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In July 1985 Steve Cram from Great Britain set a world record of 3 min. 29.67 sec. for the 1500 meter race and a world record of 3 min 46.31 sec for the one mile race. Suppose you know only that these two events occurred 11 days apart and that the 1500 meter record was set on July 16th. Use an absolute value equation to find the possible dates for the one mile record run.

2007-07-13 06:59:55 · 9 answers · asked by Anonymous in Science & Mathematics Mathematics

9 answers

Let M = date of one mile run

|M - 16| = 11

So either M - 16 = 11
M = 27

or 16 - M = 11
M = 5

So the possible dates are July 5th or 27th

2007-07-13 07:03:57 · answer #1 · answered by Dr D 7 · 1 0

Let the date of one mile run be D
|D – 16| = 11
So D – 16 = 11 or 16 – D = 11
Therefore D = 27 or D = 5

So the possible dates are July 5th or 27th

2007-07-13 22:49:40 · answer #2 · answered by Pranil 7 · 0 0

Hi,

Let x = the date of the one mile run in July
16 refers to July 16th.
| x - 16 | = 11

x - 16 = 11 or x - 16 = -11
x = 27 or x = 5

So it could have occurred on July 5th or July 27th

I hope this helps!! :-)

2007-07-13 07:04:30 · answer #3 · answered by Pi R Squared 7 · 1 0

|July 16 - record| = 11

July 16 - record = 11 or July 16 - record = -11
-record = 11 - (July 16) OR -record = -11 - (July 16)
record = (July 16) - 11 OR record = 11 + (July 16)
record = July 5 OR record = July 27

2007-07-13 07:04:29 · answer #4 · answered by Mathematica 7 · 1 0

|x - 16| = 11

x - 16 = 11 or -(x - 16) = 11

x = 27 or -x = -5 --> x = 5

2007-07-13 07:05:41 · answer #5 · answered by TychaBrahe 7 · 1 0

july 27th or july 5th.

11 = |x - 16|

2007-07-13 07:08:47 · answer #6 · answered by A Military Veteran 5 · 0 0

|16-x| = 11
x = 5 or 27

2007-07-13 07:13:49 · answer #7 · answered by ironduke8159 7 · 1 0

the only possible days are the 5th and the 27th.

2007-07-13 07:04:45 · answer #8 · answered by E_man5 2 · 0 0

why don't you just do your homework and not ask every question here? then maybe you'll actually pass your class.

2007-07-13 07:05:12 · answer #9 · answered by Andrew 4 · 2 0

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