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3 years ago

Select all diagrams that show 4 x 1/3

Mathematics

1 answer:

Marta_Voda [28]3 years ago

6 0

Answer:

Step-by-step explanation:

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Pls help being timed

Ronch [10]

A. just gave you the answer

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3 years ago

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In Aunt Peggy’s backyard there are cows and chickens. There are 20 more chickens than cows. The total number of legs is 130. How

finlep [7]

Answer:

There's 15 cows and 35 chickens.

Step-by-step explanation:

Take 20 and multiply it by 2. The answer is 40. Then subtract 40 from 130. The answer will be 90. The cow has double the amount of legs than the chicken, so the cow is equivalent to 2 chicken. Divide 90 by 3. The answer is 30. Then divide 30 by 2 the answer is 15. Add 15 and 20 and the answer is 35 chickens. Divide 30 by 2 and the answer is 15.

You will get 35 chicken and 15 cows as your answer. Happy Election Night, and I hope this helps! ;)

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3 years ago

PLEASE HELP ILL MARK BRAINLIEST!!!

ivolga24 [154]

Answer:

If I am not mistaking is 6

Step-by-step explanation:

Answer number 1

6 0

4 years ago

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A population of rabbits increases 30.8% each year. If the rabbit population is currently 75,

Katen [24]

Answer:

176

Step-by-step explanation:

175 is not enough. Good luck

3 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 :

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