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3 years ago

What is x in this equation W=2hx-11x

Mathematics

2 answers:

Illusion [34]3 years ago

6 0

Answer:

$x=\frac{W}{2h-11}$

Step-by-step explanation:

The given equation is

$W=2hx-11x$

We need to solve this equal for x.

Taking out common factors from right side of the equation.

$W=x(2h-11)$

Divide both sides by (2h-11) to isolate variable x.

$\frac{W}{2h-11}=\frac{x(2h-11)}{2h-11}$

Cancel out common factors.

$\frac{W}{2h-11}=x$

Interchange the sides.

$x=\frac{W}{2h-11}$

Therefore, the value of x for the given equation is $x=\frac{W}{2h-11}$ .

Monica [59]3 years ago

4 0

X=W/(2h-11)

2hx-11x=W
(2h-11)x=W
x=W/(2h-11)

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aliya0001 [1]

Answer:

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Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

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We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

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Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

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3 years ago

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