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Elenna [48]
3 years ago
13

6 times 7 is 42 6 times 5 is 30 if that's any help please help me out.

Mathematics
1 answer:
alexgriva [62]3 years ago
6 0

6 * 5 = 30

42 - 30 = 12

? * 2 = 12

? = 6

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pashok25 [27]

Direct computation:

Parameterize the top part of the circle x^2+y^2=1 by

\vec r(t)=(x(t),y(t))=(\cos t,\sin t)

with 0\le t\le\pi, and the line segment by

\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)

with 0\le t\le1. Then

\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)

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Using the fundamental theorem of calculus:

The integral can be written as

\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}

If there happens to be a scalar function f such that \vec F=\nabla f, then \vec F is conservative and the integral is path-independent, so we only need to worry about the value of f at the path's endpoints.

This requires

\dfrac{\partial f}{\partial x}=\sin x\implies f(x,y)=-\cos x+g(y)

\dfrac{\partial f}{\partial y}=\cos y=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\sin y+C

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f(x,y)=-\cos x+\sin y+C

which means \vec F is indeed conservative. By the fundamental theorem, we have

\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=f(2,-\pi)-f(1,0)=-\cos2-(-\cos1)=\boxed{\cos1-\cos2}

3 0
3 years ago
Need help with number 3 inequalities with variables on both sides
Volgvan

Answer:

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\displaystyle{2x-3 \leq \dfrac{x}{2}}

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\displaystyle{2x\cdot 2-3\cdot 2 \leq \dfrac{x}{2}\cdot 2}\\\\\displaystyle{4x-6 \leq x}

Add both sides by 6 then subtract both sides by x:

\displaystyle{4x-6+6 \leq x+6}\\\\\displaystyle{4x \leq x+6}\\\\\displaystyle{4x-x \leq x+6-x}\\\\\displaystyle{4x-x \leq 6}\\\\\displaystyle{3x \leq 6}

Then divide both sides by 3:

\displaystyle{\dfrac{3x}{3} \leq \dfrac{6}{3}}\\\\\displaystyle{x \leq 2}

Therefore, the answer is x ≤ 2

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Answer:

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Step-by-step explanation:

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Answer:

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I think it might be the second option
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