Let x be the mass of the paperbacks and y be the mass of the textbook.
20x + 9y = 44.4 ----------- (1)
25x + 10y = 51 -------------(2)
(1) x 10:
200x + 90y = 444 --------(1a)
(2) x 9:
225x + 90y = 459 --------(2a)
(2a) - (1a):
25x = 15
x = 0.6 -------- sub into (1)
20 (0.6) + 9y = 44.4
12 + 9y = 44.4
9y = 44.4 - 12
9y = 32.4
y = 3.6
So the paperback's mass is 0.6 pounds and textbook is 3.6 pounds
let's say that C is "x" units farther from B, that means that CB = x, and therefore AC = 1.5x.
![\bf \underset{\leftarrow ~~30~~\to}{\boxed{A}\stackrel{1.5x}{\rule[0.35em]{18em}{0.25pt}}C\stackrel{x}{\rule[0.35em]{10em}{0.25pt}}\boxed{B}} \\\\\\ AB=AC+CB\implies \stackrel{AB}{30}=\stackrel{AC}{1.5x}+\stackrel{CB}{x}\implies 30=2.5x \\\\\\ \cfrac{30}{2.5}=x\implies 12=x \\\\[-0.35em] ~\dotfill\\\\ AC=1.5(12)\implies AC=18~\hspace{8em} CB=x\implies CB=12](https://tex.z-dn.net/?f=%5Cbf%20%5Cunderset%7B%5Cleftarrow%20~~30~~%5Cto%7D%7B%5Cboxed%7BA%7D%5Cstackrel%7B1.5x%7D%7B%5Crule%5B0.35em%5D%7B18em%7D%7B0.25pt%7D%7DC%5Cstackrel%7Bx%7D%7B%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%7D%5Cboxed%7BB%7D%7D%0A%5C%5C%5C%5C%5C%5C%0AAB%3DAC%2BCB%5Cimplies%20%5Cstackrel%7BAB%7D%7B30%7D%3D%5Cstackrel%7BAC%7D%7B1.5x%7D%2B%5Cstackrel%7BCB%7D%7Bx%7D%5Cimplies%2030%3D2.5x%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B30%7D%7B2.5%7D%3Dx%5Cimplies%2012%3Dx%0A%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%5C%5C%5C%5C%0AAC%3D1.5%2812%29%5Cimplies%20AC%3D18~%5Chspace%7B8em%7D%20CB%3Dx%5Cimplies%20CB%3D12)
Answer:
because none of them say MPH
Answer: 2nd one is the right one
Answer:
31.4 in³
Step-by-step explanation:
The box is just big enough to hold the 3 balls, so it must have a length 6 times the radius of each ball, a width 2 times the radius, and a height 2 times the radius.
The volume of the box is:
V = (6r)(2r)(2r)
V = 24r³
The volume of the 3 balls is:
V = 3 (4/3 π r³)
V = 4πr³
So the volume of the air is:
V = 24r³ − 4πr³
V = (24 − 4π) r³
Since r = 1.4 inches:
V = (24 − 4π) (1.4 in)³
V ≈ 31.4 in³