Write with only positive exponents. (3x^2 y) (-2x^-2 y) I got -6x^2 y^2 But the teacher got -6y^2 I don't understand how he got that
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Originally Posted by Candy101 Write with only positive exponents. (3x^2 y) (-2x^-2 y) $\displaystyle (3x^2 y) (-2x^{-2} y)=(3)(-2)(x^{2-2})(y^{1+1})=-6y^2$
Oh I see Wht I did wrong. I forgot about the x^-2 . Which cancels out! Thanks (2x^-2 y)^3 / -2x^-2 y^-3 For this one I can't understand why he got -4y^6/x^4 I got -4x^-4 y^-6 I know I have to flip it So y can't I get y^6 / 4x^4
Originally Posted by Candy101 Oh I see Wht I did wrong. I forgot about the x^-2 . Which cancels out! Thanks (2x^-2 y)^3 / -2x^-2 y^-3 For this one I can't understand why he got -4y^6/x^4 I got -4x^-4 y^-6 I know I have to flip it So can't I got y^6 / 4x^4 Why not learn latex, so that we can easily comprehend what you're trying to say? Is this the initial expression: $\displaystyle \frac{(2x^{-2}y)^3}{-2x^{-2}y^{-3}}$
Originally Posted by Candy101 (2x^-2 y)^3 / -2x^-2 y^-3 For this one I can't understand why he got -4y^6/x^4 Why not learn to use some basic LaTeX code. [TEX]\frac{(x^{-2}y)^3}{(-2x^{-2}y^{-3})}[/TEX] gives $\displaystyle \frac{(x^{-2}y)^3}{(-2x^{-2}y^{-3})}$ $\displaystyle \frac{(x^{-2}y)^3}{(-2x^{-2}y^{-3})}=-\frac{y^6}{2x^4}$
Yeah
Originally Posted by Candy101 Oh I see Wht I did wrong. I forgot about the x^-2 . Which cancels out! Thanks (2x^-2 y)^3 / -2x^-2 y^-3 For this one I can't understand why he got -4y^6/x^4 I got -4x^-4 y^-6 I know I have to flip it So y can't I get y^6 / 4x^4 (2x^-2 y)^3= 8x^-6y^3. The denominator, 1/(-2x^-2y^-3), becomes (-1/2)x^2y^3 8(-1/2)= -4. x^-6(x^2)= x^{-4}. y^3(y^3)= y^6 So that is -4x^{-4}y^6 which, using only positive exponents, is -4y^6/x^4. Again, 8/-2 is -4, not -1/4.
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