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Mathematics 21 Online
OpenStudy (anonymous):

Can someone double check my work?

OpenStudy (anonymous):

\[1. (7x^3 + 13x - 1) + (5x^2 - 9x) - (x^3 - 2x^2 - 3x + 8)\] I got: \[6x^3 + 7x^2 + 1x - 9\]

OpenStudy (anonymous):

\[2. (-4x^5)(-5xy^6)(15x^2y)^0\] I got: \[20x^5y^6\]

OpenStudy (anonymous):

\[3. \frac{ ch^4 }{ xch^-3 }\] I got: \[\frac{ h^7 }{ x }\]

OpenStudy (jdoe0001):

did you mean \(\bf \cfrac{ ch^4 }{ xch^{-3} }\) ?

OpenStudy (anonymous):

Yes.

OpenStudy (jdoe0001):

\(\bf (7x^3 + 13x - 1) + (5x^2 - 9x) - (x^3 - 2x^2 - 3x + 8)\\\qquad \\ 7x^3 + 13x - 1 + 5x^2 - 9x - x^3 + 2x^2 + 3x - 8\\\qquad \\ (7x^3- x^3 ) (+ 13x- 9x + 3x) (+ 5x^2 + 2x^2)(- 1 - 8)\\\qquad \\ 6x^3+7x^2+7x-9\\ -------------------------\\ (-4x^5)(-5xy^6)(15x^2y)^0 \implies (-4x^5)(-5xy^6)1\\ \qquad \\ 20x^6y^6\\ -------------------------\\ \cfrac{ ch^4 }{ xch^{-3} }\implies \cfrac{\cancel{c}h^4h^3}{x\cancel{c}} \implies \cfrac{h^{4+3}}{x}\)

OpenStudy (anonymous):

\[5. (64^{12}y)^{2/3}\] I got: \[16x^8y ^{2/3}\] Also, thank you.

OpenStudy (anonymous):

\[6. (4x - 3y)(9x - 2y)\] I got: \[36x - 26xy + 6y\]

OpenStudy (anonymous):

\[7. (3x^2 + 2) + 2(x - 3)^{2}\] I got: \[-7x^2 + 18\]

OpenStudy (anonymous):

\[4. (\frac{ 3b^3n^6 }{ n^8z^2 })^{-1}\] I got: \[\frac{ n^2z^2 }{ 3b^3 }\]

OpenStudy (anonymous):

I got: \[54x^2 - 45x + 9x\]

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