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:
hello. I would love to answer but can you please show me the photo. if yes i will answer.
Step-by-step explanation:
Well done mate you’re correct.
X=-1.8
To solve first use the distributive property to change the equation to -2x+1 1/3+1=5
Then you subtract 1 from 5 so then the equation looks like this: -2x+1 1/3=4
Next subtract 1 1/3 from 4 to get 3.6
then divide it by -2 and you get -1.8
