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Bad White [126]
2 years ago
7

Activity

Mathematics
1 answer:
pashok25 [27]2 years ago
4 0

Answer:

2 8

3 27

4 64

5 125

6 204

7 343

8 512

9 729

10 1000

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Find the least common multiple of 8c4 and 6c2 .
Ksju [112]
The least common multiple is the smallest term that can be divided to both terms without any remainder. For the two terms 8c^4 and 6c^2, you can determine it into two part. First, you find the LCM for 8 and 6. You find the prime number that is common between the two. That would be 2. For the variables c^4 and c^2, the 'prime variable' is c. Therefore, the least common multiple for 8c^4 and 6c^2 is 2c.
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3 years ago
Coping strategies can help individuals build
noname [10]

Answer: Coping strategies can help individuals build <u><em>resiliency.</em></u>

Step-by-step explanation:

Resiliency is defined as the capacity to recover quickly from difficulties. Coping with something means finding ways to face and understand the negative feelings that thing causes you and how to overcome them.

The other three options are actually <em>things you would want to learn how to cope with</em>, thus not correct answers.

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3 years ago
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The Mazurskys decide to retile their mud room in 18 in. by 18 in. tiles. If their mudroom is 5 ft 6 in. by 13 ft, how many tiles
Mrac [35]

The area of one tile is 18 x 18 = 324 square inches.


Convert the dimensions of the room into inches:

5 ft 6 in. = 66 inches.

13 ft = 156 inches.

The area of the room is: 156 x 66 = 10296 square inches.


To find the number of tiles needed, divide the total area of the room by the area of one tile:

10,296 / 324 = 31.78, which means they will need 32 tiles.


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3 years ago
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Can somebody please help in this question?
Phoenix [80]

Answer:

Assume the pair of dice are fair; if not, all bets are off!

Color one die green (G) and the other one red (R). There are 36 possibilities in throwing one green and one red die, eg (G1, R1), (G1, R2), etc; (G2, R1), (G2, R2), etc; etc; (G6, R1), (G6, R2), etc.

In the set of possible numbers from throwing a die, there are only 3 possibilities that sum to to 4: (G1, R3), (G2, R2), (G3, R1).

So the answer is 3/36, or one-twelfth.

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2 years ago
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What is 1 1/3 × 2 1/5?????
Cerrena [4.2K]
77/5 <span>Decimal: (15.4) Hope this helps

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2 years ago
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