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2 years ago

I need to know what this is lolololol thanks.

Mathematics

2 answers:

kakasveta [241]2 years ago

4 0

Answer:

29/61

Step-by-step explanation:

Rewrite both mixed numbers as improper fractions. So (29/9)/(61/9)=(29/9)*(9/61)=(29/1)*1/61=29/61~~0.47540983...

Hope it helps

Jlenok [28]2 years ago

3 0

Answer:

29/61

Step-by-step explanation:

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Valerie bought a coat at a winter half-price sale. A sales tax of $6.24 was added to the sale price, bringing the cost of the co

egoroff_w [7]

Answer:

84.24 - 6.24 = p

Step-by-step explanation:

6 0

2 years ago

Pls help me, its not hard

natulia [17]

Answer:

1. y = -x + 3

2. y = x - 3

3. y = 3x + 1

4. y = -3x - 3

Step-by-step explanation:

Hello! sorry, I just saw your message on the question I answered before asking for help so here you go, these are the answers, hope this helps.

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3 0

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

-3x+3y=12

max2010maxim [7]

Answer:

Step-by-step explanation:

Reform the first graph: -3x + 3y = 12; 3y = 12 + 3x; y = 4 + x; y = x + 4
Compare y = x + 4 and y = x + 4: they are equal, thus the same slope (1) and the same y-intercept (4)

7 0

2 years ago

System of linear equations:

Vedmedyk [2.9K]

Answer:

x= 500 y= 1500

Step-by-step explanation:

Isolate x or y in the first equation

Isolating for x would be x= -y+2000

Substitute X=-y+2000 into equation 10x+20y=35000

10(-y+2000)+20y=35000

Exapnd the equation above and you should get y=1500

Same steps for finding x by isloating for y

3 0

2 years ago