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olga_2 [115]
2 years ago
10

PLEASE HELP!!!! I will award brainliest!! Look at the figure shown below. A student made the table shown below to prove that PQ

is equal to RQ. 1. Provide the missing statement and justification in the proof. 2. Using complete sentences, explain why the proof would not work without the missing step.

Mathematics
1 answer:
Rama09 [41]2 years ago
4 0

Answer:

See below.

Step-by-step explanation:

1.

Statement 8. triangle SQR is congruent to triangle TQP

Reason 8. ASA

2.

The only way to prove those two sides are congruent is to first prove that the triangles that contain those sides are congruent. Then you can use CPCTC to prove those sides congruent.

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Using <u>backward reasoning</u> we want to show that <em>"We can never get nine 0's"</em>.

Step-by-step explanation:

Basically in order to create nine 0's, the previous step had to have all 0's or all 1's. There is no other way possible, because between any two equal bits you insert a 0.

If we consider two cases for the second-to-last step:

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We obtain nine 0's if all bits in the previous step were the same, thus all bit were 0's or all bits were 1's. If the previous step contained all 0's, then we have the same case as the current iteration step. Since initially the circle did not contain only 0's, the circle had to contain something else than only 0's at some point and thus there exists a point where the circle contained only 1's.

<u>There were 9 </u><u>1's</u><u>:</u>

A circle contains only 1's, if every pair of the consecutive nine digits is different. However this is impossible, because there are five 1's and four 0's (we have an odd number of bits!), thus if the 1's and 0's alternate, then we obtain that 1's that will be next to each other (which would result in a 1 in the next step). Thus, we obtained a contradiction and thus assumption that the circle contains nine 0's after iteratins the procedure is false. This then means that you can never get nine 0's.

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I will attach the missing drawing with the answer.

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