A cellular phone company offers a contract for which the cost C, in dollars, of t minutes of telephoning is given by C=0.25(t-500)+ 56.95, where it is assumed that t is greater than or equal to 500 minutes. What times will keep cost between $92.20 and $129.95?

C = .25(t - 500) + 56.95

C = .25t - 125 + 56.95
C = .25t - 68.05
C + 68.05 = .25t
(C + 68.05) / .25 = t

(92.2 + 68.05) / .25 = t
641 = t

(129.95 + 68.05) / .25 = t
792 = t
Is this right?

Solve 92.20 = 0.25(t-500)+56.95 t≥500

0.25(t-500)=92.2-56.95
t-500=(92.2-56.95)/0.25
t=500+(92.20-56.95)/0.25=641

Similarly, solve for t in
129.95=0.25(t-500)+56.95
to get t=792.

The posted answers are right, well done!

Thank you

You're welcome!

To find the times that will keep the cost between $92.20 and $129.95, we need to solve the inequality:

92.20 ≤ C ≤ 129.95

We can substitute the expression for C from the given formula:

92.20 ≤ 0.25(t-500) + 56.95 ≤ 129.95

Let's solve this inequality step-by-step:

Step 1: Subtract 56.95 from all terms of the inequality:

92.20 - 56.95 ≤ 0.25(t-500) ≤ 129.95 - 56.95

35.25 ≤ 0.25(t-500) ≤ 73.00

Step 2: Divide all terms of the inequality by 0.25:

35.25 ÷ 0.25 ≤ (0.25(t-500)) ÷ 0.25 ≤ 73.00 ÷ 0.25

141 ≤ t - 500 ≤ 292

Step 3: Add 500 to all terms of the inequality:

141 + 500 ≤ t - 500 + 500 ≤ 292 + 500

641 ≤ t ≤ 792

Therefore, the times that will keep the cost between $92.20 and $129.95 are t values between 641 and 792 minutes.