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Thank you, Dragan !

I dont understand how to do this. Please show steps

EC=

\(\text{Let $ES=ER=30$} \\ \text{Let $DT=DR=17$} \\ \text{Let $TC=SC=32$} \\ \text{Let the perimeter is $c+d+e$} \)

\(\begin{array}{|rcll|} \hline c+d+e &=& 2*17+2*30+2*32 \\ \mathbf{c+d+e} &=& \mathbf{158} \\ \hline \end{array}\)

The perimeter of \(\triangle DEC\) is \(\mathbf{158}\)

https://web2.0calc.com/questions/help-5_28

\(\begin{array}{|rcll|} \hline \tan(A) &=& \dfrac{3}{2} \\\\ \tan(A) &=& \dfrac{x}{p} \\\\ \hline \dfrac{3}{2} &=& \dfrac{x}{p} \\ \dfrac{2}{3} &=& \dfrac{p}{x} \\ p &=& \dfrac{2}{3}*x \\ \hline \end{array} \begin{array}{|rcll|} \hline \tan(C) &=& \dfrac{2}{3} \\\\ \tan(C) &=& \dfrac{x}{q} \\\\ \hline \dfrac{2}{3} &=& \dfrac{x}{q} \\ \dfrac{3}{2} &=& \dfrac{q}{x} \\ q &=& \dfrac{3}{2}*x \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline p + x + q &=& AC \\\\ \dfrac{2}{3}*x + x+ \dfrac{3}{2}*x &=& AC \quad | \quad AC = \sqrt{13} \\\\ \dfrac{2}{3}*x + x+ \dfrac{3}{2}*x &=& \sqrt{13} \\\\ x \left( \dfrac{2}{3} +1+ \dfrac{3}{2} \right) &=& \sqrt{13} \\\\ x \left( \dfrac{4}{6} +\dfrac{6}{6}+ \dfrac{9}{6} \right) &=& \sqrt{13} \\\\ x * \dfrac{19}{6} &=& \sqrt{13} \\ \mathbf{x} &=& \mathbf{\dfrac{6}{19}* \sqrt{13}} \\ \hline \end{array}\)

The side length of the square is \(\mathbf{\dfrac{6}{19}* \sqrt{13}}\)

https://web2.0calc.com/questions/help-3_37

\(\begin{array}{|rcll|} \hline 2*\text{area}_{PQR} &=& 12*12*\sin(120^\circ) \quad | \quad \sin(120^\circ)= \sin(60^\circ)=\dfrac{\sqrt{3}} {2} \\ 2*\text{area}_{PQR} &=& 12*12*\dfrac{\sqrt{3}} {2} \\ 2*\text{area}_{PQR} &=& 12*6* \sqrt{3} \\ \text{area}_{PQR} &=& 6*6* \sqrt{3} \\ \mathbf{\text{area}_{PQR}} &=& 36* \mathbf{\sqrt{3}} \\ \hline \end{array}\)

https://web2.0calc.com/questions/help-2_42

\(\begin{array}{|rcll|} \hline 3CD &=& 12 \\ CD &=& \dfrac{12}{3} \\ \mathbf{CD} &=& \mathbf{4} \\ \hline AD &=& 3CD \\ AD &=& 3*4 \\ \mathbf{AD} &=& \mathbf{12} \\ \hline AC &=& AD+CD \\ AC &=& 12 + 4 \\ \mathbf{AC} &=& \mathbf{16} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline AB^2 + BC^2 &=& AC^2 \quad | \quad \mathbf{AC=16} \\ AB^2 + BC^2 &=& 16^2 \quad | \quad AB^2 = 12^2+BD^2 \\ 12^2+BD^2 + BC^2 &=& 16^2 \quad | \quad BC^2 = 4^2+BD^2 \\ 12^2+BD^2 + 4^2+BD^2 &=& 16^2 \\ 2BD^2 &=& 16^2 -12^2-4^2 \\ 2BD^2 &=& 256-144-16 \\ 2BD^2 &=& 96 \\ BD^2 &=& 48 \\ BD^2 &=& 16*3 \\ \mathbf{BD} &=& \mathbf{4\sqrt{3}} \\ \hline \end{array}\)

https://web2.0calc.com/questions/help-1_39

\(\begin{array}{|rcll|} \hline \cos(60^\circ) &=& \dfrac{CB}{12} \\ CB &=& 12*\cos(60^\circ) \quad | \quad \cos(60^\circ) = \dfrac12 \\ CB &=& 12*\dfrac12 \\ \mathbf{CB} &=& \mathbf{6} \\ \hline \tan{30^\circ} &=& \dfrac{XC}{CB} \\\\ \tan{30^\circ} &=& \dfrac{XC}{6} \\ XC &=& 6*\tan{30^\circ} \quad | \quad \tan{30^\circ} = \dfrac{\sin(30^\circ)} {\cos(30^\circ)}= \dfrac{\dfrac{1}{2}} {\dfrac{\sqrt{3}} {2} } = \dfrac{1}{\sqrt{3}} \\ XC &=& 6*\dfrac{1}{\sqrt{3}} \\ XC &=& 6*\dfrac{1}{\sqrt{3}}*\dfrac{\sqrt{3}}{\sqrt{3}} \\ \mathbf{XC} &=& \mathbf{2* \sqrt{3}} \\ \hline 2*\text{area}_{BXA} &=& 12*h \quad | \quad h = XC \\ 2*\text{area}_{BXA} &=& 12*XC \quad | \quad \mathbf{XC=2* \sqrt{3}} \\ 2*\text{area}_{BXA} &=& 12*2* \sqrt{3} \\ \mathbf{\text{area}_{BXA}} &=& \mathbf{12*\sqrt{3}} \\ \hline \end{array}\)

Without using a calculator, find the largest prime divisor of \(5^{12}\) minus 2 times \(10^6\) plus \(2^{12}\).

\(\begin{array}{|rcll|} \hline && 5^{12} - 2* 10^6 + 2^{12} \\ &=& 5^{12} - 2* \left(5*2\right)^6 + 2^{12} \\ &=& 5^{12} - 2* 5^6*2^6 + 2^{12} \\ &=& \left(5^{6}- 2^{6} \right)^2 \\ &=& \Big( (5^{3}- 2^{3}) (5^{3}+ 2^{3}) \Big)^2 \\ &=& \Big( (125-8) (125+8) \Big)^2 \\ &=& \left( 117*133 \right)^2 \\ &=& \Big( (3^2*13)*(7*19) \Big)^2 \\ \hline \end{array} \)

The largest prime divisor is \(\mathbf{19}\)

Let ABCD be a trapezoid with bases AB and CD. Let AD=5 and BC=7, and let P be a point on side CD such that CP/PD=7/5.

Let X,Y be the feet of the altitudes from P to AD,BC respectively.

Show that PX=PY.

\(\text{Let area of $\triangle ADP = A_1$} \\ \text{Let area of $\triangle BCP = A_2$} \)

\(\begin{array}{|rcll|} \hline 2A_1 &=& \dfrac{5}{12}DC*h \\\\ 2A_2 &=& \dfrac{7}{12}DC*h \\\\ \hline \dfrac{2A_1}{2A_2} &=& \dfrac{\dfrac{5}{12}DC*h}{\dfrac{7}{12}DC*h} \\\\ \dfrac{A_1}{A_2} &=& \dfrac{5}{12} * \dfrac{12}{7} \\\\ \mathbf{ \dfrac{A_1}{A_2} } &=& \mathbf{ \dfrac{5}{7} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 2A_1 &=& AD*PX \\\\ 2A_2 &=& BC*PY \\\\ \hline \dfrac{2A_1}{2A_2} &=& \dfrac{ AD*PX } {BC*PY} \quad | \quad AD=5,\ DC=7 \\\\ \dfrac{A_1}{A_2} &=& \dfrac{ 5*PX } {7*PY} \quad | \quad \mathbf{ \dfrac{A_1}{A_2} = \dfrac{5}{7} } \\\\ \dfrac{5}{7} &=& \dfrac{ 5*PX } {7*PY} \\\\ \dfrac{5}{7}*\dfrac{7}{5} &=& \dfrac{PX}{PY} \\\\ 1 &=& \dfrac{PX}{PY} \quad | \quad * PY \\\\ \mathbf{PY} &=& \mathbf{PX} \\ \hline \end{array}\)

Let \(n\) be a positive integer. If \(a\equiv (3^{2n}+4)^{-1}\pmod{9}\),
what is the remainder when \(a\) is divided by \(9\)?

\(\begin{array}{|rcll|} \hline \mathbf{a} &\equiv& \mathbf{ (3^{2n}+4)^{-1}\pmod{9} } \\ && \boxed{ 3^{2n}+4 = \left(3^2\right)^n+4 = 9^n+4 } \qquad \gcd(9^n+4,9) = \gcd(4,9)=1!\\ a &\equiv& (9^n+4)^{-1}\pmod{9} \\ && \boxed{(9^n+4)^{-1}\pmod{9} = (9^n+4)^{\phi(9)-1} \quad | \quad \phi(9)=9*\left( 1-\dfrac{1}{3}\right)=6 \\ =(9^n+4)^{5} \\ \mathbf{(9^n+4)^{-1}\pmod{9} =(9^n+4)^{5}}\pmod{9} } \\ a &\equiv& (9^n+4)^{5} \pmod{9} \\\\ a &\equiv& \underbrace{\binom50 9^{5n}+\binom51 9^{4n}*4+\binom52 9^{3n}*4^2+\binom53 9^{2n}*4^3+\binom54 9^{n}*4^4}_{=0\pmod{9} }+ \binom55 4^5 \pmod{9} \\\\ a &\equiv& \binom55 4^5 \pmod{9} \\\\ a &\equiv& 4^5 \pmod{9} \\\\ a &\equiv& 1024 \pmod{9} \\\\ a &\equiv& 1024-113*9 \pmod{9} \\\\ \mathbf{a} &\equiv& \mathbf{7 \pmod{9}} \\ \hline \end{array}\)

If

\(4x\equiv 8\pmod{20}\) and \(3x\equiv 16\pmod{20}\), then what is the remainder when \(x^2\) is divided by \(20\)?

\(\begin{array}{|lrcll|} \hline (1) & 4x &\equiv& 8\pmod{20} \\ & 4x &=& 8 + 20n \qquad n\in \mathbb{Z}\\ (2) & 3x &\equiv& 16\pmod{20} \\ & 3x &=& 16+20m \qquad m\in \mathbb{Z}\\ \hline (1)-(2): & 4x-3x &=& 8-16 + 20(n-m) \\ & x &=& -8 + 20(n-m) \\ & x &\equiv& -8\pmod{20} \\ & x &\equiv& -8+20 \pmod{20} \\ &\mathbf{ x } &\equiv& \mathbf{ 12 \pmod{20} } \\ \hline & x &=& 12+20z \qquad z\in \mathbb{Z}\\ & x^2 &=& (12+20z)^2 \\ & x^2 &=& 144+ 2*12*20z + 20*20z^2 \\ & x^2 &\equiv& 144 \pmod{20} \\ & x^2 &\equiv& 144-7*20 \pmod{20} \\ & \mathbf{ x^2 } &\equiv& \mathbf{ 4 \pmod{20}} \\ \hline \end{array}\)