A cup of coffee cools so that its temperature decreases by every minutes. The coffee is initially . Which equation models the temperature , in degrees Celsius, minutes after the coffee begins to cool?

Solution available with step-by-step explanation

SAT Nonlinear Functions Question 94 | Free SAT Prep | Aniko

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Question 94·Easy·Nonlinear Functions

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A cup of coffee cools so that its temperature decreases by $12\%$ every $10$ minutes. The coffee is initially $85\,{}^{\circ}\text{C}$ . Which equation models the temperature $T$ , in degrees Celsius, $t$ minutes after the coffee begins to cool?

For exponential growth/decay word problems, immediately write the model in the form $T = a \cdot b^{t/k}$ : identify the initial value $a$ , turn a percent decrease into a decay factor $b = 1 - \text{(percent as decimal)}$ , and set $k$ to the time interval over which that percent change occurs. Then check the exponent carefully (it should count how many intervals fit into $t$ ) and, if needed, plug in one simple time value like one full interval to see which choice matches the described change.

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Identify the type of function

The temperature changes by the same percentage every 10 minutes. What kind of function models repeated percentage changes over equal time intervals?

Turn a percent decrease into a multiplier

If a quantity decreases by 12%, what fraction of the original amount remains after each interval? Express that fraction as a decimal multiplier.

Handle the "every 10 minutes" part

The percent change happens every 10 minutes. How can you write the exponent so that it counts the number of 10‑minute intervals that have passed when $t$ minutes have gone by?

Check with a simple time value

After 10 minutes, the coffee should be 12% cooler than 85°C. Plug $t = 10$ into each choice: which equation gives a temperature that is slightly less than 85°C (not way smaller, and not larger)?

Desmos Guide

Enter each candidate function

In Desmos, define four functions (you can use $x$ instead of $t$ ):

$f(x) = 85(0.12)^{x/10}$
$g(x) = 85(1.12)^{x/10}$
$h(x) = 85(0.88)^{10x}$
$p(x) = 85(0.88)^{x/10}$

Check what happens after one time interval

Either make a table for each function (click the gear icon and add a table) or just type $f(10)$ , $g(10)$ , $h(10)$ , and $p(10)$ into Desmos. Compare the values: the correct model should give a temperature that is a bit less than 85°C (a moderate decrease), not close to 0 and not greater than 85.

Confirm the decay pattern

Look at the graphs of the four functions. The correct one should start at 85 when $x=0$ and decrease gradually over time, staying positive and cooling steadily, not dropping almost immediately to near 0 and not increasing.

Step-by-step Explanation

Recognize the exponential decay form

A repeated percent change over equal time intervals is modeled by an exponential function.

The general form here is

T = a \cdot b^{t/k}

where:

$a$ is the initial temperature,
$b$ is the multiplication factor each time interval,
$k$ is the length of the time interval (in minutes),
$t$ is the total time (in minutes).

Identify the initial value

The problem says the coffee is initially $85\,{}^{\circ}\text{C}$ .

That means when $t=0$ , $T=85$ , so in the model, $a = 85$ .

So we now have

T = 85 \cdot b^{t/k}.

Convert a 12% decrease into a multiplier

"Decreases by 12%" means after each time interval, the new temperature is 88% of the previous temperature, because

100\% - 12\% = 88\% = 0.88.

So the multiplication factor each interval is $b = 0.88$ (not $0.12$ and not $1.12$ ).

Now the model looks like

T = 85 \cdot 0.88^{t/k}.

Account for the 10-minute interval in the exponent and match the choice

The 12% decrease happens every 10 minutes, so each factor of $0.88$ corresponds to a 10‑minute interval.

The number of 10‑minute intervals in $t$ minutes is $t/10$ , so $k = 10$ and the exponent should be $t/10$ .

Substitute $k = 10$ into the model:

T = 85 \cdot 0.88^{t/10}.

This matches choice D, $T = 85(0.88)^{t/10}$ .

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Step-by-step Explanation

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