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3 years ago

If (5, 4) and (7,2) are the endpoints of a diameter of a circle, what is the equation of the circle?

Mathematics

2 answers:

STatiana [176]3 years ago

8 0

Answer:

(x -6)^2 +(y -3)^2 = 2

Step-by-step explanation:

The midpoint (C) is the center of the circle:

C = ((5, 4) +(7, 2))/2 = ((5+7)/2, (4+2)/2) = (6, 3)

A circle with center (h, k) that goes through point (a, b) can be written as ...

(x -h)^2 +(y -k) ^2 = (a -h)^2 +(b -k)^2

Using the values for this problem, we get ...

(x -6)^2 +(y -3)^2 = (5 -6)^2 +(4 -3)^2

(x -6)^2 +(y -3)^2 = 2

liberstina [14]3 years ago

6 0

Answer:

Its A.

Step-by-step explanation:

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B is the correct answer.

Step-by-step explanation:

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3 years ago

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Alexeev081 [22]

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So the way that your choice expressed that is:

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3 years ago

Find the following integral

ololo11 [35]

There's nothing preventing us from computing one integral at a time:

$\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2$

$\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2$

$\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx$

Expand the integrand completely:

$x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x$

Then

$\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}$

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2 years ago

Determine the midpoint between the points (-5, 3) and (5, 9).

PSYCHO15rus [73]

Answer:

M (0, 6)

Step-by-step explanation:

By midpoint formula,

M [(x1+x2)/2 , (y1+y2)/2]

M [-5+5/2 , 3+9/2]

M (0/2 , 12/2)

M (0, 6)

4 0

2 years ago