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You drop a ball from a height of 1.5 meters. Each curved path has 74% of the height of the previous path.
a. Write a rule for the sequence using centimeters. The initial height is given by the is term n = 1.
b. What height will the ball be at the top of the sixth path?

1) A(n) = 1.5 • (0.74)^x-1; 0.33cm
2)A(n) = 0.74 • (1.5)^x-1; 5.62cm
3) A(n) = 150•(0.74)^x-1; 33.29cm
4) A(n) = 150 • (74) ^x-1; 332,850,993,600cm

Respuesta :

MrRoyal

Answer:

[tex](a)\ A(n) = 150 * (0.74)^{n-1[/tex]

[tex](b)\ A(6) = 33.29cm[/tex]

Step-by-step explanation:

Given

[tex]a = 1.5[/tex] --- initial

[tex]b = 74\%[/tex] --- rate

Solving (a): The rule of the sequence

First convert [tex]a = 1.5[/tex] to centimeters

[tex]a = 1.5 * 100[/tex]

[tex]a = 150[/tex]

The equation is then calculated using the following geometric progression formula

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

This gives:

[tex]A(n) = 150 * (74\%)^{n-1}[/tex]

Express percentage as decimal

[tex]A(n) = 150 * (0.74)^{n-1[/tex]

Solving (b): The height on the 6th path

This implies that:

[tex]n = 6[/tex]

So, we have:

[tex]A(6) = 150 * (0.74)^{6-1[/tex]

[tex]A(6) = 150 * (0.74)^5[/tex]

[tex]A(6) = 150 * 0.222[/tex]

[tex]A(6) = 33.3cm[/tex]

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