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3 years ago

Given the net, what is the surface area of the rectangular prism below?

Mathematics

1 answer:

Vera_Pavlovna [14]3 years ago

8 0

Answer: 56

Step-by-step explanation: I got this answer by taking each rectangles legth and width and multiplying those, then I added all of the final answers together to get 56

You might be interested in

The incenter is the center of the _______ circle of a triangle

blagie [28]

Answer:

Step-by-step explanation:

An incircle is a circle inside a triangle that touches all three sides of the triangle.

5 0

3 years ago

25 POINTS!! (EASY)MORE POINTS!!!!!!! BRAINLIEST PLSSSSSSSSSSSSSSSSSSSHELP. For Brainliest Please help)

Lostsunrise [7]

Answer:

A) 50

B) 20

C) 20

Step-by-step explanation:

To solve the first question, we can multiply both sides by 10.

Thus, we get:

6x = 300.

Then, divide by 6, to get x:

x = 50.

You can do the same thing to all of these questions.

I've already checked the answers, but if you don't trust them, just pluck the values in.

For example, let's check the answer for question A.

0.6*50 is supposed to equal 30.

Does it equal? Yes, so we know that the answer is correct.

3 0

3 years ago

Please help due today

Len [333]

Answer:

x = 5

Step-by-step explanation:

The diagonals of a parallelogram bisects each other, which means that each diagonal is splitter into two equal parts.

Therefore:

x - 2 = 2x - 7

Collect like terms

x - 2x = 2 - 7

-x = -5

Divide both sides by -1

x = 5

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

A 2.905x10 B 2.905x10 C 2.95x10

Dafna1 [17]

Answer:

a. 29.05 b. 29.05 c. 29.5

6 0

2 years ago