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kirill [66]
3 years ago
8

The following is an incomplete paragraph proving that ∠WRS ≅ ∠VQT, given the information in the figure where segment UV is paral

lel to segment WZ.:
Segments UV and WZ are parallel with line ST intersecting both at points Q and R respectively

According to the given information, segment UV is parallel to segment WZ while angles SQU and VQT are vertical angles. Angle VQT is congruent to angle SQU by the Vertical Angles Theorem. Because angles SQU and WRS are corresponding angles, they are congruent according to the Corresponding Angles Postulate. Finally, angle VQT is congruent to angle WRS by the _____________________.

Which Property of Equality accurately completes the proof?

Reflexive
Substitution
Subtraction
Transitive
Mathematics
1 answer:
Lorico [155]3 years ago
6 0
Answer: Transitive Property of Equality
The Subtraction Property does not apply here, and the Reflexive Property is used to show something is equal to itself (a = a).

Given that ∠VQT is congruent to ∠SQU by the Vertical Angles Theorem and ∠SQU is congruent to ∠WRS by the Corresponding Angles Postulate, ∠VQT is congruent to ∠WRS because of the Transitive Property of Equality.

Basically, Angle 1 = Angle 2, Angle 2 = Angle 3, therefore Angle 1 should also equal Angle 3.
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dangina [55]
No, she is incorrect (in my opinion) i think the answer is :
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We have

\dfrac{x^2}{x+1}=\dfrac{(x+1)^2-2(x+1)+1}{x+1}=(x+1)-2+\dfrac1{x+1}=x-1+\dfrac1{x+1}

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\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-ax-b\right)=\lim_{x\to\infty}\left(x-1+\frac1{x+1}-ax-b\right)=0

The rational term vanishes as <em>x</em> gets arbitrarily large, so we can ignore that term, leaving us with

\displaystyle\lim_{x\to\infty}\left((1-a)x-(1+b)\right)=0

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To confirm, we have

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