Tuesday
May 3, 2016

# Homework Help: MATH

Posted by jasmine20 on Thursday, December 21, 2006 at 3:02am.

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-)

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