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andre [41]
2 years ago
6

Drag the tiles to the boxes to form correct pairs. Not all tiles will be used. Match the resulting values to the corresponding l

imits.

Mathematics
1 answer:
belka [17]2 years ago
7 0

The correct solution to the limits of x in the tiles can be seen below.

  • \mathbf{ \lim_{x \to 9^+} (\dfrac{|x-9|}{-x^2-34+387}) }\mathbf{ = -\dfrac{1}{52} }
  • \mathbf{ \lim_{x \to 8^-} (\dfrac{8-x}{|-x^2-63x+568|}) }\mathbf{=\dfrac{1}{79} }
  • \mathbf{ \lim_{x \to 7^+} (\dfrac{|-x^2-17x+168| }{x-7}) }= -31
  • \mathbf{ \lim_{x \to 6^-} (\dfrac{|x-6| }{-x^2-86x+552}) }\mathbf{ =\dfrac{1}{98}}

<h3>What are the corresponding limits of x?</h3>

The limits of x approaching a given number of a quadratic equation can be determined by knowing the value of x at that given number and substituting the value of x into the quadratic equation.

From the given diagram, we have:

1.

\mathbf{ \lim_{x \to 9^+} (\dfrac{|x-9|}{-x^2-34+387}) }

So, x - 9 is positive when x → 9⁺. Therefore, |x -9) = x - 9

\mathbf{ \lim_{x \to 9^+} (\dfrac{x-9}{-x^2-34+387}) }

Simplifying the quadratic equation, we have:

\mathbf{ \lim_{x \to 9^+} (-\dfrac{1}{x+43}) }

Replacing the value of x = 9

\mathbf{ = (-\dfrac{1}{9+43}) }

\mathbf{ = -\dfrac{1}{52} }

2.

\mathbf{ \lim_{x \to 8^-} (\dfrac{8-x}{|-x^2-63x+568|}) }

  • -x²-63x+568 is positive when x → 8⁻.

Thus |-x²-63x+568| = -x²-63x+568

\mathbf{ \lim_{x \to 8^-} (\dfrac{1}{x+71}) }

\mathbf{=\dfrac{1}{8+71} }

\mathbf{=\dfrac{1}{79} }

3.

\mathbf{ \lim_{x \to 7^+} (\dfrac{|-x^2-17x+168| }{x-7}) }

  • x -7 is positive, therefore |x-7| = x - 7

\mathbf{ \lim_{x \to 7^+} (\dfrac{-x^2-17x+168 }{x-7}) }

\mathbf{ \lim_{x \to 7^+} (-x-24)}

\mathbf{ \lim_{x \to 7^+} (-7-24)}

= -31

4.

\mathbf{ \lim_{x \to 6^-} (\dfrac{|x-6| }{-x^2-86x+552}) }

  • x-6 is negative when x → 6⁻. Therefore, |x-6| = -x + 6

\mathbf{ \lim_{x \to 6^-} (\dfrac{-x+6 }{-x^2-86x+552}) }

\mathbf{ \lim_{x \to 6^-} (\dfrac{1}{x+92}) }

\mathbf{ \lim_{x \to 6^-} (\dfrac{1}{6+92}) }

\mathbf{ =\dfrac{1}{98}}

Learn more about calculating the limits of x here:

brainly.com/question/1444047

#SPJ1

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