Partial Question 3 Determine the components of the vectors [u₁, u2] and [v1. v2l so that a[1, 2]+[v1, v2] generates the line segment from (1.5, -0.5) to (5.5, -4.5). Do not include spaces in your answers. 4.0 • U₁ = -4.0 02- 1.5 V1 -0.5 V2 1/2 pts
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- The interval between the points in the vector rt=[1:0.4:5] isA vector has the coordinates of VB = = [10, 20, 30] in frame {B}. Given the following homogeneous transformation between frames {A} and {B}, find the coordinates of this vector in frame {A}, i.e., find VA. 0.866 - 0.500 0.000 11.0 HÂ 0.500 0.866 0.000 -3.0 = 0.000 0.000 1.000 9.0 -0 0 0Create an algorithm for calculating a circle and the Schreier vector.Output: the orbit G; a Schreier vector v and backward references w of the orbit 8 G relative to the set S of generators.
- Write a function linear_independence that takes a collection of vectors with integer entries (each written as a list), and returns True if this collection of vectors is linearly independent, and False otherwise. Examples: linear_independence([1,2]) should return True. linear_independence([1,3,7],[2,8,3],[7,8,1]) should returnTrue. linear_independence([1,3,7],[2,8,3],[7,8,1],[1,2,3]) should return False.Hippity hoppity, abolish loopity def frog_collision_time(frog1, frog2): A frog hopping along on the infinite two-dimensional lattice grid of integers is represented as a 4- tuple of the form (sx, sy, dx, dy) where (sx, sy) is its starting position at time zero, and (dx, dy) is its constant direction vector for each hop. Time advances in discrete integer steps 0, 1, 2, 3, ... so that each frog makes one hop at every tick of the clock. At time t, the position of that frog is given by the formula (sx+t*dx, sy+t*dy) that can be nimbly evaluated for any t. Given two frogs frog1 and frog2 that are guaranteed to initially stand on different squares, return the time when both frogs hop into the same square. If these two frogs never simultaneously arrive at the same square, return None. This function should not contain any loops whatsoever. The result should be calculated using conditional statements and integer arithmetic. Perhaps the best way to get cracking is to first solve a simpler…Hippity hoppity, abolish loopity def frog_collision_time(frog1, frog2): A frog hopping along on the infinite two-dimensional lattice grid of integers is represented as a 4-tuple of the form (sx, sy, dx, dy) where (sx, sy) is its starting position at time zero, and (dx, dy) is its constant direction vector for each hop. Time advances in discrete integer steps 0, 1, 2, 3, ... so that each frog makes one hop at every tick of the clock. At time t, the position of that frog is given by the formula (sx+t*dx, sy+t*dy) that can be nimbly evaluated for any t.Given two frogs frog1 and frog2 that are guaranteed to initially stand on different squares, return the time when both frogs hop into the same square. If these two frogs never simultaneously arrive at the same square, return None.This function should not contain any loops whatsoever. The result should be calculated using conditional statements and integer arithmetic. Perhaps the best way to get cracking is to first solve a simpler version…
- Consider a vector of numbers: 12, 62, 44, 25, 79, 85, 64, 93, 29. Iterate over the elements of this vector and find how many have an integer value as its square-root value in rstudio.The coordinates of a polygon can be represented as a list of tuples: [(x1, y1), (x2, y2), .., (xn, yn)], where (x1, y1), ., (xn, yn) are points of the polygon in a counterclockwise order. Find the area of such a polygon using using shoelace formula. It is no longer required to take absolute value. п-1 п—1 A = > xi+1Yi ) – x1Yn i=1 i=1 1 x142 + x2Y3 + + xn-1Yn + xnY1 - x2y1 – X3Y2 – ··- xn Yn-1 – x1Yn| >>> area ([(0,0), (1,0), (0,1)]) 0.5 >>> area ([(0,0), (1,0),(1,1),(0,1)]) 1.0 >> area([(0,0),(2,0), (2,2),(1,2), (0,1)]) 3.5 Write the function area(C) to find the area of a triangle with vertices at c = [(x1, y1), (x2, y2),..., (xn, yn)].Suppose a computer program needs to apply an affine transformation to a complex three-dimensional object made up of 3000 points. The transformation is composed of 8 matrices (call them M1 through M3), so for each point (x, y, z) in the object, the following operation is performed. |-|-|-|-|| y M1 M2 M3 M7 M8 Each multiplication of a matrix times a column vector involves 16 multiplications (of one number by another) and 12 additions, for a total of 28 arithmetic operations. Each multiplication of a matrix times another matrix involves 64 multiplications and 48 additions, for a total of 112 arithmetic operations. (These numbers are not made up or chosen randomly; they are facts about 4 x 4 matrix multiplication.) The most inefficient way of applying the transformation to the 3000 points would be to begin on the left, multiplying M1 by M2, then that result by M3, and so on along the list from left to right, and doing the same 8 multiplications again for each of the 3000 points. How many…
- Consider the vector: x = c(1,2,3,4). What is the value of (x+2)[(!is.na(x)) & x > 0]? What do you find?Affine transformations preserve parallel lines. True or falseI am trying to work with some generic search algorithms in a coding homework, and I am having trouble understanding some of it. Here are my questions: 1. For this linear search algorithm below I am trying to send an array of x and y cordinate points into it and search to see if a specific x, y point is in the array, but I don't know how to send the points into the linear search function. By the way, I am using a struct point type for the array of points and the point I am looking for. (The below code is an exact example from our teacher during a lecture.) void* linearSearchG(void* key, void* arr, int size, int elemSize, int(*compare)(void* a, void* b)) { for (int i = 0; i < size; i++) { void* elementAddress = (char*)arr + i * elemSize; if (compare(elementAddress, key) == 0) return elementAddress; } return NULL; } 2. I forgot what this line of code specifically does: void* elementAddress = (char*)arr + i * elemSize; I need to code this myself, so if you could explain…