how can u calculate the 12th or 13th root on excel?\nand how u wrote it as a formula?\nand if i have the 12 cells and wish to calculate the (1/12th) root how can i write it?
125764
This is relatively easy.\n\nSay that the number of which you want the 12th root is in A1. The formula =POWER(A1,1/12) will give the 12th root of that number.
427163
(12x^2 +16x) - (12x-7)- (4x-5)\n12x^2 +16x -12x +7 -4x +5\n\n12x^2+12\n\nthat is what I got....\nI think you might be able to put 12 on the other side of the = sign and turn it into 12x^2 = -12 and then since both sides have 12 in common you can divide both sides by 12 and you will have x^2 = -1 then you can add 1 to both sides and you will have x^2 +1 but I think I'm getting off track by that point....it's been a while since I had to do this stuff.
642933
Put 10 in B5, and 6 in C5, then use this formula in D5: \n=IF(B5<C5, B5, C5)\nB5<C5 - checks the values\nB5 - if the test is true (B5 is less) then D5=B5\nC5 - if the test is false (C5 is less) then D5=C5\nTry it. Let me know how it goes! \nI love Excel problems. Okay, I'm a nerd. \nHave an EXCELlent day!
396381
you know there are 4 primitive roots, because phi(phi(13))=4.  we have 12 possible numbers, the order of each number must divide 12, and we want to find the 4 that have order 12.\n\njust go through these numbers (remember, the only possible orders are 1,2,3,4,6,12 so you don't even have to try the other numbers\n\n1^1=1 so order(1)=1\n\n2^1=2\n2^2=4\n2^3=8\n2^4=3 (mod 13)\n2^6=2^2 * 2^4 = 12 (mod 13) = -1 (mod 13)\n2^12=(2^6)^2=(-1)^2 (mod 13) = 1\nso order (2)=12 and 2 is a primitive root\n\n3^1=3\n3^2=9\n3^3=1 (mod 13)\norder(3)=3\n\n4=2^2, therefore, order(4)=6\n(4^6=(2^2)^6=2^12=1)\n\n5^1=5\n5^2=25 = -1 (mod 13)\nso 5^4 = 1(mod 13)\norder(5)=4\n\n6^1=6\n6^2=36=10 (mod 13) = -3 (mod 13)\n6^3=8 (mod 13)\n6^4=9 (mod 13) = -4 (mod 13)\n6^6 = 6^2 * 6^4 = 13 (mod 13)\n6^12=1 (mod 13)\norder(6)=12 and 6 is a primitive root\n\n7^1=7= -6 (mod 13)\n7^2= -3 (mod 13)\n7^3=5 (mod 13)\n7^4=9 (mod 13) = -4 (mod 13)\n7^6=12 (mod 13)\n7^12=1 (mod 13)\norder(7)=12 7 is a primitive root\n\n8=2^3 so order(8)=4\n\n9=3^2 so order(9)=3\n\norder(10)=6\n\n11=-2 (mod 13) so it's easy to see order(11)=12\n\nso your 4 primitive roots are 2,6,7,11 (mod 13)
286691
6root2 you need a calculator to reduce\nyou can make it a single termy thing by makeing the 6 into 36 and placing it inside the root, multiply to get 72 and you have root72 otherwise you need to get a calc to find root2 as 1.41421...* 6 to get 8.485281...\n\nfor the rest I'm assuming you mean root(80/4) and root(9/16)not root80 /4 and root9 /16 \n\nroot 80/4 you separate it to root80/root4 which is root(16*5)/2 then you can take out the 16 as 4 to get 4root5/2\n\nroot9/16 is the easiest. you do the same where you split it to root9/root16 root9 is 3 and root 16 is 4 so you have 3/4 as your answer\nhelpful?
427779
In mathematics, the cube root ( ) of a number, x, is the number which, when multiplied by itself and then multiplied by itself again gives x. For example, the cube root of 8 is 2, because 2 × 2 × 2 = 8. In general though, a real number has three cube roots, one real number, and two complex numbers. \nThe cube root operation is associative with exponentiation and distributive with multiplication and division, but is not associative or distributive with addition or subtraction.\n\nFigure it out!\n\nGood Luck!
412797
use the Power function\n\n=power(100,1/7)\n(proof - the value you receive "1.93069....."^7 = 100)\n=power(x,1/y)\n\nx=the number you want to find the y root of\ny=the root power
676804
maybe something like 7^(1/3) over 7^(1/2)\n\nthat's the same as the cube root of 7 over the square root of 7.\nTo solve it, you would just subtract the exponents and you'd get 1 over 7^(1/6)\n\nI hope that is what you are looking for.
502176
if the number is positive you get real roots\nso root 36=+/-6\nif the number is negative you get imaginary roots\nso root -144=+/-12i
611709
It *is* awfully hard to write algebra expressions on Yahoo Answers, isn't it? Oh well, we do what we can.\n\nAddressing your first question, let's write it out like a long division problem.  You'll have to imagine the horizontal bar above the larger polynomial, because I can't format it properly here:\n\nx^2 + 3x - 2  )  x^4 + x^3 - 5x^2 + 13x - 6\n\nOur first step is to look at the first term in each polynomial, and divide the smaller into the larger.  x^4 / x^2 = x^2, so the first term we write in our quotient (above the bar) is x^2.  You can write it directly above the x^4, if you wish, or above the 5x^2 term, your preference.\n\nNow, we're going to multiply that "partial quotient" by the entire divisor (x^2 + 3x - 2), which gives us x^4 + 3x^3 - 2x^2.  Write that "partial product" below the first three terms of the original dividend (x^4 + x^3 - 5x^2).  Our next step is to subtract the partial product: (x^4 + x^3 - 5x^2) - (x^4 + 3x^3 - 2x^2) = -2x^3 - 3x^2.  We then drop down the next term from the dividend, the 13x term, and repeat:\n\n-2x^3 / x^2 = -2x (add that term to our quotient)\n\n-2x * (x^2 + 3x - 2) = -2x^3 - 6x^2 + 4x\n\n(-2x^3 - 3x^2 + 13x) - (-2x^3 - 6x^2 + 4x) = 3x^2 + 9x\n\nFinally, we drop down the last term from the dividend, -6, and repeat once more.  I'm sorry that I can't write this out the way your math book can... I really hope Yahoo Answers creates some sort of way of inputting formulas, or at least monospacing text entry!\n\n3x^2 / x^2 = 3 (so we add a +3 term to our quotient)\n\n3 * (x^2 + 3x - 2) = 3x^2 + 9x - 6\n\n(3x^2 + 9x - 6) - (3x^2 + 9x - 6) = 0... no remainder!\n\nSo our final quotient is x^2 - 2x + 3, with no remainder.\n===\nNow, when doing the 2nd problem, you'll want to write the larger polynomial, the dividend, as x^4 - 3x^3 + 0x^2 + 6x - 5.  We incorporate that extra 0x^2 term so that we never find ourselves trying to subtract unlike terms.\n\nSorry there's no better way to write this out on Yahoo Answers.  Hope that helped anyway!
100509
What's exactly your problem? given number compute the dozen (say 1,2 or 3)?\ndozen=int((number-1)/12)+1\nfor 1..12 it will give 1;\nfor 13..24 it will give 2 adn\nfor 25 ..36 it will yield 3;\nthat helps?
420210
I'm Ioana and I think I can help because I have  been doing this since the fift grade here in Romainia.So , because that in your equation the number below the fraction line is the same you can say 4/x.In other case you have to amplify to get the same number below the fraction line.Now you take 2 and put it at the other side of  the equal so  4/x=7-2 from were results that 4/x=5.Now 5 is a natural number so if you want to write it as a fraction (in this case u need)you say 5/1.Now try writing it like this  4 =5\n        -    -\n       x    1\nit is the same thing.Now yo multiply on the diagonaly.So u get \n5x=4  .Remember what I said earlier.When you change the position of a number to one side or other of the equal ,the mark changes too exactly the oposite.So for - you have + and for / u have multiplication.Get it ?So the final form is x=4/5.Tell me if it works out .My adress is janne1992@yahoo.com.\n\nIoana
486669
idk but cube root is easy look here is an example the cube root of 27 is 3 because 3*3=9*3=27 so 3*3*3=27 it is 3 numbers that is multiplied times itself 3 times and give u a whole number and the 3 same numbers u multiplied r called a cube root cube mean 3
438409
I accept the challenge for the fun of it.\n\n...will get back to you.\n\n\nOK.  in cell G5, based on what you describe, the formula is "=VLOOKUP(+E5,$B$2:$C$5,2)"  Note the dollar signs in the formula referring to cells B2 and C5.  This will allow you to copy cell G5 into G6, G7, G8, and so on if you have more values in column E.\n\n\nNOTE: In column B2 to B5 are the numbers 0, 32, 63, and 94 respectively.  In column C2 to C5 are the words Freshman, Sophomore, Junior, and Senior respectively.
481186
given,\ny= X^4-13x^2=36\n\nsubstitute x^2 = t\n=> t^2-13t=36\n=> t^2-13t-36 = 0\n=> t=15.3, -2.3 (use any scientific calculator or quadratic formula)\n\n=> x=sqroot(t)\n=> x = +3.91, -3.91, +1.5i, -1.5i\n\nwhere i = sqrt(-1)\n\n\nthe result to the second sum can be obtained by simple division\n\nits a bitch to type it out, so im just gonna give you the answer\n\n\n9x^3-6x^2+8x+3 divded by 3x+2 = \n\nquotient: 3x^2-4x+5, remainder: x-7
28367
-1 works\n-1*3 + 1 = -2\n-2/2 = -1 (back to where we started)\n\nAddition> Now you say they should be pos! Jeez\n\n0 works though is it technically positive?\n\nWow, try a number like .1  - The numbers get REALLY big before coming back down to 1.  (5.00315450989997E+15)\n\n\n  Function IsOddNum(ByVal intNumber As Double) As Boolean\n    If (intNumber Mod 2 = 0) Then\n      Return False\n    Else\n      Return True\n    End If\n  End Function\n\n  Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click\n    Dim num As Double = Convert.ToDouble(TextBox1.Text)\n    Dim count As Integer = 0\n    Literal1.Text = num.ToString() & "<BR>"\n    While num <> 1 And count <> 1000\n      If IsOddNum(num) Then\n        num = 3 * num + 1\n      Else\n        num = num / 2\n      End If\n      Literal1.Text &= num.ToString() & "<BR>"\n      count += 1\n    End While\n  End Sub\n\n\nAnother Note:  I just plotted the number of iterations for the numbers between 1 and 20,000.  The graph totally shows a pattern!  (Mathematicians love patterns)\n\nAnother Note, Doh, now the numbers have to be integers?  Jeez, there goes my repeating number answer.  Are you going to keep adding rules every time I get it right? LOL
278802
-13 - (-10) + (-64) = -64\n\nThe Working : \n\n-13 - (-10) + (-64)\n=-13 + 10 - 64 = -67 \n\nNote:\nthe minus and minus in between 13 and 10 changed to a plus b'coz\n\n- and - = +\n\nBest of Luck !!