heureka

877÷54 with remainder.

\(\begin{array}{rcl} \frac{877}{54} &=& 16.2407407407\\\\ 16.2407407407 - 16 &=& 0.2407407407\\ 0.2407407407 \cdot 54 &=& 13\\ \end{array}\)

13 is the remainder

What is the measure of the supplementary angle of a 76 degreese angle?

Two Angles are Supplementary when they add up to 180 degrees

\(180^{\circ} - 76^{\circ} = 104^{\circ}\)

So C is right.

The latitude value of Toronto is 43.70 degrees, and the longitude value is 79.40 degrees. The latitude value of Melbourne is -37.81 degrees, and the longitude value is 144.96 degrees. The two cities are degrees apart in latitude. The two cities are degrees apart in longitude.

\(\begin{array}{lcl} \Delta \lambda &=& 144.96^{\circ} - 79.40^{\circ} = 65.56^{\circ} \\\\ \Delta \varphi &=& 43.70^{\circ} -(-37.81^{\circ} ) = 43.70^{\circ} +37.81^{\circ} =81.51^{\circ} \end{array} \)

The two cities are degrees apart in latitude \(81.51^{\circ}\)

The two cities are degrees apart in longitude \(65.56^{\circ}\)

what is 0 F in degrees Celsius?

\(T_C=\frac59(T_F-32) \qquad T_F=0 \ °F\\ T_C=\frac59(0-32) \\ T_C=\frac59(-32) \\ T_C=\frac{5\cdot (-32)}{9} \\ T_C=-\frac{5\cdot 32}{9} \\ T_C=-\frac{160}{9} \\ T_C=-17.7\overline{7} \ °C\\\)

Ein Schiff hat 920422 km wie viel mal ist sie um die erde gefahren?

Einmal um die Erde

\(= 2\cdot \pi \cdot 6370 ~\text{km} \\ = 40023.89~\text{km}\)

\(\frac{920422 ~\text{km} } {40023.89 ~\text{km} } = 22.997\)

Das Schiff ist rund 23 mal um die Erde gefahren.

sin^3(x) cos^5(x)

they then rewrite it as:

sin x (cos^5(x) - cos^7(x))

how?

\(\begin{array}{lcl} \sin^3{(x)} \cos^5{(x)} && \\ = \sin{(x)}\cdot \sin^2{(x)}\cos^5{(x)} \qquad | \qquad \sin^2{(x)} = 1-\cos^2{(x)}\\ = \sin{(x)}\cdot [ {1-\cos^2{(x)}} ] \cdot \cos^5{(x)} \\ = \sin{(x)}\cdot [ {\cos^5{(x)}-\cos^2{(x)}\cos^5{(x)}} ] \qquad | \qquad \cos^2{(x)}\cos^5{(x)} = \cos^7{(x)}\\ = \sin{(x)}\cdot [ {\cos^5{(x)}-\cos^7{(x)}} ] \end{array}\)

New Edited:

[p or(~p and q)] = ?

\([p \text{ or }( \text{~} p \text{ and } q)] = p + \bar{p}\cdot q = p+q= p \text{ or } q \)

\(\text{true }=1 \quad \text{false }=0 \qquad \text{ and } = "\cdot" \quad \text{ or } = "+"\\ \begin{array}{|r|r|r|r||r|} \hline p & q & \bar{p} & \bar{p} \cdot q & p+ \bar{p} \cdot q & p+q = p \text{ or } q \\ \hline 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 1 & 1\\ 1 & 1 & 0 & 0 & 1 & 1\\ \hline \end{array}\)

[p or(~p and q)] = ?

\([p \text{ or } (~p \text{ and }q)] = p + \bar{p}\cdot q\)

\(\text{true }=1 \quad \text{false }=0 \qquad \text{ and } = "\cdot" \quad \text{ or } = "+"\\ \begin{array}{|r|r|r|r||r|} \hline p & q & \bar{p} & \bar{p} \cdot q & p+ \bar{p} \cdot q & p \text{ and } q \\ \hline 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 1 & 1\\ 1 & 1 & 0 & 0 & 1 & 1\\ \hline \end{array}\)

1/8+1/2

\(\begin{array}{lcl} \frac{1}{8} + \frac{1}{2}\\\\ &=& \frac{1}{8} + \frac{1}{2}\cdot \frac{4}{4}\\\\ &=& \frac{1}{8} + \frac{4}{8}\\\\ &=& \frac{5}{8}\\\\ &=& 0.625 \end{array}\)

  0 4

- 6 5

\(\boxed{~ A=\begin{pmatrix} a&b\\ c&d \end{pmatrix}\\ A^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d&-b\\ -c&a \end{pmatrix} ~}\)

\(\begin{array}{rcl} A&=&\begin{pmatrix} 0&4\\ -6&5 \end{pmatrix}\\ A^{-1}&=&\frac{1}{0\cdot 5- 4\cdot (-6)}\begin{pmatrix} 5&-4\\ -(-6)&0 \end{pmatrix}\\\\ A^{-1}&=&\frac{1}{24}\begin{pmatrix} 5&-4\\ 6&0 \end{pmatrix}\\\\ A^{-1}&=&\begin{pmatrix} \frac{5}{24}&\frac{-4}{24}\\ \frac{6}{24}&\frac{0}{24} \end{pmatrix}\\\\ A^{-1}&=&\begin{pmatrix} \frac{5}{24}&\frac{-1}{6}\\ \frac{1}{4}&0 \end{pmatrix} \end{array}\)