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tekilochka [14]

3 years ago

14

At a store, notebooks are on sale for 78% off the regular price. The cost of a notebook is represented by n. Which of the follow

ing expressions does NOT represent the sale price of the notebooks?
a) 0.22n

b) n + 0.22n

c) n – 0.78n

d) (1 – 0.78)n

1 answer:

Nitella [24]3 years ago

5 0

Answer:

Ima say D but I’m not sure srry

Step-by-step explanation:

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$\underbrace{(4x+3y^2)}_M\,\mathrm dx+\underbrace{2xy}_N\,\mathrm dy=0$

The ODE is exact if $\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$ .

$M_y=6y$
$N_x=2y$

This is not the case, so look for an integrating factor $\mu(x)$ such that

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Since $\mu$ is a function of $x$ only, you have

$\mu M_y=\mu'N+\mu N_x\implies\dfrac{\mu'}\mu=\dfrac{M_y-N_x}N\implies\mu=\exp\left(\displaystyle\int\frac{M_y-N_x}N\,\mathrm dx\right)$

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$\underbrace{(4x^3+3x^2y^2)}_{M^*}\,\mathrm dx+\underbrace{2x^3y}_{N^*}\,\mathrm dy=0$

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${M^*}_y=6x^2y$
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Now, you're looking for a solution of the form $\Psi(x,y)=C$ , since differentiating via the chain rule yields

$\dfrac{\mathrm d}{\mathrm dx}\Psi(x,y)=\Psi_x+\Psi_y\dfrac{\mathrm dy}{\mathrm dx}=0$

Matching up components, you would have

$\Psi_x=M^*=4x^3+3x^2y^2$
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Differentiate this with respect to $y$ to get

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So the solution here is

$\Psi(x,y)=x^4+x^3y^2+C_1=C\implies x^4+x^3y^2=C$

Just for a final check, take the derivative to get back the original ODE:

$\dfrac{\mathrm d}{\mathrm dx}[x^4+x^3y^2]=\dfrac{\mathrm d}{\mathrm dx}C$
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$(4x+3y^2)\,\mathrm dx+2xy\,\mathrm dy=0$

so the solution is correct.

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