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If there is a sertain number and someone takes 1/3 of the number then takes 1/4 of the number then someone takes 1/2 of the number. If the ending number is 6 than what was the starting number

$\begin{array}{rcll} n\cdot \frac13 \cdot \frac14 \cdot \frac12 &=& 6 \\ n &=& 6 \cdot 3 \cdot 4 \cdot 2 \\ n&=& 144 \end{array}$

A piece of rectangular plywood 42 cm by 35 cm is to be cut into smaller rectangular pieces each 10 cm by 7 cm. What is the maximum number of pieces it can be out into?

$\begin{array}{rcll} \frac{35}{7} \times [\frac{42}{10}] = 5 \times 4 = 20 \\\\ [\frac{35}{10}] \times \frac{42}{7} = 3 \times 6 = 18\\\\ [\dots] = \text{integer} \end{array}$

The maximum number is 20

The ratio of boys to girls at a party was 4:3 respectively. When 60 boys and 150 girls came to join in, the ratio of boys to girls became 2:3. How many boys were there at first?

b = boys

g = girls

$\small{ \begin{array}{lrcll} (1) & \frac{b}{g} &=& \frac43 \\ & 3b &=& 4g \\ & 3b - 4g &=& 0 \\\\ (2) & \frac{b+60}{g+150} &=& \frac23 \\ & 3\cdot (b+60) &=& 2\cdot (g+150)\\ & 3b+180 &=& 2g + 300 \\ & 3b - 2g &=& 300 - 180 \\ & 3b - 2g &=& 120 \\ \\ \hline \\ (1) & 3b - 4g &=& 0 \\ (2) & 3b - 2g &=& 120 \\\\ (2)-(1) & 3b - 2g - (3b - 4g) &=& 120 -0 \\ &3b - 2g - 3b + 4g &=& 120 \\ & 2g &=& 120 \\ & \mathbf{g} &\mathbf{=}& \mathbf{60} \\\\ & 3b &=& 4g \\ & 3b &=& 4\cdot 60 \\ & 3b &=& 240 \\ & \mathbf{b} &\mathbf{=}& \mathbf{80} \end{array} }$

$\lim \limits_{t\to 2} { \frac{2t^2-3t-2}{t^2+t-6} }$

l'Hospital's rule

$\begin{array}{rcll} \lim \limits_{t\to 2} { \frac{2t^2-3t-2}{t^2+t-6} } &=& \lim \limits_{x\to 2} { \frac{2\cdot 2\cdot t-3}{2\cdot t + 1} }\\ &=& \frac{2\cdot 2\cdot 2-3}{2\cdot 2 + 1}\\ &=& \frac{5}{5}\\ \mathbf{\lim \limits_{t\to 2} { \frac{2t^2-3t-2}{t^2+t-6} } }&\mathbf{=}& \mathbf{1}\\ \end{array}$

$\lim \limits_{x\to 2} { \frac{x^3-8}{12x-24} }$

l'Hospital's rule

$\begin{array}{rcll} \lim \limits_{x\to 2} { \frac{x^3-8}{12x-24} } &=& \lim \limits_{x\to 2} { \frac{3x^2}{12} }\\ &=& { \frac{3\cdot 2^2}{12} }\\ &=& { \frac{12}{12} }\\ \mathbf{\lim \limits_{x\to 2} { \frac{x^3-8}{12x-24} } }&\mathbf{=}& \mathbf{1}\\ \end{array}$

Find the square root of 60i-11 

$\begin{array}{rcll} z^2 &=& a+b\cdot i\\ z^2 &=& -11+60i \qquad a = -11 \qquad b = 60\\ z&=&\sqrt{-11+60i}\\\\ r &=& \sqrt{a^2+b^2} \\ r &=& \sqrt{(-11)^2+60^2}\\ r &=& 61\\\\ \tan{(\varphi)} &=& \frac{b}{a} \\ \tan{(\varphi)} &=& \frac{60}{-11} \qquad (\mathbf{II.})\\ \tan{(\varphi)} &=& -5.45454545455 \\ \varphi &=& \arctan{(-5.45454545455)} +180^\circ\\ \varphi &=& -79.6111421845 +180^\circ\\ \varphi &=& 100.388857815^\circ \\\\ z &=& \sqrt{r}\cdot \left[ \cos{( \frac{\varphi}{2} + k\cdot \frac{360^\circ}{2} )} + i \cdot \sin{ (\frac{\varphi}{2} + k\cdot \frac{360^\circ}{2}) } \right] \qquad k=0,1\\\\ z_0 &=& \sqrt{61}\cdot \left[ \cos{( \frac{100.388857815^\circ }{2} + 0\cdot \frac{360^\circ}{2} )} + i \cdot \sin{ (\frac{100.388857815^\circ }{2} + 0\cdot \frac{360^\circ}{2}) } \right]\\ z_0 &=& 7.81024967591\cdot \left[ \cos{( 50.1944289077^\circ )} + i \cdot \sin{ (50.1944289077^\circ) } \right]\\ z_0 &=& 7.81024967591\cdot \cos{( 50.1944289077^\circ )} + i \cdot 7.81024967591 \cdot \sin{ (50.1944289077^\circ) } \\ \mathbf{z_0} &\mathbf{=}& \mathbf{5 + 6\cdot i} \\\\ z_1 &=& \sqrt{61}\cdot \left[ \cos{( \frac{100.388857815^\circ }{2} + 1\cdot \frac{360^\circ}{2} )} + i \cdot \sin{ (\frac{100.388857815^\circ }{2} + 1\cdot \frac{360^\circ}{2}) } \right]\\ z_1 &=& 7.81024967591\cdot \left[ \cos{( 230.194428908^\circ )} + i \cdot \sin{ (230.194428908^\circ) } \right]\\ z_1 &=& 7.81024967591\cdot \cos{( 230.194428908^\circ )} + i \cdot 7.81024967591\cdot \sin{ (230.194428908^\circ) } \\ \mathbf{z_1 }&\mathbf{=}&\mathbf{ -5 -6\cdot i }\\ \end{array}$

$\boxed{~ \sec{(\alpha)} = \frac{1}{\cos{(\alpha)}} ~}$

$\begin{array}{rcll} \sec{\left(\frac{13}{12}\cdot \pi \right)} &=& \frac{1}{\cos{\left(\frac{13}{12}\cdot \pi \right)}} \\ &=& \frac{1}{\cos{ (3.40339204139 )} } \\ &=& \frac{1}{ \cos{ (3.40339204139 )} } \\ &=& \frac{1}{ 0.99823631510 } \\\\ \mathbf{\sec{ \left(\frac{13}{12}\cdot \pi \right)} } &\mathbf{=}& \mathbf{1.00176680098} \end{array}$

The answer is A

Löse $\sec{(\alpha)} = \sqrt{2}$  für $\alpha$, wenn $0^\circ \le \alpha \le 90^\circ$

Ich weiß nicht was sec hat bestimmt was mit "Tangens, Sinus, Cosinus" zu tun und die Lösung muss sich zwischen diesen Zahlen  befinden:

A) 30°

B) 60°

C) 45°

D) 90°

$\boxed{~ \sec{(\alpha)} = \frac{1}{\cos{(\alpha)}} ~}$

$\begin{array}{rcll} \sec{(\alpha)} = \frac{1}{\cos{(\alpha)}} &=& \sqrt{2} \\ \frac{1}{\cos{(\alpha)}} &=& \sqrt{2} \\ \cos{(\alpha)} &=& \frac{1}{ \sqrt{2} } \cdot \frac{\sqrt{2}}{\sqrt{2}}\\ \cos{(\alpha)} &=& \frac{\sqrt{2}}{ 2 } \\ \alpha &=& \pm\arccos{(\frac{\sqrt{2}}{ 2 })} +k\cdot 360^\circ \qquad & k \in Z\\ \alpha &=& \pm\arccos{(0.70710678119)} +k\cdot 360^\circ\\ \alpha &=& \pm 45^\circ +k\cdot 360^\circ\\ \alpha &=& \dots, -45^\circ, 45^\circ, 315^\circ, 405^\circ, \dots \\\\ \mathbf{ \alpha } &\mathbf{=}& \mathbf{45^\circ} \qquad 0^\circ \le \alpha \le 90^\circ \end{array}$

Die Antwort ist C

How long does it take for Marks car to change to its velocity from 12 m/s to 25 m/s if the acceleration is 5 m/s2

a = acceleration

v = velocity

t = time

$\begin{array}{rcll} \boxed{~ \begin{array}{rcll} a &=& \frac{ \Delta v } { \Delta t } \end{array} ~}\\\\ \Delta v &=& 25\ \frac{m}{s} - 12\ \frac{m}{s} \\ a &=& 5\ \frac{m}{s^2} \\\\ \end{array}\\\\ \begin{array}{rcll} 5\ \frac{m}{s^2} &=& \frac{ 25\ \frac{m}{s} - 12\ \frac{m}{s} } { \Delta t } \\ 5\ \frac{m}{s^2} &=& \frac{ 13\ \frac{m}{s} } { \Delta t } \\ \Delta t &=& \frac{ 13\ \frac{m}{s} } { 5\ \frac{m}{s^2} } \\ \Delta t &=& \frac{ 13 } { 5 } \ s \\ \mathbf{ \Delta t } &\mathbf{=}& \mathbf{2.6 \ s }\\ \end{array}$

A sector of a circle is inside a rectangle. The rectangle has the side lengths 25cm and 15cm. This also means that the radius of the sector is 25cm. What would be the area in the rectangle that is outside the sector?

The answer is 185cm^2 but i dont no how to get it. Please help!

 $\begin{array}{rcll} A_{\text{rectangle}} &=& 25\cdot 15 \ cm^2 \\ &=& 375 \ cm^2 \\\\ A_{\text{sector of a circle}} &=& \pi \cdot 25^2 \cdot \frac{\varphi}{360^\circ} \ cm^2 \\ A_{\text{Area in the rectangle and outside sector}} &=& A_{\text{rectangle}} - A_{\text{sector of a circle}}\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - \pi \cdot 25^2 \cdot \frac{\varphi}{360^\circ} \ cm^2\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - \pi \cdot 625 \cdot \frac{\varphi}{360^\circ} \ cm^2\\\\ \sin{(\frac{\varphi}{2})} &=& \frac{\frac{15}{2}}{25} \\ \sin{(\frac{\varphi}{2})} &=& \frac{7.5}{25} \\ \sin{(\frac{\varphi}{2})} &=& 0.3 \\ \frac{\varphi}{2} &=& \arcsin{(0.3)} \\ \frac{\varphi}{2} &=& 17.4576031237^\circ \\ \varphi &=& 2\cdot 17.4576031237^\circ \\ \varphi &=& 34.9152062474^\circ \\\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - \pi \cdot 625 \cdot \frac{34.9152062474^\circ}{360^\circ} \ cm^2\\\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - 190.432908760 \ cm^2\\\\ A_{\text{Area in the rectangle and outside sector}} &=& 184.567091240 \ cm^2\\\\ A_{\text{Area in the rectangle and outside sector}} &\approx& 185 \ cm^2\\\\ \end{array}$