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4 years ago

Solve for y when x=15 4x-3y=51

1 answer:

6 0

4x - 3y = 51, when x = 15

Plug in 15 for x.

4(15) - 3y = 51

Simplify.

60 - 3y = 51

Subtract 60 from both sides.

-3y = 51 - 60

Simplify.

-3y = -9

Divide both sides by -3.

y = -9/-3

Simplify.

y = 3

~Hope I helped!~

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Use the definition of a Taylor series to find the first four nonzero terms of the series for f(x) centered at the given value of

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Step-by-step explanation:

The Taylor series of the function f at a (or about a or centered at a) is given by

$f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$

To find the first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ you must:

In our case,

$f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k=\sum\limits_{k=0}^{4}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$

So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

$f^{(0)}\left(x\right)=f\left(x\right)=\frac{7}{x + 1}$

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$f^{(1)}\left(x\right)=\left(f^{(0)}\left(x\right)\right)^{\prime}=\left(\frac{7}{x + 1}\right)^{\prime}=- \frac{7}{\left(x + 1\right)^{2}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime }=- \frac{7}{9}$

$f^{(2)}\left(x\right)=\left(f^{(1)}\left(x\right)\right)^{\prime}=\left(- \frac{7}{\left(x + 1\right)^{2}}\right)^{\prime}=\frac{14}{\left(x + 1\right)^{3}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime }=\frac{14}{27}$

$f^{(3)}\left(x\right)=\left(f^{(2)}\left(x\right)\right)^{\prime}=\left(\frac{14}{\left(x + 1\right)^{3}}\right)^{\prime}=- \frac{42}{\left(x + 1\right)^{4}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime \prime }=- \frac{14}{27}$

$f^{(4)}\left(x\right)=\left(f^{(3)}\left(x\right)\right)^{\prime}=\left(- \frac{42}{\left(x + 1\right)^{4}}\right)^{\prime}=\frac{168}{\left(x + 1\right)^{5}}$

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$f\left(x\right)\approx\frac{\frac{7}{3}}{0!}\left(x-\left(2\right)\right)^{0}+\frac{- \frac{7}{9}}{1!}\left(x-\left(2\right)\right)^{1}+\frac{\frac{14}{27}}{2!}\left(x-\left(2\right)\right)^{2}+\frac{- \frac{14}{27}}{3!}\left(x-\left(2\right)\right)^{3}+\frac{\frac{56}{81}}{4!}\left(x-\left(2\right)\right)^{4}$

The first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ are:

$f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}$

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