heureka

The perimeter of a rectangle is 26 meters. The difference between the length and the width is 5 meters. Find the width ?

$$\small{\text{ l=length w=width: $ \begin{array}{rrcrr} &2l+2w &=&26\ m & \quad | \quad :2\\ (1)&l+w&=&13\ m \\ (2)&l-w&=&5\ m \end{array} $ }} \\\\ \text{ $\frac{(1)+(2)}{2}=\frac{l+\not{w}+l-\not{w}}{2}=l=\frac{13+5}{2}=9 \ m$ } \\\\ \text{ $ \frac{(1)-(2)}{2} =\frac{l+w-(l-w)}{2}=\frac{\not{l}+w-\not{l}+w}{2} =w=\frac{13-5}{2}=4\ m $ }$$

solve by factoring 3x^2+1=4x

$$\small{\text{ \begin{array}{rcl} $ 3x^2+1 & = & 4x \\ 3x^2 -4x + 1 & = &0 \\ (x-1)(3x-1) &=& 0 \\ 3(x-1)(x-\frac{1}{3}) &=& 0 $ \end{array} }} \\ \small{\text{ $ x_1 = 1 \qquad x_2 = \frac{1}{3} $ }} \\ \small{\text{ Check: $\quad 3*1^2+1=4*1 $ okay $ \qquad 3*\frac{1}{3}*\frac{1}{3}+1=4*\frac{1}{3} $ okay }}$$

$$\small{\text{ $x^2 = 0 \Rightarrow c-a=0 \Rightarrow (2a-3b)= 0 \Rightarrow b=\frac{2}{3}a $ and $ c=\frac{a+3b}{3} \Rightarrow c=a $ }}$$

The triangle is isosceles triangle:

$$\small{\text{ $ \sin{(\frac{1}{2}(180\ensurement{^{\circ}} - 2\alpha))}= \frac{\frac{2}{3}a}{a}=\frac{2}{3}=\sin{(90\ensurement{^{\circ}}-\alpha)}=\cos{(\alpha)} \quad \boxed{\cos{(\alpha)}=\frac{2}{3}} $ }} \\ \small{\text{ $ (\frac{2}{3}a)^2=2a^2-2a^2\cos{(\beta)} \Rightarrow \boxed{\cos{(\beta)}=\frac{7}{9}} $ }}$$

$$\small{\text{ $ \sin{(\alpha)}=\sqrt{1-\cos^2{(\alpha)}}=\frac{1}{3}\sqrt{5} \qquad \sin{(\beta)}=\sqrt{1-\cos^2{(\beta)}}= \frac{4}{9}\sqrt{2} $ }}\\ \small{\text{ $ \sin{(\alpha)} + \sin{(\beta)}= \frac{1}{3}\sqrt{5} +\frac{4}{9}\sqrt{2} $ }}$$

Prove that the expression y(y + 1) always represents an even number, where y is any positive integer.

$$\small{\text{ y and y+1 are next numbers. So if y is even, y+1 must be odd and }} \\ \small{\text{ if y is odd then y+1 must be even. \boxed{even \times odd = even} or \boxed{odd \times even = even} }}$$

$$\small{\text{ $ \begin{array}{rc|rc} y & even \quad & \quad y & odd \\ y+1 & odd \quad & \quad y+1 & even \\ \hline y(y+1) & even \times odd \quad & \quad y(y+1) & odd \times even \end{array} $ }}$$

$$\boxed{ \begin{array}{c} even \times odd = even \\ odd \times even = even \end{array} }$$

$$\small{\text{ Example: $ \begin{array}{rc|rc} 4 & even \quad & \quad 7 & odd \\ 5 & odd \quad & \quad 8 & even \\ \hline 20 & even \times odd \quad & \quad 56 & odd \times even \end{array} $ The products are always even \ }}$$

An observer at the top of a tower of height 20m sees a man due East of him at an angle of depression of 27 degrees. He sees another man due south of him at an angle of depression of 30 degrees. Find the distance between the men on the ground

$$\small{\text{ $ h=20\ m $; \quad distance $ d=\ ? $ }} \\ \\ \small{\text{ $ d^2 = \left( \dfrac {h} { \tan{(30\ensurement{^{\circ}} ) } } \right) ^2 + \left( \dfrac {h} { \tan{(27\ensurement{^{\circ}} ) } } \right) ^2 $ }} \\ \\ \small{\text{ $ d^2 = h^2 \left( \dfrac {1} { \tan^2{(30\ensurement{^{\circ}} ) } } \right) + \left( \dfrac {1} { \tan^2{(27\ensurement{^{\circ}} ) } } \right) $ }} \\ \\ \small{\text{ $ d = h \sqrt{ \dfrac {1} { \tan^2{(30\ensurement{^{\circ}} ) } } + \dfrac {1} { \tan^2{(27\ensurement{^{\circ}} ) } } } $ }} \\ \\ \small{\text{ $ d = 20\ m * \sqrt{ 3.69466131307 } $ }} \\ \\ \small{\text{ $ d= 20 * 1.92215017964 \ m $ }} \\ \\ \small{\text{ $ d= 38.4430035927\ m $ }}$$

How do you do this: $$(-4x)^3 = (-4)^3*x^3=(-64)x^3 = -64x^3$$

5ppm i decimalform:

$$1ppm = 1 * 10^{-6} = 0.000001 \quad 5ppm = 5 * 10^{-6} = 0.000005$$

k=c+273.15

K = °C + 273.15

Your answer will be in Kelvin.
Remember, the Kelvin temperature scale does not use the degree (°) symbol.

For example, if you want to know what 20°C is in Kelvin:

K = 20 + 273.15 = 293.15 K

logy x=3 logy (4x)=5

$$\small{\text{ $ \begin{array}{rclc|rclc} \log_y{ (x) } & = & 3 & | \quad y^{()} \quad & \quad \log_y {(4x)} & = & 5 \quad | \quad y^{()} \\ y^{\log_y{ (x) }} & = & y^3 & \quad & \quad y^{\log_y {(4x)}} & = &y^5 \\ x & = & y^3 & \quad & \quad 4x & = & y^5 \\ & & & & 4y^3 & = & y^5 \quad | \quad :y^3 \\ & & & & 4 & = & y^2 \quad | \quad \sqrt \\ & & & & 2 & = & y \\ x & = & 2^3 & & y & = & 2 \\ x & = & 8 & & y & = & 2 \\ \end{array} $ }}$$

$$\small{\text{ $ \log_2{(8)}=3 \qquad \log_2{(4*8)}=5 $ }}$$

r³ = 8pi

$$\small{\text{ $r^3 = 8\pi$. Ziehen der 3. Wurzel: $r=\sqrt[3]{ \dfrac{8}{\pi}}$. }}$$

Nun kann die 3. Wurzel aufgeteilt werden in 3. Wurzel aus Zähler durch 3. Wurzel aus Nenner:

$$r= \dfrac{\sqrt[3] {8} }{ \sqrt[3]{\pi }}$$

Die 3. Wurzel im Zähler kann gezogen werden. Sie ist nämlich = 2. So bleibt nur noch die 3. Wurzel im Nenner und das Ergebnis ist:

$$r= \dfrac{ 2 }{ \sqrt[3]{\pi }}$$