so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check
so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.
![\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cstackrel%7Bf%28x%29%7D%7B2x%5E3-x%5E2-5x%7D~~%20-%20~~%5B%5Cstackrel%7Bg%28x%29%7D%7B-x%5E2%2B3x%7D%5D%5Cimplies%202x%5E3-x%5E2-5x%2Bx%5E2-3x%20%5C%5C%5C%5C%5C%5C%202x%5E3-8x%5Cimplies%202%28x%5E3-4x%29%5Cimplies%20%5Cdisplaystyle%202%5Cint%5Climits_%7B-2%7D%5E%7B0%7D%20%28x%5E3-4x%29dx%20%5Cimplies%202%5Cleft%5B%20%5Ccfrac%7Bx%5E4%7D%7B4%7D-2x%5E2%20%5Cright%5D_%7B-2%7D%5E%7B0%7D%5Cimplies%20%5Cboxed%7B8%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}](https://tex.z-dn.net/?f=%5Cstackrel%7Bg%28x%29%7D%7B-x%5E2%2B3x%7D~~%20-%20~~%5B%5Cstackrel%7Bf%28x%29%7D%7B2x%5E3-x%5E2-5x%7D%5D%5Cimplies%20-x%5E2%2B3x-2x%5E3%2Bx%5E2%2B5x%20%5C%5C%5C%5C%5C%5C%20-2x%5E3%2B8x%5Cimplies%202%28-x%5E3%2B4x%29%20%5C%5C%5C%5C%5C%5C%20%5Cdisplaystyle%202%5Cint%5Climits_%7B0%7D%5E%7B2%7D%20%28-x%5E3%2B4x%29dx%20%5Cimplies%202%5Cleft%5B%20-%5Ccfrac%7Bx%5E4%7D%7B4%7D%2B2x%5E2%20%5Cright%5D_%7B0%7D%5E%7B2%7D%5Cimplies%20%5Cboxed%7B8%7D%20~%5Chfill%20%5Cboxed%7B%5Cstackrel%7B%5Ctextit%7Btotal%20area%7D%7D%7B8~~%20%2B%20~~8~~%20%3D%20~~16%7D%7D)
Expplanation and Answer:
y=mx−7
Swap sides so that all variable terms are on the left hand side.
mx−7=y
Add 7 to both sides.
mx=y+7
Divide both sides by m.
m
mx
=
m
y+7
Dividing by m undoes the multiplication by m.
x=
m
y+7
66.5 divided by 3.5 is 19.
Answer:C
Explanation:none
The total number of outcomes of rolling a die = 6
The total number of outcomes of tossing a coin = 2
The total number of possible outcomes of executing both experiments is 6*2=12, which is also the number of leaves (end of the branch of a tree) on a tree diagram.
