Answer:
Angle <u>A</u><u> </u>= 15°
use angle properties of triangle to solve this.
<h3>
Answer: E) x^5</h3>

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Explanation:
We simply take half of the exponent 10 to get 5. This applies to square roots only.
So the rule is 
A more general rule is
![\sqrt[n]{a^b} = a^{b/n}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Eb%7D%20%3D%20a%5E%7Bb%2Fn%7D)
If n = 2, then we're dealing with square roots like with this problem. In this case, a = x and b = 10.
it takes 5 seconds to reach a height of 40 ft. and It takes 2.5 seconds to reach maximum height.
Let h(t) represent the height of the ball at time t.
Given that the height is given by:
h(t) = -16t² + 80t + 40
1) For the ball to reach a height 0f 40 ft, h(t) = 40, hence:
40 = -16t² + 80t + 40
16t² - 80t = 0
t(t - 5) = 0
t = 0 or t = 5
Hence it takes 5 seconds to reach a height of 40 ft.
2) The maximum height is at h'(t) = 0,
h'(t) = -32t + 80
-32t + 80 = 0
32t = 80
t = 2.5
It takes 2.5 seconds to reach maximum height.
Find out more at: brainly.com/question/11535666
x = total amount split between Adam and Tom.
since we know the total amount split between both in a 18 : 17 ratio is "x", let's divide "x" by (18 + 17) and distribute accordingly to get the amount of each.
![\stackrel{Adam~received}{18\cdot \cfrac{x}{18+17}}\qquad \qquad \stackrel{Tom~received}{17\cdot \cfrac{x}{18+17}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{since we know that Adam got "5" more}}{ \stackrel{Adam}{18\cdot \cfrac{x}{18+17}}~~ = ~~\stackrel{Tom}{17\cdot \cfrac{x}{18+17}~~ + ~~5} }\qquad \implies \qquad \cfrac{18x}{35}~~ + ~~\cfrac{17x}{35}+5](https://tex.z-dn.net/?f=%5Cstackrel%7BAdam~received%7D%7B18%5Ccdot%20%5Ccfrac%7Bx%7D%7B18%2B17%7D%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7BTom~received%7D%7B17%5Ccdot%20%5Ccfrac%7Bx%7D%7B18%2B17%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bsince%20we%20know%20that%20Adam%20got%20%225%22%20more%7D%7D%7B%20%5Cstackrel%7BAdam%7D%7B18%5Ccdot%20%5Ccfrac%7Bx%7D%7B18%2B17%7D%7D~~%20%3D%20~~%5Cstackrel%7BTom%7D%7B17%5Ccdot%20%5Ccfrac%7Bx%7D%7B18%2B17%7D~~%20%2B%20~~5%7D%20%7D%5Cqquad%20%5Cimplies%20%5Cqquad%20%5Ccfrac%7B18x%7D%7B35%7D~~%20%2B%20~~%5Ccfrac%7B17x%7D%7B35%7D%2B5)
![\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{35}}{35\left( \cfrac{18x}{35} \right)~~ = ~~35\left( \cfrac{17x}{35}+5 \right)}\implies 18x~~ = ~~17x+175\implies \boxed{x =175} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{Tom~received}{17\cdot \cfrac{x}{18+17}}\implies \cfrac{17(175)}{35}\implies \blacktriangleright 85 \blacktriangleleft](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B35%7D%7D%7B35%5Cleft%28%20%5Ccfrac%7B18x%7D%7B35%7D%20%5Cright%29~~%20%3D%20~~35%5Cleft%28%20%5Ccfrac%7B17x%7D%7B35%7D%2B5%20%5Cright%29%7D%5Cimplies%2018x~~%20%3D%20~~17x%2B175%5Cimplies%20%5Cboxed%7Bx%20%3D175%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7BTom~received%7D%7B17%5Ccdot%20%5Ccfrac%7Bx%7D%7B18%2B17%7D%7D%5Cimplies%20%5Ccfrac%7B17%28175%29%7D%7B35%7D%5Cimplies%20%5Cblacktriangleright%2085%20%5Cblacktriangleleft)