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2 years ago

SOMEBODY ANSWER THIS PLZ PLZ PLZ!!!

Mathematics

1 answer:

Luden [163]2 years ago

8 0

Answer:

D. 18

Step-by-step explanation:

This is the combination of 3, 3 and 2 options.

<u>The number of combinations is:</u>

3*3*2 = 18

Correct choice is D

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Solve the inequality 5 + 1/x > 16/x

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x < 0 or x > 3

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3 years ago

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What is the reference angle for a 240 degree angle?

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2 years ago

Hello! Please help asap

ValentinkaMS [17]

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3 years ago

Carson needs to lease out a music studio to record his new album. The studio charges an initial studio-use fee plus an hourly fe

Naily [24]

Answer:

The y-intercept of the equation is 100 and represents the initial studio-use fee.

Step-by-step explanation:

In this equation, our t variable (time) is the equivalent of the x-variable on a graph. This is because it is the variable that we 'change' to see its impact on y. We see how the amount of hours affects the price. So our P variable (price) is the equivalent of y on a graph. The y-intercept is where the line crosses the y-axis on a graph. At this point, x=0.

Since P is our y, and t is our x, to find the y-intercept, we simply need to make t = 0.

P = 50(0) + 100

P = 100

Therefore the y-intercept is 100.

In this context, t represents time, so even though the studio has been used for 0 hours, the price is still 100. This is because the 100 represents the initial studio-use fee, and using it for certain amounts of time adds onto the initial fee of $100. The hourly fee is represented by 50t so it costs $50 more for each hour of use.

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5 0

3 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

3 years ago