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3 years ago

Literal Equations: C=2πr solve for r

1 answer:

6 0

$\sf C = 2\pi r\\\\ Divide~both~sides~by~2\pi\\\\\boxed{\bf{r = \frac{C}{2\pi}}}$

<span>Hope this helped! If you have any other questions or would like further explanation just let me know!! :)</span>

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What is 11/5 in a decimal

Vesnalui [34]

You can divide it and get 2.2

5 0

3 years ago

Read 2 more answers

Which two expressions have the same value?<br> 1. 23-82<br> 2. 23 - (8 x 2)<br> 3. (23 -8) x 2

Elza [17]

Answer:

Any expression doesn't have the same value

Step-by-step explanation:

Subtract 82 from 23.

−59

Simplify each term.

Multiply 8 by 2.

23−1⋅16

Multiply −1 by 16.

23−16

Subtract 16 from 23.

Subtract 8 from 23.

15⋅2

Multiply 15 by 2.

<h2>Hope it is helpful.....</h2>

6 0

2 years ago

Consider the graph of the function f(x) = 25

trasher [3.6K]

Considering it's horizontal asymptote, the statement describes a key feature of function g(x) = 2f(x) is given by:

Horizontal asymptote at y = 0.

<h3>What are the horizontal asymptotes of a function?</h3>

They are the limits of the function as x goes to negative and positive infinity, as long as these values are not infinity.

Researching this problem on the internet, the functions are given as follows:

$f(x) = 2^x$ .

$g(x) = 2f(x) = 2(2)^x$

The limits are given as follows:

$\lim_{x \rightarrow -\infty} g(x) = \lim_{x \rightarrow -\infty} 2(2)^x = \frac{2}{2^{\infty}} = 0$

$\lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} 2(2)^x = 2(2)^{\infty} = \infty$

Hence, the correct statement is:

Horizontal asymptote at y = 0.

More can be learned about horizontal asymptotes at brainly.com/question/16948935

#SPJ1

3 0

1 year ago

The vertex is at (-1,-3) move it to (8,10) on the coordinate grid.

givi [52]

Right 9 spaces and up 13 spaces

5 0

3 years ago

Please help I’m lost it has to be 3 or 2 or more answers by the way and I will except answers from Expert, Ace, or Genius or any

deff fn [24]

The correct answer is only A, because you need the tenths place to be greater than the number on the left, and not just the hundredths place.

P.S. Even though I'm a beginner, I took Multi-Variable calculus in 12th grade.
P.P.S. STOP PUTTING ON RANDOM ANSWERS ON QUESTIONS JUST SO YOU CAN GET THE POINTS!!!!!!

4 0

2 years ago