Houghton Mifflin Harcourt Algebra 1, 2015
HM
Houghton Mifflin Harcourt Algebra 1, 2015 View details
Unit 1 Assessment Readiness

Exercise 2 Page 79

Practice makes perfect
a We know that the distance from Town A to Town B on a map is 6 inches and the scale of the map is 1 in: 4 mi. Using this information we can write the ratio that compares the actual distance and the distance on the map. actual distance/map distance = 4/1 If we let d represent the actual distance from A to B, we can write a second ratio. actual distance/map distance = d/6 Since these ratios are comparing the same objects, we can form a proportion. 4/1 = d/6 Let's solve the proportion for d.
4/1 = d/6
4=d/6
4*6=d
24=d
d=24
The actual distance from Town A to Town B is 24 miles.

Therefore, the statement is true.

b Now, we are asked for the distance on the map from Town B to Town C. Since the scale of the map is the still 1 in: 4 mi, and we know that the actual distance between these two towns is 12 miles we can write the ratios just like we did before. actual distance/map distance = 4/1 If we let m represent the distance from B to C on the map, we can write a second ratio. actual distance/map distance = 12/m Since these ratios are comparing the same objects, we can form a proportion. 4/1 = 12/m Let's solve the proportion for m.
4/1 = 12/m
4=12/m
4* m=12
4m=12
4m/4=12/4
m=3
The distance on the map from Town B to Town C is 3 inches and not 36 inches. Therefore, the statement is false.
c Since we calculated that the actual distance from Town A to Town B is 24 miles, and we know the actual distance from Town B to Town C is 12 miles, we can calculate the distance from Town A to Town C by adding them.
24+12
36
The distance between Town A and Town C is 36 miles and not 4.5 miles, so the statement is false.