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Mathematical Formulae
(a + b)2 = a2 + b2 + 2ab
(a - b)2 = a2 + b2 - 2ab
(a + b)2 + (a - b)2 = 2(a2 + b2)
(a + b)2 - (a - b)2 = 4ab
a2 - b2 = (a + b) (a - b)
(a + b)3 = a3 + b3 + 3ab (a + b)
(a - b)3 = a3 - b3 - 3ab (a - b)
a3 + b3 = (a + b) (a2 - ab + b2)
a3 - b3 = (a - b) (a2 + ab + b2)
logax = y →ay = x
logaxp = p logax
logaxy = loga x + logay
why cant we find the value of 1 upon zero
ln(4x+19)=ln(2x+9)
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Here we will use the expansion method Firstly lim x-0 log a (1+x)/x firstly using log property we get: lim x-0 log a (1+x)-logx then we change the base of log i.e lim x-0 {l
sir/madam, i abdulla working as a maths teacher want to join ur esteemed organisation as a tutor how can i proceed i have created an account even pls guide me, thanks abdulla
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