A sphere with a radius of 21cm JUST fits in a cube-shaped box. Calculate the amount of empty space in the box rounded to the nearest tenth of a cubic centimetre. How do you solve this?

Since its a cube and a sphere just fits in side it we know that both the sphere and the cube are 21cm wide and since its a cube the dimensions of the cube is 21*21*21 giving you 9,261 cm³ then you find the volume of the V=4/3π10³≈4849.04826 now all you have to do is subtract your two totals to get 4411.95174

Step-by-step explanation:

Brₐinliest plz

35,295.6 cm^3 to the nearest tenth.

Step-by-step explanation:

The diameter of the sphere = 2*21 = 42.

The volume of the box = 42^3 ( because the dimensions of the box = the diameter of the sphere).

The volume of the sphere = 4/3 π r^3

= 4/3 π 21^3

So the empty space = 42^3 - 4/3 π 21^3

= 35295.615 cm^3.

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841