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Mathematics 13 Online
OpenStudy (oswaldmurphy):

The table below represents the volume of a liquid sample as a function of its temperature: http://prntscr.com/bs66hd (graph) The average rate of change of f(x) between x = 11 to x = 21 is _____ml per °C and represents the rate of change of volume per degree rise in temperature.

OpenStudy (oswaldmurphy):

@mathstudent55 @jim_thompson5910 @AloneS

jimthompson5910 (jim_thompson5910):

Using the table, when x = 11, what is the value of f(x) ?

OpenStudy (oswaldmurphy):

f(11)?

jimthompson5910 (jim_thompson5910):

look at the row that has x = 11. Then look directly across from x = 11 to find the value ______

OpenStudy (oswaldmurphy):

95?

jimthompson5910 (jim_thompson5910):

yes

jimthompson5910 (jim_thompson5910):

when x = 21, the value of f(x) is what?

OpenStudy (oswaldmurphy):

f(21)

jimthompson5910 (jim_thompson5910):

again, use the table

jimthompson5910 (jim_thompson5910):

you don't have a function f(x) like you did in the last problem

OpenStudy (oswaldmurphy):

100,

jimthompson5910 (jim_thompson5910):

no

OpenStudy (oswaldmurphy):

Is the final answer 5?

OpenStudy (oswaldmurphy):

Oh 105 sorry typo.

jimthompson5910 (jim_thompson5910):

when the temp is 21, the volume is _______

jimthompson5910 (jim_thompson5910):

yes 105

jimthompson5910 (jim_thompson5910):

so we have this info temp = 11 volume = 95 temp = 21 volume = 105

OpenStudy (oswaldmurphy):

Oh so the difference between the two is 10?

jimthompson5910 (jim_thompson5910):

change in volume = 105 - 95 = 10 change in temp = 21-11 = 10 divide the two changes (change in volume)/(change in temp) = 10/10 = 1

OpenStudy (oswaldmurphy):

So is that the final answer?

jimthompson5910 (jim_thompson5910):

the rate of change is 1 mL per degree Celsius basically that means each time the temperature goes up one degree Celsius, the volume goes up 1 mL

jimthompson5910 (jim_thompson5910):

yeah it's just 1

OpenStudy (oswaldmurphy):

Thank you again! I might have 1 or two more. Ill tag you :D

jimthompson5910 (jim_thompson5910):

no problem

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