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
3 is being added each time, so:
-25,-22,-19,-16,-13,-10,-7,-4,-1,2,5,8
I hope this helps!
~cupcake
Answer:
Step-by-step explanation:
the bottom left is the answer. I hope this helps.
Answer:
6 times 3 times 4
18 times 4
= 72
Step-by-step explanation:
hope this helps good luck
I draw the two triangles, see the picture attached.
As you can see, angle 1 and 2 are vertically opposite angles because they are formed by the same two crossing lines and they face each other.
Angles <span>ABQ and QPR, as well as angles BAQ and QRP, are alternate interior angles because they are formed by </span><span>two parallel lines crossed by a transversal, and they are inside the two lines on opposite sides of the transversal.</span>
Hence, Allison's correct claims are:
1 = 2 because they are vertically opposite angles. BAQ = QRP because they are alternate interior angles. Therefore Allison, in order to prove her claim, can use the AA similarity theorem: if two angles of a triangle are congruent to two angles of the other triangle, then the two triangles are similar.