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jeyben [28]
3 years ago
9

Helpppppp mathhhhhhh

Mathematics
1 answer:
lorasvet [3.4K]3 years ago
6 0
Use sin
Sin35=x/4.1
X=4.1×sin35
=2.351
=2.4m
Answer is 2.4
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£9.50

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(1-i)^2 find 4 th root​
sergiy2304 [10]

By de Moivre's theorem,

1 - i = \sqrt2\,e^{-i\pi/4} \implies (1-i)^2 = 2\,e^{-i\pi/2}

\implies \sqrt[4]{(1 - i)^2} = \sqrt[4]{2}\,e^{i(2\pi k-\pi/2)/4} = \sqrt[4]{2}\,e^{i(4k-1)\pi/8}

where k\in\{0,1,2,3\}. The fourth roots of (1-i)^2 are then

k = 0 \implies \sqrt[4]{2}\,e^{-i\pi/8}

k = 1 \implies \sqrt[4]{2}\,e^{i3\pi/8}

k = 2 \implies \sqrt[4]{2}\,e^{i7\pi/8}

k = 3 \implies \sqrt[4]{2}\,e^{i11\pi/8}

or more simply

\boxed{\pm\sqrt[4]{2}\,e^{-i\pi/8} \text{ and } \pm\sqrt[4]{2}\,e^{i3\pi/8}}

We can go on to put these in rectangular form. Recall

\cos^2(x) = \dfrac{1 + \cos(2x)}2

\sin^2(x) = \dfrac{1 - \cos(2x)}2

Then

\cos\left(-\dfrac\pi8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac{1 + \cos\left(\frac\pi4\right)}2} = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}

\sin\left(-\dfrac\pi8\right) = -\sin\left(\dfrac\pi8\right) = -\sqrt{\dfrac{1 - \cos\left(\frac\pi4\right)}2} = -\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}

\cos\left(\dfrac{3\pi}8\right) = \sin\left(\dfrac\pi8\right) = \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}

\sin\left(\dfrac{3\pi}8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}

and the roots are equivalently

\boxed{\pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} - i\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right) \text{ and } \pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} + i \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right)}

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Please help me! (Tangent Ratios) (30 points)
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Answer:

Step-by-step explanation:

A = 48/20

B = 20/48

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