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xxTIMURxx [149]
3 years ago
7

What is the value of X ? A-17 B-26 C-39 D-41

Mathematics
1 answer:
Hatshy [7]3 years ago
4 0

Answer:

D.

Step-by-step explanation:

It's a right triangle so

x^2=40^2+9^2

x = 41

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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>

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2 years ago
When comparing answers on a group assignment, Jenny and Jeremy found that their answers did not match. Part 1: Who do you believ
vitfil [10]
PART 1:

Jeremy gives the correct answer. 

The value of 0.41 [with a bar over the digit 4 and 1] shows that the digit 4 and 1 are reoccurring = 0.414141414141414141....

Jenny's assumption of 41/100 will give a decimal equivalency of 0.41 [without a bar over digit 4 and 1]. This value is not a reoccurring decimal value. 

PART 2:

The long division method is shown in the picture below

PART 3:

As mentioned in PART 1, the result of converting 41/100 into a decimal is 0.41 [non-reoccuring decimal] while converting 41/99 into a decimal is 0.41414141... [re-occuring decimal]. The conjecture in PART 1 is correct



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<span>
Write two others names for fg
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