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...10.. .....3/10 ] Now we have accomplished the identity matrix on the

**LHS**, and we are left with [1... -3/2........ -1/2 ] [1/5....-3/10..... 1/10] [-2...1 Answers · Education & Reference · 12/03/2014

**LHS**is -, RHS is + (3x-2) - (5x-2) - (x+1) - x = 4+3+6+2 Now...1 Answers · Education & Reference · 11/03/2014

**LHS**: cscx - sinx = (1/sinx) - sinx = (1 - sin^2 x) / sinx = cos^2 x / sinx = (cosx/sinx) * cosx = cosx cotx = RHS1 Answers · Education & Reference · 05/03/2014

...also equal] [ check, whether the answer is right or wrong if x=3/5 then,

**LHS**=(32)^x =(32)^3/5 =(2)^(5 x (3/5) ) [x sign ...2 Answers · Education & Reference · 27/02/2014

..., we will use u-substitution. u = x^2 + 1 and du = 2x dx. So the

**LHS**becomes, ∫(-0.5)(u^(-0.5)) du = -0.5 * u ^(1/2) / (1/2) = -(x^2 + 1...1 Answers · Education & Reference · 26/02/2014

... is very important to know. The coefficients of

**LHS**variables compose Matrix A and the RHS values...1 Answers · Education & Reference · 25/02/2014

Just write z= x+iy and equate

**LHS**and RHS to get answer.1 Answers · Education & Reference · 23/02/2014

...and a = -b - c ..........................(3) then (a + b + c)^3 = 0 too Expanding

**LHS**results in a^3 + 3a^2b + 3a^2c + 3ab^2 +6abc + 3ac^2 + b^3 + 3b^2c +3bc^2...1 Answers · Education & Reference · 22/02/2014

Just multiply out the brackets to demonstrate that the

**LHS**=RHS. The bit about negative areas having no ...2 Answers · Education & Reference · 22/02/2014

...b2y) / a2 we cross-multiply first a2(c1 - b1y) = a1(c2 - b2y) we expand both

**LHS**and RHS a2c1 -a2b1y = a1c2 - a1b2y we gather like terms -a2b1y + a1b2y...1 Answers · Education & Reference · 18/02/2014