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Nesterboy [21]
3 years ago
8

What is the solution to the inequality |x-4|< 3? –7 < x < –1 1 < x < 7 x < –7 or x < –1 x > 1 or x <

7
Mathematics
1 answer:
balandron [24]3 years ago
5 0

|x-4| < 3\iff x-4 < 3\ \wedge\ x-4 > -3\ \ \ \ |+4\\\\x < 7\ \wedge\ x > 1\\\\Answer:\ \boxed{1 < x < 7}

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The answer would be 4. It would be 4 because 4x16=64. And 0-4x3=-12.
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Please help me on this problem !
Rashid [163]

Answer:

x = 5 , y = 5\sqrt{3}

Step-by-step explanation:

Using the sine and cosine ratios in the right triangle and exact values.

sin30° = \frac{1}{2} , cos30° = \frac{\sqrt{3} }{2} , then

sin30°= \frac{opposite}{hypotenuse} = \frac{x}{10} = \frac{1}{2} ( cross- multiply )

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and

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3 years ago
Identify the oblique asymptote of f(x) = quantity x plus 4 over quantity 3 x squared plus 5 x minus 2.
faltersainse [42]
f(x) = \frac{x + 4}{3x^{2} + 5x - 2}

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                3x² + 6x - x - 2 = 0
3x(x) + 3x(2) - 1(x) - 1(2) = 0
          3x(x + 2) - 1(x + 2) = 0
                  (3x - 1)(x + 2) = 0
3x - 1 = 0       or       x + 2 = 0
    + 1 + 1                     - 2  - 2 
     3x = 1          or          x = -2
      3     3                       1     1
       x = ¹/₃         or         x = -2

      f(x) = 3x² + 5x - 2
     f(¹/₃) = 3(¹/₃)² + 5(¹/₃) - 2
     f(¹/₃) = 3(¹/₉) + 1²/₃ - 2
     f(¹/₃) = ¹/₃ - ¹/₃
     f(¹/₃) = 0
(x, f(x)) = (¹/₃, 0)

      f(x) = 3x² + 5x - 2
     f(-2) = 3(-2)² + 5(-2) - 2
     f(-2) = 3(4) - 10 - 2
     f(-2) = 12 - 12
     f(-2) = 0
(x, f(x)) = (-2, 0)

    Vertical Asymptotes: ¹/₃ or -2
Horizontal Asymptotes: 0
      Oblique Asymptote: No Asymptotes
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