Question

(1.1x-0.9y)^2-(0.9x-1.1y)^2​

Answer 1

Answer:

$(1.1x-0.9y)^{2}-(0.9x-1.1y)^{2}\\=4(x-y)(0.1x+0.1y)$

Step-by-step explanation:

$(1.1x-0.9y)^{2}-(0.9x-1.1y)^{2}\\=[(1.1x-0.9y)+(0.9x-1.1y)][(1.1x-0.9y)-(0.9x-1.1y)]\\=(1.1x-0.9y+0.9x-1.1y)(1.1x-0.9y-0.9x+1.1y)\\=[(1.1+0.9)x-(0.9+1.1)y][(1.1-0.9)x-(0.9-1.1)y]\\=(2x-2y)(0.2x+0.2y)\\=2(x-y)[2(0.1x+0.1y)]\\=4(x-y)(0.1x+0.1y)$

Therefore,

$(1.1x-0.9y)^{2}-(0.9x-1.1y)^{2}\\=4(x-y)(0.1x+0.1y)$

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Answer 2

0.4(x+y)(x-y)

Step-by-step explanation:

Given:

$(1.1x-0.9y)^2-(0.9x-1.1y)^2\\\\ We\ know\ that\ (a-b)^2=a^2+b^2-2ab\\\\So,\\\\(1.1x)^2+(0.9y)^2-2(1.1x)(0.9y)-[(0.9x)^2+(1.1y)^2-2(0.9x)(1.1y)]\\\\(1.1x)^2+(0.9y)^2-2(1.1x)(0.9y)-(0.9x)^2-(1.1y)^2+2(0.9x)(1.1y)]\\\\1.21x^2+0.81y^2-1.98xy-0.81x^2-1.21y^2+1.98xy\\\\1.21x^2+0.81y^2-0.81x^2-1.21y^2\\\\0.4x^2-0.4y^2\\\\0.4(x^2-y^)\\\\0.4(x+y)(x-y)$

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https://brainly.in/question/11958007