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Lera25 [3.4K]
3 years ago
11

ATTENTION WILL GIVE BRAINLIEST IF YOU HAVE IT RIGHT, IF YOU DONT PUT A RIGHT ANSWR I WILL REPORT YOU AND BLOCK YOU. 1. Write an

expression with (−1) as its base that will produce a positive product, and explain why your answer is valid. 2. Write an expression with (−1) as its base that will produce a negative product, and explain why your answer is valid.
Mathematics
2 answers:
WITCHER [35]3 years ago
8 0

Answer: -1^0 because any number to the 0 is 1

-1^1 because it stays the same number which is negative

Llana [10]3 years ago
8 0

Answer:

Hey there!

For question number 1.

(-1)^2, because this equals (-1)(-1), or 1.

For question number 2.

(-1)^3, becuase this equals (-1)(-1)(-1), or -1.

Let me know if this helps :)

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3 0
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Read 2 more answers
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Treat \mathcal C as the boundary of the region \mathcal S, where \mathcal S is the part of the surface z=2xy bounded by \mathcal C. We write

\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r

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so the line integral is equivalent to

\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S
=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv


where \mathbf s(u,v) is a vector-valued function that parameterizes \mathcal S. In this case, we can take

\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)

with 0\le u\le1 and 0\le v\le2\pi. Then

\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv

and the integral becomes

\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv
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