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Karo-lina-s [1.5K]
2 years ago
7

Edward has a hot dog stand. Last week he spent $13 for hot dogs and

Mathematics
1 answer:
xxTIMURxx [149]2 years ago
5 0
His profit is $12
25-13=12
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Cuanto es la tercera parte de 39​
KengaRu [80]

Answer:

13

Step-by-step explanation:

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3y''-6y'+6y=e*x sexcx
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From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

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y_1=e^x\cos x
y_2=e^x\sin x

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the fundamental solutions. We have

W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

and so

u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx
u_1=\dfrac13\ln|\cos x|

u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
u_2=\dfrac13x

Therefore the particular solution is

y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x

so that the general solution to the ODE is

y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x
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3 years ago
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3 years ago
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kakasveta [241]

Answer:

A = ((a+b)/2)h

Step-by-step explanation:

a and b are base (the top and bottom)

h is height

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3 years ago
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Fudgin [204]

Answer:

<h3>              D. (2, 7)</h3>

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