Question

jds regents preparation, llc presents- algebra iexponential equations08 2016 1711.the table below shows the temperature, $t(m)$, of a cup of hot chocolate that is allowed to chill over several minutes, $m$.| time, m (minutes) | temperature, t(m) (°f) || ---- | ---- || 0 | 160 || 2 | 108 || 4 | 78 || 6 | 66 || 8 | 41 |which expression best fits the data for $t(m)$?1) $150(0.85)^m$2) $150(1.15)^m$3) $150(0.85)^{m-1}$4) $150(1.15)^{m-1}$08 2016 2412.milton has his money invested in a stock portfolio. the value, $v(x)$, of his portfolio can be modeled with the function $v(x)=30,000(0.78)^x$, where $x$ is the number of years since he made the investment.which statement describes the rate of change of the value of his portfolio?1) it decreases 78% per year.2) it decreases 22% per year.3) it increases 78% per year.4) it increases 22% per year.

Explanation:

Question 11

Step1: Identify initial value

At $m=0$, $T(0)=160$. The general exponential decay form is $T(m) = a(1-r)^m + C$, but here we test the options against the data. First, check the difference between initial temp and room temp (implied, since it's chilling): $160-10=150$ (the decaying part).

Step2: Test option 1 with m=2

Substitute $m=2$ into $150(0.85)^m$:

$$\begin{align*} 150(0.85)^2 &= 150 \times 0.7225 \\ &= 108.375 \approx 108 \end{align*}$$

This matches $T(2)=108$.

Step3: Verify with m=4

Substitute $m=4$ into $150(0.85)^4$:

$$\begin{align*} 150(0.85)^4 &= 150 \times (0.7225)^2 \\ &= 150 \times 0.52290625 \\ &= 78.4359375 \approx 78 \end{align*}$$

This matches $T(4)=78$.

Step1: Recognize decay function form

The function is $v(x)=30,000(0.78)^x$, which follows $v(x)=a(1-r)^x$ (exponential decay).

Step2: Calculate rate of change

Here, $1-r=0.78$, so solve for $r$:
$r = 1 - 0.78 = 0.22$
This means the value decreases by 22% per year.

Answer:

1. $150(0.85)^m$

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Question 12