Question 214676
Any linear function can be written in the form 


{{{f(t)=mt+b}}} where 'm' is the slope and represents the change of the function and 'b' is the initial value of the function.



In our case, the change is $30 (since the total cost goes up $30 a month) while the initial fee is $20. So this means {{{m=30}}} and {{{b=20}}} giving us the function {{{C(t)=30t+20}}}



If you're unsure of the answer, simply plug in values of 't'



Examples: When {{{t=0}}}, the number of months is zero. So this represents the start of the plan. Plug this into C(t) to get {{{C(0)=30(0)+20=20}}}. So no matter what, you're going to pay $20 up front (which is exactly what the problem states).


Now one month later, the value of 't' is now {{{t=1}}}. If you simply add $30 to the down payment of $20, you get $50. In other words, you've paid $50 for the first month. If we plug in t=1, we get {{{C(1)=30(1)+20=30+20=50}}} which is exactly what we just got.



So you may be asking why we're going through all the trouble of using functions when we could just add. Using functions will save us time for larger values of 't'. For instance, let's say you want to know what the total cost will be 20 months from now. Are you going to add 20 sets of $30 to 20 to find the answer? You could, but why do that when you can simply use the function. Also, you can also answer other questions that the previous technique would have a hard time doing so.


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To find the cost of 7 months of service, just plug in t=7:



{{{C(t)=30t+20}}} Start with the given function.



{{{C(7)=30(7)+20}}} Plug in {{{t=7}}}



{{{C(7)=210+20}}} Multiply



{{{C(7)=230}}} Add



So 7 months of service will cost $230