Answers

  • Yhi
  • Ambitious
2015-03-07T16:56:03+08:00

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I think you got confused how to use the squared thing..so I guess you mean this for number 1? 3 x^{2} y + 5x y^{3} + 9/3xy ? So this is my answer if you got wrong.

1. 3x^2y + 5xy^3 + 9 / 3xy
= 3x^2y / 3xy + 5xy^3 / 3xy + 9/3xy
= x + 5y^2/3 + 3/xy
Answer: 3x + 5y^2 + 3/xy

2. a^2b^2c-6abc^2 + 5a^3b^5 / 2abc^2
= a^2b^2c/2abc^2 - 6abc^2 / 2abc^2 + 5a^3b^5 / 2abc^2
= ab / 2c - 3 + 5a^2b^4 / 2c^2
answer: ab/2c - 3 + 5a^2b^4/2c^2

3. (3x2 + 2x + 1) + (5x2 - 7x)  =  
=  3x2 + 2x + 1 + 5x2 - 7x
=  8x2 - 5x + 1

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yeah I saw it too
the final answer in no. 1 must x ONLY(on the first term) because you have to cancel it.
In no.3 you must multiply it because you must distribute + to the second factor.
2015-03-07T17:16:41+08:00

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 1.)\ Solution; \\  \\ \frac{3x^{2}y+5xy^{3}+9}{3xy} \\ \\  \frac{3x^{2}y+5xy^{3}+9}{3xy} \\ \\  \frac{\not3x^{2}\not{y}}{\not3x\not{y}}+\frac{5\not{x}y^{3}}{3\not{x}y}+\frac{9}{3xy} \\ \\ \\ \boxed{\boxed{x+ \frac{5y^{2}}{3}+ \frac{3}{xy}}} \\ \\ \\ \\ 2.)Solution; \\ \\   \frac{a^{2}b^{2}c-6abc^{2}+5a{3}b{5}}{2abc^{2}} \\ \\  \frac{a^{2}b{2}c}{2abc^{2}}-\frac{6\not{a}\not{b}\not{c^{2}}}{2\not{a}\not{b}\not{c^{2}}}+  \frac{5a^{3}b^{5}}{2abc{2}}

\boxed{\boxed{ \frac{ab}{2c}-3+ \frac{5a^{2}b^{4}}{2c^2}}} \\ \\ \\ 3.)\ Solution; \\ \\ (3x^2+2x+1)+(5x^2-7x) \\ \\ (3x^2+2x+1)(5x^2-7x) \\ \\ (5x^2-7x) (3x^2+2x+1) \\ \\ 3x^2+2x+1 \\ \times5x^2-7x \\ ----------- \\ 15x^4+10x^3+5x^2 \\ .\ \ \ \ \ -21x^3-14x^2-7x \\ ------------ \\ \boxed{\boxed{15x^4-11x^3-9x^2-7x}} \\ \\ \\ \\ Hope\ it\ Helps :) \\ Domini
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