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Ths is my whole code

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\begin{document}



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\begin{center}

\large

%OSNOVNA \v{S}KOLA VUK KARAD\v{Z}I\'{C}

%\medskip

%LESKOVAC

\vfill

{\bf\Large ZBIRKA RE\v{S}ENIH ZADATAKA NAMENJENA ZA UPIS U 7. RAZRED SPECIJALNOG MATEMATI\v{C}KOG ODELJENJA}

\vfill

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\vspace{5mm}

{\LARGE AUTOR: Milo\v{s} Vu\v{c}kovi\'{c}}

\vspace{15mm}


{\LARGE\bfseries OSNOVNA \v{S}KOLA VUK KARAD\v{Z}I\'{C}}

\vfill


LESKOVAC $2018/2019$

\vfill

%\begin{tabular}{rl}
%Mentor: & Prof. dr Predrag Stanimirovi\' c \\
%Mentor: & dr Jelena Milo\v{s}evi\'{c} \\ 
%\noalign{\vspace{2mm}}

%\noalign{\vspace{2mm}}
%Oblast: & Numeri\v cka matematika i optimizacija \\
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%Ni\v{s}%, decembar 2018.

\end{center}

\newpage
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\tableofcontents

\newpage
\section{ Racionalni izrazi}




\begin{zad} (2015)
Ako je $a=\frac{1}{2}+\frac{1}{3}\cdot \frac{1}{4}- \frac{1}{5}:\frac{1}{6}$, onda je:
\begin{enumerate}
    \item[A)]$|a|-a<1-a<|a|<a+1$
    \item[B)]$1-a<|a|-a<|a|<a+1$
    \item[C)]$a+1<|a|<1-a<|a|-a$
    \item[D)]$|a|<a+1<|a|-a<1-a$
    \item[E)]$a+1<|a|<|a|-a<1-a $
    \item[N)] ne znam.
\end{enumerate}

\bigskip

{\bf{Re\v{s}enje:}}
\bigskip
\newline  
$$a=\frac{1}{2}+\frac{1}{3}\cdot \frac{1}{4}- \frac{1}{5}:\frac{1}{6}=\frac{6}{12}+\frac{1}{12}-\frac{6}{5}=$$
$$\frac{7}{12}-\frac{6}{5}=\frac{35}{60}-\frac{72}{60}=-\frac{37}{60}$$
Jednostavnom proverom utvr\dj{}ujemo da je jedino ta\v{c}na nejednakost $ E)a+1<|a|<|a|-a<1-a  $, to jest $-\frac{37}{60}+1<|-\frac{37}{60}|<|-\frac{37}{60}|+\frac{37}{60}<1+\frac{37}{60}$
$$\frac{23}{60}< \frac{37}{60}<\frac{37}{60}+\frac{37}{60}<\frac{97}{60} $$ 
$$ \frac{23}{60}<\frac{37}{60}<\frac{74}{60}<\frac{97}{60}$$
Dakle, odgovor je pod E). 
\end{zad}


\begin{zad}(2014)
Vrednost izraza $\frac{2014:38-2014:(2014:53-2014:106)}{2014:38-2014:53-2014:106}$ je:

\begin{enumerate}
    \item[A)] manja od $-15$
    \item[B)] izme\dj{}u $-15$ i $-5$
    \item[C)] izme\dj{}u $-5$ i $5$
    \item[D)] izme\dj{}u $5$ i $15$
    \item[E)] ve\'{c}a od $15$
    \item[N)] ne znam.
\end{enumerate}

\bigskip

{\bf{Re\v{s}enje:}}
\bigskip
\newline  
$$\frac{2014:38-2014:(2014:53-2014:106)}{2014:38-2014:53-2014:106}=\frac{53-2014\cdot(38-19)}{53-38-19}=\frac{53-2014:19}{53-57}=$$ 
$$\frac{53-106}{-4}=\frac{\cancel{-}53}{\cancel{-}4}=\frac{53}{4}=13\frac{1}{4}.$$
Dakle, vidimo da je ta\v{c}an odgovor pod D) izme\dj{}u $5$ i $15$. 
\end{zad}



\begin{zad} (2013)
Vrednost izraza $\frac{2,4 \cdot 0,25 - 0,75:2,5}{2,4-\frac{1}{4}+\frac{3}{4}-2,5}$ je :
\begin{enumerate}
 \item[A)] $\frac{30}{4}$
    \item[B)] $\frac{57}{4}$
    \item[C)] $\frac{1}{2}$
    \item[D)] $\frac{1}{3}$
    \item[E)] $\frac{3}{4}$
    \item[N)] ne znam.
\end{enumerate}
\bigskip

{\bf{Re\v{s}enje:}}
\bigskip
\newline  
$$\frac{2,4 \cdot 0,25 - 0,75:2,5}{2,4-\frac{1}{4}+\frac{3}{4}-2,5}=\frac{\frac{24}{\cancel{10}\quad 2}\cdot \frac{\cancel{25}\quad 5}{100}-\frac{\cancel{75}\quad 3 }{10\cancel{0}}\cdot \frac{1\cancel{0}}{\cancel{25}\quad 1}}{\frac{\cancel{24}\quad 12}{\cancel{10}\quad 5}-\frac{1}{4}+\frac{3}{4}-\frac{\cancel{25}\quad 5}{\cancel{10}\quad 2}}=\frac{\frac{12}{1}\cdot \frac{\cancel{5}\quad 1}{\cancel{100}\quad 20}-\frac{3}{10}}{\frac{12}{5}+\frac{2}{4}-\frac{5}{2}}=$$
$$ \frac{\frac{12}{1}\cdot \frac{1}{20}-\frac{3}{10}}{\frac{48}{20}+\frac{10}{20}-\frac{50}{20}}=\frac{\frac{12}{20}-\frac{3}{10}}{\frac{8}{20}}=\frac{\frac{12}{20}-\frac{6}{20}}{\frac{8}{20}}=\frac{\frac{6}{20}}{\frac{8}{20}}=\frac{6 \cdot \cancel{20}}{\cancel{20} \cdot 8}=\frac{3}{4}.$$
Dakle, odgovor je pod $E) \frac{3}{4}$.

\end{zad}

\begin{zad} (2012)
Ako je 
$$ A=\frac{1-\left(\frac{3}{2}-0,25\right):\left(\frac{5}{4}-1,125\right)}{\frac{3}{40}-0,25:\frac{5}{4}+1,125},$$
onda je $10\% $ apsolutne vrednosti broja $A$ jednako:

\begin{enumerate}
 \item[A)] $1$
    \item[B)] $\frac{36}{5}$
    \item[C)] $\frac{9}{10}$
    \item[D)] $\frac{40}{197}$
    \item[E)] $\frac{153}{14}$
    \item[N)] ne znam.
\end{enumerate}

\bigskip

{\bf{Re\v{s}enje:}}
\bigskip
\newline
$$ A=\frac{1-\left(\frac{3}{2}-0,25\right):\left(\frac{5}{4}-1,125\right)}{\frac{3}{40}-0,25:\frac{5}{4}+1,125}=\frac{1-(1,50-0,25):(1,200-1,125)}{\frac{3}{40}-\frac{25}{100}\cdot \frac{4}{5}+\frac{1125}{1000}}=\frac{1-1,25:0,075}{\frac{3}{40}-\frac{100}{500}+\frac{1125}{1000}}=$$

$$ \frac{1-\frac{125}{100}: \frac{75}{1000}}{\frac{75}{1000}-\frac{200}{1000}+\frac{1125}{1000}}=\frac{1-\frac{125}{1\cancel{0}\cancel{0}}\cdot \frac{10\cancel{0}\cancel{0}}{75}}{\frac{1\cancel{0}\cancel{0}\cancel{0}}{1\cancel{0}\cancel{0}\cancel{0}}}=1-\frac{1250}{75}=\frac{75}{75}-\frac{1250}{75}=-\frac{1175}{75}=-\frac{47}{3}.$$
$A=-\frac{47}{3}$, pa je $10\%$ apsolutne vrednosti od $A$, $10\% \cdot |A|=\frac{10}{100}\cdot |-\frac{47}{3}|=\frac{1}{10}\cdot \frac{47}{3}=\frac{47}{30}$
\end{zad}

\begin{zad} (2011)
Ako je $a=4\cdot \left(-\frac{1}{2}+\frac{1}{3}\right)\cdot 1,8, \quad b=4\cdot \left(-\frac{1}{2}\right)+\frac{1}{3}\cdot 1,8, \quad c=4\cdot\left(-\frac{1}{2}+\frac{1}{3}\cdot 1,8\right),$

\begin{enumerate}
 \item[A)] $a=b<c$
    \item[B)] $b<a<c$
    \item[C)] $a<b<c$
    \item[D)] $a<c<b$
    \item[E)] $c<a<b$
    \item[N)] ne znam.
\end{enumerate}

\bigskip
{\bf{Re\v{s}enje:}}
\bigskip
\newline
$a=4\cdot \left(-\frac{1}{2}+\frac{1}{3}\right)\cdot 1,8= 4\cdot \left(-\frac{3}{6}+\frac{2}{6}\right)\cdot \frac{18}{10}=\frac{4}{1}\cdot \left(-\frac{1}{6}\right)\cdot \frac{18}{10}=-\frac{4}{6}\cdot \frac{18}{10}=-\frac{2}{3}\cdot \frac{18}{10}=-\frac{36}{30}=-\frac{6}{5}$.
\newline
Dakle, $a=-\frac{6}{5}$.



\end{zad}




\begin{zad} (2010)
Vrednost izraza $(1\cdot 2 - 2\cdot 3 + 3\cdot 4 - 4 \cdot 5): \left( \frac{1}{1\cdot 2}+ \frac{1}{2 \cdot 3}+ \frac{1}{3 \cdot 4}+ \frac{1}{4 \cdot 5 } \right) $ je: 

\begin{enumerate}
 \item[A)] $-15$
    \item[B)] $-\frac{48}{5}$
    \item[C)] $-\frac{1}{5}$
    \item[D)] $0$
    \item[E)] $15$
    \item[N)] ne znam.
\end{enumerate}

\end{zad}


\begin{zad} (2009)
Vrednost izraza $\frac{\left(6-\frac{4}{3}\right):0,3}{\left(\left(\frac{3}{20}-1,9\right)\cdot4\right):\frac{1}{5}} $ je: 

\begin{enumerate}
 \item[A)] $-\frac{7}{216}$
    \item[B)] $-2$
    \item[C)] $-\frac{4}{9}$
    \item[D)] $\frac{35}{81}$
    \item[E)] $\frac{4}{9}$
    \item[N)] ne znam.
\end{enumerate}

\end{zad}



\begin{zad} (2008)
Neka je $A=\frac{\left(-3\frac{3}{4}-2\frac{2}{3}-1\frac{1}{2}\right)\cdot 3\frac{3}{5}}{5-15\frac{1}{8}:2\frac{1}{5}}. $ Tada je: 

\begin{enumerate}
 \item[A)] $A<0$
    \item[B)] $0\le A<1$
    \item[C)] $1 \le A <2$
    \item[D)] $2\le A <3$
    \item[E)] $A\ge 3$
    \item[N)] ne znam.
\end{enumerate}

\end{zad}




\begin{zad} (2007)
Vrednost izraza $\frac{0,4+8\cdot \left(5-0,8\cdot \frac{5}{8}\right)- 5:2\frac{1}{2}}{\left(1\frac{7}{8}\cdot 8 - \left(8,9-2,6:\frac{2}{3}\right)\right)\cdot 34 \frac{2}{5}} $ je: 

\begin{enumerate}
 \item[A)] ve\'{c}a od $12$ ali manja od $16$;
    \item[B)] manja od $4$;
    \item[C)] ve\'{c}a od $8$ ali manja od $12$;
    \item[D)] ve\'{c}a od $4$ ali manja od $8$;
    \item[E)] ve\'{c}a od $16$;
    \item[N)] ne znam.
\end{enumerate}

\end{zad}


\begin{zad} (2006)
Ako je $a=\left(\frac{1}{2}-\frac{3}{4}\right)\cdot \left(\frac{5}{6}-\frac{7}{8}\right), $ onda je $96\%$ broja $a$ jednako: 

\begin{enumerate}
 \item[A)] -$0,01$;
    \item[B)] $0,01$;
    \item[C)] -$0,1$;
    \item[D)] $0,1$;
    \item[E)] $1$;
    \item[N)] ne znam.
\end{enumerate}

\end{zad}


\begin{zad} (2005)
Vrednost izraza $\frac{39\frac{1}{5}:18\frac{2}{3}+7\frac{1}{2}:1\frac{12}{13}}{1\frac{13}{20}:2\frac{1}{5}}$ je: 

\begin{enumerate}
 \item[A)] $2\frac{1}{2}$;
    \item[B)] $8$;
    \item[C)] $6$;
    \item[D)] $12$;
    \item[E)] $16$;
    \item[N)] ne znam.
\end{enumerate}

\end{zad}


\begin{zad} (2004)
Vrednost izraza $\frac{3}{4}-\frac{3}{4}:\left(-\frac{3}{4}\right)+\frac{3}{4}\cdot\frac{3}{4}-\frac{3}{4}$ je: 

\begin{enumerate}
 \item[A)] $\frac{25}{16}$;
    \item[B)] $0$;
    \item[C)] $\frac{9}{8}$;
    \item[D)] $\frac{3}{16}$;
    \item[E)] -$\frac{3}{16}$;
    \item[N)] ne znam.
\end{enumerate}

\end{zad}

\newpage

\section{Apsolutna vrednost}

\begin{zad2} (2018)
Zbir re\v{s}enja jedna\v{c}ine $\frac{1}{1+\frac{1}{|x+1|}}=\frac{2017}{2018}$ je: 
\begin{enumerate}
    \item[A)] $-2018$
    \item[B)] $-2$
    \item[C)] $0$
    \item[D)] $2$
    \item[E)] $2018$
\end{enumerate}
\end{zad2}
%\vspace{5cm}

\begin{zad2} (2017)
Broj re\v{s}enja jedna\v{c}ine $$ \left| |1-|x||-2\right|=3$$je: 
\begin{enumerate}
    \item[A)] $0$
    \item[B)] $1$
    \item[C)] $2$
    \item[D)] $4$
    \item[E)] $8$
    \item[N)] ne znam
\end{enumerate}
\end{zad2}



\begin{zad2}(2006)
Celih brojeva $x$ koji zadovoljavaju nejednakost $\frac{|x-1|-2}{2}<1$ ima:


\begin{enumerate}
    \item[A)] $1$
    \item[B)] $3$
    \item[C)] $6$
    \item[D)] $7$
    \item[E)] $9$ 
    \item[N)] ne znam.
\end{enumerate}

\end{zad2}


\begin{zad2}(2005)
Zbir svih re\v{s}enja jedna\v{c}ine $|2-|x+1||=3$ je:


\begin{enumerate}
    \item[A)] $-4$
    \item[B)] $2$
    \item[C)] $-2$
    \item[D)] $4$
    \item[E)] jedna\v{c}ina nema re\v{s}enja  
    \item[N)] ne znam.
\end{enumerate}

\end{zad2}



\begin{zad2}(2004)
Zbir svih re\v{s}enja jedna\v{c}ine $\frac{5+|x-1|}{2}=5$ je:


\begin{enumerate}
    \item[A)] $-4$
    \item[B)] $2$
    \item[C)] $6$
    \item[D)] $0$
    \item[E)] $10$  
    \item[N)] ne znam.
\end{enumerate}

\end{zad2}

\newpage

\section{Deljjivost brojeva}

\begin{zad3} (2018)
Iz kesice sa bombonama Vasilije je sestri dao tre\'{c}inu, bratu \v{c}etvrtinu, majci \v{s}estinu i ocu osminu od ukupnog broja bombona koje je imao. Ako su ukupan broj bombona i broj bombona koji je dobio svaki \v{c}lan njegove porodice dvocifreni brojevi, koliki je zbir cifara broja bombona koje je dobio otac? 
\begin{enumerate}
    \item[A)] $3$
    \item[B)] $6$
    \item[C)] $7$
    \item[D)] $8$
    \item[E)] $9$
\end{enumerate}
\end{zad3}
%\vspace{5cm}

\begin{zad3} (2017)
\v{C}etvorocifreni broj $\overline{abc1}$ je tri puta ve\'{c}i od \v{c}etvorocifrenog broja $\overline{2abc}$. Zbir cifara $a$, $b$ i $c$ je:
\begin{enumerate}
    \item[A)] $17$
    \item[B)] $18$
    \item[C)] $19$
    \item[D)] $20$
    \item[E)] $21$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}

\begin{zad3} (2016)
Koliko ima prirodnih brojeva manjih od $2016$ koji se zavr\v{s}avaju cifrom $5$ i jednaki su proizvodu \v{c}etiri me\dj{}usobno razli\v{c}ita prosta broja?
\begin{enumerate}
    \item[A)] $2$
    \item[B)] $4$
    \item[C)] $6$
    \item[D)] $8$
    \item[E)] $10$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}


\begin{zad3} (2016)
Ana, Bane i Ceca poga\dj{}aju nepoznati \v{s}estocifreni broj, znaju\'{c}i da su njegove cifre $1$, $2$, $3$, $4$, $5$ i $6$. Oni daju slede\'{c}e prognoze za taj broj:
\newline
Ana : $1$ $2$ $3$ $4$ $5$ $6$;
\newline
Bane : $2$ $4$ $5$ $1$ $6$ $3$;
\newline
Ceca : $4$ $6$ $3$ $2$ $1$ $5$;
\newline
Ako se zna da je Ana pogodila ta\v{c}no mesto za tri cifre, Bane tako\dj{}e za $3$ cifre, a Ceca samo za jednu cifru, nepoznati broj je deljiv sa: 
\begin{enumerate}
    \item[A)] $18$
    \item[B)] $45$
    \item[C)] $15$, ali ne sa $45$
    \item[D)] $24$
    \item[E)] $12$, ali ne sa $24$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}





\begin{zad3} (2015)
Koliko ima trocivrenih brojeva $\overline{abc}$ takvih da va\v{z}i: 
$$\overline{ aa } \cdot \overline{bc} \cdot \overline{abc}= \overline{abcabc}?$$
(cifre $a$, $b$, $c$ ne moraju da budu razli\v{c}ite. ) 
\begin{enumerate}
    \item[A)] $0$
    \item[B)] $1$
    \item[C)] $2$
    \item[D)] $3$
    \item[E)] $4$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}


\begin{zad3} (2014)
Mira je na $10$ listi\'{c}a napisala $10$ uzastopnih prirodnih brojeva (na svakom po jedan), pa je jedan listi\'{c} izgubila. Kada je sabrala brojeve na listi\'{c}ima koji su joj ostali zbir je bio $2014$. Na listi\'{c}u koji je Mira izgubila je broj:
\begin{enumerate}
    \item[A)] $211$
    \item[B)] $219$
    \item[C)] $221$
    \item[D)] $226$
    \item[E)] $231$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}

\begin{zad3} (2013)
Proizvod $2013$ prirodnih brojeva jednak je $2013$. Najmanja mogu\'{c}a vrednost zbira tih $2013$ brojeva je: 
\begin{enumerate}
    \item[A)] $2013$
    \item[B)] $2085$
    \item[C)] $2105$
    \item[D)] $2205$
    \item[E)] $4025$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}

\begin{zad3} (2012)
Koliko ima \v{s}estocifrenih brojeva oblika $\overline{a17a6b}$ koji su deljivi sa $18$?
\begin{enumerate}
    \item[A)] $1$
    \item[B)] $2$
    \item[C)] $3$
    \item[D)] $4$
    \item[E)] vi\v{s}e od $4$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}

\begin{zad3} (2011)
Koliko postoji razli\v{c}itih \v{s}estocifrenih brojeva $\overline{a2012b}$ koji su deljivi sa $12$?
\begin{enumerate}
    \item[A)] $3$
    \item[B)] $6$
    \item[C)] $7$
    \item[D)] $9$
    \item[E)] $10$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}


\begin{zad3} (2010)
na $2010$-om mestu posle zapete u decimalnom zapisu broja $\frac{5}{7}$ je cifra: 
\begin{enumerate}
    \item[A)] $1$
    \item[B)] $4$
    \item[C)] $5$
    \item[D)] $7$
    \item[E)] $8$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}



\begin{zad3} (2009)
Proizvod $2009$ prirodnih brojeva jednak je $2009$. Najmanja mogu\'{c}a vrednost zbira tih $2009$ brojeva je : 
\begin{enumerate}
    \item[A)] $2009$
    \item[B)] $2061$
    \item[C)] $2097$
    \item[D)] $2301$
    \item[E)] $4017$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}


\begin{zad3} (2008)
Celih brojeva $x$ za koje je $\frac{12x+24}{x}$ prirodan broj ima: 
\begin{enumerate}
    \item[A)] $16$
    \item[B)] $6$
    \item[C)] $8$
    \item[D)] $12$
    \item[E)] $14$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}

\begin{zad3} (2007)
Zbir svih prirodnih brojeva koji pri deljenju sa $7$ daju koli\v{c}nik jednak ostatku je: 
\begin{enumerate}
    \item[A)] $180$
    \item[B)] $80$
    \item[C)] $140$
    \item[D)] $48$
    \item[E)] $168$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}


\begin{zad3} (2006)
Celih brojeva $x$ koji zadovoljavaju nejednakost $\frac{|x-1|-2}{2}<1$ ima: 
\begin{enumerate}
    \item[A)] $1$
    \item[B)] $3$
    \item[C)] $6$
    \item[D)] $7$
    \item[E)] $9$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}

\begin{zad3} (2005)
Prostih brojeva $p$ za koje va\v{z}i $\frac{334}{2004}<\frac{5}{p}<\frac{802}{2005}$ ima:
\begin{enumerate}
    \item[A)] $2$
    \item[B)] $4$
    \item[C)] $3$
    \item[D)] $5$
    \item[E)] vi\v{s}e od $5$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}


\begin{zad3} (2005)
Neka su $a$ i $b$ prirodni brojevi, takvi da je $a-b=-2$. Najmanja vrednost razlike $2005a - 2000 b$ je:
\begin{enumerate}
    \item[A)] $3995$
    \item[B)] $-4005$
    \item[C)] $4005$
    \item[D)] $-4010$
    \item[E)] $-3995$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}


\begin{zad3} (2004)
Neka je broj $n$ zbir petocifrenih brojeva $\overline{a2004}$ i $\overline{2004b}$, i, gde su $a$ i $b$ cifre. Takvih brojeva $n$ koji su deljivi sa $15$ ima ukupno: 
\begin{enumerate}
    \item[A)] $4$
    \item[B)] $2$
    \item[C)] $6$
    \item[D)] $7$
    \item[E)] $3$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad3}



\newpage

\section{ Trougao}

\begin{zad4} (2018)
Neka je $D$ ta\v{c}ka na stranici $AC$ trougla $ABC$ pri \v{c}emu je $|CD|=2\cdot |AD|$ i neka je $\sphericalangle BAC=45\degree$, a $\sphericalangle BDC=60\degree$. Tada je razlika uglova $\sphericalangle BCA$ i $\sphericalangle CBA$ jednaka:
\begin{enumerate}
    \item[A)] $15\degree$
    \item[B)] $30\degree$
    \item[C)] $45\degree$
    \item[D)] $60\degree$
    \item[E)] $90\degree$
\end{enumerate}
\end{zad4}
%\vspace{5cm}

\begin{zad4} (2017)
Simetrala katete $AC$ pravouglog trougla $ABC$ se\v{c}e hipotenuzu $AB$ u ta\v{c}ki $E$. Obim trougla $ABC$ je $80cm$, obim trougla $AEC$ je $50cm$, a obim trougla $BCE$ je $64cm$. Povr\v{s}ina trougla $ABC$ je: 
\begin{enumerate}
    \item[A)] $240cm^2$
    \item[B)] $120cm^2$
    \item[C)] $360cm^2$
    \item[D)] $180cm^2$
    \item[E)] $480cm^2$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad4}

\begin{zad4} (2016)
Trougao $ABC$ je tupougli sa tupim uglom kod temena $B$. Simetrale spolja\v{s}njih uglova kod temena $A$ i $C$ seku prave $BC$ i $AB$, redom u ta\v{c}kama $D$ i $E$. Ako je $AD=AC=CE$, tada je ugao kod temena $B$ jednak:
\begin{enumerate}
    \item[A)] $135 \degree$
    \item[B)] $120 \degree$
    \item[C)] $100 \degree$
    \item[D)] $105 \degree$
    \item[E)] $108 \degree$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad4}

\begin{zad4} (2016)
Trougao $ABC$ ima stranice du\v{z}ina $BC=17cm$, $CA=12cm$ i $AB=23cm$. Ta\v{c}ka $D$ je sredi\v{s}te stranice $AB$, a ta\v{c}ka $E$ je podno\v{z}je normale iz temena $A$ na simetralu unutra\v{s}njeg ugla datog trougla iz temena $C$. Du\v{z}ina du\v{z}i $DE$ je: 
\begin{enumerate}
    \item[A)] $1,5cm$
    \item[B)] $2cm$
    \item[C)] $2,5cm$
    \item[D)] $3cm$
    \item[E)] $3,5cm$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad4}


\begin{zad4} (2015)
Povr\v{s}ina jednakostrani\v{c}nog trougla $ABC$ je $1cm^2$. Ta\v{c}ke $E$, $F$, $G$ pripadaju stranicama $AB$, $BC$, $CA$, redom pri \v{c}emu je $AE:EB=BF:FC=CG:GA=3:1$. Povr\v{s}ina trougla $EFG$ je:

\begin{enumerate}
    \item[A)] $\frac{3}{8}cm^2$
    \item[B)] $\frac{7}{16}cm^2$
    \item[C)] $\frac{3}{4}cm^2$
    \item[D)] $\frac{5}{8}cm^2$
    \item[E)] $\frac{9}{16}cm^2$
    \item[N)] Ne znam.
\end{enumerate}
\end{zad4}




\end{document}

I have problem here because my command "\renewcommand{\contentsname}{Sadr\v{z}aj}" doesnt work and I cant change name than it. Can someone help me, i tried many things here and doesnt work.

1
  • 1
    Too much irrelevant code (for me to read). Please reduce it to a minimal working example. For example: your color definitions, math symbols and so on are not relevant.
    – campa
    Jul 8 '19 at 6:18
1

Add

\addto\captionsenglish{%
  \renewcommand{\contentsname}%
    {Sadr\v{z}aj}}

to the preamble of your document, instead of

\renewcommand{\contentsname}{Sadr\v{z}aj}

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