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Lapatulllka [165]
3 years ago
9

1. In the above figure, AE = 2, CE = 3, and DE = 4. What is the length of BE? A. 6 B. 8 C. 5 D. 12

Mathematics
1 answer:
babunello [35]3 years ago
4 0

Answer:

<u>The answer that you are looking for is A.6. </u><-----------------------------

<u>Hope this helps..</u>


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Answer:

2

Step-by-step explanation:

sin 17° = x/7

(sin 17°) 7 = (x/7)7

x = sin 17° (7)

x = 0.292371704722737 (7)

x = 2.046601933059157

x ≈ 2

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Step-by-step explanation:

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The radius of a circle is 1 mm. What is the circle's circumference? ​
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C≈6.28mm

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(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 <em>µ(x, y)</em> 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 <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

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}

The modified DE,

\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0

is now exact:

\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}

Integrate both sides of the first condition with respect to <em>x</em> :

F(x,y) = -e^{-x}y - \dfrac1x + g(y)

Differentiate both sides of this with respect to <em>y</em> :

\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C

Then the general solution to the DE is

F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}

5 0
3 years ago
What is the missing term in the factorization?
wel

Answer:

100x^2-36=4(5x+3)(5x-3)

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