I put the solution on the paper
28 men are needed to paint the room in 3 hours
<h3><u>Solution:</u></h3>
Given that it takes 12 hours for 7 men to paint a room
We are asked to find number of men required to paint the room in 3 hours
Recognize, "paint the room" is 1 task. One job.
7 men -------- 12 hours ------ 1 job
(7/7) = 1 men ------- 12 x 7 (84) ------- same 1 job
The one men is rate is 84 hours to do the job
We can express this as 1/84 jobs per hour, the one-person rate
Now lets find how many men needed to paint the room in 3 hours
Let the required number of men for 3 hours be "a"
The rates of each person is simply additive.

corresponds to rate x hours = jobs and "a" is a variable for how many men

Thus 28 men are needed to paint the room in 3 hours
R=(3V4<span>Home: Kyle's ConverterKyle's CalculatorsKyle's Conversion Blog</span>Volume of a Sphere CalculatorReturn to List of Free Calculators<span><span>Sphere VolumeFor Finding Volume of a SphereResult:
523.599</span><span>radius (r)units</span><span>decimals<span> -3 -2 -1 0 1 2 3 4 5 6 7 8 9 </span></span><span>A sphere with a radius of 5 units has a volume of 523.599 cubed units.This calculator and more easy to use calculators waiting at www.KylesCalculators.com</span></span> Calculating the Volume of a Sphere:
Volume (denoted 'V') of a sphere with a known radius (denoted 'r') can be calculated using the formula below:
V = 4/3(PI*r3)
In plain english the volume of a sphere can be calculated by taking four-thirds of the product of radius (r) cubed and PI.
You can approximated PI using: 3.14159. If the number you are given for the radius does not have a lot of digits you may use a shorter approximation. If the radius you are given has a lot of digits then you may need to use a longer approximation.
Here is a step-by-step case that illustrates how to find the volume of a sphere with a radius of 5 meters. We'll u
π)⅓
9514 1404 393
Answer:
$7.14
Step-by-step explanation:
Let p, d, q represent the numbers of pennies, dimes, and quarters in the collection, respectively.
p + d + q = 45 . . . . . . . . there are 45 coins in the collection
2p +5 = q . . . . . . . . . . . . 5 more than twice the number of pennies
p + 4 = d . . . . . . . . . . . . . 4 more than the number of pennies
Substituting the last two equations into the first gives ...
p +(p +4) +(2p +5) = 45
4p = 36 . . . . . . . . . . . . . subtract 9
p = 9 . . . . . . . . . . . divide by 4
d = 9 +4 = 13
q = 2(9) +5 = 23
The value of the collection is ...
23(0.25) +13(0.10) +9(0.01) = 5.75 +1.30 +0.09 = 7.14
The coin collection is worth $7.14.
1.04 kilograms
1 gram = .001 kilogram
2x520= 1040
1040 x .001 = 1.04 kilograms