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A regular pyramid has a regular polygon base and a what over the center of the base?
vertex
A regular pyramid has a base and a vertex over the center of the base?
Regular Polygon...
Quadratic equation x2-k-x equals 0?
2
How do you figure out if a parabola is stretched or shrunk?
Assuming the parabola is of the form y = ax^2 + bx + c or y= a(x-h)^2 - k, you look to the 'a' coefficient to determine whether the parabola has undergone a vertical "stretch" or "shrink." If a>1, then it's a stretch. If 0<a<1, then it's a shrink. If, by the way, the a is negative, this test still works... just ignore the negative sign. So if for example a = -2/3, it's a shrink, but if a = -3 it's a stretch. (Incidentally, the negative sign makes the parabola "reflect" over the x-axis.)
What is a platonic solid and how many are there and what are their names?
A Platonic solid is the 3-D shape equivalent of a polygon: it is a three dimensional figure whose sides are congruent, regular polygons, with identical vertices. Unlike the 2-dimensional case (in which there are infinitely many polygons) there are only 5 Platonic solids: * The tetrahedron, which has 4 triangular sides. * The cube (or hexahedron), which has 6 square sides. * The octahedron, which has 8 triangular sides. * The dodecahedron, which has 12 pentagonal sides. * The icosahedron, which has 20 triangular sides. Here is how the 5 Platonic solids were found, and how we know there aren't any more: Think about the sum of the angles at a vertex (by the definition of a Platonic solid, all vertices are identical). In the plane, angles around a vertex add up to 360 degrees, but we don't want the vertex to lie flat - otherwise, we'd end up with a huge flat sheet instead of a polyhedron. We also want at least 3 polygons around a vertex, because otherwise the result will become a flat figure without volume. If the sides are triangles, we can have 3 triangles around a vertex (getting the tetrahedron), 4 triangles around a vertex (getting the octahedron), or 5 triangles around a vertex (getting the icosahedron). We can't have 6 or more, because then the sum of angles wouldn't be less than 360. If the sides are squares, we can have 3 squares around a vertex, getting the cube. 4 squares around a vertex would mean the sum of angles is 360, and 5 or more is even more impossible. Finally, we can take 3 pentagons around a vertex, getting the dodecahedron; more pentagons will give us an angle sum of over 360. We can't use any shapes with more than 6 sides, because their angles are larger and we can't even fit 3 around a vertex. Even 3 hexagons will give an angle sum of 360 degrees, and anything more than that is even worse.
How do you find the vertex of an absolute value function?
There are three main types of vertices for an absolute value function. There are some vertices which are carried over from the function, and taking its absolute value makes no difference. For example, the vertex of the parabola y = 3*x^2 + 15 is not affected by taking absolute values. Then there are some vertices which are reflected in the x-axis because of the absolute value. For example, the vertex of the absolute value of y = 3*x^2 - 15, that is y = |3*x^2 - 15| will be the reflection of the vertex of the original. Finally there are points where the function is "bounced" off the x-axis. These points can be identified by solving for the roots of the original equation. -------------- The above answer considers the absolute value of a parabola. There is a simpler, more common function, y = lxl. In this form, the vertex is (0,0). A more general form is y = lx-hl +k, where y = lxl has been translated h units to the right and k units up. This function has its vertex at (h,k). Finally, for y = albx-hl + k, where the graph has been stretched vertically by a factor of a and compressed horizontally by a factor of b, the vertex will be at (h/b,ak). Of course, you can always find the vertex by graphing, especially since you might not remember the 2nd or 3rd parts above.
A regular pyramid has a base and vertex over the center of the base?
A regular pyramid has a regular polygon base and a vertex over the center of the base. By:Cherrylvr :)
What is the highest point over your head known as?
Apex, or vertex
How does the formula -b over 2a give the x value of the vertex given ax2 bx c?
From the quadratic formula, the zeros (where y=0) is -b/(2a) +/- sqrt(b^2 -4ac)/(2a). The expression inside the square root is known as the descriminant. Being that the parabola is symmetric, one of the zeros will be at the vertex + sqrt(b^2 -4ac)/(2a) and the other zero is at vertex - sqrt(b^2 -4ac)/(2a). The vertex is halfway between these two values, so the x coordinate of the vertex is -b/(2a). Or if you use Calculus, you can take the derivative of f(x) = ax^2 + bx + c. f'(x) = 2ax + b. Set equal to zero and solve for x. x = -b/(2a).
What is the equation for the axis of symmetry of a parabola?
x=-b/2a [negative B over 2A]
Can the graph of a polynomial function have no x-intercept?
Yes, over the real set of numbers. For example, the graph of y=x2+1 is a regular parabola with a vertex that is one unit above the origin. Because the vertex is the lowest point on the graph, and 1>0, there is no way for it to touch the x-axis.NOTE: But if we're considering imaginary numbers, the values "i" and "-i" would be the zeroes. I'm pretty sure that all polynomial functions have a number of zeroes equal to their degree if we include imaginary numbers.
How is parabola used in basketball?
A thrown basketball follows a path that can be approximated by a parabola. The approximation ignores air resistance and any curve imparted by spin on the ball. Over the distances involved, both are likely to be negligible.
A has a regular polygon base and a vertex over the center of the base?
regular pyramid
A regular pyramid has a base and a vertex over the center of the base?
Regular Polygon...
A regular pyramid has a regular polygon base and a what over the center of the base?
vertex
A regular pyramid has a regular polygon as its base and vertex over the base's center?
Yes.
Does a regular pyramid have a regular polygon as its base and a vertex over the base's center?
Yes
