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Zepler [3.9K]
2 years ago
15

Pls help urgently extra points and mark brainlist

Mathematics
2 answers:
son4ous [18]2 years ago
8 0

Answer:

6 to the power of one

Step-by-step explanation:

bdhsbabasjbsbesjwushbe

andreyandreev [35.5K]2 years ago
7 0

Answer:

Step-by-step explanation:

6^1 = 6 x 1 = 6

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What I use the Pythagorean theorem for this problem?
andrew-mc [135]
No you would not use the Pythagorean Theorem, You would take the two sides and multiply them together, and divide it by 2.
8 0
3 years ago
What is the value of x in the equation 2(x+3) = 4(x-1)?<br> own
11111nata11111 [884]

Step-by-step explanation:

Solving

2x + 6 = 4x - 4

Bringing like terms on one side

6 + 4 = 4x - 2x

10 = 2x

10 / 2 = x

5 = x

8 0
3 years ago
A pack of 9 toilet rolls costs £4.23
OleMash [197]
 a)  9 toilet rolls cost $ 4.23
       1 toilet roll costs  $ x

          9x = 4.23 * 1
           x= 4.23 / 9
            x = 0.47
So in first case one toilet roll costs $ 0.47.

b) 4 toilet rolls cost $1.96
    1 toilet roll costs  $ x
     
    4x = 1.96 * 1
     x = 1.96 / 4
     x = 0.49
So in second case one toilet roll costs $ 0.49.

0.47 < 0.49
The toilet roll in first case is cheaper than the toilet roll in second case.

Hope this heps!


8 0
3 years ago
Give the domain and range.
Ymorist [56]

Answer:Graphs of inverse functions have a domain and range just like any other graph of a function. The domain of an inverse function is the range of the original, and the range of an inverse function is the domain of an original.

Step-by-step explanation:

7 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
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