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
We use this formula:
<span>Area = ½ • side 1 • sine (A) • side 2
</span>Area = <span>½ • 6 • sine (74) • 7
</span><span>Area = <span>21 • sine (74)
Area = 21*0.96126
Area = </span></span><span><span><span>20.18646
</span>
</span>
</span>
Area = 20.2 (rounded)
Source:
http://www.1728.org/triang.htm
We are given with the following:
Effective annual interest, i = 0.0425
Future worth, F = $25000
Number of years, n = 18
We use the formula to solve for the present worth of the money:
P = F / (1 + i)^n
P = 25000 / (1 + 0.0425)^18
P = 11818.73
The closest answer is:
<span>b.
$11,820</span>
Answer:
A
Step-by-step explanation:
