## Engage NY Eureka Math 8th Grade Module 4 Lesson 30 Answer Key

### Eureka Math Grade 8 Module 4 Lesson 30 Exercise Answer Key

Mathematical Modeling Exercise

(1) If t is a number, what is the degree in Fahrenheit that corresponds to tÂ°C?

(2) If t is a number, what is the degree in Fahrenheit that corresponds to ( – t)Â°C?

Answer:

â†’ Instead of trying to answer these questions directly, letâ€™s try something simpler. With this in mind, can we find out what degree in Fahrenheit corresponds to 1Â°C? Explain.

â†’ We can use the following diagram (double number line) to organize our thinking.

Answer:

â†’ At this point, the only information we have is that 0Â°C = 32Â°F, and 100Â°C = 212Â°F. We want to figure out what degree of Fahrenheit corresponds to 1Â°C. Where on the diagram would 1Â°C be located? Be specific.

Provide students time to talk to their partners about a plan, and then have them share. Ask them to make conjectures about what degree in Fahrenheit corresponds to 1Â°C, and have them explain their rationale for the numbers they chose. Consider recording the information, and have the class vote on which answer they think is closest to correct.

â†’ We need to divide the Celsius number line from 0 to 100 into 100 equal parts. The first division to the right of zero will be the location of 1Â°C.

Now that we know where to locate 1Â°C on the lower number line, we need to figure out what number it corresponds to on the upper number line representing Fahrenheit. Like we did with Celsius, we divide the number line from 32 to 212 into 100 equal parts. The number line from 32 to 212 is actually a length of 180 units (212 – 32 = 180). Now, how would we determine the precise number in Fahrenheit that corresponds to 1Â°C?

Provide students time to talk to their partners and compute the answer.

â†’ We need to take the length 180 and divide it into 100 equal parts.

\(\frac{180}{100}\) = \(\frac{9}{5}\) = 1 \(\frac{4}{5}\) = 1.8

â†’ If we look at a magnified version of the number line with this division, we have the following diagram:

â†’ Based on your computation, what number falls at the intersection of the Fahrenheit number line and the red line that corresponds to 1Â°C? Explain.

Since we know that each division on the Fahrenheit number line has a length of 1.8, then when we start from 32 and add 1.8, we get 33.8. Therefore, 1Â°C is equal to 33.8Â°F.

Revisit the conjecture made at the beginning of the activity, and note which student came closest to guessing 33.8Â°F. Ask the student to explain how he arrived at such a close answer.

â†’ Eventually, we want to revisit the original two questions. But first, letâ€™s look at a few more concrete questions. What is 37Â°C in Fahrenheit? Explain.

Provide students time to talk to their partners about how to answer the question. Ask students to share their ideas and explain their thinking.

â†’ Since the unit length on the Celsius scale is equal to the unit length on the Fahrenheit scale, then 37Â°C means we need to multiply (37 Ã— 1.8) to determine the corresponding location on the Fahrenheit scale. But, because 0 on the Celsius scale is 32 on the Fahrenheit scale, we will need to add 32 to our answer. In other words, 37Â°C = (32 + 37 Ã— 1.8)Â°F = (32 + 66.6)Â°F = 98.6Â°F.

Exercises

Determine the corresponding Fahrenheit temperature for the given Celsius temperatures in Exercises 1â€“5.

Exercise 1.

How many degrees Fahrenheit is 25Â°C?

Answer:

25Â°C = (32 + 25 Ã— 1.8)Â°F = (32 + 45)Â°F = 77Â°F

Exercise 2.

How many degrees Fahrenheit is 42Â°C?

Answer:

42Â°C = (32 + 42 Ã— 1.8)Â°F = (32 + 75.6)Â°F = 107.6Â°F

Exercise 3.

How many degrees Fahrenheit is 94Â°C?

Answer:

94Â°C = (32 + 94 Ã— 1.8)Â°F = (32 + 169.2)Â°F = 201.2Â°F

Exercise 4.

How many degrees Fahrenheit is 63Â°C?

Answer:

63Â°C = (32 + 63 Ã— 1.8)Â°F = (32 + 113.4)Â°F = 145.4Â°F

Exercise 5.

How many degrees Fahrenheit is tÂ°C?

Answer:

tÂ°C = (32 + 1.8t)Â°F

### Eureka Math Grade 8 Module 4 Lesson 30 Problem Set Answer Key

Question 1.

Does the equation tÂ°C = (32 + 1.8t)Â°F work for any rational number t? Check that it does with t = 8 \(\frac{2}{3}\) and t = – 8 \(\frac{2}{3}\).

Answer:

(8 \(\frac{2}{3}\))Â°C = (32 + 8 \(\frac{2}{3}\) Ã— 1.8)Â°F = (32 + 15.6)Â°F = 47.6Â°F

( – 8 \(\frac{2}{3}\))Â°C = (32 + ( – 8 \(\frac{2}{3}\)) Ã— 1.8)Â°F = (32 – 15.6)Â°F = 16.4Â°F

Question 2.

Knowing that tÂ°C = (32 + \(\frac{9}{5}\) t)Â°F for any rational t, show that for any rational number d, dÂ°F = (\(\frac{5}{9}\) (d – 32))Â°C.

Answer:

Since dÂ°F can be found by (32 + \(\frac{9}{5}\) t), then d = (32 + \(\frac{9}{5}\) t), and dÂ°F = tÂ°C. Substituting d = (32 + \(\frac{9}{5}\) t) into dÂ°F, we get

dÂ°F = (32 + \(\frac{9}{5}\) t)Â°F

d = 32 + \(\frac{9}{5}\) t

d – 32 = \(\frac{9}{5}\) t

\(\frac{5}{9}\) (d – 32) = t.

Now that we know t = \(\frac{5}{9}\) (d – 32), then dÂ°F = (\(\frac{5}{9}\) (d – 32))Â°C.

Question 3.

Drake was trying to write an equation to help him predict the cost of his monthly phone bill. He is charged $35 just for having a phone, and his only additional expense comes from the number of texts that he sends. He is charged $0.05 for each text. Help Drake out by completing parts (a)â€“(f).

a. How much was his phone bill in July when he sent 750 texts?

Answer:

35 + 750(0.05) = 35 + 37.5 = 72.5

His bill in July was $72.50.

b. How much was his phone bill in August when he sent 823 texts?

Answer:

35 + 823(0.05) = 35 + 41.15 = 76.15

His bill in August was $76.15.

c. How much was his phone bill in September when he sent 579 texts?

Answer:

35 + 579(0.05) = 35 + 28.95 = 63.95

His bill in September was $63.95.

d. Let y represent the total cost of Drakeâ€™s phone bill. Write an equation that represents the total cost of his phone bill in October if he sends t texts.

Answer:

y = 35 + t(0.05)

e. Another phone plan charges $20 for having a phone and $0.10 per text. Let y represent the total cost of the phone bill for sending t texts. Write an equation to represent his total bill.

Answer:

y = 20 + t(0.10)

f. Write your equations in parts (d) and (e) as a system of linear equations, and solve. Interpret the meaning of the solution in terms of the phone bill.

Answer:

(y = 35 + t(0.05)

y = 20 + t(0.10)

35 + (0.05)t = 20 + (0.10)t

15 + (0.05)t = (0.10)t

15 = 0.05t

300 = t

y = 20 + 300(0.10)

y = 50

The solution is (300,50), meaning that when Drake sends 300 texts, the cost of his bill will be $50 using his current phone plan or the new one.

### Eureka Math Grade 8 Module 4 Lesson 30 Exit Ticket Answer Key

Use the equation developed in class to answer the following questions:

Question 1.

How many degrees Fahrenheit is 11Â°C?

Answer:

11Â°C = (32 + 11 Ã— 1.8)Â°F

11Â°C = (32 + 19.8)Â°F

11Â°C = 51.8Â°F

Question 2.

How many degrees Fahrenheit is – 3Â°C?

Answer:

– 3Â°C = (32 + ( – 3) Ã— 1.8)Â°F

– 3Â°C = (32 – 5.4)Â°F

– 3Â°C = 26.6Â°F

Question 3.

Graph the equation developed in class, and use it to confirm your results from Problems 1 and 2.

Answer:

When I graph the equation developed in class, tÂ°C = (32 + 1.8t)Â°F, the results from Problems 1 and 2 are on the line, confirming they are solutions to the equation.