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3 years ago

8

A shipping container is in the shape of a right rectangular prism with a length of 2.5 feet, a width of 11 feet, and a height of

8 feet. The container is completely filled with contents that weigh, on average, 0.36 pound per cubic foot. What is the weight of the contents in the container, to the nearest pound?

1 answer:

adelina 88 [10]3 years ago

5 0

Answer:

You can hold 188.19 or a rounded 188 pounds per cubic feet

Step-by-step explanation:

First we need to find the area which is 271

I found this using the formula 2(WL+HL+HW)

If you can fit 271 square feet then you need to divide that by 4 to find cubic feet.

if you do this you get 67.75 cubic feet

next divide 67.75 by 0.36

you get 188.19

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When I double my number and add eight, I get my number plus twelve. What is my number?

Wewaii [24]

Let say the number = x

double my number and add eight, I get my number plus twelve

double x and add eight, I get x plus twelve

2x + 8 = x + 12 (subtract x from each side)

2x - x + 8 = 12

x + 8 = 12 (subtract 8 from each side)

x = 12 - 8

x = 4

Your number is 4

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3 years ago

1 Which of these expressions is equivalent to 32 + 48? A 8(4 + 8) B 16 (2 + 3) C 4(8 + 16) D 16 (16 + 32)

Maksim231197 [3]

Answer:

B)16(2+3) is equivalent to 32+48

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2 years ago

Each day Shelia doubles the number of push-ups she did the day before. On the fifth day, she does 2x2x2x2x2 push-ups.Use an expo

mel-nik [20]

Answer:

It's 2^5

Step-by-step explanation:

2*2*2*2*2=32

2^5=32

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3 years ago

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Prove the diagonal connecting these vertex angles is perpendicular to the diagonal connecting the non-vertex angles

Volgvan

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3 years ago

The angle,2Θ, lies in the third quadrant such that cos2Θ=-2/5. Determine an exact value for tanΘ . Show your work including any

horrorfan [7]

Answer:

$tan(\theta)=\frac{\sqrt{21}}{3}$

Step-by-step explanation:

1. Approach

One is given the following information:

$cos(2\theta)=-\frac{2}{5}$

One can rewrite this as:

$cos(2\theta)=-0.4$

Also note, the problem says that the angle ( $2\theta$ ) is found in the third quadrant.

Using the trigonometric identities ( $cos(2\theta)=2(cos^2(\theta))-1$ ) and ( $cos(2\theta)=1-2(sin^2(\theta))$ ) one can solve for the values of ( $cos(\theta)$ ) and ( $sin(\theta)$ ). After doing so one can use another trigonometric identity ( $tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$ ). Substitute the given information into the ratio and simplify.

2. Solve for $(cos(\theta))$

Use the following identity to solve for ( $cos(\theta)$ ) when given the value ( $cos(2\theta)$ ).

$cos(2\theta)=2(cos^2(\theta))-1$

Substitute the given information in and solve for ( $cos(\theta)$ ).

$cos(2\theta)=2(cos^2(\theta))-1$

$-0.4=2(cos^2(\theta))-1$

Inverse operations,

$-0.4=2(cos^2(\theta))-1$

$0.6=2(cos^2(\theta))$

$0.3=cos^2(\theta)$

$\sqrt{0.3}=cos(\theta)$

Since this angle is found in the third quadrant its value is actually:

$cos(\theta)=-\sqrt{0.3}$

3. Solve for $(sin(\theta))$

Use the other identity to solve for the value of ( $sin(\theta)$ ) when given the value of ( $cos(2\theta)$ ).

$cos(2\theta)=1-2(sin^2(\theta))$

Substitute the given information in and solve for ( $sin(\theta)$ ).

$cos(2\theta)=1-2(sin^2(\theta))$

$-0.4=1-2(sin^2(\theta))$

Inverse operations,

$-0.4=1-2(sin^2(\theta))$

$-1.4=-2(sin^2(\theta))$

$0.7=sin^2(\theta)$

$\sqrt{0.7}=sin(\theta)$

Since this angle is found in the third quadrant, its value is actually:

$sin(\theta)=-\sqrt{0.7}$

4. Solve for $(tan(\theta))$

One can use the following identity to solve for $(tan(\theta))$ ;

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

Substitute the values on just solved for and simplify,

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

$tan(\theta)=\frac{-\sqrt{0.7}}{-\sqrt{0.3}}$

$tan(\theta)=\frac{\sqrt{0.7}}{\sqrt{0.3}}$

$tan(\theta)=\frac{\sqrt{\frac{7}{10}}}{\sqrt{\frac{3}{10}}}$

Rationalize the denominator,

$tan(\theta)=\frac{\sqrt{\frac{7}{10}}}{\sqrt{\frac{3}{10}}}$

$tan(\theta)=\frac{\sqrt{\frac{7}{10}}}{\sqrt{\frac{3}{10}}}*\frac{\sqrt{\frac{3}{`0}}}{\sqrt{\frac{3}{10}}}$

$tan(\theta)=\frac{\sqrt{\frac{7}{10}*\frac{3}{10}}}{\sqrt{\frac{3}{10}*\frac{3}{10}}}$

$tan(\theta)=\frac{\sqrt{\frac{21}{100}}}{\frac{3}{10}}$

$tan(\theta)=\frac{\frac{\sqrt{21}}{10}}{\frac{3}{10}}$

$tan(\theta)=\frac{\sqrt{21}}{10}*\frac{10}{3}$

$tan(\theta)=\frac{\sqrt{21}}{3}$

4 0

3 years ago