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madam [21]
3 years ago
12

Write an equation of the line in slope-intercept form.

Mathematics
1 answer:
aalyn [17]3 years ago
6 0

Answer:

The answer would be y=-1/3x+1.

Step-by-step explanation:

To find the slope, you must use the formula y2-y1/x2-x1. That would make it 1-2/0-(-3) which the slope is -1/3. Now, using the formula y-y1=m(x-x1) which would be y-2=-1/3(x-(-3) if you plug it in. Solving the equation, you would get y=-1/3x+1.

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HELP I CANT FIGURE IT OUT!<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7B18%7D%20%20-%20%20%5Cfrac%7B1%7D%7B4%7D%20"
bija089 [108]

<em>Answer:</em>

<em>1/36</em>

<em>Step-by-step explanation:</em>

<em>²⁾5/18 - ⁹⁾1/4 =</em>

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8 0
3 years ago
Find the area of the part of the paraboloid z = 9 - x^2 - y^2 that lies above the xy-plane.
svetoff [14.1K]

Parameterize this surface (call it S) by

\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(9-u^2)\,\vec k

with 0\le u\le3 and 0\le v\le2\pi. Take the normal vector to S to be

\vec r_u\times\vec r_v=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k

Then the area of S is

\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^{2\pi}\int_0^3u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv

=\displaystyle2\pi\int_0^3u\sqrt{1+4u^2}\,\mathrm du=\boxed{\frac{37\sqrt{37}-1}6\pi}

3 0
4 years ago
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