x2+3y2+4z2+2√3xy+4√3yz+4xz factorise
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question the diagonals of the rhombus are 16cm 24 cm . find the area of the rhombus
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Step-by-step explanation:
x²+3y²+4z²+2(3)½xy+4(3)½yz+4xz
(x)²+((3)½y)+(2z)²+2[(3)½y)(x)+(2z)((3)½y)+(2z)(x)
By applying the identity
a²+b²+c²+2(ab+bc+ca)=(a+b+c)²
Therefore,
x²+3y²+4z²+2(3)½xy+4(3)½yz+4xz=(x+(3)½y+2z)²
I don't have the under root sign.
Sorry for the inconveniency
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