Mathematics: What is the value of n in the equation 0 = 18 + 128n2

What is the value of n in the equation 0 = 18 +

fgcherubin

12.03.2022

Consider the equation:

0= 18+128n^2

Subtracting '18' from both the sides of the equation, we get

0-18 = 18+128n^2-18

-18 = 128n^2

Dividing by '128' from both the sides of the equation, we get

$\frac{-18}{128}= \frac{128n^2}{128}$

$\frac{-9}{64}= {n^2}$

$n=\sqrt{\frac{-9}{64}}$

$n = \frac{3i}{8}$ and $n= \frac{-3i}{8}$

So, the values of 'n' are $n = \frac{3i}{8}$ and $n= \frac{-3i}{8}$ .

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devin3634

15.06.2020

A. 6/7

Step-by-step explanation:

you remove the a so that there are only 7 letters left. if you count everything except the k and a there are 6 letters

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