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Solve 2 cos 2𝑥° = 1 for 0 ≤ 𝑥 ≤ 180.

\(\begin{array}{|rcll|} \hline 2 \cos (2x) &=& 1 \quad & | \quad : 2 \\ \cos (2x) &=& \frac{1} {2} \quad & | \quad \pm\arccos() \\ 2x &=& \pm\arccos( \frac{1} {2} ) +360\cdot k \qquad k \in \mathbb{Z} \\ 2x &=& \pm 60^{\circ} + 360^{\circ} \cdot k \quad & | \quad : 2 \\ x &=& \pm 30^{\circ} + 180^{\circ} \cdot k \\ \mathbf{x_1} & \mathbf{=} & \mathbf{30^{\circ}} \\\\ x_2 &=& -30^{\circ} + 180^{\circ} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{150^{\circ}} \\ \hline \end{array} \)

help

11+ 22+ 33+ 44+ 55+ 66+ 77+ 88+ 99+ 110+ 121+ 132+ 143+ 154+ 165+ 176+ 187+ 198+ 209+ 220+ 231+ 242+ 253+ 264+ 275+ 286+ 297+ 308+ 319+ 330+ 341+ 352+ 363+ 374+ 385+ 396+ 407+ 418+ 429+ 440+ 451+ 462+ 473+ 484+ 495+ 506+ 517+ 528+ 539+ 550+ 561+ 572+ 583+ 594+ 605+ 616+ 627+ 638+ 649+ 660+ 671+ 682+ 693+ 704+ 715+ 726+ 737+ 748+ 759+ 770+ 781+ 792+ 803+ 814+ 825+ 836+ 847+ 858+ 869+ 880+ 891+ 902+ 913+ 924+ 935+ 946+ 957+ 968+ 979+ 990+ 1001+ 1012+ 1023+ 1034+ 1045+ 1056+ 1067+ 1078+ 1089+ +1100+ 1111+ 1122+ 1133+ 1144+ 1155+ 1166+ 1177+ 1188+ 1199+ 1210+ 1221+ 1232+ 1243+ 1254+ 1265+ 1276+ 1287+ 1298+ 1309+ 1320+ 1331

Arithmetic Sequence:

\(\begin{array}{|rcll|} \hline a_n &=& a_1+(n-1)\cdot d \quad & | \quad a_1 = 11 \qquad d = 11 \qquad a_n = 1331 \\ 1331 &=& 11 + (n-1)\cdot 11 \\ 1320 &=& (n-1)\cdot 11 \\ 120 &=& n-1 \\ \mathbf{121} & \mathbf{=} & \mathbf{n} \\\\ s_n &=& \frac{(a_1+a_n)}{2} \cdot n \\ s_{121} &=& \frac{(11+1331)}{2} \cdot 121 \\ s_{121} &=& \frac{1342}{2} \cdot 121 \\ s_{121} &=& 671 \cdot 121 \\ \mathbf{s_{121}} & \mathbf{=} & \mathbf{81191} \\ \hline \end{array} \)

A sample of 4 different calculators is randomly selected from a group containing

20 that are defective and

37 that have no defects.

Find the probability that at least one of the calculators is defective. Round to the nearest thousandth.

\(\begin{array}{|rcll|} \hline && \frac{ \binom{20}{1} \cdot \binom{37}{3} } {\binom{57}{4} } \\ &=& \frac{ 20 \cdot \frac{37}{3} \cdot \frac{36}{2} \cdot \frac{35}{1} } { \frac{57}{4} \cdot \frac{56}{3} \cdot \frac{55}{2} \cdot \frac{54}{1} } \\ &=& 4\cdot \frac{ 20 \cdot 37 \cdot 36 \cdot 35 } { 57 \cdot 56 \cdot 55 \cdot 54 } \\ &=& 0.39340776183 \\ &\approx& 0.393\quad (39.3\ \% ) \\ \hline \end{array} \)

Find the next three terms in the geometric sequence: 

6075,4050,2700,1800

\(\text{ratio } r = \frac{4050}{6075} = \frac{2700}{4050} = \frac{1800}{2700} = \frac23\)

\(a_4 = 1800 \\ a_5 = a_4 \cdot r = 1800\cdot \frac23 = 1200 \\ a_6 = a_5 \cdot r = 1200\cdot \frac23 = 800 \\ a_7 = a_6 \cdot r = 800\cdot \frac23 = 533.\bar{3} \\\)

What is 5 and 1/3 divided by 1 and 7/9

\(\begin{array}{|rcll|} \hline && \dfrac{5 + \frac13 } { 1 + \frac79 } \\\\ &=& \dfrac{ \frac{3\cdot5 + 1}{3} } { \frac{9\cdot 1 + 7}{9} } \\\\ &=& \dfrac{ \frac{16}{3} } { \frac{16}{9} } \\\\ &=& \frac{16}{3} \cdot \frac{9}{16} \\\\ &=& \frac{9}{3} \\\\ &=& 3 \\ \hline \end{array}\)

Let v be the vector from initial point P1 to terminal point P2.

Write v in terms of i and j. P1= (-1,2), P2= (-9,-5)

\(\begin{array}{|rcll|} \hline \vec{v} &=& \vec{P_2} - \vec{P_1} \\ \vec{v} &=& \binom{-9}{-5} - \binom{-1}{2} \\ \vec{v} &=& \binom{-9-(-1)}{-5-2} \\ \vec{v} &=& \binom{-9+1}{-5-2} \\ \vec{v} &=& \binom{-8}{-7} \\ \vec{v} &=& -8\cdot i -7\cdot j \\ \hline \end{array} \)

how to you tell when the next leap year is?

1700 is divisible by 4 but not a leap year
1800 is divisible by 4 but not a leap year
1900 is divisible by 4 but not a leap year

Leap year:
We have a leap year if  the year  is divisible by 4
and ( the year is not divisible by 100 or the year y is divisible by 400 )

Example:
y= 2020
2020 : 4        remainder 0  okay
2020 : 100    remainder \(\ne\) 0 okay
-> 2020 is a leap year

  what is the remainder of 1320559350 divided by 97 ?

\(\begin{array}{|rcll|} \hline 1320\ 55\ 93\ 50 \pmod{97} \\\\ 1320 \pmod{97} &=& 59 \\ 59\ 55 \pmod{97} &=& 38 \\ 38\ 93 \pmod{97} &=& 13 \\ 13\ 50 \pmod{97} &=& \color{red}89 \\ \hline \end{array} \)

\(1320\ 55\ 93\ 50 \pmod{97} = \color{red}89 \)

Help

Let \(\angle{TUV} = \varphi\)

\(\begin{array}{|rcll|} \hline \tan(\varphi) &=& \frac{24\cdot \sin(38^{\circ})}{12-24\cdot \cos(38^{\circ})} \\ \tan(\varphi) &=& \frac{24\cdot 0.61566147533}{12-24\cdot 0.78801075361} \\ \tan(\varphi) &=& \frac{14.7758754078}{-6.91225808656} \quad & | \quad \text{II. Quadrant}\\ \tan(\varphi) &=& -2.13763363908 \\ \varphi &=& -64.9294774677^{\circ} + 180^{\circ} \\ \varphi &=& 115.070522532^{\circ} \\ \hline \end{array} \)

TU = ?

\(\begin{array}{|rcll|} \hline \frac{ \sin(38^{\circ}) } {TU} &=& \frac{ \sin(\varphi) } {24} \\ \frac{ \sin(38^{\circ}) } {TU} &=& \frac{ \sin(115.070522532^{\circ}) } {24} \\ TU &=& 24\cdot \frac{\sin(38^{\circ})}{\sin(115.070522532^{\circ}) } \\ TU &=& 24\cdot \frac{0.61566147533}{0.90578692079 } \\ TU &=& 16.3127497965 \\ \hline \end{array} \)

\(TU \approx 16.3\ mm\)

Can you prove sin4x=(4sinxcosx)(2cos^2x-1)?

\(\begin{array}{|rcll|} \hline \sin(4x) &=& [4\sin(x)\cos(x)][2\cos^2(x)-1] \\ && \cos(2x) = \cos^2(x) - \sin^2(x) = \cos^2(x) - [1- \cos^2(x)] = 2\cos^2(x)-1 \\ \sin(4x) &=& [4\sin(x)\cos(x)][\cos(2x)] \\ && \sin(2x) = 2\cdot \sin(x)\cos(x) \\ \sin(4x) &=& 2\sin(2x)\cos(2x) \\ && \sin(4x)=\sin(2x+2x) = \sin(2x)\cos(2x) + \cos(2x)\sin(2x) = 2\sin(2x)\cos(2x) \\ \sin(4x) &=& \sin(4x) \ \checkmark \\ \hline \end{array}\)