Wikipedia
Homogeneous coordinate system - A homogeneous coordinate system is a coordinate system in which there is an extra dimension, used most commonly in computer science to specify whether the given coordinates represent a vector (if the last coordinate is zero) or a point (if the last coordinate is non-zero). A homogeneous..
Homogeneous function - In mathematics, a homogeneous function is a function with multiplicative scaling behaviour: if the argument is multiplied by a factor, then the result is multiplied by some power of this factor. More precisely, if is a function between two vector space s over a field F, then ƒ is said to be homogeneous of degree k if for all nonzero and . When the vector spaces involved are over the real numbers, a slightly more general form of homogeneity is often used, requiring only that () hold for all α > 0. Homogeneous functions can also be defined for vector spaces with the origin deleted, a fact that is used in the definition of sheaves on projective space in algebraic geometry. More generally, if S ⊂ V is any subset that is invariant under scalar multiplication by elements of the field (a 'cone'), then a homogeneous function from S to W can still be defined by (). Examples Linear functions Any linear function..
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Homogeneous system
A system is called homogeneous if physical properties and chemical composition are identical throughout the system. A pure gas or consistent mixture of gases e.g., an oxygen cylinder, or a pure liquid or solid in a container are examples of homogeneous ..
Homogeneous Equations (Constant = 0)
Consider the homogeneous equations a 1 x + b 1 y + c 1 z = 0 a 2 x + b 2 y + c 2 z = 0 a 3 x + b 3 y + c 3 z = 0 The homogenous system of equations is always consistent because x = 0, y = 0, z = 0 satisfies all the equations in the s..
Consider the homogeneous equations a 1 x + b 1 y + c 1 z = 0 a 2 x + b 2 y + c 2 z = 0 a 3 x + b 3 y + c 3 z = 0 The homogenous system of equations is always consistent because x = 0, y = 0, z = 0 satisfies all the equations in the s..Which of the following linear systems matches with the linear system i..
Which of the following linear systems matches with the linear system in the graph? => 3 y = -2, y = 1 or 3 y = 2 x , y = - x - 1 or 3 y = 2 x , y = x - 1 or None of the above..
Consistency of a system of linear equation
If a system of linear equations has at least one solution, then the system is called consistent, otherwise it is called inconsistent. Solve the system of linear equations (1) by using method of elimination as studied earlier Mul..
If a system of linear equations has at least one solution, then the system is called consistent, otherwise it is called inconsistent. Solve the system of linear equations (1) by using method of elimination as studied earlier Mul.. Homogeneous Linear Equations?of their disguise and start hunting prey. Predator interaction creates interesting animation effects, which is convincingly similar to flocks, herds or schools observed in nature. The simulation is quite simplistic, it uses linear motion and basic distance equations to enforce boundaries and to animate the fish. This of course introduces obvious artifacts, such as jerky animation, oscillations, and poor collision detection with external objects. You will notice that the aquarium fails ...
IKA UTTD Closed Homogenizer SystemWorlds first closed homogenizing system developed by IKA. Disperse, stir and grind using a single drive unit with this tube disperser closed homogenizing system. The system is easy to operate and offers continuously variable speed control, timer function with digital display and antilocking function. Hermetically sealable disposable sample tubes provide protection and security for infectious sample materials, toxic substances and high-odor substances. Complete system includes workstation with tube drive, two stirring tubes, two dispersing tubes, two grinding/milling tubes and carrying case. Replacement tubes are available separately.
Question : 3x+y+z=0
x-y+z=0
-2x-4y+z=0
a) (2c, -c, c)
b) (2c, c, -c)
c) (c, 2c, -c)
d) (-c/2, c?2, c)
Can anyone help?
Answer : The row reduced echelon form of the coefficient matrix is [1 0 1/2] [0 1 -1/2] [0 0 0]. If we let z = c, the first row stands for the equation x + (1/2)c = 0, or x = -(1/2)c; the second row stands for y - (1/2)c = 0, or y = (1/2)c. Therefore, answer d is the correct response.
Answer : The row reduced echelon form of the coefficient matrix is [1 0 1/2] [0 1 -1/2] [0 0 0]. If we let z = c, the first row stands for the equation x + (1/2)c = 0, or x = -(1/2)c; the second row stands for y - (1/2)c = 0, or y = (1/2)c. Therefore, answer d is the correct response.
Question : Solve the linear homogeneous recurrence relation a(n) = 2a(n-1) a(n-2) with a(o) = 4 and a(1) = 1.
Answer : Here's the general way to solve these: Assume that a(n) = r^n for some constant r. Then, r^n = 2r^(n-1) - r^(n-2) ==> r^n - 2r^(n-1) + r^(n-2) = 0 ==> r^(n-2) * (r - 1)^2 = 0. Ignoring the first factor (since this gives nothing), we get a double root of r = 1. So, the solution may be written a(n) = 1^n * (A + Bn) = A + Bn for some constants A and B. We use the initial values to find A and B. 4 = a(0) = A + B * 0 ==> A = 4. 1 = a(1) = A + B ==> B = -3. So, the solution is a(n) = 4 - 3n. I hope this helps!
Answer : Here's the general way to solve these: Assume that a(n) = r^n for some constant r. Then, r^n = 2r^(n-1) - r^(n-2) ==> r^n - 2r^(n-1) + r^(n-2) = 0 ==> r^(n-2) * (r - 1)^2 = 0. Ignoring the first factor (since this gives nothing), we get a double root of r = 1. So, the solution may be written a(n) = 1^n * (A + Bn) = A + Bn for some constants A and B. We use the initial values to find A and B. 4 = a(0) = A + B * 0 ==> A = 4. 1 = a(1) = A + B ==> B = -3. So, the solution is a(n) = 4 - 3n. I hope this helps!