Answer: 05%
Step-by-step explanation:
60-57=3
3/60=0.05
0.05x100=5
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
Answer is 264
You just multiply at three numbers
<h2>33 km</h2>
3.3cm = 0.000033km (divide by 100,000.
So, 0.000033km/yr.
0.000033 × 1,000,000
=33km
Plug in (7, 12) for x and y in the equation.
12 = 2(7) + 1
12 = 14 + 1
12 ≠ 15
Therefore, (7, 12) is not on the straight line equation.
