Suppose that the height (in centimeters) of a candle is a linear function of
the amount of time (in hours) it has been burning.
Let x be the amount of time in hours
Let y be the heoght of a candle in centimeters
The two points are then as (9,24.5) and (23,17.5).
[tex] \mathrm{Find\:the\:line\:}\mathbf{y=mx+b}\mathrm{\:passing\:through\:}\left(9,\:24.5\right)\mathrm{,\:}\left(23,\:17.5\right) [/tex]
[tex] \mathrm{Slope\:between\:two\:points}:\quad \mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1} [/tex]
[tex] \left(x_1,\:y_1\right)=\left(9,\:24.5\right),\:\left(x_2,\:y_2\right)=\left(23,\:17.5\right) [/tex]
[tex] m=\frac{17.5-24.5}{23-9} [/tex]
[tex] m=-0.5 [/tex]
[tex] \mathrm{Plug\:the\:slope\:}-0.5\mathrm{\:into\:}y=mx+b [/tex]
[tex] y=\left(-0.5\right)x+b [/tex]
[tex] \mathrm{Plug\:in\:}\left(9,\:24.5\right)\mathrm{:\:}\quad \:x=9,\:y=24.5 [/tex]
[tex] 24.5=\left(-0.5\right)\cdot \:9+b [/tex]
[tex] 24.5=\left(-0.5\right)\cdot \:9+b [/tex]
[tex] -4.5+b=24.5 [/tex]
[tex] b=29 [/tex]
[tex] \mathrm{Construct\:the\:line\:equation\:}\mathbf{y=mx+b}\mathrm{\:where\:}\mathbf{m}=-0.5\mathrm{\:and\:}\mathbf{b}=29 [/tex]
[tex] y=-0.5x+29 [/tex]
Now plug in x=21, we get
[tex] y=-0.5*21+29=18.5 [/tex]
Thus the height of the candle after 21 hours is 18.5 centimeters.
