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miskamm [114]
3 years ago
11

Fred begins walking toward John house at 3 mi/h. John leaves his house at the same time and walks toward Fred's house on the sam

e path at a rate of 2 mi/h. How long will it be before they meet if the distance between the houses is 4 miles?
Mathematics
1 answer:
Tcecarenko [31]3 years ago
7 0
Let us assume as at when they meet, the time elapsed is x hours.

Fred  3 mi/h
John  2 mi/h

Distance = speed * time

Fred = 3x
John = 2x

Houses are 4miles apart. 
So:            3x + 2x  = 4    5x = 4

x = 4/5 hour =  0.8 hour or 0.8*60 = 48 minutes

Time elapsed is 0.8 hour or 48 minutes.
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88

Step-by-step explanation:

6 + 5 = 11

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44 · 2 = 88

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Explain how you could use either of two operation to solve exercise 10.
wel
You want to know the factor by which 3 2/3 is multiplied to get 7 1/3.

1. You can estimate that it is 2 from 7/3 ≈ 2, then check by multiplication to see if that is right.
.. 2*(3 2/3) = 6 4/3 = 7 1/3 . . . . 2 is the correct factor.

2. You can divide 7 1/3 by 3 2/3 to see what the factor is.
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3. You could see how many times you can subtract 3 2/3 from 7 1/3.
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4. You could add 3 2/3 to see how many times it takes to get 7 1/3.
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We have shown methods using multiplication, division, subtraction, addition. Take your pick.
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HELP PLEASE!!!❗️❗️❗️
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