1/3(6x-12y)
2x - 4y
the answer is A
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:
Step-by-step explanation:
First you do the parentheses so the -3 x 8n= -24 and -3 x -5= 15
so it will look like this. -24n+15-2n=8n-21
Now combine like terms the -24n-2n.
-26n+15=8n-21 now subtract 8n now it will look like this
-34n+15=-21 now -15 from -21
-34n=-36
now if you want n by its self divide -34 and -36
