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kobusy [5.1K]
3 years ago
13

Subject differential equationDay. Month.Year. ..(y-secx) dx + tanxdy...o​

Mathematics
1 answer:
Colt1911 [192]3 years ago
8 0

(y-\sec x)\,\mathrm dx+\tan x\,\mathrm dy=0

Divide both side by \mathrm dx and rearrange terms to get a linear ODE;

\tan x\dfrac{\mathrm dy}{\mathrm dx}+y=\sec x

Multiply both sides by \cos x:

\sin x\dfrac{\mathrm dy}{\mathrm dx}+\cos x\,y=1

The left side can be condensed as the derivative of a product:

\dfrac{\mathrm d}{\mathrm dx}(\sin x\,y)=1

Integrate both sides, then solve for y:

\sin x\,y=x+C\implies\boxed{y(x)=x\csc x+C\csc x}

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Contact [7]

Answer:

Pedro pagó $448

Step-by-step explanation:

Sea P el precio inicial de un objeto.

Si aplicamos un descuento del X%, entonces el nuevo precio del objeto es:

NP = P*(1 - X%/100%)

y lo que estamos ahorrando es:

P - NP

En este caso, primero tenemos un descuento del 30%, entonces:

NP = P*(1 - 30%/100%) = P*(1 - 0.3)

Luego tenemos otro descuento, esta vez del 20%, entonces:

NP' = NP*(1 - 20%/100%) = P*(1 - 0.3)*(1 - 20%/100%) = P*(1 - 0.3)*(1 - 0.2)

Lo que Pedro ahorra es igual a $352

entonces:

P - NP' = $352

P -  P*(1 - 0.3)*(1 - 0.2) = $352

P*(1 - (1 - 0.3)*(1 - 0.2)) = $352

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P = $352/(1 - 0.56) = $800

Esto significa que el precio original era $800.

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2 years ago
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