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Leno4ka [110]
3 years ago
13

PLEASE!!!!! HELP ME!!!!!!

Mathematics
1 answer:
kati45 [8]3 years ago
4 0

Answer:

15 = a_1 r^4 (1)

1 = a_1 r^5 (2)

If we divide equations (2) and (1) we got:

\frac{r^5}{r^4}= \frac{1}{15}

And then r= \frac{1}{15}

And then we can find the value a_1 and we got from equation (1)

a_1 = \frac{15}{r^4} = \frac{15}{(\frac{1}{15})^4} =759375

And then the general term for the sequence would be given by:

a_n = 759375 (\frac{1}{15})^n-1 , n=1,2,3,4,...

And the best option would be:

C) a1=759,375; an=an−1⋅(1/15)

Step-by-step explanation:

the general formula for a geometric sequence is given by:

a_n = a_1 r^{n-1}

For this case we know that a_5 = 15, a_6 = 1

Then we have the following conditions:

15 = a_1 r^4 (1)

1 = a_1 r^5 (2)

If we divide equations (2) and (1) we got:

\frac{r^5}{r^4}= \frac{1}{15}

And then r= \frac{1}{15}

And then we can find the value a_1 and we got from equation (1)

a_1 = \frac{15}{r^4} = \frac{15}{(\frac{1}{15})^4} =759375

And then the general term for the sequence would be given by:

a_n = 759375 (\frac{1}{15})^n-1 , n=1,2,3,4,...

And the best option would be:

C) a1=759,375; an=an−1⋅(1/15)

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surface area (S) of a right rectangular solid is:

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where:

L = length

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you have:

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formula becomes:

S = 2*7*a + 2*7*4 + 2*a*4

simplify:

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combine like terms:

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-----

answer is:

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-----

to prove, substitute any value for a in equation 2:

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S = 22*15 + 56

S = 330 + 56

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-----

since a = 15, then W = 15 because W = a

go back to equation 1 and substitute 15 for W:

S = 2*L*W + 2*L*H + 2*W*H (equation 1)

where:

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W = width

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you have:

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equation 1 becomes:

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S = 22*a + 56

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