mqm dobrze to zadanie ? plis pomocy na za 30 minut !!

[latex]1cfrac{1}{4} cdot (-5) = cfrac{5}{4} cdot (-5) = -cfrac{25}{4} = -6cfrac{1}{4}[/latex] [latex](-cfrac{3}{4}) cdot (-2cfrac{2}{3}) = cfrac{3}{4} cdot cfrac{8}{3} = 2[/latex] [latex]-2cfrac{3}{4} - (-1cfrac{3}{4}) = -2cfrac{3}{4} + 1cfrac{3}{4} = -1[/latex] [latex]1cfrac{1}{4} - (-3cfrac{3}{4}) = 1cfrac{1}{4} + 3cfrac{3}{4} = 5[/latex] [latex]-2cfrac{3}{4}-1cfrac{3}{4} = -3cfrac{6}{4} = -4cfrac{2}{4} = -4cfrac{1}{2}[/latex] [latex]1cfrac{1}{2} - 2cfrac{3}{4} = -1cfrac{1}{4}[/latex] [latex]-6cfrac{1}{4} <-4cfrac{1}{2} <-1cfrac{1}{4} <-1 < 2 <5[/latex] Hasło: K O M T U R

[latex]K: 1frac14cdot(-5)=frac54cdot(-5)=-frac{25}4=-6frac14\\ U: (-frac34)cdot(-2frac23)=(-frac34)cdot(-frac83)=frac11cdotfrac21=2\\ T: -2frac34-(-1frac34)=-2frac34+1frac34=-1\\ R: 1frac14-(-3frac34)=1frac14+3frac34=4frac44=5\\O: -2frac34-1frac34=-3frac64=-4frac24=-4frac12\\ M: 1frac12-2frac34=1frac24-2frac34=-1frac14\\\ -6frac14 extless -4frac12 extless -1frac14 extless -1 extless 2 extless 5\\ ~quad Kqquad Oqquad Mqquad Tqquad Uquad R[/latex] Komtur był zwierzchnikiem domu zakonnego w niektórych zakonach rycerskich (np.: krzyżacy lub templariusze) . U nas pojęcie znane głównie z "Krzyżaków" H. Sienkiewicza :)