What's the answer and how do you do it?

Mathematics

2 answers:

Monica [59]3 years ago

7 0

The vertical asymptotes for a rational expression, occur at the values of "x" that make the denominator 0, and therefore the expression undefined.

what are those anyway? simple, just set the denominator to 0, solve for "x", and that will spit them out.

x + 4 = 0
x = -4

there, if "x" ever becomes -4, then the fraction turns to $\bf \cfrac{1}{x+4}\implies \cfrac{1}{-4+4}\implies \stackrel{und efined}{\cfrac{1}{0}}$

thus, that's where the vertical asymptote is at, x = -4.

kykrilka [37]3 years ago

4 0

A vertical asymptote is when y gets higher and higher or lower and lower without stopping around a certain point. In a fraction one way this can happen is when there is a point where the denominator is 0, as when it gets close to 0 it gets bigger and bigger, and anything over 0 is undefined. Therefore the answer is A as 1/4-4 is 1/0 and an asymptote

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Nutka1998 [239]

Answer:

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8 0

3 years ago

The sum of five consecutive positive integers is 2020. Find the largest of these numbers.

lawyer [7]

Answer:

406

Step-by-step explanation:

Five consecutive positive integers = x, (x+1), (x+2), (x+3), (x+4)

x + (x+1) + (x+2) + (x+3) + (x+4) = 2020

5x + 10 = 2020

5x = 2020-10

5x = 2010

x = 2010÷5

x = 402

The largest of the number is = x+4

x+4 = 402 + 4 = 406

8 0

3 years ago

(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 :

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