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2 years ago

A carton has a length of feet, width of feet, and height of feet. What is the volume of the carton?

Mathematics

1 answer:

noname [10]2 years ago

6 0

Answer:

L times W times H

Step-by-step explanation:

just multiply the length times the width times the height

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the density of pure gold 1,204 pounds per cubic foot? how many cubic feet is 481.6lbs of pure gold?

Nadusha1986 [10]

Answer:

Volume of Gold = 0.4 cubic feet.

Step-by-step explanation:

First we have to make sure the units are same as the units the answer is required in.

Answer should be given in cubic feet.

Density is given in pounds per cubic foot. And Mass is given in pounds. So units are in order. Therefore, we can go ahead with the calculation.

The Density of Gold = [Mass of Gold/Volume of Gold]

By substituting values,

$1204$ ponds per cubic foot = [ $481.6$ pounds / Volume of gold]

Making Volume of God the subject of the equation,

Volume of Gold = [ $481.6$ pounds / tex]1204[/tex] ponds per cubic foot ]

Volume of Gold = 0.4 cubic feet.

5 0

3 years ago

Write the equation of a circle with center M(2,6) and a radius of 4

ycow [4]

Answer:

(x -2)^2 + (y-6)^2 = 16

Step-by-step explanation:

The equation of a circle is of the form

(x-h)^2 + (y-k)^2 = r^2

where (h,k) is the center and r is the radius

(x- 2)^2 + (y-6)^2 = 4^2

(x -2)^2 + (y-6)^2 = 16

7 0

3 years ago

The total number of running yards in a football game was less than 100. The inequality x < 100 represents the situation.

Aleks [24]

Answer:

sorry I'm late but it's b I'm on unit test on edge

4 0

3 years ago

1. Write the equation of the line best fit

Pavel [41]

Answer:

y=3/4x+2

Step-by-step explanation:

I found this by using desmos graphing calculator

8 0

3 years ago

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

3 years ago