Ashish had a box of 117 books.

Mathematics

1 answer:

il63 [147K]2 years ago

4 0

117 - 13 = 114 books in each box

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Need math help please!!

s2008m [1.1K]

Just by looking at it, 2 seems to be the correct answer. I mean if you think about it, when the triangle is split in two smaller triangles, side LN is longer than LP so that means x needs to be a small number. I could either be 2 or 1. Well if you look at the big picture, side LM is longer than LN so it’s gonna need to be bigger. If you say x=1 then you are saying 2(1) +4 = 6 (therefore LM is 6) but if you say x=2 then you are saying 2(2) +4 = 8 which makes more sense

If x=2 then you can use the Pythagorean’s theorem

a2 + b2 = c2
4^2 + b^2 = 5^2
16 + b^2 = 25
25-16= 9
b^2= 9
b= 3

Now that we know what NP is we can do Pythagorean’s theorem to figure out what side LP is (we already know it but you can do it again to check lol)

So the answer is 2

5 0

3 years ago

Is this equation a linear function? If so give the intersepts 12x+3y=6

Katarina [22]

Dear Mendozaaj55, 12x + 3y = 6 is a linear function. To find the y-intercept set x = 0 and solve --> 3y = 6 --> y = 2. Therefore, the y-intercept is (0,2). To find the x-intercept, set y = 0 and solve. 12x = 6 ---> x = .5---> the x-intercept is the coordinate (.5, 0). Best.

7 0

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 .