solve using identities (x/2 + 3y/4) *(x/2 + 3y/4)
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(x/2+3y/4)*(x/2+3y/4)=(x/2+3y/4)^2
(a+b)^2= a^2+b^2+2ab
(x/2+3y/4)^2= (x/2)^2+(3y/4)^2+2*x/2*3y/4
= x^2/4+9y^2/16+3xy/4
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this is wrong
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here (x/2+3y/4) * (x/2+3y/4)
we can use ( a+b) (a+b) = a^2+b^2
(x/2)^2+(3y/4)^2
x^2/4 + 9y^2/16
4x^2 + 9y^2
-----------------
16
we can use ( a+b) (a+b) = a^2+b^2
(x/2)^2+(3y/4)^2
x^2/4 + 9y^2/16
4x^2 + 9y^2
-----------------
16
make me BRAINLIEST
This is wrong. (a+b)(a+b)=a^2+b^2+2ab. First of all, multiply "a" with "a" and "b" both. Then multiply "b" with "a" and "b" both. As a result, a^2+b^2+2ab is formed.
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