Asked by kelly
In 1920 the record for a certain race was 46.5 seconds. in 1980 it was 45.3 seconds let R(t) = the record in the race and t=the number of years since 1920
R(t) = 46.5 - t*0.02
what is the predicted record in 2003? 44.84
what is the predicted record in 2006?44.78
what year will the record be 44.58 seconds? 2015
help me figure this out, first I subtract 46.5 - 45.3 = 1.2 next 1980- 1920 = 60 1.2/60 = 0.02
are these correct
R(t) = 46.5 - t*0.02
what is the predicted record in 2003? 44.84
what is the predicted record in 2006?44.78
what year will the record be 44.58 seconds? 2015
help me figure this out, first I subtract 46.5 - 45.3 = 1.2 next 1980- 1920 = 60 1.2/60 = 0.02
are these correct
Answers
Answered by
Damon
t = year - 1920
r = m t + 46.5
in 1980 t = 1980 - 1920 = 60
r = m (60) + 46.5 = 45.3
so
m = (45.3-46.5)/60 = -0.02
so
r = -0.02 t + 46.5 Our equation agreed
in 2003:
t = 2003 - 1920 = 83 years
r = -0.02 (83) + 46.5 = 44.84 agree
now for example in what year will it be 44 seconds?
44 = -0.02 t + 46.5
0.02 t = 46.5 - 44 = 2.5
so
t = 125
125 = year - 1920
so year = 2045 (that is just an example so you can do the real problem.)
r = m t + 46.5
in 1980 t = 1980 - 1920 = 60
r = m (60) + 46.5 = 45.3
so
m = (45.3-46.5)/60 = -0.02
so
r = -0.02 t + 46.5 Our equation agreed
in 2003:
t = 2003 - 1920 = 83 years
r = -0.02 (83) + 46.5 = 44.84 agree
now for example in what year will it be 44 seconds?
44 = -0.02 t + 46.5
0.02 t = 46.5 - 44 = 2.5
so
t = 125
125 = year - 1920
so year = 2045 (that is just an example so you can do the real problem.)
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