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svetlana [45]
3 years ago
5

the 43rd term of an arithmetic sequence is -671 and the 52nd term is -806. find the first term and the common difference.

Mathematics
1 answer:
Bond [772]3 years ago
3 0

Answer:

First term = -41.

Common difference = -15.

Step-by-step explanation:

nth term: an = a1 + (n - 1)d  where a = first term and d = the common difference.

-671 = a1 + (43 - 1)d

-806 = a1 +(52 - 1)d

-671  = a1 + 42d

-806 = a1 + 51d

Subtracting ( to eliminate a1):

-671 - (-806) = 42d - 51d

-9d =  135

d = -15

Substitute for d in the first equation:

-671 = a1 + 42*-15

-671 = a1 - 630

a1 = -671 + 630 = -41.

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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}
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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:

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This gives

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

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

The function for number of systems is:

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<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}

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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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