2 years ago

Helppppppppppppp 20 points

Mathematics

2 answers:

Over [174]2 years ago

5 0

The answer to that is 4 3/7

Kobotan [32]2 years ago

3 0

Answer:

4 3/7

Step-by-step explanation:

Divide 31 by 7 to the closest whole number. 28/7 is 4 and you have 3/7 left, so you get 4 3/7

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Which algebraic expression could NOT represent the phrase below?

skad [1K]

Answer:

Step-by-step explanation:

because 4 and 3 are grouped together you would add them first. this would change the outcome of the problem.

6 0

3 years ago

Read 2 more answers

Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G

Kaylis [27]

$\Rightarrow$

Suppose first that $H\subset G$ is a normal subgroup. Then by definition we must have for all $a\in H$ , $xax^{-1} \in H$ for every $x\in G$ . Let $a\in G$ and choose $(ab)\in aH$ ( $b\in H$ ). By hypothesis we have $aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H$ , i.e. $aba^{-1}=c$ for some $c\in H$ , thus $ab=ca \in Ha$ . So we have $aH\subset Ha$ . You can prove $Ha\subset aH$ in the same way.

$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .