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
To find the area you do base times height. Base is the number across,and height is the number that is up and down.So in this case you will multiply 9 and 5 and 9 and 7 and get 45 and 72.Then you will add 45 and 72 together and get 117. So the area is 117. Answer 117. Hope this helps :D.
The negitives are far from the 0 so its less than the non negitives
Answer:
Step-by-step explanation:
Let's retype that as y = b^x. The inverse of this function is log y = x
b
which you could also write as y = ln x
b
If you had y = e^x, the inverse function would be y = ln x.
Note that if you let ln x be the input to y = e^x, the result would be just 'x,' which confirms that these two functions are inverses of one another.
