How many solutions does this nonlinear system have?
402617
No solutions.\nFrom the first equation you have\ny=x^2+2x+9/2\n\nUsing the second:\nx^2+2x+3=x^2+2x+9/2\n3=9/2 , so there is no solution
587075
Would be happy to help, but some information is missing.  I am assuming that you are asking about the region bounded by the following four equations:\n{2x + y < 25}  {x + y < ?}  {x < 12}  {x > 0}\n\nNeed to know what ? is in order to give an exact answer.
293371
The equation has 0 real solutions, but 2 complex solutions.\n\nThere is a very fundamental theorem of algebra which states that any polynomial equation of degree n has n roots. In your case, the degree is 2, so the equation has to have 2 roots.\n\nHowever, roots can be either real, complex, or repeated roots. For example, x^2=0 appears to only have one solutions (x=0) but x=0 is a repeated root, occurring twice. For polynomials of degree 2, you can find out about the solutions by using the discriminant. An equation of the form ax^2+bx+c=0 has discriminant b^2-4ac. The discriminant can be one of three things:\n\n>0 This means that both roots are REAL roots.\n=0 This means that there is ONE REAL root, and it is a repeated root.\n<0 This means that both roots are COMPLEX roots and there is no real solution.\n\nThe methods for general degree n polynomials are much more difficult, if not impossible, to classify.
593300
8 squirts.\n8 works.\nmaybe even 9.
970181
1) This has no elementary solution\n2) This one, however, does have a solution, with the substitution e^x = tanu.  You are at the answer arctan(e^x)\n\nSteve
593616
It's not hard if you draw the graph of the functions y=x and y=5*sin(Pi*x), and see how many times they cross.\n\nNow, you know that sin(pi*x) is always less than 1, so there are only solutions when -5<x<5.  And you know that at x=5, sin(pi*x) =0. (Same at x=-5)\n\nIf you plot the graph of sin(pi*x), you see that it's positive if 0<x<1, 2<x<3, and 4<x<5.  Since the sine curve goes up and down, you can see that there are two solutions in each interval.  Counting these up gives 6 solutions in each interval, or 6 solutions.  Now, look at x<0-- there are 6 solutions there also, but the x=0 is included in both sets, so you have to ignore one of them-- there are 11 solutions to the equation x=5sin(pi*x)
316686
the process u follow is correct, but actually it is wrong. fundamentally wat u did is ok, but technically it is wrong. because this becomes a quadratic equation which has more than 1 root. so this equation has more than 1 answers for which it is true.
840437
Well, I can see that one possible solution is r = 0, and my calculator says the other answer is r = 0.0809394063, so hopefully that helps...
279194
It easily follows that the general solution to this differential equation is y=Ae^{kt} for some constant A [You may differentiate to double check your answers.] \n\nSo given the initial condition, i.e. y(t=60)=0.5*A, we obtain k=(ln(0.5))/(60).
665576
x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4
531201
the solution is 1. see 6^1,5^2,4^3,3^4,2^5,1^6.
613913
I know it has at least 4. You can probably find the definitive number in one of the links below.
779994
If you are solving a system of linear equations (presumbale, two equations with two variables), then there are an infinite number of solutions if the graphs coincide, i.e. are the same.\n\nFor example, if I ask you to solve the system\n\n4x - 2y = 12\ny = 2x - 6\n\nby graphing, you should discover that ythe graphs are the same.  As such, this system has an infinite number of solutions.  Any ordered pair that makes the first equation true (and there are an infinite number of them) will also make the second equation true.\n\n\nI really have no idea what you are asking, but that seemed reasonable.
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)
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).
939642
Counterexample.  Let u(x), v(x) be members of {sin x, cos x}.  Both of those are solutions, but you do not get zero by adding one to the other, or by doubling one of them.
274355
Depends on the system you are trying to solve. check your limits of integration. from -infinity to 0 plus 0 to infinity.