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mylen [45]
3 years ago
6

Say the slope value of the next line -5x + 3y=30

Mathematics
1 answer:
OLEGan [10]3 years ago
3 0
<h3>Answer:</h3>

\large\boxed{C).\,m= 5/3}

<h3>Step-by-step explanation:</h3>

In this question, we're finding the slope of the line.

What this means is that we have to make the equation into slope intercept form.

Slope intercept form is: y = mx + b

In order to get the equation to slope intercept form, we have to get y by itself and solve from there.

Solve:

-5x + 3y=30\\\\\text{Add 5x to both sides to get 3y by itself}\\\\3y=5x+30\\\\\text{Divide both sides by 3}\\\\y = 5/3x+30\\\\\text{The slope of the equation would be 5/3}

When you're done solving, you would see that answer choice C). m = 5/3 would be the correct answer.

<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3><h3 />
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a restaurant has an all-you-can-eat buffet. They charge $13.95 per person. what is the constant of proportionality?​
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staurant has an all-you-can-eat buffet. They charge $13.95 per person. what is the constant of proportionality?​

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3n *power 2 +10n+7 how to factories and pls tell with solution
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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

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