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3 years ago

Subtract 8 from me. Then add 21. If you divide me by 4 and add 21 you get 29. What number am I?

Mathematics

1 answer:

enyata [817]3 years ago

3 0

Answer:

Step-by-step explanation:

x - 8 + 21 ÷ 4 + 21 = 29

x - 8 + 21 ÷ 4 = 29 - 21

= 8

x - 8 + 21 = 8 x 4

= 32

x - 8 = 32 - 21

= 11

x = 11 + 8

= 19

Therefore the number is 19. A bit confusing to me at first but I solved it :)

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$\bold{Heya!}$

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>

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

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

$\sf{(1) = 1$

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

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

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

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