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 #1
avatar+18281 
+1

Right \(\triangle{ABC}\) has AB = 3, BC = 4, and AC = 5.
Square XYZW is inscribed in triangle ABC with X and Y on \(\overline{AC}\),
W on \(\overline{AB}\), and
Z on \(\overline{BC}\).
What is the side length of the square?

 

Let s is the side length oft the square \( = \overline{XY} = \overline{YZ} = \overline{ZW} = \overline{WX}\)

Let h = \(\overline{BT}\)

Let A the area of \(\triangle{ABC}\)

 

 

h = ?

\(\begin{array}{|rcll|} \hline A &=& \frac{\overline{AB} \cdot \overline{BC} }{2} \\ A &=& \frac{3\cdot 4}{2} \\ \mathbf{A} &\mathbf{=}& \mathbf{6} \\\\ A &=& \frac{\overline{AC}\cdot h}{2} \\ A &=& \frac{5\cdot h}{2} \quad & | \quad \mathbf{A=6} \\ 6 &=& \frac{5\cdot h}{2} \\ \mathbf{h} &\mathbf{=}& \mathbf{ \frac{12}{5} } \\ \hline \end{array}\)

 

\(\mathbf{\overline{BW} =\ ?}\)

\(\begin{array}{|rcll|} \hline \frac{ \overline{BW} } {s} &=& \frac{ \overline{AB} } { \overline{AC} } \\ \frac{ \overline{BW} } {s} &=& \frac{ 3 } { 5 } \\ \mathbf{ \overline{BW} } & \mathbf{=} & \mathbf{ \frac{3}{5}s } \\ \hline \end{array}\)

 

\(\mathbf{\overline{BZ} =\ ?} \)

\(\begin{array}{|rcll|} \hline \frac{ \overline{BZ} } {s} &=& \frac{ \overline{BC} } { \overline{AC} } \\ \frac{ \overline{BZ} } {s} &=& \frac{ 4 } { 5 } \\ \mathbf{\overline{BZ}} &\mathbf{=}& \mathbf{\frac{4}{5}s } \\ \hline \end{array}\)

 

s = ?

\(\begin{array}{|rcll|} \hline A_{\triangle{ZBW}} = \frac{ \overline{BW}\cdot \overline{BZ} } {2} &=& \frac{s\cdot(h-s)} {2} \\ \overline{BW}\cdot \overline{BZ} &=& s\cdot(h-s) \quad & | \quad \mathbf{ \overline{BW} =\frac{3}{5}s } \quad \mathbf{ \overline{BZ} =\frac{4}{5}s } \quad \mathbf{h=\frac{12}{5}} \\ \frac{3}{5}s \cdot \frac{4}{5}s &=& s\cdot(\frac{12}{5}-s) \\ \frac{12}{25}s &=& \frac{12}{5}-s \\ s+\frac{12}{25}s &=& \frac{12}{5} \\ s \cdot \left(1+\frac{12}{25} \right) &=& \frac{12}{5} \\ s \cdot \left(\frac{25+12}{25} \right) &=& \frac{12}{5} \\ s \cdot \left(\frac{37}{25} \right) &=& \frac{12}{5} \\ s &=& \frac{25}{37} \cdot \frac{12}{5} \\ s &=& \frac{5}{37} \cdot 12 \\ s &=& \frac{60}{37} \\ \mathbf{s} &\mathbf{=}& \mathbf{1.\overline{621}} \\ \hline \end{array}\)

 

The side length oft the square is \(\mathbf{1.\overline{621}}\)

 

 

laugh

heureka 11 hours ago
 #1
avatar+18281 
+2

Let triangle ABC be a triangle such that AB=13, BC=14, and AC=15. 

Meanwhile, D is a point on BC such that AD bisects angle A.

Find the area of triangle ADC

 

Let area of triangle \(\text{ADC} = A_\text{ADC} \)

Let area of triangle \( \text{ADB} = A_\text{ADB}\)
Let area of triangle \(\text{ABC} = A\)

 

Let \(A_\text{ADB} = A - A_\text{ADC}\)

 

\(\begin{array}{|rcll|} \hline \frac { A_\text{ADB} } { A_\text{ADC} } &=& \frac{ \overline{AD}\cdot 13 \cdot \sin(\frac{A}{2}) \cdot \frac12 } { \overline{AD}\cdot 15 \cdot \sin(\frac{A}{2}) \cdot \frac12 } \\ \frac { A_\text{ADB} } { A_\text{ADC} } &=& \frac{ 13 } { 15 } \quad & | \quad A_\text{ADB} = A - A_\text{ADC} \\ \frac { A - A_\text{ADC} } { A_\text{ADC} } &=& \frac{ 13 } { 15 } \\ \frac { A } { A_\text{ADC} } - 1 &=& \frac{ 13 } { 15 } \\ \frac { A } { A_\text{ADC} } &=& 1+ \frac{ 13 } { 15 } \\ \frac { A } { A_\text{ADC} } &=& \frac{ 28 } { 15 } \\ \frac { A_\text{ADC} } { A } &=& \frac{ 15 } { 28 } \\\\ \mathbf{ A_\text{ADC } } & \mathbf{=} & \mathbf{ \frac{ 15 } { 28 } A } \\ \hline \end{array} \)

 

Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is

\( A = \sqrt{s(s-a)(s-b)(s-c)}\)

 

where s is the semiperimeter of the triangle; that is,
\(s=\frac{a+b+c}{2}\).

 

\(\begin{array}{|rcll|} \hline s &=& \frac{a+b+c}{2} \quad & | \quad a= 14, \ b= 15, \ c= 13 \\ s &=& \frac{15+15+13}{2} \\ s &=& \frac{42}{2} \\ \mathbf{ s }& \mathbf{=} & \mathbf{21} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline A &=& \sqrt{s(s-a)(s-b)(s-c)} \quad & | \quad s= 21, \ a= 14, \ b= 15, \ c= 13 \\ A &=& \sqrt{21(21-14)(21-15)(21-13)} \\ A &=& \sqrt{21\cdot 7 \cdot 6 \cdot 8 } \\ A &=& \sqrt{3\cdot 7 \cdot 7 \cdot 2\cdot 3 \cdot 2\cdot 4 } \\ A &=& \sqrt{4^2\cdot 7^2 \cdot 3^2 } \\ A &=& 4\cdot 7 \cdot 3 \\ \mathbf{ A }& \mathbf{=} & \mathbf{84} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{ A_\text{ADC } } & \mathbf{=} & \mathbf{ \frac{ 15 } { 28 } A } \quad & | \quad \mathbf{ A } \mathbf{=} \mathbf{84} \\ A_\text{ADC } & = & \frac{ 15 } { 28 } \cdot 84 \\ A_\text{ADC } & = & 15 \cdot 3 \\\\ \mathbf{ A_\text{ADC } }& \mathbf{=} & \mathbf{45} \\ \hline \end{array}\)

 

laugh

heureka Jul 27, 2017
 #1
avatar+18281 
+1

Find dy/dx

Yx+XX+Xy= ab

 

\(\begin{array}{|rcll|} \hline y^x+x^x+x^y &=& a^b \\ y^x+x^x+x^y -a^b &=& 0 \\ \hline \end{array} \)

 

Formula:

\(\begin{array}{|rcll|} \hline \frac{dy}{dx} = - \dfrac{F_x}{F_y} \\ \hline \end{array} \)

 

\(\mathbf{F_x =\ ?} \begin{array}{|rcll|} \hline F_x &=& \frac{d }{dx}y^x + \frac{d }{dx}x^x + \frac{d }{dx}x^y + \frac{d }{dx}a^b \\\\ h &=& y^x \quad & | \quad \ln() \\ \ln(h) &=& x\cdot \ln(y) \\ \frac{h'}{h} &=& \ln(y) \\ h' &=& h\cdot \ln(y) \\ h' &=& y^x\cdot \ln(y) \\ \mathbf{ \frac{d }{dx}y^x } & \mathbf{=} & \mathbf{ y^x\cdot \ln(y) } \\\\ h &=& x^x \quad & | \quad \ln() \\ \ln(h) &=& x\cdot \ln(x) \\ \frac{h'}{h} &=& x\cdot \frac{1}{x} + 1\cdot \ln(x) \\ h' &=& h \cdot \Big( 1+\ln(x) \Big) \\ \mathbf{ \frac{d }{dx}x^x } & \mathbf{=} & \mathbf{ x^x \cdot \Big( 1+ \ln(x) \Big) } \\\\ h &=& x^y \quad & | \quad \ln() \\ \ln(h) &=& y\cdot \ln(x) \\ \frac{h'}{h} &=& \frac{y}{x} \\ h' &=& h \cdot \frac{y}{x} \\ h' &=& x^y \cdot \frac{y}{x} \\ \mathbf{ \frac{d }{dx}x^y } & \mathbf{=} & \mathbf{ x^y \cdot \frac{y}{x} } \\\\ \mathbf{ \frac{d }{dx}a^b } & \mathbf{=} & \mathbf{ 0 } \\ \hline \end{array} \)

 

\(\mathbf{F_y =\ ?} \begin{array}{|rcll|} \hline F_y &=& \frac{d }{dy}y^x + \frac{d }{dy}x^x + \frac{d }{dy}x^y +\frac{d }{dy}a^b \\\\ h &=& y^x \quad & | \quad \ln() \\ \ln(h) &=& x\cdot \ln(y) \\ \frac{h'}{h} &=& \frac{x}{y} \\ h' &=& h\cdot \frac{x}{y} \\ h' &=& y^x\cdot \frac{x}{y} \\ \mathbf{ \frac{d }{dy}y^x } & \mathbf{=} & \mathbf{ y^x\cdot \frac{x}{y} } \\\\ \mathbf{ \frac{d }{dy}x^x } & \mathbf{=} & \mathbf{ 0 } \\ h &=& x^y \quad & | \quad \ln() \\ \ln(h) &=& y\cdot \ln(x) \\ \frac{h'}{h} &=& \ln(x) \\ h' &=& h \cdot \ln(x) \\ h' &=& x^y \cdot \ln(x) \\ \mathbf{ \frac{d }{dy}x^y } & \mathbf{=} & \mathbf{ x^y \cdot \ln(x) } \\\\ \mathbf{ \frac{d }{dy}a^b } & \mathbf{=} & \mathbf{ 0 } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \frac{dy}{dx} &=& - \dfrac{F_x}{F_y} \\\\ \frac{dy}{dx} &=& - \dfrac{ y^x\cdot \ln(y) + x^x \cdot \Big( 1+ \ln(x) \Big) + x^y \cdot \frac{y}{x} + 0 } { y^x\cdot \frac{x}{y} + 0 + x^y \cdot \ln(x) + 0 } \\\\ \frac{dy}{dx} &=& - \dfrac{ y^x\cdot \ln(y) + x^x \cdot \Big( 1+ \ln(x) \Big) + x^y \cdot \frac{y}{x} } { y^x\cdot \frac{x}{y} + x^y \cdot \ln(x) } \\\\ \frac{dy}{dx} &=& - \dfrac{ y^x\cdot \ln(y) + x^x \cdot \Big( 1+ \ln(x) \Big) + x^{y-1} \cdot y } { y^{x-1}\cdot x + x^y \cdot \ln(x) } \\\\ \mathbf{ \frac{dy}{dx} } & \mathbf{=} & \mathbf{ - \dfrac{ y^x\cdot \ln(y) + x^x \cdot \Big( 1+ \ln(x) \Big) + x^{y-1} \cdot y } { x^y \cdot \ln(x) + y^{x-1}\cdot x } } \\ \hline \end{array}\)

 

laugh

heureka Jul 25, 2017
 #1
avatar+18281 
+1

lim n→∞{(1− 3/n)^3n +3*n'te√3n}

 

\(\lim \limits_{x\to 0} { \Big( \left( 1- \frac{3}{n} \right)^{3n} +3 \cdot \sqrt[n]{3n} \Big) }= \ ? \)

 

\(\begin{array}{|rcll|} \hline && \lim \limits_{x\to 0} { \Big( \left( 1- \frac{3}{n} \right)^{3n} +3 \cdot \sqrt[n]{3n} \Big) } \\ &=& \lim \limits_{x\to 0} {\Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) } + \lim \limits_{x\to 0} {\Big( 3 \cdot \sqrt[n]{3n} \Big) } \\ \hline \end{array} \)

 

\(y=\lim \limits_{x\to 0} {\Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) }= \ ?\)

\(\begin{array}{|rcll|} \hline \ln(y) &=& \ln\Big( \lim \limits_{x\to 0} \Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) \Big) \\ &=& \lim \limits_{x\to 0} \Big( \ln \Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) \Big) \\ &=& \lim \limits_{x\to 0} \Big( 3n\ln \left( 1- \frac{3}{n} \right) \Big) \\ &=& \lim \limits_{x\to 0} \Big( \frac{3\ln \left( 1- \frac{3}{n} \right)} {n^{-1}} \Big) \\\\ && \text{Regel von de l’Hospital anwenden} \\ && \text{Die 1. Ableitung von } 3\cdot \ln(1-3\cdot n^{-1}) \text{ lautet } 3 \cdot \frac{ [(-3)(-1)\cdot n^{-2} ] } {1-\frac{3}{n} } \\\\ &=& \lim \limits_{x\to 0} \Big( \frac{3 \cdot \frac{ [(-3)(-1)\cdot n^{-2} ] } {1-\frac{3}{n} } } {(-1)\cdot n^{-2}} \Big) \\ &=& \lim \limits_{x\to 0} \Big( \frac{ 3 \cdot[(-3)(-1)\cdot n^{-2} ] } { (1-\frac{3}{n})(-1)\cdot n^{-2} } \Big) \\ &=& \lim \limits_{x\to 0} \Big( \frac{ 3 \cdot(-3)} { (1-\frac{3}{n}) } \Big) \\ &=& \lim \limits_{x\to 0} \Big( \frac{ -9 } { (1-\frac{3}{n}) } \Big) \\ \ln(y) &=& \frac{ -9 } { 1-0 } \\ \ln(y) &=& -9 \\ e^{\ln(y)} = y &=& e^{-9}\\ y &=& \lim \limits_{x\to 0} {\Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) }= e^{-9} \\ \hline \end{array}\)

 

 

\(\lim \limits_{x\to 0} {\Big( 3 \cdot \sqrt[n]{3n} \Big) } = \ ?\)

\(\begin{array}{|rcll|} \hline && \lim \limits_{x\to 0} {\Big( 3 \cdot \sqrt[n]{3n} \Big) } \\ &=& \lim \limits_{x\to 0} {\Big( 3 \cdot (3n)^{\frac{1}{n}} \Big) } \\ &=& \lim \limits_{x\to 0} {\Big( 3 \cdot e^{\ln \left((3n)^{\frac{1}{n}} \right) } \Big) } \\ &=& \lim \limits_{x\to 0} {\Big( 3 \cdot e^{ \frac{\ln(3n)} {n} } \Big) } \\ &=& 3 \cdot e^{ \lim \limits_{x\to 0} {\Big( \frac{\ln(3n)} {n} } \Big) } \\ && \text{Regel von de l’Hospital anwenden} \\ && \text{Die 1. Ableitung von } \frac{\ln(3n)} {n} \text{ lautet } \frac{ \frac{3}{3n} } {1} = \frac{3}{3n} = \frac{1}{n} \\\\ &=& 3 \cdot e^{ \lim \limits_{x\to 0} {\Big(\frac{1}{n}} \Big) } \\ &=& 3 \cdot e^0 \\ &=& 3 \cdot 1 \\ &=& 3 \\ \lim \limits_{x\to 0} {\Big( 3 \cdot \sqrt[n]{3n} \Big) } &=& 3 \\ \hline \end{array} \)

 

Somit erhalten wir insgesamt:

 

\(\lim \limits_{x\to 0} { \Big( \left( 1- \frac{3}{n} \right)^{3n} +3 \cdot \sqrt[n]{3n} \Big) } = e^{-9} + 3\)

 

laugh

heureka Jul 21, 2017
 #2
avatar+18281 
+1

Compute

\(1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.\)

 

Let \(r =\frac12\)

 

\(\begin{array}{|rcll|} \hline s_n &=& 1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} \\\\ && r = \frac12 \\\\ \hline s_n &=& 1\cdot r + 2\cdot r^2 + 3\cdot r^3 + \dots + n \cdot r^n \\ rs_n &=& \qquad \quad 1\cdot r^2 + 2\cdot r^3 + \dots + (n-1) \cdot r^n + n\cdot r^{n+1} \\ \hline s_n -r\cdot s_n &=& r+r^2+r^3+ \dots +r^n-n\cdot r^{n+1} \\ s_n\cdot(1-r) &=& \underbrace{( r+r^2+r^3+ \dots +r^n )}_{=S_n} -n\cdot r^{n+1} \\ \hline && S_n = r+r^2+r^3+ \dots +r^n \\ && rS_n = \quad r^2+r^3 + \dots + r^{n+1} \\ \hline && S_n -r\cdot S_n = r - r^{n+1} \\ && S_n\cdot (1-r) = r - r^{n+1} \\ && S_n = \frac{r - r^{n+1}}{1-r} \\ \hline s_n\cdot(1-r) &=& \underbrace{( r+r^2+r^3+ \dots +r^n )}_{=S_n} -n\cdot r^{n+1} \\ s_n\cdot(1-r) &=& \frac{r - r^{n+1}}{1-r} -n\cdot r^{n+1} \\ s_n &=& \frac{1}{1-r} \cdot \Big( \frac{r - r^{n+1}}{1-r} -n\cdot r^{n+1} \Big) \\\\ && \frac{1}{1-r} = 2 \\\\ s_n &=& 2 \cdot \Big( 2\cdot(r - r^{n+1}) -n\cdot r^{n+1} \Big) \\ s_n &=& 2 \cdot ( 2\cdot r - 2\cdot r^{n+1} - n\cdot r^{n+1} ) \\ s_n &=& 2 \cdot r\cdot ( 2 - 2\cdot r^n - n\cdot r^n ) \\ \\ \mathbf{s_n} & \mathbf{=} & \mathbf{2 \cdot r\cdot \Big( 2 - (2+n)\cdot r^n \Big)} \quad & | \quad r=\frac12 \\ \\ s_n & = & 2 \cdot \frac12 \cdot \Big( 2 - (2+n)\cdot (\frac12)^n \Big) \\ s_n & = & 2 - (2+n)\cdot \frac{1}{2^n} \\ \\ \mathbf{s_n} & \mathbf{=} & \mathbf{2 - \frac{2+n}{2^n} } \\ \hline \end{array}\)

 

laugh

heureka Jul 21, 2017