Envision Math 2.0: Grade 8, Volume 1
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4. Construct Functions to Model Linear Relationships
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Exercise 14 Page 188

Practice makes perfect

We want to find the thaw rate of a turkey in hours per pound for refrigerator thawing and for cold water thawing. We will find these thaw rates one at a time.

Thaw Rate for Refrigerator Thawing

We are given that we need 1 day to thaw 4 pounds of a turkey in the refrigerator. Recall that a rate is a ratio that compares two quantities measured in different units. Let's find the thaw rate by writing a ratio of these two values. Since we want to find the thaw rate in hours per pound, we will express the number of days in hours.

1 day/4 pounds = 24 hours/4 pounds = 6 hours per pound Therefore, the thaw rate for refrigeration thawing is equal to 6 hours per pound.

Thaw Rate for Cold Water Thawing

We are given that we need 30 minutes to thaw 1 pound of a turkey in cold water. Let's find the thaw rate by writing a ratio of these two values. Since we want to find the thaw rate in hours per pounds, we will express the number of minutes in hours. 30 minutes/1 pound = 0.5 hours/1 pound = 0.5 hours per pound Therefore, the thaw rate for cold water thawing is equal to 0.5 hours per pound.

We are asked to find the linear function that represents the time t in hours it takes to thaw a turkey in the refrigerator as a function of the weight w in pounds of a turkey. The desired linear function has to follow a specific format. y= m x+ b In this case, y is represented by t and x is represented by w. t= m w+ b

For an equation in this form, m is the slope and b is the y-intercept. To find the desired linear function we will calculate the slope and the y-intercept. We can do this by using any two points lying on the line that is the graph of the linear function. Let's find these points by going through the Part A!

Fact From the Part A Point (w,t)
We need 1 day (24 hours) to thaw each 4 pounds of a turkey in the refrigerator. ( 4, 24)
The thaw rate for refrigeration thawing is equal to 6 hours per pound. ( 1, 6)
Therefore, we have the points ( 4, 24) and ( 1, 6) that we can use to calculate m. We will start by substituting the points into the Slope Formula.
m = y_2-y_1/x_2-x_1
m=24- 6/4- 1
â–Ľ
Simplify right-hand side
m=18/3
m=6
Now that we know the slope, we can write a partial version of the equation. t= 6 w+ b To complete the equation, we also need to determine the y-intercept b. Since we know that the obtained points will satisfy the equation, we can substitute one of them into the equation to solve for b. Let's use ( 1, 6).
t=6w+b
6=6( 1)+b
â–Ľ
Solve for b
6=6+b
6-6=6+b-6
0=b
b=0
We can now complete the equation. t= 6w+ 0 ⇔ t =6w