`5(x+3)=20`

Solve the equations and check your answer in the case: (x+5)/(2)+(x)/(3)=20

Factorise : ( 5x -3) ^(2) - ( 5x -3) - 20

Without expanding at any stage, show that Delta=|{:(x^2+x+2,2x^2+3x+1,3x^2+5x+3),(20x^2+20x+59,40x^2+60x+20,60x^2+100x+70),(2x^2+2x+6,4x^2+6x+2,6x^2+10x+7):}|=x^2A+xB+C , where A and B are determinants of order 3 not involving x….

Solve 3x + 2(x + 2) = 20 - (2x - 5)

Solve : 3x + 2 (x+2) = 20 - (2x -5)

3x + 2 (x + 2) = 20- (2x-5)

Factorize the following : (1) 3x-9 (2) 5x-15 x^2 (3) 20 a^(12)b^2-15 a^8b^4 (4) 72 x^6y^7-96 x^7y^6

If (x+5) is a factor of p(x)=x^3-20x+5k then k=?

If (x+5) is a factor of p(x)=x^3-20x+5k , then k=

Supply the two missing figures in order indicated by x and y in the given equation,the fractions being in their lowest terms.5(1)/(x)xy(3)/(4)=20( a) 3,1 (b) 3,3 (c) 4,1 (d) 5,3