Question 70228
One approach to solving this problem is to treat the given input data, ie, heart-rate and time, as the coordinates of points on a graph.  Instead of the usual (x, y) you can use (t, h) where t is time and h is heart rate.
With the two given points you can find the slope of this linear relationship using:
{{{m = (y[2]-y[1])/(x[2]-x[1])}}} or rather: {{{m = (h[2]-h[1])/(t[2]-t[1])}}}
Once you have found the slope, you can start writing the equation in the slope-intercept form (h = mt+b) which will be just perfect for this problem.
Let's find the slope using the two given point (t, h) which are: (2, 84) and (3, 86)
{{{m = (86-84)/(3-2)}}}
{{{m = 2/1}}}
{{{m = 2}}}
You can now write the start of the equation as:
{{{h = 2t+b}}} where b is the h-intercept. Substitute the t- and h-coordinates from either one of the two given points and solve for b. Using (2, 84) we get:
{{{84 = 2(2)+b}}} Simplify and solve for b.
{{{b = 80}}}
Now you can write the final equation for this linear relationship:
{{{h = 2t+80}}}
To find Jenny's heart rate (h) after 10 minutes (t=10) on the treadmill, make the appropriate substitutions into your final equation and solve for h.
{{{h = 2(10)+80}}}
{{{h = 20+80}}}
{{{h = 100}}}
Jenny's heart rate will be 100 bpm after 10 minutes on the treadmill.