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Gwar [14]
3 years ago
7

Pls help with these

Mathematics
1 answer:
Studentka2010 [4]3 years ago
7 0

Answer:

(0,-11)

(1,-2)

(2,1)

(3,-2)

(4-11)

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Hunter-Best [27]

Let's simplify the following expression:

\text{ }\frac{\text{ \lparen2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{ }\frac{\text{ 6x}^5\text{y}^{-4}\text{z}^{-5}}{\text{ \lparen2x}^6\text{y}^3\text{z}^{-3})^2}

We get,

\text{ }\frac{(\text{2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{  }\frac{6x^5y^{-4}z^{-5}}{(2x^6y^3z^{-3})^2}\text{ }\frac{(2x^4-y^6z^{-4})^2(6x^5y^{-4}z^{-5})}{(2x^2z^3)(2x^6y^3z^{-3})^2}\text{ }\frac{(4x^8\text{ - 4x}^4y^6z^{-4}\text{ + y}^{12}z^{-8})(6x^5y^{-4}z^{-5})}{(2x^2z^3)(4x^{12}y^6z^{-6})}\text{ }\frac{24x^{13}y^{-4}z^{-5}\text{ - 24x}^9y^2z^{-9}\text{ + 6x}^5y^8z^{-13}}{8x^{14}y^6z^{-3}}

4 0
1 year ago
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