-11 and 1/6 as a decimal

Mathematics

1 answer:

Nikolay [14]3 years ago

7 0

-11.1666667 should be the answer

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◆ COMPLEX NUMBERS ◆

$given \: expression \: is \: - \\ \\ - 3 {i}^{4} + 2 {i}^{3} + 2 {i}^{2} + \sqrt{ - 9} \\ \\ we \: know \: that \: , \: \\ i = \sqrt{ - 1} \\ \\ {i}^{2} = - 1 \\ \\ {i}^{3 } = - i \\ \\ {i}^{4} = 1 \\ \\ \\ - 3 {i}^{4} + 2 {i}^{3} + 2 {i}^{2} + \sqrt{9} i \\ = - 3 {i}^{4} + 2 {i}^{3} + 2 {i}^{2} + 3i \\ \\ using \: the \: above \: properties \: of \: iota \: (i) \\ \\ = - 3(1) + 2( - i) + 2( - 1) + 3i \\ \\ = - 3 - 2i - 2 + 3i \\ \\ = (- 5 + i) \: \: \: ans.$

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What's the correct answer for this?

11Alexandr11 [23.1K]

$\bf \qquad \qquad \textit{inverse proportional variation}\\\\
\textit{\underline{y} varies inversely with \underline{x}}\qquad \qquad y=\cfrac{k}{x}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\$

$\bf \textit{\underline{N} is inversely proportional to \underline{A}}\qquad N=\cfrac{k}{A}
\\\\\\
\textit{we also know that }
\begin{cases}
A=3\\
N=16
\end{cases}\implies 16=\cfrac{k}{3}\implies 16\cdot 3=k
\\\\\\
48=k\qquad thus\qquad \boxed{N=\cfrac{48}{A}}\\\\
-------------------------------\\\\
\textit{if \underline{A} is 4, what is \underline{N}?}\qquad N=\cfrac{48}{4}$

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2 years ago