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Troyanec [42]
3 years ago
12

Solve for y.8^3-3y=256^4y

Mathematics
1 answer:
aliina [53]3 years ago
3 0
8^{3-3y}=256^{4y}\\\\(2^3)^{3-3y}=(2^8)^{4y}\\\\2^{3(3-3y)}=2^{8\cdot4y}\\\\2^{9-9y}=2^{32y}\iff9-9y=32y\\\\-9y-32y=-9\\-41y=-9\ \ \ \ /:(-41)\\\\y=\frac{9}{41}
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4 is the correct answer i believe! :) sorry if wrong.
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What is (2-2i√3)4 equivalent to?
maxonik [38]

a. (2 - 2i√3)⁴ in polar form is 256(cos(-4π/3) + isin(-4π/3)) = 256cis(-4π/3)

b. (2 - 2i√3)⁴ in rectangular form is -128 + 128√3

To answer the question, we need to know what complex numbers are

<h3>What are complex numbers?</h3>

Complex numbers are numbers of the form z = x + iy

<h3>a. Complex numbers in polar form</h3>

Complex numbers in polar form z = r(cosθ + isinθ) where

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Given that z = (2 - 2i√3)⁴ =

So,

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So, converting to polar form

r = √(x² + y²)

= √[2² + (-2√3)²]

= √[4 + 4(3)]

= √[4 + 12]

= √16

= 4

θ = tan⁻¹(y/x)

θ = tan⁻¹(-2√3/2)

θ = tan⁻¹(-√3)

θ = -π/3

So, z = r(cosθ + isinθ)

= 4(cos(-π/3) + isin(-π/3))

<h3>Powers of complex numbers</h3>

A complex number z raised to power n is zⁿ = rⁿ(cosnθ + isin(nθ)]

z⁴ = (2 - 2i√3)⁴

= r⁴(cos4θ + isin4θ)

= 4⁴(cos(4 × -π/3) + isin(4 × -π/3))

= 256(cos(-4π/3) + isin(-4π/3))

= 256cis(-4π/3)

(2 - 2i√3)⁴ in polar form is 256(cos(-4π/3) + isin(-4π/3)) = 256cis(-4π/3)

<h3>b. Complex numbers in rectangular form</h3>

The complex number z =  r(cosθ + isinθ) in rectangular form is z = x + iy where

  • x =  rcosθ and
  • y =  rsinθ

Given that z⁴ = 256(cos(-4π/3) + isin(-4π/3)) in rectangular form,

x = rcosθ

= 256(cos(-4π/3)

= 256cos(-4 × 60°)

= 256cos(-240)

= 256cos(240)

= 256 × -1/2

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y =  rsinθ

= 256sin(-4π/3)

= 256sin(-4 × 60°)

= 256sin(-240)

= -256sin240

= -256 × -√3/2

= 128√3

So, z⁴ = x + iy

= -128 + 128√3

So, (2 - 2i√3)⁴ in rectangular form is -128 + 128√3

Learn more about complex numbers in polar form here:

brainly.com/question/9678010

#SPJ1

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Answer:

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Step-by-step explanation:

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