Question #f28b6

A candle is 10 in. tall after burning for 2 hours. After 3 hours, it is 8.5 in. tall. How tall will the candle be after 6 hours?

We can make a linear equation to model the situation. Since time is the independent variable and height is the dependent variable, we can let #(x,y)# be #("time", "height")#, therefore the question gives us the points #(2,10)# and #(3,8.5)#, which we can use to find the slope, m:

#m=(10-8.5)/(2-3)=-1.5/1=-3/2#

Consider point-slope form:
#(y-y_1)=m(x-x_1)#
where #m# is the slope of the line and #(x_1,y_1)# is a point of the line.

Now, we can write a linear equation using point-slope form:
#y-10=-3/2(x-2)#

Since we want to know the height of the candle after 6 hours, we simply plug #x=6# into the above equation and solve for #y#:
#y-10=-3/2(6-2)#
#y=-3/2(4)+10#
#y=-6+10#
#y=4#

Therefore, the height of the candle after #6# hours, will be #4# in.

Sometimes it is a good idea to 'scribble' a quick sketch to assist in visualisation of what the question is asking.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let the required height be #h#

What the target is:

Reference height - #x# = h

where #x# = burn rate x 4 hours
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Determine the burn rate")#

#("EB-EC inches")/("3-2 hours") -> (10-8.5)/(3-2) =1.5/1=3/2#

#" "3/2 # inches in 1 hour

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine "x)#

# x=3/2 ("inches")/("hour")xx 4" hours"#

#x=3/(cancel(2)^1)xxcancel(4)^2" "("inches")/(cancel("hour"))xx cancel(" hours")#
#" "color(red)( uarr)#
#color(red)("manipulating units the same way as numbers")#

#x=3xx2" inches "=" "6" inches"#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine "h)#

#h=10 -x" inches "=" "10-6" "=" "4" inches"#