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Sliva [168]
3 years ago
10

Figure RST on the graph is reflected over the y-axis creating figure R’S’T’. Figure R’S’T’ is translated down 2 units and left 1

unit creating R’’S’’T’’. On a coordinate plane, triangle R S T has points (1, 1), (2, 3), (1, 4). Triangle R prime S prime T prime has points (negative 1, 1), (negative 2, 3), (negative 1, 4). Triangle R double-prime S double-prime T double-prime has points (negative 2, negative 1), (negative 3, 1), (negative 2, 2). Tyler knows that RST Is congruent to R’S’T’ because reflections produce congruent figures. He also knows that R’S’T’ Is congruent to R’’S’’T’’ because translations produce congruent figures. What can Tyler determine based on these two statements? Check all that apply. RST Is congruent to R’’S’’T’’ RS Is congruent to ST Is congruent to TR Angle R is congruent to angle S is congruent to angle T Angle R is congruent to angle R prime is congruent to angle R double-prime TS Is congruent to T’S’ Is congruent to T’’S’’
Mathematics
2 answers:
oee [108]3 years ago
4 0

For those who like the short answer, it is A and D

Crazy boy [7]3 years ago
3 0

Answer:

RST Is congruent to R’’S’’T’’

Angle R is congruent to angle R prime is congruent to angle R double-prime

TS Is congruent to T’S’ Is congruent to T’’S’’

Step-by-step explanation:

we know that

A reflection and a translation are rigid transformation that produce congruent figures

If two or more figures are congruent, then its corresponding sides and its corresponding angles are congruent

In this problem

Triangles RST, R'S'T and R''S''T'' are congruent

That means

Corresponding sides

RS≅R'S'≅R''S''

ST≅S'T'≅S''T''

RT≅R'T'≅R''T''

Corresponding angles

∠R≅∠R'≅∠R''

∠S≅∠S'≅∠S''

∠T≅∠T'≅∠T''

therefore

RST Is congruent to R’’S’’T’’

Angle R is congruent to angle R prime is congruent to angle R double-prime

TS Is congruent to T’S’ Is congruent to T’’S’’

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1,436.8

Step-by-step explanation:

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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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You <em>multiply</em> the number of people by the number of each slices each person gets. Which is 100; 50x2.

Each cake has 20 slices so you <em>divide</em> 100 by 20. Which equals 5.

It would take 5 cakes to feed all 50 people 2 slices each.

<em>Hope this helps</em>. :)

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