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GaryK [48]
3 years ago
15

In the picture is de question

Mathematics
1 answer:
valentina_108 [34]3 years ago
8 0
2+(x+4)= ?
So it really is: 2+1(x+4)=0
You distribute the 1 to the x+4
2+ 1x +4= 0
Combine like terms
1x+ 6=0
Subtract six on both sides
1x=-6
Divide by 1 on both sides
x=-6
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Answer:

y= $25-$10.50

Step-by-step explanation:

5 0
3 years ago
Which of the following statements is true
konstantin123 [22]
Hey!

The answer is option c).

√9 is a rational number as √9 = 3..

while √10 is an irrational number..

Hope it helps^^

5 0
3 years ago
(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
Translate the phrase into an algebraic expression.The product of 4 and b
Korvikt [17]

Answer:

the answer is 4×b=4b

Step-by-step explanation:

product=×

4×b=4b

6 0
2 years ago
What is 49 divided by 3 plus 87
ozzi

Answer: 103.3

Step-by-step explanation:

Divide 49/3 and u get 16.3333333333. Then add 16.3333333333 and 87 and get 103.333333333. Round 103.333333333 to the nearest tenth and get 103.3

Hope this helps! #BaconSquad

4 0
3 years ago
Read 2 more answers
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