Solve for square root of y=0.5

<img src="https://tex.z-dn.net/?f=%5Cbegin%7Bequation%7D%5Ctext%20%7B%20Question%3A%20Find%20the%20value%20of%20%7D%20%5Cfrac%7B

ankoles [38]

$\bold{Heya!}$

Your answer to this is:

$\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0}$

<h2>→ <u>EXPLANATION :-</u></h2>

<u />

<u /> $\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0} \: (1)$

$\sf{Apply \: rule:} \: (a) = a$

$\sf{(1) = 1$

$\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0} \: ^. \: 1$

$\sf{\frac{1}{1 \: + \:1} = \frac{1}{2}$

$\sf{\frac{2^1^0^0}{1 \: + \: x^2^0^0} ^.^ \: 1 \: = \: \frac{2^1^0^0}{1 \: + \: x^2^0^0}$

$\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0}$

Hopefully This Helps ! ~

#LearnWithBrainly

$\underline{Answer :}$

<em>Jaceysan ~</em>

7 0

2 years ago

At a sporting good store the price of a basketball at $650 during a sale this prices marked down 45% what is the sale price of a

Ne4ueva [31]

The multiplier for decreasing by 45% is 0.55
So you times the multiplier by the price to get your answer:

650 x 0.55 = $357.50

4 0

3 years ago

Find the number of ways of arranging the numbers

Doss [256]

First of all, note that all integers are either 0,1, or 2 modulo 3 (if you're not familiar with this terminology, it means that every integer is either a multiple of 3, or it is 1 or 2 away from a multiple of 3).

So, we can think of our numbers as

$\begin{array}{c|c}x&x\mod 3\\0&0\\1&1\\2&2\\3&0\\4&1\\5&2\\6&0\\7&1\\8&2\\9&0\end{array}$

In order to make sure that the sum of any three adjacent numbers is divisible by 3, we have to make sure that any group of 3 three adjacent numbers contains a 0, a 1 and a 2. This is possible only if we arrange our 9 numbers in 3 groups of 3 numbers containing 0,1 and 2 exactly once, repeating always the same pattern.

For example, we could arrange our numbers following the pattern

$0,1,2,0,1,2,0,1,2$

or

$2,0,1,2,0,1,2,0,1$

We have $3!=6$ possible patterns. Suppose for example that we choose the pattern

$0,1,2,0,1,2,0,1,2$

One possible way of following this pattern would be the arrangement

$3,1,2,6,4,5,9,7,8$

In fact, we substituted every '0' with a multiple of 3 (3, 6 or 9), every '1' with a number 1 away from a multiple of 3 (1, 4 or 7) and every '2' with a number 2 away from a multiple of 3 (2, 5 or 8).

This means that, once we fix a patter, we have 3 choices for the first 3 slots, 2 choices for the next 3 slots, and the final slot will be fixed. So, we have

$3\cdot 3\cdot 3\cdot 2 \cdot 2 \cdot 2 = 216$