- 5.5.1: Tell what operation on the first inequality gives the second one, a...
- 5.5.2: Find three values of the variable that satisfy each inequality. a. ...
- 5.5.3: Give the inequality graphed on each number line. a. b. c. d. e.
- 5.5.4: Translate each phrase into symbols. a. 3 is more than x b. y is at ...
- 5.5.5: Solve each equation for y. a. 3x + 4y = 5.2 b. 3(y 5) = 2x
- 5.5.6: Solve each inequality and show your work. a. 4.1 + 3.2x > 18 b. 7.2...
- 5.5.7: Solve each inequality and graph the solutions on a number line. a. ...
- 5.5.8: Ezra received $50 from his grandparents for his birthday. He makes ...
- 5.5.9: For each graph, tell what operation moves the two points in the ine...
- 5.5.10: Tell whether each inequality is true or false for the given value. ...
- 5.5.11: Solve each inequality. Explain the meaning of the result. On a numb...
- 5.5.12: Data collected by a motion sensor will vary slightly in accuracy. A...
- 5.5.13: You read the inequality symbols, <, , >, and , as is less than, is ...
- 5.5.14: The table gives equations that model the three vehicles distances i...
- 5.5.15: In Example B, the inequality 8 0.25x < 5 was written to represent t...
- 5.5.16: List the order in which you would perform these operations to get t...
- 5.5.17: The table shows the 2004 U.S. postal rates for letters, large envel...
- 5.5.18: Use the distributive property to rewrite each expression without us...
Solutions for Chapter 5.5: Inequalities in One Variable
Full solutions for Discovering Algebra: An Investigative Approach | 2nd Edition
Associative Law (AB)C = A(BC).
Parentheses can be removed to leave ABC.
Augmented matrix [A b].
Ax = b is solvable when b is in the column space of A; then [A b] has the same rank as A. Elimination on [A b] keeps equations correct.
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!
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.
Complete solution x = x p + Xn to Ax = b.
(Particular x p) + (x n in nullspace).
cond(A) = c(A) = IIAIlIIA-III = amaxlamin. In Ax = b, the relative change Ilox III Ilx II is less than cond(A) times the relative change Ilob III lib II· Condition numbers measure the sensitivity of the output to change in the input.
Determinant IAI = det(A).
Defined by det I = 1, sign reversal for row exchange, and linearity in each row. Then IAI = 0 when A is singular. Also IABI = IAIIBI and
A sequence of row operations that reduces A to an upper triangular U or to the reduced form R = rref(A). Then A = LU with multipliers eO in L, or P A = L U with row exchanges in P, or E A = R with an invertible E.
Four Fundamental Subspaces C (A), N (A), C (AT), N (AT).
Use AT for complex A.
Fourier matrix F.
Entries Fjk = e21Cijk/n give orthogonal columns FT F = nI. Then y = Fe is the (inverse) Discrete Fourier Transform Y j = L cke21Cijk/n.
Set of n nodes connected pairwise by m edges. A complete graph has all n(n - 1)/2 edges between nodes. A tree has only n - 1 edges and no closed loops.
Hilbert matrix hilb(n).
Entries HU = 1/(i + j -1) = Jd X i- 1 xj-1dx. Positive definite but extremely small Amin and large condition number: H is ill-conditioned.
A symmetric matrix with eigenvalues of both signs (+ and - ).
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.
Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.
Left inverse A+.
If A has full column rank n, then A+ = (AT A)-I AT has A+ A = In.
Minimal polynomial of A.
The lowest degree polynomial with meA) = zero matrix. This is peA) = det(A - AI) if no eigenvalues are repeated; always meA) divides peA).
The diagonal entry (first nonzero) at the time when a row is used in elimination.
Right inverse A+.
If A has full row rank m, then A+ = AT(AAT)-l has AA+ = 1m.
Sum V + W of subs paces.
Space of all (v in V) + (w in W). Direct sum: V n W = to}.