Task: Use the following situation to answer Parts A and B Instructions At noon, a tank contains 100 gallons of water. The table shows the input and output of water from pipes A, B, and C. The pipes begin operating simultaneously at noon. A B C Pipe Flow IN (gallons per minute) a (2) = 172 6(2) = 80 Flow OUT (gallons per minute) c (2)=30r Complete each of the 2 activities for this Task. Activity 1 of 2 Let T (2) represent the amount of water in the tank x minutes after all of the pipes A, B, and C are opened. Which of the following functions represents T (x)? O A T (2) = 100 + a(z) +b(2) - c(x) OCT (x) = 100 + a (x) +b(2) + c(2) OBT () = a (x) + b() +(2) ODT (2) = a (x) +b(x) - c(:) Activity 2 of 2 Using the equation, after how many minutes will the tank be empty?

Step-by-step explanation:

Refer to attached for the table and clearer version of the question

As per table A and B pipes feed the tank and C is the outlet

Considering initial volume of 100 gallons, the amount of water in the tank after x minutes is:

We can calculate when tank is going to be empty:

Since we got negative number, the tank is not going to be empty when all 3 pipes are open.

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

i cant answer this, theres no table. sorry man

step-by-step explanation: