The diameter of the sphere = the side of the cube
The empty volume of the cube is the difference between the volume of the cube and the volume of the sphere

∵ The basketball is shaped a sphere

∵ The box is shaped a cube

∵ The ball touching every wall of the box

∴ The diameter of the ball = the side of the box

∵ The length of one side of the cube = 9.5 inches

∴ The diameter of the ball = 9.5 inches

∵ The diameter of the ball = 2 r, where r is the radius of the ball

∴ 9.5 = 2 r

- Divide both sides by 2

∴ r = 4.75 inches

∵ The volume of the air that surrounded the ball in the box =

volume of the box - volume of the ball

∵ Volume box (cube) = s³

∵ s = 9.5 inches

∴ Volume of the box = (9.5)³

∴ Volume of the box = 857.375 inches³

∵ Volume the ball (sphere) = $\frac{4}{3}$ π r³

∵ r = 4.75

∴ Volume the ball = $\frac{4}{3}$ (π)(4.75)³

∴ Volume the ball = 448.921 inches³

Substitute the values of the volumes of the box and the ball in the rule above

∴ The volume of the air = 857.375 - 448.921

∴ The volume of the air = 408.454 inches³

408.454 cubic inches of air is surrounding the ball in the box

Learn more:

You can learn more about volume of solids in brainly.com/question/12497249

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6 0

3 years ago

To prove :One plus cot square theta into tan theta by sec square theta = cot theta

pickupchik [31]

we are given

$\frac{(1+cot^2(\theta))*tan(\theta)}{sec^2(\theta)} =cot(\theta)$

We will simplify left side and make it equal to right side

Left side:

$\frac{(1+cot^2(\theta))*tan(\theta)}{sec^2(\theta)}$

we can use trigonometric identity

$1+cot^2(\theta)=csc^2(\theta)$

we can replace it

$\frac{(csc^2(\theta))*tan(\theta)}{sec^2(\theta)}$

we know that

csc=1/sin and sec=1/cos

so, we can replace it

and we get

$\frac{cos^2(\theta)tan(\theta)}{sin^2(\theta)}$

now, we know that

tan =sin/cos

$\frac{cos^2(\theta)*sin(\theta)}{sin^2(\theta)*cos(\theta)}$

we can simplify it

and we get

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

we can also write it as

$=cot(\theta)$

Right Side:

$cot(\theta)$

we can see that

left side = right side

so,

$\frac{(1+cot^2(\theta))*tan(\theta)}{sec^2(\theta)} =cot(\theta)$ ......Answer

6 0

3 years ago

Please help me with b) and c)

Anuta_ua [19.1K]

Answer:

i dont know that answer

Step-by-step explanation:

8 0

3 years ago