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2 years ago

How do you solve ? 2(x−(3+2x)+9)=3x−8

Mathematics

1 answer:

PIT_PIT [208]2 years ago

6 0

Hey there!!

Given equation :

... 2 ( x - ( 3 + 2x ) + 9 ) = 3x - 8

Using the distributive property.

... 2 ( x - 3 - 2x + 9 ) = 3x - 8

... 2 ( -x + 6 ) = 3x - 8

Using the distributive property.

... -2x + 12 = 3x - 8

Subtracting 12 on both sides.

... -2x = 3x - 8 - 12

... -2x = 3x - 20

Subtracting 3x on both sides.

... -2x - 3x = -20

... -5x = -20

Dividing by -5 on both sides.

... x = -20 / -5

... x = 4

Hence, the answer is 4.

Hope my answer helps!

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2 years ago

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Option (4)

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3 years ago

Use the definition of a Taylor series to find the first three non zero terms of the Taylor series for the given function centere

Ket [755]

Answer:

$e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...$

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

Step-by-step explanation:

Taylor series expansions of f(x) at the point x = a

$\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...$

This expansion is valid only if $\text{f}\:^{(n)}(a)$ exists and is finite for all $n \in \mathbb{N}$ , and for values of x for which the infinite series converges.

$\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1$

$\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...$

$\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}$