Q: What is 2 1 4 x 2?

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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

2x2 = x+1 => 2x2-x-1= 0 Divide all terms by 2: x2-1/2x-1/2 = 0 Completing the square: (x-1/4)2-1/2 = 0 (x-1/4)2-1/2-1/16 = 0 (x-1/4)2-9/16 = 0 (x-1/4)2 = 9/16 Square root both sides: x-1/4 = -/+ 3/4 x = 1/4 -/+ 3/4 x = -1/2 and x = 1

x2-x-30 = (x-1/2)2 - 1/4 -30 as (a+b)2=a2+2ab+b2 with a=x then b=-1/2 = (x-1/2)2 - 121/4 = [(x-1/2)-sqrt(121/4)] [ [(x-1/2)+sqrt(121/4)] as a2-b2=(a-b)(a+b) then = (x-1/2-11/2)(x-1/2+11/2) = (x-6)(x+5)

5x2 + x - 4 = 0 a = 5, b = 1, c = -4 The equation of the axis of the symmetry of a parabola is x = - b/2a, which also is the x-coordinate of the vertex. x = - b/2a = - 1/2 Find the y-coordinate: f(x) = 5x2 + x - 4 f(-1/2) = 5(-1/2)2 + (-1/2) - 4 = 5(1/4) - 1/2 - 4 = 5/4 - 1/2 - 4 = 4/2 - 4 = 2 - 4 = -2 Thus, the vertex is (-1/2, -2).

X^2 + X = 0 halve the linear term (1) and square it then add to both sides X^2 + X + 1/4 = 1/4 factor left; gather terms right (X + 1/2)^2 = 1/4 (X + 1/2)^2 - 1/4 = 0 (-1/2,-1/4) vector and the number 1/4 was added to both sides completing the square

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Use the chain rule:d/dx √(4 - x) = d/dx (4 - x)1/2= 1/2 (4 - x)-1/2 x d/dx (4 - x)= 1/2 (4 - x)-1/2 x -1= -1/2 (4 - x)-1/2 or -1/2√(4 - x)

(-9x^2/√x) + 4= [-9x^2/x^(1/2)] + 4= (-9x^2)[x^(-1/2)] + 4= -9x^[2 + (-1/2)] + 4= -9x^(2 - 1/2) + 4= -9x^(3/2) + 4= -9√x^3 + 4= -9√[(x^2)(x)] + 4= -9x√x + 4Or,(-9x^2/√x) + 4= [(-9x^2)(√x)/(√x)(√x)] + 4= [(-9x^2)(√x)/√x^2] + 4= [-9(x)(x)(√x)/x] + 4 simplify x= -9x√x + 4

(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

16(x+5)**2=1 (x+5)**2=1/16 (x+5)=2root(1/16) (x+5)=1/4 x=1/4-5 x= 1/4-20/4 x=-19/4 16(-19/4+5)**2=1 16(-19/4+20/4)**2=1 16(1/4)**2=1 16(1/16)=1 1=1

4 x -1 x -1 = 4 4 + (-1) + (-1) = 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

2*x^2 - x - 1 = 02*x^2 - x = 1[x^2 - x/2] = 1/2[x^2 - x/2 + (1/4)^2] = 1/2 + (1/4)^2(x - 1/4)^2 = 1/2 + 1/16 = 9/16x - 1/4 = +/- sqrt(9/16) = +/- 3/4x = 1/4 +/- 3/4x = -1/2 or x = 1.

1, 2, 4, 8: 1 x 8, 2 x 4, 4 x 2, 8 x 1

1 x 4, 2 x 2, 4 x 1

1 x 4, 2 x 2, 4 x 1 = 4

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

-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