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.
Step-by-step explanation:
sqrt(-4) = sqrt(4×-1) = sqrt(4)×sqrt(-1) = 2i
yes it is right when the vertex of the first polygon is located on 3 - 2 after rotation the work vertex will be there on (2 , 3) it is right
If you divide by 8, you can put the equation into intercept form. That form is ...
... x/a + y/b = 1
where <em>a</em> and <em>b</em> are the x- and y-intercepts, respectively.
Here, your equation would be
... x/(-2) + y/(-4) = 0
The graph with those intercepts is not shown with your problem statement here. See the attachment for the graph.
