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2 years ago

7

You were given a free ticket for a boat ride for you and 2 people of your choice. Sadly,5 of your friends want to

1 answer:

givi [52]2 years ago

6 0

✨Jokes on you I only have two friends✨

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13. Which equation is equivalent to 8x + 4y = -24?

7nadin3 [17]

Answer:

2x + y = - 6

Step-by-step explanation:

Given

8x + 4y = - 24 ( divide each term by 4 )

2x + y = - 6 ← is an equivalent equation

4 0

2 years ago

Solve: 9x - 2(4x + 5) = 2x - (4 - x) - 12 A 16 B 9 C 3 D 13

Margarita [4]

Step-by-step explanation:

4x+2x+x+9x=16x

16x=2+5+4=11

x=11

7 0

2 years ago

please help me i have a huge science test that i have to study for after this so PLEASE HELP ME GET THE ANSWER

enot [183]

Usually, quadratic equations equal 0, and to find the solution you need to make x equal the value that, when added or subtracted to the rest of its bracket, makes 0. This means that your answer is (x + 5)(x - 2) = 0, because 5 - 5 = 0 and 2 - 2 = 0. I hope this helps! Let me know if you would like me to explain further :)

4 0

2 years ago

Brian has $30 in his piggy bank. He earns $10 a week at his job.

sleet_krkn [62]

Answer:

B. 2 weeks

Step-by-step explanation:

1st week

Brian will have $40

Sarah will have $30

2nd week

Brian will have $50

Sarah will have $50

3 0

2 years ago

Read 2 more answers

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