7th root of z-8 I think it would only reduce to the 7th root  of z-2^3?
103512
Actually, there's no compelling reason to rewrite the 7th root of z-8 as the 7th root of z-2^3. Here's why:\n\n(1) z-8 is a binomial, so you have to consider the expression as a whole: (z-8). Even if you had the cube root of z-8, you couldn't rewrite it as z-2^3 and then pull the 2 outside the radical to get 2*cube root of (z-1). That's not allowed. If z and 8 were multiplied that's a different story; the cube root of 8z does equal 2 times the cube root of z. However, addition and subtraction means if anything gets simplified the whole chunk has to be simplified. For instance, the cube root of (z-8)^7 does equal z-8.\n\n(2) In order to simplify a seventh root, you need a value in the radical that can be generated by multiplying something by itself 6 times, such as 2*2*2*2*2*2*2=2^7=128. You could simplify 7th root of (128z) as 2*7th root of z.\n\nLong story short, writing 8 as 2^3 really isn't simplifying any more than writing z as z^1 is considered simplifying z. You cannot reduce or simplify the expression at all.
423594
(3x^3-7x^2+x-20)/(x-3) = 3x^2+2x+7 with a remainder of 1\n         \n         3x^2+2x+7 \n     _______________\nx-3 | 3x^3-7x^2+ x-20\n      -3x^3+9x^2\n    __________\n          2x^2+ x\n         -2x^2+6x\n         ________\n              7x - 20\n             -7x + 21\n             ________\n                   1\n\nI tried to show what this would look like as a division problem but it keeps jumbling up.
1037006
I am guessing 168 Ft. This doesn't seem right to me, and I'm not sure if I did it the correct way:\n\n                       7 \n*                      7 \n=             ----------\n                      49 \n ------------------------------         \n                       24 \n*                     24 \n=             ----------\n                     576 \n-------------------------------         \n                      49 \n*                    576 \n=             ----------\n                   28224 \n-------------------------------         \n                   28224 \nsquare root\n                     168
502365
Not as stated.\nIt should be;\n\nThe integral of z-squared d z\nbetween one and the cube root of three\ntimes the cosine\nof 3 pi over nine\nis the log of the cube root of e\n\n\nThe integral of z-squared d z\nbetween one and the cube root of three = 2/3\n\ncos(3pi/9) = cos(pi/3) = 1/2\n\nlog (natural) cube root of e = 1/3
294113
In general the binomial expansion of\n(a+b) to the 8th degree is:\n\na^8+8a^7b+28a^6b^2+56a^5b^3+\n\n70a^4b^4+56a^3b5+28a^2b^6+\n\n8ab^7+b^8.\n\nHere your "a" term is 1 and your "b" term is 2x^3.\nDo the substitution and the rest is tedious arithmetic.
233017
the number of roots for any equation is the degree of the equation. here it is 3 (coz the max power or exponent is 3)
896031
Why not?  If you divide top and bottom of this expression by x^7 you get\n(3 + 4/(x^5))/(9 - 1/(x^4) + 2/(x^7))\n\nEvery term except the 3 and the 9 now has a power of x as denominator, and therefore approaches 0 as x goes to infinity,\ntherefore it approaches 3/9.
353472
2)(x-7)^2 + (y-6)^2 + (z-9)^2 = 36\n3)a)1 b) 3 c) 8\nFor 2 it is 36 because you use the absolute value of the smallest number of the center.  The smallest value is 6.\n6^2 = 36
926034
If by the whole you mean where the finction has a hole in the graph (a removeable discontinuity), then that is when the denominator equals 0.  so there is a hole at -3.  And, yes, I belive that the roots are both -3.
859279
This would be a special case of the Fermat-Catalan Conjecture.\n\nThis conjecture is that there are only a finite number of solutions of x^p + y^q = z^r for 1/p + 1/q + 1/r < 1.  Ten solutions are known, but all of them have the three exponents different (except for 1^p + 2^3 = 3^2) and one exponent equal to 2.\n\nYour conjecture would be the special case that there are no solutions with p > 2 and q = r.  It is already known that there are no solutions with p = q and r > 2 (Darmon and Merel, 1997).
837237
Hi\nI think you're referring to one of my answers.\nhttp://answers.yahoo.com/question/index;_ylt=ApZ1XohHNUdL5xkBg8bXQEfsy6IX?qid=20061124162535AAURdJ5\n\nHy is correct, this was done in order to eliminate the x^3 term in the quartic equation.  It made the subsequent solution much easier to handle, since the actual values of a,b,c,d or p,q,and r didn't matter:  the solution wasn't looking for an answer to a specific quartic but for a generalization about quartics in general.\n\nConverting a quartic into reduced form is also done when finding the roots of any quartic equation by an old method (hardly used anymore since computers).  After placed into reduced form, the 'cubic resolvant' is formed:\n\nz^3 + 2pz^2 + (p^2 - 4r)z - q^2 = 0\n\nFinding the three roots (z1,z2,z3) of this equation (using Cardano's or other methods), will yield the four roots of the reduced form quartic:\n\nx1 = (√z1 + √z2 + √z3) / 2\nx2 = (√z1 - √z2 - √z3) / 2\nx3 = (-√z1 + √z2 - √z3) / 2\nx4 = (-√z1 - √z2 + √z3) / 2
758804
Look at the denominator: e^-6x - 7;  as x gets very large, e^-6x gets very small and soon becomes insignificant with respect to -7, so the limit will be -3/7.
37863
There are a few cases that need to be considered here.\n\n1. If n = 0, then n^2 would smaller and 2^n would be larger. (coz 1 is always greater than 0 ;) )\n\n2. If n = 2, 2^n would be smaller and n^2 would be larger.\n\n3. If n < 0 and with decreasing negative values, 2^n would become smaller and smaller and n^2 would become larger and larger.(This case is valid if and only if the instance size can be negative)\n\n4. If n > 26, then n^2 is smaller and 2^n is larger.\n\n5. I haven't considered 'n' as a fraction here coz that will mighty complicate things. \n\nOverall, it was a nice question! Keep posting such questions!
68796
Let's consider base N and a number k (the divisor)\nand take the powers of N: 1, N, N^2, N^3, ...\n\nThere's an infinity of such numbers, however their remainders when divided by k can only be smaller than k, therefore some must be repeating.\n\nLet N^a and N^b (a<b) be two powers with the same remainder, that means k | N^b-N^a.\nThen k | N^a*(N^(b-a)-1). Assuming that k is not divisible by the prime factors of N, it follows that k and N^a are mutual primes, so k | N^(b-a)-1.\n\nLet c=b-a; c>0 because a<b. Then k | N^c-1.\nBut N^c-1 is nothing else than the digit (N-1) repeated c times.\nIf you choose the minimal such number c, then (N^c-1)/k gives exactly the period of 1/k in base N.\n\n\nExample: N=10, k=7\n\npowers of N: 1, 10, 10^2=100, 10^3=1000, 10^4=10000, 10^5=100000, 10^6=1000000, ...\nremainders when divided by 7: 1, 3, 2, 6, 4, 5, 1, ...\nWe notice a repeating remainder: 1, for 1=10^0 and 10^6. Then 7 | 10^6-1=999999\n999999/7=142857\n1/7=0.(142857)
1187142
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.
222795
It is an easy problem.\n\nPut it as the\n\nProposition. There does not exist a positive integer n such that 2^n+1 is divisible by 7.\nProof. Re-write the comparison 2^n+1=0 mod 7 as 2^n=-1 mod 7 or else 2^n=6 mod 7.\n\nFurther, it is clear that 2^0=1 mod 7, 2^1=2 mod 7, 2^2=4 mod 7, 2^3=1 mod 7, hence, one observes the periodicity of remainders (periodicity with respect to n of length 3). Consequently it suffices to consider only 3 types of powers of 2: namely, for\n\n1) n=3*k, 2) n=3*k+1, 3) n= 3*k+2 (k is a positive integer).\n\nThese cases imply respectively:\n\n1) 2^(3*k)=1 mod 7;\n2) 2^(3*k+1)=2 mod 7;\n3) 2^(3*k+2)=4 mod 7;\n\nThese cases exhaust all possibilities. In neither of the cases the remainder 6 is attained (or, what*s the same, -1). The proposition has thereby been proved.
894556
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