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3 years ago

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One survey estimates that, on average, the retail value of a mid-sized car decreases by 8% annually. If the retail value of a ca

r is V dollars today, which expression represents the car’s value 1 year later?
A. 0.08V
B. 0.92V
C. 1.08V
D. V-0.08

2 answers:

nexus9112 [7]3 years ago

8 0

$\bf \qquad \textit{Amount for Exponential Decay}\\\\
A=I(1 - r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\to &V\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
t=\textit{elapsed time}\to &1\\
\end{cases}
\\\\\\
A=V(1-0.08)^1\implies A=V(0.92)\implies A=0.92V$

AURORKA [14]3 years ago

5 0

Answer:

sorry i just need points and this was 3 1.5 years ago....

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

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Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

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The positive value of $k$ will result in exactly one real root is approximately 0.028.

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Answer:

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Step-by-step explanation:

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Square root both sides to get y by itself:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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