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vesna_86 [32]
3 years ago
14

Help ! ASAP doing an exam

Mathematics
1 answer:
Alona [7]3 years ago
8 0

Answer:

ok ill explain if want after test

Step-by-step explanation:

angle 2 and 4 are 50

angle 1 and 3 are 130

angle 5 and 7 are 50

angle 8 and 5 are 130

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(a) If G is a finite group of even order, show that there must be an element a = e, such that a−1 = a (b) Give an example to sho
Dahasolnce [82]

Answer:

See proof below

Step-by-step explanation:

First, notice that if a≠e and a^-1=a, then a²=e (this is an equivalent way of formulating the problem).

a) Since G has even order, |G|=2n for some positive number n. Let e be the identity element of G. Then A=G\{e} is a set with 2n-1 elements.

Now reason inductively with A by "pairing elements with its inverses":

List A as A={a1,a2,a3,...,a_(2n-1)}. If a1²=e, then we have proved the theorem.

If not, then a1^(-1)≠a1, hence a1^(-1)=aj for some j>1 (it is impossible that a^(-1)=e, since e is the only element in G such that e^(-1)=e). Reorder the elements of A in such a way that a2=a^(-1), therefore a2^(-1)=a1.

Now consider the set A\{a1,a2}={a3,a4,...,a_(2n-1)}. If a3²=e, then we have proved the theorem.

If not, then a3^(-1)≠a1, hence we can reorder this set to get a3^(-1)=a4 (it is impossible that a^(-1)∈{e,a1,a2} because inverses are unique and e^(-1)=e, a1^(-1)=a2, a2^(-1)=a1 and a3∉{e,a1,a2}.

Again, consider A\{a1,a2,a3,a4}={a5,a6,...,a_(2n-1)} and repeat this reasoning. In the k-th step, either we proved the theorem, or obtained that a_(2k-1)^(-1)=a_(2k)

After n-1 steps, if the theorem has not been proven, we end up with the set A\{a1,a2,a3,a4,...,a_(2n-3), a_(2n-2)}={a_(2n-1)}. By process of elimination, we must have that a_(2n-1)^(-1)=a_(2n-1), since this last element was not chosen from any of the previous inverses. Additionally, a_(2n1)≠e by construction. Hence, in any case, the statement holds true.

b) Consider the group (Z3,+), the integers modulo 3 with addition modulo 3. (Z3={0,1,2}). Z3 has odd order, namely |Z3|=3.

Here, e=0. Note that 1²=1+1=2≠e, and 2²=2+2=4mod3=1≠e. Therefore the conclusion of part a) does not hold

7 0
3 years ago
Frank is trying to factor y^2+6y-27 . He has determined that one factor is (y + 9). What is the other factor? A. y – 27 B. y – 6
beks73 [17]
The other factor is (y - 3).  If you multiply the two terms together, you will end up with y^2+6y-27:

(y - 3)(y + 9) \\ y^{2} +9y-3y-27 \\ y^2+6y-27
5 0
3 years ago
Read 2 more answers
How is the point (2, 9) in the scatterplot described?
AnnZ [28]
Im not sure but u can search up the name and see online i would help if i could i hope the advise i gave helps just seach up if no one can help im so sorry i wanted to help but idk this stuff yet sorry
7 0
3 years ago
Triangle ABC has vertices A(–2, –3), B(–5, –3), and C(–4, –2). The triangle is rotated 90° counterclockwise around the origin.
Ulleksa [173]
The rule for this rotation is (x, y) converted into (-y, x). So, to find the answer you would make the y negative and switch the values in the ordered pair. A would be (3, -2), B would be (3, -5), and C would be (2, -4). So, your answer would be B. Hope this helps!
4 0
3 years ago
Read 2 more answers
The table shoes the results if a 100-meter race including the time in seconds for the first three finishers. The first-place fin
Luden [163]
For the answer to the question above asking <span>how much faster than the second-place finisher? 1st place got 13.89 and 2nd place got 14.02.
Based on your question the
1st place is    13.89
2nd place is  14.02
Just get the difference

The 1st place is 0.13 milliseconds faster than the second. and that was too close.</span>
6 0
3 years ago
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