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OpenStudy (anonymous):
OpenStudy (anonymous):
Again! Haha.
OpenStudy (anonymous):
I only have a couple more
OpenStudy (anonymous):
Does this mean no?
OpenStudy (anonymous):
It means I'll do what I can. But Help Monkeys also need some rest :-)
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OpenStudy (anonymous):
sorry, I hate to bug
OpenStudy (anonymous):
Let's call the year 1920 the year t=0. This means that 1990 is the year t=70.
\[R(0)=46.4\]\[R(70)=44.3\]You have two points, (0, 46.4) and (70, 44.3). The first step is almost always finding slope. Find the slope using the equation
\[m=\frac{y_2-y_1}{x_2-x_1}\]
\[m=\frac{(46.4)-(44.3)}{(0)-(40)}\]\[m=-0.03\]
Then find the y-intercept. The y-intercept happens where the t-coordinate is 0, so b=46.4.
\[y=-0.03t+46.4\]
That's your linear function. Let me know if you need more on the next two parts.
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
there was one part missing
in what year will the predicted record be 43.76 seconds?
OpenStudy (anonymous):
c) 2008
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OpenStudy (anonymous):
You want to know what t is when R is 43.76.
\[43.76=-0.03t+46.4\]Solve for t.
OpenStudy (anonymous):
Did anyone get what the seconds would be for 2003 and 2006?
OpenStudy (anonymous):
Great, I was forgotten
OpenStudy (anonymous):
2003: 43.91
2006: 43.82
OpenStudy (anonymous):
so what did R(t)=?
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OpenStudy (anonymous):
Which suestion?
OpenStudy (anonymous):
In the beginning of the series of questions, it ask R(t)=