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.
Answer:
The answer is C: X = 3
Step-by-step explanation:
N^4
1 + 1 + 4 - 2 = 4
Mark brainliest
Rate of change of profit for this period is $2750 per month
<em><u>Solution:</u></em>
Given that,
Profit of $6500 in January and $17,500 in May
<em><u>To find: Rate of change</u></em>
Since,
January is the first month of the year (1) while May is the fifth month (5)
<em><u>Therefore, we get two points</u></em>
(1, 6500) and (5, 17500)
Using these points we can find the rate of change in profit for this time period
<em><u>The rate of change using the following formula:</u></em>

Here from the points,

<em><u>Therefore, rate of change is given as:</u></em>

Thus rate of change of profit for this period = $2750 per month