Alright, so we have the difference = 4500-2800=1700. To find the percent increase, we do 1700/2800=around 0.61 = around 61% percent
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:
The coordinates of H′ are (-2,0).
Step-by-step explanation:
It is given that square EFGH stretches vertically by a factor of 2.5 with respect to the x-axis to create rectangle E′F′G′H′.
If a figure stretches vertically by a factor of 2.5 with respect to the x-axis, then the x-coordinates remains same and the points which lie on the axis are also remains the same after stretch.
It is given that the coordinates of H are (-2,0). This point lie on the x-axis, therefore it will remains the same after stretch.
Therefore the coordinates of H′ are (-2,0).
A) and translate figure A 11 units left and 5 units down
