Temat:
MatematykaAutor:
bridgetUtworzono:
1 rok temuOdpowiedź:
[tex]1.\\\\a)\ \ -1\frac{5}{9}:1,4=-\frac{14}{9}:\frac{14}{10}=-\frac{\not14^1}{9}\cdot\frac{10}{\not14_{1}}=-\frac{10}{9}=-1\frac{1}{9}\\\\b)\ \ 1\frac{1}{8}:(-0,3)=\frac{9}{8}:(-\frac{3}{10})=-\frac{\not9^3}{\not8_{4}}\cdot\frac{\not10^5}{\not3_{1}}=-\frac{3}{4}\cdot5=-\frac{15}{4}=-3\frac{3}{4}[/tex]
[tex]2.\\\\-2,5:a=-1\frac{2}{3}\\\\a=-2,5:(-1\frac{2}{3})\\\\a=\frac{25}{10}:\frac{5}{3}\\\\a=\frac{\not5^1}{2}\cdot\frac{3}{\not5_{1}}\\\\a=\frac{3}{2}\\\\Sprawdzenie\\\\L=-2,5:\frac{3}{2}=-\frac{25}{10}\cdot\frac{2}{3}=-\frac{5}{\not2_{1}}\cdot\frac{\not2^1}{3}=-\frac{5}{3}=-1\frac{2}{3}\\\\P=-1\frac{2}{3}\\\\L=P[/tex]
[tex]3.\\\\-3,6:(-2\frac{1}{4})-2\frac{2}{3}\cdot1,5=\frac{36}{10}:\frac{9}{4}-\frac{8}{3}\cdot\frac{15}{10}=\frac{\not18^2}{5}\cdot\frac{4}{\not9_{1}}-\frac{\not8^4}{\not3_{1}}\cdot\frac{\not3^1}{\not2_{1}}=\frac{8}{5}-4=\frac{8}{5}-\frac{20}{5}=\\\\=-\frac{12}{5}=-2\frac{2}{5}\\\\Odp.B[/tex]
Autor:
amazonf7od
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