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Q: What is 2x squared plus 6 equals 2?

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(2x + 1)(2x + 1) or (2x + 1)2 so x = -0.5

They intersect at the point of: (-3/2, 11/4)

2(x - 2)(x - 4)

(x - 3)(x + 5)

3(2x - 1)(x - 2)

2x + 52 = 7 2x + 25 = 7 2x = 7 - 25 2x = -18 x = -18/2 x = -9

(2x + 1)(x + 2)

Add the x's squared together to get 2x^2+x. You cannot add anymore so the answer is 2x^2+x

If: y = x^2 -2x +4 and y = 2x^2 -4x +4 Then: 2x^2 -4x +4 = x^2 -2x +4 Transposing terms: x^2 -2x = 0 Factorizing: (x-2)(x+0) => x = 2 or x = 0 Therefore by substitution points of intersect are at: (2, 4) and (0, 4)

If: 2x+y = 5 and x2-y2 = 3 Then the solutions work out as: (2, 1) and ( 14/3, -13/3)

If: y = 4x^2 -2x -1 and y = -2x^2 +3x+5 Then: 4x^2 -2x-1 = -2x^2 +3x+5 or 6x^2 -5x -6 = 0 Factorizing the above: (3x+2)(2x-3) = 0 meaning x = -2/3 or x = 3/2 By substitution points of contact are at: (-2/3, 19/9) and (3/2, 5)

x^2+x^2=2x^2

48

It works out that line 3x-y = 5 makes contact with the curve 2x^2 +y^2 = 129 at (52/11, 101/11) and (-2, -11)

2x2 + 2x - 8 = 2(x2 + x - 4). This can take any value greater than or equal to -8.5

That factors to (x - 2)(x + 4) x = 2, -4

x^2+2x^2=3x^2

8x-7 plus 2x equals 6 plus 5x plus 2?

Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C

104

If: y = 4x^2 -2x -1 and y = -2x^2 +3x +5 Then: 4x^2 -2x -1 = -2x^2 +3x +5 => 6x^2 -5x -6 = 0 Solving the above quadratic equation: x = -2/3 or x = 3/2 Therefore by substitution the points of intersection are: (-2/3, 19/9) and (3/2, 5)

If: y = 4x^2 -2x -1 and y = -2x^2+3x+5 Then: 4x^2 -2x -1 = -2x^2+3x+5 => 6x^2-5x-6 = 0 Solving the above quadratic equation: x = -2/3 or x = 3/2 Therefore the points of intersection by substitution are: (-2/3, 19/9) and (3/2, 5)

If: y = 4x^2 -2x -1 and y = -2x^2 +3x +5 Then: 4x^2-2x-1 = -2x^2+3x+5 =>6x^2-5x-6 = 0 Solving the above quadratic equation: x = -2/3 or x = 3/2 Therefore by substitution the points of intersection are: (-2/3, 19/9) and (3/2, 5)

If: y = -2x^2 +3x +5 and y = 4x^2 -2x -1 Then: 4x^2 -2x -1 = -2x^2 +3x +5 So it follows: 6x^2 -5x -6 = 0 Using the quadratic equation formula: x = -2/3 or x = 3/2 Therefore points of intersection by substitution are at: (-2/3, 19/9) and (3/2, 5)

If: y = 4x^2 -2x -1 and y = -2x^2 +3x +5 Then: 4x^2 -2x -1 = -2x^2 +3x +5 Transposing terms: 6x^2 -5x -6 = 0 Factorizing: (3x+2)(2x-3) = 0 => x = -2/3 or x = 3/2 By substitution the points of contact are at: (-2/3, 19/9) and (3/2, 5)