In Exponential form 2(X x 4x^-5/2)^1/2
The distance from the lighthouse to the boat will be given by:
tan θ=opposite/adjacent
θ=angle of depression=23
opposite= height of the lighthouse
adjacent=distance from the boat to the bottom of the light house
hence
tan 23=210/a
thus solving for a we obtain:
a=210/tan23
hence
a=494.73 ft
![\bf \textit{sum of all interior angles in a polygon}\\\\ S=180(n-2)~~ \begin{cases} n=\textit{number of sides}\\[-0.5em] \hrulefill\\ n=8 \end{cases}\implies S=180(8-2) \\\\\\ S=180(6)\implies S=1080](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bsum%20of%20all%20interior%20angles%20in%20a%20polygon%7D%5C%5C%5C%5C%20S%3D180%28n-2%29~~%20%5Cbegin%7Bcases%7D%20n%3D%5Ctextit%7Bnumber%20of%20sides%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20n%3D8%20%5Cend%7Bcases%7D%5Cimplies%20S%3D180%288-2%29%20%5C%5C%5C%5C%5C%5C%20S%3D180%286%29%5Cimplies%20S%3D1080)
now, we know 7 of those angles are already 835°, so the last one must be 1080 - 835 = 245°.
Answer:
1 ≥ t ≤ 3
Step-by-step explanation:
Given
h(t) = -16t² + 64t + 4
Required
Determine the interval which the bar is at a height greater than or equal to 52ft
This implies that
h(t) ≥ 52
Substitute -16t² + 64t + 4 for h(t)
-16t² + 64t + 4 ≥ 52
Collect like terms
-16t² + 64t + 4 - 52 ≥ 0
-16t² + 64t - 48 ≥ 0
Divide through by 16
-t² + 4t - 3 ≥ 0
Multiply through by -1
t² - 4t + 3 ≤ 0
t² - 3t - t + 3 ≤ 0
t(t - 3) -1(t - 3) ≤ 0
(t - 1)(t - 3) ≤ 0
t - 1 ≤ 0 or t - 3 ≤ 0
t ≤ 1 or t ≤ 3
Rewrite as:
1 ≥ t or t ≤ 3
Combine inequality
1 ≥ t ≤ 3
I should be any more points for this but 26 per cubic meter he only has 1500 but that’ an answer of 14,000 600 m³