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wie löse ich

sqrt(x-2016)+sqrt(y-56)=11

x+y=2193

1. Lösung

x = 2016, y = 177

$\begin{array}{rcll} \sqrt{2016-2016} + \sqrt{177-56} \stackrel{?} = 11 \\ \sqrt{0} + \sqrt{121} \stackrel{?} = 11 \\ \sqrt{0} + 11 \stackrel{?} = 11 \\ 11 =11 \checkmark\\\\ 2016 + 177 \stackrel{?} = 2193 \\ 2193 = 2193 \checkmark\\ \end{array}$

2. Lösung
x = 2137, y = 56

$\begin{array}{rcll} \sqrt{2137-2016} + \sqrt{56-56} \stackrel{?} = 11 \\ \sqrt{121} + \sqrt{0} \stackrel{?} = 11 \\ 11 + \sqrt{0} \stackrel{?} = 11 \\ 11 =11 \checkmark\\\\ 2137 + 56 \stackrel{?} = 2193 \\ 2193 = 2193 \checkmark\\ \end{array}$

wie löse ich

sqrt(x-2016)+sqrt(y-56)=11

x+y=2193

Three resistors are connected in parallel.
Their values are 50 ohms,75 ohms and 150 ohms.
Calculate the total resistance (RT) ?

$R_1 = 50 ~Ohms\\ R_2 = 75 ~Ohms\\ R_3 = 150~ Ohms$

$\begin{array}{|rcll|} \hline \frac{1}{R_T} &=& \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \\\\ \frac{1}{R_T} &=& \frac{1}{50} + \frac{1}{75} + \frac{1}{150} \\ \frac{1}{R_T} &=& 0.02 + 0.01333333333 + 0.00666666667 \\ \frac{1}{R_T} &=& 0.04 \\ R_T &=& \frac{1}{0.04} \\ R_T &=& 25~Ohms \\ \hline \end{array}$

wie löse ich

sqrt(x-2016)+sqrt(y-56)=11

x+y=2193

Wir haben:

$\begin{array}{|lrcrl|} \hline (1) & \sqrt{x-2016}+\sqrt{y-56} &=& 11\\ (2) & x+y &=& 2193\\ \hline \end{array}$

Auflösung der 2. Gleichung nach x:

$\begin{array}{|lrcll|} \hline (2) & x+y &=& 2193 \quad & | \quad -y \\ & \mathbf{ x } & \mathbf{=} & \mathbf{2193 - y} \\ \hline \end{array}$

Auflösung der 1. Gleichung nach y:

$\begin{array}{|rcll|} \hline \sqrt{x-2016} +\sqrt{y-56}&=& 11 & | \quad -\sqrt{y-56} \\ \sqrt{x-2016} &=& 11 \quad -\sqrt{y-56} & | \quad \text{beide Seiten quadrieren}\\ x-2016 &=& (11 \quad -\sqrt{y-56})^2 \\ x-2016 &=& 11^2 -2\cdot 11 \cdot \sqrt{y-56} +(y-56) & | \quad + 2016\\ x &=& 11^2 -2\cdot 11 \cdot \sqrt{y-56} + y-56+ 2016\\ x &=& 121 -2\cdot 11 \cdot \sqrt{y-56} + y+1960\\ x &=& -2\cdot 11 \cdot \sqrt{y-56} + y+ 1960+121\\ x &=& -2\cdot 11 \cdot \sqrt{y-56} + y+ 2081 & | \quad +2\cdot 11 \cdot \sqrt{y-56}\\ 2\cdot 11 \cdot \sqrt{y-56} + x &=& y+ 2081 d & | \quad - x\\ 2\cdot 11 \cdot \sqrt{y-56} &=& y+ 2081 - x & | \quad x =2193 - y\\ 2\cdot 11 \cdot \sqrt{y-56} &=& y+ 2081 - (2193 - y) \\ 2\cdot 11 \cdot \sqrt{y-56} &=& y+ 2081 - 2193 + y \\ 11 \cdot \sqrt{y-56} &=& y - 56 & | \quad \text{beide Seiten quadrieren}\\ 121 \cdot (y-56) &=& (y-56)^2 \\ 121 \cdot (y-56) &=& y^2- 112y + 3136 \\ 121y - 6776 &=& y^2- 112y + 3136 & | \quad - 121y + 6776\\ 0 &=& y^2- 112y + 3136 - 121y + 6776\\ 0 &=& y^2- 233y + 9912\\ \mathbf{ y^2- 233y + 9912 } & \mathbf{=} & \mathbf{0} \\\\ y_{1,2} &=& \frac{233\pm\sqrt{233^2-4\cdot 9912} } { 2 } \\ y_{1,2} &=& \frac{233\pm\sqrt{14641} } { 2 } \\ y_{1,2} &=& \frac{233\pm 121 } { 2 } \\\\ y_1 &=& \frac{233 + 121 } { 2 } \\ \mathbf{ y_1} &\mathbf{ =}& \mathbf{ 177} \\\\ y_2 &=& \frac{233 - 121 } { 2 } \\ \mathbf{ y_2} &\mathbf{ =}& \mathbf{ 56} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline x_1 & = & 2193 - y_1 \\ x_1 & = & 2193 - 177 \\ \mathbf{ x_1} & \mathbf{ =} & \mathbf{2016} \\\\ x_2 & = & 2193 - y_2 \\ x_2 & = & 2193 - 56 \\ \mathbf{ x_2} & \mathbf{ =} & \mathbf{2137} \\ \hline \end{array}$

2sec(x+60)=5sec(x-20)

$\begin{array}{|rcll|} \hline 2\cdot \sec(x+60^{\circ}) &=& 5\cdot \sec(x-20^{\circ}) \\ \text{We substitute } x = y-20^{\circ} \\\\ 2\cdot \sec(y-20^{\circ}+60^{\circ}) &=& 5\cdot \sec(y-20^{\circ}-20^{\circ}) \\ 2\cdot \sec(y+40^{\circ}) &=& 5\cdot \sec(y-40^{\circ}) \quad | \quad \sec(\phi) = \frac{1}{\cos(\phi)}\\ \frac{ 2 } { \cos(y+40^{\circ}) } &=& \frac{ 5 } { \cos(y-40^{\circ}) } \\ \frac{ \cos(y-40^{\circ}) } { \cos(y+40^{\circ}) } &=& \frac{ 5 } { 2 } \quad | \quad \cos(y-40^{\circ}) = \cos(y)\cos(40^{\circ})+\sin(y)\sin(40^{\circ}) \\ \frac{ \cos(y)\cos(40^{\circ})+\sin(y)\sin(40^{\circ}) } { \cos(y+40^{\circ}) } &=& \frac{ 5 } { 2 } \quad | \quad \cos(y+40^{\circ}) = \cos(y)\cos(40^{\circ})-\sin(y)\sin(40^{\circ}) \\ \frac{ \frac{\cos(y)\cos(40^{\circ})+\sin(y)\sin(40^{\circ})} {\cos(y)\cos(40^{\circ})} } { \frac{\cos(y)\cos(40^{\circ})-\sin(y)\sin(40^{\circ})} {\cos(y)\cos(40^{\circ})} } &=& \frac{ 5 } { 2 } \\ \frac{ 1+\tan(y)\tan(40^{\circ}) } {1-\tan(y)\tan(40^{\circ}) } &=& \frac{ 5 } { 2 } \\\\ 2\cdot [~ 1+\tan(y)\tan(40^{\circ}) ~] &=& 5\cdot [~ 1-\tan(y)\tan(40^{\circ}) ~] \\ 2 + 2\cdot \tan(y)\tan(40^{\circ}) &=& 5 - 5 \cdot \tan(y)\tan(40^{\circ}) \\ 7\cdot \tan(y)\tan(40^{\circ}) &=& 3 \\ \tan(y) &=& \frac37\cdot \frac{1}{\tan(40^{\circ})} \quad | \quad y=x+20^{\circ}\\ \tan(x+20^{\circ}) &=& \frac37\cdot \frac{1}{\tan(40^{\circ})} \\ x+20^{\circ} &=& \arctan{\left( \frac37\cdot \frac{1}{\tan(40^{\circ})} \right)} \pm n\cdot 180^{\circ}\\ x &=& -20^{\circ} + \arctan{\left( \frac37\cdot \frac{1}{\tan(40^{\circ})} \right)} \pm n\cdot 180^{\circ}\\ x &=& -20^{\circ} + \arctan{\left( 0.51075153968 \right)} \pm n\cdot 180^{\circ}\\ x &=& -20^{\circ} + 27.0557431286^{\circ} \pm n\cdot 180^{\circ}\\\\ \mathbf{x} &\mathbf{=}& \mathbf{7.0557431286^{\circ} \pm n\cdot 180^{\circ} } \qquad n \in N\\ \hline \end{array}$

If log_(8)3=P and log_(3)5=Q, express log_(10)5 in terms of P and Q.

Your answer should no longer include any logarithms.

$\begin{array}{|rcll|} \hline \log _b \left( x \right) &=& \dfrac{ \log _c \left( x \right) } { \log _c \left( b \right) } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline &P &=& \log_8 \left( 3 \right) \\\\ &P &=& \dfrac{ \log _{10} \left( 3 \right) } { \log _{10} \left( 8 \right) } \\\\ &P &=& \dfrac{ \log _{10} \left( 3 \right) } { \log _{10} \left( 2^3 \right) } \\\\ &P &=& \dfrac{ \log _{10} \left( 3 \right) } { 3\cdot \log _{10} \left( 2 \right) } \\ \hline (1) & \log _{10} \left( 2 \right) &=& \dfrac{ \log _{10} \left( 3 \right) } {3P} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline &Q &=& \log_3 \left( 5 \right) \\\\ &Q &=& \dfrac{ \log _{10} \left( 5 \right) } { \log _{10} \left( 3 \right) } \\\\ \hline (2) & \log _{10} \left( 3 \right) &=& \dfrac{ \log _{10} \left( 5 \right) } {Q} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \log _{10} \left( 5 \right) &=& \log _{10} \left( \frac{10}{2} \right) \\\\ \log _{10} \left( 5 \right) &=& \log _{10} \left( 10 \right)- \log _{10} \left( 2 \right) \quad &| \quad \log _{10} \left( 10 \right) = 1\\\\ \log _{10} \left( 5 \right) &=& 1- \log _{10} \left( 2 \right) \quad &| \quad \log _{10} \left( 2 \right) = \dfrac{ \log _{10} \left( 3 \right) } {3P}\\\\ \log _{10} \left( 5 \right) &=& 1- \dfrac{ \log _{10} \left( 3 \right) } {3P} \quad &| \quad \log _{10} \left( 3 \right) = \dfrac{ \log _{10} \left( 5 \right) } {Q}\\\\ \log _{10} \left( 5 \right) &=& 1- \dfrac{ \dfrac{ \log _{10} \left( 5 \right) } {Q} } {3P} \\\\ \log _{10} \left( 5 \right) &=& 1- \dfrac{ \log _{10} \left( 5 \right) } {3PQ} \\\\ \log _{10} \left( 5 \right) + \dfrac{ \log _{10} \left( 5 \right) } {3PQ} &=& 1 \\\\ \log _{10} \left( 5 \right)\cdot \left( 1+ \dfrac{ 1 } {3PQ} \right ) &=& 1 \\\\ \log _{10} \left( 5 \right)\cdot \left( \dfrac{ 3PQ+1 } {3PQ} \right ) &=& 1 \\\\ \mathbf{\log _{10} \left( 5 \right)} &\mathbf{=}& \mathbf{\dfrac{3PQ}{ 3PQ+1 } }\\ \hline \end{array}$

35th term of Arithmetic sequence 3,7,11,15,19

$\small{\text{We have $t_x=t_1=3$ and $t_y=t_{2}=7$ and we want $t_z=t_{35}=?$ }}\\\\ \small{\text{$\boxed{~~t_z = t_x\cdot \left( \dfrac{y-z}{y-x}\right) + t_y\cdot \left(\dfrac{z-x}{y-x}\right) ~~} $}}\\\\ \small{\text{$ \begin{array}{rcl} t_{35} &=& t_1\cdot \left( \dfrac{2-35}{2-1}\right) + t_{2}\cdot \left(\dfrac{35-1}{2-1}\right) \\\\ t_{35} &=& 3\cdot \left( \dfrac{-33}{1}\right) + 7\cdot \left(\dfrac{34}{1}\right) \\\\ t_{35} &=& -3\cdot 33 + 7\cdot 34 \\ t_{35} &=& -99 + 238 \\ \mathbf{t_{35}} & \mathbf{=} & \mathbf{139}\\ \hline \\ \end{array} $}}$

Find the remainder when $1^3 + 2^3 + 3^3 + \dots + 100^3$ is divided by 6.

$\text{The sum of } ~ 1^3 + 2^3 + 3^3 + \dots + n^3 = \left[ \frac{ n*(n+1) } {2} \right]^2\\ \text{The sum of } ~ 1^3 + 2^3 + 3^3 + \dots + 100^3 = \left[ \frac{ 100*(100+1) } {2} \right]^2$

$\begin{array}{|rcll|} \hline && 1^3 + 2^3 + 3^3 + \dots + 100^3 \pmod 6 \\ &\equiv & \left[ \frac{ 100*(100+1) } {2} \right]^2 \pmod 6 \\ &\equiv & \left[ \frac{ 100*101 } {2} \right]^2 \pmod 6 \\ &\equiv & \left[ 50*101 \right]^2 \pmod 6 \\ &\equiv & 5050^2 \pmod 6 \qquad &| \qquad 5050 \pmod 6 = 4 \\ &\equiv & 4^2 \pmod 6 \\ &\equiv & 16 \pmod 6 \\ &\equiv & 4 \pmod 6 \\ \hline \end{array}$

What is the remainder when $13^{51}$ is divided by 5?

$\begin{array}{|rcll|} \hline && 13^{51} \pmod {5} \qquad &| \qquad 13 \pmod 5 \equiv 3 \\ &\equiv& 3^{51} \pmod {5} \qquad &| \qquad 3^4 \equiv 1 \pmod 5 \\ &\equiv & 3^{4\cdot 12+3} \pmod {5} \\ &\equiv & 3^{4\cdot 12} \cdot 3^3 \pmod {5} \\ &\equiv & (3^{4})^{12} \cdot 3^3 \pmod {5} \qquad &| \qquad 3^4 \equiv 1 \pmod 5 \\ &\equiv & (1)^{12} \cdot 3^3 \pmod {5} \qquad &| \qquad (1)^{12}=1\\ &\equiv & 1 \cdot 3^3 \pmod {5} \\ &\equiv & 27 \pmod {5}\\ &\equiv & 2 \pmod {5}\\ \hline \end{array}$

What is the remainder when $2^{2005}$ is divided by 7?

$\begin{array}{|rcll|} \hline && 2^{2005} \pmod {7} \qquad &| \qquad 2^6 \equiv 1 \pmod 7 \\ &\equiv & 2^{6\cdot 334+1} \pmod {7} \\ &\equiv & 2^{6\cdot 334} \cdot 2 \pmod {7} \\ &\equiv & (2^{6})^{334} \cdot 2 \pmod {7} \qquad &| \qquad 2^6 \equiv 1 \pmod 7 \\ &\equiv & (1)^{334} \cdot 2 \pmod {7} \qquad &| \qquad (1)^{334}=1\\ &\equiv & 1 \cdot 2 \pmod {7}\\ &\equiv & 2 \pmod {7}\\ \hline \end{array}$