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Anestetic [448]
3 years ago
12

PLEASE HELPPPP!!

Mathematics
1 answer:
geniusboy [140]3 years ago
7 0

Answer:

qwsedfgh

Step-by-step explanation:

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Find the value of x , if log₂ [log₅ (log₃x)] = 0
Lina20 [59]
\log_2 (\log_5( \log_3x)) = 0\\
D:x>0 \wedge \log_3x>0 \wedge \log_5(\log_3x)>0\\
D:x>0\wedge x>1 \wedge\log_3x>1\\
D:x>1 \wedge x>3\\
D:x>3\\
\log_5(\log_3x)=1\\
\log_3x=5\\
x=3^5\\
x=243
3 0
3 years ago
Problem Solving
Murrr4er [49]

Answer:

31 games

Step-by-step explanation:

round 1 has 32 teams ÷2=16games

round 2 has 16 teams ÷2=8games

round 3 has 8 teams ÷2=4games

round 4 has 4 teams ÷ 2=2games

round 5 has 2 teams ÷2=1games

16+8+4+2+1

6 0
3 years ago
Pls help answer this
Paladinen [302]

Answer: Third option

Step-by-step explanation:

For this exercise it is important to remember the following:

1. By definition, the Associative property of addition states that it does not matter how you grouped the numbers, you will always obtained the same sum.

2. The rule for the  Associative property of addition is the following (given three numbers "a", "b" and "c"):

a + (b + c) = (a + b) + c

Knowing the information shown before, you can identify in the picture attached that the option that illustrates the Associative property of addition is the third one. This is:

(11+8)+1=11+(8+1)

As you can notice that you will always get the same result:

(19)+1=11+(9)\\\\20=20

8 0
3 years ago
What is the volume of this cube?
Charra [1.4K]

Answer:

50

Step-by-step explanation:

im not that sure sorry

8 0
3 years ago
#2 <br> #3 <br> Need help for my discrete math class
miss Akunina [59]

For (2), start with the base case. When n = 2, we have

(n + 1)! = (2 + 1)! = 3! = 6

2ⁿ = 2² = 4

6 > 4, so the case of n = 2 is true.

Now assume the inequality holds for n = k, so that

(k + 1)! > 2ᵏ

Under this hypothesis, we want to show the inequality holds for n = k + 1. By definition of factorial, we have

((k + 1) + 1)! = (k + 2)! = (k + 2) (k + 1)!

Then by our hypothesis,

(k + 2) (k + 1)! > (k + 2) 2ᵏ = k•2ᵏ + 2ᵏ⁺¹

and k•2ᵏ ≥ 2•2² = 8, so

k•2ᵏ + 2ᵏ⁺¹ ≥ 8 + 2ᵏ⁺¹ > 2ᵏ⁺¹

which proves the claim.

Unfortunately, I can't help you with (3). Sorry!

3 0
3 years ago
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