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2 years ago

14

Sams annual college tuition will be $230 more than ryan's annual college tuition after sams college tuition increase by 10% next

year. If ryan's current tuition is $22,200, what sams current tuition

1 answer:

Alik [6]2 years ago

3 0

Answer:

Sounds like Sam better pull it together

Step-by-step explanation:

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The width of a rectangle is 2 cm less than 8 times the length. The perimeter is 68 cm. What are the width and length of the rect

eduard

Answer:

length=4

width=30

Step-by-step explanation:

length=x

width=8x-2

(8x-2)*2+2x=68

16x-4+2x=68

18x=72

x=4 (Length)

4*8=32, 32-2=30(width)

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2 years ago

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Step-by-step explanation:

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What is the value of x if f(x) = 76

GaryK [48]

Answer:

The roots (zeros) are the x values where the graph intersects the x-axis. To find the roots (zeros), replace f(x) with 0 and solve for x

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3 years ago

Clarissa helps her mom put the 200-Newton lawn mower in the back of her mom’s truck. They lift the mower up from the ground to t

Anarel [89]

Answer:

220 N * m or 220 J

Step-by-step explanation:

We have that the work formula would be the applied force multiplied by the distance traveled, that is:

W = F * d

In this case the F is 200 N and the distance is 1.1 meters, therefore we replace:

W = 200 * 1.1

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Therefore the work carried out by Clarissa and her mom is 200 joules.

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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