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evablogger [386]
2 years ago
13

Can anyone help me with this?

Mathematics
1 answer:
avanturin [10]2 years ago
5 0

The simplified expression of (x^0 y^2/3 z^-2y^)^2/3 divided by (x^2 z^1/2)^-6 is x^(12) y^(10/9) z^(-1/3)

<h3>How to simplify the expression?</h3>

The algebraic statement is given as:

(x^0 y^2/3 z^-2y^)^2/3 divided by (x^2 z^1/2)^-6

Rewrite the algebraic statement as:

[(x^0 y^2/3 z^-2y)^2/3]/[(x^2 z^1/2)^-6]

Evaluate the like factors

[(x^0 y^(2/3+1) z^-2)^2/3]/[(x^2 z^1/2)^-6]

Evaluate the sum

[(x^0 y^5/3 z^-2)^2/3]/[(x^2 z^1/2)^-6]

Expand the exponents

[(x^(0*2/3) y^(5/3 * 2/3)z^(-2*2/3)]/[(x^(2*-6) z^(1/2*-6)]

Evaluate the products

[(x^0 y^(10/9) z^(-4/3)]/[(x^(-12) z^(-3)]

Apply the quotient law of indices

x^(0+12) y^(10/9) z^(-4/3+3)

Evaluate the sum of exponents

x^(12) y^(10/9) z^(-1/3)

Hence, the simplified expression of (x^0 y^2/3 z^-2y^)^2/3 divided by (x^2 z^1/2)^-6 is x^(12) y^(10/9) z^(-1/3)

Read more about simplified expression at:

brainly.com/question/723406

#SPJ1

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Two positive, consecutive, odd integers have a product of 143. set up an equation to then solve for the integers.
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Answer:

Step-by-step explanation:

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ANSWER

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EXPLANATION

The given expression is;

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\sin(x)

and

\cos(x)

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\frac{ \cos^{2} (x) }{ \sin^{2} (x) }  \times ( \frac{1}{ \cos(x) } )-  \cos(x)

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Answer:

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Step-by-step explanation:

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