He goes 1.5k/h. To get that you divide the km by the hour. Then multiply the hour by 1.5 to get the km. (Ex. 2hrs x 1.5km = 3km)
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:
Rect Area= Lw
3D Rect Area= 2(wh+lw+lh)
Step-by-step explanation:
Answer:
Let total students who prefer lemonade is 2a and who prefer ice tea is a.
Students in 6th grade=39
2a+a=39
3a=39
a=39/3
a=13
students who prefer lemonate is 13
and those who prefer ice tea is 2a=26
The answer is C. 4x^5 + 3 + 2/x^5
