heureka

6q^2 - 7qr - 24r^2

\( 6q^2 - 7qr - 24r^2 = 6(q-\frac83r) (q+\frac32r) \)

completely factor this expression?

6m^2n - 12mn^2 - 10mn + 20n^2

\(\begin{array}{|rcll|} \hline && 6m^2n - 12mn^2 - 10mn + 20n^2 \\ &=& 2n ( 3m^2 - 6mn - 5m + 10n) \\ &=& 2n (m-2n) (3m-5) \\ \hline \end{array}\)

A rectangular sheet of paper measures 12 inches by 9 inches.

One corner is folded onto the diagonally opposite corner and the paper is creased.

What is the length in inches of the crease? Express your answer as a mixed number.

Let a = 12 inches
Let b = 9 inches
Let x = crease = ?

\(\begin{array}{|rcll|} \hline x &=& \frac{b}{a}\cdot \sqrt{a^2+b^2} \\ &=& \frac{9}{12}\cdot \sqrt{12^2+9^2} \\ &=& \frac{3}{4}\cdot \sqrt{225} \\ &=& \frac{3}{4}\cdot 15 \\ &=& \frac{45}{4} \\ &=& 11 \frac14\ \text{ inches}\\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & b^2 &=& p\cdot a \\ & p &=& \frac{b^2}{a} \\\\ (2) & x^2 &=& p\cdot(p+a) \\ & x^2 &=& \frac{b^2}{a} \cdot(\frac{b^2}{a}+a) \\ & x^2 &=& \frac{b^2}{a} \cdot \frac{a}{a} \cdot(\frac{b^2}{a}+a) \\ & x^2 &=& \frac{b^2}{a^2} \cdot a \cdot(\frac{b^2}{a}+a) \\ & x^2 &=& \frac{b^2}{a^2} \cdot(b^2+a^2) \\ & x^2 &=& \frac{b^2}{a^2} \cdot(a^2+b^2) \\ & \mathbf{x} & \mathbf{=} & \mathbf{ \frac{b}{a} \cdot \sqrt{a^2+b^2} } \\ \hline \end{array}\)

Thank you, CPhill

heureka

If A2 = 2A + 3I, find A³ and A⁴ in linear form kA + sI 

\(\begin{array}{|lrcll|} \hline 1.& A^3 &=& A * A^2 \quad & | \quad A^2 = 2A + 3I \\ 2.& A^3 &=& A *(2A + 3I) \\ 3.& A^3 &=& 2A^2 + 3AI \quad & \text{ or } \quad 3AI =A^3-2A^2\\\\ 4.& A^3 &=& \underbrace{2A^2}_{=2(2A + 3I)} + \underbrace{A^3 - 2A^2}_{=3AI} \\ & A^3 &=& 2(2A + 3I) + 3AI \quad & | \quad A*I = A \\ & A^3 &=& 2(2A + 3I) + 3A \\ & A^3 &=& 4A + 6I + 3A \\ 5. & A^3 &=& 7A + 6I \\ \hline \end{array}\)

cos(pi/7) + (cos(pi/7))^2 - 2(cos(pi/7))^3 = ?

step by step

\(\begin{array}{|l|rcll|} \hline \mathbf{1. \text{ Substitution } } \\ x = \frac{\pi}{7} && \cos(\frac{\pi}{7}) + \cos^2(\frac{\pi}{7}) - 2\cos^3(\frac{\pi}{7}) \\ &=& \cos(x) + \cos^2(x) - 2\cos^3(x) \\ \hline \end{array}\)

\(\mathbf{2. \cos^2(x) \text{ to } \cos(2x) \text{ and } \cos^3(x) \text{ to } \cos(3x) }\)

\(\begin{array}{|rcll|} \hline \text{Formula}: \\ \cos(2x) &=& 2\cos^2(x) - 1 \qquad \text{ or } \qquad \cos^2(x) = \frac12\Big(1+\cos(2x)\Big) \\ \cos(3x) &=& 4\cos^3(x) - 3\cos(x) \qquad \text{ or } \qquad \cos^3(x) = \frac14\Big(3\cos(x)+\cos(3x)\Big) \\ \\ && \cos(x) + \cos^2(x) - 2\cos^3(x) \\ &=& \cos(x) + \frac12\Big(1+\cos(2x)\Big) - \frac24\Big(3\cos(x)+\cos(3x)\Big) \\ &=& \frac12\Big( 2\cos(x) + 1 + \cos(2x)-3\cos(x) - \cos(3x) \Big) \\ &=& \frac12\Big( 1- \cos(x)+\cos(2x)-\cos(3x) \Big) \\ \hline \end{array} \)

3. Change minus to plus

\(\begin{array}{|l|rclcl|} \hline \text{Formula}: \\ \cos(x) = -\cos(\pi-x) \\\\ & \cos(\frac{\pi}{7}) &=& -\cos(\pi-\frac{\pi}{7}) \\ & &=& -\cos(\frac67 \pi) \\ & \mathbf{\cos(x)} &\mathbf{=}&\mathbf{ -\cos(6x)} \\\\ & \cos(\frac{3\pi}{7}) &=& -\cos(\pi-\frac{3\pi}{7}) \\ & &=& -\cos(\frac47 \pi) \\ & \mathbf{\cos(3x) } &\mathbf{=}& \mathbf{-\cos(4x)} \\\\ \hline \end{array} \\ \begin{array}{|rcll|} \hline && \frac12\Big( 1- \cos(x)+\cos(2x)-\cos(3x) \Big) \\ &=& \frac12\Big( 1+ \cos(6x)+\cos(2x)+\cos(4x) \Big) \\ &=& \frac12\Big( 1+\cos(2x)+\cos(4x)+ \cos(6x) \Big) \\ \hline \end{array} \)

\(\begin{array}{|l|rcll|} \hline \mathbf{4. \text{ Substitution } } \\ y = 2x && \frac12\Big( 1+\cos(2x)+\cos(4x)+ \cos(6x) \Big) \\ &=& \frac12\Big( 1+\cos(y)+\cos(2y)+ \cos(3y) \Big) \\ \hline \end{array} \)

5. complex numbers \(1+\cos(y)+\cos(2y)+ \cos(3y) = \ ?\)

\(\begin{array}{|rcll|} \hline C &=& 1+\cos(y)+\cos(2y)+ \cos(3y) \\ S &=& \qquad \sin(y)+\sin(2y)+ \sin(3y) \\ \mathbf{T} &\mathbf{=}& \mathbf{ C+i\cdot S } \\\\ T &=& 1+\cos(y)+\cos(2y)+ \cos(3y)+i\cdot \Big( \sin(y)+\sin(2y)+ \sin(3y) \Big)\\ &=& 1+\underbrace{\cos(y)+i\cdot \sin(y)}_{=e^{iy}} +\underbrace{\cos(2y)+i\cdot \sin(2y)}_{=e^{i2y}} +\underbrace{ \cos(3y)+i\cdot \sin(3y)}_{=e^{i3y}} \\ &=& 1+ e^{iy} + e^{i2y} + e^{i3y} \\ &=& 1+ e^{iy} + (e^{iy})^2 + (e^{iy})^3 \qquad \text{geometric series } r=e^{iy} \\ T &=& 1+ r + r^2 + r^3 \\ \hline T &=& 1+ r + r^2 + r^3 \\ r\cdot T &=& \qquad r + r^2 + r^3 + r^4 \\ \hline r\cdot T - T &=& r^4 - 1 \\ T\cdot (r-1) &=& r^4 - 1 \\ T &=& \frac{r^4 - 1} {r-1} \\ &=& \frac{(e^{iy})^4 - 1} {e^{iy}-1} \\ \mathbf{T} & \mathbf{=} & \mathbf{\frac{e^{i4y} - 1} {e^{iy}-1}} \\ \hline \end{array}\)

\(\begin{array}{|l|rclcl|} \hline \text{Formula}: \\ e^{i\phi} - 1 = 2i\cdot \sin(\frac{\phi}{2})\cdot e^{i\frac{\phi}{2}} \\\\ & \mathbf{T} & \mathbf{=} & \mathbf{\frac{e^{i4y} - 1} {e^{iy}-1}} \\ & &=& \frac{ 2i\cdot \sin(2y) \cdot e^{i2y} } { 2i\cdot \sin(\frac{y}{2})\cdot e^{ i\frac{y}{2} }} \qquad | \qquad y = 2x \\ & &=& \frac{ \sin(4x) \cdot e^{i4x} } { \sin(x)\cdot e^{ ix }} \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot e^{i4x-ix} \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot e^{i3x} \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot \Big( \cos(3x) + i\cdot \sin(3x) \Big) \\ & &=& \underbrace{\frac{ \sin(4x) } { \sin(x)} \cdot \cos(3x)}_{=C} + i\cdot \underbrace{ \frac{ \sin(4x) } { \sin(x)} \cdot \sin(3x) }_{=S} \\\\ \hline \end{array}\)

6. Solution

\(\begin{array}{|rcll|} \hline & C &=& 1+\cos(y)+\cos(2y)+ \cos(3y) \qquad | \qquad y = 2x \\ & &=& 1+\cos(2x)+\cos(4x)+ \cos(6x) \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot \cos(3x) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{Formula}: \\ \sin(x) = \sin(\pi-x) \\\\ & \cos(\frac{4\pi}{7}) &=& \sin(\pi-\frac{4\pi}{7}) \\ & &=& \sin(\frac37 \pi) \\ & \mathbf{\sin(4x)} &\mathbf{=}&\mathbf{ \sin(3x)} \\\\ & \cos(\frac{6\pi}{7}) &=& \sin(\pi-\frac{6\pi}{7}) \\ & &=& \sin(\frac17 \pi) \\ & \mathbf{\sin(6x)} &\mathbf{=}&\mathbf{ \sin(x)} \\\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \cos(\frac{\pi}{7}) + \cos^2(\frac{\pi}{7}) - 2\cos^3(\frac{\pi}{7}) \quad & | \quad x=\frac{\pi}{7} \\ &=& \cos(x) + \cos^2(x) - 2\cos^3(x) \\ &=& \frac12\Big( 1- \cos(x)+\cos(2x)-\cos(3x) \Big) \\ &=& \frac12\Big( 1+\cos(2x)+\cos(4x)+ \cos(6x) \Big) \\ &=& \frac12\Big( \frac{ \sin(4x) } { \sin(x)} \cdot \cos(3x) \Big) \quad & | \quad \mathbf{\sin(4x)= \sin(3x)}\\ &=& \frac12\Big( \frac{ \sin(3x) } { \sin(x)} \cdot \cos(3x) \Big) \quad & | \quad \sin(3x)\cos(3x)=\frac{\sin(6x)}{2} \\ &=& \frac12\Big( \frac{ \sin(6x) } { 2 \sin(x)} \Big) \quad & | \quad \mathbf{\sin(6x)=\sin(x)} \\ &=& \frac12\Big( \frac{ \sin(x) } { 2 \sin(x)} \Big) \\ &=& \frac12\Big( \frac12 \Big) \\ &=& \mathbf{\frac14} \\ \hline \end{array}\)

In right triangle ABC, M and N are midpoints of legs AB and BC, respectively.

Leg AB is 6 units long, and

leg BC is 8 units long.

How many square units are in the area of triangle APC?

another approach

Let \(BC = 8\)
Let \(AB = 6\)

\(\text{Let } A = \text{area}_{ABC} = \frac{AB\cdot BC}{2} = \frac{6\cdot 8}{2} = 24\)

\(\text{Let } A_1 = \text{area}_{BPN} = \text{area}_{NPC} . \quad ( \text{The base and the height is equal} ) \\ \text{Let } A_2 = area_{BCP} = area_{CAP} . \quad ( \text{The base and the height is equal} ) \\ \text{Let } A_3 = area_{APC} = \ ? \\ \text{Let area}_{BAN} = \frac{ AB\cdot \frac{BC}{2} } {2} = \frac{ AB\cdot BC } {4}\\ \text{Let area}_{BMC} = \frac{ BC\cdot \frac{AB}{2} } {2} = \frac{ AB\cdot BC } {4}\)

\(\begin{array}{|rcll|} \hline \text{area}_{BAN} &=& \text{area}_{BMC} = \frac{ AB\cdot BC } {4} \quad & | \quad \text{area}_{BAN} = A_1 +2A_2 \qquad \text{area}_{BMC} = A_2 +2A_1 \\ A_1 +2A_2 &=& A_2 +2A_1 \\ 2A_2-A_2 &=& 2A_1-A_1 \\ A_2 &=& A_1 \\\\ A_1 +2A_2 &=& \frac{ AB\cdot BC }{4} \quad & | \quad A_2 = A_1 \\ 3A_1 &=& \frac{ AB\cdot BC }{4} \\ A_1 &=& \frac{ AB\cdot BC }{12} \quad & | \quad AB = 6 \qquad BC = 8 \\ A_1 &=& \frac{ 6\cdot 8 }{12} \\ \mathbf{A_1} & \mathbf{=} & \mathbf{4} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline A &=& 2A_1+2A_2+A_3 \quad & | \quad A_2 = A_1 \\ A &=& 2A_1+2A_1+A_3 \\ A &=& 4A_1+A_3 \\ A_3 &=& A-4A_1 \quad & | \quad A = 24 \qquad A_1 = 4\\ A_3 &=& 24-4\cdot 4 \\ A_3 &=& 24-16 \\ \mathbf{A_3 } & \mathbf{=} & \mathbf{8} \\ \hline \end{array} \)

The area of triangle APC is 8 square units

In right triangle ABC, M and N are midpoints of legs AB and BC, respectively.

Leg AB is 6 units long, and

leg BC is 8 units long.

How many square units are in the area of triangle APC?

Centroid of a Triangle: The point where the three medians of the triangle intersect.

So P is the centroid of the triangle ABC.

\(\begin{array}{lcll} \text{Let } AB = 6 \\ \text{Let } BC = 8 \\\\ \text{Let } \vec{B} = \binom {0}{0} \\ \text{Let } \vec{A} = \binom {0}{AB} \\ \text{Let } \vec{C} = \binom {BC}{0} \\ \end{array} \)

\(\vec{P} \text{ the centroid of the triangle ABC}\ =\ ?\)

\(\begin{array}{|rcll|} \hline \vec{P} &=& \frac13 (\vec{A}+\vec{B}+\vec{C}) \\ \vec{P} &=& \frac13 \Big( \binom {0}{AB}+\binom {0}{0}+\binom {BC}{0} \Big) \\ \vec{P} &=& \frac13 \binom {BC}{AB} \quad & | \quad AB = 6 \qquad BC = 8 \\ \vec{P} &=& \frac13 \binom {8}{6} \\ \vec{P} &=& \binom {\frac{8}{3}}{2} \\ \hline \end{array} \)

\(\text{Area}_{\text{APC}} = \ ? \)

\(\begin{array}{|rcll|} \hline \text{Area}_{\text{APC}} &=& \frac12 | (\vec{C}-\vec{P}) \times (\vec{A}-\vec{P}) | \\ &=& \frac12 | \Big(\binom {BC}{0}- \binom {\frac{8}{3}}{2} \Big) \times \Big(\binom {0}{AB}- \binom {\frac{8}{3}}{2} \Big) | \quad & | \quad AB = 6 \qquad BC = 8 \\ &=& \frac12 | \Big(\binom {8}{0}- \binom {\frac{8}{3}}{2} \Big) \times \Big(\binom {0}{6}- \binom {\frac{8}{3}}{2} \Big) | \\ &=& \frac12 | \binom {8-\frac{8}{3}}{0-2} \times \binom {0-\frac{8}{3}}{6-2} | \\ &=& \frac12 | \binom {\frac{16}{3}}{-2} \times \binom {-\frac{8}{3}}{4} | \\ &=& \frac12 \cdot \Big( \frac{16}{3} \cdot 4 - (-2)\cdot (-\frac{8}{3}) \Big) \\ &=& \frac12 \cdot \Big( \frac{64}{3} - \frac{16}{3} \Big) \\ &=& \frac12 \cdot \frac{48}{3} \\ &=& \frac{48}{6} \\ \text{Area}_{\text{APC}} &=& 8 \text{ square units } \\ \hline \end{array}\)

Der NVP Wert eines Kabels beträgt 0,65.Wie groß ist die Siganlgeschwindigkeit und die Signallaufzeit?
Die Geschwindigkeit habe ich errechnet . 195000 km/s ,
aber mit der Signallaufzeit habe ich meine Probleme

Die Formel lautet. T=1/v in ns/m

1.  Signalgeschwindigkeit v

\(\begin{array}{|rcll|} \hline \text{NVP } &=& \frac{v}{c_0} \quad &| \quad c_0 = 300000\ \frac{km}{s} \qquad \text{NVP } = 0,65 \\ 0,65 &=& \frac{v}{300000\ \frac{km}{s} } \\ v &=& 0,65 \cdot 300000\ \frac{km}{s} \\ v &=& 195000\ \frac{km}{s} \\ \hline \end{array} \)

Die Signalgeschwindigkeit beträgt rund 195000 \(\frac{km}{s}\)

2. Die Signallaufzeit T

\(\begin{array}{|rcll|} \hline T &=& \frac{1}{v} \quad &| \quad v = 195000\ \frac{km}{s}\\ &=& \frac{1}{ 195000\ \frac{km}{s}} \\ &=& \frac{1}{ 195000 } \cdot \frac{s}{km } \\ &=& \frac{1}{ 195000 } \cdot \frac{s}{km } \cdot \frac{1\ km}{1000\ m} \\ &=& \frac{1}{ 195000\cdot 1000 } \cdot \frac{s}{m } \\ &=& \frac{1}{ 195000000 } \cdot \frac{s}{m } \\ &=& \frac{1}{ 0,195\cdot 10^{9} } \cdot \frac{s}{m } \cdot \frac{1\ ns}{10^{-9} s}\\ &=& \frac{1}{ 0,195\cdot 10^{9}\cdot 10^{-9} } \cdot \frac{ns}{m } \\ &=& \frac{1}{ 0,195 } \cdot \frac{ns}{m } \\ &=& 5,12820512821\cdot \frac{ns}{m } \\ \hline \end{array}\)

Die Signallaufzeit beträgt rund 5,1282  \(\frac{ns}{m}\)

We have a right triangle triangle ABC where the legs AB and BC have lengths 6 and 3sqrt3, respectively.

Medians AM and CN meet at point P.

What is the length of CP?

another approach

See labeling CPhill :

\(\begin{array}{rcll} \text{Let } \overline{AB} &=& 6 \\ \text{Let } \overline{BC} &=& 3 \sqrt{3} \\ \text{Let } \overline{CP} &=& ? \\\\ \text{Let } \vec{A} &=& \binom{0}{6} \\ \text{Let } \vec{B} &=& \binom{0}{0} \\ \text{Let } \vec{C} &=& \binom{ 3 \sqrt{3} }{0} \\ \end{array}\)

Point P is the barycenter of triangle ABC

\(\begin{array}{|rcll|} \hline \vec{P} &=& \frac13 \cdot( \vec{A} + \vec{B} + \vec{C} ) \\ \vec{P} &=& \frac13 \cdot\Big( \binom{0}{6} + \binom{0}{0} + \binom{ 3 \sqrt{3} }{0} \Big) \\ \vec{P} &=& \frac13 \cdot \binom{ 0+0+3\sqrt{3} } { 6+0+0 } \\ \vec{P} &=& \frac13 \cdot \binom{ 3\sqrt{3} } { 6 } \\ \vec{P} &=& \binom{ \sqrt{3} } { 2 } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \overline{CP} &=& |\vec{C}-\vec{P}| \\ &=& |\binom{ 3 \sqrt{3} }{0}-\binom{ \sqrt{3} } { 2 }| \\ &=& | \binom{ 3 \sqrt{3}-\sqrt{3} }{0-2} | \\ &=& | \binom{ 2\sqrt{3} }{-2} | \\ &=& \sqrt{ (2\sqrt{3})^2 + (-2)^2 } \\ &=& \sqrt{ 4\cdot 3 + 4 } \\ &=& \sqrt{ 16 } \\ \mathbf{\overline{CP} } & \mathbf{=} & \mathbf{4} \\ \hline \end{array}\)