heureka

Vielen Dank,

asinus

Thank you

Find the coefficient of x^3 y^3 z^2 in the expansion of (x+y+z)^8.

Trinomial Coefficient:

\(\begin{array}{|rcll|} \hline \dbinom{8}{3,3,2} &=& \dfrac{8!}{3!3!2!} \\\\ &=& \dfrac{3!4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{3!3!2!} \\\\ &=& \dfrac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{3!2!} \quad & | \quad 3! = 6 \quad 2! = 2 \\\\ &=& \dfrac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{6\cdot 2} \\\\ &=& \dfrac{4\cdot 5 \cdot 7 \cdot 8}{2} \\\\ &=& 2\cdot 5 \cdot 7 \cdot 8 \\\\ &=& 10 \cdot 56 \\\\ &\mathbf{=}& \mathbf{ 560 } \\ \hline \end{array}\)

The trinomial coefficient of \(x^3 y^3 z^2\) in the expansion of \((x+y+z)^8\) is 560

Bitte die Gleichung ist so weit als möglich vereinfachen.

\(\Large h=\dfrac{ \dfrac{2t}{c}+\dfrac{2}{g}-\sqrt{\left(\dfrac{2t}{c}+\dfrac{2}{g} \right)^2-4\cdot \dfrac{1}{c^2}\cdot t^2}}{2\cdot \dfrac{1}{c^2}}\)

\(\begin{array}{|rcll|} \hline h &=& \dfrac{ \dfrac{2t}{c}+\dfrac{2}{g}-\sqrt{\left(\dfrac{2t}{c}+\dfrac{2}{g} \right)^2-4\cdot \dfrac{1}{c^2}\cdot t^2}}{2\cdot \dfrac{1}{c^2}} \\\\ &=& \dfrac{ 2\cdot \left( \dfrac{t}{c}+\dfrac{1}{g} \right) -\sqrt{\left(2\cdot \left(\dfrac{t}{c}+\dfrac{1}{g}\right) \right)^2-4\cdot \dfrac{1}{c^2}\cdot t^2}}{2\cdot \dfrac{1}{c^2}} \\\\ &=& \dfrac{ 2\cdot \left( \dfrac{t}{c}+\dfrac{1}{g} \right) -\sqrt{4\cdot \left(\dfrac{t}{c}+\dfrac{1}{g}\right)^2-4\cdot \dfrac{t^2}{c^2}}}{2\cdot \dfrac{1}{c^2}} \\\\ &=& \dfrac{ 2\cdot \left( \dfrac{t}{c}+\dfrac{1}{g} \right) -2\cdot \sqrt{\left(\dfrac{t}{c}+\dfrac{1}{g}\right)^2-\dfrac{t^2}{c^2}}}{2\cdot \dfrac{1}{c^2}} \\\\ &=& \dfrac{ \dfrac{t}{c}+\dfrac{1}{g} - \sqrt{\left(\dfrac{t}{c}+\dfrac{1}{g}\right)^2-\dfrac{t^2}{c^2}}}{\dfrac{1}{c^2}} \\\\ &=& c^2 \cdot \left(\dfrac{t}{c}+\dfrac{1}{g} - \sqrt{\left(\dfrac{t}{c}+\dfrac{1}{g}\right)^2-\dfrac{t^2}{c^2}} \right) \\\\ &=& c \cdot \left[ c \cdot \left( \dfrac{t}{c}+\dfrac{1}{g} - \sqrt{\left(\dfrac{t}{c}+\dfrac{1}{g}\right)^2-\dfrac{t^2}{c^2}} \right) \right] \\\\ &=& c \cdot \left( \dfrac{ct}{c}+\dfrac{c}{g} - c\cdot \sqrt{\left(\dfrac{t}{c}+\dfrac{1}{g}\right)^2-\dfrac{t^2}{c^2}} \right) \\\\ &=& c \cdot \left( t +\dfrac{c}{g} - \sqrt{c^2\cdot\left[ \left(\dfrac{t}{c}+\dfrac{1}{g}\right)^2-\dfrac{t^2}{c^2} \right] } \right) \\\\ &=& c \cdot \left( t +\dfrac{c}{g} - \sqrt{ c^2\cdot \left(\dfrac{t}{c}+\dfrac{1}{g}\right)^2-\dfrac{c^2t^2}{c^2} } \right) \\\\ &=& c \cdot \left( t +\dfrac{c}{g} - \sqrt{ c^2\cdot \left(\dfrac{tg+c}{cg}\right)^2-t^2 } \right) \\\\ &=& c \cdot \left( t +\dfrac{c}{g} - \sqrt{ c^2\cdot \dfrac{\left(tg+c\right)^2}{c^2g^2}-t^2 } \right) \\\\ &=& c \cdot \left( t +\dfrac{c}{g} - \sqrt{ \dfrac{\left(tg+c\right)^2}{g^2}-t^2 } \right) \\\\ &=& c \cdot \left( t +\dfrac{c}{g} - \sqrt{ \dfrac{t^2g^2+2tgc+c^2}{g^2}-t^2 } \right) \\\\ &=& c \cdot \left( t +\dfrac{c}{g} - \sqrt{\dfrac{t^2g^2}{g^2} + \dfrac{2tgc}{g^2} + \dfrac{c^2}{g^2}-t^2 } \right) \\\\ &=& c \cdot \left( t +\dfrac{c}{g} - \sqrt{ t^2 + \dfrac{2tc}{g} + \dfrac{c^2}{g^2}-t^2 } \right) \\\\ &=& c \cdot \left( t +\dfrac{c}{g} - \sqrt{ \dfrac{2tc}{g} + \dfrac{c^2}{g^2} } \right) \\\\ \mathbf{h} & \mathbf{=} & \mathbf{ c \cdot \left( t +\dfrac{c}{g} - \sqrt{ \dfrac{c}{g} \left( 2t + \dfrac{c}{g} \right) } \right) } \\\\ \hline \end{array}\)

\(\ldots\) man könnte noch weiter vereinfachen:

\(\begin{array}{|rcll|} \hline h & = & c \cdot \left( t +\dfrac{c}{g} - \sqrt{ \dfrac{c}{g} \left( 2t + \dfrac{c}{g} \right) } \right) \\\\ & = & c \cdot \left( t +\dfrac{c}{g} - \sqrt{ \dfrac{c}{g} } \sqrt{ 2t + \dfrac{c}{g} } \right) \\\\ & = & c \cdot \left( t +\dfrac{c}{g} - \left(\dfrac{c}{g} \right)^{\dfrac{1}{2} } \sqrt{ 2t + \dfrac{c}{g} } \right) \\\\ & = & c \cdot \left( t +\dfrac{c}{g}\cdot \left(1 - \left(\dfrac{c}{g} \right)^{-\dfrac{1}{2} } \sqrt{ 2t + \dfrac{c}{g} } \right) \right) \\\\ & = & c \cdot \left[ t +\dfrac{c}{g}\cdot \left(1 - \left(\dfrac{g}{c} \right)^{\dfrac{1}{2} } \sqrt{ 2t + \dfrac{c}{g} } \right) \right] \\\\ & = & c \cdot \left[ t +\dfrac{c}{g}\cdot \left(1 - \sqrt{ \dfrac{g}{c} } \sqrt{ 2t + \dfrac{c}{g} } \right) \right] \\\\ & = & c \cdot \left[ t +\dfrac{c}{g}\cdot \left(1 - \sqrt{ \dfrac{g}{c} \left( 2t + \dfrac{c}{g} \right) } \right) \right] \\\\ & = & c \cdot \left[ t +\dfrac{c}{g}\cdot \left(1 - \sqrt{ \dfrac{2gt}{c} + \dfrac{gc}{gc} } \right) \right] \\\\ \mathbf{h} & \mathbf{=} & \mathbf{ c \cdot \left[ t +\dfrac{c}{g}\cdot \left(1 - \sqrt{ 1+ \dfrac{g}{c}\cdot 2t } \right) \right] } \\\\ \hline \end{array}\)

When x is divided by each of 4, 5, and 6, remainders of 3, 4, and 5 (respectively) are obtained.

What is the smallest possible positive integer value of x?

\(\begin{array}{|lrclcrcl|} \hline & x &\equiv& 3 \pmod 4 &\text{or} & x &\equiv& -1 \pmod 4 \\ & x &\equiv& 4 \pmod 5 &\text{or} & x &\equiv& -1 \pmod 5 \\ & x &\equiv& 5 \pmod 6 &\text{or} & x &\equiv& -1 \pmod 6 \\\\ \Rightarrow & x &\equiv& -1 \pmod{\text{lcm}(4,5,6)} \\ & x &\equiv& -1 \pmod{60} \\ & x &\equiv& -1+60 \pmod{60} \\ & \mathbf{x} & \mathbf{\equiv} & \mathbf{59 \pmod{60}} \\ \hline \end{array}\)

The smallest possible positive integer value x is 59

In how many ways can 2 postcards be mailed into 3 mail boxes.

The answer is 9. It's solved by doing 3^2. 

Could anyone please explain why it is solved like this?

\(\text{ Let postcard $1 = PC_1$ } \\ \text{ Let postcard $2 = PC_2$ }\)

\(\text{The subset $PC_1 = $ { Box $1,\ $ Box $2,\ $ Box $3$ } } \\ \text{The subset $PC_2 = $ { Box $1,\ $ Box $2,\ $ Box $3$ } } \)

The cartesian product of  the subsets is \(PC_1 \times PC_2\) :

\(\begin{array}{|rcll|} \hline && PC_1 \times PC_2 \\ &=& 3\times 3 \\ &=& 3^2 \\ &=& 9 \\ \hline \end{array}\)

\(\begin{array}{|r|c|c|c|c|} \hline & PC_2: & \text{Box}_1 & \text{Box}_2 & \text{Box}_3 \\ PC_1 & & \\ \hline \text{Box}_1 & & (1,1) & (1,2) & (1,3) \\ \text{Box}_2 & & (2,1) & (2,2) & (2,3) \\ \text{Box}_3 & & (3,1) & (3,2) & (3,3) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline PC_1 \times PC_2 &=& \{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3) \} \\ \hline \end{array}\)

Please help with this geometry question

In rectangle \(ADEH\), points B and C trisect \(\overline{AD}\),
and points G and F trisect \(\overline{HE}\).
In addition, \(\overline{AH}=\overline{AC}=2\).
What is the area of quadrilateral \(WXYZ\) ?

\(\text{Let $\overline{AH}=\overline{AC}=2$ } \\ \text{Let $\overline{AB}=\overline{BC}=\dfrac{\overline{AC}}{2} = 1$ } \\ \text{Let $\overline{AB}=\overline{XZ} = 1$ } \\ \text{Let $\overline{BX}=\dfrac{\overline{AH}}{2} = 1$ } \)

\(\text{Let triangle $XWZ = $ triangle $BYC$ } \)

\(\begin{array}{|rcll|} \hline \mathbf{A_{\text{quadrilateral } WXYZ}} &\mathbf{=} & \mathbf{ A_{XWZ} + A_{XYZ} } \quad | \quad A_{XWZ} = A_{BYC} \\\\ &=& A_{BYC} + A_{XYZ} \quad | \quad A_{BYC} =\dfrac{\overline{BC}\cdot x}{2} \quad A_{XYZ} =\dfrac{\overline{XZ}\cdot (1-x)}{2} \\\\ &=& \dfrac{\overline{BC}\cdot x}{2} + \dfrac{\overline{XZ}\cdot (1-x)}{2} \quad | \quad \overline{BC}=\overline{XZ}=1 \quad \\\\ &=& \dfrac{x}{2} + \dfrac{1-x}{2} \\\\ &=& \dfrac{x+1-x}{2} \\\\ &\mathbf{=} & \mathbf{ \dfrac{1}{2} } \\ \hline \end{array}\)

If \(y>0\), find the range of all possible values of  \(y\)  such that  \(\lceil{y}\rceil\cdot\lfloor{y}\rfloor=42.\)

\lceil{y}\rceil\cdot\lfloor{y}\rfloor=42. 

Express your answer using interval notation.

\(6 \lt y \lt 7\)

Source:

If the surface area of a cylinder with radius of 4 feet is 48pi square feet,

what is its volume?

\(\text{Let $V$ the volume of a cylinder.} \\ \text{Let $r$ the radius of a cylinder.} \\ \text{Let $h$ the height of a cylinder.} \\ \text{Let $S$ the surface area of a cylinder.}\)

\(\begin{array}{|lrcll|} \hline (1) & V &=& \pi r^2h \\ \\ \hline \\ & S &=& 2\pi r^2 + 2\pi rh \quad & | \quad -2\pi r^2 \\ (2) & S-2\pi r^2 &=& 2\pi rh \\ \\ \hline \\ \dfrac{(1)}{(2)} & \dfrac{V}{S-2\pi r^2} &=& \dfrac{\pi r^{2}h}{2\pi rh} \\\\ & \dfrac{V}{S-2\pi r^2} &=& \dfrac{\not{\pi} r^{\not{2}}\not{h}}{2\not{\pi} \not{r}\not{h}} \\\\ & \dfrac{V}{S-2\pi r^2} &=& \dfrac{r}{2} \quad & | \quad \cdot \left( S-2\pi r^2 \right) \\\\ & V &=& \dfrac{r}{2}\cdot \left(S-2\pi r^2 \right) \quad & | \quad S=48\pi,\quad r=4 \\\\ & V &=& \dfrac{4}{2}\cdot \left(48\pi-2\pi 4^2 \right) \\\\ & V &=& 2\cdot \left(48\pi-32\pi \right) \\\\ & V &=& 2\cdot \left(16\pi \right) \\\\ & \mathbf{V} &\mathbf{=}& \mathbf{32\pi \ feet^3 } \\ \hline \end{array}\)

Auf einer Burg in einem Gebirge in Thüringen wirft an einem Wintertag (0°C) ein Kind einen Stein in einen Brunnen. Der Stein fällt ohne die Schachtwände zu berühren nach unten. Nach 6,52 Sekunden laut Stoppuhr hört man das Platschen des Steins auf der Wasseroberfläche. Der Schall braucht bei 0°C für 331,5 m Weg eine Sekunde.

Wie tief ist es im Brunnen von oben bis zur Wasseroberfläche?