For the first one, I define it as \(-{1\over3} + {2 {5\over6}} - {3\over8} \cdot 1 {2\over3}.\) Now you calculate \(-{1\over3}+{2 {5\over6}}\), which is \(-{2\over6}+{2 {5\over6}}\). This is equal to \({2 {3\over6}}=2{1\over2}\).
Now we have to solve \({3\over8}\cdot1{2\over3}\). This is equal to \({3\over8}\cdot{5\over3}\), or \(5\over8\). Now all you need to do is calculate \(2{1\over2}-{5\over8}\), which is just \(2{4\over8}-{5\over8}=1{12\over8}-{5\over8}=1{7\over8}.\)
(The latex takes some time this took around 5 minutes lol)
For the second one, I define it as \({{-2{1\over2}}\over{-3{1\over3}}}+{({-{1\over6}})}\). Now we calculate the division, which is \({{5\over2}\over{10\over3}}.\) Since division is multiplying by a reciprocal, it is just \({5\over2}\cdot{3\over10}={3\over4}.\) Now you have to do \({3\over4}+(-{1\over6}),\) which becomes \({9\over12}+(-{2\over12})={7\over12}.\)
For the third one, I define it as \(-{1\over4}+(-{4\over5})\cdot{2{1\over6}}-{5\over16}.\) First we have to calculate the multiplication, which is \({-{4\over5}}\cdot{2{1\over6}}={-{4\over5}}\cdot{13\over6}={-{26\over15}}.\)Now you do the addition and subtraction, which is \({-{1\over4}}-{26\over15}-{5\over16}={-{60\over240}}-{416\over240}-{75\over240}={-{551\over240}}.\)
I really hope this helped, it took me around 20 minutes to solve this and do the latex.
HOPE THIS HELPED!!!!!!!!!!!!!!!!!!!!!!!!!!!
