Ask question

Ask question

All categories

3 years ago

15

The space program has 6 astronauts in outer space and 10 astronauts on Earth. What is the ratio of the number of astronauts in o

uter space to the total number of astronauts?

1 answer:

natulia [17]3 years ago

4 0

6:16

explanation:
6+10=16
6 is the amount in outer space
ration is 6:16

You might be interested in

ect the best answer for the question. 3. Rewrite the equation x + y + 3 = –2x + 4y + 18 into standard form (ax + by = c). A. 3x

trapecia [35]

Answer:

B

Step-by-step explanation:

To rewrite the equation x + y + 3 = -2x + 4y + 18, first combine like terms.

x + y + 3 = -2x + 4y + 18

3x - 3y + 3 = 18

3x - 3y = 15

This is the standard form.

It can also be simplified by dividing each term by 3 to x - y = 5.

5 0

3 years ago

Wich one is cheer to buy 20 for 4.40 or 24 for 4.80

lawyer [7]

4.4/20=0.22 per item 4.8/24=0.2 per item Answer: 24 for 4.80

6 0

3 years ago

Translate the sentence into an inequality.<br><br> Four subtracted from c is less than - 20 .

11Alexandr11 [23.1K]

Answer:

$\displaystyle -4 + c < -20$

Step-by-step explanation:

You could either do what I did in the above answer, or you could do this:

$\displaystyle c - 4< -20$

It does not matter how you write it, as long as you understand the concept!

I am joyous to assist you anytime.

3 0

3 years ago

yKpoI14uk [10]

Answer:

-6 This is your answer it help you

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 .

5 0

3 years ago