heureka

Hi Melody and Chris, 

thank you.

What's

\((x+\frac{1}{x})^4 ~ \text{ ?}\)

\(\begin{array}{rcll} && (x+\frac{1}{x})^4\\ &=& \binom40 \cdot x^4 + \binom41 \cdot x^3 \cdot \frac1x + \binom42 \cdot x^2 \cdot \frac1{x^2} + \binom43 \cdot x \cdot \frac1{x^3} + \binom44 \cdot \frac1{x^4} \\ &=& 1 \cdot x^4 + 4 \cdot x^3 \cdot \frac1x + 6 \cdot x^2 \cdot \frac1{x^2} + 4 \cdot x \cdot \frac1{x^3} + 1 \cdot \frac1{x^4} \\ &=& x^4 + 4 \cdot x^2 + 6 + \frac4{x^2} + \frac1{x^4} \\ \end{array} \)

1. Days between A and B

\(\begin{array}{|rcll|} \hline d &=& 2\cdot 365,25\ \text{days} - 687\ \text{days} \\ d &=& 43.5\ \text{days} \\ \hline \end{array}\)

2. Angle \(\alpha\) = ACB:

\( \begin{array}{|rcll|} \hline \alpha &=& \frac{ 43.5\ \text{days} } {365.25\ \text{days} } \cdot 360^{\circ}\\ \alpha &=& 42.8747433265^{\circ}\\ \hline \end{array}\)

3. Angle \(\beta \) = ABC = BAC ( isosceles triangle )

\( \begin{array}{|rcll|} \hline 2\cdot \beta + \alpha &=& 180^{\circ} \\ 2\cdot \beta &=& 180^{\circ} - \alpha \\ \beta &=& 90^{\circ} - \frac{ \alpha } {2} \\ && \beta = 68.5626283368^{\circ}\\ \hline \end{array} \)

4. \(b = \overline{BA}:\)

\( \begin{array}{|rcll|} \hline \frac{ \sin(\alpha) } {b} &=& \frac{ \sin(\beta) } {1\ \text{AU}} \\ b &=& \frac{ \sin(\alpha) } { \sin(\beta) } \cdot 1\ \text{AU} \qquad & | \qquad \beta = 90^{\circ} - \frac{ \alpha } {2} \\ b &=& \frac{ \sin(\alpha) } { \sin(90^{\circ} - \frac{ \alpha } {2}) } \cdot 1\ \text{AU} \\ b &=& \frac{ \sin(\alpha) } { \cos(\frac{ \alpha } {2}) } \cdot 1\ \text{AU} \\ b &=& \frac{ \sin(\alpha) } { \cos(\frac{ \alpha } {2}) } \qquad & | \qquad b \text{ in AU}\\ && b = 0.70212731514\ \text{AU}\\ \hline \end{array}\)

5. Angle \(\delta\) = BDA:
 \( \begin{array}{|rcll|} \hline \delta + (\beta -17.1^{\circ}) + (\beta+42.8^{\circ}) &=& 180^{\circ} \\ \delta + 2\cdot \beta -17.1^{\circ} + 42.8^{\circ} &=& 180^{\circ} \\ \delta + 2\cdot \beta + 25.7^{\circ} &=& 180^{\circ} \\ \delta + 2\cdot \beta &=& 180^{\circ} - 25.7^{\circ}\\ \delta &=& 180^{\circ} - 25.7^{\circ} - 2\cdot \beta \qquad & | \qquad \beta = 90^{\circ} - \frac{ \alpha } {2} \\ \delta &=& 180^{\circ} - 25.7^{\circ} - 2\cdot ( 90^{\circ} - \frac{ \alpha } {2} ) \\ \delta &=& 180^{\circ} - 25.7^{\circ} - 180^{\circ} + \alpha \\ \delta &=& \alpha - 25.7^{\circ} \\ && \delta = 17.1747433265^{\circ} \\ \hline \end{array} \)

6. \( a = \overline{AD}:\)

\(\begin{array}{|rcll|} \hline \frac{ \sin(\delta) } {b} &=& \frac{ \sin(\beta+42.8^{\circ} ) } {a} \\ a &=& \frac{ \sin(\beta+42.8^{\circ} ) } { \sin(\delta) } \cdot b \qquad & | \qquad \beta = 90^{\circ} - \frac{ \alpha } {2} \qquad \delta = \alpha - 25.7^{\circ} \\ a &=& \frac{ \sin(90^{\circ} - \frac{ \alpha } {2}+42.8^{\circ} ) } { \sin(\alpha - 25.7^{\circ}) } \cdot b \\ a &=& \frac{ \sin(90^{\circ} - [~\frac{ \alpha } {2}-42.8^{\circ} ~]) } { \sin(\alpha - 25.7^{\circ}) } \cdot b \\ a &=& \frac{ \cos(\frac{ \alpha } {2}-42.8^{\circ} ) } { \sin(\alpha - 25.7^{\circ}) } \cdot b \qquad & | \qquad b = \frac{ \sin(\alpha) } { \cos(\frac{ \alpha } {2}) }\\ a &=& \frac{ \cos(\frac{ \alpha } {2}-42.8^{\circ} ) } { \sin(\alpha - 25.7^{\circ}) } \cdot \frac{ \sin(\alpha) } { \cos(\frac{ \alpha } {2}) }\\ && a = 2.30537070178\ \text{AU}\\ \hline \end{array} \)

7. \(x = \overline{CD}\)

\(\begin{array}{|rcll|} \hline x^2 &=& 1^2 + a^2 - 2\cdot 1 \cdot a \cdot \cos(17.1^{\circ}) \qquad & | \qquad x \text{ in AU}\\ x^2 &=& 1 + a^2 - 2 \cdot a \cdot \cos(17.1^{\circ}) \qquad & | \qquad a = 2.30537070178\ \text{AU}\\ x^2 &=& 1 + 2.30537070178^2 - 2 \cdot 2.30537070178 \cdot \cos(17.1^{\circ}) \\ x^2 &=& 1.90781964606\\ x &=& 1.38123844649\ \text{AU} \\ && x = 1.38123844649\ \text{AU} \cdot 93000000 \frac{ \text{miles} } {\text{AU}} \\ && x = 128455175.524\ \text{miles} \\ \hline \end{array} \)

The distance of mars from the sun is 1.38123844649 AU or 128,455,175.524 miles

Skylar has 40% more pennies than nickels. If the value of Skylar's pennies and nickels is $2.56, how many nickels does Skylar have?

p = pennies

n = nickels

Skylar has 40% more pennies than nickels:

\(\begin{array}{|rcll|} \hline p &=& n\cdot(1+40\ \%) \\ &=& n\cdot(1+\frac{40}{100}) \\ &=& n\cdot(1+0.4) \\ &=& n\cdot 1.4 \\ \mathbf{p} & \mathbf{=} & \mathbf{1.4\cdot n} \\ \hline \end{array}\)

The value of Skylar's pennies and nickels is $2.56:

\(\begin{array}{|rcll|} \hline p\cdot 1\ \text{Cent} + n \cdot 5\ \text{Cent} &=& 256\ \text{Cent}\\ p + 5\cdot n &=& 256 \qquad & | \qquad p = 1.4\cdot n\\ 1.4\cdot n + 5\cdot n &=& 256 \\ 6.4\cdot n &=& 256 \qquad & | \qquad : 6.4 \\ n &=& \frac{256} {6.4} \\ \mathbf{ n } & {=} & \mathbf{ 40 }\\ \hline \end{array}\)

Skylar does have 40 nickels.

pennies?

\(\begin{array}{|rcll|} \hline p &=& 1.4\cdot n\\ p &=& 1.4\cdot 40\\ \mathbf{ p } & {=} & \mathbf{ 56 }\\ \hline \end{array}\)

and Skylar does have 56 pennies.

if (x-7) is a factor of \(2x^2-11x+k\) what is the value of k

I literally have no idea what to do here.

Because a factor is x-7, therefore one root of the equation of \(2x^2-11x+k=0\), must be 7.

The roots are:

\(\begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\ x_{1,2} &=& \frac{ b \pm \sqrt{b^2-4ac} } { 2a } \\ \hline \end{array}\)

So we have:

\(\begin{array}{|rcll|} \hline 2x^2 -11x + k &=& 0 \qquad \qquad a = 2 \qquad b=-11 \qquad c = k \\\\ x_{1,2} &=& \frac{ 11 \pm \sqrt{11^2-4\cdot 2 \cdot k} } { 2\cdot 2 } \\\\ \mathbf{x_{1,2}} &\mathbf{=}& \mathbf{\frac{ 11 \pm \sqrt{121-8k} } { 4 } }\\ \hline \end{array}\)

We can to equal:
\(\begin{array}{|rcll|} \hline \mathbf{x_\text{root}} &\mathbf{=}& \mathbf{\frac{ 11 \pm \sqrt{121-8k} } { 4 } } \qquad x_\text{root} = 7\\\\ \frac{ 11 \pm \sqrt{121-8k} } { 4 } &=& 7 \qquad & | \qquad \cdot 4 \\ 11 \pm \sqrt{121-8k} &=& 7 \cdot 4 \\ 11 \pm \sqrt{121-8k} &=& 28 \qquad & | \qquad -11 \\ \pm \sqrt{121-8k} &=& 28 -11 \\ \pm \sqrt{121-8k} &=& 17 \qquad & | \qquad \text{square both sides} \\ 121-8k &=& 17^2 \\ 121-8k &=& 289 \qquad & | \qquad \cdot (-1) \\ 8k &=& 121-289 \qquad & | \qquad +121 \\ 8k &=& -168 \qquad & | \qquad :8 \\ \mathbf{k} &\mathbf{=}& \mathbf{-21} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 2x^2 -11x -21 &=& (x-7)\cdot(2x+3) \\ \hline \end{array}\)

the value of k is -21

in einem park soll ein rechteckiges beet mit kantensteinen von 1m eingefaßt werden. Das Bett ist 15m lang und 8m breit. Wie viele Steine braucht man?

Je nachdem wie die Frage gemein ist.

Entweder liegen die Steine innerhalb der Begrenzung 15m und 8m:

dann haben wir 42 Steine

Oder die Steine liegen außerhalb der Begrenzung 15m und 8m:

dann haben wir 50 Steine.

Solve for x.  Plese show all real and non-real solutions and show how you got to your answer.  If you end up with a radial in the numerator, rationalize the numerator.

\(\begin{array}{|rcll|} \hline \frac{4\sqrt{3}}{3}{x}^{2}+\frac{4\sqrt{3}}{3}x &=& -\frac{4\sqrt{3}}{3} \qquad & | \qquad +\frac{4\sqrt{3}}{3}\\ \frac{4\sqrt{3}}{3}{x}^{2}+\frac{4\sqrt{3}}{3}x +\frac{4\sqrt{3}}{3} &=& 0 \qquad & | \qquad :\frac{4\sqrt{3}}{3}\\ x^2+ x + 1 &=& 0 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\ x_{1,2} &=& \frac{ b \pm \sqrt{b^2-4ac} } { 2a } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline x^2+ x + 1 &=& 0 \qquad a = 1 \qquad b=1 \qquad c = 1 \\ x_{1,2} &=& \frac{ 1 \pm \sqrt{1^2-4\cdot 1 \cdot 1} } { 2\cdot 1 } \\ x_{1,2} &=& \frac{ 1 \pm \sqrt{-3} } { 2 } \qquad & | \qquad \sqrt{-1} = i\\ x_{1,2} &=& \frac{ 1 \pm i \cdot \sqrt{3} } { 2 } \\\\ x_{1} &=& \frac{ 1 + i \cdot \sqrt{3} } { 2 } \\ \mathbf{ x_{1} }&\mathbf{=}& \mathbf{ \frac12 + i \cdot \frac{ \sqrt{3}}{2} }\\\\ x_{2} &=& \frac{ 1 - i \cdot \sqrt{3} } { 2 } \\ \mathbf{ x_{2} } &\mathbf{=} & \mathbf{ \frac12 - i \cdot \frac{ \sqrt{3}}{2} }\\ \hline \end{array}\)

Neil Armstrong is on the moon (acceleration due to gravity is 1.6 m/s2) and throws a golf ball straight up with an upward with a speed of 10 m/s. How high does the golf ball go?

\(\begin{array}{|rcl|} \hline v_{\text{up}} &=& v_0 \\ v_{\text{down}} &=& g_{\text{moon}}\cdot t \\ \hline v &=& v_{\text{up}} - v_{\text{down}}\\ v &=& v_0 - g_{\text{moon}}\cdot t\\ \text{high max, if } v = 0 \\ 0 &=& v_0 - g_{\text{moon}}\cdot t\\ g_{\text{moon}}\cdot t &=& v_0 \\ \color{red}\mathbf{t} & \color{red}\mathbf{=} & \color{red}\mathbf{\frac{v_0}{g_{\text{moon}}} } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline h_{\text{up}} &=& v_0\cdot t \\ h_{\text{down}} &=& \frac{g_{\text{moon}}} {2} \cdot t^2 \\ \hline h &=& h_{\text{up}} - h_{\text{down}}\\ h &=& v_0\cdot t - \frac{g_{\text{moon}}} {2} \cdot t^2 \qquad &| \qquad \color{red}\mathbf{t} & \color{red}\mathbf{=} & \color{red}\mathbf{\frac{v_0}{g_{\text{moon}}} }\\ h_{\text{max}} &=& v_0\cdot \left( \frac{v_0}{g_{\text{moon}}} \right) - \frac{g_{\text{moon}}} {2} \cdot \left(\frac{v_0}{g_{\text{moon}}} \right)^2 \\ h_{\text{max}} &=& \frac{v_0^2}{g_{\text{moon}}} - \frac{v_0^2}{2\cdot g_{\text{moon}}} \\ \color{red}\mathbf{h_{\text{max}} } &\color{red}\mathbf{=}& \color{red}\mathbf{ \frac{v_0^2}{2\cdot g_{\text{moon}}} }\\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline h_{\text{max}} & = & \frac{v_0^2}{2\cdot g_{\text{moon}}} \qquad & | \qquad v_0 = 10\ \frac{m}{s} \qquad g_{\text{moon}} = 1.6\frac{m}{s^2} \\ h_{\text{max}} & = & \frac{ (10\ \frac{m}{s})^2}{2\cdot 1.6\frac{m}{s^2} } \\ h_{\text{max}} & = & \frac{10^2}{2\cdot 1.6} \cdot \frac{m^2}{s^2}\cdot \frac{s^2}{m}\\ h_{\text{max}} & = & \frac{100}{2\cdot 1.6} \ m\\ \mathbf{h_{\text{max}} }& \mathbf{=} & \mathbf{31.25 \ m }\\ \hline \end{array}\)

The golf ball does go high 31.25 m.

The parabola y = x^2 is reflected about the x-axis, then translated, so it has a turning point of (-3,5). What is the quadratic equation to represent this parabola?

\(\begin{array}{rcr} x_v &=& -3\\ y_v &=& 5 \end{array} \)