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The graph of the logarithm to base 2 crosses the x axis (horizontal axis) at 1 and passes through the points with coordinates (2, 1), (4, 2), and (8, 3). For example, log 2 (8) = 3, because 2 3 = 8. The graph gets arbitrarily close to the y axis, but does not meet or intersect it.
The roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
The program is solvable in polynomial time if the graph has all undirected or all directed edges. Variants include the rural postman problem. [3]: ND25, ND27 Clique cover problem [2] [3]: GT17 Clique problem [2] [3]: GT19 Complete coloring, a.k.a. achromatic number [3]: GT5 Cycle rank; Degree-constrained spanning tree [3]: ND1
By the fundamental theorem of algebra, if the monic polynomial equation x 2 + bx + c = 0 has complex coefficients, it must have two (not necessarily distinct) complex roots. Unfortunately, the discriminant b 2 − 4c is not as useful in this situation, because it may be a complex number. Still, a modified version of the general theorem can be ...
How to Solve It suggests the following steps when solving a mathematical problem: . First, you have to understand the problem. [2]After understanding, make a plan. [3]Carry out the plan.
If y 2 = x 3 − x − 1, then the field C(x, y) is an elliptic function field. The element x is not uniquely determined; the field can also be regarded, for instance, as an extension of C(y). The algebraic curve corresponding to the function field is simply the set of points (x, y) in C 2 satisfying y 2 = x 3 − x − 1.
When the task is to find the solution that is the best under some criterion, this is an optimization problem. Solving an optimization problem is generally not referred to as "equation solving", as, generally, solving methods start from a particular solution for finding a better solution, and repeating the process until finding eventually the ...
Solve the problem using the usual simplex method. For example, x + y ≤ 100 becomes x + y + s 1 = 100, whilst x + y ≥ 100 becomes x + y − s 1 + a 1 = 100. The artificial variables must be shown to be 0. The function to be maximised is rewritten to include the sum of all the artificial variables.