i am new and i have not received any help with my algebra homework please help me as soon as possible please 
1.) which function should be used to when evaluating the function at x = 3?a.)b.)2.) a computer training institute has 625 students paying $400 as the course fee. their research shows that for every $20 reduction in their fee, they will attract another 50 students. which table represents this situation?a.) b.)c.)d.)please please explain and solve please
1) p(x) = 1/x2 + 1 if x ≤ 3 it is clear that x= 3 is included, so you have to use this function.2) loo k at table d, and compare the given information with the data written inside the table.
given:- function f(x) = x3 – 12x2 + 36x + 17theorem:- let f be a differentiable real function defined on an open interval (a,b).(i) if f’(x) > 0 for all , then f(x) is increasing on (a, b)(ii) if f’(x) < 0 for all , then f(x) is decreasing on (a, b)algorithm:-(i) obtain the function and put it equal to f(x)(ii) find f’(x)(iii) put f’(x) > 0 and solve this inequation.for the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.here we have,f(x) = x3 – 12x2 + 36x + 17⇒ ⇒ f’(x) = 3x2 – 24x + 36for f(x) lets find critical point, we must have⇒ f’(x) = 0⇒ 3x2 – 24x + 36 = 0⇒ 3(x2 – 8x + 12) = 0⇒ 3(x2 – 6x – 2x + 12) = 0⇒ x2 – 6x – 2x + 12 = 0⇒ (x – 6)(x – 2) = 0⇒ x = 2, 6clearly, f’(x) > 0 if x < 2 and x > 6and f’(x) < 0 if 2 < x < 6thus, f(x) increases on (–∞, 2) ∪ (6, ∞)and f(x) is decreasing on interval x ∈ (2, 6) One difficulty is that it's far from clear what the ideal dropout rate should be. Probably not zero, for two reasons:

 1. Even the best students sometimes find that their interests change over time, in ways nobody could have predicted. Dropping out may become a quite sensible choice, in which case it's harmful for the university to discourage it.
2. It's reasonable for a university to give someone a chance even if it's not certain that they will succeed. The only way to get a really low drop-out rate is to admit just the applicants that are obviously destined for success. The top few schools can get away with this, but if everyone tried it, then many talented candidates would be shut out from graduate school.

 So some attrition is OK, and some is bad but may be a necessary consequence of policies that are on the whole good. The question then becomes how you distinguish these kinds from needless and damaging attrition, and then how you minimize that kind.

 Advisors play a key role here, because some are much better than others at being supportive or motivational. However, at least in mathematics, most advisors don't supervise very many students, so the numbers often aren't large enough to see patterns clearly.

 I'm sure people have studied this problem, and perhaps identified best practices for addressing it. However, in my experience any studies or solutions are not especially influential (at least in the few math departments I've been in). Most discussions are at the stage of trying to figure out whether there is a real problem and if so why, rather than what to do about it. y = 3x

It can be observed that x = −1, y = −3 and x = 1, y = 3 are solutions of the above equation. Therefore, the solution table is as follows.



| x | -1 | 1 |
| y | -3 | 3 |

The graph of the above equation is constructed as follows.

![](https://www.sarthaks.com/?qa=blob&qa_blobid=15681428413536225923) students are given **form 1098-t** to show either payments paid to or billed from an educational facility. if students paid qualified tuition, or if someone paid on the student's behalf, the student (or if the student is a dependent, the taxpayer) may be able to claim an education credit on their **tax return**. we know that- - - - - (i)taking 3 as common, we get3; solving only bracket term first,here a = 2a, b = 3b; putting values in eq.i}applying }}ans: - 3}note: must memorize cubes upto 12 given : let a , b and c be three students whose chances of solving a problem is given i.e , p(a) = , p(b) = and p(c) = . p() = , p() = and p()= to find: the probability that the problem is solved .here , p(the problem is solved) = 1 - p(the problem is not solved)= 1- p( )= 1 – [p( ) p( ) p( )]= 1- []= 1 - = therefore , the probability that the problem is solved is . hi,

lots of great explanations below, but i fear they might be difficult to truly conceptualize. 

let's start with a pie!

if there's a pie in front of us, that is the market. we'll call it 100 units of pie

we cut the pie in half - one piece (50 pie units) for you, and one piece (50 pie units) for me

scenario 1: we do nothing, we both have 50% market share.

scenario 2: i eat half my pie. my pie slice is now 25 units, yours is still 50 units. as a result, you have 67% of the market share and i have 33%. my "revenues" have fallen and my market share has fallen. your revenues haven't changed but you've gained market share

scenario 3: not only do i eat half my pie, but you also eat 20% of yours. still, i ate more, so even though your "revenues" are down, you've still gained market share.

scenario 4: we both get given more pie (yay). but, i get a whole other 50 units, and you only get 5 more. your "revenues" are up, but you've still lost market share!

i've created the below image for you to visualize this better (click on it to blow it up)...hope this helps! Definition used:  

Congruence.  

Let m be an integer greater than 1. If xand yare integers, then x is congruent to y modulo m if x — y is divisible by m. It can be represented as \(\displaystyle{x}={y}{\left({b}\ mod\ {m}\right)}\). This relation is called as congruence modulo m.  

Calculation:   

Consider the values and congruence classes modulo m,  

\(\displaystyle{\left[{13}\right]}^{{{6}}}\) and m = 15.  

It is evident that [x]=[y] if \(\displaystyle{\left[{x}={y}{\left({b}\ mod\ {m}\right)}\right]}\). Since \(\displaystyle{13}=-{2}{\left({b} \ mod \ {15}\right)}\).  

Then,  

[13]=[-2]  

\(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[-{2}\right]}^{{{6}}}\)  

\(\displaystyle={\left[{\left(-{2}\right)}^{{{6}}}\right]}\)  

=[64]  

Since, by the principal or congruence \(\displaystyle{64}={4}{\left({b}\ mod \ {15}\right)}\). Therefore, \(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[{4}\right]}\) in \(\displaystyle{Z}\_{{{15}}}\).  

Therefore, the value of the expression is [4]. 18 sq.units explanation: x2 = 12y ∴ 4b = 12 ∴ b = 3 area of triangle = `1/2 xx "ab" xx "os"` = `1/2 xx 4"a" xx "a"` = `1/2(12)(3)` = 18 sq. units hi @sam\_cat - there was another topic on this query. i can't find it, so i've tweaked yours to make it more searchable moving forwards. 

this is to do with the direct debit figure. we look at your prices for the two year plan, and the annual consumption of the property, getting an annual spend and a monthly cost (if you divide that figure by 12). the tariff that you're on now just displays your current direct debit, which will of course take into account your current credit balance. 

tim