Melody

Thank you Chris  :D

More for me to think about :)))

See if this helps 

$$\\59-3|6-9^2|\\
=59-3*|6-81|\\
=59-3*|-75|\\
=59-3*+75\\
=59-225\\
=5-166\\$$

If you added 100 to f(x) then you would know.

$$\\11x^2+13x-28+100=0\\
11x^2+13x+72=0\\
\triangle=b^2-4ac\\
\triangle=13^2-4*11*72\\
\triangle<0\\
$therefore the graph will never reach as low as $ f(x)=-100$$

Suggestion:

x-3<1

x<4

the web2 calc handles this easily

[2nd][sec](1.2934292)

$${asec}{\left({\mathtt{1.293\: \!429\: \!2}}\right)} = {\mathtt{39.363\: \!427\: \!756\: \!281^{\circ}}}$$

on othe calcs you often have to do this

$$\\sec \theta = 1.2934292\\
cos\theta = 1/1.2934292\\
\theta= cos^{-1}(1/1.2934292)\\$$

arc cos is the same as inverse cos

I haven't tried that yet either.   

Thank you Heureka   

21*sqrt(31.4)

$${\mathtt{21}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{31.4}}}} = {\mathtt{117.674\: \!976\: \!099\: \!423\: \!959\: \!1}}$$