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Ket [755]
3 years ago
13

If α, β are the zeroes of the polynomials f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1) =

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
4 0

Answer:

f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)

Step-by-step f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)explanation:

f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(xf(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1) + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(xf(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1) + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)

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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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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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Hope this helps! If you have any questions, just ask!
4 0
3 years ago
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Answer:

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