Asked by Jamese
A international phone call cost a fixed amount for the first minute and a certain rate per each additional minute. If a 7 minute phone call cost $10 and a $4 call cost &6.40, find the fixed charge and the rate per each additional minute.
Answers
Answered by
Shell
Variables,
X=first, (fixed) rate
y=second, rate(/minute)
t=time of call
z=total cost
Equation,
z=x+((t-1)*y)
Values,
7minutes is $10, 4minutes is $6.40
Plug them in, isolate one variable, set them equal to each other, and solve for the remaining variable,
10=x+((7-1)*y), 6.4=x+((4-1)*y)
10=x+(6*y), 6.4=x+(3*y)
10=x+(6*y), 6.4=x+(3*y)
-6y= -6y, -3y= -3y
10-6y=x, 6.4-3y=x
10-6y= 6.4-3y
-6.4=-6.4
3.6-6y=-3y
+6y=+6y
3.6=3y
(3.6)/3=(3y)/3
1.2=y
pick one original equation, plug in your new value, and solve for the remaining variable
10-6y=x
10-6(1.2)=x
10-7.2=x
2.8=x
Fixed first minute is $2.80, with $1.20 per each additional minute
I gave you the equations so that you could have them to help you on later problems. And I hope the directions helped.
X=first, (fixed) rate
y=second, rate(/minute)
t=time of call
z=total cost
Equation,
z=x+((t-1)*y)
Values,
7minutes is $10, 4minutes is $6.40
Plug them in, isolate one variable, set them equal to each other, and solve for the remaining variable,
10=x+((7-1)*y), 6.4=x+((4-1)*y)
10=x+(6*y), 6.4=x+(3*y)
10=x+(6*y), 6.4=x+(3*y)
-6y= -6y, -3y= -3y
10-6y=x, 6.4-3y=x
10-6y= 6.4-3y
-6.4=-6.4
3.6-6y=-3y
+6y=+6y
3.6=3y
(3.6)/3=(3y)/3
1.2=y
pick one original equation, plug in your new value, and solve for the remaining variable
10-6y=x
10-6(1.2)=x
10-7.2=x
2.8=x
Fixed first minute is $2.80, with $1.20 per each additional minute
I gave you the equations so that you could have them to help you on later problems. And I hope the directions helped.
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