heureka

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Nombre de usuarioheureka
Puntuación21824
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Preguntas 11
Respuestas 4113

 #3
avatar+21824 
+8

Find the value of \(\mathbf{ {\color{red}x}}\) if

 
\(\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}}}=\dfrac{67}{96}\)

 

\(\begin{array}{|rcll|} \hline \dfrac{67}{96} &=& \dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}}} \\ \left( 1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}} \right)67 &=& 96 \\\\ 67+\dfrac{67}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}} &=& 96 \\\\ \dfrac{67}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}} &=& 96-67 \\\\ \dfrac{67}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}} &=& 29 \\\\ \left( 2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}} \right)29 &=& 67 \\\\ 58+\dfrac{29}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}} &=& 67 \\\\ \dfrac{29}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}} &=& 67-58 \\\\ \dfrac{29}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}} &=& 9 \\\\ \left( 3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}} \right)9&=& 29 \\\\ 27+\dfrac{9}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}&=& 29 \\\\ \dfrac{9}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}&=& 29-27 \\\\ \dfrac{9}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}&=& 2 \\\\ \left( 4+\dfrac{1}{\mathbf{ {\color{red}x}}} \right)2&=& 9 \\\\ 8+\dfrac{2}{\mathbf{ {\color{red}x}}}&=& 9 \\\\ \dfrac{2}{\mathbf{ {\color{red}x}}}&=& 9-8 \\\\ \dfrac{2}{\mathbf{ {\color{red}x}}}&=& 1 \\\\ \mathbf{2} & \mathbf{=} & \mathbf{ {\color{red}x} } \\ \hline \end{array}\)

 

laugh

08-feb-2019
 #2
avatar+21824 
+8

Consider the vectors

\( \mathbf{\vec{v}} = \dbinom{1}{3}\) and  \(\mathbf{\vec{w}} = \dbinom{3}{2}\)
Can you write \( \mathbf{\vec{u}} = \dbinom{1}{2}\)
as a linear combination of  \(\vec{v}\) and \(\vec{w}\) ?
If no, answer with 0.
If yes, find coefficients a and b such \(a \dbinom{1}{3} + b \dbinom{3}{2} = \dbinom{1}{2}\)
and answer with \(\dfrac{a}{b}\).

 

\(\mathbf{\vec{v}_{\perp} = \ ?} \)

\(\begin{array}{|rcll|} \hline \vec{v} \cdot \vec{v}_{\perp} &=& 0 \\ \dbinom{1}{3} \cdot \underbrace{\dbinom{-3}{1}}_{=\vec{v}_{\perp}} &=& -3+3 = 0 \\ \hline \end{array}\)

 

\(\mathbf{\vec{w}_{\perp} = \ ?} \)

\(\begin{array}{|rcll|} \hline \vec{w} \cdot \vec{w}_{\perp} &=& 0 \\ \dbinom{3}{2} \cdot \underbrace{\dbinom{-2}{3}}_{=\vec{w}_{\perp}} &=& -6+6 = 0 \\ \hline \end{array}\)

 

\(\mathbf{a,b = \ ?}\)

\(\begin{array}{|lrclccccc|} \hline & & & & & I. & & II. \\ & a\vec{v} + b\vec{w} &=& \vec{u} & | & \cdot \vec{v}_{\perp}& | & \cdot \vec{w}_{\perp}& | \\ \hline I. & \underbrace{a\vec{v}\vec{v}_{\perp}}_{=0} + b\vec{w}\vec{v}_{\perp} &=& \vec{u}\vec{v}_{\perp} \\ & b\vec{w}\vec{v}_{\perp} &=& \vec{u}\vec{v}_{\perp} \\ & \mathbf{b} &\mathbf{=}& \mathbf{\dfrac{ \vec{u}\vec{v}_{\perp} } { \vec{w}\vec{v}_{\perp} } } \\\\ & b &=& \dfrac{ \dbinom{1}{2}\dbinom{-3}{1} } { \dbinom{3}{2}\dbinom{-3}{1} } \\\\ & b &=& \dfrac{ -3+2 } { -9+2 } \\\\ & b &=& \dfrac{ -1 } { -7} \\\\ & \mathbf{b} &\mathbf{=}& \mathbf{\dfrac{ 1 } { 7}} \\ \hline II. & a\vec{v}\vec{w}_{\perp} + \underbrace{b\vec{w}\vec{w}_{\perp}}_{=0} &=& \vec{u}\vec{w}_{\perp} \\ & a\vec{v}\vec{w}_{\perp} &=& \vec{u}\vec{w}_{\perp} \\ & \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{ \vec{u}\vec{w}_{\perp} } { \vec{v}\vec{w}_{\perp} } } \\\\ & a &=& \dfrac{ \dbinom{1}{2}\dbinom{-2}{3} } { \dbinom{1}{3}\dbinom{-2}{3} } \\\\ & a &=& \dfrac{ -2+6 } { -2+9 } \\\\ & \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{ 4 } { 7 }} \\ \\ \hline & \dfrac{a}{b} &=& \dfrac{\dfrac{ 4 } { 7 }}{\dfrac{ 1 } { 7}} \\\\ & &=& \dfrac{ 4 } { 7 } \cdot \dfrac{ 7 } { 1} \\\\ & \mathbf{\dfrac{a}{b}} &\mathbf{=}& \mathbf{4} \\ \hline \end{array}\)

 

 

laugh

08-feb-2019
 #2
avatar+21824 
+8

A square $DEFG$ varies inside equilateral triangle $ABC,$
so that $E$ always lies on side $\overline{AB},$ $F$ always lies on side $\overline{BC},$
and $G$ always lies on side $\overline{AC}.$
The point $D$ starts on side $\overline{AB},$ and ends on side $\overline{AC}.$
The diagram below shows the initial position of square $DEFG,$ an intermediate position, and the final position.

 

Show that as square DEFG varies, the height of point D above line BC remains constant.

 

\(\text{Let $\overline{DH}=h $ the height of point $D$ above line $\overline{BC}$ } \\ \text{Let ${\color{red}s} $ the side of the square $DEFG$ } \\ \text{Let $\overline{DF}=\sqrt{2}\color{red}s$ } \\ \text{Let $\angle EFD = 45^{\circ}$ } \\ \text{Let $\angle BFE = \alpha$ } \\ \text{Let $\angle CFG = 90^{\circ}-\alpha$ } \\ \text{The equilateral triangle $ABC$ has the angles $60^{\circ}$, $60^{\circ}$, $60^{\circ}$ and the sides $a$, $a$, $a$. }\)

 

In the rectangular triangle DHF the following applies:

\(\begin{array}{|rcll|} \hline \sin(\alpha + 45^{\circ}) &=& \dfrac{h}{\sqrt{2}\color{red}s} \\\\ \sqrt{2}\sin(\alpha + 45^{\circ}) &=& \dfrac{h}{\color{red}s} \quad | \quad \sqrt{2}\sin(\alpha + 45^{\circ}) = \sin(\alpha) + \cos(\alpha) \\\\ \sin(\alpha) + \cos(\alpha) & = & \dfrac{h}{\color{red}s} \\\\ \mathbf{h}& \mathbf{=} & \mathbf{{\color{red}s}\sin(\alpha) + {\color{red}s}\cos(\alpha)} \\ \hline \end{array}\)

 

\(\text{Let $\overline{BC}=a $ is the side of the equilateral triangle $ABC$. }\)

\(\begin{array}{|rcll|} \hline a &=& \overline{BM} + \overline{MF} +\overline{FN} + \overline{NC} \\ && \overline{BM} = \dfrac{s\sin(\alpha)}{\tan(60^{\circ})} \\ && \overline{MF} = s\cos(\alpha) \\ && \overline{FN} = s\sin(\alpha) \\ && \overline{NC} = \dfrac{s\cos(\alpha)}{\tan(60^{\circ})} \\ a &=& \dfrac{ {\color{red}s} \sin(\alpha)}{\tan(60^{\circ})} + {\color{red}s}\cos(\alpha) +{\color{red}s}\sin(\alpha) + \dfrac{{\color{red}s}\cos(\alpha)}{\tan(60^{\circ})} \\ a &=& {\color{red}s}\sin(\alpha) +{\color{red}s}\cos(\alpha) + \Big({\color{red}s}\sin(\alpha) +{\color{red}s}\cos(\alpha)\Big) \dfrac{1}{ \tan(60^{\circ}) } \\ a &=& \Big({\color{red}s}\sin(\alpha) +{\color{red}s}\cos(\alpha)\Big) \left( 1+\dfrac{1}{ \tan(60^{\circ}) } \right) \quad | \quad \mathbf{h={\color{red}s}\sin(\alpha) + {\color{red}s}\cos(\alpha)} \\ a &=& h \left( 1+\dfrac{1}{ \tan(60^{\circ}) } \right) \quad | \quad \tan(60^{\circ}) = \sqrt{3} \\ a &=& h \left( 1+\dfrac{1}{ \sqrt{3} } \right) \\ a &=& h \left( \dfrac{ \sqrt{3}+1}{ \sqrt{3} } \right) \\ \mathbf{h}& \mathbf{=} & \mathbf{a \left( \dfrac{ \sqrt{3} }{ 1+\sqrt{3}} \right)} \\ \hline \end{array}\)

 

So the height of point D above line BC remains constant.

 

laugh

08-feb-2019