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
Use the formula a^2+b^2=c^ which is 7.21
Step-by-step explanation:
Refer to the attachment.
I hope it helps:)
So, for the area of a circle, there should be a line in the middle. Correct?
Well, that line is the diameter. Then cut that in half, so if the diameter is 6, the radius is 3.
Then multiply the the radius by the radius, so 3x3 once you get that, multiply the answer by pi. Pi= 3.14, so if you do 3x3x3.14 you should get 28.26. Hope this helps.
