We have been given that you drop a ball from a window 50 metres above the ground. The ball bounces to 50% of its previous height with each bounce. We are asked to find the total distance traveled by up and down from the time it was dropped from the window until the 25th bounce.
We will use sum of geometric sequence formula to solve our given problem.
, where,
a = First term of sequence,
r = Common ratio,
n = Number of terms.
For our given problem 
, 
and 
.
Therefore, the ball will travel 100 meters and option B is the correct choice.
Answer:
Step-by-step explanation:
Here are a few doubles facts:
5+5=10
2+2=4
3+3=6
A double is simply a pair of identical numbers added together. There's a pair of doubles you can <em>subtract </em>1 from to get 6+7, and there's a pair you can <em>add</em> 1 to get the same answer. What are those pairs?
Hint: If you take the example 3+4, you can either <em>subtract 1</em> from the double 4+4 or <em>add 1</em> to the double 3+3 to obtain your answer.
Answer:
9
Step-by-step explanation:
Using the property that
a
m
a
n
=
a
m
−
n
, we have
3
4
2
2
=
3
4
−
2
=
3
2
=
9
Note that if we evaluated the numerator and denominator first, we would arrive at the same result:
3
4
3
2
=
81
9
=
9
18. The perimeter is simply √3 + √3 + √3 + √3 or 4√3cm, since the perimeter is just all sides added together. You could add the decimal numbers together using a calculator, which I'm not sure if you're supposed to do in your class.
The area is just width times length, so √3 • √3 = 3cm².
19. The perimeter is 2√5 + 2(9 - √5).
This can also be written as 2√5 + 18 - 2√5, which leaves you with a perimeter of 18ft.
The area would be √5 • (9 - √5), which leaves you with (9√5 - 5)ft².
20. The formula for the perimeter (or circumference) of a circle is π times the diameter of the circle. Using the radius of the circle, 1/π, the diameter is 2/π, so
π • 2/π = 2. The circumference of the circle is 2 inches.
The area of the circle is calculated with the equation πr², so
π(1/π)² = π • 1/(π²) = π/(π²) = π. The area is simply π in².
