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4. 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?
An) - 50 (0.52)* ), 13.52 cm
O A(n) - 0.5 (0.52)*-1, 0.14 cm
A(n) = 50 (52)-1; 135,200 cm
O A(n) = 0.52. (0.5)*-1, 013 cm
305
90%

Respuesta :

calculista

Answer:

Part a) The rule of the sequence is [tex]A(n)=50(0.52^{n-1})[/tex]

Part b) The height of the ball will be [tex]13.52\ cm[/tex]

Step-by-step explanation:

Part a) Write a rule for the sequence using centimeters. The initial height is given by the term n = 1.

we know that

In a Geometric Sequence each term is found by multiplying the previous term by a constant called the common ratio (r)

In this problem we have a geometric sequence

Let

n-----> the number of path

a1 ----> is the initial height

r -----> the common ratio

we have

[tex]a1=0.5\ m=0.5*100=50\ cm[/tex]

[tex]r=52\%=52/100=0.52[/tex]

The rule for the sequence is equal to

[tex]A(n)=a1(r^{n-1})[/tex]

substitute

[tex]A(n)=50(0.52^{n-1})[/tex]

Part b) What height will the ball be at the top of the third path?

For n=3      

substitute in the equation

[tex]A(3)=50(0.52^{3-1})[/tex]  

[tex]A(3)=50(0.52^{2})[/tex]

[tex]A(3)=13.52\ cm[/tex]

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