the sequence of numbers when the first 2 numerals are 0 then 1 followed by the addition of the past t2 numbers example-0,1,1,2,3,5,8,13 etc
It doesn't. It varies inversely as that square root.Here's how I did it:T = time for a planet's revolutionR = mean distance from the sunK = some konstant, any konstantV = planet's linear speedKepler's 3rd law: T2/R3 = K . . . . . T2 = K R3But T = orbital circumference/V , so (2 pi R/V)2 = K R3 , and (2 pi/V)2 = K RDivide ' 1 ' by each side:(V/2 pi)2 = 1/K R or V2 = 1/K R (new konstant)V = K sqrt( 1/R )qed
t2 + 3t - 10
The triangular numbers are 1, 3, 6, 10, 15, ... and are calculated as: t1 = 1 t2 = 1 + 2 t3 = 1 + 2 + 3 tn = 1 + 2 + ... + n The formula for the sum tn = 1 + 2 + ... + n is: tn = n(n+1)/2
46, 51, 57, 64 ----------------------------------------------------------------------------------------- What numbers would you like to be missing? Whatever numbers you specify, I can give you a formula for the general term of the pattern so that the numbers already specified, and the wanted numbers appear for the terms 1, 2, ..., n However, the pattern given looks like it is: 36, 37, 39, 42, w, x, y, z, 72, 81, 91, 102, for which the simplest formula for the general term is t{n} = (x² -x + 72)/2 which means the missing terms t{5-8} are: w = t5 = (5² - 5 + 72)/2 = 46 x = t6 = (6² - 6 + 72)/2 = 51 y = t7 = (7² - 7 + 72)/2 = 57 z = t8 = (8² -8 + 72)/2 = 64 The formula is derived from the fact that the difference between one term and the next is one more than the difference between the previous term and that term: t2 = t1 + 1 t3 = t2 + 2 = t2 + 1 + 1 t4 = t2 + 3 = t2 + 2 + 1 → t5 = t4 + 3 + 1 = t4 + 4 = 42 + 4 = 46 → t6 = t5 + 4 + 1 = t5 + 5 = 46 + 5 = 51 → t7 = t6 + 5 + 1 = t6 + 6 = 51 + 6 = 57 → t8 = t7 + 6 + 1 = t7 + 7 = 57 + 7 = 64 → t9 = t8 + 7 + 1 = t8 + 8 = 64 + 8 = 72 and the sequence continues.
/t=/t1=/t2=/t3... vt=v1+v2+v3... rt=r1+r2+r3... series formula....
B( T2/R3 = constant )
the sequence of numbers when the first 2 numerals are 0 then 1 followed by the addition of the past t2 numbers example-0,1,1,2,3,5,8,13 etc
Planets orbit the sun in proportion to the radius of their orbit. The Radius Cubed is proportional to the Time Squared. R3 = kT2 Therefore, R3/T2 = k (k will always be the same number)
The answer depends on the context. It could refer to the nth term in a sequence of numbers: T1, T2, ...
12 hole ocarina music is just the same as any other music. You read the dots, and play the notes. Of course you will need to know the proper finger positions for each note, and know the name of each note. First things first, to hold your ocarina, have your left palm facing you, your four fingers on the holes on the top, and your left thumb over the hole on the left side on the bottom of the ocarina. Then, place your right hand over the other side, covering the top holes with your fingers, and your thumb covering the other bottom hole. For the purpose of this answer, I will be referring to the fingers and holes L1 L2 L3 L4 ___________ ,--- o o o o ---.... | ---. |__ __o o o o___---` \_/ R1 R2 R3 R4 T1 T2 o o Now, I will list the primary note names as well as their finger positions: B - All fingers down (including the tiny little holes above L3 and R3) C - Everything but the tiny little holes D - Everything but R4 and the tiny little holes E - Everything but R3, R4, and the tiny little holes F - Everything but R2, R3, R4, and the tiny little holes. G - T1, T2, L1, L2, L3, L4 A - T1, T2, L1, L2, L4 B - T1, T2, L1, L4 C - T1, T2, L4 D - T2, L4 If you want any of the sharps and/or flats of these notes, just experiment. You now know the main notes on an ocarina, and are completely free to practise to your heart's content.
Type your answer here... it is a T2 hyperintense foci
Air
If you start on the G on the second line on the staff and then go to the G above the staff, the fingerings go like this: 0-T12-T2-T0-T12-T2-T2-T0-T2-T2-T12-T0-T2-T12-0 (G, A, B, C, D, E, F#, G, F#, E, D, C, B, A, G)
On the golf score board, the designation "T2" means tied for second place.
win the stick t2 tournament, as strange as that may seem
It doesn't. It varies inversely as that square root.Here's how I did it:T = time for a planet's revolutionR = mean distance from the sunK = some konstant, any konstantV = planet's linear speedKepler's 3rd law: T2/R3 = K . . . . . T2 = K R3But T = orbital circumference/V , so (2 pi R/V)2 = K R3 , and (2 pi/V)2 = K RDivide ' 1 ' by each side:(V/2 pi)2 = 1/K R or V2 = 1/K R (new konstant)V = K sqrt( 1/R )qed