Answer:
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
Step-by-step explanation:
Given the data in the question;
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is?
dA/dt = rate in - rate out
first we determine the rate in and rate out;
rate in = 3pound/gallon × 5gallons/min = 15 pound/min
rate out = A pounds/1000gallons × 5gallons/min = 5Ag/1000pounds/min
= 0.005A pounds/min
so we substitute
dA/dt = rate in - rate out
dA/dt = 15 - 0.005A
Therefore, If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
Answer:
hope this helps
Step-by-step explanation:
solution:
x^2 + 3x + 5 = x + 13
x^2 + 2x -8 = 0
(x+4)(x-2) = 0
x = -4 or x = 2
if x = -4, y = 9
if x = 2 , y = 15
Answer:
1/4=slope
-8=y intercept
Step-by-step explanation:
the equation in slope intercept form would be y=1/4x-8
The answer would be $175
Multiply the most by the number of tickets
