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Digiron [165]
3 years ago
9

Rearrange this question and make the m as the subject

Mathematics
1 answer:
balu736 [363]3 years ago
3 0
M is equal to y minus c over x
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Levart [38]

we have been asked to find the sum of the series

\sum _{n=1}^5\left(\frac{1}{3}\right)^{n-1}

As we know that a geometric series has a constant ratio "r" and it is defined as

r=\frac{a_{n+1}}{a_n}=\frac{\left(\frac{1}{3}\right)^{\left(n+1\right)-1}}{\left(\frac{1}{3}\right)^{n-1}}=\frac{1}{3}

The first term of the series is a_1=\left(\frac{1}{3}\right)^{1-1}=1

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Plugin the values we get

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On simplification we get

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Simplify 2x - 8x - 6y + 9 -14 + x + 11y.
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We can see that 2x - 8x + x = -5x
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