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3 years ago

Find the missing length indicated

Mathematics

1 answer:

liq [111]3 years ago

7 0

Tan<span>θ = opposite/adjacent
Tan</span><span>θ = 64/48
arctan(64/48) = 53
90-53 = 37
48*Tan(37) = 36
64+36 = 100</span>

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Degree 4; zeros: -4+5i, -1 multiplicity 2

Kazeer [188]

Answer:

$f(x) = [x^{2} + 8x + 41](x + 1)^{2}$

Step-by-step explanation:

If -4 + 5i is a root of a polynomial, then it's conjugate -4 - 5i will also be a root of the same polynomial.

Therefore, the polynomial has 4 degree, and zeroes are (- 4 + 5i), (- 4 - 5i) and -1 with multiplicity 2.

Hence, the polynomial will be

$f(x) = (x + 4 - 5i)(x + 4 + 5i)(x + 1)^{2}$

⇒ $f(x) = [x^{2} + 8x + (4 - 5i)(4 + 5i)] (x + 1)^{2}$

⇒ $f(x) = [x^{2} + 8x + 41](x + 1)^{2}$ (Answer)

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Step-by-step explanation:

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Step-by-step explanation:

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Dmitry_Shevchenko [17]

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2 years ago

Read 2 more answers

(sinx + icosx)^n = cos( nπ\2 - nx) + isin( nπ\2 - nx)

Alenkasestr [34]

$\sin x+i\cos x=i(\cos x-i\sin x)=ie^{-ix}=e^{i\frac\pi2}e^{-ix}=e^{i\left(\frac\pi2-x\right)}$

By DeMoivre's theorem,

$(\cos y+i\sin y)^n=(e^{iy})^n=e^{iny}=\cos(ny)+i\sin(ny)$

which would here mean that

$(\sin x+i\cos x)^n=\left(e^{i\left(\frac\pi2-x\right)}\right)^n=e^{i\left(\frac{n\pi}2-nx\right)}=\cos\left(\dfrac{n\pi}2-nx\right)+i\sin\left(\dfrac{n\pi}2-nx\right)$

as required.

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3 years ago