Ask question

Ask question

All categories

3 years ago

9

Reggie and jay go for a walk morning.reggie walks 2 1/4 miles. Jay walks 1 3/8 miles less than Reggie. What is the total distanc

e they walk every morning

1 answer:

jek_recluse [69]3 years ago

7 0

Jay walks 2 1/4 miles which if you double it so the denominators are e same for both 2 1/4 and 1 1/8 then your answer will be 18/8 (Jay) and 9/8 (Reggie). Meaning that Reggie walks have the distance Jay walks.

You might be interested in

Find the product of (x − 5)^2.

Alex17521 [72]

(x - 5)^2 =
(x - 5)(x - 5) =
x^2 - 5x - 5x + 25 =
x^2 - 10x + 25 <===

8 0

3 years ago

|2.5–y|=3 please help asap

never [62]

Answer:

-0.5 and 5.5

Step-by-step explanation:

removing absolute value will yield two equation: 2.5-y=3 and 2.5-y=-3

solve for y in both case.

8 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 .

5 0

3 years ago

Is 10 + 0.01 rational or irrational

Lady bird [3.3K]

Answer:

its a rational number because 10+0.01 is 10.01 and thats rational

3 0

3 years ago

Read 2 more answers

Rewrite 4/7 and 3/5 so they have common denominator

Vladimir [108]

20/35 and 21/35. The common denominator is 35.

6 0

3 years ago