2x-4y = 16 -4y = -2x+16 y = 1/2x-4 Any equation that has a slope of 1/2 but a different intercept of -4.
Considering a general quadratic equation y=ax^2 + bx + c, the x coordinate of the vertex is found from the formula x= -b/2a and the y coordinate is found from putting that x coordinate back into the original quadratic equation which in this case I am assuming is y= -2x^2 + 16x -15. So, the x coordinate of the vertex is x=-16/(2*-2) = 4 To find the y coordinate we plug 4 back into y= -2x^2 + 16x -15 so we have y= -2 * 4^2 + 16*4 - 15. Following the order of operations we get y= -2 *16 + 64 - 15= 17 Therefore the vertex is at (4, 17).
You could graph this Polynomial by using substution to solve for two points..which will define a line. If x=16, then y=0, If x=0, then y=8. Graph this line and you have the solution set for the equation.
In the form y = ax² + bx + c the axis of symmetry is given by the line x = -b/2a The axis of symmetry runs through the vertex, and the vertex is given by (-b/2a, -b²/4a + c). For y = 2x² + 4x - 10: → axis of symmetry is x = -4/(2×2) = -4/4 = -1 → vertex = (-1, -4²/(4×2) - 10) = (-1, -16/8 - 10) = (-1, -12)
g(x) = √(x - 16) The graph of g(x) = √(x - 16) has the same shape as the graph of f(x) = √x. However, it is shifted horizontally to the right 16 units. The graph of the function f(x)=square root(x) is made up of half a parabola (in the first quadrant) with directrix (16, 0), which opens rightward. The domain is [16,∞) and range [0, ∞).
2x-4y = 16 -4y = -2x+16 y = 1/2x-4 Any equation that has a slope of 1/2 but a different intercept of -4.
In the equation y x-5 2 plus 16 the standard form of the equation is 13. You find the answer to this by finding the value of X.
Answer this question… What is the line of symmetry of the graph of the equation ? A. x = -2 B. x = -4 C. x = -16 D. x = -8
160 x 0.1 = 16
A parabola with vertex (h, k) has equation of the form: y = a(x - h)² + k → vertex (k, h) = (3, 5), and a point on it is (-1, 6) → 6 = a(-1 - 3)² + 5 → 6 = a(-4)² + 5 → 1 = 16a → a = 1/16 → The coefficient of the x² term is 1/16
Considering a general quadratic equation y=ax^2 + bx + c, the x coordinate of the vertex is found from the formula x= -b/2a and the y coordinate is found from putting that x coordinate back into the original quadratic equation which in this case I am assuming is y= -2x^2 + 16x -15. So, the x coordinate of the vertex is x=-16/(2*-2) = 4 To find the y coordinate we plug 4 back into y= -2x^2 + 16x -15 so we have y= -2 * 4^2 + 16*4 - 15. Following the order of operations we get y= -2 *16 + 64 - 15= 17 Therefore the vertex is at (4, 17).
It represents an algebraic equation in the variable, k.
This is slightly out of context but you will have an equation of a graph we call it x and y because they are the axises we use on a graph. Usually in an equation we try to find the value of y. An equation that works in this case is y=x-7 but there are many others. But if we use the example i have given when x=16 this means y=16-7 which is 9 if you were to plot this point on a graph it would be 16 across the x axis and 9 up Hope this helps
It is 1/16.
x2 -y2 =16 This is an equation that describes your problem. We can write this equation as (1/16)x2 -(1/16)y2 =1 You may recognize this as the equation whose graph is a hyperbola. So there are an infinite number of solutions.
Sparse vs. Dense GraphsInformally, a graph with relatively few edges is sparse, and a graph with many edges is dense. The following definition defines precisely what we mean when we say that a graph ``has relatively few edges'': Definition (Sparse Graph) A sparse graph is a graph in which .For example, consider a graph with n nodes. Suppose that the out-degree of each vertex in G is some fixed constant k. Graph G is a sparse graph because .A graph that is not sparse is said to be dense:Definition (Dense Graph) A dense graph is a graph in which .For example, consider a graph with n nodes. Suppose that the out-degree of each vertex in G is some fraction fof n, . E.g., if n=16 and f=0.25, the out-degree of each node is 4. Graph G is a dense graph because .
You are describing a circle, with its center at the origin and a radius of 4 (the square root of 16)