It had to develop into a polynomial solution.
210673
Given that f(x) is a polynomial of degree EIGHT such that\n\nf(m) = 1/m for m = 1,2,3,4,5,...,9\n\nFind f(10)  \n\n\n*Answer: 2/10\n\n*Solution:\n\nConsider another polynomial, g, defined as g(x) = x f(x) - 1\n\ng(1)=0, g(2)=0, …, g(9)=0\n\nSo g(x) = A(x-1)(x-2)…(x-9)\n\ng(0) = 0 f(0) - 1 = -1\n\nalso, g(0) = A(-1)(-2)…(-9), so A = 1/9!\n\ng(10) = A(9)(8)…(1) = (1/9!)(9!) = 1\n\ng(10) = 1 = 10 f(10) - 1, so f(10) = 2/10\n\n\n**Or try doing some research on "Rabin's method"
668072
i think you need to give more informations but if you re talking about equation it is not so informal than you think\nyou just have to concentrate and try to figure the way you re supposed to resolve it
91108
Eventually the liquid would boil
426790
It didn't disappear.  Expand (x + b/2a)^2 and you will find it.
574692
if y=x^2 (quadratic model) the pay will increase gradually at first and then greatly at the end\n\nif y=x (linear) there will be a linear increase in the pay
31766
This would most definitely be a polynomial function, since it starts with x cubed. Anything that has x cubed or higher in it, like x to the fifth power, would be a polynomial function. An equation with x squared would be a quadratic function. An equation with just x would be a linear equation.
298472
You need the first volume of CCl4 to solve the problem.
172000
its a good idea to standardize your solutions everyday.  Depending on the accuracy you are trying to achieve this is the procedure i would follow.
380122
Yes, many people do know how to solve that problem.  They learned it the same you should: read your book, go to class, talk to TAs, and do your own homework.
589182
Here is my proof. Suppose P(x) is a polynomial in odd degree, for example 4x^5 + 3x^4 - 5x^3 + 2x^2 - 6x + 9. For large enough values of x, the term in x^5 will be so much larger than any of the other terms that it will swamp the smaller powers of x, and x^5 alone will determine the value of P. \n\nNow, if P is of odd degree and x is very large and positive, P will be positive; if x is very large and negative, then P will be negative. Therefore, somewhere between these two values of P there must be at least one point where P = 0. QED.
185638
Here's what I came up with below.
189126
It's the standard form of a quadratic function.
238896
It's hard to show it clearly here but the solution is actually:\n\n[-b±√(b²-4ac)]/2a\n\nthere will usually be two solutions!\nThe site below has it laid out more clearly...\n\n("invited" to leave school early.. 1974)
371758
Add all the polynomials together and you get:\nx^4 y^4
463871
D ( 2 or 0)\n\nThis is because the coefficients of the given cubic polynomial are real numbers and so any real zero needs be accompanied by its complex conjugate and so as stated  earlier MUST come in pairs so either there are 2 or there cannot be any.
611474
The answer is b, multiple solutions, if you work the problem so that it's in reduced row echelon form you will see that there are many possible solutions to the linear equations.
291631
b)\nbecause when you plug each value in for x you get a true equation