If TWO^2 = THREE where the alphabets are single-digit integers then find T + W + O

Mathematics

1 answer:

Simora [160]2 years ago

4 0

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Anita does the same number of jumping jacks each day over a period of several days. On the fifth day, she has completed a total

poizon [28]

65 is the answer. m = slope = delta y / delta x
delta y = 520 - 325
delta x = 8 - 5

Be well.

7 0

3 years ago

Help 69 points

worty [1.4K]

Answer:

a) No Solution

Step-by-step explanation:

36 - 7p = -7(p - 5)

36 - 7p = -7p + 35

1 - 7p = -7p

1 = 0p

so a) No Solution

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2 years ago

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 .