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jasenka [17]
3 years ago
7

How do I solve this? x+1/x-2 > 0

Mathematics
1 answer:
Eduardwww [97]3 years ago
8 0

Answer:

x ∈ (-oo ; - 1) ∪ (+ 2 ; +oo)

Step-by-step explanation:

\frac{x+1}{x-2}>0\\ \\x+1>0=>x>-1\\\\x-2>0=>x>2

x ∈ (+ 2 ; +oo)

x+1x

x ∈ (-oo ; - 1)

x ∈ (-oo ; - 1) ∪ (+ 2 ; +oo)

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Find the complex fourth roots of 81(cos(3pi/8) + i sin(3pi/8))
BartSMP [9]
By using <span>De Moivre's theorem:
</span>
If we have the complex number ⇒ z = a ( cos θ + i sin θ)
∴ \sqrt[n]{z} =  \sqrt[n]{a} \ (cos \  \frac{\theta + 360K}{n} + i \ sin \ \frac{\theta +360k}{n} )
k= 0, 1 , 2, ..... , (n-1)


For The given complex number <span>⇒ z = 81(cos(3π/8) + i sin(3π/8))
</span>

Part (A) <span>find the modulus for all of the fourth roots
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<span>∴ The modulus of the given complex number = l z l = 81
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∴ The modulus of the fourth root = \sqrt[4]{z} =  \sqrt[4]{81} = 3

Part (b) find the angle for each of the four roots

The angle of the given complex number = \frac{3 \pi}{8}
There is four roots and the angle between each root = \frac{2 \pi}{4} =  \frac{\pi}{2}
The angle of the first root = \frac{ \frac{3 \pi}{8} }{4} =  \frac{3 \pi}{32}
The angle of the second root = \frac{3\pi}{32} +  \frac{\pi}{2} =  \frac{19\pi}{32}
The angle of the third root = \frac{19\pi}{32} +  \frac{\pi}{2} =  \frac{35\pi}{32}
The angle of the  fourth root = \frac{35\pi}{32} +  \frac{\pi}{2} =  \frac{51\pi}{32}

Part (C): find all of the fourth roots of this

The first root = z_{1} = 3 ( cos \  \frac{3\pi}{32} + i \ sin \ \frac{3\pi}{32})
The second root = z_{2} = 3 ( cos \  \frac{19\pi}{32} + i \ sin \ \frac{19\pi}{32})

The third root = z_{3} = 3 ( cos \  \frac{35\pi}{32} + i \ sin \ \frac{35\pi}{32})
The fourth root = z_{4} = 3 ( cos \  \frac{51\pi}{32} + i \ sin \ \frac{51\pi}{32})
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4 years ago
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