The answer you eliminated is the right answer which is A
The other rest are wrong because 5x5=25 and 5x5x5=125 if you times both of thos you should get 3125 and A 5x5x5x5x5=3125 Btw i used a calculator.
The answer is x= -21
3x-3=6(x-10)
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
To find a,construct a line that parallel to AB and DE cutting c, which you could refer to the photo((sorry for my terrible diagram ^^;;
from the photo,we can see that
a= <BCG + <GCE
<BCG=44° (alt.<, AB//FG)
<GCE=20° (alt<, FG//DE)
thus,a= 44°+20° =64°
hope it helps!
6c^2-5d+8
6 (5)^2-5(4)+8
30^2-20+8
900-20=8
880-8=872
ANSWER:872