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.
Pretty sure 200feet but its kinda easy
Answer:
A) 2C
Step-by-step explanation:
The relevant rule of logarithms is ...
log(x²) = 2·log(x)
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We know that 64 = 8². So, ...
log(64) = log(8²) = 2·log(8)
We are given that log(8) = C, so 2·log(8) = 2C
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Here, all logarithms are to the base 9. That does not change the relations shown.
