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Free_Kalibri [48]
2 years ago
15

A ball inside of a cube has a volume of 113 cubic inches. If each side of the cube measures 12.6 inches, what is the volume of a

ir inside the cube
Mathematics
1 answer:
Fantom [35]2 years ago
4 0

The volume of air inside the cube is 1887.376 cubic inches

<h3>What is Volume?</h3>

The volume of a shape or an object is the amount of space in the object

<h3>How to determine the volume of air inside the cube?</h3>

The given parameters are:

  • Volume of the ball = 113 cubic inches
  • The side length of the cube = 12.6 inches

The volume of the cube is calculated using

Volume = (Side length)^3

Substitute 12.6 for Side length

Volume = 12.6^3

Expand the exponent

Volume = 12.6 * 12.6 * 12.6

Evaluate the product

Volume = 2000.376

The volume of air inside the cube is calculated using

Volume of the air = Volume of the cube - Volume of the ball

So, we have

Air = 2000.376 - 113

Evaluate

Air = 1887.376

Hence, the volume of air inside the cube is 1887.376 cubic inches

Read more about volume at:

brainly.com/question/1972490

#SPJ1

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goldfiish [28.3K]

Answer:

k=32

Step-by-step explanation:

Given the points:

A = (4,-1)$, $B = (6,2)$, and $C = (-1,2)$.

The first step is to find the <u>Centroid</u> of the triangle.

Centroid, X

=\left(\dfrac{x_1+x_2+x_3}{3} ,\dfrac{y_1+y_2+y_3}{3} \right)\\=\left(\dfrac{4+6+(-1)}{3} ,\dfrac{-1+2+2}{3} \right)\\=\left(\dfrac{9}{3} ,\dfrac{3}{3} \right)=(3,1)

Next, let P be a point (x,y)

Using the <u>distance formula, </u>\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}<u />

<u />PA^2=(x-4)^2+(y-(-1))^2\\PB^2=(x-6)^2+(y-2)^2\\PC^2=(x-(-1))^2+(y-2)^2\\PX^2=(x-3)^2+(y-1)^2\\<u />

On Substitution into: PA^2 + PB^2 + PC^2 = 3PX^2 + k

(x-4)^2+(y-(-1))^2+(x-6)^2+(y-2)^2+(x-(-1))^2+(y-2)^2=3[(x-3)^2+(y-1)^2]+k

Let us simplify the LHS first

\\LHS: x^2-8x+16+y^2+2y+1+x^2-12x+36+y^2\\-4y+4+x^2+2x+1+y^2-4y+4\\=3x^2-18x+3y^2-6y+62

Also, the Right Hand Side

RHS:3[(x-3)^2+(y-1)^2]+k\\=3[x^2-6x+9+y^2-2y+1]+k\\=3x^2-18x+27+3y^2-6y+3+k\\=3x^2+3y^2-18x-6y+30+k

Therefore:

3x^2-18x+3y^2-6y+62=3x^2+3y^2-18x-6y+30+k\\k=3x^2-18x+3y^2-6y+62-3x^2-3y^2+18x+6y-30\\k=3x^2-3x^2+3y^2-3y^2-18x+18x+62-30\\k=32

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Answer:

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Step-by-step explanation:

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Stepwise proof is given in two columns

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