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Degger [83]
3 years ago
12

Find the equation for the circle with a diameter whose endpoints are (-5, -3) and (3,2).

Mathematics
1 answer:
densk [106]3 years ago
3 0

Answer:

  (x +1)^2 +(y +0.5)^2 = 22.25

Step-by-step explanation:

The center of the circle is the midpoint of the diameter, so is ...

  (h, k) = ((-5, -3) +(3, 2))/2 = (-5+3, -3+2)/2 = (-1, -1/2)

The circle formula is then ...

  (x -h)^2 +(y -k)^2 = r^2

  (x -(-1))^2 +(y -(-1/2))^2 = r^2

We can find r^2 by substituting one of the points for (x, y).

  (3 +1)^2 +(2 +1/2)^2 = r^2 = 16 +6.25 = 22.25

Then the circle equation is ...

  (x +1)^2 +(y +0.5)^2 = 22.25

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From the CENTER, it would be radius of the circle

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Stretching across the entire circle is diameter, while from the center is radius.

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Your answer would be -2/3.

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Jannelle needs 280,000 points on a computer game so far she has 96,675 points how many more dose she need use mental math
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Describe, in words, the process you would use to find w.
Otrada [13]

Answer:

w = 52°

Step-by-step explanation:

We know that the sum of the angle of a triangle is equal to 180 degrees

In this figure,

w+w+76 = 180

As w = w (Vertically opposite angles)

So,

2w = 180-76

2w = 104

w = 52

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7 0
3 years ago
Solve the given differential equation by using an appropriate substitution. the de is homogeneous. (y2 yx) dx − x2 dy = 0
Schach [20]

The solution for the given differential equation is lnlxl = \frac{-x}{y} +C

Given,

(y^{2} +y^{x} )dx - x^{2} dy =0

Here,

x = vy, dx = vdy + ydv

y = ux, dy = udx + xdu

Then,

((ux)^{2} +(ux)x)dx-x^{2} (udx+xdu)=0\\=u^{2} x^{2} dx+ux^{2} dx-x^{2} udx-x^{3} du=0\\=u^{2} x^{2} dx=x^{3} du\\=\frac{x^{2} }{x^{3} } dx=\frac{1}{u^{2} } du

Now,

\int\limits^a_b {\frac{1}{x} } \, dx =\int\limits^a_b {u^{-2} } \, du

lnlxl=\frac{u^{-1} }{(-1)} +C

lnlxl=\frac{-1}{u} +C

y = ux

u = \frac{y}{x}

That is lnlxl =\frac{-x}{y} +C

Learn more about differential equation here: brainly.com/question/21852102

#SPJ4

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1 year ago
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