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2 years ago

Which multiplication expression is eqeul to 3/5 ÷ 1/2

Mathematics

2 answers:

eduard2 years ago

8 0

Answer:

Lets solve = 3/5 ÷ 1/2

3/5 ÷ 1/2

= 3/5 × 2/1

= 3 × 2/5 × 1

= 6/5

In mixed fraction its

$1 \frac{1}{5}$

____________________

Now lets solve "a" option

3/5 × 2

= 3 × 2/5 × 1

= 6/5

In mixed fraction

$1 \frac{1}{5}$

So option a = 3/5 x 2/1 is ur answer

STatiana [176]2 years ago

3 0

<h3>Answer:</h3>

is $\sf \frac{3}{2} \times \frac{2}{1} [B]$

<h3>Step-by-step explanation:</h3>

$\sf \frac{ 3}{5} \div \frac{1}{2}$

$\sf = \frac{3}{2} \times \frac{2}{1} [B]$

<h3>Conclusion:</h3>

Which multiplication expression is equal to 3/5 1/2 the answer is $\sf \frac{3}{2} \times \frac{2}{1} [B]$ .

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It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor <em>µ(x, y)</em> such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$