$\text{The domain}\\2x-6\neq0\ \wedge\ 6x-18\neq0\to x\neq3\\\\\dfrac{x^2}{2x-6}=\dfrac{9}{6x-18}\\\\\dfrac{x^2}{2(x-3)}=\dfrac{9}{6(x-3)}\ \ \ |\cdot6\\\\\dfrac{3x^2}{x-3}=\dfrac{9}{x-3}\iff3x^2=9\ \ \ |:3\\\\x^2=3\to x=\pm\sqrt3$

4 0

3 years ago

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If you have 2 squares with the same side lengths are they still considerd scaled copies of each other?

Mekhanik [1.2K]

Yes as long as one side on a square is the same on another they are the copies of each other

7 0

3 years ago

The points A(12,7), B(12,−3), C(−2,−3), and D(−2,7) form rectangle ABCD. Which point is halfway between A and B? CLEAR CHECK (12

Flura [38]

We have that
coordinates are
A(12,7), B(12,−3), C(−2,−3), and D(−2,7)

i use the formula of midpoint between A and B
Xm=(x1+x2)/2 Ym=(y1+y2)/2
A(12,7) B(12,−3)
Xm=(12+12)/2=12
Ym=(7+(-3))/2=2

the point (12,2) is halfway between A and B

4 0

3 years ago