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Sonja [21]
3 years ago
8

3. write an equation of the circle whose center is (-5,2) and whose radius length is 4 express your answer in standard form. I N

EED MY WORK TO BE SHOWN PLEASE HURRY I AM BEING TIMED!!!!!
Mathematics
1 answer:
marin [14]3 years ago
6 0
Circle = (x-h)^2 + (y-k)^2 = r^2

Center is (h,k) h = -5, k = 2
Radius is 4, r = 4

(x - -5)^2 + (y - 2)^2 = 4^2
(x + 5)^2 + (y - 2)^2 = 16
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It is not normally distributed as it has it main concentration in only one side.

Step-by-step explanation:

So, we are given that the class width is equal to 0.2. Thus we will have that the first class is 0.00 - 0.20, second class is 0.20 - 0.40 and so on(that is 0.2 difference).

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Class interval            frequency.

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4 0
3 years ago
PLEASE HELP! I'll give 5 out of 5 stars, give thanks, and give as many points as I can.
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Answer:

\boxed{\boxed{x=\dfrac{\pi}{3}\ \vee\ x=\pi\ \vee\ x=\dfrac{5\pi}{3}}}

Step-by-step explanation:

\cos(3x)=-1\iff3x=\pi+2k\pi\qquad k\in\mathbb{Z}\\\\\text{divide both sides by 3}\\\\x=\dfrac{\pi}{3}+\dfrac{2k\pi}{3}\\\\x\in[0,\ 2\pi)

\text{for}\ k=0\to x=\dfrac{\pi}{3}+\dfrac{2(0)\pi}{3}=\dfrac{\pi}{3}+0=\boxed{\dfrac{\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=1\to x=\dfrac{\pi}{3}+\dfrac{2(1)\pi}{3}=\dfrac{\pi}{3}+\dfrac{2\pi}{3}=\dfrac{3\pi}{3}=\boxed{\pi}\in[0,\ 2\pi)\\\\\text{for}\ k=2\to x=\dfrac{\pi}{3}+\dfrac{2(2)\pi}{3}=\dfrac{\pi}{3}+\dfrac{4\pi}{3}=\boxed{\dfrac{5\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=3\to x=\dfrac{\pi}{3}+\dfrac{2(3)\pi}{3}=\dfrac{\pi}{3}+\dfrac{6\pi}{3}=\dfrac{7\pi}{3}\notin[0,\ 2\po)

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