The approximate amount of water that remains in the tub after the 6 spherical balls are placed in the tub are 3479.12 in³.
<h3>What is the approximate amount of water that remains in the tub?</h3>
The first step is to determine the volume of the cylinder.
Volume of the tub = πr²h
Where:
- r = radius = diameter / 2 = 18/2 = 9 inches
- h = height
- π = 3.14
3.14 x 9² x 20 = 5086.8 in³
The second step is to determine the volume of the 6 balls.
Volume of a sphere= 4/3πr³
r = diameter / 2 = 8/2 = 4 inches
6 x (3.14 x 4/3 x 4³) = 1607.68 in³
Volume that remains in the tub = 5086.8 in³ - 1607.68 in³ = 3479.12 in³
To learn more about the volume of a sphere, please check: brainly.com/question/13705125
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Answer: Choice A. 1/64
Work Shown:
f(x) = 4^x
f(-3) = 4^(-3)
f(-3) = 1/(4^3) ... see note below
f(-3) = 1/64
Note: I'm using the rule that a^(-b) = 1/(a^b)
Dimensions of the room in cm = 2.54 x 12 by 15 x 2.54 by 2.54 x 8.5 = 30.48 by 38.1 by 21.59
Volume of the room in cubic cm = 30.48 x 38.1 x 21.59 cubic cm = 25,072.21 cubic cm
Given that the density of air at room temperature is

, thus the mass of air in the room = 25,072.21 x 0.00118 = 29.59 g = 0.0296 kg
Given that the lethal dose of HCN is approximately 300 mg HCN per kilogram of air when inhaled, thus the <span>amount of HCN that gives the lethal dose in the small laboratory room is given by 300 x 0.0296 =
8.88 mg.</span>
Answer: a
Step-by-step explanation:
Answer:
36.97+9.48
=$46.45
Step-by-step explanation:
hope it helps