SOLUTION: 1. if it takes john 900 seconds to run 1 mile then he is running at a rate of 4 mph. show work 2. Three consecutive odd counting numbers have the property that the sum of the lar

Question 338228: 1. if it takes john 900 seconds to run 1 mile then he is running at a rate of 4 mph.
show work
2. Three consecutive odd counting numbers have the property that the sum of the largest and twice the smallest is 64. what is the middle number.
Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Only one question per post. Which one do you want us to do? Don't write to me, repost.

John

My calculator said it, I believe it, that settles it

Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!
1. if it takes john 900 seconds to run 1 mile then he is running at a rate of 4 mph.
show work

AMP Parsing Error of [1mi/900sec]: Invalid function '': opening bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 70. . AMP Parsing Error of [(60sec)/(1min)]: Invalid function ')/(1\min)': opening bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 70. . Cancel the seconds: AMP Parsing Error of [1mi/900cross(sec)]: Invalid function ')': opening bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 70. . AMP Parsing Error of [(60cross(sec))/(1min)]: Invalid function '))/(1\min)': opening bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 70. . Cancel the minutes: AMP Parsing Error of [1mi/900cross(sec)]: Invalid function ')': opening bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 70. . AMP Parsing Error of [(60cross(sec))/(1cross(min))]: Invalid function '))/(1\cross(min))': opening bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 70. . All that's left is

2. Three consecutive odd counting numbers have the property that the sum of the largest and twice the smallest is 64. what is the middle number.

Smallest of the three consecutive odd integers = n Middle of the three consecutive odd integers = (n+2) Largest of the three consecutive odd integers = (n+4) largest + 2*smallest = 64 (n+4) + 2*n = 64 n + 4 + 2n = 64 3n + 4 = 64 3n = 60 n = 20 But 20 is even, not odd. If the problem had asked for three consecutive even counting numbers, they would be Smallest = n = 20 Middle = (n+2) = 20+2 = 22 Largest = (n+4) = 20+4 = 24 The largest, 24 plus twice the smallest, 2*20 or 40, is 64, since 24+40 = 64. But they're even, not odd. Thus, there is no solution, since they must be odd. Only such even consecutive counting numbers exist, but no odd ones. Edwin

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