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3 years ago

8

En una fabrica de zapatillas 15 obreros confeccionan unaa cierta cantidad de pares en 20 dias si se aumenta un 20% el numero de

obreros ¿cuantos dias tardaran en confeccionar la misma cantidad de pares?

1 answer:

Lady bird [3.3K]3 years ago

8 0

Answer:

sorry but i do not the answer

Step-by-step explanation:

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Danielle makes the claim that when the polynomial x^2-3x-10 is divided by x-5, the remainder is 0. Use what you have learned abo

Mumz [18]

Answer:

<h3>Daniel is correct</h3>

Step-by-step explanation:

Given the polynomial P(x) = x^2-3x-10

To check that the remainder is zero if divided by x - 5, we will first have to equate x - 5 to zero and get x;

x - 5 = 0

x = 5

Then find P(5)

P(5) = 5^2 - 3(5) - 10

P(5) = 25 - 15 - 10

P(5) = 25-25

P(5) = 0

<em>This shows that x - 5 is a factor of the polynomial since P(5) gave us zero according to the factor theorem</em>

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3 years ago

Explain why the expression 2y-y can be written as y

harina [27]

Answer:

2y is just 2 times y, so subtracted it would be 1y and 1y is just y.

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3 years ago

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A large bowl contains 10 small balls and each ball has a color and a number. The 10 balls are denoted by B1, B2, B3, B4, R1, R2,

nasty-shy [4]

Answer:

Size of N(S) = 90, or S=90

Step-by-step explanation:

So since there are 10 items:

Each single item has 9 possibilities to be paired with, since it is drawn without replacement, and 1 item can't be drawn again.

So : 10 items x 9 possibilites = array of 90

SO ANSWER : 90

Hope I helped :)

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3 years ago

9. Jonathan can rent an unfurnished apartment for

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The furnished apartment would be $760 altogether whereas the furnished apartment would be $1,800

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3 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 .

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3 years ago