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3 years ago

Evaluate the expression with the given variable replacements. 2x3 (if x = 3)

Mathematics

1 answer:

MArishka [77]3 years ago

5 0

Answer:

if the expression is 2x^3 then answer is 54

Step-by-step explanation:

2x^3

substitute in x=3

2* (3)^3

= 2*3*3*3

= 2*27

= 54

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Find the complex fourth roots of 81(cos(3pi/8) + i sin(3pi/8))

BartSMP [9]

By using De Moivre's theorem:

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 ⇒ z = 81(cos(3π/8) + i sin(3π/8))


Part (A) find the modulus for all of the fourth roots

∴ The modulus of the given complex number = l z l = 81

∴ 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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3 years ago

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DerKrebs [107]

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8. If k = log, 3 then log, 48 = 16 (1) 2k +3 (3) k +8 (2) 3k +1 (4) k +4

vekshin1

assuming you means k = log_2(3) [as log(2)3 is the same thing as 3log(2) due to multiplication being commutative]

given log(ab) = log(a) + log(b)

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siniylev [52]

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