## Engage NY Eureka Math Precalculus Module 4 Lesson 13 Answer Key

### Eureka Math Precalculus Module 4 Lesson 13 Example Answer Key

Example
The Statue of Liberty is 151 feet tall and sits on a pedestal that is 154 feet above the ground. An observer who is 6 feet tall wants to stand at the ideal viewing distance in front of the statue.
a. Sketch the statue and observer. Label all appropriate measurements on the sketch, and define them in context.

h: height of the observer at eye level
a: height of pedestal
b: height of the pedestal and statue
x: horizontal distance between viewer and base of the statue
y: viewing angle
w: angle formed between line of sight from viewer to the base of the statue and a line segment parallel to x from the viewer’s eye level to the statue

b. How far back from the statue should the observer stand so that his or her viewing angle (from the feet of the statue to the tip of the torch) is largest? What is the value of the largest viewing angle?
h = 6
a = 151
b = 154 + 151 = 305
y = tan-1($$\frac{b-h}{x}$$) – tan-1($$\frac{a-h}{k}$$)
y = tan-1($$\frac{299}{x}$$) – tan-1($$\frac{148}{x}$$)

The ideal viewing distance is approximately 210.4 feet from the base of the statue, which produces a maximum viewing angle of approximately 19.7°.

c. What would be your best viewing distance from the statue?
Answers will vary. For example, for a viewer whose height is 5.5 feet:
y = tan-1 ($$\frac{299.5}{x}$$) – tan-1 ($$\frac{148.5}{x}$$), which has a maximum y value of 19.7° when x = 210.9. Therefore, the best viewing distance is 210.9 feet from the base of the statue.

d. If there are 66 meters of dry land in front of the statue, is the viewer still on dry land at the best viewing distance?
Since 66 meters is approximately 216.5 feet, the observer would be on land at our best viewing angle with about 6 feet between him or her and the water.

### Eureka Math Precalculus Module 4 Lesson 13 Exercise Answer Key

Exercise
Hanging on a museum wall is a picture with base a inches above a viewer’s eye level and top b inches above the viewer’s eye level.
a. Model the situation with a diagram.

b. Determine an expression that could be used to find the ideal viewing distance x that maximizes the viewing angle y.
y = tan-1 ($$\frac{b}{x}$$) – tan-1 ($$\frac{a}{x}$$)

c. Find the ideal viewing distance, given the a and b values assigned to you. Calculate the maximum viewing angle in degrees.
Answers will vary. An example of an appropriate response is shown.
For a = 1 and b = 60, y = tan-1 ($$\frac{60}{x}$$) – tan-1 ($$\frac{1}{x}$$)

The ideal viewing distance is approximately 7.8 inches, with a maximized viewing angle of 75.3°.

d. Complete the table using class data, which indicates the ideal values for x given different assigned values of a and b. Note any patterns you see in the data.

It appears that x is the geometric mean of a and b.

### Eureka Math Precalculus Module 4 Lesson 13 Problem Set Answer Key

Question 1.
Consider the situation of sitting down with eye level at 46 in. Find the missing distances and heights for the following:
a. The bottom of the picture is at 50 in. and the top is at 74 in. What is the optimal viewing distance?
$$\sqrt{4 \cdot 28}$$ = 4$$\sqrt{7}$$ ≈ 10.583

b. The bottom of the picture is at 52 in. and the top is at 60 in. What is the optimal viewing distance?
$$\sqrt{6 \cdot 14}$$ = 2$$\sqrt{21}$$ ≈ 9.165

c. The bottom of the picture is at 48 in. and the top is at 64 in. What is the optimal viewing distance?
$$\sqrt{2 \cdot 18}$$ = 6
6 in. away

d. What is the height of the picture if the optimal viewing distance is 1 ft. and the bottom of the picture is hung at 47 in.?
12 = $$\sqrt{1 \cdot x}$$
x = 144
12 ft. above eye level, so 15 ft. 10 in. tall

Question 2.
Consider the situation where you are looking at a painting a inches above your line of sight and b inches below your line of sight.
a. Find the optimal viewing distance if it exists.
It is not humanly possible to maximize the optimal viewing distance. The closer you get to the painting, the larger the angles become. Eventually, the picture fully encompasses the field of view of the person.

b. If the average standing eye height of Americans is 61.4 in., at what height should paintings and other works of art be hung?
The center of the artwork should be hung at eye level so that viewers can get the largest image of the picture they want by getting closer to the image without having to worry about finding the optimal viewing distance.

Question 3.
The amount of daylight per day is periodic with respect to the day of the year. The function y = -3.016 cos⁡($$\frac{2 \pi x}{365}$$) + 12.25 gives the number of hours of daylight in New York, y, as a function of the number of days since the winter solstice (December 22), which is represented by x.
a. On what days will the following hours of sunlight occur?
i. 15 hours, 15 minutes
15.25 = -3.016 cos⁡($$\frac{2 \pi x}{365}$$) + 12.25
3 = -3.016 cos⁡($$\frac{2 \pi x}{365}$$)
-0.995 = cos⁡($$\frac{2 \pi x}{365}$$)
cos-1 (-0.995) = cos-1⁡(cos($$\frac{2 \pi x}{365}$$))
($$\frac{2 \pi x}{365}$$) = 3.0416 and 3.2416
x ≈ 177 and 188
On the 177th and 188th days

ii. 12 hours
On the 86th and 279th days

iii. 9 hours, 15 minutes
On the 6th and 359th days

iv. 10 hours
On the 42nd and 323rd days

v. 9 hours
This will never happen. The function does not go that low.

b. Give a function that will give the day of the year from the solstice as a function of the hours of daylight.
y = $$\frac{365}{2 \pi}$$ ⋅ cos-1⁡(-$$\frac{x-12.25}{3.016}$$)

c. What is the domain of the function you gave in part (b)?
This function is only accurate for as long as the argument to the inverse cosine stays between -1 and 1, so |x – 12.25|≤3.016, which tells us 9.234 ≤ x ≤ 15.266.

d. What does the domain tell you in the context of the problem?
The hours of sunlight for the year vary between about 9 hours, 14 minutes and 15 hours, 16 minutes.

e. What is the range of the function? Does this make sense in the context of the problem? Explain.
The range of the function is 0 to 182.5. This only covers half of the year because if the entire year was covered, the inverse would not be a function. We can find other dates with the same amount of daylight by subtracting the number of the day from 365.

Question 4.
Ocean tides are an example of periodic behavior. At a particular harbor, data was collected over the course of
24 hours to create the following model: y = 1.236 sin⁡($$\frac{\pi}{3}$$ x) + 1.798, which gives the water level, y, in feet above the MLLW (mean lower low water) as a function of the time, x, in hours.
a. How many periods are there each day?
Four

b. Write a function that gives the time in hours as a function of the water level. How many other times per day will have the same water levels as those given by the function?
y = $$\frac{3}{\pi}$$ sin-1($$\frac{x-1.798}{1.236}$$)
There are potentially eight times per day that have the same water level (with the exception of peaks and troughs, which only occur four times per day).

### Eureka Math Precalculus Module 4 Lesson 13 Exit Ticket Answer Key

Question 1.
The pedestal that the Statue of Liberty sits on is 89 ft. tall with a foundation fashioned in the shape of an eleven-point star making up the rest of the height. The front point of the star juts out about 145 ft. from the front of the statue and stands about 35.2 ft. tall.
How far from the Statue of Liberty does someone whose eye-height is 6 ft. need to stand in order to see the base of the statue without being obscured by the foundation? Include a diagram and appropriate work to justify your answer.

Note that the larger triangle is similar to the smaller triangle, so the angles are equal. This tells us that
tan-1⁡($$\frac{89 + 35.2-6}{x + 145}$$) = tan-1⁡($$\frac{35.2-6}{x}$$)
$$\frac{118.2}{x + 142}$$ = $$\frac{29.2}{x}$$
118.2x = 29.2x + 4234
89x = 4234
x ≈ 47.6
The person will have to stand about 47.6 ft. away from the point of the star.

## Engage NY Eureka Math Precalculus Module 4 Lesson 14 Answer Key

### Eureka Math Precalculus Module 4 Lesson 14 Example Answer Key

Example 1.
A designer wants to test the safety of a wheelchair ramp she has designed for a building before constructing it, so she creates a scale model. To meet the city’s safety requirements, an object that starts at a standstill from the top of the ramp and rolls down should not experience an acceleration exceeding 2.4 $$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$ .
a. A ball of mass 0.1 kg is used to represent an object that rolls down the ramp. As it is placed at the top of the ramp, the ball experiences a downward force due to gravity, which causes it to accelerate down the ramp. Knowing that the force applied to an object is the product of its mass and acceleration, create a sketch to model the ball as it accelerates down the ramp.

b. If the ball rolls at the maximum allowable acceleration of 2.4 $$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$, what is the angle of elevation for the ramp?
Framp = mball×aball = 0.1 kg×2.4 $$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$ = 0.24 N parallel to the ramp and directed down the ramp
F = mball×agravity = 0.1 g×9.8 $$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$ = 0.98 N directed down toward the ground
sin (θ) = $$\frac{F_{\text {ramp }}}{F}$$ = $$\frac{0.24 \mathrm{~N}}{0.98 \mathrm{~N}}$$
sin-1 (sin (θ)) = sin-1($$\frac{0.24 \mathrm{~N}}{0.98 \mathrm{~N}}$$) ≈ 14.2°
The angle of elevation for the ramp is approximately 14.2°.

c. If the designer wants to exceed the safety standards by ensuring the acceleration of the object does not exceed 2.0 $$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$ , by how much will the maximum angle of elevation decrease?
Framp = mball×aball = 0.1 kg×2.0 $$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$ = 0.2 N parallel to the ramp and directed down the ramp
F = mball×agravity = 0.1 g×9.8 $$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$ = 0.98 N directed down toward the ground
sin (θ) = $$\frac{F_{\text {ramp }}}{F}$$ = $$\frac{0.24 \mathrm{~N}}{0.98 \mathrm{~N}}$$
sin-1 (sin (θ)) = sin-1 ($$\frac{0.24 \mathrm{~N}}{0.98 \mathrm{~N}}$$) ≈ 11.8°
The maximum angle of elevation would be approximately 11.8°, which is 2.4 degrees less than the maximum angle permitted to meet the safety specification.

d. How does the mass of the ball used in the scale model affect the value of θ? Explain your response.
It doesn’t. The mass of the ball is a common factor in the numerator and denominator in the ratio used to calculate θ.

Example 2.
The declination of the sun is the path the sun takes overhead the earth throughout the year. When the sun passes directly overhead, the declination is defined as 0°, while a positive declination angle represents a northward deviation and a negative declination angle represents a southward deviation. Solar declination is periodic and can be roughly estimated using the equation
δ = -23.44°(cos(($$\frac{360}{365}$$)(N + 10))), where N represents a calendar date (e.g., N = 1 is January 1, and δ is the declination angle of the sun measured in degrees).
a. Describe the domain and range of the function.
D: 1 ≤ N ≤ 365 where N is a counting number
R: -23.44° ≤ δ ≤ 23.44°

b. Write an equation that represents N as a function of δ.
δ = -23.44°(cos(($$\frac{360}{365}$$)(N + 10)))
$$\frac{\delta}{-23.44^{\circ}}$$ = (cos(($$\frac{360}{365}$$)(N + 10)))
cos-1 ($$\frac{\delta}{-23.44^{\circ}}$$) = ($$\frac{360}{365}$$)(N + 10)
-10+(365/360) cos-1 ($$\frac{\delta}{-23.44^{\circ}}$$) = N

c. Determine the calendar date(s) for the given angles of declination:
i. 10°
N = -10+($$\frac{360}{365}$$) cos-1 ($$\frac{10^{\circ}}{-23.44^{\circ}}$$) ≈ 107 and 238
N = 107 corresponds to a calendar date of April 17, and N = 238 corresponds to August 26.

ii. -5.2°
N = -10+($$\frac{360}{365}$$) cos-1 ($$\frac{-5.2^{\circ}}{-23.44^{\circ}}$$) ≈ 68 and 277
N = 68 corresponds to a calendar date of March 9, and N = 277 corresponds to October 4.

iii. 25°
No date will correspond to this angle because it lies outside of the domain of the function.

d. When will the sun trace a direct path above the equator?
When the sun passes directly overhead, the declination is 0°. This means that
N = -10+($$\frac{360}{365}$$) cos-1 ($$\frac{0^{\circ}}{-23.44^{\circ}}$$) ≈ 81 and 264, which correspond to the calendar dates March 22 and September 21.

### Eureka Math Precalculus Module 4 Lesson 14 Exercise Answer Key

Exercise 1.
A vehicle with a mass of 1,000 kg rolls down a slanted road with an acceleration of 0.07 $$\frac{0.24 \mathrm{~N}}{0.98 \mathrm{~N}}$$ . The frictional force between the wheels of the vehicle and the wet concrete road is 2,800 newtons.
a. Sketch the situation.

b. What is the angle of elevation of the road?
Froad = mvehicle×avehicle+Ffr = 1000 kg×0.07 $$\frac{0.24 \mathrm{~N}}{0.98 \mathrm{~N}}$$ = 70 N + 2800 N = 2870 N
Fgravity = mvehicle×agravity = 1000 kg×9.8 $$\frac{0.24 \mathrm{~N}}{0.98 \mathrm{~N}}$$ = 9800 N
sin (θ) = $$\frac{F_{\text {road }}}{F_{\text {gravity }}}$$ = $$\frac{2870 \mathrm{~N}}{9800 \mathrm{~N}}$$
sin-1 (sin (θ)) = sin-1 ($$\frac{2870 \mathrm{~N}}{9800 \mathrm{~N}}$$) ≈ 17°
The angle of elevation for the road is approximately 17°.

c. What is the maximum angle of elevation the road could have so that the vehicle described would not slide down the road?
If the vehicle does not slide, then the frictional force must be greater than or equal to the downward force parallel to the road. So, for the maximum angle θ, Froad = Ffriction = 2800 N.
sin (θ) = $$\frac{F_{\text {road }}}{F_{\text {gravity }}}$$ = $$\frac{2870 \mathrm{~N}}{9800 \mathrm{~N}}$$
sin-1 (sin (θ)) = sin-1 ($$\frac{2870 \mathrm{~N}}{9800 \mathrm{~N}}$$) ≈ 16.6°
The angle of elevation for the road that would prohibit sliding is approximately 16.6°.

Exercises 2–3
Exercise 2.
The average monthly temperature in a coastal city in the United States is periodic and can be modeled with the equation y = -8 cos⁡((x-1)(π/6))+17.5, where y represents the average temperature in degrees Celsius and
x represents the month, with x = 1 representing January.
a. Write an equation that represents x as a function of y.
x = 1+$$\frac{6}{\pi}$$ cos-1 ($$\frac{y-17.5}{-8}$$)

b. A tourist wants to visit the city when the average temperature is closest to 25° Celsius. What recommendations would you make regarding when the tourist should travel? Justify your response.
x = 1+$$\frac{6}{\pi}$$ cos-1 ($$\frac{25-17.5}{-8}$$)) ≈ 6.32
If the tourist wants to visit when the temperature is closest to 25° Celsius, she should travel about the second week in June.

Exercise 3.
The estimated size for a population of rabbits and a population of coyotes in a desert habitat are shown in the table. The estimated population sizes were recorded as part of a long-term study related to the effect of commercial development on native animal species.

a. Describe the relationship between sizes of the rabbit and coyote populations throughout the study.
The rabbit population started at approximately 15,000 rabbits and then decreased while the coyote population increased (perhaps because of the abundance of prey for the coyotes). Over time, both the rabbit and coyote populations declined until the coyote population was about 1,800, when the rabbit population increased again. Both species’ population numbers appear to cycle, with the coyotes’ values shifted about
3 years from the rabbit’s values.

b. Plot the relationship between the number of years since the initial count and the number of rabbits. Fit a curve to the data.

Equation of curve: r = 5000 cos($$\frac{\pi n}{6}$$)+10000

c. Repeat the procedure described in part (b) for the estimated number of coyotes over the course of the study.

Equation of curve: c = 200 sin($$\frac{\pi n}{6}$$)+2000

d. During the study, how many times was the rabbit population approximately 12,000? When were these times?
r = 5000 cos($$\frac{\pi n}{6}$$)+10000
n = $$\frac{6}{\pi}$$ cos-1 ($$\frac{r-10000}{5000}$$)
n = $$\frac{6}{\pi}$$ cos-1 ($$\frac{12000-10000}{5000}$$) ≈ 2.2
Given that the function cycles every 12 years, n = 0 + 2.2 = 2.2; n = (12-2.2) = 9.8; n = (12 + 2.2) = 14.2; and n = (24-2.2) = 21.8. The rabbit population was approximately 12,000 four times at 2.2, 9.8, 14.2, and 21.8 years.

e. During the study, when was the coyote population estimate below 2,100?
c = 200 sin($$\frac{\pi n}{6}$$)+2000
n = $$\frac{6}{\pi}$$ sin-1 ($$\frac{c-2000}{200}$$)
n = $$\frac{6}{\pi}$$ sin-1 ($$\frac{2100-2000}{200}$$) = 1,5,13,17
By analyzing the graph of the coyote population estimates, the population of coyotes was less than 2,100 prior to n = 1, between n = 5 and n = 13, and from n = 17 to n = 24.

### Eureka Math Precalculus Module 4 Lesson 14 Problem Set Answer Key

Question 1.
A particle is moving along a line at a velocity of y = 3 sin⁡($$\frac{2 \pi x}{5}$$)+2 $$\frac{m}{s}$$ at location x meters from the starting point on the line for 0 ≤ x ≤ 20.
a. Find a formula that represents the location of the particle given its velocity.
y = $$\frac{5}{2 \pi}$$⋅sin-1($$\frac{x-2}{3}$$)

b. What is the domain and range of the function you found in part (a)?
The domain is -1 ≤ x ≤ 5, and the range is –$$\frac{5}{4}$$ ≤ y ≤ $$\frac{5}{4}$$.

c. Use your answer to part (a) to find where the particle is when it is traveling 5 m/s for the first time.
The particle will be located at x = 1.2 meters from the starting point on the line.

d. How can you find the other locations the particle is traveling at this speed?
In this case, the velocity is a maximum, so it will only occur once every period. All other values can be found by adding multiples of 5 to the location. If it was not a maximum, we could subtract the location from 5/2 to find another value within the same period and then add multiples of 5 to find analogous values in other periods.

Question 2.
In general, since the cosine function is merely the sine function under a phase shift, mathematicians and scientists regularly choose to use the sine function to model periodic phenomena instead of a mixture of the two. What behavior in data would prompt a scientist to use a tangent function instead of a sine function?
The tangent function has infinitely many vertical asymptotes and rapidly takes on extreme values. Since the tangent function is the ratio between the sine and cosine functions, it will probably show up when comparing the ratio of two sets of periodic data. Otherwise, the extreme values would be reasons to use the tangent function.

Question 3.
A vehicle with a mass of 500 kg rolls down a slanted road with an acceleration of 0.04 $$\frac{\mathrm{m}}{s^{2}}$$ . The frictional force between the wheels of the vehicle and the road is 1,800 newtons.
a. Sketch the situation.

b. What is the angle of elevation of the road?
Froad = ma + Ffr = 500 kg×0.04 $$\frac{\mathrm{m}}{s^{2}}$$ + 1800 N = 20 N + 1800 N = 1820 N
Fg = mg = 500 kg×9.8 $$\frac{\mathrm{m}}{s^{2}}$$ = 4900 N
sin(θ) = $$\frac{F_{\text {road }}}{F_{\mathrm{g}}}$$ = $$\frac{1820 \mathrm{~N}}{4900 \mathrm{~N}}$$
sin-1 (sin(θ)) = sin-1 ($$\frac{1820 \mathrm{~N}}{4900 \mathrm{~N}}$$) ≈ 21.8°
The angle of elevation for the road is approximately 21.8°.

c. The steepness of a road is frequently measured as grade, which expresses the slope of a hill as a ratio of the change in height to the change in the horizontal distance. What is the grade of the hill described in this problem?
We need to find the perpendicular force. We get
cos(21.8) = $$\frac{F_{\text {perp }}}{4900}$$
Fperp ≈ 4549.46.
So the grade of the hill is $$\frac{1820}{4549.46}$$ ≈ 40%.

Question 4.
Canton Avenue in Pittsburgh, PA is considered to be one of the steepest roads in the world with a grade of 37%.
a. Assuming no friction on a particularly icy day, what would be the acceleration of a 1,000 kg car with only gravity acting on it?
tan⁡(θ) = 0.37
θ ≈ 20.30
Since the acceleration due to gravity is the only acceleration on the car, the acceleration due to gravity is being transferred into an acceleration as the car goes down the hill. We can envision the force due to gravity as two separate forces, one that is parallel to the road and the second that is perpendicular to the road. The force parallel to the road is F = m⋅g⋅sin⁡(θ), and the acceleration parallel to the road is
a = g⋅sin⁡(θ)
a = 9.8 sin⁡(20.3)
a ≈ 3.4
The acceleration is 3.4 $$\frac{\mathrm{m}}{s^{2}}$$ .

b. The force due to friction is equal to the product of the force perpendicular to the road and the coefficient of friction μ. For icy roads and a non-moving vehicle, assume the coefficient of friction is μ = 0.3. Find the force due to friction for the car above. If the car is in park, will it begin sliding down Canton Avenue if the road is this icy?
The force perpendicular to the road is 9191.3 N. F = m⋅g⋅cos⁡(θ) = 1000⋅9.8⋅cos⁡(20.3) ≈ 9191.3
Thus, the force due to friction is approximately 2757.4 N. μF ≈ 2757.4.
The force parallel to the road is 3400.0 N. F = 1000⋅9.8⋅sin⁡(20.3) ≈ 3400.0
Because the force parallel to the road is greater than the force due to friction, the car will slide down the road once Canton Avenue gets this icy.

c. Assume the coefficient of friction for moving cars on icy roads is μ = 0.2. What is the maximum angle of road that the car will be able to stop on?
The car will be able to slow (and eventually stop) when the force due to friction is greater than the force parallel to the road, so we need to solve,
9800⋅sin⁡(θ) = 0.2⋅9800 cos⁡(θ)
$$\frac{\sin (\theta)}{\cos (\theta)}$$ = 0.2
tan⁡(θ) = 0.2.
So the car will be able to slow on any hill with less than a 20% grade, which corresponds to an angle of 11.3°.

Question 5.
Talladega Superspeedway has some of the steepest turns in all of NASCAR. The main turns have a radius of about 305 m and are pitched at 33°. Let N be the perpendicular force on the car and N_v and N_h be the vertical and horizontal components of this force, respectively. See the diagram below.

a. Let μ represent the coefficient of friction; recall that μN gives the force due to friction. To maintain the position of a vehicle traveling around the bank, the centripetal force must equal the horizontal force in the direction of the center of the track. Add the horizontal component of friction to the horizontal component of the perpendicular force on the car to find the centripetal force. Set your expression equal to (mv2)/r, the centripetal force.
The force due to friction is μN, and the horizontal component then is μN cos⁡(θ).
The horizontal perpendicular force would be N sin⁡(θ).
We get, $$\frac{m v^{2}}{r}$$ = N sin⁡(θ)+μN cos⁡(θ).

b. Add the vertical component of friction to the force due to gravity, and set this equal to the vertical component of the perpendicular force.
N cos⁡(θ) is the vertical component of the perpendicular force.
The force due to gravity is mg, and the vertical force of friction is μN sin⁡(θ).
We get, N cos⁡(θ) = mg + μN sin⁡(θ).

c. Solve one of your equations in part (a) or part (b) for m, and use this with the other equation to solve for v.
m = $$\frac{N \cos (\theta)-\mu N \sin (\theta)}{g}$$
$$\frac{N \cos (\theta)-\mu N \sin (\theta)}{g} \cdot \frac{v^{2}}{r}$$ = N sin⁡(θ)+μN cos⁡(θ)
$$\frac{v^{2}}{g r}$$ = $$\frac{N \sin (\theta)+\mu N \cos (\theta)}{N \cos (\theta)-\mu N \sin (\theta)}$$
We can factor out the N and cancel the common factor from here onward.
v2 = gr⋅$$g r \cdot \frac{\sin (\theta)+\mu \cos (\theta)}{\cos (\theta)-\mu \sin (\theta)}$$
v = $$\sqrt{g r \cdot \frac{\sin (\theta)+\mu \cos (\theta)}{\cos (\theta)-\mu \sin (\theta)}}$$

d. Assume μ = 0.75, the standard coefficient of friction for rubber on asphalt. For the Talladega Superspeedway, what is the maximum velocity on the main turns? Is this about how fast you might expect NASCAR stock cars to travel? Explain why you think NASCAR takes steps to limit the maximum speeds of the stock cars.
v = $$\sqrt{g r \cdot \frac{\sin (\theta)+\mu \cos (\theta)}{\cos (\theta)-\mu \sin (\theta)}}$$
= $$\sqrt{9.8 \cdot 305 \cdot \frac{\sin (33)+0.75 \cos (33)}{\cos (33)-0.75 \sin (33)}}$$
≈ 90.3
90.3 m/s is about 202 mph, which is the speed most people associate with NASCAR stock cars (although they usually go slower than this). This means that at Talladega, the race cars are able to go at their maximum speeds through the main turns.
If the cars go any faster than this, then they would drift up toward the wall, which may prompt them to oversteer and possibly spin out of control. NASCAR may limit the speeds of the cars because the racetracks themselves are not designed for cars that can go faster.

e. Does the friction component allow the cars to travel faster on the curve or force them to drive slower? What is the maximum velocity if the friction coefficient is zero on the Talladega roadway?
The friction component is in the direction toward the center of the racetrack (away from the walls), so it allows them to travel faster more safely. If μ = 0, the equation becomes v = $$\sqrt{g r \tan (\theta)}$$. Therefore, the maximum velocity is about 44.1 $$\frac{\mathrm{m}}{\mathrm{s}}$$ or 99 mph.

f. Do cars need to travel slower on a flat roadway making a turn than on a banked roadway? What is the maximum velocity of a car traveling on a 305 m turn with no bank?
They need to travel much slower on a flat roadway making a turn than when traveling on a banked roadway. The normal force does not prevent the car from spinning out of control the way it does on a banked turn.
If θ = 0, then the equation becomes v = . Therefore, the maximum velocity is about 47.3$$\frac{\mathrm{m}}{\mathrm{s}}$$ or 106 mph, which is much slower than the velocity that cars can travel on the banked turns (202 mph).

Question 6.
At a particular harbor over the course of 24 hours, the following data on peak water levels was collected (measurements are in feet above the MLLW):

a. What appears to be the average period of the water level?
It takes 13 hours to get from the first low-point to the second, and 13 hours to get from the first high-point to the second, so the average period is $$\frac{13 + 13}{2}$$ = 13.

b. What appears to be the average amplitude of the water level?
There are three areas we can examine to get amplitudes, from 1:30 to 7:30, 7:30 to 14:30, and 14:30 to 20:30. We get amplitudes of $$\frac{8.21-(-0.211)}{2}$$ = 4.2105, $$\frac{8.21-(-0.619)}{2}$$ = 4.4145, and $$\frac{7.518-(-0.619)}{2}$$ = 4.0685. We get 4.231 as the average amplitude.

c. What appears to be the average midline for the water level?
3.748

d. Fit a curve of the form y = A sin⁡(ω(x-h))+k or y = A cos⁡(ω(x-h))+k modeling the water level in feet as a function of the time.
Since it would make our curve more inaccurate to guess at what point the water levels will cross the midline, we can either start at 1:30 or 7:30 and use the cosine function. For 7:30, we get
y = 4.231 cos⁡($$\frac{2 \pi}{13}$$ (x-7.5))+3.748.

e. According to your function, how many times per day will the water level reach its maximum?
It should reach its maximum levels twice a day usually, but there is the rare possibility that it will reach its maximum only once.

f. How can you find other time values for a particular water level after finding one value from your function?
The values repeat every 13 hours, so immediately once a value is found, adding any multiple of 13 will give other values that work. Additionally, if you have the inverse cosine value for a particular water level (but have not solved for x yet), then take the opposite, solve normally, and you will have another time to which you can add multiples of 13.

g. Find the inverse function associated with the function in part (d). What is the domain and range of this function? What type of values does this function output?
y = $$\frac{13}{2 \pi}$$ cos-1($$\frac{x-3.748}{4.231}$$)+7.5
The domain is all real numbers x, such that -0.483 ≤ x ≤ 7.979, and the range is all real numbers y such that 7.5 ≤ y ≤ 14.

### Eureka Math Precalculus Module 4 Lesson 14 Exit Ticket Answer Key

Question 1.
The minimum radius of the turn r needed for an aircraft traveling at true airspeed v is given by the following formula
r = $$\frac{v^{2}}{g \tan (\theta)}$$
where r is the radius in meters, g is the acceleration due to gravity, and θ is the banking angle of the aircraft. Use
g = 9.78 $$\frac{\mathrm{m}}{s^{2}}$$ instead of 9.81 $$\frac{\mathrm{m}}{s^{2}}$$ to model the acceleration of the airplane accurately at 30,000 ft.
a. If an aircraft is traveling at 03 $$\frac{m}{s}$$, what banking angle is needed to successfully turn within 1 km?
1000 = $$\frac{103^{2}}{9.78 \cdot \tan (\theta)}$$
θ = tan-1⁡($$\frac{103^{2}}{1000 \cdot 9.78}$$)
≈ 47.328

b. Write the formula that gives the banking angle as a function of the radius of the turn available for a fixed airspeed v.
θ = tan-1⁡($$\frac{v^{2}}{r \cdot g}$$)

c. For a variety of reasons, including motion sickness from fluctuating g-forces and the danger of losing lift, many airplanes have a maximum banking angle of around 60°. Does this maximum on the model affect the domain or range of the formula you gave in part (b)?
If the angle cannot be greater than 60°, then the range caps out at 60° instead of normally being able to go up to 90°.

## EngageNY Math Precalculus Module 5 Answer Key | Precalculus Eureka Math Module 5 Answer Key

Eureka Math Precalculus Module 5 Probability and Statistics

Eureka Math Precalculus Module 5 Topic A Probability

Engage NY Math Precalculus Module 5 Topic B Random Variables and Discrete Probability Distributions

Eureka Math Precalculus Module 5 Mid Module Assessment Answer Key

Precalculus Eureka Math Module 5 Topic C Using Probability to Make Decisions

Eureka Math Precalculus Module 5 End of Module Assessment Answer Key

## EngageNY Math Precalculus Module 4 Answer Key | Precalculus Eureka Math Module 4 Answer Key

Eureka Math Precalculus Module 4 Trigonometry

Eureka Math Precalculus Module 4 Topic A Trigonometric Functions

Eureka Math Precalculus Module 4 Mid Module Assessment Answer Key

Engage NY Math Precalculus Module 4 Topic B Trigonometry and Triangles

Precalculus Eureka Math Module 4 Topic C Inverse Trigonometric Functions

Eureka Math Precalculus Module 4 End of Module Assessment Answer Key

## EngageNY Math Precalculus Module 3 Answer Key | Precalculus Eureka Math Module 3 Answer Key

Eureka Math Precalculus Module 3 Rational and Exponential Functions

Eureka Math Precalculus Module 3 Topic A Polynomial Functions and the Fundamental Theorem of Algebra

Eureka Math Precalculus Module 3 Mid Module Assessment Answer Key

Engage NY Math Precalculus Module 3 Topic B Rational Functions and Composition of Functions

Precalculus Eureka Math Module 3 Topic C Inverse Functions

Eureka Math Precalculus Module 3 End of Module Assessment Answer Key

## EngageNY Math Precalculus Module 2 Answer Key | Precalculus Eureka Math Module 2 Answer Key

Eureka Math Precalculus Module 2 Vectors and Matrices

Eureka Math Precalculus Module 2 Topic A Networks and Matrices

Engage NY Math Precalculus Module 2 Topic B Linear Transformations of Planes and Space

Precalculus Eureka Math Module 2 Topic C Systems of Linear Equations

Eureka Math Precalculus Module 2 Mid Module Assessment Answer Key

EngageNY Precalculus Math Module 2 Topic D Vectors in Plane and Space

Great Minds Eureka Math Precalculus Module 2 Topic E First-Person Video Games—Projection Matrices

Eureka Math Precalculus Module 2 End of Module Assessment Answer Key

## Engage NY Eureka Math 2nd Grade Module 5 End of Module Assessment Answer Key

### Eureka Math Grade 2 Module 5 End of Module Assessment Answer Key

Question 1.
Solve each problem with a written strategy such as a tape diagram, a number bond, the arrow way, the vertical form, or chips on a place value chart.
a. 460 + 200 = _______

460 + 200 = 660.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as tape diagram, a number bond.
460 + 200.
460 = 450 + 10.
200 + 10 = 210.
210 + 450 = 660.

b. _______ = 865 – 300

865 – 300 = 565.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as tape diagram, a number bond.
865 – 300.
865 = 860 + 5.
300 + 5 = 305.
860 – 305 = 565.

c. _______ + 400 = 598

198 + 400 = 598.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as tape diagram, a number bond.
198 + 400.
400 = 398 + 2.
198 + 2 = 200.
200 + 398 = 598.

d. 240 – 190 = _______

240 – 190 = 50.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as tape diagram, a number bond.
240 – 190.
240 = 230 + 10.
190 + 10 = 200.
200 + 230 = 430.

e. _______ = 760 – 280

760 – 280 = 480.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as tape diagram, a number bond.
760 – 280.
760 = 740 + 20.
280 + 20 = 300.
740 – 280 = 480.

f. 330 – 170 = _______

330 – 170 = 160.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as tape diagram, a number bond.
330 – 170.
330 = 300 + 30.
170 + 30 = 200.
200 – 360 = 160.

Question 2.
Use the arrow way to fill in the blanks and solve. Use place value drawings if that will help you.
a.

630 – 400 = 230.
230 + 10 = 240.
630 – 240 = 390.

Explanation:
In the above-given question,
given that,
use the arrow way to fill in the blanks.
630 – 400 = 230.
230 + 10 = 240.
630 – 240 = 390.

b.

570 – 300 = 270.
270 + 20 = 290.
570 – 280 = 290.

Explanation:
In the above-given question,
given that,
use the arrow way to fill in the blanks.
570 – 300 = 270.
270 + 20 = 290.
570 – 280 = 290.

c.

958 – 400 = 558.
558 – 40 = 518.
958 – 440 = 518.

Explanation:
In the above-given question,
given that,
use the arrow way to fill in the blanks.
958 – 400 = 558.
558 – 40 = 518.
958 – 440 = 518.

Question 3.
Solve.
Draw a place value chart with chips to model the problems. Show a written subtraction method to check your work.
a. 756 + 136 = ______

Subtraction number sentence:

756 + 136 = 892.

Explanation:
In the above-given question,
given that,
draw a place value chart with chips to model the problems.
show a written subtraction method to check your work.
756 + 136 = 892.
892 – 136 = 756.

b. 267 + 545 = ______

Subtraction number sentence:

267 + 545 = 812.

Explanation:
In the above-given question,
given that,
draw a place value chart with chips to model the problems.
show a written subtraction method to check your work.
267 + 545 = 812.
812 – 545 = 267.

Draw a place value chart with chips to model the problems. Show a written addition method to check your work.

c. 617 – 229 = ______

Check:

617 – 229 = 398.

Explanation:
In the above-given question,
given that,
draw a place value chart with chips to model the problems.
617 – 229 = 398.
398 + 229 = 617.

d. 700 – 463 = ______

Check:
700 – 463 = 437.

Explanation:
In the above-given question,
given that,
draw a place value chart with chips to model the problems.
700 – 463 = 437.
437 + 463 = 700.

Question 4.
Find the missing numbers to make each statement true. Show your strategy to solve.
a. 300 – 106 = ________

300 – 106 = 194.

Explanation:
In the above-given question,
given that,
find the missing numbers to make each statement true.
300 – 106 = 194.
194 + 106 = 300.

b. ________ = 407 – 159

407 – 159 = 248.

Explanation:
In the above-given question,
given that,
find the missing numbers to make each statement true.
407 – 159 = 248.
248 + 159 = 407.

c. 410 – 190 = 420 – ___200_____

410 – 190 = 220.

Explanation:
In the above-given question,
given that,
find the missing numbers to make each statement true.
410 – 190 = 220.
220 + 190 = 410.

d. 750 – 180 = ________ – 200

750 – 180 = 570.

Explanation:
In the above-given question,
given that,
find the missing numbers to make each statement true.
750 – 180 = 570.
770 – 200 = 570.
570 + 180 = 750.

e. 900 – ________ = 600 – 426

600 – 426 = 174.

Explanation:
In the above-given question,
given that,
find the missing numbers to make each statement true.
600 – 426 = 174.
174 + 426 = 600.

Question 5.
Martha answered the problem 456 – 378 incorrectly. She does not understand her mistake.
a. Explain to Martha what she did wrong using place value language.

Explanation:
________________________________________
________________________________________
________________________________________

456 – 378 = 78.

Explanation:
In the above-given question,
given that,
456 – 378.
4 hundred = 400.
5 tens = 50.
6 ones = 6.
400 + 50 + 6 = 456.
3 hundred = 300.
7 tens = 70.
8 ones = 8.
300 + 70 + 8 = 378.
456 – 378 = 78.

b. Model an alternative strategy for 456 – 378 to help Martha avoid making this mistake again.

456 – 378 = 78.

Explanation:
In the above-given question,
given that,
456 = 454 + 2.
378 + 2 = 380.
454 – 380 = 78.

## Engage NY Eureka Math 2nd Grade Module 5 Mid Module Assessment Answer Key

### Eureka Math Grade 2 Module 5 Mid Module Assessment Answer Key

Question 1.
Solve each problem with a written strategy such as a tape diagram, a number bond, the arrow way, the vertical form, or chips on a place value chart.
a. 220 + 30 = _____________

220 + 30 = 250.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as a tape diagram, a number bond.
220 + 30.
220 = 10 + 210.
30 + 10 = 40.
40 + 210 = 250.

b. 200 + 380 = _____________

200 + 380 = 580.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as a tape diagram, a number bond.
200 + 380.
200 = 20 + 180.
380 + 20 = 400.
400 + 180 = 580.

c. 450 + 210 = _____________

450 + 210 = 660.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as a tape diagram, a number bond.
450 + 210.
210 = 10 + 200.
450 + 10 = 460.
460 + 200 = 660.

d. 490 + 12 = _____________

490 + 12 = 502.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as a tape diagram, a number bond.
490 + 12.
12 = 10 + 2.
490 + 10 = 500.
500 + 2 = 502.

e. _____________ = 380 + 220

220 + 380 = 600.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as a tape diagram, a number bond.
220 + 380.
220 = 20 + 200.
380 + 20 = 400.
400 + 200 = 600.

f. 750 – 590 = _____________

750 – 590 = 160.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as a tape diagram, a number bond.
750 – 590.
750 = 740 + 10.
10 + 590 = 600.
600 – 580 = 160.

Question 2.
Use the arrow way to solve.
a.

342 + 100 + 142 = 542.

Explanation:
In the above-given question,
given that,
use the arrow way to solve.
342 + 100 = 400.
400 + 100 = 500.
500 + 42 = 542.

b.

600 – 100 – 10 = 490.

Explanation:
In the above-given question,
given that,
use the arrow way to solve.
600 – 100 = 500.
500 – 10 = 490.

c.

658 + 100 + 10 = 768.

Explanation:
In the above-given question,
given that,
use the arrow way to solve.
658 + 100 = 758.
758 + 10 = 768.

d. 542 + 207 = __________

542 + 100 + 107 = 749.

Explanation:
In the above-given question,
given that,
use the arrow way to solve.
542 + 100 = 642.
642 + 100 = 742.
742 + 7 = 749.

e. 430 + 361 = __________

430 + 361 = 791.

Explanation:
In the above-given question,
given that,
use the arrow way to solve.
430 + 100 = 530.
530 + 261 = 791.

f. 660 – 190 = __________

660 – 190 = 470.

Explanation:
In the above-given question,
given that,
use the arrow way to solve.
660 – 100 = 560.
560 – 100 = 460.
460 + 10 = 470.

Question 3.
Solve each by drawing a model of a place value chart with chips and using the vertical form.
a. 328 + 259 = ________

328 + 259 = 587.

Explanation:
In the above-given question,
given that,
solve each by drawing a model of a place value chart.
328 + 259.
3 hundred = 300.
2 tens = 20.
8 ones = 8.
300 + 20 + 8 = 328.
2 hundred = 200.
5 tens = 50.
9 ones = 9.
200 + 50 + 9 = 259.
328 + 259 = 587.

b. 575 + 345 = ________

575 + 345 = 920.

Explanation:
In the above-given question,
given that,
solve each by drawing a model of a place value chart.
575 + 345.
5 hundred = 500.
7 tens = 70.
5 ones = 5.
500 + 70 + 5 = 575.
3 hundred = 300.
4 tens = 40.
5 ones = 5.
300 + 40 + 5 = 345.
575 + 345 = 920.

Circle True or False for each number sentence. Explain your thinking using pictures, words, or numbers.

c. 466 + 244 = 600 + 100 (True / False)

466 + 244 = 710.

Explanation:
In the above-given question,
given that,
circle true or false for each number sentence.
466 + 244.
466 + 244 = 710.
710 is not equal to 700.

d. 690 + 179 = 700 + 169 (True / False)

690 + 179 = 869.

Explanation:
In the above-given question,
given that,
circle true or false for each number sentence.
690 + 179 = 869.
700 + 169 = 869.
869 is equal to 869.

e. 398 + 6 = 400 + 5 (True / False)

398 + 6 = 404.

Explanation:
In the above-given question,
given that,
circle true or false for each number sentence.
398 + 6 = 404.
400 + 5 = 405.
404 is not equal to 405.

f. 724 – 298 = 722 – 300 (True / False)

724 – 298 = 426.

Explanation:
In the above-given question,
given that,
circle true or false for each number sentence.
724 – 298 = 426.
722 – 300 = 422.
426 is not equal to 422.

Question 4.
Solve each problem with two written strategies such as a tape diagram, a number bond, the arrow way, the vertical form, or chips on a place value chart.
a. 299 + 436 = _______

299 + 436 = 735.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as a tape diagram, a number bond.
299 + 436.
299 = 4 + 295.
436 + 4 = 440.
440 + 295 = 735.

b. 470 + 390 = _______

470 + 390 = 860.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as a tape diagram, a number bond.
470 + 390.
470 = 10 + 460.
390 + 10 = 400.
400 + 460 = 860.

c. 268 + 122 = _______

268 + 122 = 490.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as a tape diagram, a number bond.
268 + 122.
268 = 8 + 260.
122 + 8 = 130.
130 + 260 = 390.

d. 330 – 190 = _______

330 – 190 = 140.

Explanation:
In the above-given question,
given that,
solve each problem with a written strategy such as a tape diagram, a number bond.
330 – 190.
330 = 10 + 320.
190 + 10 = 200.
200 – 320 = 140.

## Engage NY Eureka Math 2nd Grade Module 5 Lesson 18 Answer Key

### Eureka Math Grade 2 Module 5 Lesson 18 Problem Set Answer Key

Question 1.
Use the arrow way and counting on to solve.
a. 300 – 247

300 – 247 = 53.

Explanation:
In the above-given question,
given that,
use the arrow way and counting on to solve.
300 – 247.
3 hundred = 300.
2 hundred = 200.
4 tens = 40.
7 ones = 7.
200 + 40 + 7 = 247.
300 – 247 = 53.

b. 600 – 465

600 – 465 = 135.

Explanation:
In the above-given question,
given that,
use the arrow way and counting on to solve.
600 – 465.
600 – 100 = 500.
500 – 35 = 465.
600 – 465 = 135.

Question 2.
Solve vertically, and draw a place value chart and chips. Rename in one step.
a. 507 – 359

507 – 359 = 148.

Explanation:
In the above-given question,
given that,
solve vertically, and draw a place value chart and chips.
507 – 359.
5 hundred = 500.
7 ones = 7.
500 + 7 = 507.
3 hundred = 300.
5 tens = 50.
9 ones = 9.
300 + 50 + 9 = 359.
507 – 359 = 148.

b. 708 – 529

708 – 529 = 179.

Explanation:
In the above-given question,
given that,
solve vertically, and draw a place value chart and chips.
708 – 529.
7 hundred = 700.
8 ones = 8.
700 + 8 = 708.
5 hundred = 500.
2 tens = 20.
9 ones = 9.
500 + 20 + 9 = 529.
708 – 529 = 179.

Question 3.
Choose a strategy to solve, and explain why you chose that strategy.

 a. 600 – 437 Explanation:

600 – 437 = 163.

Explanation:
In the above-given question,
given that,
600 – 437.
600 – 100 = 500.
500 – 100 = 400.
400 + 37 = 437.
600 – 437 = 163.

 b. 808 – 597 Explanation:

808 – 597 = 211.

Explanation:
In the above-given question,
given that,
808 – 597.
8 hundred = 800.
8 ones = 8.
800 + 8 = 808.
5 hundred = 500.
9 tens = 90.
7 ones = 7.
500 + 90 + 7 = 597.
808 – 597 = 211.

Question 4.
Prove the student’s strategy by solving both problems to check that their solutions are the same. Explain to your partner why this way works.

800 – 543 = 257.
799 – 542 = 257.

Explanation:
In the above-given question,
given that,
the student’s strategy is the same.
800 – 543 = 257.
799 – 542 = 257.
257 + 542 = 799.

Question 5.
Use the simplifying strategy from Problem 4 to solve the following two problems.
a. 600 – 547

600 – 547 = 53.

Explanation:
In the above-given question,
given that,
600 – 547.
600 – 100 = 500.
500 + 47 = 547.
600 – 547 = 53.

b. 700 – 513

700 – 513 = 187.

Explanation:
In the above-given question,
given that,
700 – 513.
700 – 100 = 600.
600 – 100 = 500.
500 + 13 = 513.
700 – 513 = 187.

### Eureka Math Grade 2 Module 5 Lesson 18 Exit Ticket Answer Key

Choose a strategy to solve, and explain why you chose that strategy.

Question 1.

 1. 400 – 265 Explanation:

400 – 265 = 135.

Explanation:
In the above-given question,
given that,
400 – 265.
400 – 100 = 300.
300 – 35 = 265.
265 + 35 = 300.
400 – 265 = 135.

Question 2.

 2. 507 – 198 Explanation:

507 – 198 = 309.

Explanation:
In the above-given question,
given that,
507 – 198.
507 – 100 = 407.
407 – 100 = 307.
307 – 100 = 207.
207 – 9 = 198.
507 – 198 = 309.

Question 1.
Use the arrow way and counting on to solve.
a. 700 – 462

700 – 462 = 238.

Explanation:
In the above-given question,
given that,
700 – 462.
700 – 100 = 600.
600 – 100 = 500.
500 – 100 = 400.
400 + 62 = 462.
700 – 462 = 238.

b. 900 – 232

900 – 232 = 668.

Explanation:
In the above-given question,
given that,
900 – 232.
900 – 100 = 700.
700 – 100 = 600.
600 + 68 = 668.
900 – 232 = 668.

Question 2.
Solve vertically, and draw a place value chart and chips. Rename in one step.
a. 907 – 467

907 – 467 = 440.

Explanation:
In the above-given question,
given that,
907 – 467.
9 hundred = 900.
7 ones = 7.
900 + 7 = 907.
4 hundred = 400.
6 tens = 60.
7 ones = 7.
400 + 60 + 7 = 467.
907 – 467 = 440.

b. 803 – 667

803 – 667 = 136.

Explanation:
In the above-given question,
given that,
803 – 667.
8 hundred = 800.
3 ones = 3.
800 + 3 = 803.
6 hundred = 600.
6 tens = 60.
7 ones = 7.
600 + 60 + 7 = 667.
803 – 667 = 136.

Question 3.
Choose a strategy to solve, and explain why you chose that strategy.

 a. 700 – 390 Explanation:

700 – 390 = 310.

Explanation:
In the above-given question,
given that,
700 – 390.
7 hundred = 700.
3 hundred = 300.
9 tens = 90.
300 + 90 = 390.
700 – 390 = 310.

 b. 919 – 657 Explanation:

919 – 657 = 262.

Explanation:
In the above-given question,
given that,
919 – 657.
9 hundred = 900.
1 ten = 10.
9 ones = 9.
900 + 10 + 9 = 919.
6 hundred = 600.
5 tens = 50.
7 ones = 7.
600 + 50 + 7 = 657.
919 – 657 = 262.

Question 4.
Explain why 300 – 186 is the same as 299 – 185.
Explanation:

300 – 186 = 114.
299 – 185 = 114.

Explanation:
In the above-given question,
given that,
yes both the problems get the same answer.
300 – 186 = 114.
299 – 185 = 114.

Question 5.
Solve 500 – 278 using the simplifying strategy from Problem 4.
Solution:

500 – 278 = 222.

Explanation:
In the above-given question,
given that,
5 hundred = 500.
2 hundred = 200.
7 tens = 70.
8 ones = 8.
200 + 70 + 8 = 278.
500 – 278 = 222.

## Engage NY Eureka Math 2nd Grade Module 5 Lesson 20 Answer Key

### Eureka Math Grade 2 Module 5 Lesson 20 Problem Set Answer Key

Question 1.
399 + 237 = _________

399 + 237 = 636.

Explanation:
In the above-given question,
given that,
solve the problem using two different strategies.
399 + 237.
I solved using the number bond.
my friend solved using the arrow way.
3 hundreds = 300.
9 tens = 90.
9 ones = 9.
300 + 90 + 9 = 399.
2 hundred = 200.
3 tens = 30.
7 ones = 7.
200 + 30 + 7 = 237.
399 + 237 = 636.

Question 2.
400 – 298 = ________

400 – 298 = 102.

Explanation:
In the above-given question,
given that,
solve the problem using two different strategies.
400 – 298.
I solved using the number bond.
my friend solved using the place value strategies.
4 hundred = 400.
2 hundred = 200.
9 tens = 90.
8 ones = 8.
200 + 90 + 8 = 298.
400 – 298 = 102.

Question 3.
548 + 181 = _________

548 + 181 = 729.

Explanation:
In the above-given question,
given that,
solve the problem using two different strategies.
548 + 181.
I solved using the number bond.
my friend solved using the place value strategies.
5 hundred = 500.
4 tens = 40.
8 ones = 8.
500 + 40 + 8 = 548.
1 hundred = 100.
8 tens = 80.
1 one = 1.
100 + 80 + 1 = 181.
548 + 181 = 729.

Question 4.
360 + ______ = 754

754 – 360 = 394.

Explanation:
In the above-given question,
given that,
solve the problem using two different strategies.
394 + 360.
I solved using the number bond.
my friend solved using the place value strategies.
3 hundred = 300.
6 tens = 60.
300 + 60 = 360.
3 hundred = 300.
9 tens = 90.
4 ones = 4.
300 + 90 + 4 = 394.
394 + 360 = 754.

Question 5.
862 – ______ = 690

862 – 172 = 690.

Explanation:
In the above-given question,
given that,
solve the problem using two different strategies.
862 – 172.
I solved using the number bond.
my friend solved using the place value strategies.
8 hundred = 800.
6 tens = 60.
2 ones = 2.
800 + 60 + 2 = 862.
1 hundred = 100.
7 tens = 70.
2 ones = 2.
100 + 70 + 2 = 172.
862 – 172 = 690.

### Eureka Math Grade 2 Module 5 Lesson 20 Exit Ticket Answer Key

Solve each problem using two different strategies.

Question 1.
299 + 156 = _________

299 + 156 = 455.

Explanation:
In the above-given question,
given that,
solve the problem using two different strategies.
299 + 156.
1st strategy is the number bond.
2nd strategy solved using the place value strategies.
2 hundred = 200.
9 tens = 90.
9 ones = 9.
200 + 90 + 9 = 299.
1 hundred = 100.
5 tens = 50.
6 ones = 6.
100 + 50 + 6 = 156.
299 + 156 = 455.

Question 2.
547 + ______ = 841

547 + 294 = 841.

Explanation:
In the above-given question,
given that,
solve the problem using two different strategies.
547 + 294.
1st strategy is the number bond.
2nd strategy solved using the place value strategies.
5 hundred = 500.
4 tens = 40.
7 ones = 7.
500 + 40 + 7 = 547.
2 hundred = 200.
9 tens = 90.
4 ones = 4.
200 + 90 + 4 = 294.
547 + 294 = 841.

### Eureka Math Grade 2 Module 5 Lesson 20 Homework Answer Key

Question 1.
456 + 244 = _________

456 + 244 = 700.

Explanation:
In the above-given question,
given that,
solve the problem using two different strategies.
456 + 244.
1st strategy is the number bond.
2nd strategy solved using the place value strategies.
4 hundred = 400.
5 tens = 50.
6 ones = 6.
400 + 50 + 6 = 456.
2 hundred = 200.
4 tens = 40.
4 ones = 4.
200 + 40 + 4 = 244.
456 + 244 = 700.

Question 2.
698 + ______ = 945

698 + 247 = 945.

Explanation:
In the above-given question,
given that,
solve the problem using two different strategies.
698 + 247.
1st strategy is the number bond.
2nd strategy solved using the place value strategies.
6 hundred = 600.
9 tens = 90.
8 ones = 8.
600 + 90 + 8 = 698.
698 + 247 = 945.

Circle a strategy to solve, and explain why you chose that strategy.

Question 3.
257 + 160 = _____
a. Arrow way or vertical form

 b. Solve Explanation: __________________________ __________________________ __________________________ __________________________

257 + 160 = 317.

Explanation:
In the above-given question,
given that,
solve the problem using arrow way or vertical form.
2 hundred = 200.
5 tens = 50.
7 ones = 7.
200 + 50 + 7 = 257.
1 hundred = 100.
6 tens = 60.
100 + 60 = 160.
257 – 160 = 317.

Question 4.
754 – 597 = _____

754 – 597 = 157.

Explanation:
In the above-given question,
given that,
solve the problem using arrow way or vertical form.
7 hundred = 700.
5 tens = 50.
4 ones = 4.
700 + 50 + 4 = 754.
5 hundred = 500.
9 tens = 90.
7 ones = 7.
500 + 90 + 7 = 597.
754 – 597 = 157.

a. Number bond or arrow way

 b. Solve Explanation: __________________________ __________________________ __________________________ __________________________

754 – 597 = 157.

Explanation:
In the above-given question,
given that,
solve the problem using arrow way or number bond.
7 hundred = 700.
5 tens = 50.
4 ones = 4.
700 + 50 + 4 = 754.
5 hundred = 500.
9 tens = 90.
7 ones = 7.
500 + 90 + 7 = 597.
754 – 597 = 157.

## Engage NY Eureka Math 2nd Grade Module 5 Lesson 19 Answer Key

### Eureka Math Grade 2 Module 5 Lesson 19 Problem Set Answer Key

Question 1.
Explain how the two strategies to solve 500 – 211 are related.

__________________________________________________________________________
_____the __two___strategies____are___same._______________________________________________________
_____because while we are solving._____________________________________________________________________
______the__two___models___are__different.
__________________________________________________________________________
__________________________________________________________________________

500 – 211 = 289.

Explanation:
In the above-given question,
given that,
the two strategies are the same.
why because the answer for the two strategies is the same.
5 hundred = 500.
2 hundred = 200.
1 ten = 10.
1 ones = 1.
200 + 10 + 1 = 211.
500 – 211 = 289.

Question 2.
Solve and explain why you chose that strategy.

 a. 220 + 390 = ________ 220 = 210 + 10. 390 + 10 = 400. 400 + 210 = 610 Explanation: __220_=__210__+__10________________ _____390__+__10_=__400______________ ___400__+__210__=__610_______________ __________________________ b. 547 – 350 = ________ Explanation: __________________________ __________________________ __________________________ __________________________ c. 464 + 146 = ________ Explanation: __________________________ __________________________ __________________________ __________________________ d. 600 – 389 = ________ Explanation: __________________________ __________________________ __________________________ __________________________

220 + 390 = 610.

Explanation:
In the above-given question,
given that,
I choose the number bond strategy.
220 = 10 + 210.
390 + 10 = 400.
400 + 210 = 610.

547 – 350 = 197.

Explanation:
In the above-given question,
given that,
I choose the number bond strategy.
547 = 10 + 537.
350 + 10 = 360.
360 + 537 = 897.

464 + 146 = 610.

Explanation:
In the above-given question,
given that,
I choose the number bond strategy.
146 = 6 + 140.
464 + 6 = 470.
470 + 140 = 610.

600 – 389 = 211.

Explanation:
In the above-given question,
given that,
I choose the number bond strategy.
600 = 1 + 599.
389 + 1 = 390.
390 – 601 = 211.

### Eureka Math Grade 2 Module 5 Lesson 19 Exit Ticket Answer Key

Solve and explain why you chose that strategy.

 1. 400 + 590 = ________ Explanation: __________________________ __________________________ __________________________ __________________________ 2. 775 – 497 = ________ Explanation: __________________________ __________________________ __________________________ __________________________

400 + 590 = 990.

Explanation:
In the above-given question,
given that,
I choose the number bond strategy.
590 = 10 + 580.
400 + 10 = 410.
580 + 410 = 990.

775 – 497 = 278.

Explanation:
In the above-given question,
given that,
I choose the number bond strategy.
497 = 5 + 492.
775 + 10 = 780.
780 – 492 = 278.

### Eureka Math Grade 2 Module 5 Lesson 19 Homework Answer Key

Question 1.
Solve and explain why you chose that strategy.

 a. 340 + 250 = ________ Explanation: __________________________ __________________________ __________________________ __________________________ b. 490 + 350 = ________ Explanation: __________________________ __________________________ __________________________ __________________________ c. 519 + 342 = ________ Explanation: __________________________ __________________________ __________________________ __________________________ d. 610 + _________ = 784 Explanation: __________________________ __________________________ __________________________ __________________________ e. 700 – 456 = ________ Explanation: __________________________ __________________________ __________________________ __________________________ f. 904 – 395 = ________ Explanation: __________________________ __________________________ __________________________ __________________________

250 + 340 = 590.

Explanation:
In the above-given question,
given that,
I choose the number bond strategy.
250 = 10 + 240.
340 + 10 = 350.
350 + 240 = 590.

350 + 490 = 840.

Explanation:
In the above-given question,
given that,
I choose the number bond strategy.
350 = 10 + 340.
490 + 10 = 500.
500 + 340 = 840.

784 – 610 = 174.

Explanation:
In the above-given question,
given that,
I choose the number bond strategy.
610 = 6 + 604.
784 + 6 = 790.
790 – 610 = 174.

700 + 456 = 244.

Explanation:
In the above-given question,
given that,
I choose the number bond strategy.
700 = 4 + 696.
456 + 4 = 500.
500 + 700 = 244.