3 years ago

I need help with this Pythagorean Theorem.

Mathematics

1 answer:

sasho [114]3 years ago

6 0

The answer is C= 11.42

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X plus 1/3 equals nine enter your answer in the box in simplest form

LekaFEV [45]

$= > x + \frac{1}{3} = 9$

$\text{Add} \: - \frac{1}{3} \: \text{ on both sides ,}$

$= > x + \frac{1}{3} - \frac{1}{3} = 9 - \frac{1}{3} \\ \\ \\ \\ = > x = \frac{(9 \times 3) - 1}{3} \\ \\ \\ \\ = > x = \frac{27 - 1}{3} \\ \\ \\ \boxed{ \bold{= > x = \frac{26}{3} }} \\ \\ \\ or , \boxed{ \bold{ \: x = 8. \bar{6}}}$

Hence,

$Value \: \: \: of \: \: x \: \: is \: \frac{26}{3} \: \: \: or \: \: \: 8. \bar{6}$

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I need help with this Pythagorean Theorem. ​

I need help with this Pythagorean Theorem.