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sweet-ann [11.9K]
2 years ago
12

Please help!! It’s engenuity.

Mathematics
1 answer:
MakcuM [25]2 years ago
4 0

9514 1404 393

Answer:

  (c)  1/(36·a^4·b^10)

Step-by-step explanation:

The applicable rules of exponents are ...

  (a^b)/(a^c) = a^(b-c)

  (a^b)^c = a^(bc)

  a^-1 = 1/a

__

  \left(\dfrac{(2a^{-3}b^4)^2}{(3a^5b)^{-2}}\right)^{-1}=\dfrac{(3a^5b)^{-2}}{(2a^{-3}b^4)^2}=\dfrac{3^{-2}a^{-10}b^{-2}}{2^2a^{-6}b^8}=\dfrac{1}{3^22^2}a^{-10-(-6)}b^{-2-8}\\\\=\boxed{\dfrac{1}{36a^4b^{10}}}

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Which equation represents 4x – 3y = 15 when solved for y
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-4x      -4x

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What are the fourth roots of 6+6√(3i) ?
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Answer:

Step-by-step explanation:

The genral form of a complex number in rectangular plane is expressed as z = x+iy

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r is the modulus = √x²+y²

∅ is teh argument = arctan y/x

Given thr complex number z = 6+6√(3)i

r = √6²+(6√3)²

r = √36+108

r = √144

r = 12

∅ = arctan 6√3/6

∅ = arctan √3

∅ = 60°

In polar form, z = 12(cos60°+isin60°)

z = 12(cosπ/3+isinπ/3)

To get the fourth root of the equation, we will use the de moivres theorem; zⁿ = rⁿ(cosn∅+isinn∅)

z^1/4  = 12^1/4(cosπ/12+isinπ/12)

When n = 1;

z1 =  12^1/4(cosπ/3+isinn/3)

z1 = 12^1/4cis(π/3)

when n = 2;

z2 = 12^1/4(cos2π/3+isin2π/3)

z2 = 12^1/4cis(2π/3)

when n = 3;

z2 = 12^1/4(cosπ+isinπ)

z2 = 12^1/4cis(π)

when n = 4;

z2 = 12^1/4(cos4π/3+isin4π/3)

z2 = 12^1/4cis(4π/3)

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3 years ago
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