Simplifying
18x + -44 = 13y + -38
Reorder the terms:
-44 + 18x = 13y + -38
Reorder the terms:
-44 + 18x = -38 + 13y
Solving
-44 + 18x = -38 + 13y
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '44' to each side of the equation.
-44 + 44 + 18x = -38 + 44 + 13y
Combine like terms: -44 + 44 = 0
0 + 18x = -38 + 44 + 13y
18x = -38 + 44 + 13y
Combine like terms: -38 + 44 = 6
18x = 6 + 13y
Divide each side by '18'.
x = 0.3333333333 + 0.7222222222y
Simplifying
x = 0.3333333333 + 0.7222222222y
A) zeroes
P(n) = -250 n^2 + 2500n - 5250
Extract common factor:
P(n)= -250 (n^2 - 10n + 21)
Factor (find two numbers that sum -10 and its product is 21)
P(n) = -250(n - 3)(n - 7)
Zeroes ==> n - 3 = 0 or n -7 = 0
Then n = 3 and n = 7 are the zeros.
They rerpesent that if the promoter sells tickets at 3 or 7 dollars the profit is zero.
B) Maximum profit
Completion of squares
n^2 - 10n + 21 = n^2 - 10n + 25 - 4 = (n^2 - 10n+ 25) - 4 = (n - 5)^2 - 4
P(n) = - 250[(n-5)^2 -4] = -250(n-5)^2 + 1000
Maximum ==> - 250 (n - 5)^2 = 0 ==> n = 5 and P(5) = 1000
Maximum profit =1000 at n = 5
C) Axis of symmetry
Vertex = (h,k) when the equation is in the form A(n-h)^2 + k
Comparing A(n-h)^2 + k with - 250(n - 5)^2 + 1000
Vertex = (5, 1000) and the symmetry axis is n = 5.
Answer:
cosine = adjacent/hypotenuse
cos A = 20/29 (choice: yellow)
Step-by-step explanation:
Ooh, fun
what I would do is to make it a piecewise function where the absolute value becomse 0
because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up
so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points
we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5
A.

B.
sepearte the integrals
![\int\limits^{-5}_{-4} {x^2+x-12} \, dx = [\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-5}_{-4}=(\frac{-125}{3}+\frac{25}{2}+60)-(\frac{64}{3}+8+48)=\frac{23}{6}](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E%7B-5%7D_%7B-4%7D%20%7Bx%5E2%2Bx-12%7D%20%5C%2C%20dx%20%3D%20%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%2B%5Cfrac%7Bx%5E2%7D%7B2%7D-12x%5D%5E%7B-5%7D_%7B-4%7D%3D%28%5Cfrac%7B-125%7D%7B3%7D%2B%5Cfrac%7B25%7D%7B2%7D%2B60%29-%28%5Cfrac%7B64%7D%7B3%7D%2B8%2B48%29%3D%5Cfrac%7B23%7D%7B6%7D)
next one
![\int\limits^{-4}_3 {-x^2-x+12} \, dx=-1[\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-4}_{3}=-1((-64/3)+8+48)-(9+(9/2)-36))=\frac{343}{6}](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E%7B-4%7D_3%20%7B-x%5E2-x%2B12%7D%20%5C%2C%20dx%3D-1%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%2B%5Cfrac%7Bx%5E2%7D%7B2%7D-12x%5D%5E%7B-4%7D_%7B3%7D%3D-1%28%28-64%2F3%29%2B8%2B48%29-%289%2B%289%2F2%29-36%29%29%3D%5Cfrac%7B343%7D%7B6%7D)
the last one you can do yourself, it is

the sum is

so the area under the curve is
The y-intercept, in this case, represents when x is at 0, or when she just began observing.