3 years ago

The sum of five consecutive integers is five. What is the product of the five integers?

Mathematics

1 answer:

nadya68 [22]3 years ago

3 0

Answer:Let n = the smallest integer

Then the other integers are n+1, n+2, n+3 and n+4

Sum = n + n+1 + n+2 + n+3 + n+4 = 5n + 10 = 5(n + 2)

So the sum is always divisible by 5, for any value of n

Step-by-step explanation:

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Given f(x)=(x+5)(x-2), what are the x-intercepts?

Yuri [45]

Answer:

The intercepts are -5 and 2

Step-by-step explanation:

Using the zero product property

f(x)=(x+5)(x-2)

0 =(x+5)(x-2)

0 = (x+5) 0=(x-2)

x = -5 x = 2

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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 .