Answer:
Step-by-step explanation:
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.
Let x = adult tickets
Let y = children's tickets
x + y = 800
8x + 4y = 4,400
From the first eq, x = 800 - y, plug this in to the second eq
8(800-y) + 4y = 4,400
6400 - 8y + 4y = 4,400
2000 = 4y
500 = y
x = 800 - 500
x = 300
300 adult tickets were sold.
F(x) = (x - 4) (x^2 + 4) would be your answer.
Answer:
1 gato come 5 ratos em 1 dia X 5 = 5 gatos come 25 ratos por dia
5 gatos come 25 ratos por dia X 5 = 25 gatos come 125 ratos por dia
