C&A Help Desk received the following requests during the last five weeks:

Mathematics

1 answer:

Nana76 [90]2 years ago

4 0

The answer would be 21 weeks, the either go up or go down

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the building manager put new door knobs on 4 doors in each apartment there are 3 apartments on each floor and 3 floors in the ap

barxatty [35]

(4) x (3) x (3)= (12) x (3)= 36

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2 years ago

Cos(0.5)+cos(0.5+2(3.14159265))+cos(0.5+4(3.14159265))

aleksandr82 [10.1K]

Answer:

2.63274768567

Step-by-step explanation:

First, note that 3.1415... is p, so 2(3.1415...) is 2pi and 4(3.1415...) is 4pi.

2pi and 4pi are full rotations, which will lead to the same cosine (i.e. cos(x) = cos (x+2pi) = cos(x+4pi) = ... = cos (x+2k*pi)).

So, the expression equals cos0.5 + cos 0.5 + cos0.5 = 3cos0.5 = 3(0.87758256189) = 2.63274768567

I hope this helps! :)

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2 years ago

Solve the given equation by using the quadratic formula: 2x^2-3x-4=0

kaheart [24]

Let's solve your equation step-by-step.2x2−3x−4=0Step 1: Use quadratic formula with a=2, b=-3, c=-4.x=−b±√b2−4ac2ax=−(−3)±√(−3)2−4(2)(−4)2(2)x=3±√414x=34+14√41 or x=34+−14√41

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3 years ago

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Marinette Dupain-Cheng

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2 years ago

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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 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 :

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