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To use the sum and difference formulas, we need to find  cos(a),  cos(B),  sin(a),  sin(B),  tan(a)  and  tan(B).

Using the information given, we can find these values with the Pythagorean theorem.

\(\text{1)}\qquad\sin(a+B)\,=\,\sin( a)\cos (B)+\cos(a)\sin(B)\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,(\frac{5}{13})(-\frac12)+(-\frac{12}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,-\frac{5}{26}-\frac{12\sqrt3}{26}\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,\frac{-5-12\sqrt3}{26}\\~\\ \text{2)}\qquad\cos(a+B)\,=\,\cos(a)\cos(B)-\sin(a)\sin(B)\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,(-\frac{12}{13})(-\frac{1}{2})-(\frac{5}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,\frac{12}{26}-\frac{5\sqrt3}{26}\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,\frac{12-5\sqrt3}{26}\\~\\ \text{3)}\qquad\sin(a-B)\,=\,\sin(a)\cos(B)-\cos(a)\sin(B)\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,(\frac{5}{13})(-\frac12)-(-\frac{12}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,-\frac{5}{26}+\frac{12\sqrt3}{26}\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,\frac{12\sqrt3-5}{26}\\~\\ \text{4)}\qquad\tan(a-B)\,=\,\frac{\tan(a)-\tan(B)}{1+\tan(a)\tan(B)}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}--\frac{\sqrt3}{1}}{1+(-\frac{5}{12})(-\frac{\sqrt3}{1})}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}+\sqrt3}{1+\frac{5\sqrt3}{12}}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}+\sqrt3}{1+\frac{5\sqrt3}{12}}\cdot\frac{12}{12}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-5+12\sqrt3}{12+5\sqrt3}\)

Sorry..

The periodic time is 0.08 second. (I found it by 4*0.02 as you did) 

Frequency = 1/periodic time 

1/0.08 = 12.5HZ Is that correct ?? 

To clear:

1 Amplitude = 1/4 periodic time = 1/4 complete oscillation

so 4 amplitudes = periodic time of 1 complete oscillation 

Let \(z\) and \(w\) be complex numbers satisfying \(|z| = 5\), \(|w| = 2\), and \(z\overline{w} = 6+8i\). 

Find the value of \(|z+w|^2\), \(|zw|^2\), \(|z-w|^2\),  \(\left|\dfrac{z}{w}\right|^2\) .

\(\begin{array}{|rcll|} \hline && \mathbf{|z+w|^2} \\ &=& |z|^2+|w|^2+2*\Re{(z\overline{w})} \\ &=& 5^2+2^2+2*6 \quad | \quad \Re{(z\overline{w})} = \Re{(6+8i)} = 6 \\ &=& \mathbf{41} \\ \hline && \mathbf{|zw|^2} \\ &=&|z|^2|w|^2 \\ &=& 5^2*2^2 \\ &=& 25*4 \\ &=& \mathbf{100} \\ \hline && \mathbf{|z-w|^2} \\ &=& |z|^2+|w|^2-2*\Re{(z\overline{w})} \\ &=& 5^2+2^2-2*6 \quad | \quad \Re{(z\overline{w})} = \Re{(6+8i)} = 6 \\ &=& \mathbf{17} \\ \hline && \mathbf{\left|\dfrac{z}{w}\right|^2} \\ &=& \dfrac{|z|^2}{|w|^2} \\ &=& \dfrac{5^2}{2^2} \\ &=& \dfrac{25}{4} \\ &=& \mathbf{6.25} \\ \hline \end{array}\)

Not clear (to me!) exactly what you mean.  Do you mean one period is 0.02s, or the time between the mid-line and the maximum (or minimum) is 0.02s, in which case the period would be 4*0.02s or 0.08s (or do you mean something else?)?  Either way the frequency is given by frequency = 2pi/period.

10 / 81 =0.12345679012345679012345679012346..........etc.

4k - 8 = 36

4k = 36 + 8

4k = 44

k = 44/4

k = 11

Thanks Ginger for showcasing these questions by   Hectictar,   Anthrax   and   Alan.

And thanks Alan, Hectictar and Anthrax for supplying these great answers in the first place

As soon as I get a moment I will be taking a proper look at each of them.  

1)

A, B, C, D, E, F, G, H, I, J all represent a different integer from 0 to 9.

If ABCDE-FGHIJ=66995, what is the value of each letter.

(ABCDE and FGHIJ are 5 digits integers, not multiplication)

\(ABCDE=83497 \\ FGHIJ=16502 \\ ABCDE-FGHIJ=66995\)

Here, a well-presented solution by Anthrax.

Here is an excellent presentation for a solution by Hectictar.

a² - 2ab + b²   =   (a - b)²

So....

123456789²  -  2(123456789)(123456787)  +  123456787²   =   (123456789 - 123456787)²

And....

(123456789 - 123456787)²   =   ( 2 )²   =  4

Don't know the first one but.

33 is an answer to question two.

10/33 = 0.303030303030303

If you want p > 5 to be  a prime, how many primes above 5 do you want tested?

Yea, I guess I can

okay, thank you!

The conjugate of  a - bi  =   a + bi

Can you take it from here, Cupcake   ????

Not that great at these....but..here's what I think

Expected value for Mark  = (3/6)(1)  = .50

Expected value for Evan  = (2/6)(2)  = (1/3)(2)  ≈.66

Expected value for Cayla = (1/6)(3)  = .50

Not fair....Evan has the advantage

%%time c=0 for n in range(1,1001): if((n ** 100) - 1) % 1000==0: c=c+1 print(n, end=", ") print("Total = ", c)

n=1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, 31, 33, 37, 39, 41, 43, 47, 49, 51, 53, 57, 59, 61, 63, 67, 69, 71, 73, 77, 79, 81, 83, 87, 89, 91, 93, 97, 99, 101, 103, 107, 109, 111, 113, 117, 119, 121, 123, 127, 129, 131, 133, 137, 139, 141, 143, 147, 149, 151, 153, 157, 159, 161, 163, 167, 169, 171, 173, 177, 179, 181, 183, 187, 189, 191, 193, 197, 199, 201, 203, 207, 209, 211, 213, 217, 219, 221, 223, 227, 229, 231, 233, 237, 239, 241, 243, 247, 249, 251, 253, 257, 259, 261, 263, 267, 269, 271, 273, 277, 279, 281, 283, 287, 289, 291, 293, 297, 299, 301, 303, 307, 309, 311, 313, 317, 319, 321, 323, 327, 329, 331, 333, 337, 339, 341, 343, 347, 349, 351, 353, 357, 359, 361, 363, 367, 369, 371, 373, 377, 379, 381, 383, 387, 389, 391, 393, 397, 399, 401, 403, 407, 409, 411, 413, 417, 419, 421, 423, 427, 429, 431, 433, 437, 439, 441, 443, 447, 449, 451, 453, 457, 459, 461, 463, 467, 469, 471, 473, 477, 479, 481, 483, 487, 489, 491, 493, 497, 499, 501, 503, 507, 509, 511, 513, 517, 519, 521, 523, 527, 529, 531, 533, 537, 539, 541, 543, 547, 549, 551, 553, 557, 559, 561, 563, 567, 569, 571, 573, 577, 579, 581, 583, 587, 589, 591, 593, 597, 599, 601, 603, 607, 609, 611, 613, 617, 619, 621, 623, 627, 629, 631, 633, 637, 639, 641, 643, 647, 649, 651, 653, 657, 659, 661, 663, 667, 669, 671, 673, 677, 679, 681, 683, 687, 689, 691, 693, 697, 699, 701, 703, 707, 709, 711, 713, 717, 719, 721, 723, 727, 729, 731, 733, 737, 739, 741, 743, 747, 749, 751, 753, 757, 759, 761, 763, 767, 769, 771, 773, 777, 779, 781, 783, 787, 789, 791, 793, 797, 799, 801, 803, 807, 809, 811, 813, 817, 819, 821, 823, 827, 829, 831, 833, 837, 839, 841, 843, 847, 849, 851, 853, 857, 859, 861, 863, 867, 869, 871, 873, 877, 879, 881, 883, 887, 889, 891, 893, 897, 899, 901, 903, 907, 909, 911, 913, 917, 919, 921, 923, 927, 929, 931, 933, 937, 939, 941, 943, 947, 949, 951, 953, 957, 959, 961, 963, 967, 969, 971, 973, 977, 979, 981, 983, 987, 989, 991, 993, 997, 999, Total = 400 >>Wall time: 11 ms

ab^4  = 12           a^5b^5  = 7776

Using the first equation    a =  12/ b^4         sub this into the second equation

(12/b^4)^5 * b^5  = 7776

248832 * b^5 / b^20  = 7776

248832 / b^15  = 7776

b^15 =   248832 / 7776 

b^15  =  32      

b^15  =  2^5    

b =  (2^5)^(1/15)   =  ∛2   

And a   =  12 / (2^1/3)^4      =  12 / (2^4/3)  =  12 / (2 ∛2)   =   6/∛2  =  3*(2)^(2/3)

So....assuming real solutions....a  = 3*(2)^(2/3)

Point C(3.6, -0.4) divides in the ratio 3 : 2. If the coordinates of A are (-6, 5), the coordinates of point B are . 

I'm assuming that  C divides  AB  in the ratio of  3 : 2

We have 5 equalline segments on AB    and   C are 3 of these

We can find the x coordinate of B, thusly :

[  -6 + (3/5) ( x coordinate of B -   - 6 ] =  3.6       add 6 to both sides

(3/5)(x coordinate of B  + 6 )  =  9.6      multiply both side by (5/3)

x coordinate of B + 6  =  16   subtract  6 form both sides

x coordinate of B  = 10

Similarly, we can find the y coordinate of B thusly.....

[ 5 + (3/5)( y coordinate of B - 5)  =  -0.4        subtract 5 from both sides

(3/5) (y coordinate of B - 5)  = -5.4       multiply both sides by 5/3

y coordinate of B - 5  = -9       add 5 to both sides

y coordinate of B  =  -4

So....B  =  ( 10 , - 4)   ⇒  "C"

See the graph, here : https://www.desmos.com/calculator/hfo8kkx0lp

Thank you! I copied this from my homework doc. Theres a lot more sh!t like this. I might post a question with doc links instead 😅

Thanks Omi,   

I am really sorry that one one has added to this thread yet.

I know there have been many great answers in the last few days.

Anyway, here is one of my own that I was pleased with.

It does not involve calculus or any high level maths so some less advanced  (in maths) people will be able to understand it.

If you have questions, just ask.

This was the question:

How many zeroes does 66! end in when written in base 12?

What does "1920" have to do with the problem????......LOL!!!!