- 49.1: When the product of 0.2 and 0.3 is subtracted from the sum of 0.2 a...
- 49.2: Four fifths of a dollar is how many cents? Draw a diagram toillustr...
- 49.3: The rectangular, 99-piece Nano jigsaw puzzle is only 2.6 incheslong...
- 49.4: Find the perimeter of the puzzle described in problem 3.
- 49.5: Compare: a. 0.31 0.301 b. 31% 30.1%
- 49.6: 0.67 + 2 + 1.33
- 49.7: 12(0.25)
- 49.8: 0.07 3.5
- 49.9: 0.5 12
- 49.10: 8 0.14
- 49.11: (0.012)(1.5)
- 49.12: Find each unknown number:n - 618 4 38
- 49.13: Find each unknown number:45 x100
- 49.14: Find each unknown number:5 m = 1.37
- 49.15: Find each unknown number:m 714 = 15
- 49.16: Write the decimal number one and twelve thousandths.
- 49.17: 5 710 4 910
- 49.18: 52 53
- 49.19: How much money is 40% of $25.00?
- 49.20: There are 24 hours in a day. James sleeps 8 hours each night.a. Eig...
- 49.21: What factors do 12 and 18 have in common (that is, the numbersthat ...
- 49.22: What is the average of 1.2, 1.3, and 1.7?
- 49.23: Jan estimated that 49% of $19.58 is $10. She rounded 49%to 50% and ...
- 49.24: a. How many 34 s are in 1? b. Use the answer to part a to find the ...
- 49.25: Refer to the number line shown below to answer parts ac.a. Which po...
- 49.26: Multiply and divide as indicated:2 3 2 5 72 5 7
- 49.27: We can find the number of quarters in three dollars bydividing $3.0...
- 49.28: Use a ruler to find the length of each side of this squareto the ne...
- 49.29: A paper-towel tube is about 4 cm in diameter. The circumferenceof a...
- 49.30: Sam was given the following division problem:2.50.5Instead of multi...
Solutions for Chapter 49: Dividing by a Decimal Number
Full solutions for Saxon Math, Course 1 | 1st Edition
Column picture of Ax = b.
The vector b becomes a combination of the columns of A. The system is solvable only when b is in the column space C (A).
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 n-l - An).
When random variables Xi have mean = average value = 0, their covariances "'£ ij are the averages of XiX j. With means Xi, the matrix :E = mean of (x - x) (x - x) T is positive (semi)definite; :E is diagonal if the Xi are independent.
Cross product u xv in R3:
Vector perpendicular to u and v, length Ilullllvlll sin el = area of parallelogram, u x v = "determinant" of [i j k; UI U2 U3; VI V2 V3].
Eigenvalue A and eigenvector x.
Ax = AX with x#-O so det(A - AI) = o.
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.
Ellipse (or ellipsoid) x T Ax = 1.
A must be positive definite; the axes of the ellipse are eigenvectors of A, with lengths 1/.JI. (For IIx II = 1 the vectors y = Ax lie on the ellipse IIA-1 yll2 = Y T(AAT)-1 Y = 1 displayed by eigshow; axis lengths ad
A = L U. If elimination takes A to U without row exchanges, then the lower triangular L with multipliers eij (and eii = 1) brings U back to A.
Hypercube matrix pl.
Row n + 1 counts corners, edges, faces, ... of a cube in Rn.
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.
Linearly dependent VI, ... , Vn.
A combination other than all Ci = 0 gives L Ci Vi = O.
Ln = 2,J, 3, 4, ... satisfy Ln = L n- l +Ln- 2 = A1 +A~, with AI, A2 = (1 ± -/5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = O.
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 pivot row j is multiplied by eij and subtracted from row i to eliminate the i, j entry: eij = (entry to eliminate) / (jth pivot).
In each column, choose the largest available pivot to control roundoff; all multipliers have leij I < 1. See condition number.
Reduced row echelon form R = rref(A).
Pivots = 1; zeros above and below pivots; the r nonzero rows of R give a basis for the row space of A.
Reflection matrix (Householder) Q = I -2uuT.
Unit vector u is reflected to Qu = -u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q-1 = Q.
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.
Singular Value Decomposition
(SVD) A = U:E VT = (orthogonal) ( diag)( orthogonal) First r columns of U and V are orthonormal bases of C (A) and C (AT), AVi = O'iUi with singular value O'i > O. Last columns are orthonormal bases of nullspaces.
Tridiagonal matrix T: tij = 0 if Ii - j I > 1.
T- 1 has rank 1 above and below diagonal.