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spayn [35]
4 years ago
8

Determine the missing information in the paragraph proof. Given: Lines a and c intersect at point S, creating 4 angles. Prove: C

orresponding angles are congruent. Lines a and c intersect at point S. Line c is diagonal and also contains points R, Q, P. Line a is horizontal and also contains points U, T. We are given that lines a and c intersect at point S. Translate line a down line c until point S reaches point Q. Call the new line through Q line b. Because translations preserve orientation, lines a and b are ________. Because translations preserve angle measure, ∠RSU ≅ ∠RQU'. For the same reason, ∠RST ≅ ∠RQT', ∠PSU ≅ ∠PQU', and ∠PST ≅ ∠PQT'. Each of the angle pairs are corresponding angles. Therefore, corresponding angles are congruent. parallel perpendicular congruent reflected
Mathematics
2 answers:
Alenkasestr [34]4 years ago
9 0

Answer:

©The answer is Parallel

Step-by-step explanation:

iren [92.7K]4 years ago
6 0

Answer:

Parallel

Step-by-step explanation:

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When we approach limits, we are finding values that are infinitesimally approaching this x-value. Essentially, we consider the approximate location that this root or limit appears. This is essential when it comes to taking Calculus, and finding the limit or rate of change of a function.

When we are attempting limits questions, there are several tests we attempt first.

1. Evaluate the limit by substituting the value of the x-value as it approaches the value (direct evaluation of a limit)
2. Rearrangement of the function, such that we can evaluate the limit.
3. (TRIGONOMETRIC PROPERTIES)
\lim_{x \to 0} (\frac{sinx}{x}) = 1
\lim_{x \to 0} (\frac{tanx}{x}) = 1
4. Using L'Hopital's Rule for indeterminate limits, such as 0/0, -infinity/infinity, or infinity/infinity.

For example:

1) \lim_{x \to 0}\frac{\sqrt{x} - 5}{x - 25}

We can do this using the first and second method.
<em>Method 1: Direct evaluation:</em>

Substitute x = 0 to the function.
\frac{\sqrt{0} - 5}{0 - 25}
= \frac{-5}{-25}
= \frac{1}{5}

<em>Method 2: Rearranging the function
</em>

We can see that x - 25 can be rewritten as: (√x - 5)(√x + 5)
By rewriting it in this form, the top will cancel with the bottom easily, and our limit comes out the same.

\lim_{x \to 0}\frac{(\sqrt{x} - 5)}{(\sqrt{x} - 5)(\sqrt{x} + 5)}
= \lim_{x \to 0}\frac{1}{(\sqrt{x} + 5)}}
= \frac{1}{5}

Every example works exactly the same way, and by remembering these criteria, every limit question should come out pretty naturally.
8 0
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