Q: What is the answer to x-2 5?

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3x4 plus 5x3 plus x2 - 5 divided by x 2 =[(3x4) + (5x3) + (x2 - 5)]/x2 =(12 + 15 + x2 -5)/x2 =(27 - 5 + x2)/x2 =(22 + x2)/x2

x2 + 6x - 5 is a quadratic expression that can not be factored.

Implicit: x2 + 2y = 5 Explicit : y = (5 - x2)/2

x1:y1 = x2:y2 4:-2 = x2:5 x2 = (4*5)/-2 x2 = -10

x4 + 7x2 - 60 = x4 + 12x2 - 5x2 - 60 = x2(x2 + 12) - 5(x2 +12) = (x2 - 5)(x2 +12)

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3x4 plus 5x3 plus x2 - 5 divided by x 2 =[(3x4) + (5x3) + (x2 - 5)]/x2 =(12 + 15 + x2 -5)/x2 =(27 - 5 + x2)/x2 =(22 + x2)/x2

Final solution, x = 2 ----- WORK ----- (5, -3) (X2, 1) (X1, Y1) (X2, Y2) D = SQRT( [X1 - X2]2 + [Y1 - Y2]2 ) 5 = SQRT( [5 - X2]2 + [-3 - 1]2 ) 5 = SQRT( [5 - X2]2 + (-4)2 ) 5 = SQRT( [5 - X2]2 + 16 ) Since The square root of 25 is 5, we know that -> [5 - X2]2 + 16 = 25 Subtract 16 from both sides, [5 - X2]2 = 9 Which means that 5-X2 = 3 Because 32 is 9 5 - X2 = 3 subtract 5 from both sides -X2 = -2 multiply both sides by -1 X2 = 2 There you go... X2 = 2 (5, -3) (2, 5)

x2 + 6x + 5 = (x + 1)(x + 5)

x2 + 6x - 5 is a quadratic expression that can not be factored.

Implicit: x2 + 2y = 5 Explicit : y = (5 - x2)/2

10x + 5 unless you know what x & y are * * * * * Perhaps this answer will be of help: x3 + x2 + 5x + 5 = (x + 1)(x2 + 5). If you are willing to go to complex roots, then, x3 + x2 + 5x + 5 = (x + 1)(x2 + 5). = (x + 1)(x + i√5)(x - i√5); in which case, x = -1 or ±i√5.

x2 + 10x = 0 x2 + 10x + 25 = 25 (x + 5)2 = 25 x + 5 = +-5 x1 = 0 x2 =10

x1:y1 = x2:y2 4:-2 = x2:5 x2 = (4*5)/-2 x2 = -10

Given: x2 + 10x - 16 Let: x2 + 10x - 16 = 0 x2 + 10x - 16 = 0 ∴ x2 + 10x + 25 = 16 + 25 ∴ (x + 5)2 = 41 ∴ x = -5 ± √41 ∴ x2 + 10x - 16 = (x + 5 + √41)(x + 5 - √41)

x4 + 7x2 - 60 = x4 + 12x2 - 5x2 - 60 = x2(x2 + 12) - 5(x2 +12) = (x2 - 5)(x2 +12)

x2 - 6x + 5 = (x - 1)(x - 5)

If you have an equation of x2 plus x2, it is the same as saying 2 times x plus 2 times x. This can be written as 2x + 2x or x2 + x2. It is the same, then, as 4x. For example, if x equals 5: 2(5) + 2(5) = 10 + 10 = 20. 20 also equals 4(5).