SOLUTION: Ineed your help once again I try own my own and my answer are looking craazt help is needed. Thank you so much
Steve traveled 200 miles at a certain speed. Had he gone 10mph fa
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Steve traveled 200 miles at a certain speed. Had he gone 10mph fa
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Question 172638This question is from textbook
: Ineed your help once again I try own my own and my answer are looking craazt help is needed. Thank you so much
Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle
can you show me this one too.
The Hudson River flows at a rate of 3 miles per hour. A patrol boat travels 60 miles upriver, and returns in a total time of 9 hours. What is the speed of the boat in still water?
This question is from textbook
You can put this solution on YOUR website! I set up the equations you solve
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d=rt........in both instances distance is equal
lets call the rate of the slow trip r and time t
lets call the rate of the fast trip r+10 and time(t-1)
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so 200=rt...eq 1
...200=(r+10)(t-1)..eq 2
:change eq 1 to t=200/r and substitute it into eq 2 and solve for r...this will involves a quadratic equation again
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throw out negative value
so mph and =50mph
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Quadratic equation (in our case ) has the following solutons:
For these solutions to exist, the discriminant should not be a negative number.
First, we need to compute the discriminant : .
Discriminant d=8100 is greater than zero. That means that there are two solutions: .
Quadratic expression can be factored:
Again, the answer is: 40, -50.
Here's your graph:
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Let s= speed of the boat in still water
The current =3 mi/hr
we know that d=rt or d/r=t for this problem we must break this up into 2 parts
(distance upriver)/(rate going upriver)
+ (distance downriver)/(rate going downriver) = 9 hrs
:the we know the distance is 60 each way
the rate going up stream is 60/s-3
the rate going down stream is 60/s+3
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so we have again a quadratic equation
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multiply each term by (s+3)(s-3)
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drop the negative value
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so mph