Answer:C. 36
Step-by-step explanation: height is 12 base is 9
12^2+9^2= 144+81=225 sq rt of 225 =15
So 12+9+15=36
Answer:
<u>D) (f o g)(x) = 10x² - 60x + 93</u>
Step-by-step explanation:
f(x) = 10x² + 3
g(x) = x - 3
⇒ (f o g)(x)
⇒ f(x - 3)
⇒ f(x - 3) = 10(x - 3)² + 3
⇒ f(x - 3) = 10(x² - 6x + 9) + 3
⇒ f(x - 3) = 10x² - 60x + 90 + 3
⇒ <u>(f o g)(x) = 10x² - 60x + 93</u>
F(g(x)) = [(-7x-8)/(x-1) - 8} / [(-7x - 8)/(x-1) + 7] =
[(-7x - 8 - 8(x-1)) / (x-1)] / [(-7x - 8 + 7(x-1)) / (x-1)] = (-15x) / (-15) = x.
g(f(x)) = [-7*(x-8)/(x+7) - 8] / [(x-8)/(x+7) - 1] =
[(-7x + 56 -8*(x+7)) / (x+7)] / [(x - 8 - (x + 7)) / (x+7)] = (-15x) / (-15) = x.
So since f(g(x)) = g(f(x)) = x we can conclude that f and g are inverses.
Answer:
X=30; Z=80; Y=70
Step-by-step explanation:
The straight lines equal 180 and you subtract 180 to 150 and 180 to 100 to get 30 and 80 and the inside of a triangle is 180 and you subtract 30 and 80 to 180 to get 70.
let's first off notice that, on the 2), the sector is really half of the whole circle, and on 3) the sector is one quarter of the whole circle.
now, on 2) AB is the diameter of 4 units, therefore it has a radius of 2, or half that.
![\bf \boxed{2} \\\\\\ \stackrel{\textit{area of a circle}}{A=\pi r^2}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=2 \end{cases}\implies A=\pi 2^2\implies A=4\pi \\\\\\ \stackrel{\textit{half of that}}{A=2\pi}\implies A=\stackrel{\textit{rounded up}}{A=6.3~ft^2} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cboxed%7B2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20circle%7D%7D%7BA%3D%5Cpi%20r%5E2%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D2%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%202%5E2%5Cimplies%20A%3D4%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bhalf%20of%20that%7D%7D%7BA%3D2%5Cpi%7D%5Cimplies%20A%3D%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7BA%3D6.3~ft%5E2%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \boxed{3} \\\\\\ \stackrel{\textit{area of a circle}}{A=\pi r^2}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=20 \end{cases}\implies A=\pi 20^2\implies A=400\pi \\\\\\ \stackrel{\textit{one quarter of that}}{A=100\pi }\implies \stackrel{\textit{rounded up}}{A=314.2~in^2}](https://tex.z-dn.net/?f=%5Cbf%20%5Cboxed%7B3%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20circle%7D%7D%7BA%3D%5Cpi%20r%5E2%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D20%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%2020%5E2%5Cimplies%20A%3D400%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bone%20quarter%20of%20that%7D%7D%7BA%3D100%5Cpi%20%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7BA%3D314.2~in%5E2%7D)
