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9966 [12]
3 years ago
14

+{-x2y+[-(x2y-2xy2+y2)+(-xy2-3y2+x2y)]-(10y2-x2y)}

Mathematics
1 answer:
Kay [80]3 years ago
6 0

Answer:

3xy² -  14y²      

Step-by-step explanation:

I hope that this is the problem

   - x²y + [ - (x²y - 2xy² + y²) + (xy² - 3y² + x²y)] - (10y² - x²y)

= - x²y + [ - x²y + 2xy² - y² + xy² - 3y² + x²y] - 10y² + x²y

Now combine like terms in the [  ].

= - x²y  + [ -x²y + x²y + 2xy² + xy² - y² - 3y² ] - 10y² + x²y

= - x²y  + [ 0 + 3xy² - 4y²] - 10y² + x²y

= - x²y + 3xy² - 4y² -10y² + x²y                      Now combine like terms

=   (-x²y + x²y) + 3xy² + (-4y² - 10y²)

=       0 + 3xy² - 14y²

=  3xy² -  14y²       or    y²(3x - 14)

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3 years ago
7 1/6 = 5 2/5 + p<br><br> p = ???
Crank

\huge\textbf{Hey there!}

\mathbf{7\dfrac{1}{6} = 5\dfrac{2}{5} + p}

\rightarrow\mathbf{7 \dfrac{1}{6} = 5\dfrac{2}{5} + p}

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\text{SUBTRACT }\rm{\dfrac{27}{5}}\text{ to BOTH SIDES}

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\text{NEW EQUATION: }\rm{p = \dfrac{43}{6} - \dfrac{27}{5}}

\large\textsf{SIMPLIFY IT!}

\mathbf{p = \dfrac{53}{30} \approx p = 1\dfrac{23}{30}}

\huge\boxed{\textsf{Therefore, your answer is: \boxed{\mathsf{p = 1\dfrac{23}{30}}}}}\huge\checkmark

\huge\textbf{Good luck on your assignment \&}\\\huge\textbf{enjoy your day!}

~\frak{Amphitrite1040:)}

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