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Solve for variable 'a' in a beastly equation (numerical or analytical ok)AA = RootOf(2 * Z * b * R * sin(Z) + R * d * tan(theta) - d * R * tan(theta) * cos(2*Z) + 2 * b * tan(theta) * (-b^2 * (1+cos(2*Z)) / (-1+cos(2*Z)))^(1...
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Mathematica中 Solve::解方程的时候输入Solve函数不能得出结果In[12]:=Solve[x^2+2x-7==0,x]Solve::ivar :3 is not a vaid variable.>>请教为什么这样 教材上是这么写的In[1]:= Solve[x^2 + 2x - 7 == 0,x]Out[1]={{x→-1-2根下2}
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The mathematical statement x < 3, read as "x is less than 3," indicates that the variable x can be any number less than (or to the left of) 3. Remember, we are considering the real numbers and not just integers, so do not think of the values of x for x < 3 as only 2, 1...
integration of ((x/(b+x^a)); or integration of (exp(x)/(b+exp(a*x)); or integration of (exp(a*x)/(b+exp(x)); where, a and b are constants. x is variable. please find the solution for any one of the question.
I personally try not to useNas a variable name, as it is a built-in function name. Reply | Flag 0 Evan Cooch Posted3 months ago So, after starting over, and Solve[h[N] == 0, N] doesnowseem to work (perhaps Remove["Global`*"] did the trick, but even though Mathematica did re...
Aquadratic equationis a polynomial equation in one unknown that contains the second degree, but no higher degree, of the variable. Thestandard formof a quadratic equation is ax2+ bx + c = 0, when a ≠ 0. Anincomplete quadratic equationis of the form ax2+ bx + c = 0, and either ...
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