Vertex

Answers: 1. Segment A (Position-Time on Ramp): The graph is a curve. On any graph that curves, the slope or steepness of the graph changes from one point on the graph to another. Since the slope is constantly changing, the velocity is non-constant. Also, as the graph lies above the x-axis and its slope is increasing, the velocity of the object is also increasing (speeding up) in a positive left direction. Segment B (Position-Time on Flat Surface): The graph is a straight line, which represents constant velocity. The slope is negative, which indicates that the object is moving at a constant velocity left. Including the part in which the car slows down to a stop, segment B would cross the x-axis with a curved line. Since it is a curved line, the slope will constantly be changing which results in a non-constant velocity. Since the line crosses the x-axis and the slope is decreasing, the velocity of the object will decrease (slow down), cross the observer and continue slowing down in the left direction. 2. Acceleration on ramp: -0.70m/s^2 right Acceleration on flat surface: 0m/s^2 right No they are not similar. As the car rolls down the ramp, it accelerates and when the car is on a flat surface, it continues at a constant velocity (no acceleration). Therefore, it is entirely different as one has acceleration and the other does not. However, if the segment continued until the ball came to a stop, the velocity would slowly decrease and then the acceleration on a flat surface…

by itself. I subtracted 2W to both sides. Then I divided 2 on both sides. I found that L equals -W+20. Next, to find out what the W was, I plugged in the L value of the perimeter into LW, which is the area formula. Then I distributed the W times the L value and I got -W2+20W. To find the x of my vertex, I plugged -W2+20W into -b/2a. I got 10, which was my W. I plugged my W value back into the perimeter formula, which was 2W+2L=40. I multiplied 2 and 10 and got 20. Then I…

for A. Equation 2a can be written as y=-0.16(x)(x-10) and y=-0.16(x-5)2+4. Those two equations are equal because they are just in different form, I just expanded the first equation and then used completing the square to find the second equation. I would prefer to use the 1st equation in this case because I find it easier to find the x intercepts and then solve for A. If I knew the point of the vertex I would prefer the 2nd equation because that would be easier. The first equation is the…

esteems from tests of costly lighting estimations at vertices is less processor serious than playing out the lighting figuring for every pixel as in Phong shading. Nonetheless, exceedingly restricted lighting impacts, (for example, specular features, e.g. the gleam of thought about light the surface of an apple) won't be rendered accurately, and if a feature lies amidst a polygon, yet does not spread to the polygon's vertex, it won't be clear in a Gouraud rendering; on the other hand, if a…

converting to a common denominator: x? + + = +. After this, I converted the left-side to square form plus a bit of simplifying on the right: = From this point, I began to square root both sides and remembered to put the sign on the right: x. Once I did that, I solved for x and simplified as necessary: x= By evaluating the answer, it finally resulted in the Quadratic Formula: Problem 2: 2x?-x+2=0 Rewriting in vertex form: Before I rewrote the quadratic equation of 2x?-x+2=0 in…

guards required to have vision of each point inside of the polygon. Begin with the process of vertex guards. The most visually optimal way to prove the result _n 3 _ is by first triangulating the polygon. Every vertex of the polygon can be attached in such a way that the entire polygon is made up of solely triangles. This can be shown in _gure 2.2 below. Now that the polygon is separated into triangles, we know from the definition of a convex 8 polygon that each triangle created has now become…

x. Problem 2. Analyze the quadratic equation below. Rewrite it in vertex form, and state the domain, range, maximum or minimum value as well as where it occurs. Find the axis of symmetry, interval(s) where it is increasing and decreasing, and its x and y-intercepts, if any. Determine the types of solution(s), whether it opens up or down, and last, but not least, solve the equation 〖2x〗^2-x+2=0 Rewriting a quadratic equation to a vertex form helps the individual find the axis of symmetry, and…

Here vertex 0 corresponds to the depot and vertices i = 1…n corresponds to customers. A nonnegative cost, cij, is associated with each edge (i,j) ∈ E and represents the travel cost spent to go from vertex i to vertex j. If G is a directed graph, the cost matrix c is asymmetric, and the corresponding problem is called asymmetric CVRP (ACVRP) on the other hand if cij = cji for all (i, j) ∈ E, then the problem is called symmetric CVRP (SCVRP), and the edge set E is generally replaced by a set of…

able to accomplish this task. To find a perpendicular line, the slope needs to be inversed and opposite. A quadratic model can be identified as through visual graphs or number patterns. The common quadratic equation is x=( -b +/- √b ^2 – 4ac)/ 2a. The standard form for a quadratic function is f(x) = a(x - h)2 + k, the bottom of this function forms a vertex with the points of (h, k) and forms the graph of a quadratic which is called a parabola. In a real life situation, a quadratic model could…

in a different way. Escher used 27 out of these 28. Tessellations have two definitions. One is, “A plane tessellation is an infinite set of polygons fitting together to cover the whole plane just once, so that every side of the polygon belongs also to another polygon. No two of the polygons have common interior points. (Cornell)” and the other is, “ A tessellation is called regular if its faces are regular and equal. The same number of polygons meet at each vertex. We denote with n, k (called…