0
What else can I help you with?
Is LeBron James in 2k2?
no
What is k squared plus k squared?
k2 + k2 = 2k2
2k2 plus 7k - 30?
It is a quadratic expression that can be factored as: (2x-5)(x+6)
Which basketball video game is Michael Jordan in?
I know he isin NBA 2k2 while playing for the Wizards
What are the highist mountains' in the world?
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
What is the difference between 100ohm and 100E?
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.
Is NBA 2K9 on PSP?
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
What is the smallest number that divides 1152 to give a perfect square?
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.
How do you find the radius of a circle if you know three points on the circumference?
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...
What is the nth term and sum of n terms of 6 13 24 39?
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
Solve -8k2 -12k plus 92?
Presumably this is a quadratic equation in the form of -8k2-12+92 = 0 which will have two solutions. First divide all terms by -4 to bring the equation at its lowest terms remembering that a - divided into a - is equal to a + 2k2+3k-23 = 0 Use the quadratic equation formula to factorise the equation: (2k-5.446221994)(k+4.223110997) = 0 Therefore the solutions are: k = 2.723110997 or k = -4.223110997 to nine decimal places respectively.
Can you prove n2 - n is even for any natural number n?
There are two cases here: one where n is even and one where n is odd. Let's consider the case where n is even: If n is even then n2 has to be even (since multiple of an even number must always be even.) In this case, we are subtracting an even number from and even number, the result must be even. This proves than n2 - n is even when n is even. Now let's consider the case where n is odd: If n is odd, then n2 must be odd. This is because an odd number times an odd number is always odd. (You could think if this as an odd number times and even number and then adding an odd number. For example, say that n is odd. n-1 is then even, and n2 = n(n-1) + n. n(n-1) must be even, since it is a multiple of an even number. And even number plus and odd number then has to be odd.) Now we know we have and odd number minus and odd number, which has to be even. So this proves that n2 - n is even when n is odd. Since we have proved that n2 - n is even for both when n is even and when n is odd, and there are no other cases, n2 - n must be even for any natural number n. or Let n be a natural number. Then n can be even or odd. We want to show that n2 - n = 2m where m is any positive nteger (by the def. of even). Case 1: Let n be even. Then n = 2k (def. of even), where k is any positive integer. Then, n2 - n = (2k)2 - (2k) = 2(2k2 - k); let 2k2 - k = m = 2m Therefore, n2 - n is even. Case 2: Let n be odd. Then n = 2k +1 (def. of odd), where k is any positive integer. Then, n2 - n = (2k - 1)2 - (2k - 1) = 4k2 - 4k + 1 - 2k + 1 = 4k2 - 6k + 2 = 2(2k2 - 3k + 1); let 2k2 -3k + 1 = m = 2m Therefore, n2 - n is even. Therefore, for any natural number n, n2 - n is even.
