A water tank is being filled by pumps at a constant rate. The volume of water in the tank V, in gallons, is given by the equation: V(t) = 65t+280 ,where t is the time, in minutes the pump has been on. At what rate, in gallons per minute, is the water being pumped into the tank?

65 gallons per minute

Step-by-step explanation:

The total volume of the tank at any given time is given by the equation:

V(t) = 65t + 280

In order to find the rate of change of volume, we can simply differentiate this equation with respect to time. This will give us the rate of change of the volume or the rate at which water is being pumped into the tank.

Differentiating the above equation we get:

V'(t) = 65

So we can see that the rate at which water is being pumped into the tank is 65 gallons per minute

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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