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:
Limit is 2
Step-by-step explanation:
The last term approaches zero as n approaches infinity so the fuction approaches 2.
Answer:
I think the answer may be 465
Step-by-step explanation:
105+115=220
150+95=245
245+220=465
These are vertical angles so they are equal so
6a + 11 = 2a + 83
now solve for a
Step One
Calculate the number of feet traveled in 1 rotation.
Formula
C = π*d
P = 3.14
d = 9 inches = 9/12 feet = 3/4 of a foot.
C = 3.14 * 3/4 = 2.355 So that means that every time the tire turns around 1 complete turn, the distance traveled on the ground is 2.355 feet.
Step Two
Figure out the number of revolutions.
1 revolution = 2.355 feet
x revolutions = 300 feet.
1/x = 2.355/ 300 Cross multiply
2.355 feet * x = 1 rev * 300 feet
2.355 x = 300 rev Divide by 2.355
x = 300 / 2.355
x = 127.39 revolutions. <<<< Answer
