Question

In a library, bookcases A,B,C and D are lined up in order. Bookcase A is 1.38 meters from bookcase B and 6.8 meters from bookcase C. Bookcase D is 0.925 meters farther from bookcase C than the distance between bookcases B and C. How far apart are bookcases C and D? (Show your work).

Answer 1

Answer:

Bookcases C and D are 6.345 meters apart

Explanation:

The given scenario can be modeled using the attached diagram

Assume that the distance between cases B and C is x meters

We are given that:

Distance between cases A and B is 1.38 m

Distance between cases A and C is 6.8 m

We know that:

Distance between A & C = Distance between A & B + Distance between  B & C

This means that:

6.8 = 1.38 + x

x = 6.8 - 1.38 = 5.42 m

Now, we are given that bookcase D is 0.925 meters farther from bookcase C than the distance between bookcases B and C

This means that:

Distance between cases C & D = Distance between cases B & C + 0.925

Distance between cases C & D = x + 0.925

Distance between cases C & D = 5.42 + 0.925

Distance between cases C & D = 6.345 meters

Hope this helps :)

Answer 2

Answer: Bookcases C and D are 6.345 (or six and three hundred forty five thousandth) meters apart

Step-by-step explanation:Since book case A is 1.38 meters from bookcase B and 6.8 meters from bookcase C, we must subtract 1.38 from 6.8 which would give you 5.42, and since bookcase D is 0.925 meters farther from bookcase C then you would want to add 5.42 by 0.925 which would give you your answer which would be 6.345. In other words subtract 1.38-6.8=5.42, then add 5.42+0.925=6.345 and that's your explanation simplified!