Question 46365: Could someone please help us figure this out!
Time on a treadmill:
After 2 minute, Jeremy has a heart rate of 82. After 3 minutes he has a heart rate of 86. Assume there is a linear equation that gives his heart "h" in terms of time on the treadmill "t". Find the equation and use it to predict his heart rate after 10 minutes on the treadmill.
We are getting confused. Can anyone help?
Found 2 solutions by mszlmb, Earlsdon:
Answer by mszlmb(115)   (Show Source):
You can put this solution on YOUR website! OK so let x be time in minutes and let y be heart rate [per minute].
Let's make a line that goes through (2,82) and (3,86). First off, you know
that the slope must be 4/1 because the slop is difference in y divided by
difference in x, or (Y1-Y2)/(X1-X2).
The difference in y is 4 and in x is 1, so slope must equal 4/1, or 4. Our equation so far is:
Y=MX+B we know it's that format because it is a line.
Y=4X+B we know m=4 because m is the slope.
What is b? to find out, let's replace y and x with actual numbers. These can be 82 and 2 or 86 and 3 [respectively],
you will get the same answer:
82=4(2)+B
82=8+B
74=B
let's try the other pair:
86=3(4)+B
86=12+B
74=B
Our equation now yields:
Y=4X+74.
Let's replace Y (heartrate) with h and X (minutes) with m as instructed to get
h=4m+74
wooooo! ok now let's assume 10 minutes, or m=10
h=4(10)+74
h=40+74
h=114
his heartrate will be 114 if he excercises for 10 minutes.
A.)H=4M+74
B.)114 beats per minute
Answer by Earlsdon(6294)  (Show Source):
You can put this solution on YOUR website! If you were to graph the two given points (2, 82) and (3, 86) on a graph which was labeled t (for time) on the horizontal axis and h (for heart rate) on the vertical axis, you will find that they do indeed generate a straight line when connected.
But you can generate the corresponding linear equation when you have two points that lie on the line, as you do in this case. Only instead of using (x,y) as you would normally, you would use (t, h) for time and heart rate.
Let's find the slope, m, of the line through these two points (2, 82) and (3, 86).
Now you can use the slope-intercept form of a linear equation to write:
 We are just replacing x with t and y with h.
 Now we need to find the value of b, so substitute the t and h values from either one of the two points (2, 82) or (3, 86)
 Subtract 8 from both sides.
 Now you write the final equation.
 This is the linear equation to predict Jeremy's heart rate after t minutes on the treadmill.
Ok, to find Jeremy's heart rate after 10 minutes on the treadmill, substitute t = 10 and solve for h.
 This is Jeremy's predicted heart rate after 10 minutes on the treadmill.
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