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Sever21 [200]
3 years ago
10

Mr. Garcia needs at least 60 paintbrushes for his art classes. He has 22 paintbrushes already and will buy more paintbrushes of

8. Which inequality can be used to find how many packages of paintbrushes, p, Mr. Garcia needs to buy in order to have at least 60 paintbrushes?
Mathematics
1 answer:
elena-s [515]3 years ago
4 0

Answer:

60≥22+8x

Step-by-step explanation:

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Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.
Tasya [4]

Answer:

\frac{dy}{dx}=-[(\frac{5x+24}{36x-6x^2})]

Step-by-step explanation:

Given function:

y =\ln(\frac{(6 - x)^{\frac{3}{2}}}{x^{\frac{2}{3}}})

we know

\ln(\frac{A}{B}) = ln(A) - ln(B)

thus,

y = \ln((6 - x)^{\frac{3}{2}}) - \ln(x^{\frac{2}{3}})

or

also,

ln(Aⁿ) = n × ln(A)

thus,

y = (\frac{3}{2})\times\ln(6 - x) - (\frac{2}{3})\times\ln(x)

therefore,

\frac{dy}{dx}=[(\frac{3}{2})\times\frac{1}{(6-x)}\times(0 - 1)] - [ (\frac{2}{3})\times\frac{1}{x}\times1]

or

\frac{dy}{dx}=-(\frac{3}{2(6-x)}) - (\frac{2}{3x})

or

\frac{dy}{dx}=-[(\frac{3(3x)+2\times2(6-x)}{2(6-x)\times(3x)})]

or

\frac{dy}{dx}=-[(\frac{5x+24}{36x-6x^2})]

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3 years ago
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lora16 [44]

In cylindrical coordinates, we have r^2=x^2+y^2, so that

z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}

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1 \le r \le \sqrt2 means our region is between two cylinders with radius 1 and \sqrt2. In spherical coordinates, the inner cylinder has equation

x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)

This cylinder meets the sphere when

x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)

which occurs at

\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi

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Putting everything together, we have

\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3

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Answer: 9.5

Step-by-step explanation:

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