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.
She must at least score a 38
To get the average you need to add 45 + 32 + 37 together than divide it by the amount of numbers added, in this case 3.
45 + 32 + 37 = 114
114÷3 = 38
R = speed in still water
r + c = speed with current
r - c = speed against current
3.5(r + c) = 70
r + c = 20
c = 20 - r
4(r + c) = distance from pier to pier
5(r - c) = distance from pier to pier
4(r + c) = 5(r - c)
4(r + 20 - r) = 5(r - 20 + r)
80 = 10r - 100
180 = 10r
18 = r
The speed in still water is 18 km/hour
= (16x3 - 8x2 + 4x4) / 2x
= 4x * (4x2 - 2x + x3) / 2x
= 2 * (4x2 - 2x + x3)
= 8x2 - 4x + 2x3
Answer A
N=#
-6(n)
I think this is the answer
