Could I have the choices?
Answer:
Step-by-step explanation:
In general, remember that the converse of 
is 
, therefore in this case 
so
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
Congruent sides, so isosceles triangle. When we have an isosceles triangle the bisector of angle A is a perpendicular bisector of BC with foot D. So x=ADB=90 degrees.
Angle B is 47 degrees and angle y=BAD is complementary, the other acute angle in a right triangle.
y = 90 - 47 = 43
Answer: x=90, y=43
