POMOCY!! potrzebuje na jutro! DAJE NAJ!! <3 zad10 d) f). zad 4 c)

[tex]x=-48:0,6+1 \frac{2}{3}*54-1 \frac{5}{7}: 1 \frac{1}{14}=-48:\frac{6}{10}+\frac{5}{3}*54- \frac{12}{7}:\frac{15}{14}=-48*\frac{10}{6}+5*18-\frac{12}{7}*\frac{14}{15}=-80+90-\frac{4*2}{5}=10-\frac{8}{5}=\frac{50-8}{5}=\frac{42}{5}[/tex]

zatem liczba przeciwna do x to [tex]-\frac{42}{5}[/tex]

[tex]y=1 \frac{1}{3}*(-3 \frac{3}{5})+(\frac{3}{5}-2 \frac{4}{15})*0,5=\frac{4}{3}*(-\frac{18}{5})+(\frac{3}{5}-\frac{34}{15})*\frac{1}{2}=-\frac{4*6}{5}+(\frac{9-34}{15})*\frac{1}{2}=-\frac{24}{5}-\frac{25}{15}*\frac{1}{2}=-\frac{24}{5}-\frac{5}{3}*\frac{1}{2}=-\frac{24}{5}-\frac{5}{6}=-\frac{144}{30}-\frac{25}{30}=-\frac{169}{30}[/tex]

Zatem liczba odwrotna do y to [tex]-\frac{30}{169}[/tex]

Sawanna1viz Sawanna1viz · Answer 2

Odpowiedź:

zad 10 d)

[tex](2\frac{3}{4}-3,5)*(-2^{2})=(\frac{11}{4}-\frac{7}{2})*4=(\frac{11}{4}-\frac{14}{4})*4=(-\frac{3}{4})*4=-\frac{12}{4}=-3[/tex]

zad 10 f)

(0,75-(-³/₄)²)²=[tex](\frac{3}{4}-(-\frac{3}{4})^{2})^{2}=(\frac{3}{4}-\frac{9}{16})^{2}=(\frac{12}{16}-\frac{9}{16})^{2} =[/tex][tex](\frac{3}{16})^{2}=\frac{9}{256}[/tex]

zad 4 c)

x=-48:0,6+1²/₃*54-1⁵/₇:1¹/₁₄=

-80+⁵/₃*54-¹²/₇:¹⁵/₁₄=

-80+²⁷⁰/₃-¹²/₇*¹⁴/₁₅=

-80+90-¹⁶⁸/₁₀₅=-80+90-1,6=10-1,6=8,4

y=[tex]1\frac{1}{3}*(-3\frac{3}{5})+(\frac{3}{5}-2\frac{4}{15})*0,5=[/tex]

[tex]\frac{4}{3}*(-\frac{18}{5})+(\frac{9}{15}-\frac{34}{15})*0,5=[/tex]

[tex]-\frac{24}{5}+(-\frac{25}{15})*\frac{1}{2}=[/tex]

[tex]-\frac{24}{5}+(-\frac{25}{30})=[/tex]

[tex]-\frac{144}{30}-\frac{25}{30}=[/tex][tex]-\frac{169}{30}[/tex]=

[tex]-5\frac{19}{30}=-5,63[/tex]

Szczegółowe wyjaśnienie: