The equation is -5x = 4
x = -0.8
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:
E
Step-by-step explanation:
Answer:
P = 5
Step-by-step explanation:
2/5(10p) - 2/5(15)= 14
20p/5 - 30/5= 14
4p - 6= 14
4p = 20
p=5
