# MATH

posted by
**jasmine20** on
.

This problem makes no sence to me i tried it but i know i am completely wrong someone please help.

The cost for a long-distance telephone call is $0.36 for the first minute and $0.21 for each additional minute or portion thereof. Write an inequality representing the number of minutes a person could talk without exceeding $3.

my answer and solving way:

.36x+.21 = 3

-.21 .21

.36x = 2.79

---- ----

.36 .36

x = 7.75

i think o.36+0.21x=3.00

so 0.21x=2.64. then you just find x

let he talks for X minutes

out of this 1 minute is @ 0.36$/min

and (X-1) minutes @ 0.21$/min

therefore

=> 0.36+0.21(X-1)<=3.0

=> 0.36+0.21X-0.21 <=3.0

=> 0.15+0.21X <=3.0

=> 0.21X <=2.85

=> X <= ( 2.85/0.21 )

=> X <14 minutes

precisely X= 13.57 minutes

Let x = number of minutes.

0.21 x + 0.36 = 3.00

0.21 (x-1) = 2.64

x = 13.57 minutes IF the person were charged for the unused part of the last minute. However, the caller would be charged for 1 + 13 minutes, or $3.09. That exceeds the $3 limit, so talking would actually have to stop at 13 minutes, when the charge would be $2.88.

I am having a little trouble with maths because It is really hard 8-)