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aev [14]
2 years ago
6

URGENT: (TIMED) Determine the equation of the circle graphed below.

Mathematics
1 answer:
Shtirlitz [24]2 years ago
5 0

Answer:

(x-3)^2 + (y-2)^2 = 16

Step-by-step explanation:

This is because the centre is at (3,2) and the radius of the circle is 4 and you square all the terms to find the equation of the circle.

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Helppp
Darya [45]

Answer:

w= -35  so i would go with the first one

Step-by-step explanation:

5 0
3 years ago
Find the missing side, please
AVprozaik [17]

Answer:

a. √442, b. √540, c. 8

Step-by-step explanation:

remember that a^2 + b^2 = c^2

4 0
2 years ago
Evalute 2 (a - 5) + 2, if a = 15<br> O 14<br> 0 27<br> O 22<br> 07
Mrac [35]
I’m pretty sure it’s 22
4 0
3 years ago
Read 2 more answers
A particle moves with velocity vector
asambeis [7]

Answer:

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}

Step-by-step explanation:

We are given that velocity vector of a particle

\vec{v(t)}=t\hat{i}-2\hat{j}+t^2\hat{k}

When t=0 then the particle is at the point (0,-1,1).

We have to find the position of particle  at time t.

We know that

Velocity =\frac{Displacement }{time}=\frac{ds}{dt}

Therefore,\vec{v}=\frac{\vec{ds}}{dt}

\int{ds}=\int (t\hat{i}-2\hat{j}+t^2\hat{k})dt

Integrate on both sides then we get

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}+C

\int x^n dx=\frac{x^{n+1}}{n+1}+C

Substitute the value of point at time t=0 then we get

C=-\hat{j}+\hat{k}

Substitute the value of C then we get

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}-\hat{j}+\hat{k}

Therefore, the position of particle at time t

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}

8 0
2 years ago
Sam was in 4th grade 146 weeks ago. What fade is she in now?
hichkok12 [17]
She is in 6th grade. there are 52 weeks in a year, and 146/52 is approximately equal to 2.8. If she was starting 4th grade in the beginning of the year, then she would now be nearing the end of 6th grade.
4 0
3 years ago
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