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Vlad1618 [11]
3 years ago
7

Christine runs 10 miles in 75 minutes. At the same rate, how many miles would she ru I miles

Mathematics
1 answer:
Len [333]3 years ago
7 0

Answer:

\frac{46}{5}  miles  [or, 9.2 miles]

Step-by-step explanation:

<u>Complete Question:</u>

Christine runs 10 miles in 75 minutes. At the same rate, how many miles would she ru n in 69 minutes?

<u>Solution:</u>

We need to find the unit rate first and then find our answer. To get unit rate, we need to know the number of miles she runs in "1 minute". So, we will divide number of miles by number of minutes, that is given. Hence:

\frac{10}{75}=\frac{2}{15}

That means in 1 minute, Christine runs 2/15th of a mile.

Now, we want to know how many miles she will run in 69 minutes. We simply multiply unit rate (2/15) by the number of minutes (which is 69). Thus we have:

\frac{2}{15}*69=\frac{2*23}{5}=\frac{46}{5}

In decimal, doing division, that is 9.2 miles

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14.       1.5, 10    <- Answer
15.       5,1          <- Answer

Proof 14 

Solve the following system:
{2 x - y = -7 | (equation 1)
4 x - y = -4 | (equation 2)
Swap equation 1 with equation 2:
{4 x - y = -4 | (equation 1)
2 x - y = -7 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{4 x - y = -4 | (equation 1)
0 x - y/2 = -5 | (equation 2)
Multiply equation 2 by -2:
{4 x - y = -4 | (equation 1)
0 x+y = 10 | (equation 2)
Add equation 2 to equation 1:
{4 x+0 y = 6 | (equation 1)
0 x+y = 10 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 3/2 | (equation 1)
0 x+y = 10 | (equation 2)
Collect results:
Answer: {x = 1.5               
                y = 10

Proof 15. 

Solve the following system:
{5 x + 7 y = 32 | (equation 1)
8 x + 6 y = 46 | (equation 2)
Swap equation 1 with equation 2:
{8 x + 6 y = 46 | (equation 1)
5 x + 7 y = 32 | (equation 2)
Subtract 5/8 × (equation 1) from equation 2:{8 x + 6 y = 46 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Divide equation 1 by 2:
{4 x + 3 y = 23 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Multiply equation 2 by 4/13:
{4 x + 3 y = 23 | (equation 1)
0 x+y = 1 | (equation 2)
Subtract 3 × (equation 2) from equation 1:
{4 x+0 y = 20 | (equation 1)
0 x+y = 1 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 5 | (equation 1)
0 x+y = 1 | (equation 2)
Collect results:
Answer:  {x = 5                           y = 1
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