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3 years ago

1-6n=-n+16 what dose n equal to?

Mathematics

1 answer:

Setler [38]3 years ago

7 0

Answer:

-3

Step-by-step explanation:

1 - 6n = -n + 16

1 - 5n = 16

-5n = 15

n = -3

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Three times a number subtracted from the product of fifteen and the reciprocal of a number

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To write the problem out as an equation, let 'n' equal your number.

(15 x 1/n) - 3n

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Write each expression in exponential form 2 x 2 x 2 x 2 =

jok3333 [9.3K]

Answer:

2^4

Step-by-step explanation:

there are four 2's being multiplied to each other.

You can combine them and say 2 x 2 x 2 x 2 = 2 ^4

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-7÷x=19 need your help

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Answer:

the answer is 26

Step-by-step explanation:

you add 7 to -7 and 19. the positive and negative 7 cancels each other out. them you add 19 and 7 which is 26. that means x is 26.

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3 years ago

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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