SAT Nonlinear Functions Question 213 | Free SAT Prep | Aniko

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Question 213·Hard·Nonlinear Functions

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$T(h)=740(0.88)^{\tfrac{2}{3}h}$

The function $T$ models the temperature, in degrees Celsius, of a cooling liquid $h$ hours after it is removed from a heat source. According to the model, the temperature decreases by $12\%$ every $k$ minutes. What is the value of $k$ ?

For exponential decay questions that say something like "decreases by p% every k units of time," recognize that the base of the exponent is $1-p$ (written as a decimal), and the exponent counts how many such time intervals have passed. Express the number of intervals in terms of the variable given (for example, $\frac{60h}{k}$ intervals in $h$ hours if each interval is $k$ minutes), then match this exponent to the exponent in the model and solve the resulting simple equation for $k$ . This approach is faster and more reliable than trying to reason about the percentage change per hour in your head.

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Connect 0.88 to the 12% decrease

Focus on the base of the exponent, $0.88$ . How is that related to a 12% decrease from 100%? What does this base represent in terms of repeated percentage changes?

Relate the exponent to the number of k-minute intervals

In models like $T(h)=740(0.88)^{\text{exponent}}$ , the exponent counts how many times the 12% decrease happens. If the decrease happens every $k$ minutes, how many such intervals occur in $h$ hours?

Match the given model to the "every k minutes" model

Write an expression for $T(h)$ assuming a 12% decrease every $k$ minutes (using the number of $k$ -minute intervals in $h$ hours), then compare its exponent with the exponent $(2/3)h$ in the given function. What equation does that give you for $k$ ?

Solve the equation for k

Once you have an equation relating $60$ , $k$ , and the fraction $2/3$ , solve it carefully. Watch out for inverting fractions incorrectly when you cross-multiply.

Desmos Guide

Use the exponent to find the time for one decay interval

In Desmos, graph the line $y = (2/3)x$ and the horizontal line $y = 1$ . The $x$ -coordinate of their intersection is the number of hours it takes for the exponent $(2/3)h$ to equal 1, meaning one full 12% decrease has occurred.

Convert that time from hours to minutes

Take the $x$ -value you found (in hours) and type that value times 60 into Desmos (for example, if it shows $x = 1.5$ , type 1.5*60). The output is the number of minutes for one decay interval; use that value as $k$ .

Step-by-step Explanation

Interpret the exponential model

The function is

T(h)=740(0.88)^{\tfrac{2}{3}h}.

In an exponential decay model of the form $\text{initial} \times (\text{decay factor})^{\text{number of intervals}}$ :

The base $0.88$ is the decay factor per interval.
$0.88 = 1 - 0.12$ , so it represents a 12% decrease each decay interval.
The exponent $(2/3)h$ is the number of such intervals that have happened after $h$ hours.

Express the number of k-minute intervals in h hours

If the liquid cools by 12% every $k$ minutes, then in $h$ hours:

There are $60h$ minutes total (because $1$ hour $=60$ minutes).
The number of $k$ -minute intervals in $h$ hours is $\dfrac{60h}{k}$ .

So the model would also look like

T(h) = 740(0.88)^{\frac{60h}{k}}.

Here, $\dfrac{60h}{k}$ is the number of 12%-drop intervals.

Match the exponents to find a relationship for k

We now have two expressions for $T(h)$ :

Given in the problem:

T(h) = 740(0.88)^{\tfrac{2}{3}h}

Written in terms of $k$ minutes:

T(h) = 740(0.88)^{\tfrac{60h}{k}}

For these to model the same situation, the exponents must be equal for all $h$ :

\frac{60h}{k} = \frac{2}{3}h.

Since this holds for every $h>0$ , we can divide both sides by $h$ to get:

\frac{60}{k} = \frac{2}{3}.

Solve for k

Now solve

\frac{60}{k} = \frac{2}{3}.

Cross-multiply:

60 \cdot 3 = 2k \\[0.7em] 180 = 2k.

Divide both sides by 2:

k = 90.

So the temperature decreases by 12% every $90$ minutes, and the correct answer is $90$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation