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3 years ago

Solve this equation 3x^2 − 16x = x^2 − 30

Mathematics

2 answers:

ValentinkaMS [17]3 years ago

3 0

3x^2-16x=x^2-30

-x^2 -x^2

2x^2-16x=-30

+30 +30

2x^2-16x+30=0

2x-6=0 or x-5=0

x=3 or x=5

A. 3,5

If you need an explanation, I can provide one. Hope it helped! If it didn't tell me what went wrong or what confused you.

vitfil [10]3 years ago

3 0

Answer:

Step-by-step explanation:

2x^2-16x+30=0

x^2 -8x+15=0

(x-3)(x-5)=0

x=3, x=5

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3 years ago

1. A-Tunes requires a membership fee of $12 a month and $1 for each download.

matrenka [14]

<h3>Answer:</h3>

After 16 downloads the charges will equal each other.

<h3>Step-by-step explanation:</h3>

First set up an equation

x = Downloads

12 + 1x = 5 + 1.5x

Solve

12 + 1x = 5 + 1.5x <- Rewrite your equation

-5 -5 <- Subtract 5 from both sides

8 + 1x = 1.5x <- Simplify

-1x -1x <- Subtract 1 from both sides

8 = 0.5x <- Simplify

8 = 1/2x <- Convert 0.5 to a fraction to make the next step easier

16 = x <- Multiply both sides by 2 to divide by

a fraction

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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:

<u>Taylor series</u> 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}}$