A telephone company offers a monthly cellular phone plan for \( \$ 24.99 \). It includes 350 anytime minutes plus \( \$ 0.20 \) per minute for additional minutes. The following function is used to compute the monthly cost for a subscriber, where \( x \) is the number of anytime minutes used. \( C(x)=\left\{\begin{array}{ll}24.99 & \text { if } 0350\end{array}\right. \) Compute the monthly cost of the cellular phone for use of the following anytime minutes. \( \begin{array}{lll}\text { (a) } 210 & \text { (b) } 425 & \text { (c) } 351\end{array} \) (a) \( C(210)=\$ \square \) (Round to the nearest cent as needed.)

To compute the monthly cost of the cellular phone for the given anytime minutes, we will evaluate the function \( C(x) \) for each case. ### Step 1: Evaluate \( C(210) \) Since \( 210 \) is less than \( 350 \), we will use the first part of the piecewise function: \[ C(x) = 24.99 \quad \text{if } 0 < x \leq 350 \] Thus, for \( x = 210 \): \[ C(210) = 24.99 \] ### Step 2: Evaluate \( C(425) \) Since \( 425 \) is greater than \( 350 \), we will use the second part of the piecewise function: \[ C(x) = 0.20x - 45.01 \quad \text{if } x > 350 \] Now, substituting \( x = 425 \): \[ C(425) = 0.20(425) - 45.01 \] Calculating \( 0.20 \times 425 \): \[ 0.20 \times 425 = 85 \] Now, substituting back into the equation: \[ C(425) = 85 - 45.01 = 39.99 \] ### Step 3: Evaluate \( C(351) \) Since \( 351 \) is also greater than \( 350 \), we will again use the second part of the piecewise function: \[ C(x) = 0.20x - 45.01 \quad \text{if } x > 350 \] Now, substituting \( x = 351 \): \[ C(351) = 0.20(351) - 45.01 \] Calculating \( 0.20 \times 351 \): \[ 0.20 \times 351 = 70.2 \] Now, substituting back into the equation: \[ C(351) = 70.2 - 45.01 = 25.19 \] ### Summary of Results - (a) \( C(210) = 24.99 \) - (b) \( C(425) = 39.99 \) - (c) \( C(351) = 25.19 \) Thus, the final answers are: - (a) \( C(210) = \$24.99 \) - (b) \( C(425) = \$39.99 \) - (c) \( C(351) = \$25.19 \)