Question

Check my work?

Answer 1

C is just a variable, and has been predefined.
F(x) = 12c + 100
F(x) = 12(5) + 100
F(x) = 60 + 100
F(x) = 160

Answer 2

can functions work this way?

Answer 3

correct.

Answer 4

You forgot to say please

Answer 5

F(x) = 12c+100. F(x) is, therefore, constant. It always equals 12c+100, no matter what x is. Did you mean to write F(x) = 12x+100?

Answer 6

I suppose C and X are one in he same.

Answer 7

Then you must write it as F(x)=12x+100.

In that case, F(5)=12(5)+100 = 160. So, the function F evaluated at x=5 is 160. The function F evaluated at x=0 is F(0) = 100. As you can see, F changes as x changes.

Answer 8

You may also write it as F(c)=12c+100.

Answer 9

Okay, so because C is defined, and X has no set value, my equation will be wrong?

Answer 10

Well, you don't have an equation. You have a function. And, how is C defined in F(x)=12x+100?

Answer 11

Either C or X need to both be on either side of the equals sign?

Answer 12

It wasn't. C = number of charms purchased. 100 is the initial price of a bracelet.
so, F(x) = 12c + 100. But X will never change. So C needs to be equal to X, so, my function needs to be F(x) = 12x + 100

Answer 13

Yes. That is the general idea. You can give the "input" to the function F any label -- generally, x is used. So, if you call the input c, then F(c)=12c+100 would probably be correct. The idea is that F is a function that depends on what the input is. In this case, it is a function that is 12 times the value of its input plus an additional 100.

Answer 14

Yes. That's right. Also, I would use "C" as the input, since they use C=number of charms purchased. So, F(C) = 12C+100.

Answer 15

Thank you!