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tino4ka555 [31]
3 years ago
10

One week Alaina ran 12 miles in 131.25 minutes. The next, Alaina ran 12 miles in 119.5 minutes. If she ran in a constant pace du

ring each run, about how much faster did she run each mile in he second week than in the first week?
Mathematics
2 answers:
vlabodo [156]3 years ago
6 0
Ou should not increase to more than 12 miles next week<span>; ... </span>25 miles<span> a </span>week<span>, usually a 7 </span>milerun<span> ... before I </span>run<span> a half. Using that method</span>
Ronch [10]3 years ago
5 0

Answer:

0.98 minutes

Step-by-step explanation:

First, we have to determine how much time it took for Alaina to run one mile on the first and the second week:

First week:

12 miles → 131.25 minutes

1 mile     →     x

x=(131.25 minutes*1 mile)/12 miles= 10.93 minutes

Second week:

12 miles → 119.5 minutes

1 mile     →     x

x= (119.5 minutes*1 mile) /12 miles= 9.95 minutes

Now, we have to find the difference between the time it took to run 1 mile in the first and the second week:

10.93-9.95= 0.98 minutes

She run 0.98 minutes faster in the second week.

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Now, onto Part B. We can use <u>another</u> equation!

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Now, we must solve the given equation, first by taking 100.7 from each side:

192.7 - 2.3x = 100.7

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\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]

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Using the new aug. matrix, combine row 1 and 3 times row 3 :

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\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 5 & 4 & 1 & 0 & 3 \end{array} \right]

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Combine -5 times row 2 and 7 times row 3 :

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\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 3 & -3 & -15 & 21 \end{array} \right]

• Multiply row 3 by 1/3 :

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• Eliminate the column 2 and 3 entries in row 1.

Combine row 1, -2 times row 2, and -1 times row 3 :

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