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
Answer:

Step-by-step explanation:
1) The other curve is
then the common points of both curves are x-intercepts, the roots of 

2). Then those intersection points are the upper and the lower limits. Plugging in to this formula for they belong to the interval [-1,1]:


The answer is B: (-1/4, 1/2)
60 mph = 1 mile per minute
10:20 AM -> 1:00 PM = 2 hours and 40 minutes
2 hours 40 minutes = 160 minutes
Boston -> Stamford = 160 miles
Answer:
333.3 meters per minute
Step-by-step explanation:
<u>The best way to solve this problem is using </u><u>dimensional anaysis</u><u>. First, we write out our starting units, that being 20km/1hr. We have to keep in mind that we want to change the kilometers to meters and the hours to minutes.</u>

<u>We know that there are 1000 meters in 1 kilometer. We add this to the dimensional analysis as 1000m/1km. We write it as this because we want the kilometers to cancel each other out. We only want the meters.</u>

<u>We also know that 1 hour is 60 minutes. We add this to the analysis as well so that the hours cancel each other.</u>

<u>We now solve this expression. Since both the kilometers and the hours cancel out, we have meters per minute as our unit. All that's left are the numbers.</u>
= (20*1000*1)/(1*1*60) m/min
= 333.3 meters per minute