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3 years ago

Solve for x 6(x-1)=9(x+2)

2 answers:

8 0

<span>Simplifying 6(x + -1) = 9(x + 2) Reorder the terms: 6(-1 + x) = 9(x + 2) (-1 * 6 + x * 6) = 9(x + 2) (-6 + 6x) = 9(x + 2) Reorder the terms: -6 + 6x = 9(2 + x) -6 + 6x = (2 * 9 + x * 9) -6 + 6x = (18 + 9x) Solving -6 + 6x = 18 + 9x Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '-9x' to each side of the equation. -6 + 6x + -9x = 18 + 9x + -9x Combine like terms: 6x + -9x = -3x -6 + -3x = 18 + 9x + -9x Combine like terms: 9x + -9x = 0 -6 + -3x = 18 + 0 -6 + -3x = 18 Add '6' to each side of the equation. -6 + 6 + -3x = 18 + 6 Combine like terms: -6 + 6 = 0 0 + -3x = 18 + 6 -3x = 18 + 6 Combine like terms: 18 + 6 = 24 -3x = 24 Divide each side by '-3'. x = -8 Simplifying x = -8</span>

soldi70 [24.7K]3 years ago

4 0

6(x-1)=9(x+2)
6x-6=9x+18
6x=9x+24
-3x=24
-x=8
x=-8

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$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

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

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

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

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