math

rame said:
(sqrt2x-5y)^2

What do you need to do? - there is no "question.

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
 
Another thing:\displaystyle Another \ thing:

Is that: (2x5y)2 or (2x5y)2?\displaystyle Is \ that: \ (\sqrt2 x-5y)^2 \ or \ (\sqrt{2x}-5y)^2?

If you dont start using grouping symbols, unless you marry the boss daughter, youre doom.\displaystyle If \ you \ don't \ start \ using \ grouping \ symbols, \ unless \ you \ marry \ the \ boss' \ daughter, \ you're \ doom.
 
BigGlenntheHeavy said:
Is that: (2x5y)2 or (2x5y)2?\displaystyle Is \ that: \ (\sqrt2 x-5y)^2 \ or \ (\sqrt{2x}-5y)^2?
Huh?
 


Another possibility: (2x5y)2\displaystyle \text{Another possibility:} \ (\sqrt{2x - 5y})^2

 


I love it!\displaystyle \text{I love it!}


\(\displaystyle \text{We have: }\;(\text{sqrt}2x\!\!-\!\!5y)}^2\)
. . . \(\displaystyle \text{which could equal: }\;\begin{Bmatrix} \left(\sqrt{2}\!\cdot\!x - 5y\right)^2 \\ \\[-3mm] \left(\sqrt{2x} - 5y\right)^2 \\ \\[-3mm] \left(\sqrt{2x-5}\cdot y\right)^2 \\ \\[-3mm] \left(\sqrt{2x-5y}\,\right)^2 \end{array}\)

. . and no instructions . . .\displaystyle \text{and no instructions . . .}



If I had a nickel . . .
.
 
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