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
X/5120=20/100
x/5120=1/5
5120/5=1024
1024+5120=6144
6144 is the answer.
Answer:
The answer is A
Step-by-step explanation:
Hello there!
(Equation #1) 32 + 12 = 44
(Equation #2) 6 - 3 = 3, 3 x 8 = 24
These both equations don't have the same answer, even though it says that they are equal with each other, if you could correct these these 2 equations I am willing to answer again.
Hope this helped!!
Answer:
3t-21 hope I helped you! >w<
