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3 years ago

The radius is 9 cm. what is the approximate circumference of the circle

Mathematics

1 answer:

Mumz [18]3 years ago

4 0

Answer:

56.5487 (this is exact, just round it if you want)

Step-by-step explanation:

The formula to find the circumference is 2, pi, r (radius).

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Use the fundamental identities and appropriate algebraic operations to simplify the following expression. (18 +tan x) (18-tan x)

andrezito [222]

Answer:

a) $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)=325$

b) The lowest point of $y=\cos \left(x\right)$ , $0\leq x\leq 2\pi$ is when x = $\pi$

Step-by-step explanation:

a) To simplify the expression $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)$ you must:

Apply Difference of Two Squares Formula: $\left(a+b\right)\left(a-b\right)=a^2-b^2$

$a=18,\:b=\tan \left(x\right)$

$\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)=18^2-\tan ^2\left(x\right)=324-\tan ^2\left(x\right)$

$324-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Apply the Pythagorean Identity $1+\tan ^2\left(x\right)=\sec ^2\left(x\right)$

From the Pythagorean Identity, we know that $1=-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Therefore,

$324[-\tan ^2\left(x\right)+\sec ^2\left(x\right))]\\324[+1]\\325$

b) According with the below graph, the lowest point of $y=\cos \left(x\right)$ , $0\leq x\leq 2\pi$ is when x = $\pi$

3 0

4 years ago

What is the value of the expression?<br> 1/3-5/6<br> Enter your answer in simplest form.

algol [13]

Answer:

(1/3 - 5/6) / 5/6

You want to make 1/3 and 5/6 have the same denominator so you multiply both the numerator and denominator of 1/3 by 2 to get 2/6. You plug that back into your equation to get: (2/6 - 5/6) / 5/6. 2/6 - 5/6 is -3/6. -3/6 divided by 5/6 is -3/6 multiplied by 6/5 which is -3/5.

5 0

3 years ago

Read 2 more answers

(4x+1)(3x-5) / 10x^2 +2x • A / 15x2 -22x -5 = 64 x^2 -4 / 5x + 1

AfilCa [17]

Answer:

4x^3

Step-by-step explanation:

5 0

3 years ago

The range of f(x) = |x| is y ≥ 0. If a < 0 and b ≠ 0 for g(x) = a|x| + b, what is the range of function g?

kifflom [539]