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3 years ago

BEING TIMED !! In circle O, the length of arc AB is 2(pi) and the radius is 5. Find the

Mathematics

1 answer:

inn [45]3 years ago

8 0

Answer:

31.4

Step-by-step explanation:

3.14 x 2 x 5 = 31.4

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Find the length of side x in simplest radical form with a rational a<br> denominator.

geniusboy [140]

Step-by-step explanation:

the answer is

$x = \sqrt{3}$

5 0

3 years ago

Janet drove 300 miles in 4.5 hours. Write and equation to find the rate at which she was traveling.

Kobotan [32]

If you would like to know the rate at which she was travelling, you can calculate this using the following steps:300 miles ... 4.5 hoursx miles = ? ... 1 hour300 * 1 = 4.5 * x300 = 4.5 * xx = 300 / 4.5x = 200 / 3x = 66.67 miles per hour Janet was travelling at the rate 66.67 miles per hour.

Read more on Brainly.com - brainly.com/question/1927497#readmore

7 0

3 years ago

688×29=<br>how to do this step by stwp

laila [671]

688

x 29

6192

1376 x

19952

So you start by taking 9 from 29 and multiplying it by 688. when you get the answer put it below the line. then multiply the 2 from the 29 and whne you get the answer put a zero on the end and then add the answer you got from 9x688 (6192) and 2x688 (13760) and then thats youre answer. hope it helps!

8 0

3 years ago

What is the slope of the line that crosses through points (0,0) and (1,2). How can Find the answer to the problem?

qwelly [4]

Answer:

The correct answer is m = 2/1 because they first point it reaches is (1,2) and so you insert those numbers (x,y) into the rise and run equation (y/x) and get the answer 2/1.

Hope this helps!

(Hint: M = Slope)

5 0

3 years ago

CALC- limits<br> please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

3 0

3 years ago