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Norma-Jean [14]
4 years ago
13

Combine like terns: 16y+8×-7×+8y-2y^2

Mathematics
2 answers:
mihalych1998 [28]4 years ago
8 0
-2Y^2+24Y+X. Is the simplified version of this. <span />
creativ13 [48]4 years ago
6 0
16<em>y + 8</em>y<em>
</em>8<em>x</em> - 7<em>x</em>
-2<em>y</em>²

-2<em>y</em>² + 24<em>y</em> + <em>x</em> 
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3 years ago
∫∫(x+y)dxdy ,d là miền giới hạn bởi x²+y²=1
igor_vitrenko [27]

It looks like you want to compute the double integral

\displaystyle \iint_D (x+y) \,\mathrm dx\,\mathrm dy

over the region <em>D</em> with the unit circle <em>x</em> ² + <em>y</em> ² = 1 as its boundary.

Convert to polar coordinates, in which <em>D</em> is given by the set

<em>D</em> = {(<em>r</em>, <em>θ</em>) : 0 ≤ <em>r</em> ≤ 1 and 0 ≤ <em>θ</em> ≤ 2<em>π</em>}

and

<em>x</em> = <em>r</em> cos(<em>θ</em>)

<em>y</em> = <em>r</em> sin(<em>θ</em>)

d<em>x</em> d<em>y</em> = <em>r</em> d<em>r</em> d<em>θ</em>

Then the integral is

\displaystyle \iint_D (x+y)\,\mathrm dx\,\mathrm dy = \iint_D r^2(\cos(\theta)+\sin(\theta))\,\mathrm dr\,\mathrm d\theta \\\\ = \int_0^{2\pi} \int_0^1 r^2(\cos(\theta)+\sin(\theta))\,\mathrm dr\,\mathrm d\theta \\\\ = \underbrace{\left( \int_0^{2\pi}(\cos(\theta)+\sin(\theta))\,\mathrm d\theta \right)}_{\int = 0} \left( \int_0^1 r^2\,\mathrm dr \right) = \boxed{0}

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3 years ago
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Artist 52 [7]

Step-by-step explanation:

<em><u>1</u></em><em><u>,</u></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em>

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3 years ago
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Tom [10]
2 divided by 18 will equal 9
so just multiply that by the minutes (3)
9×3=27 minutes
3 0
3 years ago
Read 2 more answers
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