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3 years ago

Erica was deducted 450 points while she was playing a

Mathematics

1 answer:

Over [174]3 years ago

6 0

Answer:

Step-by-step explanation:

hope this helps

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Please answer I will give 20 and brainiest

Mariulka [41]

Answer:

Option 4

Step-by-step explanation:

Coordinances for a three-dimensional coordinate system are (x, y, z).

The letter labeling each axis is always on the positive side.

The fourth option is correct because you move 3 units away from the 'x', 1 unit away from the 'y', then 2 units toward the 'z'.

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3 years ago

Two lines segment that both have a midpoint at point M

AVprozaik [17]

Answer:

No se

Step-by-step explanation:

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3 years ago

(01.02 MC)Which of the following expressions shows how to rewrite 4 − 5 using the additive inverse and displays the expression c

n200080 [17]

Answer:

I think the answer is 1. The expression 4 plus negative 5 is written on top. A number line from negative 10 to positive 10 is shown, with numbers labeled at intervals of 1. An arrow is shown from point 0 to 4. Another arrow points from 4 to negative. So the answer is A

hope this helps sorry if it wrong

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3 years ago

Find the value of X?

Molodets [167]

Answer:

x = 58

Step-by-step explanation:

The exterior angle is equal to the sum of the opposite interior angles

90 = 32+x

Subtract 32 from each side

90-32 = x

58 =x

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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 :

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