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Radda [10]
2 years ago
6

The volume of a cubical box is

Mathematics
1 answer:
creativ13 [48]2 years ago
8 0

The sides length of the cubical box is 3.8 cm if the volume of a cubical box is 54.872 cm³ option second is correct.

<h3>What is a cube?</h3>

It is defined as three-dimensional geometry that has six square faces and eight vertices.

We have a volume of a cubical box is 54.872 cm³

V = 54.872 cm³

As we know the volume of the cube:

V = side³

54.872 = side³

Taking cube root on both sides:

side = 3.8 cm

Thus, the sides length of the cubical box is 3.8 cm if the volume of a cubical box is 54.872 cm³ option second is correct.

Learn more about the cube here:

brainly.com/question/15420947

#SPJ1

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NEED HELP ASAP 20POINTS The energy (and cost) needed to operate an air conditioner is proportional to the volume of a space that
insens350 [35]

Answer:

The daily cost of cooling the large warehouse with the same model of air conditioner is \$1,953

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube

In this problem

The scale factor is equal to 2.5

so

The scale factor elevated to the cube is 2.5^{3} =15.625

To estimate the daily cost of cooling the large warehouse with the same model of air conditioner, multiply $125.00 by the scale factor elevated to the cube

so

\$125(15.625)=\$1,953

4 0
3 years ago
Look at the figure. Which trigonometric ratio should you use to find x?
Elis [28]

Answer:

B.  Sine.

Step-by-step explanation:

We need to find the value of x which is opposite to the given angle  (50 degree).  Also we are given the length of the hypotenuse.

Sine = opposite / hypotenuse so we would use the sine.

Its worth committing this mnemonic to memory, which helps to find the required trig ratio:

SOH-CAH-TOA.

SineOpposite/Hypotenuse- CosineAdjacent/Hypotenuse-TangentOpposite/Adjacent.

4 0
3 years ago
Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len
Rama09 [41]

The length of a curve <em>C</em> parameterized by a vector function <em>r</em><em>(t)</em> = <em>x(t)</em> i + <em>y(t)</em> j over an interval <em>a</em> ≤ <em>t</em> ≤ <em>b</em> is

\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt

In this case, we have

<em>x(t)</em> = exp(<em>t</em> ) + exp(-<em>t</em> )   ==>   d<em>x</em>/d<em>t</em> = exp(<em>t</em> ) - exp(-<em>t</em> )

<em>y(t)</em> = 5 - 2<em>t</em>   ==>   d<em>y</em>/d<em>t</em> = -2

and [<em>a</em>, <em>b</em>] = [0, 2]. The length of the curve is then

\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt

=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt

=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt

=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt

=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}

5 0
3 years ago
4.72n - 0.1 = 8 + 0.67n
levacccp [35]
4.72n - 0.1 = 8 + 0.67n

4.72n - 0.67n = 8 + 0.1

4.05n = 8.1

4.05n/4.05 = 8.1/4.05 

n = 2
5 0
3 years ago
15 points PLEASE HELP , and be careful when answering !!
Aleks04 [339]
18.84 because the formula to find the volume of a cone is pie*r^2*h/3 and your radius is 3 and the height is 2, so if you plug those in the equation and calculate it this should be your answer
6 0
3 years ago
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