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2 years ago

How are rational expressions related to rational numbers?

1 answer:

4 0

A rational expression is the same way that any integer is also a rational number. You can stick to any polynomial in the numerator of a fraction and put "1" in the denominator.

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Which of the following is irrational?

dedylja [7]

Answer:

$\sqrt{2}$

Step-by-step explanation:

the other 3 are perfect squares and have rational answers.

$\sqrt{1.44} =1.2\\\sqrt{1}= 1\\\sqrt{4} =2$

6 0

1 year ago

Read 2 more answers

(x^2y+e^x)dx-x^2dy=0

klio [65]

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 µ(x, y) 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 µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

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

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

5 0

3 years ago

C= m over 3 π d to the second power , for m

julia-pushkina [17]

Solve for m:C = 1/3 π d^2 m
C = 1/3 d^2 m π is equivalent to 1/3 d^2 m π = C:1/3 π d^2 m = C
Divide both sides by (π d^2)/3:Answer: m = (3 C)/(π d^2)

6 0

2 years ago

How do you use electromagnetic spectrums to learn more about stars and planets?

klemol [59]

Answer: Telescopes use lenses or mirrors to collect and focus waves from the electromagnetic spectrum, including visible light, allowing us to look at celestial objects. By studying the electromagnetic waves given off by objects such as stars, galaxies, and black holes, astronomers can better understand the universe.

Step-by-step explanation:

8 0

2 years ago

Show an example of commutative property of addition

vaieri [72.5K]

For example, if we use 5+3=8.
If you change it to 3+5, it still equals 8.
Therefore, 3+5=5+3 because at the end, they both have equal value aka both end up with 8 using the same numbers. (goes by the rule a+b=b+a)

5 0

3 years ago

Read 2 more answers