Grade 9 · Algebra & functions

Exponential functions in context (SAT) practice

Exponential functions in context (SAT) is a grade 9 math skill aligned to Common Core standard HSF.LE.A.2: construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 10 exponential functions in context (sat) problems our math games drill.

CCSS HSF.LE.A.210 questions in the bank
Sample questions

Try 8 for free

Question 1easy

The function gg is defined by g(t)=840(0.5)t/4g(t) = 840(0.5)^{t/4}. The function gg models the mass, in milligrams, of a medicine in a patient's bloodstream, where tt is the number of hours after an injection. According to the model, what is the estimated mass, in milligrams, of the medicine at the time of the injection?

At the time of injection, t=0t = 0. So g(0)=840(0.5)0=840g(0) = 840(0.5)^0 = 840 milligrams.

Question 2easy

The function hh is defined by h(t)=125(1.18)th(t) = 125(1.18)^t. The function hh models the number of members in a club, where tt is the number of years after the club was founded. According to the model, what is the estimated number of members when the club was founded?

When the club was founded, t=0t = 0. So h(0)=125(1.18)0=125h(0) = 125(1.18)^0 = 125 members.

Question 3easy

The function NN is defined by N(t)=680(0.72)t/8N(t) = 680(0.72)^{t/8}. The function NN models the number of particles detected in a radiation beam, where tt is the number of millimeters the beam has traveled into a shielding material. According to the model, what is the estimated number of particles in the beam at the surface of the shielding material?

At the surface, the beam has traveled 00 millimeters, so t=0t = 0. Then N(0)=680(0.72)0=680N(0) = 680(0.72)^0 = 680 particles.

Question 4easy

The function SS is defined by S(t)=450(0.85)t/3S(t) = 450(0.85)^{t/3}. The function SS models the height, in centimeters, of a snowpack, where tt is the number of days after a winter storm. According to the model, what is the estimated height, in centimeters, of the snowpack 33 days after the storm?

When t=3t = 3, the exponent is 3/3=13/3 = 1. So S(3)=450(0.85)1=382.5S(3) = 450(0.85)^1 = 382.5 centimeters.

Question 5easy

The function II is defined by I(t)=275(1.04)tI(t) = 275(1.04)^t. The function II models the balance, in thousands of dollars, of an investment account, where tt is the number of months after the account was opened. According to the model, what is the estimated balance, in thousands of dollars, of the account 11 month after it was opened?

When t=1t = 1, I(1)=275(1.04)1=2751.04=286I(1) = 275(1.04)^1 = 275 \cdot 1.04 = 286 thousand dollars.

Question 6easy

The function LL is defined by L(w)=180(0.88)w/2L(w) = 180(0.88)^{w/2}. The function LL models the light output, in lumens, from a bulb, where ww is the number of weeks the bulb has been in use. According to the model, what is the estimated light output, in lumens, after 22 weeks of use?

When w=2w = 2, the exponent is 2/2=12/2 = 1. So L(2)=180(0.88)1=158.4L(2) = 180(0.88)^1 = 158.4 lumens.

Question 7medium

The function DD is defined by D(t)=1,500(0.6)t/5D(t) = 1{,}500(0.6)^{t/5}. The function DD models the amount of a drug, in micrograms, in a patient's bloodstream, where tt is the number of hours after a dose is given. Which statement best describes what the factor 0.60.6 represents in this model?

Every 55 hours, the exponent increases by 11, so the amount is multiplied by 0.60.6. That means 60%60\% of the previous amount remains after each 55-hour interval.

Question 8medium

The function RR is defined by R(t)=900(0.5)t/12R(t) = 900(0.5)^{t/12}. The function RR models the intensity of background radiation, in millisieverts, measured tt months after a person enters an underground shelter. Which statement best describes what the 1212 in the expression t/12t/12 represents?

The exponent is t/12t/12, so when tt increases by 1212, the exponent increases by 11 and the intensity is multiplied by 0.50.5. The 1212 is the number of months for one full application of the factor 0.50.5.

Drill it inside a game.

The free placement test finds your level, then every match serves exponential functions in context (sat) questions at exactly the right difficulty.