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2 years ago

What is the surface area of a box with a length of 6 ft, a width of 3 ft, and a height of 2½ ft?

Mathematics

1 answer:

NikAS [45]2 years ago

4 0

Answer:

Step-by-step explanation:

Remark

The surface area is the area of all six sides when the object is a box.

Formula

SA = 2*L*w + 2*L*h + 2*w*h

Givens

L = 6 ft

w= 3 ft

h = 2.5 feet

Substitute and Solution

SA = 2*6*3 + 2*6*2.5 + 2*3*2.5

SA = 36 + 30 + 15

SA = 81

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Three robots ran the same distance. The first robot ran 150 meters per minute.

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2 years ago

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Pyramid A has a triangular base where each side measures 4 units and a volume of 36 cubic units. Pyramid B has the same height,

omeli [17]

Answer:

The volume of pyramid B is 81 cubic units

Step-by-step explanation:

Given

Pyramid A

$s = 4$ -- base sides

$V = 36$ -- Volume

Pyramid B

$s = 6$ --- base sides

Required

Determine the volume of pyramid B [Missing from the question]

From the question, we understand that both pyramids are equilateral triangular pyramids.

The volume is calculated as:

$V = \frac{1}{3} * B * h$

Where B represents the area of the base equilateral triangle, and it is calculated as:

$B = \frac{1}{2} * s^2 * sin(60)$

Where s represents the side lengths

First, we calculate the height of pyramid A

For Pyramid A, the base area is:

$B = \frac{1}{2} * s^2 * sin(60)$

$B = \frac{1}{2} * 4^2 * \frac{\sqrt 3}{2}$

$B = \frac{1}{2} * 16 * \frac{\sqrt 3}{2}$

$B = 4\sqrt 3$

The height is calculated from:

$V = \frac{1}{3} * B * h$

This gives:

$36 = \frac{1}{3} * 4\sqrt 3 * h$

Make h the subject

$h = \frac{3 * 36}{4\sqrt 3}$

$h = \frac{3 * 9}{\sqrt 3}$

$h = \frac{27}{\sqrt 3}$

To calculate the volume of pyramid B, we make use of:

$V = \frac{1}{3} * B * h$

Since the heights of both pyramids are the same, we can make use of:

$h = \frac{27}{\sqrt 3}$

The base area B, is then calculated as:

$B = \frac{1}{2} * s^2 * sin(60)$

Where

$s = 6$

So:

$B = \frac{1}{2} * 6^2 * sin(60)$

$B = \frac{1}{2} * 36 * \frac{\sqrt 3}{2}$

$B = 9\sqrt 3$

So:

$V = \frac{1}{3} * B * h$

Where

$B = 9\sqrt 3$ and $h = \frac{27}{\sqrt 3}$

$V = \frac{1}{3} * 9\sqrt 3 * \frac{27}{\sqrt 3}$

$V = \frac{1}{3} * 9 * 27$

$V = 81$

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2 years ago

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I need trigometry help

Alex Ar [27]

A^2=b^2+c^2-2bc Cos(A)
(270)^2=(255)^2+(442.85)^2-2(255)(442.85)Cos(A)
A=Cos^(-1)(270^2-255^2_(442.85)^2)/(-2*255*442.85)=33.5435
solve for angle A approximately is 33.54 degrees.
sinA/a=SinB/b
Sin33.54/270=SinB/255
B=Sin^(-1)(sin33.54/270*255)approximately is 31.45.
180-31.45-33.54=115.01 is Angle C.

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2 years ago

Imagine the United States only had two postage stamps, a 3 cent stamp and a 7 cent stamp. If you can put any number of stamps on

cricket20 [7]

Answer:

not possible: 1, 2, 4, 5, 8, 11
possible: all others

Step-by-step explanation:

Obviously, all multiples of 3 and 7 can be paid.

1 cent can be added by adding a 7-cent stamp and taking away two 3-cent stamps. 2 cents can be added by adding three 3-cent stamps and taking away one 7-cent stamp. Hence any value more than 7 +2(3) = 13 cents can always be accommodated.

Amounts that are impossible:

1, 2, 4, 5, 8, 11 cents

All other positive integer amounts are possible.

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3 years ago

The quotient of w and 5 is at least 25

const2013 [10]

I think the answer is 5 because 25 divided by 5=5

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3 years ago

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