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2 years ago

Write ln(x^3y^2) in terms of u and v (picture provided)

Mathematics

1 answer:

Katena32 [7]2 years ago

6 0

Remember the properties of logarithms. If you have ln81, you can pull out the power of two (9^2) and have 2ln9. Also, you could do ln9 + ln9 because 9*9=81 and ln9*9 = ln9 + ln9

With these properties, you can get 3lnx+2lny, which is 3u+2v according to the answer.

Hope I helped!

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2 years ago

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2 years ago

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MrRissso [65]

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2 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

2 years ago

Match each value with its formula for ABC.

Ivahew [28]

For the triangle ABC use the sine theorem:

$\dfrac{a}{\sin A}= \dfrac{b}{\sin B}= \dfrac{c}{\sin C}$ .

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2. From $\dfrac{b}{\sin B}= \dfrac{c}{\sin C}$ you have $\dfrac{b}{c} \cdot \sin C=\sin B$ .

3. From $\dfrac{a}{\sin A}= \dfrac{c}{\sin C}$ you have $\dfrac{c}{a} \cdot \sin A=\sin C$ .

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5. From $\dfrac{a}{\sin A}= \dfrac{b}{\sin B}$ you have $\dfrac{\sin B}{\sin A} \cdot a=b$ .

6. From $\dfrac{b}{\sin B}= \dfrac{c}{\sin C}$ you have $\dfrac{\sin C}{\sin B} \cdot b=c$ .

5 0

2 years ago

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