After a valve is adjusted, the water level in a tank begins to approach a stable height. The equation gives the water level , in centimeters, as a function of time , in minutes, where . Which choice is the best interpretation of 0.97 in this context?

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Super-Hard SAT Math Question 184 | Free SAT Prep | Aniko

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Question 184·200 Super-Hard SAT Math Questions·Advanced Math

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After a valve is adjusted, the water level in a tank begins to approach a stable height. The equation

h(t)=40-15(0.97)^{t-5}

gives the water level $h(t)$ , in centimeters, as a function of time $t$ , in minutes, where $t \ge 5$ . Which choice is the best interpretation of 0.97 in this context? So⁠urcе: anікο.ai

For exponential models with a vertical shift (like $k+a\,b^{t-h}$ or $k-a\,b^{t-h}$ ), interpret the base $b$ as the per-step multiplier of the distance from the horizontal asymptote $y=k$ , not necessarily the per-step multiplier of the function value itself. A quick move is to rewrite the “gap” as $k-h(t)$ (or $h(t)-k$ ) and compare the gap at $t$ and $t+1$ . © Аnікo

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Look at what the function approaches

Because $0.97<1$ , the exponential part shrinks over time. What value does $h(t)$ get closer to as $t$ increases? Frοm aniкo.aі

Focus on the difference from that value

Rewrite the expression for how far the water level is from the stable height, such as $40-h(t)$ .

Compare consecutive minutes

Find a relationship between $40-h(t+1)$ and $40-h(t)$ to see what factor is applied each minute.

Desmos Guide

Graph the function and the stable height

Enter $h(t)=40-15(0.97)^{t-5}$ and also enter $y=40$ . Notice that the graph of $h$ approaches the line $y=40$ as $t$ increases. Anікo ‍- Freе‍ SAT Р⁠rep

Define the gap below 40

Enter $d(t)=40-h(t)$ . This represents how far below 40 cm the water level is.

Check the factor from one minute to the next

Make a table for $d(t)$ at several integer values (for example, $t=5,6,7,8$ ). Compute (mentally or by adding another expression) the ratio $d(t+1)/d(t)$ for consecutive rows. The ratio stays constant at 0.97, showing what 0.97 multiplies each minute.

Step-by-step Explanation

Identify the stable (long-term) height

As $t$ increases, $(0.97)^{t-5}$ gets closer to 0 (since $0.97<1$ ).

So $h(t)=40-15(0.97)^{t-5}$ gets closer to $40$ . That means 40 cm is the stable height the water level approaches.

Write the difference from the stable height

The amount by which the water level is below 40 cm is аni‌k⁠ο.аі⁠ ЅAТ ⁠Quе‍stіοn Вank

40-h(t)=40-\bigl(40-15(0.97)^{t-5}\bigr)=15(0.97)^{t-5}.

See what happens after 1 more minute

Compare the below-40 amount at times $t$ and $t+1$ :

\begin{aligned} 40-h(t) &= 15(0.97)^{t-5} \\ 40-h(t+1) &= 15(0.97)^{(t+1)-5}=15(0.97)^{t-5}(0.97). \end{aligned}

So,

40-h(t+1)=0.97\bigl(40-h(t)\bigr).

Choose the interpretation that matches this factor

This shows that each minute after $t=5$ , the amount by which the water level is below 40 cm is multiplied by 0.97.

Correct choice: Each minute after $t=5$ , the amount by which the water level is below 40 centimeters is multiplied by 0.97.

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Step-by-step Explanation

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