Answer:
A(t) = 300 -260e^(-t/50)
Step-by-step explanation:
The rate of change of A(t) is ...
A'(t) = 6 -6/300·A(t)
Rewriting, we have ...
A'(t) +(1/50)A(t) = 6
This has solution ...
A(t) = p + qe^-(t/50)
We need to find the values of p and q. Using the differential equation, we ahve ...
A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50
0 = 6 -p/50
p = 300
From the initial condition, ...
A(0) = 300 +q = 40
q = -260
So, the complete solution is ...
A(t) = 300 -260e^(-t/50)
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The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.
J)x^2 +2x +yx+2y-5x-10
x^2-3x+yx-10
k)5y^2(1-2y)
l)a(x+y+z)
The Answer to your question is 77/8
Since the farmer sells an average of 16.25 bushels a day and it has been 7 days, you would multiply 16.25 by 7 to find the amount sold in a week aka the change in the total bushels the farmer has to sell which is a change of 113.75 or 113 3/4.
Answer: y = 64
Step-by-step explanation:
y <> x² ----------------------- 1
y = kx² ----------------------- 2
36 = 3²k
36 = 9k
K = 36/9
= 4.
To find y when x = 4, we substitute for x in equation 2
y = kx²
y = 4²k
y = 16 x 4
y = 64
