kat145234 kat145234
  • 25-02-2014
  • Mathematics
contestada

using special products what is (wx-y)^2

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priyambaksi
priyambaksi priyambaksi
  • 25-02-2014
i think that supposed to be (wx - y^2)^2?

If so, multiply using FOIL:

First: wx * wx = w^2x^2

Outside: wx * -y^2 = -wxy^2

Inside: -y * wx = -wxy^2

Last: -y^2 * -y^2 = y^4

Combine w^2x^2 - 2wxy^2 + y^4
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isaacrosado04
isaacrosado04 isaacrosado04
  • 29-10-2020

Answer:

I believe it is wx^2 - 2ywx + y^2

Step-by-step explanation:

seperated into (wx-y)(wx-y) and used FOIL

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