The answer is 26 trust me I will be ur friend
Answer:
4(3x+2)
Step-by-step explanation:
Find the GCF (greatest common factor).
GCF=2
Factor out the GCF.
![2(\frac{2x}{x}+\frac{10x}{2}+\frac{8}{2})](https://tex.z-dn.net/?f=2%28%5Cfrac%7B2x%7D%7Bx%7D%2B%5Cfrac%7B10x%7D%7B2%7D%2B%5Cfrac%7B8%7D%7B2%7D%29)
Simplify.
2(x+5x+4)
Collect the like terms x and 5x.
2(6x+4)
Factor out the number 2, because 2*3=6 and 2*2=4.
2*2(3x+2)
Simplify to get the answer.
4(3x+2)
Answer:
1368
Step-by-step explanation:
4.
first one, since the lower bound is y=0
so what we do is just inetgrate from x=0 to x=8
![\int\limits^8_9 { \frac{16x}{x^2+1} } \, dx](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E8_9%20%7B%20%5Cfrac%7B16x%7D%7Bx%5E2%2B1%7D%20%7D%20%5C%2C%20dx%20)
find an antiederivitive
use u subsitution
u=x²+1
du=2x dx
so factor out the 8
![8 \int\limits^8_9 { \frac{2x}{x^2+1} } \, dx](https://tex.z-dn.net/?f=%208%20%5Cint%5Climits%5E8_9%20%7B%20%5Cfrac%7B2x%7D%7Bx%5E2%2B1%7D%20%7D%20%5C%2C%20dx%20)
![8 \int\limits^8_9 { \frac{1}{u} } \, du](https://tex.z-dn.net/?f=%208%20%5Cint%5Climits%5E8_9%20%7B%20%5Cfrac%7B1%7D%7Bu%7D%20%7D%20%5C%2C%20du%20)
and we know that the antideritivitve of 1/u is ln|u|
8ln|x²+1| is an antideritivive
now
![[8ln|x^2+1|]^8_0](https://tex.z-dn.net/?f=%5B8ln%7Cx%5E2%2B1%7C%5D%5E8_0)
![8ln|8^2+1|-8ln|0^2+1|](https://tex.z-dn.net/?f=8ln%7C8%5E2%2B1%7C-8ln%7C0%5E2%2B1%7C)
![8ln|65|+8ln|1|](https://tex.z-dn.net/?f=8ln%7C65%7C%2B8ln%7C1%7C)
![8ln65+0](https://tex.z-dn.net/?f=8ln65%2B0)
the aera is 8ln65
A is the answer
6.
find where they intersect to find the area bounded
f(x)=g(x) at x=-6 and x=2
and g(x) is on top (we can see that because g(0)>f(0))
so we integrate from -6 to 2
![\int\limits^2_{-6} {g(x)-f(x)} \, dx](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E2_%7B-6%7D%20%7Bg%28x%29-f%28x%29%7D%20%5C%2C%20dx%20)
![\int\limits^2_{-6} {8x+48-(x^2+12x+36)} \, dx](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E2_%7B-6%7D%20%7B8x%2B48-%28x%5E2%2B12x%2B36%29%7D%20%5C%2C%20dx%20)
![\int\limits^2_{-6} {8x+48-x^2-12x-36)} \, dx](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E2_%7B-6%7D%20%7B8x%2B48-x%5E2-12x-36%29%7D%20%5C%2C%20dx%20)
![\int\limits^2_{-6} {-x^2-4x+12)} \, dx](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E2_%7B-6%7D%20%7B-x%5E2-4x%2B12%29%7D%20%5C%2C%20dx%20)
this is easy
use reverse power rules
remember that
![\int\limits^a_b {f(x)+g(x)} \, dx = \int\limits^a_b {f(x)} \, dx + \int\limits^a_b {g(x)} \, dx](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5Ea_b%20%7Bf%28x%29%2Bg%28x%29%7D%20%5C%2C%20dx%20%3D%20%5Cint%5Climits%5Ea_b%20%7Bf%28x%29%7D%20%5C%2C%20dx%20%2B%20%5Cint%5Climits%5Ea_b%20%7Bg%28x%29%7D%20%5C%2C%20dx%20)
the antiderivitive is
![\frac{-x^3}{3} -2x^2+12x](https://tex.z-dn.net/?f=%5Cfrac%7B-x%5E3%7D%7B3%7D%20-2x%5E2%2B12x)
so
![[\frac{-x^3}{3} -2x^2+12x]^2_{-6}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B-x%5E3%7D%7B3%7D%20-2x%5E2%2B12x%5D%5E2_%7B-6%7D)
=
![(\frac{-2^3}{3} -2(2)^2+12(2))-(\frac{-(-6)^3}{3} -2(-6)^2+12(-6))](https://tex.z-dn.net/?f=%28%5Cfrac%7B-2%5E3%7D%7B3%7D%20-2%282%29%5E2%2B12%282%29%29-%28%5Cfrac%7B-%28-6%29%5E3%7D%7B3%7D%20-2%28-6%29%5E2%2B12%28-6%29%29)
=
![({-8}{3} -8+24)-(72 -72-72)](https://tex.z-dn.net/?f=%28%7B-8%7D%7B3%7D%20-8%2B24%29-%2872%20-72-72%29)
=
![({-8}{3}+ 16)-(-72)](https://tex.z-dn.net/?f=%28%7B-8%7D%7B3%7D%2B%2016%29-%28-72%29)
=
![{-8}{3}+ 16+72](https://tex.z-dn.net/?f=%7B-8%7D%7B3%7D%2B%2016%2B72)
=
![{-8}{3}+ 88](https://tex.z-dn.net/?f=%7B-8%7D%7B3%7D%2B%2088)
=
![\frac{-8}{3} + \frac{264}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B-8%7D%7B3%7D%20%2B%20%5Cfrac%7B264%7D%7B3%7D%20)
=
![\frac{256}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B256%7D%7B3%7D%20)
answer is C