You must substitute the (x, y) and check that the POI makes both equations ___.

Mathematics

1 answer:

Mademuasel [1]2 years ago

6 0

False because they cannot work

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Determine whether the graph of the equation is symmetric with respect to the y-axis, the x-axis, the origin, more than one of th

Murrr4er [49]

$\bf x^2y^2+3xy=1\\\\
-------------------------------\\\\
\stackrel{\stackrel{\textit{test for x-symmetry}}{y=-y}}{x^2(-y)^2+3x(-y)=1}\implies x^2y^2-3xy=1\impliedby 
\begin{array}{llll}
\textit{function differs}\\
\textit{from original}\\
\textit{no dice}
\end{array}\\\\
-------------------------------\\\\$

$\bf \stackrel{\stackrel{\textit{test for y-symmetry}}{x=-x}}{(-x)^2y^2+3(-x)y=1}\implies x^2y^2-3xy=1\impliedby 
\begin{array}{llll}
\textit{function differs}\\
\textit{from original}\\
\textit{no dice}
\end{array}\\\\
-------------------------------\\\\
\stackrel{\stackrel{\textit{test for origin-symmetry}}{x=-x~~y=-y}}{(-x)^2(-y)^2+3(-x)(-y)=1}\implies x^2y^2+3xy=1\impliedby 
\begin{array}{llll}
origin\\symmetry
\end{array}$

so, recall, the function has symmetry when the yielded resulting function resembles the original function, after negativizing the variable(s).

Also recall that minus*plus is minus, and minus*minus is plus.

7 0

3 years ago

Answers

Flura [38]

Answer:

<em>c</em><em> </em>

Step-by-step explanation:

because the coordinate of x is negative

7 0

3 years ago

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nirvana33 [79]