Rotation as you could rotate it 360 degrees back into its original place
The answer is 257.92 dollars. Comment if you need more help :)
Step-by-step explanation:
Let the height above which the ball is released be H
This problem can be tackled using geometric progression.
The nth term of a Geometric progression is given by the above, where n is the term index, a is the first term and the sum for such a progression up to the Nth term is
To find the total distance travel one has to sum over up to n=3. But there is little subtle point here. For the first bounce ( n=1 ), the ball has only travel H and not 2H. For subsequent bounces ( n=2,3,4,5...... ), the distance travel is 2×(3/4)n×H
a=2H..........r=3/4
However we have to subtract H because up to the first bounce, the ball only travel H instead of 2H
Therefore the total distance travel up to the Nth bounce is
For N=3 one obtains
D=3.625H
let's firstly convert the mixed fraction to improper fraction and then divide.
![\bf \stackrel{mixed}{2\frac{1}{2}}\implies \cfrac{2\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{5}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{5}{2}\div 4\implies \cfrac{5}{2}\div \cfrac{4}{1}\implies \cfrac{5}{2}\cdot \cfrac{1}{4}\implies \cfrac{5}{8}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B2%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B5%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B5%7D%7B2%7D%5Cdiv%204%5Cimplies%20%5Ccfrac%7B5%7D%7B2%7D%5Cdiv%20%5Ccfrac%7B4%7D%7B1%7D%5Cimplies%20%5Ccfrac%7B5%7D%7B2%7D%5Ccdot%20%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B5%7D%7B8%7D)
<span><span><span><span><span><span>(<span>x+2</span>)</span><span>(<span>x<span>−2</span></span>)</span></span><span>(<span>x+1</span>)</span></span><span>(<span>x<span>−1</span></span>)</span></span></span><span>=0</span></span> true. x=−2,2,−1,1
hope this helps
