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

Drag The Tiles To The Boxes To Form Correct Pairs Not All Tiles Will Be Used Match The Resulting Values To The Corresponding Limits class=

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

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

What are the corresponding limits of x?

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.

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

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

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

Simplifying the quadratic equation, we have:

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

Replacing the value of x = 9

[tex]\mathbf{ = (-\dfrac{1}{9+43}) }[/tex]

[tex]\mathbf{ = -\dfrac{1}{52} }[/tex]

2.

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

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

Thus |-x²-63x+568| = -x²-63x+568

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

[tex]\mathbf{=\dfrac{1}{8+71} }[/tex]

[tex]\mathbf{=\dfrac{1}{79} }[/tex]

3.

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

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

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

[tex]\mathbf{ \lim_{x \to 7^+} (-x-24)}[/tex]

[tex]\mathbf{ \lim_{x \to 7^+} (-7-24)}[/tex]

= -31

4.

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

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

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

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

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

[tex]\mathbf{ =\dfrac{1}{98}}[/tex]

Learn more about calculating the limits of x here:

https://brainly.com/question/1444047

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