It means year 2002.
no
k2 + k2 = 2k2
It is a quadratic expression that can be factored as: (2x-5)(x+6)
I know he isin NBA 2k2 while playing for the Wizards
1Everest 2K2 / Qogir / Godwin Austen 3Kangchenjunga 4Lhotse 5Makalu 6Cho Oyu 7Dhaulagiri I 8Manaslu 9Nanga Parbat 10Annapurna I 11Gasherbrum I / Hidden Peak / K5 12Broad Peak / K3
It is the same, you can use ohm, µ, R or E to represent Ohm, like 2E2 or 2R2 = 2.2 Ohm and 2K2 = 2.2 Kilo Ohm also 2M2 will be 2.2 Mega Ohm.
You can't get nba 2k9 for psp even you can't get nba 2k, 2k1, 2k2, 2k3, 2k4, 2k5, 2k6, 2k7, or 2k8 on psp. But NBA 2K10 is gonna COME OUT for PSP on October 6, 2009
The smallest integer is 2.There is no smallest number. Dividing 1152 by any number of the form x = 1/(2k2) where k is an integer will result in a perfect square. Since there is no limit to how large k can be, there is no limit to how small x can be.
First lets look at the equation for a circle... (x - h)2 + (y - k)2 = r2 What do I know? The center must be constant so H and K are two variables I need to solve for... and I know I need to solve for R as well! take your points (i'm making a simple circle with center (0,0) with radius of 1...) 1,0 0,1 -1,0 I have my 3 points.... I will make myself 3 equations: eqn #1a: (1 - h)2 +(0 - k)2 = r2 eqn #2a: (0 - h)2 +(1 - k)2 = r2 eqn #3a: (-1 - h)2 +(0 - k)2 = r2 Ok so now I have some equations to play with... eqn #1b: h2 - 2h +1 + k2 = r2 eqn #2b: h2 + k2-2k +1 = r2 eqn #3b: h2 + 2h +1 + k2= r2 The rest is subsitution... We want r's value so lets see. Lets take eqn #1b and #2b... h2 - 2h +1 + k2 = h2 + k2-2k +1 Balance it out cancle and such try to make it = 0 - 2h = -2k h = k so now I know I can replace h with k or k with h It doesn't matter which one I make to the other if we had 2h = k I would put 2h in its just simpler... keep in mind either way works. eqn #1c: k2 - 2k +1 + k2 = r2 eqn #2c: k2 + k2-2k +1 = r2 eqn #3c: k2 + 2k +1 + k2= r2 collect terms eqn #1c: 2k2 - 2k +1 = r2 eqn 1 is the same as eqn 2! eqn #2c: 2k2 - 2k +1 = r2 eqn #3c: 2k2 + 2k +1= r2 We really have two equations left now it doesn't always work out this way. Set eqn 2 and 3 equal to each other via the r 2k2 - 2k +1 = 2k2 + 2k +1 do some cancling -2k = 2k at this point I know k = 0 its the only viable answer.... take any one of our origonal equations.... h2 - 2h +1 + k2 = r2 all the parts with k or h = 0; remember h = k = 0 1 = r2 1 = r If I had time I would do a more complex equation for yea... but this kinda gets the point across...
The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6
It mean what you don't what does it mean.
Mean is the average.