Let's work through the problem step-by-step.
### Part A: Finding the Rule for the Sequence
First, let's convert the initial height from meters to centimeters because the answers need to be in centimeters.
0.5 meters = 50 centimeters
The height of the ball after each bounce is 52% of the height of the previous bounce. To write a rule for this sequence, we can use the geometric sequence formula:
\[ A(n) = A(1) \cdot r^{(n-1)} \]
where:
- \( A(n) \) is the height after the nth bounce,
- \( A(1) \) is the initial height,
- \( r \) is the common ratio (52% or 0.52),
- \( n \) is the bounce number.
So, the formula becomes:
\[ A(n) = 50 \cdot 0.52^{(n-1)} \]
Let's compare this with the given options:
1. \( A(n) = 50 \cdot (52)^{x-1}; 135,200 cm \)
2. \( A(n) = 0.5 \cdot (0.52)^{x-1}; 0.14 cm \)
3. \( A(n) = 0.52 \cdot (0.5)^{x-1}; 0.13 cm \)
4. \( A(n) = 50 \cdot (0.52)^{x-1}; 13.52 cm \)
The correct sequence rule is:
\[ A(n) = 50 \cdot (0.52)^{x-1} \]
So the correct answer is:
\[ A(n) = 50 \cdot (0.52)^{x-1}; 13.52 cm \]
### Part B: Finding the Height at the Top of the Third Path
Now, let's use the rule \( A(n) = 50 \cdot 0.52^{(n-1)} \) to find the height at the top of the third path.
For the third path, \( n = 3 \):
\[ A(3) = 50 \cdot 0.52^{(3-1)} \]
\[ A(3) = 50 \cdot 0.52^{2} \]
\[ A(3) = 50 \cdot 0.2704 \]
\[ A(3) = 13.52 \, \text{cm} \]
So, at the top of the third path, the height of the ball will be 13.52 cm.
Therefore, the formula is:
\[ A(n) = 50 \cdot (0.52)^{x-1}; 13.52 \, \text{cm} \]
7.
You drop a ball from a height of 0.5 meter. Each curved path has 52% of the height of the previous path.
a. Write a rule for the sequence using centimeters. The initial height is given by the term n = 1.
b. What height will the ball be at the top of the third path?
A(n) = 50 ∙ (52)x − 1; 135,200 cm
A(n) = 0.5 ∙ (0.52)x − 1; 0.14 cm
A(n) = 0.52 ∙ (0.5)x − 1; 0.13 cm
A(n) = 50 ∙ (0.52)x − 1; 13.52 cm
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