 6.6.1: Rewrite each expression using only positive exponents. a. 23 b. 52 ...
 6.6.2: Insert the appropriate symbol (<, =, or >) between each pair of num...
 6.6.3: Find the exponent of 10 that you need to write each expression in s...
 6.6.4: The population of a town is currently 45,647. It has been growing a...
 6.6.5: Juan says that 63 is the same as 63.Write an explanation of how Jua...
 6.6.6: Use the properties of exponents to rewrite each expression without ...
 6.6.7: APPLICATION Suppose the annual rate of inflation is about 4%. This ...
 6.6.8: APPLICATION The population of Japan in 2004 was about 1.3 108. Japa...
 6.6.9: Decide whether each statement is true or false. Use expanded form t...
 6.6.10: A large ball of string originally held 1 mile of string. Abigail cu...
 6.6.11: Suppose 36(1 + 0.5)4 represents the number of bacteria cells in a s...
 6.6.12: APPLICATION Camila received a $1,200 prize for one of her essays. S...
 6.6.13: MiniInvestigation In the last few lessons, you have worked with eq...
 6.6.14: APPLICATION A capacitor is charged with a ninevolt battery. The eq...
 6.6.15: Set your calculator in scientific notation mode for this problem. a...
Solutions for Chapter 6.6: Zero and Negative Exponents
Full solutions for Discovering Algebra: An Investigative Approach  2nd Edition
ISBN: 9781559537636
Solutions for Chapter 6.6: Zero and Negative Exponents
Get Full SolutionsDiscovering Algebra: An Investigative Approach was written by and is associated to the ISBN: 9781559537636. Since 15 problems in chapter 6.6: Zero and Negative Exponents have been answered, more than 14264 students have viewed full stepbystep solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. Chapter 6.6: Zero and Negative Exponents includes 15 full stepbystep solutions. This textbook survival guide was created for the textbook: Discovering Algebra: An Investigative Approach, edition: 2.

Companion matrix.
Put CI, ... ,Cn in row n and put n  1 ones just above the main diagonal. Then det(A  AI) = ±(CI + c2A + C3A 2 + .•. + cnA nl  An).

Cramer's Rule for Ax = b.
B j has b replacing column j of A; x j = det B j I det A

Cyclic shift
S. Permutation with S21 = 1, S32 = 1, ... , finally SIn = 1. Its eigenvalues are the nth roots e2lrik/n of 1; eigenvectors are columns of the Fourier matrix F.

Dimension of vector space
dim(V) = number of vectors in any basis for V.

Dot product = Inner product x T y = XI Y 1 + ... + Xn Yn.
Complex dot product is x T Y . Perpendicular vectors have x T y = O. (AB)ij = (row i of A)T(column j of B).

Four Fundamental Subspaces C (A), N (A), C (AT), N (AT).
Use AT for complex A.

GramSchmidt orthogonalization A = QR.
Independent columns in A, orthonormal columns in Q. Each column q j of Q is a combination of the first j columns of A (and conversely, so R is upper triangular). Convention: diag(R) > o.

Hessenberg matrix H.
Triangular matrix with one extra nonzero adjacent diagonal.

Hilbert matrix hilb(n).
Entries HU = 1/(i + j 1) = Jd X i 1 xj1dx. Positive definite but extremely small Amin and large condition number: H is illconditioned.

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 .

Indefinite matrix.
A symmetric matrix with eigenvalues of both signs (+ and  ).

lAII = l/lAI and IATI = IAI.
The big formula for det(A) has a sum of n! terms, the cofactor formula uses determinants of size n  1, volume of box = I det( A) I.

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.

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.

Pivot.
The diagonal entry (first nonzero) at the time when a row is used in elimination.

Schur complement S, D  C A } B.
Appears in block elimination on [~ g ].

Standard basis for Rn.
Columns of n by n identity matrix (written i ,j ,k in R3).

Trace of A
= sum of diagonal entries = sum of eigenvalues of A. Tr AB = Tr BA.

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

Vector v in Rn.
Sequence of n real numbers v = (VI, ... , Vn) = point in Rn.