Ask your own question, for FREE!
Mathematics 9 Online
2792000217:

Michelle rents a movie for a flat fee of $1.50 plus an additional $1.25 for each night she keeps the movie. Choose the cost function that represents this scenario if x equals the number of nights Michelle has the movie. (2 points) c(x) = 1.50 + 1.25x c(x) = 1.50x + 1.25 c(x) = 2.75 c(x) = (1.50 + 1.25)x These ones are soo confusing because I'm not the best at math

mxddi3:

ok so you see how there is a $1.50 fee?

2792000217:

Yea

mxddi3:

that is the price she is going to start at--it's our constant do you know what a constant is?

2792000217:

Its a number by itself

mxddi3:

yes, that's a good start. the constant is a number that is not going to change because,as you said, it is by itself. now, we are going to need this 1.50 in our equation, so which option can we eliminate just from this ?

2792000217:

c(x) = 2.75

mxddi3:

yes, good job! ok now, it says she has to pay $1.25 for every night, and x is the number of nights she has it. do you know how to write $1.25 per night with a variable?

2792000217:

I think so 1.25(x) I might be wrong

mxddi3:

no, you're exactly right! (you don't need the parentheses though so you could remove them) and now you can remove another option. do you know which one?

2792000217:

c(x) = 1.50x + 1.25

mxddi3:

so we are now left with c(x) = 1.50 + 1.25x and c(x) = (1.50 + 1.25)x so as you remember, we had 1.50 and 1.25x, so looking at these two options, do you know which shows the data we found?

2792000217:

c(x) = 1.50 + 1.25x

mxddi3:

good job!!! do you understand how you got this?

2792000217:

Well for the first time yes Thank you soo much

mxddi3:

of course, glad to help !

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!