What is the cube root of 27225?
471374
Dear student,\nThe cube root of 27225 is between 30.0831 and 30.0832.\nbest regards,\nMarcello
961314
easy the number is found as such \n\n21 = 3*7\n\n27 = 3^3 \n\n33 = 3 *11\n\n55 = 5 *11\n\nIn order that a number must be divisble by these, it must contain \nall factors. if you look for the samllest avoid repetition \n\n3^3 , 5, 11  =1385\n\nand you add 5 \n\nso the number is 1390
977047
The cube root of x is x^(1/3), so the cube root of a^2 is a^(2/3)
1045150
well, the only single digit cubes are 1 and 8. To take three and get a sum of 10, you need 1,1 and 8. Arranging these palidromically is left as an exercise for the reader
1175698
cube root of (-125)=-5 \n\n3*sqrt(-125)=\n3*sqrt(25*5*-1)=\n3*5*i*sqrt(5)=\n15i*sqrt(5)\n\nnot sure which you were asking. For future reference, if you are talking cube or higher roots, use the n^(1/y) notation. A lot less confusing. sqrt is just for square roots.
1115504
the cube root can be found using a calculator\nmake sure you have a scientific one though\ni had do use it alot in geometry\n\nits alot like a square root just three instead of 2
782506
(  4 * (2)^(1/2)    ,    45 degrees)\n\n\n(   - 4 * (2)^(1/2)   , 225 degrees)\n\n\n\nand that big jumble on the start of each is 4 times the square root of 2, or the square root of 32, AKA (32)^(1/2). Math is hard on keyboards.
494551
256 - the highest number that can be expressed is 255 (calculation shown below), but 0 is also a valid combination:\n\n1 1 1 1 1 1 1 1\n\n=> 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128\n\n= 255\n\nThis is in a sequence of powers of 2 = 2^0 + 2^1 +...+ 2^7
331432
27..444...oh umm...I don't know, google it, now, back to counting...1 2 34 65 R HG
772626
The number is 41312432.\n\nThe "4" is in the 1st anf fifth space. 4_ _ _ _4_ _\nThe "1" is in the 2nd and 4rth space_1_ 1_ _ _ _\nThe "3" is in the 3rd and 7th space _ _ 3_ _ _3_\nThe "2" is in the 5th and 8th space_ _ _ _ 2 _ _2\n\nDon't know any way to show steps other than the way I did here...\n\nBTW the number 23421314 also satisfies the requirements.
133233
(1+1+1)! = 6 3 factorial is 6\n2+2+2 = 6\n3*3-3 = 6\nsquare root(4) + square root(4) + square root (4) = 6\n5 + 5/5 = 6\n6+6-6 = 6\n7 - 7/7 = 6\ncube root(8) + cube root(8) + cube root (8) = 6\n9+9 / square root(9) = 6
707330
Lets call the numbers 2a-27, 2a-25, 2a-23, ..., 2a+23, 2a+25, 2a+27. Thats 28 numbers, and since all the constant terms cancel out, their sum is 28*2a = 56a = 7a*2^3.\nSince we already have a factor of 2^3 in there, all we need for the entire thing to be a cube is a = 7*7*cube.\nThe smallest value of a gives the smallest p and the smallest q, so a = 7*7 = 49.\nThat gives p = 71, q = 125, average = 98, ie B.
973342
3x^2=27\nx^2=9\nx= +- 3 ==> positive root : x=3
942305
Okay. You have a few things that you have to consider.\n\nNegative exponents: moves it to the denominator\nFractional exponents: the denominator tells you what the root is, the numerator tells you what the root is rasied to\n\nSo: (x+1)^-2/3 rewritten would be:\n\n1/[cuberoot(x+1)^2]\n\nwhen x is seven you have\n1/[cuberoot(8)^2]\n1/[2^2]\n1/4
985663
First, if you are just going to have 3 numbers from 0 to 27 inclusively, you have 28 numbers all in all and it is with repitition so you will have just raise the 28 to the 3rd power \n\nso you will get 21,952   \n\nand then , if you will have a correct answer on the first time you will get \n\n1/21952
157569
The required bond angle alpha = AOB; and using Pythagoras' theorem, AB = 2*sqrt(2) is the diagonal of a face of the cube.\nOE is the bisector of AÔB.\nFrom right-angled triangle OEB, tan(alpha/2) = sqrt(2) and therefore alpha = 2tan(^-1)sqrt(2) is approx. 109° 28'
977457
1 to 64k.. so there exists about 40 cubes (1^3, 2^3...  39^3, 40^3)..  and sum of 4 cubes. that's like. "40 choose 4" (combinatorics) so that comes out to be about 91390 combinations of 4 cubes and then you just find the cube roots of each. each should have its own cube root. and 91 thousand isn't too bad for an average computer to handle. i suggest you program a computer to do it for you. it wouldn't be hard, just basic programming language.\n\nif you cant do that, then i suggest you try to narrow down what you're looking for...