help please
in 1920 the record for a certain race was 45.9 seconds. In 1960 it was 44.7 sec. let R(t) = the record in the race and t=the number of years since 1920.
a. find a linear function that fits the data
b. use the function in (a) to predict the record in 2003 and in 2006
c. find the year when the record will be 43.26 sec.
round to nearest hundredth

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- dumbcow

use 2 points (t,R)
(0,45.9) and (40,44.7)
slope = (44.7-45.9)/(40-0) = -1.2/40 = -.03
y_intercept is when t=0: (0,45.9)
R(t) = -.03t + 45.9

- anonymous

hi dumbcow what is the ans to the other two questions b and c

- dumbcow

use R(t) and substitute in t=83 and t=86
for part c
43.26 = -.03t +45.9
solve for t

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## More answers

- anonymous

what year will the record be 43.26 sec

- dumbcow

get t by itself
1) subtract 45.9 on both sides
2) divide by -.03 on both sides

- anonymous

I'm very confuse now

- dumbcow

do you know how to solve linear equations?
like
4x -2 = 10
solve for x

- anonymous

no i do not

- dumbcow

well that would help you with this question and class im guessing...
t=88
this means 88 years after 1920
=2008

- anonymous

that helped me thank you very much

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