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:
I think the first mistake was made in step 1
Answer is D.
28:44 is the ratio but you have to simplify it. Divide both by 4. 28 divided by 4 = 7; 44 divided by 4 = 11
Answer:
that would be nice for you to have to say about me nila rc
Answer:
Building B, it is around 16 35 feet taller than building A ( C )
Step-by-step explanation:
<u>Calculating the height of Building A </u>
Height = opposite ( ? )
∅ = 73°
base/adjacent = 30 ft
∴ Height = Tan 73° * base
= 3.2709 * 30 = 98.127 ft
<u>Calculating the height of building B </u>
∅ = 73°
base/adjacent = 35 ft
∴ height = Tan 73° * base
= 3.2709 * 35 = 114.4815
Therefore we can say Building B is Taller by approximately 16.35 feet
