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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.
Figure 1. Plots of quadratic function y = ax 2 + bx + c, varying each coefficient separately while the other coefficients are fixed (at values a = 1, b = 0, c = 0). A quadratic equation whose coefficients are real numbers can have either zero, one, or two distinct real-valued solutions, also called roots.
A formula for computing the trigonometric identities for the one-third angle exists, but it requires finding the zeroes of the cubic equation 4x 3 − 3x + d = 0, where is the value of the cosine function at the one-third angle and d is the known value of the cosine function at the full angle.
The smaller piece, at the bottom, has width a-b and height b. Now the smaller piece can be detached, rotated, and placed to the right of the larger piece. In this new arrangement, shown in the last diagram below, the two pieces together form a rectangle, whose width is a + b {\displaystyle a+b} and whose height is a − b {\displaystyle a-b} .
Animation demonstrating the smallest Pythagorean triple, 3 2 + 4 2 = 5 2. A Pythagorean triple consists of three positive integers a, b, and c, such that a 2 + b 2 = c 2.Such a triple is commonly written (a, b, c), a well-known example is (3, 4, 5).
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 . An exponential equation is one which has the form a x = b {\displaystyle a^{x}=b} for a > 0 {\displaystyle a>0} , [ 43 ] which has solution
Let ABC be a triangle with side lengths a, b, and c, with a 2 + b 2 = c 2. Construct a second triangle with sides of length a and b containing a right angle. By the Pythagorean theorem, it follows that the hypotenuse of this triangle has length c = √ a 2 + b 2, the same as the hypotenuse of the first triangle.
The third formula shown is the result of solving for a in the quadratic equation a 2 − 2ab cos γ + b 2 − c 2 = 0. This equation can have 2, 1, or 0 positive solutions corresponding to the number of possible triangles given the data.