![\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}](https://tex.z-dn.net/?f=%5Chuge%5Cunderline%7B%5Cunderline%7B%5Cboxed%7B%5Cmathbb%20%7BANSWER%3A%7D%7D%7D%7D)
◉ ![\large\bm{ -4}](https://tex.z-dn.net/?f=%5Clarge%5Cbm%7B%20-4%7D)
![\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}](https://tex.z-dn.net/?f=%5Chuge%5Cunderline%7B%5Cunderline%7B%5Cboxed%7B%5Cmathbb%20%7BSOLUTION%3A%7D%7D%7D%7D)
Before performing any calculation it's good to recall a few properties of integrals:
![\small\longrightarrow \sf{\int_{a}^b(nf(x) + m)dx = n \int^b _{a}f(x)dx + \int_{a}^bmdx}](https://tex.z-dn.net/?f=%5Csmall%5Clongrightarrow%20%5Csf%7B%5Cint_%7Ba%7D%5Eb%28nf%28x%29%20%2B%20m%29dx%20%3D%20n%20%5Cint%5Eb%20_%7Ba%7Df%28x%29dx%20%2B%20%20%5Cint_%7Ba%7D%5Ebmdx%7D)
![\small\sf{\longrightarrow If \: a \angle c \angle b \Longrightarrow \int^{b} _a f(x)dx= \int^c _a f(x)dx+ \int^{b} _c f(x)dx }](https://tex.z-dn.net/?f=%5Csmall%5Csf%7B%5Clongrightarrow%20If%20%5C%3A%20a%20%5Cangle%20c%20%5Cangle%20b%20%5CLongrightarrow%20%5Cint%5E%7Bb%7D%20_a%20%20f%28x%29dx%3D%20%5Cint%5Ec%20_a%20f%28x%29dx%2B%20%20%5Cint%5E%7Bb%7D%20_c%20%20f%28x%29dx%20%7D)
So we apply the first property in the first expression given by the question:
![\small \sf{\longrightarrow\int ^3_{-2} [2f(x) +2]dx= 2 \int ^3 _{-2} f(x) dx+ \int f^3 _{2} 2dx=18}](https://tex.z-dn.net/?f=%5Csmall%20%5Csf%7B%5Clongrightarrow%5Cint%20%5E3_%7B-2%7D%20%5B2f%28x%29%20%2B2%5Ddx%3D%202%20%5Cint%20%5E3%20_%7B-2%7D%20f%28x%29%20dx%2B%20%5Cint%20f%5E3%20_%7B2%7D%202dx%3D18%7D)
And we solve the second integral:
![\small\sf{\longrightarrow2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot(3 - ( - 2)) }](https://tex.z-dn.net/?f=%5Csmall%5Csf%7B%5Clongrightarrow2%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20%2B%202%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20%3D%202%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20%2B%202%20%5Ccdot%283%20-%20%28%20-%202%29%29%20%7D)
![\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} 2dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot5 = 2 \int^3_{-2} f(x)dx10 = }](https://tex.z-dn.net/?f=%5Csmall%20%5Csf%7B%5Clongrightarrow%202%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20%2B%202%20%5Cint%20%5E3_%7B-2%7D%202dx%20%20%3D%202%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20%2B%20%20%202%20%5Ccdot5%20%3D%202%20%5Cint%5E3_%7B-2%7D%20f%28x%29dx10%20%3D%20%7D)
Then we take the last equation and we subtract 10 from both sides:
![\sf{{\longrightarrow 2 \int ^3_{-2} f(x)dx} + 10 - 10 = 18 - 10}](https://tex.z-dn.net/?f=%5Csf%7B%7B%5Clongrightarrow%202%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%7D%20%2B%2010%20-%2010%20%3D%2018%20-%2010%7D)
![\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 8}](https://tex.z-dn.net/?f=%5Csmall%20%5Csf%7B%5Clongrightarrow%202%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20%20%3D%208%7D)
And we divide both sides by 2:
![\small\longrightarrow \sf{\dfrac{2 { \int}^{3} _{2} }{2} = \dfrac{8}{2} }](https://tex.z-dn.net/?f=%5Csmall%5Clongrightarrow%20%5Csf%7B%5Cdfrac%7B2%20%20%7B%20%20%5Cint%7D%5E%7B3%7D%20_%7B2%7D%20%20%7D%7B2%7D%20%20%3D%20%20%5Cdfrac%7B8%7D%7B2%7D%20%7D)
![\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx=4}](https://tex.z-dn.net/?f=%5Csmall%20%5Csf%7B%5Clongrightarrow%202%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%3D4%7D)
Then we apply the second property to this integral:
![\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 4}](https://tex.z-dn.net/?f=%5Csmall%20%5Csf%7B%5Clongrightarrow%202%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20%2B%202%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20%2B%202%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20%3D%204%7D)
Then we use the other equality in the question and we get:
![\small\sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx = 8 + 2 \int ^3_{-2} f(x)dx = 4}](https://tex.z-dn.net/?f=%5Csmall%5Csf%7B%5Clongrightarrow%202%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20%20%3D%20%202%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20%20%3D%208%20%2B%20%202%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20%20%3D%204%7D)
![\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =4}](https://tex.z-dn.net/?f=%5Csmall%5Clongrightarrow%20%5Csf%7B2%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20%3D4%7D)
We substract 8 from both sides:
![\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx -8=4}](https://tex.z-dn.net/?f=%5Csmall%5Clongrightarrow%20%5Csf%7B2%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20-8%3D4%7D)
• ![\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =-4}](https://tex.z-dn.net/?f=%5Csmall%5Clongrightarrow%20%5Csf%7B2%20%5Cint%20%5E3_%7B-2%7D%20f%28x%29dx%20%3D-4%7D)
Answer:
2
Step-by-step explanation:
if you multiply 2 by 3=6
- 2×2=4
- 3×3=9
Answer:
x= -8 , y = 5
x= 25/4 , y = 1/4
Step-by-step explanation:
substitute first eqn into the second eqn:
(7 - 3y)^2 -y^2 = 39
49 - 42y + 9y^2 - y^2 = 39
8y^2 - 42y +10 =0
4y^2 - 21y + 5 = 0
(4y-1) (y-5) = 0
y= 1/4 , 5
when y=1/4
x = 7- 3/4
=25/4
when y= 5
x = 7- 15
= -8
Answer:
A = 12 ft
Step-by-step explanation:
Perimeter ( of SQUARE) B = 64
then each side = 64/4 = 16
AREA of SQUARE C = 400
side X side = 400
then side = sqrt 400
Now use Pyhtagorean Theorem to find leg of right triangle that is
side A Side B hypotenuse Side C
A^2 + 16^2 = (sqrt 400)^2
A^2 = 400 - 256
A^2 = 144 then A = 12 Ft
Answer:
85500000
Step-by-step explanation: