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:
19 days
Step-by-step explanation:
Given
--- initial
-- rate
Required
Days when the fish gets to 30
The function is exponential and as such it follows;
Where x represents the number of days and y the number of fishes
Because the fishes decreases;
So, we have:
Express as decimal
In
So, we have:
Divide both sides by 150
Take log of both sides
Apply law of logarithm
Make x the subject
<em>Hence, it takes approximately 19 days</em>
The answer is b
Because you have to add da x and dat is 2x den add da numbers which is 4
Answer:
2
----------
5w^11
Step-by-step explanation:
8/20 reduces to 2/5, and (w^(-2)) / w^9 becomes 1 / w^11.
Overall, #20 reduces to:
2
----------
5w^11
Recall the rule of exponents:
w^a
----------- = w^(a-b)
w^b
and note that
w^(-2)
------------ = w^(-2-9) = w^(-11) (same as above)
w^9
