it's actually
$y=\frac{4}{x}+\sqrt{x+0.2-5x}$ (see algebra problem)
$$y=\frac{4}{x}+\sqrt{x+0.2-5x}$$ if $x=\frac45$ what is y?
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If $x$ is a constant, either write it down, or use a calculator – Aseed Apr 06 '16 at 03:34
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Since $\frac{4}{5} = 0.8$, $$ y = \frac{4}{x} + \sqrt{x + 0.2} - 5x = \frac{4}{4/5} + \sqrt{0.8+0.2} - 5 \cdot \frac{4}{5} = 5 + \sqrt{1} - 4 = 2 $$
Edit: Since you've modified the question, $$ y = \frac{4}{x} + \sqrt{x + 0.2 - 5x} = \frac{4}{4/5} + \sqrt{0.8 + 0.2 - 4} = 5 \pm \sqrt{3}i $$
Henricus V.
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Actually it's not $\sqrt{x+\frac{2}{10}} - 5x$ but $\sqrt{x+\frac{2}{10} - 5x}$ (but when you answered the question's formatting was misleading) – Aseed Apr 06 '16 at 03:41
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$f(x) = \frac{4}{x} + \sqrt{x+.2-5x}$
$f(4/5) = \frac{4}{4/5}+ \sqrt{\frac{4}{5}+0.2-5(\frac{4}{5})} = 5 + \sqrt{3}i$
user41915
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but im so hopeless at maths & you guys are so good! & doesnt it feel good to help someone in need? – user329008 Apr 06 '16 at 03:44
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@user329008 Here's a website you can use to solve your math problems: http://www.wolframalpha.com/ – Henricus V. Apr 06 '16 at 04:00