Performing laplace transform of the equation.
sY(s) - y(0) + 6Y(s) = 1/(s-4)
(s+6)Y(s) - 2 = 1/(s-4)
Y(s) = 2/(s+6) + 1/(s-4)(s+6), by partial fraction decomposition
Y(s) = 2/(s+6) + 1/10 * (1/(s-4) + 1/(s+6))
Y(s) = 0.1/(s-4) + 2.1/(s+6)
Performing inverse laplace transform,
y(t) = 0.1e^4t + 2.1e^(-6t)
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Answer 3.1
Here's the formula:
V=
H=3
3 · 
≈ 3.08333
rounded to : 3.1
(P.S. I also Used a calculator)
Answer:
I think the answer is 19.
Primary equation: A(x)= (5y)(3X)
Secondary equation: 5y+3X=1000
y=200-(3X)/5
A(x)=3X(1000-3X)
A(x)=3000X-9X²
Now, find the derivative of A(x) to find the max... here's the work for that, or you could guess and check.
A'(x)=3000-18X
Set derivative equal to 0
0=3000-18X
166.6666666=X
Now test the intervals
(0,166.6666) (166.66666, 1000)
1st derivative is + 1st derivative is -
Plug the X value back into the secondary equation
5y+3(166.666666)=1000
5Y=500
Y=5
Answer:
X= 166.6666666666
Y=5
Please note, this is entry level calculus, and your teacher may expect you to use a different, longer route such as guess and check.
Answer:
192 servings
Step-by-step explanation:
