Engage NY Eureka Math Precalculus Module 3 Lesson 17 Answer Key
Eureka Math Precalculus Module 3 Lesson 17 Example Answer Key
Example 1.
Find the domain and range for the following functions:
a. f: R → R given by f(x) = x2
Answer:
Domain: All real numbers
Range: All real numbers greater than or equal to 0
b. g: R → R given by g(x) = \(\sqrt{x – 2}\)
Answer:
Domain: Since x – 2 ≥ 0, x ≥ 2
Range: \(\sqrt{x – 2}\) ≥ 0, so g(x) ≥ 0
c. f(g(x))
Answer:
f(g(x)) = f(\(\sqrt{x – 2}\)) = (\(\sqrt{x – 2}\))2 = |x – 2|
Domain: x – 2 ≥ 0, so x ≥ 2
Range: |x – 2| ≥ 0, so f(g(x)) ≥ 0
d. g(f(x)) = g(x2) = \(\sqrt{x^{2} – 2}\)
Answer:
Domain: x2 – 2 ≥ 0, so x ≤ – \(\sqrt{2}\) or x ≥ \(\sqrt{2}\)
Range: \(\sqrt{x^{2} – 2}\) ≥ 0, so g(f(x)) ≥ 0
Example 2.
According to the Global Wind Energy Council, a wind turbine can generate about 16,400 kWh of power each day. According to the Alternative Fuels Data Center, an average electric car can travel approximately 100 miles on 34 kWh of energy. An environmental nonprofit organization is interested in analyzing how wind power could offset the energy use of electric vehicles.
a. Write a function that represents the relationship between the number of wind turbines operating in a wind farm and the amount of energy they generate per day (in kilowatt – hours). Define the input and output.
Answer:
E(t) = 16400t
t: number of turbines operating daily
E(t): energy (in kilowatt – hours) produced daily by the turbines
b. Write a function that represents the relationship between the energy expended by an electric car (in kilowatt – hours) and the number of miles driven.
Answer:
The relationship between miles driven and energy used is directly proportional, so m = kE.
Since the car drives 100 miles using 34 kWh, 100 = k(34) and k ≈ 2.9.
m(E) = 2.9E
E: amount of energy expended by an average electric car (in kilowatt – hours)
m(E): miles driven by the electric car
c. Write a function that could be used to determine the number of miles that an electric car could drive based on the number of wind turbines operating daily at a wind farm. Interpret this function in context.
Answer:
m(E(t)) = m(16400t) = 2.9(16400t) = 47560t
For every turbine operating in a wind farm daily, an average electric car can drive 47,560 miles.
d. Determine an appropriate domain and range for part (c). Explain why your domain and range are reasonable in this context.
Answer:
Domain: Whole numbers—The domain represents the number of turbines, which can only be represented with whole numbers.
Range: Whole numbers—Given the function m(E(t)) = 47560t, the outputs are found by multiplying a whole number by an input that is a whole number, which always produces a whole number. The range represents miles driven, and these values cannot be negative.
e. How many miles of driving could be generated daily by 20 wind turbines in a day?
Answer:
m(E(20)) = 20 × 47560 = 951 200. 951,200 miles of driving can be generated.
Eureka Math Precalculus Module 3 Lesson 17 Exercise Answer Key
Exercise 1.
Find the domain and range for the following functions:
a. f: R → R given by f(x) = x + 2
Answer:
Domain: All real numbers
Range: All real numbers
b. g: R → R given by g(x) = \(\sqrt{x – 1}\)
Answer:
Domain: x ≥ 1
Range: \(\sqrt{x – 1}\) ≥ 0, so g(x) ≥ 0
c. f(g(x))
Answer:
f(g(x)) = f(\(\sqrt{x – 1}\)) = \(\sqrt{x – 1}\) + 2
Domain: x – 1 ≥ 0, so x ≥ 1
Range: Since \(\sqrt{x – 1}\) ≥ 0, \(\sqrt{x – 1}\) + 2 ≥ 2, so f(g(x)) ≥ 2
d. g(f(x))
Answer:
g(f(x)) = g(x + 2) = \(\sqrt{(x + 2) – 1}\) = \(\sqrt{x + 1}\)
Domain: \(\sqrt{x + 1}\) ≥ 0, so x ≥ – 1
Range: \(\sqrt{x + 1}\) ≥ 0, so g(f(x)) ≥ 0
Exercises 2–3
Exercise 2.
A product safety commission is studying the effect of rapid temperature changes on the equipment of skydivers as they descend. The commission has collected data on a typical skydiver during the part of the dive when she has reached terminal velocity (maximum speed) to the time the parachute is released. They know that the terminal velocity of a diver is approximately 56 m/s and that, given the altitude of skydivers at terminal velocity, the temperature decreases at an average rate of 6.4 \(\frac{{ }^{\circ} \mathrm{C}}{\mathbf{k m}}\).
a. Write a function that represents the altitude of a skydiver experiencing terminal velocity if she reaches this speed at a height of 3,000 m.
Answer:
h(s) = 3000 – 56t
s: number of seconds spent descending at terminal velocity
h(s): altitude of the skydiver (in meters)
c. Write a function that represents the relationship between the altitude of the skydiver and the temperature if the temperature at 3,000 m is 5.8°C.
Answer:
t(h) – t(h1) = m(h – h1)
(h1,t(h1)) = (3000,5.8)
m = – 6.4 \(\frac{{ }^{\circ} \mathrm{C}}{\mathbf{k m}}\) = – 0.0064 \(\frac{{ }^{\circ} \mathrm{C}}{\mathbf{k m}}\)
t(h) – 5.8 = – 0.0064(h – 3000)
t(h) – 5.8 = – 0.0064h + 19.2
t(h) = – 0.0064h + 25
h: altitude of the skydiver (in meters)
t(h): temperature corresponding to the altitude of the skydiver
c. Write a function that could be used to determine the temperature, in degrees Celsius, of the air surrounding a skydiver based on the time she has spent descending at terminal velocity. Interpret the equation in context.
Answer:
t(h(s)) = t(3000 – 56t) = – 0.0064(3000 – 56t) + 25 ≈ 5.8 + 0.36t
A skydiver begins the portion of her dive at terminal velocity experiencing an air temperature of approximately 5.8°C, and the temperature increases by approximately 0.36°C for each second of descent until she deploys the parachute.
d. Determine an appropriate domain and range for part (c).
Answer:
Domain: Nonnegative real numbers
Range: Real numbers greater than or equal to 5.8
e. How long would it take a skydiver to reach an altitude where the temperature is 8°C?
Answer:
For a temperature of 8°C, 8 = 5.8 + 0.36t
0.36t = 2.2, so t ≈ 6.1 sec.
Exercise 3.
A department store manager is planning to move some cement spheres that have served as traffic barriers for the front of her store. She is trying to determine the relationship between the mass of the spheres and their diameter in meters. She knows that the density of the cement is approximately 2,500 kg/m3.
a. Write a function that represents the relationship between the volume of a sphere and its diameter. Explain how you determined the equation.
Answer:
V(d) = \(\frac{1}{6}\) πd3
The volume of a sphere is equal to
\(\frac{4}{3}\) πr3 = \(\frac{4}{3}\) π(\(\frac{d}{2}\))3 = \(\frac{1}{6}\) πd3
b. Write a function that represents the relationship between the mass and the volume of the sphere. Explain how you determined the function.
Answer:
m(V) = 2500V
Mass = density × volume = 2500 × V
c. Write a function that could be used to determine the mass of one of the cement spheres based on its diameter. Interpret the equation in context.
Answer:
m(V(d)) = m(\(\frac{1}{6}\) πd3 ) = 2500(\(\frac{1}{6}\) πd3 ) = \(\frac{2500}{6}\) πd3
The numerical value of the mass of one of the cement spheres is equal to approximately 1,300 times the value of the cubed diameter (measured in meters).
d. Determine an appropriate domain and range for part (c).
Answer:
Domain: Nonnegative real numbers
Range: Nonnegative real numbers
e. What is the approximate mass of a sphere with a diameter of 0.9 m?
Answer:
m(V(0.9)) = \(\frac{2500}{6}\) π(0.9)3 ≈ 950. The mass of the sphere is approximately 950 kg.
Eureka Math Precalculus Module 3 Lesson 17 Problem Set Answer Key
Question 1.
Find the domain and range of the following functions:
a. f: R → R by f(x) = – x2 + 2
Answer:
Domain: x: all real numbers
Range: f(x) ≤ 2
b. f: R → R by f(x) = \(\frac{1}{x + 1}\)
Answer:
Domain: x ≠ – 1
Range: f(x) ≠ 0
c. f: R → R by f(x) = \(\sqrt{4 – x}\)
Answer:
Domain: x ≤ 4
Range: f(x) ≥ 0
d. f: R → R by f(x) = |x|
Answer:
Domain: x: all real numbers
Range: f(x) ≥ 0
e. f: R → R by f(x) = 2x + 2
Answer:
Domain: x: all real numbers
Range: f(x) > 0
Question 2.
Given f: R → R and: R → R, for the following, find f(g(x)) and g(f(x)), and state the domain.
a. f(x) = x2 – x, g(x) = x – 1
Answer:
f(g(x)) = x2 – 3x + 2, Domain: x: all real numbers
g(f(x)) = x2 – x – 1, Domain: x: all real numbers
b. f(x) = x2 – x, g(x) = \(\sqrt{x – 2}\)
Answer:
f(g(x)) = x – 2 – \(\sqrt{x – 2}\), Domain: x ≥ 2
g(f(x)) = \(\sqrt{x^{2} – x – 2}\), Domain: x ≤ – 1 or x ≥ 2
c. f(x) = x2, g(x) = \(\frac{1}{x – 1}\)
Answer:
f(g(x)) = \(\frac{1}{(x – 1)^{2}}\), Domain: x ≠ 1
g(f(x)) = \(\frac{1}{x^{2} – 1}\), Domain: x ≠ ±1
d. f(x) = \(\frac{1}{x + 2}\), g(x) = \(\frac{1}{x – 1}\)
Answer:
f(g(x)) = \(\frac{x – 1}{2 x – 1}\), Domain: x ≠ 1 and x ≠ \(\frac{1}{2}\)
g(f(x)) = \(\frac{x + 2}{ – x – 1}\), Domain: x ≠ – 2 and x ≠ – 1
e. f(x) = x – 1, g(x) = log2(x + 3)
Answer:
f(g(x)) = log2(x + 3) – 1, Domain: x > – 3
g(f(x)) = log2(x + 2), Domain: x > – 2
Question 3.
A company has developed a new highly efficient solar panel. Each panel can produce 0.75 MW of electricity each day. According to the Los Angeles power authority, all the traffic lights in the city draw 0.5 MW of power per day.
a. Write a function that represents the relationship between the number of solar panels installed and the amount of energy generated per day (in MWh). Define the input and output.
Answer:
E(n) = 0.75n
n is the number of panels operating in one day.
E is the total energy generated by the n panels.
b. Write a function that represents the relationship between the number of days and the energy in megawatts consumed by the traffic lights. (How many days can one megawatt provide?)
Answer:
D(E) = \(\frac{1}{0.5}\) E = 2E
E: the energy in megawatts
D(E): the number of days per megawatt
c. Write a function that could be used to determine the number of days that the traffic lights stay on based on the number of panels installed.
Answer:
D(E(n)) = 2E(n) = 2 × 0.75n = 1.5n
d. Determine an appropriate domain and range for part (c).
Answer:
Domain: whole numbers
Range: whole numbers (whole number multiples of 1.5)
e. How many days can 20 panels power all the lights?
Answer:
D(E(20)) = 1.5 × 20 = 30. It takes 30 days.
Question 4.
A water delivery person is trying to determine the relationship between the mass of the cylindrical containers he delivers and their diameter in centimeters. The density of the bottles is 1 g/cm^3. The height of each bottle is approximately 60 cm.
a. Write a function that represents the relationship between the volume of the cylinder and its diameter.
Answer:
V(d) = 15πd2
b. Write a function that represents the relationship between the mass and volume of the cylinder.
Answer:
m(V) = 1V
c. Write a function that could be used to determine the mass of one cylinder based on its diameter. Interpret the equation in context.
Answer:
D(V(d)) = 1 × 15πd2 = 15πd2
The numerical value of the mass of one of the cylindrical water containers is equal to approximately 47.1 times the value of the squared diameter.
d. Determine an appropriate domain and range for part (c).
Answer:
Domain: nonnegative real numbers
Range: nonnegative real numbers
e. What is the approximate mass of a cylinder with a diameter of 30 cm?
Answer:
D(V(60)) = 15π(30)2 ≈ 42411 g ≈ 42.4.
The mass is approximately 42.4 kg.
Question 5.
A gold mining company is mining gold in Northern California. Each mining cart carries an average 500 kg of dirt and rocks that contain gold from the tunnel. For each 2 metric tons of material (dirt and rocks), the company can extract an average of 10 g of gold. The average wholesale gold price is $20/g.
a. Write a function that represents the relationship between the mass of the material mined in metric tons and the number of carts. Define the input and output.
Answer:
V(n) = 0.5n
n is the number of carts.
V(n) is the total mass of dirt and rock carried out by the n carts in metric tons.
b. Write a function that represents the relationship between the amount of gold and the materials. Define the input and output.
Answer:
G(V) = 0.000005V
V: the amount of material in metric tons
G(V): the mass of gold in metric tons
c. Write a function that could be used to determine the mass of gold in metric tons as a function of the number of carts coming out from the mine.
Answer:
G(V(n)) = 0.000 002 5n
d. Determine an appropriate domain and range for part (c).
Answer:
Domain: whole numbers
Range: positive real numbers
e. Write a function that could be used to determine the amount of money the gold is worth in dollars and the amount of gold extracted in metric tons.
Answer:
C(V(n)) = 20 000 000 × 0.000 002 5n = 50n
f. How much gold can 40,000 carts of material produce?
Answer:
G(V(40000)) = 0.1 metric ton
g. How much, in dollars, can 40,000 carts of material produce?
Answer:
C(V(40000)) = 50 × 40000 = 2 000 000 The cost is $2 000 000.
Question 6.
Bob operates hot air balloon rides for tourists at the beach. The hot air balloon rises, on average, at 100 feet per minute. At sea level, the atmospheric pressure, measured in inches of mercury (inHg), is 29.9 inHg. Using a barometric meter, Bob notices that the pressure decreases by 0.5 inHg for each 500 feet the balloon rises.
a. Write a function that represents the relationship between the height of the hot air balloon and the time spent to reach that height.
Answer:
H(t) = 100t
t is the number of minutes.
H(t) is the height of the hot air balloon at t time.
b. Write a function that represents the relationship between the height of the hot air balloon and the atmospheric pressure being applied to the balloon.
Answer:
P(H) = 29.9 – \(\frac{0.5 H}{500}\)
H: the height of the hot air balloon
P(H): the reading on the barometer at the height of the hot air balloon
c. Write a function that could be used to determine the pressure on the hot air balloon based on the time it spends rising.
Answer:
P(H(t)) = 29.9 – \(\frac{0.5 H(t)}{500}\) = 29.9 – 0.1t
d. Determine an appropriate domain and range for part (c).
Answer:
Domain: nonnegative real numbers
Range: nonnegative real numbers
e. What is the reading on the barometer 10 minutes after the hot air balloon has left the ground?
Answer:
P(H(10)) = 29.9 – 0.1t = 28.9 The reading is .9 inHg.
Eureka Math Precalculus Module 3 Lesson 17 Exit Ticket Answer Key
Question 1.
Timmy wants to install a wooden floor in a square room. The cost to install the floor is $24 per 4 square feet.
a. Write a function to find the area of the room.
Answer:
A(x) = x2
x: a side of a square
A(x): the area of the square room
b. Write a function for the installation cost per square foot.
Answer:
C(A) = 6A
A: the area of the floor
C(A): installation cost per square foot
c. Write the function to find the total cost to install the floor.
Answer:
C(A(x)) = 6x2
d. Show how the function in part (c) is the result of a composition of two functions.
Answer:
A(x) = x2
C(A(x)) = C(x2) = 6(x2) = 6x2
e. How much does it cost to install a wood floor in a square room with a side of 10 feet?
Answer:
C(A(10)) = 6(10)2 = 600. The cost is $600.00.