solve the following system of linear inequation graphically: `x+ygeq1,\ 7x+9ylt=63 ,\ xlt=6,\ ylt=5,\ xgeq0,\ ygeq0`
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finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). general systems thursday, september 29, 2016 3:50 pm a fruit store sell two types of fruit, apples and oranges. apples are $2 and oranges are $3.today 50 fruit was sold and the store made $110. 110 = 2▯ + 3▯ 50 = ▯ + ▯ ▯ = 40 ▯ = 10 general ▯▯ ▯+ ▯ ▯▯ ▯⋯+ ▯ ▯ = ▯ ▯ a linear equation is an equation in the form of: • x's are variables and a's are constants. • solution set for a linear equation is the set of all solutions. • set of linear equations where a commonsolution is desired is called a system of linear equations. graphs • 1 variables is a point • 2 variables is a line • 3 variables is a plane • 4+ variables is a hyperplane examples of systems 2 equations, 2 unknowns • no solutions • one ?solution 23cq


a) equation for loop abgefa is (35.5)i1 + (6..25 = 21 v
b) equation for loop cbgedc is ( 9.25i2 - ( 6.25i3 = +15 v 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) dear student
red is coded as 2076
which means
r=20
e=7
d=6
now lets us number the alphabets as follows
a=1
b=2
c= 3
d= 4
e= 5




r=18


y=25
z=26

now according to the question
r=20=18+2
e=7=5+2
d=6=4+2

therefore we can say that
g = 7+2 =9
a=1+2=3


gag will be coded as 939


regards
â€‹â€‹â€‹â€‹â€‹â€‹â€‹ let a (x, y) be the point on the curve y = f(x). then slope of the tangent to the curve at point a is `"dy"/"dx"`. according to the given condition, `"dy"/"dx" = "x" + 3"y" - 1` ∴ `"dy"/"dx" - 3"y" = "x - 1"` ....(1) this is the linear differential equation of the form `"dy"/"dx" + "py" = "q"`, where p = - 3 and q = x - 1 ∴ i.f. = `"e"^(int "p dx") = "e"^(int - 3"dx") = "e"^(- 3"x")` ∴ the solution of (1) is given by `"y" * ("i.f.") = int "q" * ("i.f.") "dx" + "c"` ∴ `"y" * "e"^(- 3"x") = int ("x - 1") * "e"^(-3"x") "dx" + "c"` ∴ `"e"^(- 3"x") * "y" = ("x - 1") int "e"^(- 3"x") - int ["d"/"dx" ("x - 1") * int "e"^(- 3"x") "dx"] "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = ("x - 1") * "e"^(- 3"x")/-3 - int 1 * "e"^(- 3"x") * "y"/-3 "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1") * "e"^(- 3"x") + 1/3 int "e"^(- 3"x") "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1") "e"^(- 3"x") + 1/3 * "e"^(- 3"x")/-3 + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1")"e"^(- 3"x") - 1/9 "e"^(- 3"x") + "c"\_1` ∴ 9y = - 3(x - 1) - 1 + 9`"c"\_1 * e^("3x")` ∴ 9y + 3(x - 1) + 1 = `9"c"\_1 * e^("3x")` ∴ 9y + 3x - 3 + 1 = `9"c"\_1 * e^("3x")` ∴ 3(x + 3y) = 2 + `9"c"\_1 * e^("3x")` ∴ 3(x + 3y) = 2 + `"c" * e^("3x")` where c = 9c1 ....(2) this is the general equation of the curve. but the required curve is passing through the origin (0, 0). ∴ by putting x = 0 and y= 0 in (2), we get 0 = 2 + c ∴ c = - 2 ∴ from (2), the equation of the required curve is 3(x + 3y) = 2 - 2`e^("3x")` i.e. 3(x + 3y) = 2(1 - `e^("3x")`). the given equations arex2 + y2 = 16 …(i)and y2 = 6x …(ii)on solving the equation (i) and (ii),or x2 + 6x = 16or x2 + 6x – 16 = 0or (x + 8)(x – 2) = 0or x = 2 or – 8 is not possible solution∴ when x = 2, y = ± = ± = ±equation (i) is a circle with centre (0, 0) and meets axes at (±4,0), (0,±4)equation (ii) represents a parabola with axis as x - axis.points of intersection are a () and c (2,)these are shown in the graph below:area bounded by the circle and parabola= 2[area (oado) + area (adba)]area of circle = π(r) 2= π(4)2= 16π sq. unitsthus, required areathe area of the circle x2 + y2 = 16 which is exterior the parabola y2 = 6x is >> [x,y] = ismember(a,b);
>> [~,z] = sort(y(x));
>> v = find(x);
>> r = reshape(v(z),2,[]).'
r =
 2 9
 4 8
 5 7 :roll\_eyes:

[[image]](https://aws1.discourse-cdn.com/business6/uploads/dynamobim/original/3x/6/3/6393f7d9277c8a0ff2b48f07b1e2352f69f463b0.png "image")

t1 = [9, 9, 99, 9, 6, 7, 8];

t9 = dscore.list.chop(t1, 3);

a = t9;

t2 = get.valueatindex(a, 0);

t4 = [3, 2];

t10 = dscore.list.ofrepeateditem(t4, 3);

a1 = t10;

t3 = get.valueatindex(a1, 0);

t11 = t2<1l> + t3<1l>;

a2 = t9;

t5 = get.valueatindex(a2, 1);

a3 = t10;

t6 = get.valuea… slope of the line ⇒ slope of the lineequation of line :⇒ 3(y - 5) = 2(x - 2)⇒ 2x - 3y + 11 = 0 …equation (i)putting x = x + 3 and y = y + 2 in the above equation we get2(x + 3) - 3(y + 2) + 11 = 0⇒ 2x - 3y + 11 = 0since it again gives the same equation as (i) so it is a point on this line.hence proved