All categories

2 years ago

8-6x=-28 I need help plzzz

Mathematics

2 answers:

Schach [20]2 years ago

3 0

The answer is 6. 8-6(6) = -28.

matrenka [14]2 years ago

3 0

Answer:

Step-by-step explanation:

We need to solve the equation .The given equation is ,

→ 8 - 6x = -28

→ -6x = -28-8

→ -6x = -36

→ x = -36/-6

→ x = 6

<h3>• Hence the value of x is 6 .</h3>

You might be interested in

Which of the following methods would be the easiest to use to solve 12x2 – 48 = 0?

Katarina [22]

$\bf \stackrel{\textit{solving for \underline{x}}}{12x^2-48=0\implies 12x^2=48\implies x^2=\cfrac{48}{12}\implies x^2=4} \\\\\\ x=\pm\sqrt{4}\implies x=\pm 2$

7 0

2 years ago

Read 2 more answers

Which term is a word for an initial amount of money that is borrowed or invested?

ozzi

Answer:

principal is the correct answer

Step-by-step explanation:

principal is the correct answer

5 0

2 years ago

Read 2 more answers

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

1 year ago

Marissa bought 2 and a half gallons of orange juice to make punch how many quarts of orange juice did. Marissa buy

Nookie1986 [14]

The answer is as follows:
There are 2 pints in a quart.
There are 2 quarts in a half gallon.
There are 2 half gallons in a gallon.
Therefore, we can conclude that there are 10 quarts of orange juice.

7 0

2 years ago

the area of a rectangle is 188 meters squared .find the length of the rectangle if the width is 4 meters, using the formula A=lw

sergiy2304 [10]

A = LW
A = 188
W = 4

188 = 4L
188/4 = L
47 = L....so the length is 47 meters

5 0

2 years ago