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anonymous
 5 years ago
Help plz,
Set up an equation to solve the following problem, and then solve it.
On a sales trip, Irwin drove 330 miles averaging a certain speed. The return trip was at an average speed that is 11mph faster. Total time for the round trip was 11 hours. Find Irwin’s average speed on each part of the trip.
anonymous
 5 years ago
Help plz, Set up an equation to solve the following problem, and then solve it. On a sales trip, Irwin drove 330 miles averaging a certain speed. The return trip was at an average speed that is 11mph faster. Total time for the round trip was 11 hours. Find Irwin’s average speed on each part of the trip.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, its not hard , what dont we know , what we do we need to find , assign them to variables

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Use distance equal rate times time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let x equal the "certain speed"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so assume 330 miles for one way , that means in total he drove 660miles

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we know the total time was 11 but time = distance / speed time for first section = 330 / x times for second section = 330/ ( x+11) so ( 330/x) + [330 / (x+11) = 11

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{330}{x} + \frac{330}{x+11} =11\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it has two solutions x=6, x=55 but because we are dealling with real world ( and yes, it says "speed", not velocity ) so a negative value is not allowed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so x=55 means for first half average was 55mph, and for second half average was 66mph ( ie 55 +11 )

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Actually the average speed for the entire trip was 660/11 or 60 mi. So wouldn't that made the avg speed for the first part 54.5 mph and the second part 65.5?

radar
 5 years ago
Best ResponseYou've already chosen the best response.0The avg of 54.5 + 65.5 is 120/2=60

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0firstly where did u magically get 54.5 and 65.5 from

radar
 5 years ago
Best ResponseYou've already chosen the best response.0x + (x+11) 660  =  = 60 2 11

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes there is , there are alot of different ways to get two numbers to sum to 120 , why did you pick those two numbers? , theres no calculations to suggest why you chose thoise two

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh I think i see what you are doing now

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0bit of a dangerous way to think

radar
 5 years ago
Best ResponseYou've already chosen the best response.0x= first avg speed (certain speed) x+11 is the second avg speed (return trip Their sum divided by 2 equals the total avg speed 660/11 is also the total avg speed. Take it from there

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Just sayin your way seems correct just wondering why they dont come out the same. Did you round off or take a root of or whatever/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, the reason why they dont come out correct is because your way of thinking is wrong :P

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Well they came out close.lol gotta run

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0:\, the average of a sum is not the sum of the avergaes

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Gotcha, I see the error of my logic.
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