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KIM [24]
3 years ago
7

Use Green's Theorem to evaluate the line integral along the given positively oriented curve.

Mathematics
1 answer:
Vlad [161]3 years ago
6 0

By Green's theorem, the line integral is equivalent to the area integral

\displaystyle\int_C(3y+7e^{\sqrt x})\,\mathrm dx+(8x+9\cos(y^2))\,\mathrm dy=\int_0^1\int_{x^2}^{\sqrt x}\frac{\partial(8x+9\cos(y^2))}{\partial x}-\frac{\partial(3y+7e^{\sqrt x})}{\partial y}\,\mathrm dy\,\mathrm dx

=\displaystyle5\int_0^1\int_{x^2}^{\sqrt x}\mathrm dy\,\mathrm dx=\boxed{\frac53}

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Answer:

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Step-by-step explanation:

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Is triangle mno similar to triangle pqo explain
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(x^2 - x^(1/2))/(1-x^(1/2))
Levart [38]
\frac { \left( { x }^{ 2 }-{ x }^{ \frac { 1 }{ 2 }  } \right)  }{ \left( 1-{ x }^{ \frac { 1 }{ 2 }  } \right)  }

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\\ \\ =\frac { \left( { x }^{ 2 }-\sqrt { x }  \right)  }{ \left( 1-\sqrt { x }  \right)  } \cdot \frac { \left( 1+\sqrt { x }  \right)  }{ \left( 1+\sqrt { x }  \right)  }

\\ \\ =\frac { { x }^{ 2 }+{ x }^{ 2 }\sqrt { x } -\sqrt { x } -x }{ 1+\sqrt { x } -\sqrt { x } -x }

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\\ \\ =\frac { -\sqrt { x } \left( 1+x \right) \left( 1-x \right) -x\left( 1-x \right)  }{ \left( 1-x \right)  }

\\ \\ =\frac { \left( 1-x \right) \left\{ -\sqrt { x } \left( 1+x \right) -x \right\}  }{ \left( 1-x \right)  }

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3 0
3 years ago
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I’m confused, please help
kodGreya [7K]

Answer: (a)

Step-by-step explanation:

Given

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The tangent line to y= f(x) at (4,-6) passes through the point (10,9) Compute the following, f(4) and f'(4)
Anni [7]
The solution to your problem is as follows:

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<span>i.e. y = (5/2)(x-10) + 9 
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I hope my answer has come to your help. Thank you for posting your question here in Brainly.

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