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3 years ago

The line that has a slipe of -1/2 and passes through the point (-4,-2)

Mathematics

1 answer:

zhuklara [117]3 years ago

3 0

Answer:

y+2=-1/2(x+4)

Step-by-step explanation:

y-y1=m(x-x1)

y-(-2)=-1/2(x-(-4))

y+2=-1/2(x+4)

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PLS HELP! WILL GIVE BRAINLIEST

Leviafan [203]

Answer:

The center (2, 2.5), radius $\sqrt{37}$ / 2

Step-by-step explanation:

The standard form of the equation of a circle is:

( x - h )^2 + ( y - k )^2 = r^2, where ( h, k ) is the center and r is the radius

Rewrite the given equation in the standard form:

2x^2 + 2y^2 - 8x + 10y + 2 = 0
x^2 - 4x + y^2 + 5y = -1
x^2 - 4x + 2^2 + y^2 + 5y + (5/2)^2 = -1 + 4 + 25/4
(x - 2)^2 + (y + 2.5)^2 = 37/4

The center is:

(2, 2.5)

And radius is:

$\sqrt{37}$ / 2

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3 years ago

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Answer:

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Step-by-step explanation: cant explain it

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3 years ago

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Answer:

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The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati

makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

$y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\$

And the second one would be

$y = 2-x \,\,\,\,\,, 1 \leq x \leq 2$ .

If you rotate the first region around the "y" axis you get that

$A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51$

And if you rotate the second region around the "y" axis you get that

$A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188$

And the sum would be 2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first region around the x axis you get that

$A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708$

And if you rotate the second region around the x axis you get that

$A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472$

And the sum would be 1.5708+1.0472 = 2.618

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