<img src="https://tex.z-dn.net/?f=3%20%5Csqrt%7B50%7D%20%20%5Ctimes%20%20%5Csqrt%7B22%7D%20" id="TexFormula1" title="3 \sqrt{50}

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3 years ago

\times \sqrt{22} " alt="3 \sqrt{50} \times \sqrt{22} " align="absmiddle" class="latex-formula">
A
$15 \sqrt{22}$
B
$30 \sqrt{22}$
C
$30 \sqrt{11}$
D
$15 \sqrt{11}$

Mathematics

1 answer:

wariber [46]3 years ago

6 0

Answer:I'm 97.248316478692% sure it's C.

Step-by-step explanation:

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The set of numbers in the Collatz Conjecture are 1, 4, 2, 1, 4, 2, 1, 4, 2....

<h3>How to generate the numbers in the Collatz Conjecture</h3>

To start with, we make use of the following starting point:

Starting point = 1

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If the last number is odd, multiply it by 3 and add 1.
If the last number is even, divide the number by 2.

1 is an odd number.

So, we use the first rule

So, we have:

1 * 3 + 1 = 4

4 is an even number.

So, we use the second rule

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4/2 = 2

So, the first three numbers in the Collatz Conjecture are:

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1, 4, 2, 1, 4, 2, 1, 4, 2....

The above means that there is only one set of repeating numbers in the Collatz Conjecture.

Hence, the set of numbers in the Collatz Conjecture are 1, 4, 2, 1, 4, 2, 1, 4, 2....