 3.2.1: Decide whether each expression is positive or negative without usin...
 3.2.2: List the terms of each number sequence of ycoordinates for the poi...
 3.2.3: Make a table listing the coordinates of the points plotted in 2b an...
 3.2.4: Plot the first five points represented by each recursive routine in...
 3.2.5: The direct variation y = 2.54x describes the relationship between t...
 3.2.6: APPLICATION A car is moving at a speed of 68 mi/h from Dallas towar...
 3.2.7: APPLICATION A longdistance telephone carrier charges $1.38 for int...
 3.2.8: These tables show the changing depths of two submarines as they com...
 3.2.9: Each geometric design is made from tiles arranged in a row.a. Make ...
 3.2.10: A bicyclist, 1 mi (5280 ft) away, pedals toward you at a rate of 60...
 3.2.11: Consider the expression a. Use the order of operations to find the ...
 3.2.12: Isaac learned a way to convert from degrees Celsius to Fahrenheit. ...
 3.2.13: APPLICATION Karen is a U.S. exchange student in Austria. She wants ...
 3.2.14: Draw and label a coordinate plane with each axis scaled from 10 to ...
Solutions for Chapter 3.2: Linear Plots
Full solutions for Discovering Algebra: An Investigative Approach  2nd Edition
ISBN: 9781559537636
Solutions for Chapter 3.2: Linear Plots
Get Full SolutionsDiscovering Algebra: An Investigative Approach was written by and is associated to the ISBN: 9781559537636. Since 14 problems in chapter 3.2: Linear Plots have been answered, more than 9586 students have viewed full stepbystep solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Discovering Algebra: An Investigative Approach, edition: 2. Chapter 3.2: Linear Plots includes 14 full stepbystep solutions.

Associative Law (AB)C = A(BC).
Parentheses can be removed to leave ABC.

Basis for V.
Independent vectors VI, ... , v d whose linear combinations give each vector in V as v = CIVI + ... + CdVd. V has many bases, each basis gives unique c's. A vector space has many bases!

Block matrix.
A matrix can be partitioned into matrix blocks, by cuts between rows and/or between columns. Block multiplication ofAB is allowed if the block shapes permit.

Cofactor Cij.
Remove row i and column j; multiply the determinant by (I)i + j •

Fibonacci numbers
0,1,1,2,3,5, ... satisfy Fn = Fnl + Fn 2 = (A7 A~)I()q A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].

Full column rank r = n.
Independent columns, N(A) = {O}, no free variables.

Incidence matrix of a directed graph.
The m by n edgenode incidence matrix has a row for each edge (node i to node j), with entries 1 and 1 in columns i and j .

Jordan form 1 = M 1 AM.
If A has s independent eigenvectors, its "generalized" eigenvector matrix M gives 1 = diag(lt, ... , 1s). The block his Akh +Nk where Nk has 1 's on diagonall. Each block has one eigenvalue Ak and one eigenvector.

Left inverse A+.
If A has full column rank n, then A+ = (AT A)I AT has A+ A = In.

Markov matrix M.
All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.

Multiplier eij.
The pivot row j is multiplied by eij and subtracted from row i to eliminate the i, j entry: eij = (entry to eliminate) / (jth pivot).

Normal equation AT Ax = ATb.
Gives the least squares solution to Ax = b if A has full rank n (independent columns). The equation says that (columns of A)·(b  Ax) = o.

Normal matrix.
If N NT = NT N, then N has orthonormal (complex) eigenvectors.

Permutation matrix P.
There are n! orders of 1, ... , n. The n! P 's have the rows of I in those orders. P A puts the rows of A in the same order. P is even or odd (det P = 1 or 1) based on the number of row exchanges to reach I.

Pseudoinverse A+ (MoorePenrose inverse).
The n by m matrix that "inverts" A from column space back to row space, with N(A+) = N(AT). A+ A and AA+ are the projection matrices onto the row space and column space. Rank(A +) = rank(A).

Rank one matrix A = uvT f=. O.
Column and row spaces = lines cu and cv.

Schwarz inequality
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.

Stiffness matrix
If x gives the movements of the nodes, K x gives the internal forces. K = ATe A where C has spring constants from Hooke's Law and Ax = stretching.

Unitary matrix UH = U T = UI.
Orthonormal columns (complex analog of Q).

Vector addition.
v + w = (VI + WI, ... , Vn + Wn ) = diagonal of parallelogram.