heureka

a tree casts a shadow 4 ft. long while a house casts a shadow 2 and a half feet long. if the house is 26 ft. high, how tall is the tree? Round to the nearest foot.

$$\dfrac{ \mathrm{tree} } { \mathrm{house} } = \dfrac{ 4 \mathrm{~ft.} }{ 2.5 \mathrm{~ft.} } \qquad \mathrm{house}=26 \mathrm{~ft.}\\\\ \mathrm{tree} = \mathrm{house} \cdot \dfrac{ 4 \mathrm{~ft.} }{ 2.5 \mathrm{~ft.} }\\\\ \mathrm{tree} = 26 \mathrm{~ft.} \cdot \dfrac{ 4 \mathrm{~ft.} }{ 2.5 \mathrm{~ft.} }\\\\ \mathrm{tree} = 41.6 \mathrm{~ft.} \\\\ \mathrm{tree} = 42 \mathrm{~ft.} \mathrm{~~Round~ to~ the~ nearest ~foot}$$

A pair of the identical circles are overlapping each other. Their diameter (D) is 20cm, and the distance between the centers is 10cm. In the grey area, it is possible to inscribe only two  golden rectangles - different in size.

Can you find the area of a larger one?

$$\small{\text{$\boxed{\varphi=\dfrac{1+\sqrt{5}}{2}} \qquad \Delta_c=10~cm\qquad D = 20~cm$}}\\\\ \mathrm{circle:~~} (\vec{x})^2= (\frac{D}{2})^2\\\\ \mathrm{line:~~} \binom{\frac{\Delta_c}{2}}{0}+\lambda\binom{1}{\varphi}=\vec{x}\\\\ \mathrm{golden~ rectangle:~~} \small{\text{$ \binom{ \frac{width}{2} }{ \frac{height}{2} } = \lambda\binom{1}{\varphi} ~~ \mathrm{~or~}~~\mathrm{~width~} = 2\lambda ~~ \mathrm{~and~} ~~ \mathrm{~height~} = 2\varphi\lambda $}}\\ \mathrm{ratio~}=\frac{height}{width}=\frac{2\varphi\lambda }{2\lambda } = \varphi\\\\ \boxed{\mathrm{~area~}~~= \mathrm{~width~} \cdot \mathrm{~height~} = 4 \varphi\lambda^2}$$

$$\\\mathrm{solve~~} \lambda :\\\\ \small{\text{$ \begin{array}{rcl} \left[ \binom{\frac{\Delta_c}{2}}{0}+\lambda\binom{1}{\varphi} \right]^2 =(\vec{x})^2 &=& (\frac{D}{2})^2\\\\ \left[ \binom{\frac{\Delta_c}{2}}{0}+\lambda\binom{1}{\varphi} \right]^2 &=& (\frac{D}{2})^2\\\\ \binom{\frac{\Delta_c}{2}}{0}\cdot \binom{\frac{\Delta_c}{2}}{0} + 2\lambda\binom{\frac{\Delta_c}{2}}{0}\cdot \binom{1}{\varphi} + \lambda^2 \binom{1}{\varphi}\cdot \binom{1}{\varphi} &=& (\frac{D}{2})^2\\\\ \frac{\Delta_c^2}{4}+\lambda\Delta_c+\lambda^2(1+\varphi^2) &=&\frac{D^2}{4}\\\\ (1+\varphi^2)\lambda^2 + \Delta_c\lambda - \frac{D^2-\Delta_c^2}{4}&=& 0\\\\ \mathbf{\lambda }&\mathbf{=}& \mathbf{ \dfrac{ \sqrt{\Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c } { 2(1+\varphi^2) } } \end{array} $}}$$  

$$\small{\text{$ \begin{array}{rcl} \mathrm{~area} = 4 \varphi\lambda^2 &=& 4 \varphi\ \left[ \dfrac{ \sqrt{\Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c } { 2(1+\varphi^2) } \right]^2\\\\ \mathrm{~area} &=& 4 \varphi\ \dfrac{ \left[ \sqrt{ \Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c \right]^2} { 4(1+\varphi^2)^2 }\\\\ \mathrm{~area} &=& \varphi\ \dfrac{ \left[ \sqrt{ \Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c \right]^2} { (1+\varphi^2)^2 }\\\\ \mathrm{~area} &=& \dfrac{ \varphi } { (1+\varphi^2)^2 } \left[ \sqrt{ \Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c \right]^2 \qquad |\qquad 1+\varphi^2 = 2+\varphi = \sqrt{5}\varphi\\\\ \mathrm{~area} &=& \dfrac{ \varphi } { 5\varphi^2 } \left[ \sqrt{ \Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2 \\\\ \mathrm{~area} &=& \dfrac{ 1} { 5\varphi } \left[ \sqrt{ \Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2 \qquad |\qquad \dfrac{1}{\varphi} = \varphi -1 \\\\ \mathrm{~area} &=& \dfrac{ \varphi -1 } { 5 } \left[ \sqrt{ \Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2 \end{array} $}}$$

$$\boxed{ \mathrm{~area} = \left( \dfrac{ \varphi -1 } { 5 } \right) \left[ \sqrt{ \Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2 }$$

$$\small{\text{$ \begin{array}{rcl} \mathrm{~area} &=& \left( \dfrac{ \varphi -1 } { 5 } \right) \left[ \sqrt{ \Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2 \qquad |\qquad \Delta_c = 10 \qquad D = 20 \\\\ \mathrm{~area} &=& \left( \dfrac{ \varphi -1 } { 5 } \right) \left[ \sqrt{ 100+ (2+\varphi )(400 - 100) } -10 \right]^2 \\\\ \mathrm{~area} &=& \left( \dfrac{ \varphi -1 } { 5 } \right) \left[ \sqrt{ 100+ (2+\varphi )300 } -10 \right]^2 \\\\ \mathrm{~area} &=& 0.12360679775\cdot \left[ 34.4297864737 -10 \right]^2\\\\ \mathrm{~area} &=& 0.12360679775\cdot 596.814467151\\\\ \mathrm{~area} &=& 73.7703251354 \mathrm{~cm^2} \end{array} $}}$$

determine the interest on $5000 invested at 12% per annum compound interest for 3 years, with the interest compounded a)monthly?

I.  12% per annum compound interest for 3 years:

$$\small{\text{$ \begin{array}{rcl} \$5000\cdot \left(1+\dfrac{12}{100}\right)^3\\\\ &=&\$5000\cdot 1.12^3\\\\ &=& \$7024.64 \end{array} $}}$$

II.  12% per annum with the interest compounded monthly

$$\small{\text{$ \begin{array}{rcl} \$5000\cdot \left(1+\dfrac{12}{100}\cdot \dfrac{1}{12}\right)^{36}\\\\ &=&\$5000\cdot 1.01^{36}\\\\ &=& \$7153.84 \end{array} $}}$$

Find the horizontal asymptote of  f(x) = 2 ((x+4)(5 x-1))/((8-x)(7 x + 2))

$$\small{\text{$ \lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{(x+4)(5x-1)}{(8-x)(7x+2)} =\lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{5x^2+19x-4}{-7x^2+54x+16} =\lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{ \dfrac{5x^2}{x^2}+\dfrac{19x}{x^2}-\dfrac{4}{x^2}}{ \dfrac{-7x^2}{x^2}+\dfrac{54x}{x^2}+\dfrac{16}{x^2}} =\lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{ 5 + \dfrac{19}{x}-\dfrac{4}{x^2}}{ -7 + \dfrac{54}{x}+\dfrac{16}{x^2}} = 2\cdot \dfrac{ 5 }{ -7 }=-\dfrac{10}{7} $}}$$

$$\rm{horizontal~ Asymptote:~} y =-\dfrac{10}{7} = -1.42857142857$$

BUT i don't understand the logic behind the horizontal asymptote at y=9

$$\small{\text{$ \lim \limits_{x \rightarrow \infty } \dfrac{(x+4)(x-2)} {(x-4)(x-6)} = \lim \limits_{x \rightarrow \infty } \dfrac{x^2+2x-8} {x^2-10x+24} = \lim \limits_{x \rightarrow \infty } \dfrac { \dfrac{x^2}{x^2}+\dfrac{2x}{x^2}-\dfrac{8}{x^2}} { \dfrac{x^2}{x^2}-\dfrac{10x}{x^2}+\dfrac{24}{x^2}} = \lim \limits_{x \rightarrow \infty } \dfrac { 1 + \dfrac{2}{x}-\dfrac{8}{x^2}} { 1 - \dfrac{10}{x}+\dfrac{24}{x^2}} = \dfrac { 1 } { 1 }=1 $}}\\\\$$

$$\small{\text{$ \lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{(x+4)(x-2)} {(x-4)(x-6)} = \lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{x^2+2x-8} {x^2-10x+24} = \lim \limits_{x \rightarrow \infty } 9\cdot \dfrac { \dfrac{x^2}{x^2}+\dfrac{2x}{x^2}-\dfrac{8}{x^2}} { \dfrac{x^2}{x^2}-\dfrac{10x}{x^2}+\dfrac{24}{x^2}} = \lim \limits_{x \rightarrow \infty } 9\cdot \dfrac { 1 + \dfrac{2}{x}-\dfrac{8}{x^2}} { 1 - \dfrac{10}{x}+\dfrac{24}{x^2}} = 9\cdot \dfrac { 1 } { 1 }=9 $}}\\\\$$

Geg: a=7,1 m, ß=34°, c=8,564m, alpha=56°, y=90° Wie errechne ich nun b?

$$\\ \begin{array}{rclr} a&=&7,1~m\qquad & \alpha = 56\ensurement{^{\circ}}\\ b&=& ~? ~m\qquad & \beta = 34\ensurement{^{\circ}}\\ c&=&8,564~m\qquad& \gamma = 90\ensurement{^{\circ}}\\ \end{array}$$

1. Möglichkeit:

$$\small{\text{$ \tan{(56\ensurement{^{\circ}})} =\dfrac{7,1}{b} \qquad b = \dfrac{7,1}{\tan{(56\ensurement{^{\circ}})} } = 4,789~m $}}$$

2. Möglichkeit:

$$\small{\text{$ \cos{(56\ensurement{^{\circ}})} =\dfrac{b}{8,564} \qquad b = 8,564\cdot \cos{ (56\ensurement{^{\circ}}) } = 4,789~m $}}$$

3. Möglichkeit:

$$\small{\text{$ \tan{(34\ensurement{^{\circ}})} =\dfrac{b}{7,1} \qquad b = 7,1\cdot \tan{ (34\ensurement{^{\circ}}) } = 4,789~m $}}$$

4. Möglichkeit:

$$\small{\text{$ \sin{(34\ensurement{^{\circ}})} =\dfrac{b}{8,564} \qquad b = 8,564\cdot \sin{ (34\ensurement{^{\circ}}) } = 4,789~m $}}$$

(a+b)^3 ?

Let a = x:

If  $$(x+b)^3=0$$    then   $$(x+b)(x+b)(x+b) = 0$$   is a Polynom with the roots $$x_1 = -b\qquad x_2 = -b\qquad x_3 = -b$$ .

Vieta's formulas:

If $$x^3+Ax^2+Bx+C=0$$ , then

$$\\\small{\text{$ A =-(x_1 +x_2 +x_3 )\qquad B =x_1 x_2 +x_1 x_3 +x_2 x_3 \qquad C =-(x_1 x_2 x_3 ) $}}\\\\ \small{\text{$ \begin{array}{rcl} A &=& -[ (-b) +(-b) +(-b) ] = -[3(-b)]=3b\\ B &=& (-b) (-b) +(-b) (-b) +(-b) (-b)=3b^2 \\ C &=& -[(-b) (-b) (-b) ]=-[-3b^3]=b^3 \end{array} $}}\\$$

so we have:

$$\\ (x+b)^3= x^3+Ax^2+Bx+C=0 \\ (x+b)^3= x^3+(3b)x^2+(3b^2)x+(b^3) \qquad | \qquad \textcolor[rgb]{1,0,0}{x = a}\\ (a+b)^3= a^3+(3b)a^2+(3b^2)a+(b^3)\\\\ \mathbf{(a+b)^3= a^3+3a^2b+3ab^2+b^3}$$

Write an equation for a rational function with: Vertical asymptotes at x = -2 and x = -6 x intercepts at x = 1 and x = -5 y intercept at 2

$$y=\dfrac{\left(-4.8\right)\left(x+5\right)\left(x-1\right)}{\left(x+6\right)\left(x+2\right)}$$

Using implicit differentiation, find the equation of the tangent line to the given curve at the given point: 3x2y2 − 3y −17 = 5x +14 at (1,−3)

I. implicit differentiation:

$$\small{\text{$ \begin{array}{rcl} 3x^2y^2 - 3y -17 &=& 5x +14 \\ f(x,y) &=& 3x^2y^2 - 3y-5x -31 = 0\\\\ f'(x,y) &=& -\dfrac{ \dfrac{ d( 3x^2y^2 - 3y-5x -31 ) } {dx} } { \dfrac{ d( 3x^2y^2 - 3y-5x -31 ) } {dy} }\\\\\\ f'(x,y) &=& -\dfrac{ 6xy^2-5 } { 3x^2\cdot 2y-3 }\\\\\\ f'(x,y) &=& \dfrac{ 5- 6xy^2 } { 6yx^2-3 } \end{array} $}}$$

II.  tangent line

$$\small{\text{$ \begin{array}{rcl} x_p=1 \qquad y_p = -3\\\\ f'(x_p,y_p) &=& \dfrac{y-y_p}{x-x_p}\\\\ y &=& f'(x_p,y_p)\cdot(x-x_p)+y_p \qquad | \qquad f'(x_p,y_p) = \dfrac{5-6\cdot 1\cdot (-3)^2} {6\cdot (-3)\cdot 1^2 -3 } = 2.\overline{3}\\\\ \mathbf{y} &\mathbf{=}&\mathbf{ 2.\overline{3} \cdot(x-1)-3}\\\\ \mathbf{y} &\mathbf{=}&\mathbf{\dfrac{7}{3} \cdot(x-1)-3} \end{array} $}}$$

how do i simplify 2p^2+3p+1=0

$$\small{\text{$ \mathbf{2p^2+3p+1=2\left(p+\dfrac{1}{2}\right) \left( p+1\right)=0 } $}}$$