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adell [148]
3 years ago
9

In Exercises 1-4, determine whether the dilated figure or the original figure is closer to the center of dilation. Use the given

location of the center of dilation and scale factor k.
1. Center of dilation inside the figure; k = 3
Center of ditation inside the figure, k = 1/2
3. Center of dilation outside the figure: = 120%
4. Center of dilation outside the figure; k = 0.1
Mathematics
1 answer:
melamori03 [73]3 years ago
4 0

When the Center of dilation is inside the figure

  1. The original figure is closer to the center of dilation
  2. The dilated figure is closer to the the center of dilation

When the Center of dilation is outside the figure

    3. The original figure is closer to the the center of dilation

    4. The dilated figure is closer to the center of dilation

<em>The </em><em>center of dilation</em><em> is the fixed point from which the distances in a dilation are measured</em>

<em>The </em><em>scale factor </em><em>is ratio of the side lengths of an original figure or preimage to  the side lengths of the newly formed image</em>

Center of dilation is inside the figure

  1. Where the center of dilation is inside the figure, and the scale factor is larger than 1, k = 3 > 1,  we have;

The distance of a point on the dilated figure, including the distances from the center of dilation is 3 times the distances of points on the original image from the center of dilation

Therefore, the original figure has a shorter distance to and is therefore closer to the the center of dilation than the dilated figure

     2. Where the center of dilation is inside the figure, and the scale factor is a fraction between 0 and 1 k = 1/2, we have;

The distance of a point on the dilated figure, including the distances from the center of dilation is 1/2 times the distances of points on the original image from the center of dilation

Therefore, the dilated figure has a shorter distance to and is therefore closer to the the center of dilation than the original figure

Center of dilation outside the figure

     3. Given that the center of dilation is outside the figure and the scale factor is larger than 1, k = 120% = 120/100 = 1.2 > 1, we have;

The distance of the dilated figure from the center of dilation is 120% of the distance of the original figure from the center of dilation, therefore, the original figure is closer to the the center of dilation than the dilated figure

     4. Where the center of dilation is outside the figure and the scale factor is a fraction between 0 and 1, k = 0.1 < 1

The distance of the dilated figure from the center of dilation is only 0.1 times the distance of the original figure from the center of dilation, and therefore, the dilated figure is closer to the center of dilation

Learn more about scale factors and center of dilation here;

brainly.com/question/12162455

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salantis [7]

Answer:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]:                                                                     \displaystyle \lim_{x \to c} x = c

L'Hopital's Rule

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                    \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

We are given the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}

When we directly plug in <em>x</em> = 0, we see that we would have an indeterminate form:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle  \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}

Plugging in <em>x</em> = 0 again, we would get:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}

Substitute in <em>x</em> = 0 once more:

\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

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