Cities and companies find that the cost of pollution control increases along with the percentage of pollutants to be removed in a situation. Suppose that the cost C, in dollars, of removing p\% of the pollutants from a chemical spill is given below. Complete parts (a) through (d). \[ C(p)=\frac{40,000}{100-p} \] (a) Find $C(0), C(20), C(80)$, and $C(90)$. $C(0)=\$ \square$ (Round to the nearest whole number as necessary. Simplify your answer.)

Step 1 ：Given the cost function \(C(p)=\frac{40,000}{100-p}\), we are asked to find the cost of removing certain percentages of pollutants.

Step 2 ：Substitute \(p=0\) into the function to find \(C(0)\), which gives \(C(0)=\frac{40,000}{100-0} = \$400\).

Step 3 ：Substitute \(p=20\) into the function to find \(C(20)\), which gives \(C(20)=\frac{40,000}{100-20} = \$500\).

Step 4 ：Substitute \(p=80\) into the function to find \(C(80)\), which gives \(C(80)=\frac{40,000}{100-80} = \$2000\).

Step 5 ：Substitute \(p=90\) into the function to find \(C(90)\), which gives \(C(90)=\frac{40,000}{100-90} = \$4000\).

Step 6 ：Final Answer: \(C(0)=\boxed{\$400}\), \(C(20)=\boxed{\$500}\), \(C(80)=\boxed{\$2000}\), and \(C(90)=\boxed{\$4000}\).