X+2> _ 5, x >_5-2, x>_3
-x-2>_5, x<_ -2-5, x<_-7
the last is the answer
Answer:
9(4x+3) :)
Step-by-step explanation:
Answer:
<em>Rate of Pratap in still water is 4.5 miles/hour and rate of current is 0.5 miles/hour.</em>
Step-by-step explanation:
Pratap Puri rowed 10 miles down a river in 2 hours, but the return trip took him 2.5 hours.
We know that, 
So, the <u>speed of Pratap with the current</u> will be:
miles/hour
and the <u>speed of Pratap against the current</u> will be:
miles/hour.
Suppose, the rate of Pratap in still water is
and the rate of current is
.
So, the equations will be........

Adding equation (1) and (2) , we will get......

Now, plugging this
into equation (1), we will get.....

Thus, Pratap can row at 4.5 miles per hour in still water and the rate of the current is 0.5 miles/hour.
Answer:
the answer is 67.5
Step-by-step explanation:
look at the photo
Answer:
2 hours, 150 miles
Step-by-step explanation:
The relation between time, speed, and distance can be used to solve this problem. It can work well to consider just the distance between the drivers, and the speed at which that is changing.
<h3>Separation distance</h3>
Jason got a head start of 20 miles, so that is the initial separation between the two drivers.
<h3>Closure speed</h3>
Jason is driving 10 mph faster than Britton, so is closing the initial separation gap at that rate.
<h3>Closure time</h3>
The relevant relation is ...
time = distance/speed
Then the time it takes to reduce the separation distance to zero is ...
closure time = separation distance / closure speed = 20 mi / (10 mi/h)
closure time = 2 h
Britton will catch up to Jason after 2 hours. In that time, Britton will have driven (2 h)(75 mi/h) = 150 miles.
__
<em>Additional comment</em>
The attached graph shows the distance driven as a function of time from when Britton started. The distances will be equal after 2 hours, meaning the drivers are in the same place, 150 miles from their starting spot.