Please HELP IM FALLING MATH PLEASE HELP

Mathematics

1 answer:

Ann [662]2 years ago

8 0

Answer:

Step-by-step explanation:

32+35=67

90-67=23

it won't let me type in the comments for some reason but i got 90 because andgle MLQ is a right angle so it is 90 degrees

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2/3 divided by 1/4 <br> as a fraction! HURRY!!!

a_sh-v [17]

Answer:

2 2/3

Step-by-step explanation:

7 0

2 years ago

? divided by 2/3= 3/8

NeX [460]

When you divide by a fraction all you have to is multiply by the reciprocal. The reciprocal of 3/8 is 8/3.
<span>The real problem is 2/3 TIMES 8/3. Multiply the numerators across 2 X 8 = 16. Then multiply the denominators across 3 X 3 = 9. The answer is 16/9. If you need to simplify then the answer is 1 7/9.</span>

3 0

2 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 .