( 8 2/4 + 2 1/6) - 5 2/12

Mathematics

1 answer:

Readme [11.4K]3 years ago

5 0

Answer:

5 1/2 or 5.5

Step-by-step explanation:

=8 1/2+2 1/6-5 2/12

=17/2+ 2 1/6-5 2/12

=17/2+13/6-5 2/12

=32/3- 5 2/12

=32/3-31/6

=11/2

=5 1/2

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In the third degree polynomial function f(x) =x^3 + 4x, why are 2i and -2i zeros?

IgorC [24]

Answer:

We just need to evaluate and get f(2i)=0, f(-2i)=0.

Step-by-step explanation:

Since $i^2=-1$ , then $i^3=i^2i=-i$ , and we can apply this when we evaluate $f(x) =x^3 + 4x$ for 2i and -2i.

First we have:

$f(2i) =(2i)^3 + 4(2i)=2^3i^3+8i=8(-i)+8i=0$

Which shows that 2i is a zero of f(x).

Then we have:

$f(-2i) =(-2i)^3 + 4(-2i)=(-2)^3i^3-8i=-8(-i)-8i=8i-8i=0$

Which shows that -2i is a zero of f(x).

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4 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 :

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