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:
m<D = 14
Step-by-step explanation:
Exterior angles thm:
83 + 6x - 4 = 21x + 4
6x + 79 = 21x + 4
6x - 21x = 4 - 79
-15x = -75
x = 5
m<D:
6x - 4
6(3) - 4
18 - 4
14
Answer:
R > 7
Step-by-step explanation:
R+6> 13 (subtract 6 from both sides)
R > 13 - 6
R > 7
Answer:
y10 and x2 (I think).
Step-by-step explanation:
Answer:
40
Step-by-step explanation:
S x .35 = 14
S = 14/.35
S = 40
Verify:
40 x 35% = 14