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Haley measured the distances between her house and two of her friends' houses. ellen's house is 1.41 miles away from haley's, and dirk's house is 6,547 feet away from haley's. whose house is farther away from haley's, and how many feet farther away is it, to the nearest whole foot?

Respuesta :

danielmaduroh
We have the following variables:
H = Haley's house
E= Ellen's house
D = Dirk's house

In the figure below we can see the representation of this problem.

Given that there are two measurements with different units, namely, the distance between Haley and Ellen's houses is in miles and the distance between Haley and Dirk's houses is in feet, we need to convert one of these units.

As the final question demands which house is farther away from Haley's, we will convert the distance between Haley and Ellen's houses from miles to feet, so:

[tex]1mi = 5280ft[/tex]

[tex]1.41mi(\frac{5280ft}{1mi}) = 7444.8ft[/tex]

So:
The distance between Haley and Ellen's houses = [tex]7444.8ft[/tex]
The distance between Haley and Dirk's houses = [tex]6547ft[/tex]

Then Ellen's house is farther away from haley's, and the nearest whole foot is:
[tex]7444ft[/tex]

Ver imagen danielmaduroh
JeanaShupp

Answer: Ellen's house is farther away by 898 feet.

Step-by-step explanation:

Given: The distance between Haley's house and Ellen's house = 1.41 miles

We know that 1 mile = 5,280 feet

Then, [tex]1.41\text{ miles}=1.41\times5,280\text{ feet}=7,444.8\text{ feet}[/tex]

The distance between Haley's house and Dirk's house = 6,547 feet

Clearly, 6,547<7,444.8

And [tex]7444.8-6547= 897.8\approx898[/tex]

Therefore, Ellen's house is farther away by 898 feet.

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