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3 years ago

12

How do you solve 8a squared plus 2 = 634

1 answer:

Inga [223]3 years ago

8 0

Answer:

a ≈ 8.9

Step-by-step explanation:

Set up the equation as so:

8a^2 + 2 = 634

First, subtract two from both sides:

8a^2 = 632

Then, divide by 8 to further isolate the variable.

a^2 = 79

To get rid of the squared, you have to take the square root of both sides. The square root of 79 is roughly 8.9. Ergo, a ≈ 8.9

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Alla [95]

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3 years ago

The first term of a geometric sequence is 15, and the 5th term of the sequence is <img src="https://tex.z-dn.net/?f=%5Cfrac%7B24

sladkih [1.3K]

The geometric sequence is $15,9,\frac{27}{5},\frac{81}{25}, \frac{243}{125}$

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We need to find the 2nd, 3rd and 4th term of the geometric sequence.

To find these terms, we need to know the common difference.

The common difference can be determined using the formula,

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For $n=5$ , we have,

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Now, we shall find the 2nd, 3rd and 4th terms by substituting $n=2,3,4$ in the formula $a_n=a_1(r)^{n-1}$

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Thus, the 2nd term of the sequence is 9

For $n=3$ , we have,

$a_3=15(\frac{3}{5} )^{2}$

$=15(\frac{9}{25} )$

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Thus, the 3rd term of the sequence is $\frac{27}{5}$

For $n=4$ , we have,

$a_4=15(\frac{3}{5} )^{3}$

$=15(\frac{27}{25} )$

$=\frac{81}{25}$

Thus, the 4th term of the sequence is $\frac{81}{25}$

Therefore, the geometric sequence is $15,9,\frac{27}{5},\frac{81}{25}, \frac{243}{125}$

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natima [27]

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