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denpristay [2]
3 years ago
7

Plz help me well mark brainliest if you are correct!..

Mathematics
2 answers:
koban [17]3 years ago
8 0

Answer:

8

Step-by-step explanation:

all numbers must be positive when you have the lines on the side

Rufina [12.5K]3 years ago
5 0

Answer:

8

Step-by-step explanation:

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Help please! I don't understand how to do this at all. Please help
Viktor [21]
9. \sqrt[n]{x}
    \sqrt[6]{-64}
    \sqrt[6]{64 * (-1)}
    \sqrt[6]{64}\sqrt[6]{-1}
    2i

10. \sqrt[n]{x}
      \sqrt[3]{128}
      \sqrt[3]{64 * 2}
      \sqrt[3]{64}\sqrt[3]{2}
      4\sqrt[3]{2}

11. \sqrt[n]{x}
      \sqrt[10}{1024}
      2

12. \sqrt[n]{x}
      \sqrt[7]{\frac{8\sqrt{2}}{2187}}
      \frac{\sqrt[7]{8\sqrt{2}}}{\sqrt[7]{2187}}
      \frac{\sqrt[7]{8\sqrt{2}}}{\sqrt[7]{(3)^{7}}}
      \frac{\sqrt[7]{8\sqrt{2}}}{3}
      
4 0
3 years ago
Eli, a pretty smart ape in the Salt Lake City zoo, has correctly picked the winner of the last 7 Super Bowls. Read about Eli's p
Anna [14]

Answer:

There is a 0.78% probability that Eli chooses 7 winners in a row.

Step-by-step explanation:

There were 7 Super Bowls.

In each Super Bowl, there were two teams.

So, the probability that Eli chooses 7 winners in a row is

P = \frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2} = \frac{1}{128} = 0.0078

There is a 0.78% probability that Eli chooses 7 winners in a row.

4 0
3 years ago
PLZ ANSWER ILL MAKE YOU BRAINLIEST NO FAKE LINKS
Alex787 [66]

Answer:

the answer is D

Step-by-step explanation:

7 0
3 years ago
4 x 10^4 in Standard Form?
Fofino [41]

Answer:

4 × 10^4

→ 4× 10000

→ 40000

Correct option → 40000

hope it helped :)

4 0
3 years ago
Suppose that Aces can be either high or low; that is, that {A, 2, 3, 4, 5} is a straight, and so is {10, Jack, Queen, King, Ace}
Mumz [18]

Answer:

If OR is exclusive in the question, meaning that Ace can either be high or low but not both would give you 36 possible combinations. However the question isn't exactly stated in a way that makes it easy to interpret this. Your logic is right though, and it could be an error on your sources part. The possible combos are

A 2 3 4 5

2 3 4 5 6

3 4 5 6 7

4 5 6 7 8

5 6 7 8 9

6 7 8 9 10

7 8 9 10 J

8 9 10 J Q

9 10 J Q K

10 J Q K A

8 0
3 years ago
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