1. Linear function problem

A copying service charges a uniform rate for the first one hundred copies or less and a fee for each additional copy. Nancy Taylor paid 7.00 to make 200 copies and Rose Barbi paid 9.20 for 310 copies.
a) Write two ordered pairs (x,y) where x represents the numbers of copies over one hundred and y represents the cost of the copies.
b) Write an equation in the form of y=mx+b that expresses the value of y, the total cost of the copies, in terms of x, the number of copies over one hundred.
c) What is the cost of the first one hundred copies?
d) What is the cost of each additional copy
http://www.jmap.org/JMAP/SupportFile...Chapter_10.pdf Look at Pg 407 number 15

2. Re: Linear function problem

Originally Posted by victorwen28
A copying service charges a uniform rate for the first one hundred copies or less and a fee for each additional copy. Nancy Taylor paid 7.00 to make 200 copies and Rose Barbi paid 9.20 for 310 copies.
a) Write two ordered pairs (x,y) where x represents the numbers of copies over one hundred and y represents the cost of the copies.
b) Write an equation in the form of y=mx+b that expresses the value of y, the total cost of the copies, in terms of x, the number of copies over one hundred.
c) What is the cost of the first one hundred copies?
d) What is the cost of each additional copy
what, exactly, is confusing about part (a) ?

3. Re: Linear function problem

no because like when it says over one hundred copies i think that it means the numbers of copies greater than 100. so u take the number and subtract 100.
so the cost of the first one hundred is 5 while the cost of each additional copy is .02 . am i right?

4. Re: Linear function problem

Originally Posted by victorwen28
no because like when it says over one hundred copies i think that it means the numbers of copies greater than 100. so u take the number and subtract 100.
so the cost of the first one hundred is 5 while the cost of each additional copy is .02 . am i right?
that is correct.

next time, I recommend that you post your work and results instead of just saying you are "confused". you'll find that folks are more apt to help when you actually show an attempt to solve the problem.

5. Re: Linear function problem

oh i am awfully sorry. i am quite new to this forum so i do not completely grasp it. thanks for giving me some advice. i did trial and error. may you please explain and show work to show me how to exactly grasp this problem?

6. Re: Linear function problem

Originally Posted by victorwen28
A copying service charges a uniform rate for the first one hundred copies or less and a fee for each additional copy. Nancy Taylor paid 7.00 to make 200 copies and Rose Barbi paid 9.20 for 310 copies.
a) Write two ordered pairs (x,y) where x represents the numbers of copies over one hundred and y represents the cost of the copies.
b) Write an equation in the form of y=mx+b that expresses the value of y, the total cost of the copies, in terms of x, the number of copies over one hundred.
c) What is the cost of the first one hundred copies?
d) What is the cost of each additional copy
http://www.jmap.org/JMAP/SupportFile...Chapter_10.pdf Look at Pg 407 number 15
a) It tells you that the x variable represents copies over 100, so to find the x values take the amounts that they have and subtract 100. (200-100=100 and 310-100=210) Your x values are 100 and 210 respectively. It tells you that y represents the cost of the copies... so just use the numbers they gave you.
Doing the above you should get two ordered pairs, (100, 7) and (210, 9.2).

b) In order to make an equation you will need to find the slope. Do this via your two points you found in part a. Once you have your slope take and pick an ordered pair and solve for your y-intercept. Then you have an equation in y=mx+b form (after you plug in your slope and intercept).

c and d) Use the equation found in part b to solve for these. They'll probably have to do with a y-intercept and slope.

7. Re: Linear function problem

thank you very much for your assistance. math help forum is a great place with great assistants to help! yet again, thank you!