McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
8. Differences of Squares
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Exercise 67 Page 63

Practice makes perfect
a We are given two equations that represent the distance in miles of the cars from the center of Cleveland after t hours.

Car A: & d= 65t+15 Car B: & d=60t+25 We want to determine the speeds of the cars. The speed is the ratio of the change in distance to the change in time. Since the relationship between the distance and time is linear, the slope of the lines will give us the speed. Notice that the equations have slope-intercept form, so the coefficients of t represent the speed. r|c|c &Car A & Car B [0.3em] Equation:& d= 65t+15 & d=60t+25 & ⇓ & ⇓ Speed:& 65 & 60 Car A is faster because its speed is 65 mph and the other car has a speed of 60 mph.

b Since the given equations represent the distance, the difference of of them will give the distance between the cars.

d_A &= 65t+15 - d_B & = 60t+25 d_A-d_b & = 5t-10 The expression 5t-10 represents the distance between the cars.

c We can use the expression we wrote in Part B to find how far apart the cars are after 2 12 hours. We just need to substitute 2 12 for t in the expression.
5t-10
5( 2 12)-10
â–¼
Evaluate
5(5/2)-10
25/2-10
12.5-10
2.5
The distance between the cars is 2.5 mi after 2 12 hours.