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put this solution on YOUR website!In 1920, the record for a certain race was 45.4 sec. In 1990, it was 44.0 sec.
Let R(t)= the record in the race and t= the number of the year since 1920.
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a) Find a linear function that the date
Think of the x-axis representing the year and the y-axis representing the "record-time" in seconds.
The problem, then, really gives you two data points: (remember, since 1920)
1920 we get: (0,45.4)
1990 we get: (70,44)
Slope then is
(y2-y1)/(x2-x1) = (44-45.4)/(70-0) = -1.4/70 = -0.02
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Using the slope and one (70,44) of the two points substitute it into the "point-slope" form:
y-y1 = m(x-x1)
y-44 = -0.02(x-70)
y = -0.02(x-70) + 44
y = -0.02x + 1.4+44
y = -0.02x + 45.4
Replacing y with R(t) and x with t we get:
R(t) = -0.02t + 45.4
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b) Use the function in (a) to predict the record in 2003 and in 2006.
For 2003:
t = 2003-1920 = 83
R(t) = -0.02t + 45.4
R(t) = -0.02(83) + 45.4
R(t) = -1.66 + 45.4
R(t) = -1.66 + 45.4
R(t) = 43.74 seconds
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For 2006:
t = 2006-1920 = 86
R(t) = -0.02t + 45.4
R(t) = -0.02(86) + 45.4
R(t) = -1.66 + 45.4
R(t) = -1.72 + 45.4
R(t) = 43.68 seconds
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c) Find the year when the record will be 43.56 sec
R(t) = -0.02t + 45.4
43.56 = -0.02t + 45.4
43.56-45.4 = -0.02t
-1.84 = -0.02t
92 = t
Year then = 1920+92 = 2012
