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1 year ago

What is the radius of a circle if it’s center is (1,2) and one point on the circle is (1,6)?

Mathematics

1 answer:

dem82 [27]1 year ago

3 0

because the x values are identical the radius would be found by subtracting the y values.

6-2 = 4

the radius is 4

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Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin

saw5 [17]

Treat $\mathcal C$ as the boundary of the region $\mathcal S$ , where $\mathcal S$ is the part of the surface $z=2xy$ bounded by $\mathcal C$ . We write

$\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r$

with $\mathbf f=(y+7\sin x,z^2+9\cos y,x^3)$ .

By Stoke's theorem, the line integral is equivalent to the surface integral over $\mathcal S$ of the curl of $\mathbf f$ . We have

$\nabla\times\mathbf f=(-2z,-3x^2,-1)$

so the line integral is equivalent to

$\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$
$=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

where $\mathbf s(u,v)$ is a vector-valued function that parameterizes $\mathcal S$ . In this case, we can take

$\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)$

with $0\le u\le1$ and $0\le v\le2\pi$ . Then

$\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$

and the integral becomes

$\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$
$=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi$

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2 years ago

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Alex777 [14]

Answer:

Now, I ain't tryin' to sound mean or standoff-ish but I'm pretty sure Wikipedia could answer this question for you

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3 years ago

4 red balls 4 white balls what is the probability of drawing at least 1 red ball if two balls are drawn

vodomira [7]

The probability of drawing exactly one red ball is given by:
$P(1\ red\ ball)=\frac{4C1\times4C1}{8C2}=\frac{16}{28}$
the probability of drawing two red balls is given by:
$P(2\ red\ balls)=\frac{4C2\times4C0}{8C2}=\frac{6}{28}$
The probability of drawing at least one red ball is:
P(1 red ball) + P(2 red balls) = 16/28 + 6/28 = 11/14.
The answer is $\ \frac{11}{14}$

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Norma-Jean [14]

Answer:

it would be 3.28125

Step-by-step explanation:

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