1 = (1 + 4) ÷ (2 + 3) 2 = 4 - 3 + 2 - 1 3 = (4 + 3 - 1) ÷ 2 4 = (4 + 3 + 1) ÷ 2 5 = (4 x 2 - 3) x 1 6 = 4 + 3 - 2 + 1 7 = (4 + 3) x (2 - 1) 8 = 4 + 3 + 2 - 1 9 = 4 + 3 + 2 x 1 10 = 4 + 3 + 2 + 1 11 = 4 x 2 + 3 x 1 12 = 4 x 3 x (2 - 1) 13 = 3 x 4 + 2 - 1 14 = 3 x 4 + 2 x 1 15 = 3 x 4 + 2 + 1
2/3 x 2 1/4 2/3 x 9/4 2 x 9 3 4 cross reduce 9/3=3 3/3=1 4/2=2 2/2=1 1/1 x 3/2 3/2= 1 1/2 hope i helped
Y = X+3, therefore: 3 1/2 4 1/2 5 1/2 7 1/4 7 3/4
Do you mean? 3 = 2 + X/4 multiply through by 4 4/1(3 = 2 + X/4)4/1 12 = 8 + X 4 = X -----------check in original equation 3 = 2 + 4/4 3 = 2 + 1 3 = 3 ----------checks now, if you meant 3 = (2+X)/4 some multiplying through 12 = 2 + X 10 = X you check this one
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1 = (1 + 4) ÷ (2 + 3) 2 = 4 - 3 + 2 - 1 3 = (4 + 3 - 1) ÷ 2 4 = (4 + 3 + 1) ÷ 2 5 = (4 x 2 - 3) x 1 6 = 4 + 3 - 2 + 1 7 = (4 + 3) x (2 - 1) 8 = 4 + 3 + 2 - 1 9 = 4 + 3 + 2 x 1 10 = 4 + 3 + 2 + 1 11 = 4 x 2 + 3 x 1 12 = 4 x 3 x (2 - 1) 13 = 3 x 4 + 2 - 1 14 = 3 x 4 + 2 x 1 15 = 3 x 4 + 2 + 1
(x^2+x-1/2)= x(x+1)-1/2 [x + (1 - square root of 3)/2][x + (1 + square root of 3)/2] = 0 Check it: x^2 + x/2 + (square root of 3)x)/2 + x/2 + 1/4 + (square root of 3)/4 - (square root of 3)x/2 - (square root of 3)/4 - 3/4 = 0 x^2 + x/2 + x/2 + [(square root of 3)x]/2 - [(square root of 3)x]/2 + (square root of 3)/4 - (square root of 3)/4 + 1/4 - 3/4 = 0 x^2 + x - 2/4 = 0 x^2 + x - 1/2 = 0 How to find this roots: Using the completing the square method: x^2 + x - 1/2 = 0 x^2 + x = 1/2 x^2 + x + 1/4 = 1/2 + 1/4 (x + 1/2)^2 = 3/4 x + 1/2 = (plus & minus)(square root of 3/4) x = -1/2 + (square root of 3)/2 x = - 1/2 - (square root of 3)/2
x2 + x + 1 = 0 ∴ x2 + x + 1/4 = -3/4 ∴ (x + 1/2)2 = -3/4 ∴ x + 1/2 = ± √(-3/4) ∴ x = - 1/2 ± (i√3) / 2 ∴ x = (-1 ± i√3) / 2
cube numbers next is 64 1 = 1 x 1 x 1 = 1³ 8 = 2 x 2 x 2 = 2³ 27 = 3 x 3 x 3 = 3³ then 64 = 4 x 4 x 4 = 4³
(x^2+x-1/2)= x(x+1)-1/2 [x + (1 - square root of 3)/2][x + (1 + square root of 3)/2] = 0 Check it: x^2 + x/2 + (square root of 3)x)/2 + x/2 + 1/4 + (square root of 3)/4 - (square root of 3)x/2 - (square root of 3)/4 - 3/4 = 0 x^2 + x/2 + x/2 + [(square root of 3)x]/2 - [(square root of 3)x]/2 + (square root of 3)/4 - (square root of 3)/4 + 1/4 - 3/4 = 0 x^2 + x - 2/4 = 0 x^2 + x - 1/2 = 0 How to find this roots: Using the completing the square method: x^2 + x - 1/2 = 0 x^2 + x = 1/2 x^2 + x + 1/4 = 1/2 + 1/4 (x + 1/2)^2 = 3/4 x + 1/2 = (plus & minus)(square root of 3/4) x = -1/2 + (square root of 3)/2 x = - 1/2 - (square root of 3)/2
2/3 x 2 1/4 2/3 x 9/4 2 x 9 3 4 cross reduce 9/3=3 3/3=1 4/2=2 2/2=1 1/1 x 3/2 3/2= 1 1/2 hope i helped
-1*-1 (-2+1)(-2+1) -2*-2+1*-2+1*-2+1*1 since multiplication is repeated addition then -1*-1*4+(-2)+(-2)+(1) let -1*-1=x -1*-1=-1*-1*4-2-2+1 x=x*4-4+1 x=4x-3 3=4x-x 3=3x 3/3=x 1=x thus -1*-1=x=1
Several ways - here are four methods. 1 + 2 - 3 + 4 = 4 (3 - 1) + (4 - 2) = 4 (3 - 1) x 4 ÷ 2 = 4 3 x 4 ÷ (2 + 1) = 4
Y = X+3, therefore: 3 1/2 4 1/2 5 1/2 7 1/4 7 3/4
1 x 1 2 x 2 3 x 3 4 x 4 ............. 1(2)=2
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Do you mean? 3 = 2 + X/4 multiply through by 4 4/1(3 = 2 + X/4)4/1 12 = 8 + X 4 = X -----------check in original equation 3 = 2 + 4/4 3 = 2 + 1 3 = 3 ----------checks now, if you meant 3 = (2+X)/4 some multiplying through 12 = 2 + X 10 = X you check this one