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Anna007 [38]
3 years ago
8

The distance from Mason's house to Kevin's house is about 5 × 105 inches. The distance from Mason's house to Chloe's house is ab

out 8 × 107 inches. The distance from Mason's house to Chloe's house is how many times as great as the distance from Mason's to Kevin's house?
Mathematics
1 answer:
olga nikolaevna [1]3 years ago
6 0

Answer:

Distance from Mason's house to Chloe's house is 160 as great as the distance from Mason's to Kevin's house.

Step-by-step explanation:

Let the distance from Mason's house to Chloe's house is x times as great as the distance from Mason's to Kevin's house.

Distance from Mason's house to Kevin's house= 5 \times 10^5 inches.

Distance from Mason's house to Chloe's house =8 \times 10^7 inches

ATQ

8 \times 10^7= x \times 5 \times 10^5 \\\frac{8 \times 10^7}{5 \times 10^5}=x\\160=x

So, distance from Mason's house to Chloe's house is 160 times as great as the distance from Mason's to Kevin's house.

Hence distance from Mason's house to Chloe's house is 160 as great as the distance from Mason's to Kevin's house.

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