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3 years ago

PLEASE HELP THANK YOU EVERYBODY

Mathematics

2 answers:

VikaD [51]3 years ago

6 0

Answer:

Step-by-step explanation:

Givens

L = 9

W = 4

H = 3

Formula

SA = 2*L*w + 2*L*H + 2 * H * W

Solution

SA = 2 * 9*4 + 2*9*3 + 2*3 * 4

SA = 72 + 54 + 24

SA = 150 in^2

Nadusha1986 [10]3 years ago

5 0

Answer:

150 in^2

Step-by-step explanation:

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Goryan [66]

Answer:

Im pretty sure its x< 3

Step-by-step explanation:

7 0

3 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

3 years ago

Can someone plz help me solved this problem I need help plz help me! Will mark you as brainiest!

baherus [9]

Answer: x^2 -3

That's x squared minus three

Step-by-step explanation:

When you divide 2x^3 by 2x the 2's cancel, and x cubed becomes x squared.

In 6x divided by 2x, the X's cancel and 6 ÷2=3

BTW "cancels" means that a number divided by itself equals one, so it "disappears" from the expression

7 0

4 years ago

49 = -5n – 2n<br> I think I know the answer but just wanna be sure?

Svetach [21]

Answer:

n = -7

Step-by-step explanation:

49 = -5n - 2n

combine like terms

-5 - 2 = -7

49 = -7n

divide both sides by -7

49/-7 = -7n/-7

49/-7 = n

49/-7 = -7

-7 = n

3 0

3 years ago

May i have help with this

Leya [2.2K]

Answer: D. 7^11

Step-by-step explanation:

For exponents with the same base, we just add the exponents and keep the same base. 7 is the same base; therefore 5+6=11 which means the answer is 7^11

3 0

2 years ago