heureka

exp(-0.9/(1273*8.62*10^-5)

$$\small{\text{ $ e^{-\dfrac{0.9}{(1273\cdot 8.62 \cdot 10^{-5})}} = 0.0002741717276064 $}}$$

(1) x+(b/a-1)y-b/(a+b)=0

(2) bx/a-(b/a+1)y+b^2/(a(a-b))+1=0

Können Sie mir den Lösungsweg dieses Gleichungssystems zeigen bitte!

$$\\ \small{\text{$ \begin{array}{rcl}(1) \quad x+( \frac{b}{a}-1)\cdot y- \frac{b}{(a+b)} &=& 0\\ x+( \frac{b-a}{a} ) \cdot y- \frac{b}{(a+b)} &=& 0\\ x&=& \frac{b}{a+b}- \left( \frac{b-a}{a} \right) \cdot y \end{array} $}}$$

$$\\\small{\text{$ \begin{array}{rcl}(2) \quad \frac{b\cdot x}{a}-( \frac{b}{a}+1)\cdot y+ \frac{b^2}{(a\cdot (a-b))}+1 &=& 0 \\ \frac{b}{a}\cdot x -( \frac{a+b}{a})\cdot y+ \frac{b^2}{(a\cdot (a-b))}+1 &=& 0 \\ y\cdot( \frac{a+b}{a}) &=& \frac{b}{a}\cdot x + \frac{b^2}{(a\cdot (a-b))}+1 \\ \\ y\cdot (a+b) &=& b\cdot x + \frac{b^2}{(a-b)}+a\\ \end{array} $}}$$

(1) in (2) einsetzen:

$$\\\small{\text{$ \begin{array}{rcl} y\cdot (a+b) &=& b\cdot \left[\frac{b}{a+b}- \left( \frac{b-a}{a} \right) \cdot y \right] + \frac{b^2}{(a-b)}+a\\\\ y\cdot (a+b) &=& \frac{b^2}{a+b}- b\cdot \left( \frac{b-a}{a} \right) \cdot y + \frac{b^2}{(a-b)}+a\\\\ y\cdot (a+b)+ b\cdot \left( \frac{b-a}{a} \right) \cdot y &=& \frac{b^2}{a+b}+ \frac{b^2}{(a-b)}+a\\\\ y\cdot \left[ (a+b)+ b\cdot \left( \frac{b-a}{a} \right) \right] &=& \frac{b^2}{a+b}+ \frac{b^2}{(a-b)}+a\\\\ y\cdot \left( a+b+ \frac{b^2}{a} -b\right) &=& \frac{b^2}{a+b}+ \frac{b^2}{(a-b)}+a\\\\ y\cdot \left( a+ \not{b}+ \frac{b^2}{a} -\not{b}\right) &=& \frac{b^2}{a+b}+ \frac{b^2}{(a-b)}+a\\\\ y\cdot \left( \frac{a^2+b^2}{a} \right) &=& \frac{b^2}{a+b}+ \frac{b^2}{(a-b)}+a\\\\ y &=& \left( \frac{a}{a^2+b^2} \right) \cdot \left( \frac{b^2}{a+b}+ \frac{b^2}{(a-b)}+a \right) \\\\ y &=& \left( \frac{a\cdot b^2}{a^2+b^2} \right) \cdot \left( \frac{1}{a+b}+ \frac{1}{(a-b)}+\frac{a}{b^2} \right) \\\\ y &=& \left( \frac{a\cdot b^2}{a^2+b^2} \right) \cdot \left( \frac{(a-b)\cdot b^2+(a+b)\cdot b^2 + a\cdot(a^2-b^2) }{b^2\cdot (a+b)\cdot (a-b) } \right) \\\\ y &=& \left( \frac{a\cdot \not{b^2} }{a^2+b^2} \right) \cdot \left( \frac{(a-b)\cdot b^2+(a+b)\cdot b^2 + a\cdot(a^2-b^2) }{ \not{b^2} \cdot (a+b)\cdot (a-b) } \right) \\\\ y &=& \left( \frac{ a }{a^2+b^2} \right) \cdot \left( \frac{(a-b)\cdot b^2+(a+b)\cdot b^2 + a\cdot(a^2-b^2) }{ (a+b)\cdot (a-b) } \right) \\\\ y &=& \left( \frac{ a }{a^2+b^2} \right) \cdot \left( \frac{ a\cdot b^2-b^3 + a\cdot b^2 + b^3 + a^3 - a\cdot b^2 }{ (a+b)\cdot (a-b) } \right) \\\\ y &=& \left( \frac{ a }{a^2+b^2} \right) \cdot \left( \frac{ a\cdot b^2 + a^3 }{ (a+b)\cdot (a-b) } \right) \\\\ y &=& \left( \frac{ a^2 }{\left( a^2+b^2 \right) } \right) \cdot \left( \frac{ \left( b^2 + a^2 \right) }{ (a+b)\cdot (a-b) } \right) \\\\ y &=& \left( \frac{ a^2 }{\left( a^2+b^2 \right) } \right) \cdot \left( \frac{ \left( a^2 + b^2 \right) }{ (a+b)\cdot (a-b) } \right) \\\\ y &=& \left( \frac{ a^2 }{ (a+b)\cdot (a-b) } \right) \\\\ y &=& \dfrac{ a^2 }{ a^2-b^2 } \\\\ \end{array} $}}$$

$$\\ \small{\text{$ \begin{array}{rcl} x&=& \frac{b}{a+b}- \left( \frac{b-a}{a} \right) \cdot y \\\\ x&=& \frac{b}{a+b}- \left( \frac{b-a}{a} \right) \cdot ( \frac{ a^2 }{ a^2-b^2 } ) \\\\ x&=& \frac{b}{a+b}- \left( \frac{b-a}{1} \right) \cdot ( \frac{ a }{ a^2-b^2 } ) \\\\ x&=& \frac{b}{a+b}+\left( \frac{a-b}{1} \right) \cdot ( \frac{ a }{ a^2-b^2 } ) \\\\ x&=& \frac{b}{a+b}+ \frac{ (a-b)\cdot a }{ (a+b)\cdot(a-b) } \\\\ x&=& \frac{b}{a+b}+ \frac{ a }{ a+b } \\\\ x&=& \frac{b+a}{a+b} \\\\ x&=& \frac{a+b}{a+b} \\\\ x&=& 1 \end{array} $}}$$

$$\boxed{\quad x = 1 \qquad y = \dfrac{ a^2 }{ a^2-b^2 } \quad }$$

A woman who is 5 feet 5 inches tall is standing near the space needle in Washington; she casts a 13 inch shadow at the same time that the space needle casts a 121-foot shadow. How tall is the space needle ?

intercept theorem:

$$\\ \small{\text{$ \\ \dfrac{h}{121\;\rm{foot} } = \dfrac{ 5\;\rm{foot}+5\;\rm{inch} }{13\;\rm{inch} } $}}\\ \\ \small{\text{$ \dfrac{h}{121\;\rm{foot} } = \dfrac{ 60\;\rm{inch}+5\;\rm{inch} }{13\;\rm{inch} } $}}\\ \\ \dfrac{h}{121\;\rm{foot} } = \dfrac{ 65\;\rm{inch} }{13\;\rm{inch} } $}}\\ \\ \dfrac{h}{121\;\rm{foot} } = \dfrac{ 65\;\not{\rm{inch}} }{13\;\not{\rm{inch}} } $}}\\ \\ \dfrac{h}{121\;\rm{foot} } = \dfrac{ 65 }{ 13 } $}}\\ \\ h= 121 \cdot \dfrac{ 65 }{ 13 } \;\rm{foot} $}}\\ \\ h= 605 \;\rm{foot} $}}$$

The space needle is 605 feet tall.

What is the exact equation to find pi ?

 factors of 2^145-3

Convert 65 degrees to an equivalent number of radians

$$65\ensurement{^{\circ}}\cdot \left( \dfrac{\pi}{180\ensurement{^{\circ}}} \right)\\\\ =\dfrac{65\ensurement{^{\circ}}}{180\ensurement{^{\circ}}}\cdot\pi\\\\ =\dfrac{65\not{\ensurement{^{\circ}}} }{180\not{\ensurement{^{\circ}}} }\cdot\pi\\\\ =0.36111111111\cdot\pi\\\\ =0.36111111111\cdot3.14159265359\;\rm{rad}\\\\ =1.13446401380\;\rm{rad}$$

What is a pythagorean triple 

$$\small{\text{Quater $ = 25 \qquad $Dime $ = 10 \qquad $ Nickel $= 5 $}} \\ \small{\text{In \$0.55 max. 2 Quaters, max. 5 Dimes and max. 11 Nickels }} \\\\ \left(\sum\limits_{i=0}^{2} x^{(\textcolor[rgb]{0,0,1}{25}*i}) \right) \times \left (\sum\limits_{i=0}^{5} x^{(\textcolor[rgb]{0,0,1}{10}*i)} \right) \times \left(\sum\limits_{i=0}^{11} x^{(\textcolor[rgb]{0,0,1}{5}*i)} \right) \\\\ \small{\text{$=(1+x^{25}+x^{50})$}} \times \small{\text{$(1+x^{10}+x^{20}+x^{30}+x^{40}+x^{50})$}} \\ \times \small{\text{$(1+x^{5}+x^{10}+x^{15}+x^{20}+x^{25}+x^{30}+x^{35}+x^{40}+x^{45}+x^{50}+x^{55})$}}\\ \small{\text{$ =(x^{155})+(x^{150})+2*x^{145}+2*x^{140}+3*x^{135}+4*x^{130} $}} \\ \small{\text{$+5*x^{125}+6*x^{120}+7*x^{115}+8*x^{110}+10*x^{105}+11*x^{100} $}} \\ \small{\text{$+11*x^{95}+12*x^{90}+12*x^{85}+13*x^{80}+13*x^{75}+12*x^{70} $}} \\ \small{\text{$+12*x^{65}+11*x^{60}+\textcolor[rgb]{1,0,0}{11*x^{55}}+10*x^{50}+8*x^{45}+7*x^{40} $}} \\ \small{\text{$+6*x^{35}+5*x^{30}+4*x^{25}+3*x^{20}+2*x^{15}+2*x^{10}+(x^5)+1 $}} \\$$

The coefficient from   $$\small{\text{$\textcolor[rgb]{1,0,0}{x^{55}}$}}$$  is $$\small{\text{$\textcolor[rgb]{1,0,0}{ 11 }$}}$$ . So there are 11 possibilities.

Hallo Radix,

ich kenne noch

How many different combinations of quarters, nickels, and dimes can be used to make $0.55 ?

  $$\small{\text{ $ \begin{array}{rccrlclcl} \hline \\ 1. & \$0.55 &=& & && 3\; Dimes &+& 1\; Quater \\ 2. & \$0.55 &=& 1& Nickel && &+& 2\; Quaters \\ 3. & \$0.55 &=& 1& Nickel &+& 5\; Dimes \\ 4. & \$0.55 &=& 2& Nickels &+& 2\; Dimes &+& 1\; Quater \\ 5. & \$0.55 &=& 3& Nickels &+& 4\; Dimes \\ 6. & \$0.55 &=& 4& Nickels &+& 1\; Dime &+& 1\; Quater\\ 7. & \$0.55 &=& 5& Nickels &+& 3\; Dimes \\ 8. & \$0.55 &=& 6& Nickels && &+& 1\; Quater \\ 9. & \$0.55 &=& 7& Nickels &+& 2\; Dimes \\ 10. & \$0.55 &=& 9& Nickels &+& 1\; Dime \\ 11. & \$0.55 &=& 11& Nickels \\ \\ \hline \end{array} $}}$$