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2 years ago

Function f is graphed. According to the graph, is f even, odd, or neither?

Mathematics

1 answer:

miss Akunina [59]2 years ago

3 0

Answer:

Step-by-step explanation:

f is neither even nor odd

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How should I do this question

Ivanshal [37]

Answer:

( 11 -6)

( -5 6 )

Step-by-step explanation:

Multiply the first row of N by first column of M. This will give the first element in top row of the answer :-

-3*-2 + 1*5 = 11

Then multiply the first row in N by the second column of M to give the second element of the top row.

-3*0 + 1*-6 = -6

Then do a similar process with the second row in N.

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2 years ago

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levacccp [35]

First, you try to make sure that both the denominators are the same. In order to do this, you multiply both fractions' numerator and denominator by the other fraction's denominator. Next, you combine like terms in the numerator, and then you simplify the fraction.

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2 years ago

2d + 1 = 9 so d = what

weeeeeb [17]

Answer:

d is 4 :))

hope that helped

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3 (x-k)/w = 4 solve for x

Trava [24]

Actually the above equation can be solved by following the below steps to fins the value of x Step1: 3(x-k)/w=4 Step 2: (3x-3k)/w=4 Step 3: (3x-3k)=4w Step 4: 3x=4w+3k So hence therefore by following the above steps we can conclude that the value of x is Step 5: x=(4w+3k)/3

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2 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 :

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