Answer:
Step-by-step explanation:
First off, your drawing is kinda inaccurate, because the 
looks like a right angle, but as you drew it yourself it doesn't matter too much.
We know that the sum of the angles of a triangle will always equal 
, so we have that 
.
Combining like terms on the left side gives 
.
Subtracting 
from both sides gives 
.
So, 
and we're done!
So pythagrorean theorem is the ideal way and easiest way to solve this
a^2+b^2=c^2
8.5^2+11^2=x^2
72.25+121=193.25
square root of 193.25=13.9014 or 13.9 in
9.5^2+15^2=x^2
90.25+225=315.25
square root of 315.25=17.7553 or 17.8 ft
Answer:
dont lie, thats only 5 points
-3x+5y=30
+3x+5y=30
5y=3x+30
divide 5 by everything
answer: y=3/5x+6
Check the picture below.
so, let's notice, is really just a 2x20 rectangle with a quarter of a semicircle with a radius of 11.
![\bf \stackrel{\textit{area of a circle}}{A=\pi r^2}~~ \implies A=\pi 11^2\implies A=121\pi \implies \stackrel{\textit{one quarter of that}}{\boxed{A=\cfrac{121\pi }{4}}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\underline{\textit{area of the figure}}}{\stackrel{\textit{rectangle's area}}{(2\cdot 20)}+\stackrel{\textit{circle's quart's area}}{\cfrac{121\pi }{4}}\qquad \approx \qquad 135.03\implies \stackrel{\textit{rounded up}}{135}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20circle%7D%7D%7BA%3D%5Cpi%20r%5E2%7D~~%20%5Cimplies%20A%3D%5Cpi%2011%5E2%5Cimplies%20A%3D121%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bone%20quarter%20of%20that%7D%7D%7B%5Cboxed%7BA%3D%5Ccfrac%7B121%5Cpi%20%7D%7B4%7D%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Cunderline%7B%5Ctextit%7Barea%20of%20the%20figure%7D%7D%7D%7B%5Cstackrel%7B%5Ctextit%7Brectangle%27s%20area%7D%7D%7B%282%5Ccdot%2020%29%7D%2B%5Cstackrel%7B%5Ctextit%7Bcircle%27s%20quart%27s%20area%7D%7D%7B%5Ccfrac%7B121%5Cpi%20%7D%7B4%7D%7D%5Cqquad%20%5Capprox%20%5Cqquad%20135.03%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7B135%7D%7D)
