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From Wikipedia
In mathematics, the term linear function can refer to either of two different but related concepts:
 a firstdegree polynomial function of one variable;
 a map between two vector spaces that preserves vector addition and scalar multiplication.
Analytic geometry
In analytic geometry, the term linear function is sometimes used to mean a firstdegree polynomialfunction of one variable. These functions are known as "linear" because they are precisely the functions whose graph in the Cartesian coordinate plane is a straight line.
Such a function can be written as
 f(x) = mx + b
 (yy1) = m(xx1)
 0= Ax + By + C
(called slopeintercept form), where m and b are realconstants and x is a real variable. The constant m is often called the slope or gradient, while b is the yintercept, which gives the point of intersection between the graph of the function and the yaxis. Changing m makes the line steeper or shallower, while changing b moves the line up or down.
Examples of functions whose graph is a line include the following:
 f_{1}(x) = 2x+1
 f_{2}(x) = x/2+1
 f_{3}(x) = x/21.
The graphs of these are shown in the image at right.
Vector spaces
In advanced mathematics, a linear function means a function that is a linear map, that is, a map between two vector spaces that preserves vector addition and scalar multiplication.
For example, if x and f(x) are represented as coordinate vectors, then the linear functions are those functions f that can be expressed as
 f(x) = \mathrm{M}x,
where M is a matrix. A function
 f(x) = mx + b
is a linear map if and only if b = 0. For other values of b this falls in the more general class of affine maps.
In mathematics, the factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n. For example, 5 ! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \ 0! is a special case that is explicitly defined to be 1. The factorial operation
In the mathematical discipline of linear algebra, a matrix decomposition is a factorization of a matrix into some canonical form. There are many different matrix decompositions; each finds use among a particular class of problems.
Example
In numerical analysis, different decompositions are used to implement efficient matrix algorithms.
For instance, when solving a system of linear equations Ax=b, the matrix A can be decomposed via the LU decomposition. The LU decomposition factorizes a matrix into a lower triangular matrixL and an upper triangular matrixU. The systems L(Ux)=b and Ux=L^{1}b require fewer additions and multiplications to solve, though one might require significantly more digits in inexact arithmetic such as floating point. Similarly the QR decomposition expresses A as QR with Q a unitary matrix and R an upper triangular matrix. The system Q(Rx) = b is solved by Rx = Q^{T}b = c, and the system Rx = c is solved by "back substitution". The number of additions and multiplications required is about twice that of using the LU solver, but no more digits are required in inexact arithmetic because the QR decomposition is numerically stable.
Decompositions related to solving systems of linear equations
 Applicable to: square matrixA
 Decomposition: A=LU, where L is lower triangular and U is upper triangular
 Related: the LDU decomposition is A=LDU, where L is lower triangular with ones on the diagonal, U is upper triangular with ones on the diagonal, and D is a diagonal matrix.
 Related: the LUP decomposition is A=LUP, where L is lower triangular, U is upper triangular, and P is a permutation matrix.
 Existence: An LUP decomposition exists for any square matrix A. When P is an identity matrix, the LUP decomposition reduces to the LU decomposition. If the LU decomposition exists, the LDU decomposition does too.
 Comments: The LUP and LU decompositions are useful in solving an nbyn system of linear equations Ax=b. These decompositions summarize the process of Gaussian elimination in matrix form. Matrix P represents any row interchanges carried out in the process of Gaussian elimination. If Gaussian elimination produces the row echelon form without requiring any row interchanges, then P=I, so an LU decomposition exists.
LU Reduction
Block LU decomposition
Rank factorization
 Applicable to: square, symmetric, positive definite matrix A
 Decomposition: A=U^TU, where U is upper triangular with positive diagonal entries
 Comment: the Cholesky decomposition is a special case of the symmetric LU decomposition, with L=U^T.
 Comment: the Cholesky decomposition is unique
 Comment: the Cholesky decomposition is also applicable for complex hermitian positive definite matrices
 Comment: An alternative is the LDL decomposition which can avoid extracting square roots.
 Applicable to: mbyn matrix A
 Decomposition: A=QR where Q is an orthogonal matrix of size mbym, and R is an upper triangular matrix of size mbyn
 Comment: The QR decomposition provides an alternative way of solving the system of equations Ax=b without inverting the matrix A. The fact that Q is orthogonal means that Q^TQ=I, so that Ax=b is equivalent to Rx=Q^Tb, which is easier to solve since R is triangular.
Singular value decomposition
 Applicable to: mbyn matrix A.
 Decomposition: A=UDV^H, where D is a nonnegative diagonal matrix, and U and V are unitary matrices, and V^H denotes the conjugate transpose of V (or simply the transpose, if V contains real numbers
From Yahoo Answers
Answers:http://en.wikipedia.org/wiki/Exponential_growth#Examples you can start with that , and you need to reexpress them in your own words .
Answers:In both cases, these use the basic exponential growth formula (like calculating compounding interest in a bank account): y=K((1+g)^x). Where y is the final amount after growth, K is the initial value, g is the growth rate per period, and x is the number of growth periods. For example, if you deposit $100 in a savings account for 3 years and the interest rate (growth rate) is 1% per year, after three years, you would have: y=100((1+.01)^3) or $103.03 in your account. For #1, you have to do a little algebraic manipulation but it is solvable. 1st case: y=4=K(1+g)^5  you don't know the initial value or growth rate 2nd case: y=15=K(1+g)^8  again, you don't know values for K or g at this point, you have two equations and two unknowns that can be solved by elimination or substitution. For this work, let's solve both equations for K and eliminate K since we are looking for the value of g. 1st case: K=4/(1+g)^5, and 2nd case: K=15/(1+g)^8; since K=K, you can eliminate K by writing these equations equal to each other: 4/(1+g)^5=15/(1+g)^8, now simplify and solve for g. (1+g)^8/(1+g)^5=15/4 >>> (reduce the left side since these are the same group raised to different exponents) >>> (1+g)3=15/4 now the tricky part; you have to take the cubed root of both side to be able to isolate g >>> 1+g = (15/4)^(1/3) and solve for g >>> g=(15/4)^(1/3)  1 do the math and you find that g = .5536 and if you want to check your work, solve for K and try recalculating the original answers (btw K=.4419) The second problem is easier. again, it follows y=K(1.12)^x for annual, x=1 for a monthly growth, you will have y=K(g)^12 (12 months in a year) set these equal and solve for g: (1.12)^1=(g)^12 , take the twelfth root of both sides (I hope you have a scientific calculator) >>> g = 1.12^(1/12) or g = 1.0095 or 1.01 I hope this helps. Memorize the generic formula on top and you should always be able to do these types of problems.
Answers:hypothesisa possible explaniation or sollution to the problem dependent variablethe variable that is measured in an experiment independent variablethe factor that is changed in an experiement
Answers:Dear,I searched the whole net..but couldn't find any as they are not proper mathematical terms. But I know what a Scientific Notation is. Scientific notation is a mathematical format used to write very large and very small numbers; this system avoids using a lot of zeros by using powers (exponents). Hope this helps!!
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