gibsonj338

\(5248/(2/15)\)

\(5248*15/2\)

\(78720/2\)

\(39360\)

\(-6-b/8\)

\(-48=b\)

\(b=-48\)

\(cos^-1(8.3/9.5)\)

\(cos^-1(0.8736842105263158...)\)

\(1.062724573062...\)

\(h=c/w/d\); Let h = numbr of credits per day, Let c = number of total credits, Let w = number of weeks, Let d = number of days in a week

\(h=60/3/7\)

\(h=20/7\)

\(h=2.8571428571428571...\)

You would need to work \(20/7\) credits or approximately \(3\) crredits a day in order to completete 60 credits.  As far as the number of hours in a day, that all depends on how long it would take you to complete one credit. Once you figure that out, multiply that number by 20/7 to get the number of hours for one day.

\(3/5x=30\)

\(3=30*5x\)

\(3=150x\)

\(3/150=x\)

\(x=3/150\)

\(x=1/50\)

\(x=0.02\)

\(\sqrt(155555567854)\)

\(155555567854^(1/2)\)

Usualy you would just leave it at either \(\sqrt(155555567854)\) or at \(155555567854^(1/2)\) as the number \(155555567854\) cannot be multiplied by itself evenly.  The decimal answer is \(394405.3344644314012\)...

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Everything SpawnofAngel says is right except that it is not y equals but x equals.

\((2/9)/(3/8)\)

\(2/9*8/3\)

\(16/27\)

\(0.592...\)

\(62/36\)

Dvide both the numerator and the dominator by 2

\(31/18\)

\(31/18\) cannot be simplified any further

\(3(4-2(0.7*0.5))\)

\(3(4-2(0.35))\)

\(3(4-0.7)\)

\(3(3.3)\)

\(9.9\)