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3 years ago

Solve each equation for the indicated variable. u = -3a - 3 , for a

Mathematics

1 answer:

MAXImum [283]3 years ago

8 0

Answer:

$a = \frac{u + 3}{-3}$

Step-by-step explanation:

Isolate the variable, a. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

PEMDAS is the order of operation and stands for:

Parenthesis

Exponents (& Roots)

Multiplication

Division

Addition

Subtraction

First, add 3 to both sides of the equation:

u = -3a - 3

u (+3) = -3a - 3 (+3)

u + 3 = -3a

Next, divide -3 from both sides of the equation:

(u + 3)/-3 = (-3a)/-3

a = (u + 3)/-3

$a = \frac{u + 3}{-3}$ is your answer.

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Math question. image shown below

marin [14]

Answer:

The third one is the only one that is not a function.

Step-by-step explanation:

3 0

3 years ago

Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4

anyanavicka [17]

Answer:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Step-by-step explanation:

Given

$f(x) = \sin x\\$

$c = \frac{3\pi}{4}$

Required

Find the Taylor series

The Taylor series of a function is defines as:

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

We have:

$c = \frac{3\pi}{4}$

$f(x) = \sin x\\$

$f(c) = \sin(c)$

$f(c) = \sin(\frac{3\pi}{4})$

This gives:

$f(c) = \frac{1}{\sqrt 2}$

We have:

$f(c) = \sin(\frac{3\pi}{4})$

Differentiate

$f'(c) = \cos(\frac{3\pi}{4})$

This gives:

$f'(c) = -\frac{1}{\sqrt 2}$

We have:

$f'(c) = \cos(\frac{3\pi}{4})$

Differentiate

$f"(c) = -\sin(\frac{3\pi}{4})$

This gives:

$f"(c) = -\frac{1}{\sqrt 2}$

We have:

$f"(c) = -\sin(\frac{3\pi}{4})$

Differentiate

$f"'(c) = -\cos(\frac{3\pi}{4})$

This gives:

$f"'(c) = - * -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

So, we have:

$f(c) = \frac{1}{\sqrt 2}$

$f'(c) = -\frac{1}{\sqrt 2}$

$f"(c) = -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

becomes

$f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Rewrite as:

$f(x) = \frac{1}{\sqrt 2} + \frac{(-1)}{\sqrt 2}(x - \frac{3\pi}{4}) +\frac{(-1)/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{(-1)^2/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Generally, the expression becomes

$f(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Hence:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

3 0

2 years ago

Solve using inverse operations 83+q=125; q=

Vilka [71]

Answer:

Step-by-step explanation:

83+q=125

q=125-83

q=42

3 0

3 years ago

In cirle O, OE = 12, what is the length of the circle's radius<br><br>

iren2701 [21]

Answer:

24?

Step-by-step explanation:

I'm not very good at math and I'm sorry if I'm wrong.

8 0

3 years ago

If the area of a square is 100cm2, then what is its perimeter?

Vinil7 [7]

Answer:

40cm

Step-by-step explanation:

A=s*s

A=10*10

A=100 cm^2

P=4*s

P=4*10

P=40 cm

"s" represents side

6 0

3 years ago