MATH
posted by jasmine20 .
This problem makes no sence to me i tried it but i know i am completely wrong someone please help.
The cost for a longdistance 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$/min
and (X1) minutes @ 0.21$/min
therefore
=> 0.36+0.21(X1)<=3.0
=> 0.36+0.21X0.21 <=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 (x1) = 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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