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Arada [10]
3 years ago
12

Arturo ran 17 laps that were 1/5 mile long. Axel ran 7 laps that were 1/2 mile long. Who ran further, prove it. (5 sentence resp

onse)
Mathematics
1 answer:
Lelechka [254]3 years ago
3 0

Answer:

Axel

Step-by-step explanation:

17*1/5=17/5

7*2/7=7/2

17/5=3.4

7/2=3.5

3.4<3.5

Axel ran farther

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kvv77 [185]

The question is an illustration of composite functions.

  • Functions c(n) and h(n) are \mathbf{c(n) = 5000 + 250n} and \mathbf{n(h) = 5h}
  • The composite function c(n(h)) is \mathbf{c(n(h)) = 5000 + 1250h}
  • The value of c(n(100)) is \mathbf{c(n(100)) = 130000}
  • The interpretation is: <em>"the cost of working for 100 hours is $130000"</em>

The given parameters are:

  • $5000 in fixed costs plus an additional $250
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<u>(a) Functions c(n) and n(h)</u>

Let the number of system be n, and h be the number of hours

So, the cost function (c(n)) is:

\mathbf{c(n) = Fixed + Additional \times n}

This gives

\mathbf{c(n) = 5000 + 250 \times n}

\mathbf{c(n) = 5000 + 250n}

The function for number of systems is:

\mathbf{n(h) = 5 \times h}

\mathbf{n(h) = 5h}

<u>(b) Function c(n(h))</u>

In (a), we have:

\mathbf{c(n) = 5000 + 250n}

\mathbf{n(h) = 5h}

Substitute n(h) for n in \mathbf{c(n) = 5000 + 250n}

\mathbf{c(n(h)) = 5000 + 250n(h)}

Substitute \mathbf{n(h) = 5h}

\mathbf{c(n(h)) = 5000 + 250 \times 5h}

\mathbf{c(n(h)) = 5000 + 1250h}

<u>(c) Find c(n(100))</u>

c(n(100)) means that h = 100.

So, we have:

\mathbf{c(n(100)) = 5000 + 1250 \times 100}

\mathbf{c(n(100)) = 5000 + 125000}

\mathbf{c(n(100)) = 130000}

<u>(d) Interpret (c)</u>

In (c), we have: \mathbf{c(n(100)) = 130000}

It means that:

The cost of working for 100 hours is $130000

Read more about composite functions at:

brainly.com/question/10830110

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