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emmainna [20.7K]

3 years ago

6

A cube fits perfectly in a sphere. If the cube has a side length of 10cm, what is the volume of the sphere? Round your answer to

the nearest hundredth (2 decimal places) or leave your answer in Pi form

1 answer:

Goryan [66]3 years ago

3 0

Answer:

Step-by-step explanation:

Since the cube fits perfectly in the sphere, it means that the diagonal of the cube will form the diameter of the sphere. In a cube, all sides are equal. To determine the length of its diagonal, d, we would apply Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side²

Therefore,

d² = 10² + 10² = 200

d = √200 = 14.14cm

Formula for finding the volume of a sphere is expressed as

Volume = 4/3 × π × r³

Radius of sphere = diameter/2 = 14.14/2 = 7.07cm

Volume = 4/3 × π × 7.07³ = 471.19π cm³

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Please solve i will give brainiest 100 point question ****** do the whole page please need to pass or i will fail its my final t

dmitriy555 [2]

Answer:

1. Find the difference between the areas.

<u>Area of the small rectangle</u>: $(x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14$

<u>Area of the big rectangle</u>: $(x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99$

The difference is: $11x+85$

$( x^2+20x+99)- (x^2+9x+14)=x^2+20x+99-x^2-9x-14=11x+85$

2.

You can solve this question just by looking at the graph.

a) The height is 4 meters.

$f(d)=h=-2d^2+7d+4$

To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.

$f(0)=h=-2(0)^2+7(0)+4$

$h=4$

The height is 4 meters.

b) 9 meters.

For $d=1$

$f(1)=h=-2(1)^2+7(1)+4$

$f(1)=h=-2+7+4$

$h=9$

b) The ball travels 4 meters.

But to calculate it, it is when $h=0$

$0=-2d^2+7d+4$

Using the quadratic formula:

$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$d=\frac{-7 \pm \sqrt{7^2-4\left(-2\right)4}}{2\left(-2\right)}$

$d=\frac{-7\pm\sqrt{81}}{-4}$

$d=\frac{-7\pm9}{-4}$

It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.

$d=-\frac{1}{2} \text{ or }d=4$

3.

In this question, we have to find the area of the cylinder and the sphere.

From the information given, we have

a = 5mm and d = 5mm, therefore the radius is 2.5 mm.

The volume of a cylinder:

$V=\pi r^2h$

$V=\pi (2.5)^2 \cdot 5$

$V=31.25 \pi$

$V_{c} \approx 98.17 \text{ m}^3$

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$V=\frac{4}{3} \pi r^2$

$V_{s} \approx 65.4 \text{ m}^3$

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Answer:

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Answer:

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