# Algebra help? (different problem.)?

Question:i need help with one problem. it''s not too easy for me, but i am just asking you to help me with this problem. just show me how to do it or summarize it into simpler words, just anything tht might help me understand it and get the right answer.

problem 1:
An internet service provider charges 9.95/month for the first 20 hours and then 50 cents for each additional hours. write an expression representing the charges for "h" hours use in one month when h is more than 20 hours. What is the charge for 35 hours?

When h>20...

.50(h-20) + \$9.95
.50(35-20) + \$9.95
.50(15) + \$9.95
7.50 + \$9.95
\$17.45

FYI, when h<20...
the charge will be \$9.95 regardless
charge=9.95+ h*50
(50/100) =.5
charge for 35 hrs

charge=9.95+35*.5 = 27.45

what u have is that the charge of the first 20 hours is constant and any hours more u're charged 50 cents each now if u wanted to get the charge in one month when the hours are more 20 hours u have to add the constant 20 hours charge and theextra hours "H" multiplied by 50 cents or \$0.5
you have more than 20 hours so first you pay 9.95 plus 50 cents for each additional hour (H) Y is the amount you pay.
9.95 + .50(H) = y
so.. you have 35 hours

first do 35 - 20 to get the number of additional hours which is 15.

so you pay 9.95 for first 20 hours then 50 cents for each of the 15 hours left

9.95+ .50 (H)=y plug in 15 for H
9.95 + .50 (15) = y

Y= \$17 45 your answer. its an easy question once you understand.

BTW user coolj is wrong as you can see.

This article contents is post by this website user, EduQnA.com doesn't promise its accuracy.