun 28, 2002 150 000 Mon Jul 22, 2002 44 000 Tue Oct 1, 2002 132 666 Fri Nov 29, 2002 110 777 Sat Dec 14, 2002 27 777 Sat Dec 28, 2002 26 444 Sat Jan 11, 2003 26 777 Sat Feb 15, 2003 65 222 Wed Feb 19, 2003 7 222 Mon May 5, 2003 140 333 Thu Oct 2, 2003 279 000 Mon Oct 27, 2003 46 555 Wed Nov 26, 2003 55 666 Wed Dec 3, 2003 13 555 Mon Dec 29, 2003 49 222 Wed Jan 28, 2004 55 666 Thu Mar 11, 2004 80 333 Sat Apr 24, 2004 83 777 Sun Apr 25, 2004 1 000 Wed Jul 14, 2004 148 555 Sun Aug 8, 2004 47 888 Thu Sep 23, 2004 86 777 Mon Oct 18, 2004 45 333 Wed Feb 2, 2005 200 777 Sat Feb 12, 2005 18 000 Fri Feb 25, 2005 25 222 Fri Mar 11, 2005 25 111 Mon Apr 25, 2005 84 444 Wed Apr 27, 2005 3 000 Sun Jun 26, 2005 112 666 Sat Jul 16, 2005 37 999 Wed Feb 15, 2006 398 777 Thu Jun 22, 2006 235 444 Thu Aug 31, 2006 130 444 Fri Sep 1, 2006 2 999 Sat Mar 3, 2007 340 555 Fri Jul 27, 2007 270 777 Mon Aug 13, 2007 32 555 Tue Aug 21, 2007 14 444 Tue Sep 4, 2007 26 333 Mon Sep 17, 2007 24 111 Tue Oct 16, 2007 55 999 Fri Nov 2, 2007 31 333 Sat Feb 2, 2008 172 000 Tue Apr 1, 2008 109 666 Fri Jun 27, 2008 161 222 Sat Jul 5, 2008 16 444 Thu Jul 17, 2008 21 333 Sat Sep 13, 2008 108 111 Fri Oct 24, 2008 76 222 Sun Nov 23, 2008 56 333 Wed Feb 11, 2009 149 111 Mon Mar 9, 2009 48 444 Tue Apr 21, 2009 80 111 Sun Jun 21, 2009 113 888 Wed Jul 15, 2009 44 000 Wed Jul 29, 2009 27 111 Sun Aug 16, 2009 33 555 Mon Sep 7, 2009 40 666 Wed Oct 7, 2009 56 666 Mon Nov 16, 2009 75 888 Tue Dec 8, 2009 41 333 Fri Jun 4, 2010 331 888 Sat Jun 12, 2010 14 444 Sat Sep 4, 2010 156 777 Wed Sep 8, 2010 7 555 Mon Oct 18, 2010 74 888 Fri May 20, 2011 398 000 Mon Feb 27, 2012 526 444 Sat Jun 9, 2012 191 000 Sun Jul 1, 2012 41 222 Sun Jul 22, 2012 39 666 Mon Jul 23, 2012 2 999 Wed Jul 25, 2012 3 555 Thu Jan 17, 2013 328 555 Wed Mar 20, 2013 114 333 Sat Jun 8, 2013 149 666 Wed Jul 24, 2013 85 333 Tue Sep 17, 2013 103 999 Wed Nov 13, 2013 105
* NDAPT = Number of Draws After Previous Triple, where the very first entry is from the starting date
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• WE DO NOT SELL ANY LOTTERY, WE DO NOT BUY LOTTERY NUMBERS ON BEHALF OF ANYONE; this is only an information and guide web site. | 2,589 | 7,590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2013-48 | latest | en | 0.958253 |
https://lucatrevisan.wordpress.com/2009/11/12/the-large-deviation-of-fourwise-independent-random-variables/ | 1,670,195,192,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710980.82/warc/CC-MAIN-20221204204504-20221204234504-00085.warc.gz | 394,905,545 | 33,637 | # The Large Deviation of Fourwise Independent Random Variables
Suppose ${X_1,\ldots,X_n}$ are mutually independent unbiased ${\pm 1}$ random variables. Then we know everything about the distribution of
$\displaystyle | X_1 + \ldots + X_N | \ \ \ \ \ (1)$
either by using the central limit theorem or by doing calculations by hand using binomial coefficients and Stirling’s approximation. In particular, we know that (1) takes the values ${1,\ldots, 1/\sqrt N}$ with probability ${\Theta(1/\sqrt N)}$ each, and so with constant probability (1) is at most ${O(\sqrt N)}$.
The last statement can be proved from scratch using only pairwise independence. We compute
$\displaystyle \mathop{\mathbb E} \left| \sum_i X_i \right|^2 = N$
so that
$\displaystyle \mathop{\mathbb P} \left[ \left|\sum_i X_i \right| \geq c \cdot \sqrt N \right] = \mathop{\mathbb P} \left[ \left|\sum_i X_i \right|^2 \geq c^2 \cdot N \right] \leq \frac 1 {c^2}$
It is also true that (1) is at least ${\Omega(\sqrt N)}$ with constant probability, and this is trickier to prove.
First of all, note that a proof based on pairwise independence is not possible any more. If ${(X_1,\ldots,X_N)}$ is a random row of an Hadamard matrix, then ${\sum_i X_i = N}$ with probability ${1/N}$, and ${\sum_i X_i =0}$ with probability ${1-1/N}$.
Happily, four-wise independence suffices.
Lemma 1 Suppose ${X_1,\ldots,X_n}$ are four-wise independent ${\pm 1}$ random variables. Then
$\displaystyle \mathop{\mathbb P} \left[ \left| \sum_i X_i \right| \geq \Omega(\sqrt N) \right] \geq \Omega(1)$
This is probably well known to a lot of people, but the only place where I have seen this result is a paper by Andreev et al. that I have mentioned before.
Here is a slightly simplified version of their argument. We compute
$\displaystyle \mathop{\mathbb E} \left| \sum_i X_i \right|^2 = N$
and
$\displaystyle \mathop{\mathbb E} \left| \sum_i X_i \right|^4 = 3N^2 - 2N$
So we have (from here on is our simplified version)
$\displaystyle \mathop{\mathbb E} \left( \left(\sum_i X_i \right)^2 - 2N \right)^2 = 3N^2 - 2N \leq 3N^2$
and
$\displaystyle \mathop{\mathbb P} \left[ \left| \sum_i X_i \right| \leq \frac 13 \sqrt N \right]$
$\displaystyle = \mathop{\mathbb P} \left[ \left| \sum_i X_i \right|^2 \leq \frac 19 N \right]$
$\displaystyle \leq \mathop{\mathbb P} \left[ \left| \left| \sum_i X_i \right|^2 - 2N \right| \geq \frac {17}{9} N \right]$
$\displaystyle = \mathop{\mathbb P} \left[ \left( \left| \sum_i X_i \right|^2 - 2N \right)^2 \geq \frac {289}{81} N^2 \right]$
$\displaystyle \leq \frac 3 {\frac {289}{81}} = .80408$
Thus
$\displaystyle \mathop{\mathbb P} \left[ \left| \sum_i X_i \right| \geq \frac 13 \sqrt N \right] \geq .159$
Following the same proof, but replacing each “${X_i}$” with “${a_i X_i}$” and every “${N}$” with “${\sum_i a_i^2}$”, we get the more general version
Lemma 2 Suppose ${X_1,\ldots,X_N}$ are four-wise independent ${\pm 1}$ random variables and let ${a_1,\ldots,a_N}$ be any real numbers. Then
$\displaystyle \mathop{\mathbb P} \left[ \left| \sum_i a_i X_i \right| \geq \frac 13 \sqrt {\sum_i a_i^2 } \right] \geq .159$
In particular,
$\displaystyle \mathop{\mathbb E} \left| \sum_i a_i X_i \right| \geq \frac 1 {20} \sqrt {\sum_i a_i^2 }$
A related version that was useful in our work on distinguishers for pseudorandom generators is
Corollary 3 Let ${H}$ be a four-wise independent family of functions ${h: S \rightarrow \{-1,1\}}$ and ${g: S \rightarrow {\mathbb R}}$ be any real valued function. Then
$\displaystyle \mathop{\mathbb E}_{h\sim H} \left| \sum_{x\in S} h(x)g(x) \right| \geq \frac 1{20\cdot \sqrt{|S|}} \sum_x | g(x) |$
To briefly show how this relates to our work on distinguishers, suppose ${G: \{ 0,1 \}^{n-1} \rightarrow \{ 0,1 \}^n}$ is a length-increasing function, and that we want to construct a function ${D}$, computable by a circuit as small as possible, such that
$\displaystyle \mathop{\mathbb P}_{z\in \{ 0,1 \}^{n-1}} [ D(G(z)) =1 ] - \mathop{\mathbb P}_{x\in \{ 0,1 \}^n} [D(x)=1] \geq \epsilon$
Then this is equivalent to proving
$\displaystyle \mathop{\mathbb E}_{z \in \{ 0,1 \}^{n-1}} (-1)^{D(G(z))} - \mathop{\mathbb E}_{x\in \{ 0,1 \}^n} (-1)^{D(x)} \geq 2 \epsilon \ \ \ \ \ (2)$
and, if we set
$\displaystyle g(x) := \mathop{\mathbb P}_{z\in \{ 0,1 \}^{n-1}} [G(z) = x] - \frac 1 {2^n}$
then (2) is equivalent to
$\displaystyle \sum_{x\in \{ 0,1 \}^n} (-1)^{D(x)} \cdot g(x) \geq 2 \epsilon \ \ \ \ \ (3)$
Now, fix a simple partition ${\cal P}$ of ${\{ 0,1 \}^n}$ into ${1,600 \epsilon^{2} 2^n}$ subsets each of size ${(40 \epsilon)^{-2}}$, for example based on the value of the first ${\log 1,600 \epsilon^2 2^n}$ bits.
Fix an efficiently computable four-wise independent family ${H}$ of functions ${h: \{ 0,1 \}^n \rightarrow \{ 0,1 \}}$. Then we have, for every set ${S}$ in the partition
$\displaystyle \mathop{\mathbb E}_{h\sim H} \left| \sum_{x \in S} (-1)^{h(x)} g(x) \right| \geq \frac{1}{20\sqrt {|S|}} \sum_{x\in S} |g(x)| = 2\epsilon \sum_{x\in S} |g(x)|$
and if we sum over all sets in the partition we have
$\displaystyle \mathop{\mathbb E}_{h\sim H} \sum_{S\in {\cal P}} \left| \sum_{x \in S} (-1)^{h(x)} g(x) \right| \geq 2\epsilon \sum_{x\in \{ 0,1 \}^n} |g(x)| \geq 2\epsilon$
In particular, there is a function ${h_0 \in H}$ such that
$\displaystyle \sum_{S\in {\cal P}} \left| \sum_{x \in S} (-1)^{h_0(x)} g(x) \right| \geq 2\epsilon$
To finish the argument, define the function ${c: \{ 0,1 \}^n \rightarrow \{ 0,1 \}}$ such that ${c(z)=1}$ iff ${\sum_{x\in S} (-1)^{h_0(x)} g(x) < 0}$, where ${S}$ is the set that ${z}$ belongs to. Then
1. ${c(x)}$ depends only on the first ${\log O(\epsilon^{-2} 2^n)}$ bits of the input, and so it can be computed by a circuit of size ${O(\epsilon^{-2}2^n)}$
2. for every ${S}$ we have
$\displaystyle \left| \sum_{x \in S} (-1)^{h_0(x)} g(x) \right| = \sum_{x\in S} (-1)^{c(x)} \cdot (-1)^{h_0(x)} g(x)$
So, in conclusion
$\displaystyle \sum_{x \in \{ 0,1 \}^n} (-1)^{c(x) + h_0(x)} g(x) \geq 2\epsilon$
meaning that setting ${D(x) := c(x) \oplus h(x)}$ satisfies (3), and gives us a distinguisher obtaining advantage ${\epsilon}$ and having circuit complexity ${O(\epsilon^{-2} 2^n)}$.
## 15 thoughts on “The Large Deviation of Fourwise Independent Random Variables”
1. Similar results have been obtained by Serge Vaudenay using his “decorrelation theory” [1] in the context of the design of secure block ciphers. More precisely, his paper [2] about iterated distinguishers shows that a decorrelation of order $2d$ is sufficient to resist an attack of order $d$. An open question is whether this is a necessary condition.
[1] S. Vaudenay, “Decorrelation: A Theory for Block Cipher Security”, J. Cryptology, 16(4), 249-286, 2003.
[2] S. Vaudenay, “Resistance against general iterated attacks”, Proceedings of Eurocrypt’99, pp. 255-271, LNCS 1592, Springer, 1999.
2. Hi Luca,
Lemma 1 was also proven by Bonnie Berger back in SODA ’91. See her paper, “The Fourth Moment Method”.
3. Yeah, what James said, with Wikipedia’s $Z = (\sum a_i X_i)^2$. Indeed, if the $X_i$‘s in Lemma 2 are any $(2, 4, \eta)$-hypercontractive random variables, then so too is $\sum a_i X_i$; then Paley-Zygmund gives the lower bound (with the RHS depending on $eta$).
4. Thanks, Jelani, and her calculations are much tighter, in Lemma 2 she has a factor of $\sqrt 3$ instead of 20 (Andreev, Clementi and Rolim had 96)
5. Hi Pascal, what is the similarity that you see? Vaudenay shows that if a family of permutations is close to k-wise independent under certain norms then certain classes of attackers have small advantage in distinguishing a function from the family from a truly random permutation. (And some of those conditions are necessary in order to upper bound the advantage of distinguishers from those classes.)
Here we have a completely arbitrary distribution over $n$-bit strings that is far from uniform in total variation distance (for concreteness I have talked about the output of a length-increasing generator) and we show that, within a set S, a distinguisher sampled from a 4-wise independent family of functions has typically advantage about $1/\sqrt {|S|}$ times the best possible advantage within the set S.
(So if we are willing to invest in a look-up table of size $2^n/|S|$ to know the sign of the distinguishing probability in each set of size $|S|$ into which we partition $\{0,1\}^n$, we can get advantage $1/\sqrt{|S|}$ overall. This is not an iterated attack because there are no queries to make, it’s just a case analysis.)
So here it’s the attacker, not the construction that is coming from a $k$-wise independent distribution, and we are proving a lower bound, not an upper bound, on the distinguishing probability.
6. I had a question along similar lines on sum of $k$-wise independent variables. Assume $X_1, \ldots, X_n$ are $k$-wise independent $\pm 1$ random variables. What can be said about the following expression :
$\displaystyle {\mathbb P} \left[\left|\sum_{i} X_i\right| > \frac{n^{0.5}}{100}\right]$
By Berry – Esseen theorem, if $k=n$, then it is $1-o(1)$ and the above discussion shows that $k \geq 4$ implies that it is bigger than a constant. But can we get a better bound if $4?
7. Hi Anindya,
If the $X_i$ are k-wise independent for $k = O(1/ \epsilon^2)$, then
$\displaystyle Pr \left[ \left|\sum_i X_i \right| > \sqrt{n} \right] >= 1 - O(\epsilon). \ \ \ (*)$
Note when $1/\epsilon^2 = n$, this gives a lower bound of $1 - O(1/\sqrt{n} )$, which is what Berry-Esseen would tell you.
(*) follows from Section A.5 of http://web.mit.edu/minilek/www/papers/lp.pdf (last paragraph, which refers back to an earlier argument in the paper ^^;), though I regret that recovering (*) from what’s written there might take a bit of sorting through our notation, since the paper is really about something else. You can also get (*) from “Bounded Independence Fools Halfspaces” (Diakonikolas et al., FOCS ’09) with only a slightly larger k of $O(\log^2(1/\epsilon)/\epsilon^2)$.
8. I don’t understand: if the $X_i$ are mutually independent, then from the Berry-Esseen theorem we have that for every threshold $t$
$\displaystyle \left| {\mathbb P} \left[ \left|\sum_i X_i \right| \geq t \sqrt n \right] - {\mathbb P} \left[ \left| N \right| \geq t \right] \right| \leq O \left( \frac 1 {\sqrt n} \right)$
where $N$ is a normal random variable with expectation zero and variance 1. But the probability that $|N|$ is more than $1/100$ is not $1-o(1)$, it is a constant bounded away from 1 (and from zero).
9. I think Luca is right though I intended to put $n^{0.45}$ instead of $\sqrt{n}$. In that case, Berry-Esseen does guarantee that the sum of $X_i$ exceeds $n^{0.45}$ in absolute value with $1-o(1)$ probability. However, a calculation like above does not give anything beyond a constant probability.
Jelani: Thanks for answering. I assume what you guys prove is that if $k=1/\epsilon^2$, then the difference from the Normal distribution is at most $\epsilon$. That sounds great though I have not yet read your paper.
10. What I should have actually said is, as long as the $X_i$ are $k$-wise independent for $k = \Omega(1/\epsilon^2)$ then
$$\mathrm{Pr}[|\sum_i X_i| > \epsilon \sqrt{n}] \ge 1 – O(\epsilon)$$
I forgot to type that inner $\epsilon$ above. Sorry for the confusion.
11. In general, proving concentration bounds under bounded independence has been known for a while. What I believe was not known until recently, is that bounded independence also suffices for anti-concentration bounds. More generally, the following holds:
The Berry-Esseen version of the CLT also applies under bounded independence.
In particular, in Luca’s formulation, if the X_i’ s in the sum are \Omega(1/eps^2)-wise independent, then the distance between the CDF’s is at most 1/\sqrt{n} + eps. (**)
The degree of independence required for such a statement to hold is actually Omega(1/eps^2), i.e. the above bound is optimal up to constant factors.
As Jelani pointed out, (**) follows as a special case of a result by Gopalan, Jaiswal, Servedio, Viola and myself, but the bound there on the degree of independence required is slightly weaker (up to a log^2 factor).
12. That had been incredibly imformative. I appreciate you all the material. | 3,894 | 12,320 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 112, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-49 | latest | en | 0.834333 |
https://www.sanfoundry.com/high-voltage-engineering-questions-answers-waveshape-control/ | 1,725,803,738,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651002.87/warc/CC-MAIN-20240908115103-20240908145103-00811.warc.gz | 948,885,339 | 17,404 | # High Voltage Engineering Questions and Answers – Waveshape Control
This set of High Voltage Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Waveshape Control”.
1. An impulse voltage generator has a generator capacitance C1 of 0.02 μF, load capacitance C2 of 0.001 μF, front resistance of R1=400 Ω and tail resistance of R2=2980 Ω. The tail time is __________
a) 49.68 μs
b) 54.26 μs
c) 30.92 μs
d) 76.54 μs
Explanation: Tail time is the discharging time. The capacitances C1 and C2 may be considered to be in parallel and discharging occurs through R1 and R2. The tailing time is given by the equation
t=.7 (R1+R2)(C1+C2)
t=.7(400+2980) (.02+0.001) x 10-6
t=49.68 μs.
2. An impulse voltage generator has a generator capacitance C1 of 0.02 μF, load capacitance C2 of 0.001 μF, front resistance of R1=400 Ω and tail resistance of R2=2980 Ω. The charging time is __________
a) 1.9 μs
b) 1.14 μs
c) 2.0 μs
d) 0.5 μs
Explanation: The time taken for charging is approximately three times the time constant of the circuit.
T=3.0 R1 $$\frac{C_1 C_2}{C_1 + C_2}$$
T=3.0 x 400 x $$\frac{.02 * .001}{0.02+0.001}$$ x 10-6
T=1.14 μs.
3. Wave shape control is flexible and independent.
a) True
b) False
Explanation: Wave shape control is flexible and independent. Wave shape gets changed with the change in test object. But, the circuit is simple and hence is advantageous.
4. The transient voltage may be __________
a) an oscillatory wave
b) a damped oscillatory wave
c) critically damped wave
d) an oscillatory wave or a damped oscillatory wave
Explanation: The transient voltage may be an oscillatory wave or a damped oscillatory wave. The frequency of the waves ranges from few hundred hertz to few kilo hertz. Transient voltage can also be considered as a slow rising impulse.
5. What is the time constant of the wave shaping circuit?
a) $$t=R_1 \frac{C_1 C_2}{C_1 + C_2}$$
b) $$t=2 R_1 \frac{C_1 C_2}{C_1 + C_2}$$
c) $$t=3R_1 \frac{C_1 C_2}{C_1 + C_2}$$
d) $$t=R_1 \frac{C_1 C_2}{2C_1 + C_2}$$
Explanation: The time constant of the wave shaping circuit is $$=R_1 \frac{C_1 C_2}{C_1 + C_2}$$. R1 is the resistance, C1 is the generator capacitance and C2 is the load capacitance. The circuit inductance is very small and hence is not taken into account.
6. What is the time taken for charging the capacitance in the wave shape control circuit?
a) $$t=R_1 \frac{C_1 C_2}{C_1 + C_2}$$
b) $$t=2 R_1 \frac{C_1 C_2}{C_1 + C_2}$$
c) $$t=3R_1 \frac{C_1 C_2}{C_1 + C_2}$$
d) $$t=R_1 \frac{C_1 C_2}{2C_1 + C_2}$$
Explanation: The time taken for charging the capacitance in the wave shape control circuit is $$t=3R_1 \frac{C_1 C_2}{C_1 + C_2}$$. R1 is the resistance, C1 is the generator capacitance and C2 is the load capacitance. The time taken for charging is thrice the time constant of the circuit. The circuit inductance is very small and hence is not taken into account.
7. What is the effective capacitance in wave shaping circuit?
a) $$C=\frac{C_1 C_2}{C_1 + C_2}$$
b) $$C=2 \frac{C_1 C_2}{C_1 + C_2}$$
c) $$C=\frac{C_1 C_2}{2C_1 + 2C_2}$$
d) $$C=3 \frac{C_1 C_2}{C_1 + C_2}$$
Explanation: The effective capacitance in wave shaping circuit is $$C=\frac{C_1 C_2}{C_1 + C_2}$$, C1 is the generator capacitance and C2 is the load capacitance. The circuit inductance is very small and hence is not taken into account.
8. What is the time taken for discharging in the wave shape control circuit?
a) $$t=(R_1+R_2) (C_1 + C_2)$$
b) $$t=.7(R_1+R_2) (C_1 + C_2)$$
c) $$t=3R_1 \frac{C_1 C_2}{C_1 + C_2}$$
d) $$t=R_1 \frac{C_1 C_2}{C_1 + C_2}$$
Explanation: The time taken for discharging in the wave shape control circuit is $$t=.7(R_1 + R_2)(C_1 + C_2)$$. It is also known as the tail time of the circuit. R1 and R2 is the resistance, C1 is the generator capacitance and C2 is the load capacitance. The time taken for charging is thrice the time constant of the circuit. The circuit inductance is very small and hence is not taken into account.
9. The generator capacitance is dependent on the design of the generator.
a) True
b) False
Explanation: The generator capacitance is dependent on the design of the generator. The load capacitance is also fixed for a given design of the generator and test object. So to obtain the desired wave shape only the resistance can be controlled.
10. An impulse voltage generator has a load capacitance C2 of 0.001 μF, front resistance of R1=400 Ω, tail resistance of R2=2980 Ω and tail time 49.68 μs.The generator capacitance C1 of is __________
a) 0.001 μF
b) 0.2 μF
c) 0.02 μF
d) 0.03 μF
Explanation: Tail time is the discharging time. The capacitances C1 and C2 may be considered to be in parallel and discharging occurs through R1 and R2. The tailing time is given by the equation $$t=.7 (R_1 + R_2)(C_1 + C_2)$$
Hence, the generator capacitance C1 can be found out by the following steps.
49.68 = $$.7(400+2980)(C_1 + 0.001)$$
C1 = 0.02 μF.
Sanfoundry Global Education & Learning Series – High Voltage Engineering.
To practice all areas of High Voltage Engineering, here is complete set of Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
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# 1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60%
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Joined: 15 Jan 2018
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1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% [#permalink]
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31 May 2019, 11:00
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1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% alcohol. If x > y, how many integer values can x take?
A. 10
B. 20
C. 35
D. 13
E. 30
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Re: 1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% [#permalink]
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31 May 2019, 12:39
3
x.........y
....60....
1.........3
$$\frac{x-60}{60-y}$$=3/1
x+3y=240...(1)
x=240-3y=3(80-y)
We have 2 constraints
1. x=<100
2. x>y
Also, x will be a multiple of 3
x can be 99,96.....63
At x=60, y=60....Hence x can't take value equal to 60 or lower than 60.
Total number of integer values x can take= [$$\frac{99-63}{3}$$]+1=13
DisciplinedPrep wrote:
1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% alcohol. If x > y, how many integer values can x take?
A. 10
B. 20
C. 35
D. 13
E. 30
##### General Discussion
Manager
Joined: 25 Mar 2018
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GMAT 1: 640 Q50 V26
GMAT 2: 640 Q50 V26
GMAT 3: 650 Q50 V28
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GMAT 5: 730 Q50 V40
GPA: 4
WE: Analyst (Manufacturing)
Re: 1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% [#permalink]
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31 May 2019, 21:30
2
Question is clear but OA is surprising.
The equation on which question depends is
x + 3y = 4*60
So we can start x with 61, which is integer value. We don't bother about Y. Y can take any value integer or not.
So IMHO answer is (100-61)+1 , which is 40.
Posted from my mobile device
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9983
Location: Pune, India
Re: 1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% [#permalink]
### Show Tags
13 Jun 2019, 03:51
1
DisciplinedPrep wrote:
1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% alcohol. If x > y, how many integer values can x take?
A. 10
B. 20
C. 35
D. 13
E. 30
1 part x% alcohol + 3 parts y% alcohol -> 60% alcohol
Since x > y, x must be greater than 60% and y must be less than 60% so that they average out to 60.
The maximum value of x can be 100% (the entire solution can be alcohol).
So x can take any integer value from 61 to 100.
But it seems that the intent of the question is that x and y both should be integer values.
y --- 60 --------- x
Since we are mixing 1 part x with 3 parts y, average will be 1 unit away from y and 3 units away from x.
So if y = 59, x = 63
if y = 58, x = 66
and so on...
x will take values from 63% to 99% i.e. (3*21) to (3*33)
This means (33 - 21 = )12 + 1 = 13 multiples of 3
Check this post for the scale method of weighted avgs:
https://www.veritasprep.com/blog/2011/0 ... -averages/
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Re: 1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% [#permalink]
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31 May 2019, 20:43
nick1816 wrote:
x.........y
....60....
1.........3
$$\frac{x-60}{60-y}$$=3/1
x+3y=240...(1)
x=240-3y=3(80-y)
We have 2 constraints
1. x=<100
2. x>y
Also, x will be a multiple of 3
x can be 99,96.....63
At x=60, y=60....Hence x can't take value equal to 60 or lower than 60.
Total number of integer values x can take= [$$\frac{99-63}{3}$$]+1=13
DisciplinedPrep wrote:
1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% alcohol. If x > y, how many integer values can x take?
A. 10
B. 20
C. 35
D. 13
E. 30
I don't get it. x and y are concentration in percent. 3 and 1 are the value of units.
the 2 equations will be
1x + 3y = ?
x+y=60.
So we have to find to the value of x for which it is greater than y and the first equation gives a valid number.
right?
Manager
Joined: 11 Aug 2017
Posts: 59
Re: 1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% [#permalink]
### Show Tags
31 May 2019, 23:14
nick1816 wrote:
x.........y
....60....
1.........3
$$\frac{x-60}{60-y}$$=3/1
x+3y=240...(1)
x=240-3y=3(80-y)
We have 2 constraints
1. x=<100
2. x>y
Also, x will be a multiple of 3
x can be 99,96.....63
At x=60, y=60....Hence x can't take value equal to 60 or lower than 60.
Total number of integer values x can take= [$$\frac{99-63}{3}$$]+1=13
DisciplinedPrep wrote:
1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% alcohol. If x > y, how many integer values can x take?
A. 10
B. 20
C. 35
D. 13
E. 30
Why x has to be less than or equ. 100? i could not understand. please guide??
Manager
Joined: 01 Oct 2018
Posts: 112
Re: 1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% [#permalink]
### Show Tags
12 Jun 2019, 17:51
nick1816 wrote:
x.........y
....60....
1.........3
$$\frac{x-60}{60-y}$$=3/1
x+3y=240...(1)
x=240-3y=3(80-y)
We have 2 constraints
1. x=<100
2. x>y
Also, x will be a multiple of 3
x can be 99,96.....63
At x=60, y=60....Hence x can't take value equal to 60 or lower than 60.
Total number of integer values x can take= [$$\frac{99-63}{3}$$]+1=13
DisciplinedPrep wrote:
1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% alcohol. If x > y, how many integer values can x take?
A. 10
B. 20
C. 35
D. 13
E. 30
Why x has to be less than or equ. 100? i could not understand. please guide??
Maximum value of aclhogol in unit is 100%
Re: 1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% [#permalink] 12 Jun 2019, 17:51
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After the knowledge of 2d geometry, it is time to add a new dimension to the geometry after that it is called 3d geometry. This is just an introduction but details of this chapter is in the next class. This chapter is for basics. This chapter consists of coordinate axis and coordinate planes in three dimensions, coordinates of a point, distance between two points and section formula.
Download pdf of NCERT Solutions for Class Mathematics Chapter 12 Introduction to 3 dimensional Geometry
Download pdf of NCERT Examplar with Solutions for Class Mathematics Chapter 12 Introduction to 3 dimensional Geometry
### Exercise 1
• Q1 A point is on the x-axis. What are its y-coordinates and z-coordinates? Ans: If a point is on the x-axis, then its y-coordinates and z-coordinates are zero. Q2 A point is in the XZ-plane. What can you say about its y-coordinate? Ans: If a point is in the XZ plane, then its y-coordinate is zero. Q3 Name the octants in which the following points lie: (1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (–4, 2, –5), (–4, 2, 5), (–3, –1, 6), (2, –4, –7) Ans: The x-coordinate, y-coordinate, and z-coordinate of point (1, 2, 3) are all positive. Therefore, this point lies in octant I. The x-coordinate, y-coordinate, and z-coordinate of point (4, –2, 3) are positive, negative, and positive respectively. Therefore, this point lies in octant IV. The x-coordinate, y-coordinate, and z-coordinate of point (4, –2, –5) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII. The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, –5) are positive, positive, and negative respectively. Therefore, this point lies in octant V. The x-coordinate, y-coordinate, and z-coordinate of point (–4, 2, –5) are negative, positive, and negative respectively. Therefore, this point lies in octant VI. The x-coordinate, y-coordinate, and z-coordinate of point (–4, 2, 5) are negative, positive, and positive respectively. Therefore, this point lies in octant II. The x-coordinate, y-coordinate, and z-coordinate of point (–3, –1, 6) are negative, negative, and positive respectively. Therefore, this point lies in octant III. The x-coordinate, y-coordinate, and z-coordinate of point (2, –4, –7) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII. | 660 | 2,401 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-30 | latest | en | 0.774587 |
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# solve it - simple mathematics!
Vikrama Sanjeeva
Ranch Hand
Posts: 760
A person was about to die, and before dying he wrote his Will which went as
follows ...
"I have 17 Camels, and I have three sons. Divide my Camels in such a way,
that My eldest son gets half of them, the second one gets 1/3rd of the
total and my youngest son gets 1/9th of the total number of Camels"
How will you divide, so that every son get the right number of camels which father wrote in will?
Bye,
Viki.
Scott Johnson
Ranch Hand
Posts: 518
1st son: 17 * 1/2 = 8.500 ==> round up ==> 9
2nd son: 17 * 1/3 = 5.667 ==> round up ==> 6
3rd son: 17 * 1/9 = 1.889 ==> round up ==> 2
Checking: 9 + 6 + 2 = 17
The fathers will has been satisifed because all sons received the proper fraction (plus a little more.)
Shyju Lakshman
Greenhorn
Posts: 4
Actually i have one camel which i shall add to the total . now there are 17+1 = 18 camels.
1st son ==> 18/2 = 9 camels
2nd son ==> 18/3 = 6 camels
3rd son ==> 18/9 = 2 camels
9+6+2= 17 camels.. one camel is left which is mine.. so i am happy, and the father is happy and the sons are happy
Ryan McGuire
Ranch Hand
Posts: 1083
4
According to the problem statement, that's an invalid solution. The will specifically says, "Divide up my camels in such a way that my eldest son gets half of them..." Adding your camel to the original 17 is going against the father's request.
Besides, the father may have realized that he has allocated only 17/18 of his 17 camels and wants the remaining 17/18 of a camel to be used to pay the executor of his estate (you). Mmmm... a half ton of camel jerky.
[ August 07, 2006: Message edited by: Ryan McGuire ]
Vikrama Sanjeeva
Ranch Hand
Posts: 760
Originally posted by Shyju Lakshman:
Actually i have one camel which i shall add to the total . now there are 17+1 = 18 camels.
1st son ==> 18/2 = 9 camels
2nd son ==> 18/3 = 6 camels
3rd son ==> 18/9 = 2 camels
9+6+2= 17 camels.. one camel is left which is mine.. so i am happy, and the father is happy and the sons are happy
100% | 659 | 2,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2016-44 | longest | en | 0.957755 |
geoffprewett.com | 1,713,841,324,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00481.warc.gz | 13,714,626 | 5,522 | ## Of Bezier Curves and Teapots
Bezier curves always sounded like something super-advanced, something that only mathematically-minded, skilled professionals who wrote Path Routines for operating systems wrote. After much searching and reading, I found out that one of the reasons Bezier curves are so popular is that they are so easy. Unfortunately, a lot of the articles out there are heavy on the math but light on the implementation.
A search on Bezier curves will quickly inform you that two French guys named De Casteljau and Bezier independently invented the Bezier curve in order to design cars, which have all sorts of curves which are not easily represented by normal functions. You can find plenty of discussion on De Castlejau’s algorithm, along with a nice animation cribbed from Wikipedia, but they inevitably end with “but don’t use this in real life.” And you can also find the equation for Bezier’s algorithm. But pseudocode? Nowhere.
Eventually I realized that nobody gives pseudocode because the the equation is the algorithm:
n Pi is the i-th control point B(u) is a Bernstein coefficient P(u) = Σ Bin(u) Pi i = 0
``````
def bezierCurveAt(u, points):
p = [0.0, 0.0] # assuming 2D control points
# Note that n in the summation is len(points) - 1.
# Also note that in Python, xrange(0, m) goes from 0, 1, ... m - 1.
for i in xrange(0, len(points)):
B = bernstein(i, len(points) - 1, u)
p[0] += B * points[i][0]
p[1] += B * points[i][1]
return p
def calcCurve(controlPoints, nSegments):
points = []
for i in xrange(0, nSegments + 1):
u = float(i) / float(nSegments)
points.append(bezierCurveAt(u, controlPoints))
``````
So what are these Bernstein coefficients? Well, from a coding standpoint, they are really easy:
Bin(u) = n! ui (1 - u)n-i i! (n - i)!
``````
def bernstein(i, n, u):
binomial = float(math.factorial(n)) / (float(math.factorial(i)) * float(math.factorial(n - i)))
return binomial * math.pow((1.0 - u), n - i) * math.pow(u, i)
``````
### The Theory
That is not very satisfying from an understanding standpoint, however. Over the years I have observed that there are a bunch of magic functions that show up in math. Can’t integrate a Gaussian function? No problem, use the error function, problem magically solved! When I studied Physics, we were happily deriving the equations of the spherical model of a hydrogen atom (well, the professor was happily deriving it, I was copying Greek letters as fast as possible, and reading the book after the lecture to figure out what was going on). The equations cannot be solved with the methods of solving differential equations familiar to undergraduate students. No problem, Legendre polynomials to the rescue! We spent a week deriving the hydrogen atom, but nobody bothered explaining Legendre polynomials. The book, which was quite good, said something to the effect of “conveniently, it happens that this equation can be solved with Legendre polynomials.” We, however, are not going to take that easy road.
Bernstein coefficients belong to the Bernstein polynomials, which are named after Sergei Bernstein, who created them (or discovered, depending on your philosophical perspective on mathematics) to prove that any curve can be approximated by polynomials. Basically it is the same idea as a Fourier series, which approximates any periodic (cyclical) function as a sum of sines or cosines. A Fourier series approximates a periodic function as
f(x) ≈ A + B sin(αx) + C sin(2αx) + D sin(3αx) + ... Xn sin(nαx)
(where A, B, etc. are appropriate constants). Bernstein polynomials do the same sort of thing, but for any function, not just periodic ones:
f(x) ≈ Ax0 + Bx1 + Cx2 + ... Xn xn
It is only valid over the range [0, 1], but that is not a huge problem, just scale and translate your function appropriately. Approximating a function is pretty much exactly the use case that Bezier was trying to solve—a digital approximation to the smooth curves of cars—so it is not surprising that Bernstein polynomials show up in Bezier curves. Indeed, one would almost expect it.
### Bezier Surfaces
Bezier curves turn out to be so easy that we still have time to look at Bezier surfaces, which are basically a 3D mesh, instead of a 2D line. I recently wanted a model of the famous Utah Teapot for computer graphics (the so-called “teapotahedron”), but it turns out it is defined as a series of Bezier surfaces. I was completely unable to find a teapot model that used vertices that I could just send to glDrawArrays. Since I was intimidated by the fancy name “bezier curve,” I was going to compile Mesa and modify glEvalMesh2 to spit out the vertices that glutSolidTeapot generated. Hah! Too much effort. Let’s just write a short Python script.
n m P(u, v) = Σ Σ Bin(u) Bjm(v) Pij i = 0 j = 0
``````
def bezierSurfaceAt(u, v, points, nUPoints, nVPoints):
p = [0.0, 0.0, 0.0] # assuming 3D control points
for j in xrange(0, nVPoints):
for i in xrange(0, nUPoints):
Bi = bernstein(i, nUPoints - 1, u)
Bj = bernstein(j, nVPoints - 1, v)
cp = points[j * nUPoints + i]
p[0] += Bi * Bj * cp[0]
p[1] += Bi * Bj * cp[1]
p[2] += Bi * Bj * cp[2]
return p
``````
This gives us the vertices, but we still need the normals. The normal is simply the tangent in the u direction crossed with the tangent in the v direction. Since the tangent of a curve is simply its derivative,
N(u, v) = ∂ P(u, v) ⨉ ∂ P(u, v) ∂u ∂v
It turns out that the derivative of a Bernstein polynomial is a sum of lower order Bernstein polynomials:
∂ Bin(t) = n [ Bi-1n-1(t) - Bin-1(t)], where B-1n-1(t) = Bnn-1(t) = 0 ∂t
Thus our normal is:
n m n m N(u, v) = Σ Σ n [Bi-1n-1(u) - Bin-1(u)] Bjm(v) Pij ⨉ Σ Σ Bin(u) m [Bj-1m-1(v) - Bjm-1(v)] Pij i = 0 j = 0 i = 0 j = 0
So we can calculate our normals with a straightforward implementation of the equation:
``````
def bezierNormalAt(u, v, points, nUPoints, nVPoints):
uTangent = [0.0, 0.0, 0.0]
for j in xrange(0, nVPoints):
for i in xrange(0, nUPoints):
Bi1 = bernstein(i - 1, nUPoints - 2, u)
Bi2 = bernstein(i, nUPoints - 2, u)
Bj = bernstein(j, nVPoints - 1, v)
a = nUPoints * (Bi1 - Bi2) * Bj
cp = points[j * nUPoints + i]
uTangent[0] += a * cp[0]
uTangent[1] += a * cp[1]
uTangent[2] += a * cp[2]
vTangent = [0.0, 0.0, 0.0]
for j in xrange(0, nVPoints):
for i in xrange(0, nUPoints):
Bi = bernstein(i, nUPoints - 1, u)
Bj1 = bernstein(j - 1, nVPoints - 2, v)
Bj2 = bernstein(j, nVPoints - 2, v)
a = nUPoints * Bi * (Bj1 - Bj2)
cp = points[j * nUPoints + i]
vTangent[0] += a * cp[0]
vTangent[1] += a * cp[1]
vTangent[2] += a * cp[2]
return normalized(crossProduct(uTangent, vTangent))
``````
I have packaged this up in a short Python script (teapot.py) which will generate the vertices, normals, and vertex ids for the Utah teapot. These are suitable for passing directly to glDrawElements() with GL_VERTICES. | 1,966 | 6,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-18 | latest | en | 0.838371 |
https://www.gradesaver.com/textbooks/math/trigonometry/CLONE-68cac39a-c5ec-4c26-8565-a44738e90952/chapter-5-trigonometric-identities-section-5-6-half-angle-identities-5-6-exercises-page-243/55 | 1,726,790,068,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00192.warc.gz | 719,565,180 | 12,959 | ## Trigonometry (11th Edition) Clone
$\frac{sin~x}{1+cos~x} = tan~\frac{x}{2}$
$\frac{sin~x}{1+cos~x}$ When we graph this function, we can see that it looks like the graph of $~~tan~\frac{x}{2}$ We can verify this algebraically: $\frac{sin~x}{1+cos~x} = \frac{sin~(\frac{x}{2}+\frac{x}{2})}{1+cos~(\frac{x}{2}+\frac{x}{2})}$ $\frac{sin~x}{1+cos~x} = \frac{sin~\frac{x}{2}~cos~\frac{x}{2}+cos~\frac{x}{2}~sin~\frac{x}{2}}{1+cos~\frac{x}{2}~cos~\frac{x}{2}-sin~\frac{x}{2}~sin~\frac{x}{2}}$ $\frac{sin~x}{1+cos~x} = \frac{2~sin~\frac{x}{2}~cos~\frac{x}{2}}{1-sin^2~\frac{x}{2}+cos^2~\frac{x}{2}}$ $\frac{sin~x}{1+cos~x} = \frac{2~sin~\frac{x}{2}~cos~\frac{x}{2}}{cos^2~\frac{x}{2}+cos^2~\frac{x}{2}}$ $\frac{sin~x}{1+cos~x} = \frac{2~sin~\frac{x}{2}~cos~\frac{x}{2}}{2~cos^2~\frac{x}{2}}$ $\frac{sin~x}{1+cos~x} = \frac{sin~\frac{x}{2}}{cos~\frac{x}{2}}$ $\frac{sin~x}{1+cos~x} = tan~\frac{x}{2}$ | 447 | 895 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-38 | latest | en | 0.23313 |
https://www.vedantu.com/jee-main/in-one-dimensional-motion-instantaneous-speed-physics-question-answer | 1,708,663,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474360.86/warc/CC-MAIN-20240223021632-20240223051632-00747.warc.gz | 1,085,942,786 | 27,154 | Courses
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# In one dimensional motion, instantaneous speed satisfies the condition, $0\le v<{{v}_{0}}$, whenA) The displacement in time $T$ must always take non-negative valuesB) The displacement $x$ in time $T$ satisfies $-{{v}_{0}}T< x <{{v}_{0}}T$ C) The acceleration is always a non-negative numberD) The motion has no turning points
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Hint: Let us begin by understanding one-dimensional motion and instantaneous velocity. One dimensional motion applies to objects moving in straight lines. Speed is a measure of how quickly an object is moving along this line; it is a scalar quantity. Velocity is the speed with a direction, making it a vector. If an object’s velocity changes at a constant rate with time, the object is said to be accelerating. When studying motion along one dimension, there are only two possible directions for the velocity and acceleration vectors to point in.
Formula Used:
$v=\dfrac{dx}{dt}$
Complete step by step solution:
We know that velocity is the derivative of displacement.
Mathematically, we can say that $v=\dfrac{dx}{dt}$
Transposing sides, we can say that $dx=v.dt$
Integrating both sides of the above equation, we get
\begin{align} & \int\limits_{0}^{x}{dx}={{v}_{0}}\int\limits_{0}^{T}{dt} \\ & \Rightarrow \left[ x \right]_{0}^{x}={{v}_{0}}\left[ t \right]_{0}^{T} \\ & \Rightarrow x={{v}_{0}}T \\ \end{align}
As discussed in the hint, velocity in one-dimensional motion may be positive or negative. If the velocity is negative, the displacement will be negative as well.
So, displacement cannot always take non-negative values but can vary between the negative and the positive values, like $-{{v}_{0}}T < x < {{v}_{0}}T$.
Hence we can say that option (B) is the correct answer.
Note: We have already seen that option (A) cannot be the correct answer as the displacement can vary between positive and negative values. Similarly, acceleration is the rate of change of velocity and if the velocity is negative, the acceleration would be negative as well. Hence acceleration cannot always be a non-negative number. Whenever the velocity changes from a positive value to a negative value or vice-versa, there is a change in the direction of motion or a turning point is encountered. Hence option (D) is also incorrect. | 600 | 2,411 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-10 | longest | en | 0.826387 |
https://richardvigilantebooks.com/what-are-the-conditions-for-paralleling-two-3-phase-transformer/ | 1,685,366,329,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644855.6/warc/CC-MAIN-20230529105815-20230529135815-00048.warc.gz | 545,092,190 | 10,675 | # What are the conditions for paralleling two 3 phase transformer?
## What are the conditions for paralleling two 3 phase transformer?
Condition for Parallel Operation of Three Phase Transformers The line voltage ratio of the transformers must be same. The transformers should have equal per unit leakage impedance. (You may read per unit system) The ratio of equivalent leakage reactance to equivalent resistance should be same for all the transformers.
Can you connect 2 transformers in parallel?
It is possible to connect transformers that have different polarities in parallel. The supply line polarity determines the primary polarity of the transformer. The primary polarity determines the secondary polarity of the transformer.
### What is parallel operation of transformers?
When we connect the primary windings of two transformers to a common supply voltage and the secondary windings of both the transformers to a common load, this type of connection of transformer is said to be the parallel operation of transformers.
How do two transformers work in parallel?
If two or more transformers are connected in parallel, then load sharing per centage between them is according to their rating. If all are of same rating, they will share equal loads.
#### Which condition for parallel operation of 3 phase transformer is incorrect?
Do not connect the transformers at incorrect polarity in parallel, incorrect polarity results a dead short circuit. Hence, both the transformer will be damaged.
How do you connect two transformers together?
Series Connected Secondary Transformer As the two windings are connected in series, the same amount of current flows through each winding, then the secondary current is the same at 2.5 Amps. So for a series connected secondary, the output in our example above is rated at 24 Volts, 2.5 Amps.
## Which polarity is used for parallel operation of transformer?
In perfect parallel operation of two or more transformers, current in each transformer would be directly proportional to the transformer capacity, and the arithmetic sum would equal one-half the total current. Any combination of positive and negative polarity transformers can be used.
When two transformers operating parallel the following conditions must be satisfied?
When two or more transformers run in parallel, they must satisfy the following conditions for satisfactory performance. These are the conditions for parallel operation of transformers. Same voltage ratio of transformer. Same percentage impedance.
### Can current transformers be connected in series?
A current transformers primary coil is always connected in series with the main conductor giving rise to it also being referred to as a series transformer. The nominal secondary current is rated at 1A or 5A for ease of measurement.
What is the resistance between primary and secondary in transformer?
In a transformer the coils are not electricaly connected therefore the resistance is ideally infinite.
#### Can a parallel transformer be used in a new transformer?
Parallel in a new transformer. It is sometimes more practical to parallel in a new transformer as there is minimal downtime to the operation. Three conditions must be met before you can connect transformers in parallel. 1. The transformers must have the same primary and secondary voltage ratings.
How is the impedance of Transformers in parallel related?
From the above two statements it can be said that impedance of transformers running in parallel are inversely proportional to their MVA ratings. In other words percentage impedance or per unit values of impedance should be identical for all the transformers run in parallel.
## How is CEMF produced in paralleling Transformers?
Even though the voltages induced in the secondaries of the transformers are AC, the same circulating currents flow in each of the secondary windings. Any current flowing in the secondary of the transformer must be matched by a current in the primary so that the proper CEMF is produced in the primary windings.
Why do you parallel a subtractive polarity transformer?
When making the connections, you must observe the terminal polarity of the transformers. This still allows you to parallel a subtractive-polarity transformer with an additive-polarity transformer if you ensure that the connection terminals have the same instantaneous polarity. Figure 10. Circulating currents | 798 | 4,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-23 | latest | en | 0.911427 |
http://www.komal.hu/verseny/feladat.cgi?a=honap&h=201203&t=mat&l=en | 1,532,243,128,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593051.79/warc/CC-MAIN-20180722061341-20180722081341-00536.warc.gz | 470,215,987 | 9,098 | Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# KöMaL Problems in Mathematics, March 2012
Show/hide problems of signs:
## Problems with sign 'K'
Deadline expired on April 10, 2012.
K. 331. The task is to cross a desert with a car. The width of the desert is 600 km. The capacity of the petrol tank of the car is only enough for 400 km, and the car cannot carry fuel in any other way. The car can travel with an average speed of 50--60 km/h. It needs to start out at 8 a.m. and it is to arrive on the other side by 8 p.m. What is the minimum number of such cars needed, for one of them to be able to cross the desert, while the others return back? It is possible to transfer fuel from one car to another.
(6 pont)
solution (in Hungarian), statistics
K. 332. The sum of a few positive integers is 20. Their product is the number X. What is the maximum possible value of X?
(6 pont)
solution (in Hungarian), statistics
K. 333. The three interior angles of a triangle followed by the three exterior angles, in the appropriate order, form six consecutive terms of an arithmetic sequence. Find the measures of the angles of the triangle in degrees. (In an arithmetic sequence, the difference of the consecutive terms is constant, that is, the sequence increases uniformly.)
(6 pont)
solution (in Hungarian), statistics
K. 334. Alma measured and tabulated the distances (in steps) between fruit trees in the garden. How many steps is it from the walnut tree to the pear tree?
Apple Pear Walnut Almond 5 12 35 Apple 0 13 40
(6 pont)
solution (in Hungarian), statistics
K. 335. A peculiar calculator only has four buttons on it: (eight plus root seven), (addition), (reciprocal) and (equals) (see the figure).
The calculator always carries out the operations with exact values, and it can also store the current value as a constant if the button is pressed twice. That is, in subsequent calculations whenever the button is pressed, the number is incremented by that value any number of times. (E.g. if the buttons are pressed, it will display ). Prove that the result of the following sequence of operations is 1:
... (56 times)
... (15 times)
(6 pont)
solution (in Hungarian), statistics
K. 336. The concave 16-sided polygon in the figure is divided into four congruent squares and eight congruent rhombuses. The smaller angle of the rhombuses is 45 degrees. Given that the longer diagonal of the rhombuses is 10.7 cm, find the area of the 16-sided polygon.
(6 pont)
solution (in Hungarian), statistics
## Problems with sign 'C'
Deadline expired on April 10, 2012.
C. 1115. Show that n2(n2-1)(n2-n-2) is divisible by 48 for all natural numbers n.
(5 pont)
solution (in Hungarian), statistics
C. 1116. Each side of a convex quadrilateral is divided into eight equal parts. Each dividing point is connected to the corresponding dividing point on the opposite side, as shown in the figure. The small quadrilaterals obtained in this way are coloured black and white in a chessboard pattern. Prove that the total area of the black quadrilaterals equals the total area of the white quadrilaterals.
(5 pont)
solution (in Hungarian), statistics
C. 1117. We have drawn a rectangle on squared paper, such that (its sides are lattice lines and) it consists of n small lattice squares. Prove that if the half of the number of lattice points on the boundary of the rectangle is added to the number of lattice points in its interior, and 1 is subtracted from the sum then the result will be n.
(5 pont)
solution (in Hungarian), statistics
C. 1118. Solve the equation on the set of real numbers.
(5 pont)
solution (in Hungarian), statistics
C. 1119. Two identical set squares(of 30, 60 and 90 degrees) are fixed together along their shorter legs of length 10 cm, as shown in the figure. Then an isosceles set square is placed on top. Is there enough room for a tennis ball of radius 3.2 cm in the interior of the tetrahedron obtained in this way?
(5 pont)
solution (in Hungarian), statistics
## Problems with sign 'B'
Deadline expired on April 10, 2012.
B. 4432. There is a 98×98 chessboard on the monitor of a computer. With the mouse, you can select a rectangle bounded by lattice lines of the chessboard. If you click on the rectangle, each chessboard field inside it will change to the opposite colour. What is the minimum number of clicks needed to make the whole chessboard have the same colour.
(5 pont)
solution (in Hungarian), statistics
B. 4433. Solve the equation (1+x)8+(1+x2)4=82x4.
(3 pont)
solution (in Hungarian), statistics
B. 4434. Prove that every natural number not divisible by 10 can be multiplied by an appropriate natural number, such that the product is a palindromic number in decimal notation.
(4 pont)
solution (in Hungarian), statistics
B. 4435. T is the foot of the altitude drawn from vertex A of an acute-angled triangle ABC. The midpoint of side BC is F. The centres of the squares drawn over the sides AB and AC on the outside are K and L, respectively. Prove that KTFL is a cyclic quadrilateral.
(Suggested by Sz. Miklós, Herceghalom)
(4 pont)
solution (in Hungarian), statistics
B. 4436. Let x, y, z denote positive integers, such that . Prove that .
(Suggested by J. Mészáros, Jóka)
(4 pont)
solution (in Hungarian), statistics
B. 4437. Given the centres of the circumscribed circle and two escribed circles of a triangle, construct the triangle.
(4 pont)
solution (in Hungarian), statistics
B. 4438. The angle bisectors of a triangle ABC intersect the opposite sides at the points A1, B1 and C1, respectively. For what triangles is it true that ?
(Matlap, Cluj-Napoca -- Kolozsvár)
(3 pont)
solution (in Hungarian), statistics
B. 4439. Given two different points B and C in the plane, determine the locus of those points A for which the altitude drawn from vertex A of the triangle ABC is the geometric mean of the line segments BC+AC and BC-AC.
(3 pont)
solution (in Hungarian), statistics
B. 4440. On a winter's day, an absent minded mathematician went on a walk with his old poodle along a long straight alley. He was so absorbed in his thoughts that when he finally thought of his dog, he had to realize that it was not near him. In the snowfall, he did not see further than 5 metres. He did not see the dog in front of him, he did not see it behind either, and he did not know which direction it had gone. After a little thinking, he went to look for his poodle. The old poodle can only walk half as fast as its master. The mathematician chose a searching strategy, such that the following condition should hold for the smallest possible value of the constant c: if the dog is at a distance x from him then he needs to cover a distance of at most cx to find it. What is this smallest value of c?
(5 pont)
solution (in Hungarian), statistics
B. 4441. The areas of the faces of a tetrahedron are a, b, c, d. The angle enclosed by the faces of areas a and b is , the angle of b and c is , and the angle of c and a is . Prove that d2=a2+b2+c2-2ab.cos -2bc.cos -2ca.cos .
(5 pont)
solution (in Hungarian), statistics
## Problems with sign 'A'
Deadline expired on April 10, 2012.
A. 557. Show that the positive integers can be coloured with three colours in such a way that the equation x+y=z2 has no solution (x,y,z) consisting of distinct numbers with the same colour.
(Kolmogorov's Cup, 2011; a problem by F. Petrov and I. Bogdanov)
(5 pont)
statistics
A. 558. Prove that there exists a constant C>0 for which the following statement holds: if n is a positive integer and are sets such that every two of them has at least two, and every three of them has at most three elements in common, then N<Cn2.
(Proposed by: Z. Gyenes, Budapest)
(5 pont)
statistics
A. 559. The incircle of triangle ABC is k. The circle kA touches k and the segments AB and AC at at A', AB and AC respectively. The circles kB, kC and the points B', C' are defined analogously. The second intersection point of the circles A'B'AB and A'C'AC, other than A', is K. The line A'K meets k at R, other than A'. Prove that R lies on the radical axis of the circles kB and kC.
(Kolmogorov's Cup, 2011; a problem by F. Ivlev)
(5 pont)
statistics | 2,184 | 8,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-30 | latest | en | 0.915814 |
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