heureka

During a 4th of july weekend 32 vehicles became trapped on sunshine bridge while it was being repaved. A recent city ordinance decreed that only cars with only 4 wheels and trucks with 6 wheels could be on the bridge at a time. If there were 148 tires that needed to be replaced due to damage, how many cars and trucks would be involved in the accident?

x = cars

y = trucks

$\begin{array}{lrcll} (1)& x + y &=& 32 \\ & x &=& 32 - y \\\\ (2) &4x+6y &=& 148 \\ &4\cdot (32-y) +6y &=& 148 \\ & 4\cdot 32 -4y + 6y &=& 148 \\ & 4\cdot 32 + 2y &=& 148 \quad & | \quad : 2\\ & 2\cdot 32 + y &=& 74\\ & 64 + y &=& 74 \quad & | \quad -64\\ & y &=& 74 -64\\ & \mathbf{ y }& \mathbf{=} & \mathbf{ 10 }\\\\ & x &=& 32 - y \\ & x &=& 32 - 10 \\ & \mathbf{ x }& \mathbf{=} & \mathbf{ 22 } \end{array}$

22 cars and 10 trucks

(2^-4 + 4^-3)/2^-3

$\begin{array}{rcll} \dfrac{ 2^{-4} + 4^{-3} } { 2^{-3} } &=& 2^3\cdot ( 2^{-4} + 4^{-3} ) \\\\ &=& 2^3\cdot [ 2^{-4} + (2\cdot 2)^{-3} ] \\\\ &=& 2^3\cdot [ 2^{-4} + (2^2)^{-3} ] \\\\ &=& 2^3\cdot [ 2^{-4} + 2^{2\cdot (-3)} ]\\\\ &=& 2^3\cdot [ 2^{-4} + 2^{-6} ]\\\\ &=& 2^3\cdot 2^{-4} + 2^3\cdot 2^{-6} \\\\ &=& 2^{3-4} + 2^{3-6} \\\\ &=& 2^{-1} + 2^{-3} \\\\ &=& \frac{1}{2} + \frac{1}{2^3} \\\\ &=& \frac{1}{2} + \frac{1}{8} \\\\ &=& \frac{1}{2}\cdot \frac{4}{4} + \frac{1}{8} \\\\ &=& \frac{4}{8} + \frac{1}{8} \\\\ &=& \frac{4+1}{8} \\\\ &=& \frac{5}{8} \\\\ &=& 0.625 \\\\ \end{array}$

the multiplicative inverse of -3

multiplicative inverse or reciprocal for a number $x$, denoted by $\frac{1}{x}$ or $x^{−1}$,

is a number which when multiplied by $x$ yields the multiplicative identity, $1$.

$\begin{array}{rcl } x\cdot \underbrace{\frac{1}{x}}_{\text{multiplicative inverse for x}} &=& 1 \\\\ &\text{ or }& \\\\ x\cdot\underbrace{ x^{-1} }_{\text{multiplicative inverse for x}} &=& 1\\ \end{array}$

$\begin{array}{lrcll} x=-3 & \\ & (-3)\cdot \underbrace{\frac{1}{(-3)}}_{\text{multiplicative inverse for }-3} &=& 1 \\\\ & &\text{ or }& \\\\ & (-3)\cdot\underbrace{ (-3)^{-1} }_{\text{multiplicative inverse for }-3} &=& 1\\ \end{array}$

the multiplicative inverse of $-3 ~\text{ is } ~ -\frac{1}{3} ~ \text{ or } ~ (-3)^{-1}$

Komplementieren

FRAGE: wie komme ich von den beiden Zahl 173 und 451 auf das komplement 827.

Habe keine Ahnung, wie das gehen soll

Herzlichen Dank für die Mühe.

warum ich das Frage? Darum.

Beispiel für dreistellige Rechnung: Um 173 von 451 zu subtrahieren, addiert man das Komplement 827 und erhält 1278. Nach Streichung der führenden 1 erhält man das korrekte Ergebnis.

$\begin{array}{rcl} 451-173 &=& 451 + \underbrace{( 1000-173)}_{=827} -1000\\ &=& 451 + 827 -1000 \\ &=& 1278 - 1000\\ &=& 278 \end{array}$

$\begin{array}{rcl} 1&0&0&0 \\ -&1&7&3\\ \hline &1~ \text{Ergänzung zu }~ 9 =8 & 7~ \text{Ergänzung zu }~ 9 =2 & 3~ \text{Am Ende Ergänzung zu }~ 10 =7 \\ & 8 & 2 & 7 \\ \hline \end{array}$

Hallo Guest

The denominator must not 0 (null).

The denominator is 0, if

 $\begin{array}{rcll} x+1 &=& 0 \quad & | \quad -1\\ x+1-1 &=& 0-1 \\ x+0 &=& -1 \\ x &=& -1 \end{array}$

so $x=-1$ is not permissible or $D = R ~ \backslash ~ \{-1\}$

$\dfrac{-2x+3}{x+1} > 1$

Hallo Guest

The denominator must not 0 (null).

The denominator is 0, if

 $\begin{array}{rcll} x+1 &=& 0 \quad & | \quad -1\\ x+1-1 &=& 0-1 \\ x+0 &=& -1 \\ x &=& -1 \end{array}$

so $x=-1$ is not permissible or $D = R ~ \backslash ~ \{-1\}$

$\dfrac{-2x+3}{x+1} > 1$

l = 2*r*SIN(a/2)  die die formel gerne nach a umgestellt.

Wir haben einen Kreissektor mit dem Sektorwinkel a.

Der Kreis hat den Rasius r = 22,5 mm

und die Kreissehne beträgt l = 28 mm.

$\begin{array}{rcll} l &=& 2\cdot r \cdot \sin{( \frac{a}{2} )} \\ \end{array}$

Wir stellen die Gleichung nach $a$ um:

$\begin{array}{rcll} l &=& 2\cdot r \cdot \sin{( \frac{a}{2} )} \quad & | \quad :2\cdot r\\ \frac{l}{2\cdot r} &=& \sin{( \frac{a}{2} )} \\ \sin{( \frac{a}{2} )} &=& \frac{l}{2\cdot r} \quad & | \quad \arcsin{()} \\ \frac{a}{2} &=& \arcsin{( \frac{l}{2\cdot r} )} \quad & | \quad \cdot 2 \\ a &=& 2\cdot \arcsin{( \frac{l}{2\cdot r} )} \\ \end{array}$

Jetzt setzen wir die Zahlen ein:

$\begin{array}{rcll} a &=& 2\cdot \arcsin{( \frac{28\ mm}{2\cdot 22,5 mm} )} \\ a &=& 2\cdot \arcsin{( 0,62222222222 )} \\ a &=& 2\cdot 38,4785951333^{\circ} \\ a &=& 76,9571902666^{\circ} \end{array}$

Der Sektorwinkel $a$ beträgt rund $77^{\circ}$

Find the value of cos(5*180/2-19*180/3)=? 

$\begin{array}{rcll} \cos{\left(5\cdot \frac{180}{2}-19\cdot \frac{180}{3} \right)} &=& \cos{\left(450 - 1140 \right)} \\ &=& \cos{ (-690 )} \\ &=& \cos{ (-690 +360 )} \\ &=& \cos{ (-330 )} \\ &=& \cos{ (-330 +360 )} \\ &=& \cos{ (30 )} \\ &=& \frac{ \sqrt{3} } {2} \\ &=& 0.86602540378 \\ \end{array}$

how do i solve: ^2log(-2x+3) - ^2log(x+1)>0

rearrange:

$\begin{array}{rcll} ^2log(-2x+3) - ^2log(x+1) &>& 0 \\ \log_{2}{ (-2x+3) } -\log_{2}{ (x+1) }&>& 0 \\ \log_{2}{ \left( \frac{-2x+3}{x+1} \right) } &>& 0 \quad & | \quad 2^{()}\\ \frac{-2x+3}{x+1} & > & 2^0 \quad & | \quad 2^0 = 1\\ \frac{-2x+3}{x+1} & > & 1\\ \end{array}$

$\boxed{~ \begin{array}{rcll} \dfrac{-2x+3}{x+1} & > & 1 \qquad D = R ~ \backslash ~ \{-1\}\\ \end{array} ~}$

$\small{ \begin{array}{ccc|ccc} \hline \text{denominator } x+1 > 0 && & \text{denominator } x+1< 0 && \\ \\ \hline -2x+3~ >~ 1\cdot(x+1) &&& -2x+3 {\color{red}~<~} 1\cdot(x+1)&& \\ \\ \begin{array}{ccc|c} x+1 &>& 0 & -1 \\ -2x+3 &>& x+1 & +2x-1\\ \hline \end{array} &&& \begin{array}{ccc|c} x+1 &<& 0 & -1 \\ -2x+3 &<& x+1 & +2x-1\\ \hline \end{array} \\\\ \begin{array}{ccc|c} x &>& -1 & \\ 2 &>& 3x & :3\\ \hline \end{array} &&& \begin{array}{ccc|c} x &<& -1 & \\ 2 &<& 3x & :3\\ \hline \end{array} \\\\ \begin{array}{ccc|c} x &>& -1 & \\ \frac23 &>& x & \\ \hline \end{array} &&& \begin{array}{ccc|c} x &<& -1 & \\ \frac23 &<& 3x & \\ \hline \end{array} \\\\ \mathbf{ -1 < x < \frac23 }&& & \text{no solution}&& \end{array} }$

The size of an interior angle of a n-sided polygon is 14 times of its exterior angle.Find the value of n

HOW TO SOLVE?

The Exterior Angles of a Polygon add up to $360^{\circ} \qquad sum_e = 360^{\circ}$
Sum of Interior Angles $= (n-2) × 180^{\circ}\qquad sum_i = (n-2) \cdot 180^{\circ}$

$\begin{array}{rcll} sum_i &=& 14 \cdot sum_e \\ (n-2) \cdot 180^{\circ} &=& 14 \cdot 360^{\circ} \quad & | \quad :180 \\ n-2 &=& 14 \cdot \frac{ 360^{\circ} } { 180^{\circ} } \\ n-2 &=& 14 \cdot 2 \\ n-2 &=& 28 \quad & | \quad +2 \\ \mathbf{ n } & \mathbf{=} & \mathbf{30} \\ \end{array}$