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Kruka [31]
2 years ago
8

A bouncing ball reaches a height of 54 inches at its first peak, 36 inches at its second peak, and 24 inches at its third peak.

Which formula represents this scenario?
f(x) = 54(two-thirds) Superscript x
f(x) = 54(two-thirds) Superscript x minus 1
f(x) = Two-thirds(54)x
f(x) = Two-thirds(54)x – 1
Mathematics
2 answers:
Olegator [25]2 years ago
6 0

Answer: B

Step-by-step explanation:

Just took the test

sesenic [268]2 years ago
4 0

Answer:

f(x) = 54(two-thirds) Superscript x minus 1

Step-by-step explanation:

Given that:

First peak : 36 / 54=2/3

Second peak : 24 / 36 = 2/3

The common ratio here is 2/3 ; which mean each bounce height is 2/3 of previous height

Modeling this using geometric progression :

An=a1r^(n-1)

An = nth term of a geometric progression

a1=first term

r=common ratio = 2/3

n = nth term

a1=54

Substituting into the above formular :

An=54(2/3)^(n-1)

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Read 2 more answers
Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{
SVETLANKA909090 [29]

y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\

\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\

This shows that y' = y is true when y = e^x

-----------------------

  • Note 1: A more general solution is y = Ce^x for some constant C.
  • Note 2: It might be tempting to say the general solution is y = e^x+C, but that is not the case because y = e^x+C \to y' = e^x+0 = e^x and we can see that y' = y would only be true for C = 0, so that is why y = e^x+C does not work.
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hodyreva [135]

Answer: that the solution is true

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Answer:

Step-by-step explanation:

I will assume that the scale factor is 2.5

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