25320+11310=36,630 or 12,430+24,200=36,630
Consider right triangle with vertices B - base of the hill, S - top of the statue and Y - you. In this triangle angle B is right and angle Y is 13.2°. If h is a height of the statue, then the legs YB and BS have lengths 77 ft and 16+h ft.
You have lengths of two legs and measure of one acute angle, then you can use tangent to find h:
ft.
Answer: the height of the statue is 2.0565 ft.
First, lets create a equation for our situation. Let

be the months. We know four our problem that <span>Eliza started her savings account with $100, and each month she deposits $25 into her account. We can use that information to create a model as follows:
</span>

<span>
We want to find the average value of that function </span>from the 2nd month to the 10th month, so its average value in the interval [2,10]. Remember that the formula for finding the average of a function over an interval is:

. So lets replace the values in our formula to find the average of our function:
![\frac{25(10)+100-[25(2)+100]}{10-2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B25%2810%29%2B100-%5B25%282%29%2B100%5D%7D%7B10-2%7D%20)



We can conclude that <span>the average rate of change in Eliza's account from the 2nd month to the 10th month is $25.</span>
The paraboloid meets the x-y plane when x²+y²=9. A circle of radius 3, centre origin.
<span>Use cylindrical coordinates (r,θ,z) so paraboloid becomes z = 9−r² and f = 5r²z. </span>
<span>If F is the mean of f over the region R then F ∫ (R)dV = ∫ (R)fdV </span>
<span>∫ (R)dV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] rdrdθdz </span>
<span>= ∫∫ [θ=0,2π, r=0,3] r(9−r²)drdθ = ∫ [θ=0,2π] { (9/2)3² − (1/4)3⁴} dθ = 81π/2 </span>
<span>∫ (R)fdV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] 5r²z.rdrdθdz </span>
<span>= 5∫∫ [θ=0,2π, r=0,3] ½r³{ (9−r²)² − 0 } drdθ </span>
<span>= (5/2)∫∫ [θ=0,2π, r=0,3] { 81r³ − 18r⁵ + r⁷} drdθ </span>
<span>= (5/2)∫ [θ=0,2π] { (81/4)3⁴− (3)3⁶+ (1/8)3⁸} dθ = 10935π/8 </span>
<span>∴ F = 10935π/8 ÷ 81π/2 = 135/4</span>