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Mathematics 16 Online
OpenStudy (anonymous):

Can someone explain level sets and help me figure this question out?

OpenStudy (anonymous):

Question:

OpenStudy (anonymous):

@wio can you help? been waiting for an hour now for someone to help.

OpenStudy (anonymous):

Let \(z=1\) and mess with the equation to try to find out what sort of equation it is.

OpenStudy (anonymous):

Also \(z=1\) works as well.

OpenStudy (anonymous):

What do you mean by mess with the equation?

OpenStudy (anonymous):

Try to get it into a recognizable form.

OpenStudy (anonymous):

like \[ ax^2+by^2=c^2 \]This is an ellipse

OpenStudy (anonymous):

\[ y=ax^2+bx+c \]This is a parabola

OpenStudy (anonymous):

\[ y=mx+b \]A line

OpenStudy (anonymous):

So the first one: \[ xy+z^2=1\\ \]Well \[ xy=1-z^2 \]And \(1-z^2\) is a constant so:\[ xy=C\implies y=\frac Cx \]

OpenStudy (anonymous):

This is not on the list.

OpenStudy (anonymous):

Next one \[ z=\frac{1}{x+y-1} \] \[ \frac 1z=x+y-1 \] \[ \frac{z+1}{z}=x+y \]This ends up being: \[ y+x=C\implies y=-x+C \]This is a line

OpenStudy (anonymous):

Since the slope will always be \(-1\) it's parallel lines.

OpenStudy (anonymous):

ok I see thanks

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