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riadik2000 [5.3K]
3 years ago
14

Edwin Edwards and Karen Davis owned EEE, Inc., which owned three convenience stores, all of which sold gasoline. Reid Ellis deli

vered to the three convenience stores $26,675.02 worth of gasoline for which he was not paid. Ellis proved that Edwards and Davis owned the business, ran it, and in fact personally ordered the gasoline. He claimed that they were personally liable for the debt owed him by EEE, Inc. Decide. [Ellis v.
Edwards, 348 S.E.2d 764 (Ga. App.)]
Mathematics
1 answer:
Ulleksa [173]3 years ago
5 0

Question Completion:

Note: This is a Law question, not Mathematics.

Answer:

Decision:

Edwards and Davis are not personally liable for the debts of EEE, Inc.

Step-by-step explanation:

That Edwin Edwards and Karen Davis owned EEE Inc. does not force them to be personally liable for the debts of EEE Inc.  EEE Inc. is a separate legal entity.  There is the legal separation of ownership and the liabilities of Edwards and Davis are limited to the capital they contributed or undertook to contribute to the corporation.  The corporate veil is only lifted when it is ascertained that EEE Inc. was not run as a legal entity separate from the owners.  We can also conclude that there is no evidence of abuse of the corporate form of EEE Inc. with comingling of assets or for the purpose of promoting fraud or injustice or evasion of tort or contractual responsibility on the part of Edwards and Davis.  Therefore, Edwin Edwards and Karen Davis are not personally liable for the debts of EEE Inc. as claimed by Reid Ellis.

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Solve the initial value problem 2ty" + 10ty' + 8y = 0, for t > 0, y(1) = 1, y'(1) = 0.
Eva8 [605]

I think you meant to write

2t^2y''+10ty'+8y=0

which is an ODE of Cauchy-Euler type. Let y=t^m. Then

y'=mt^{m-1}

y''=m(m-1)t^{m-2}

Substituting y and its derivatives into the ODE gives

2m(m-1)t^m+10mt^m+8t^m=0

Divide through by t^m, which we can do because t\neq0:

2m(m-1)+10m+8=2m^2+8m+8=2(m+2)^2=0\implies m=-2

Since this root has multiplicity 2, we get the characteristic solution

y_c=C_1t^{-2}+C_2t^{-2}\ln t

If you're not sure where the logarithm comes from, scroll to the bottom for a bit more in-depth explanation.

With the given initial values, we find

y(1)=1\implies1=C_1

y'(1)=0\implies0=-2C_1+C_2\implies C_2=2

so that the particular solution is

\boxed{y(t)=t^{-2}+2t^{-2}\ln t}

# # #

Under the hood, we're actually substituting t=e^u, so that u=\ln t. When we do this, we need to account for the derivative of y wrt the new variable u. By the chain rule,

\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm dy}{\mathrm du}

Since \frac{\mathrm dy}{\mathrm dt} is a function of t, we can treat \frac{\mathrm dy}{\mathrm du} in the same way, so denote this by f(t). By the quotient rule,

\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac ft\right]=\dfrac{t\frac{\mathrm df}{\mathrm dt}-f}{t^2}

and by the chain rule,

\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\mathrm df}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm df}{\mathrm du}

where

\dfrac{\mathrm df}{\mathrm du}=\dfrac{\mathrm d}{\mathrm du}\left[\dfrac{\mathrm dy}{\mathrm du}\right]=\dfrac{\mathrm d^2y}{\mathrm du^2}

so that

\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm dy}{\mathrm du}}{t^2}=\dfrac1{t^2}\left(\dfrac{\mathrm d^2y}{\mathrm du^2}-\dfrac{\mathrm dy}{\mathrm du}\right)

Plug all this into the original ODE to get a new one that is linear in u with constant coefficients:

2t^2\left(\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm d y}{\mathrm du}}{t^2}\right)+10t\left(\dfrac{\frac{\mathrm dy}{\mathrm du}}t\right)+8y=0

2y''+8y'+8y=0

which has characteristic equation

2r^2+8r+8=2(r+2)^2=0

and admits the characteristic solution

y_c(u)=C_1e^{-2u}+C_2ue^{-2u}

Finally replace u=\ln t to get the solution we found earlier,

y_c(t)=C_1t^{-2}+C_2t^{-2}\ln t

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Answer:

Step-by-step explanation:

Let t represent the time it took Emanuel to drive home from college. If the total round trip took 11 hours, it means that the time it took

Emanuel to drive from home back to college would be (11 - t) hours.

Emanuel drove home from college traveling an average speed of 70 mph.

Distance = speed × time

Distance covered by driving from college to home is

70 × t = 70t

He drove back to the college the following week at an average speed of 62.7 mph.

Distance covered by driving back to college from home is

62.7(11 - t) = 689.7 - 62.7t

Since the distance travelled is the same, then

689.7 - 62.7t = 70t

70t + 62.7 = 689.7

132.7t = 689.7

t = 689.7/132.7

t = 5.19 hours.

Therefore, the time that it took Emanuel to drive from home back to college is

11 - 5.19 = 5.81 hours

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