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:
Step-by-step explanation:
Substitute the value of the variable into the equation and simplify.
Solve it out.
Multiply / divide / add / subtract everything out.
Send a picture of it ok then i answer
Answer:
Step-by-step explanation:
<u>Given:</u>
<u>Inscribed angle measure is half the intercepted arc:</u>
<u>The angle between the diameter and tangent is right angle:</u>
<u>According to angle addition postulate:</u>
- m∠YAC = m∠BAC - m∠BAY
- m∠YAC = 90° - 26° = 64°
Correct choice is B
Answer:
i think it is c
Step-by-step explanation:
not 100% sure tho
