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Leviafan [203]
3 years ago
8

Solve for x. −23(3x−4)+3x=56

Mathematics
2 answers:
Neporo4naja [7]3 years ago
7 0
Open bracket
-69x+92+3x=56
Collect like term
-66x=-36
Divide both side by -66
x=6\11
solniwko [45]3 years ago
7 0

Answer: 6/11

Step-by-step explanation:

Open bracket

-69x+92+3x=56

Collect like term

-66x=-36

Divide both side by -66

x=6\11

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Rewrite the following integral in spherical coordinates.​
lora16 [44]

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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

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which occurs at

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

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The volume element transforms to

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\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

4 0
2 years ago
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yKpoI14uk [10]

Answer:

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67-61 degrees-47

side-angle-side

SAS

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