ssrs barcode font download list = {{a, b, c}, {d, e, f}, {g, h, i}}; TableForm[list, TableSpacing {0, 0}] abc def ghi in Software

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list = {{a, b, c}, {d, e, f}, {g, h, i}}; TableForm[list, TableSpacing {0, 0}] abc def ghi
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TableForm[list,TableSpacing {1, 3}] a b c d e f g h i TableForm[list,TableSpacing {3, 1}] abc
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3 lines between rows, 1 space between columns. 1 line between rows, 3 spaces between columns.
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Lists can be expressed as single columns with ColumnForm.
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ColumnForm[list] presents list as a single column of objects. ColumnForm[list, horizontal] specifies the horizontal alignment of each row. Acceptable values of horizontal are Left (default), Center, and Right. ColumnForm[list, horizontal, vertical] allows vertical alignment of the column. Acceptable values of vertical are Above, Center, and Below (default).
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list = {a, bb, ccc} ColumnForm[list] a bb ccc ColumnForm[list, Right] a bb ccc
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SOLVED PROBLEMS
3.32 Construct a 3 3 matrix whose entries are consecutive integers, increasing as we go to the right and down.
SOLUTION
list = Table[3i + j, {i, 0, 2}, {j, 1, 3}] {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}} list//MatrixForm 1 2 3 4 5 6 7 8 9
Lists
3.33 The Hilbert matrix is a square matrix whose element in position (i, j) is i + 1 1 . Construct the j Hilbert matrix of order 5.
SOLUTION
a[i_, j_] = 1/(i + j 1); hilbert = Array[a,{5, 5}]; hilbert//MatrixForm 1 1 2 1 3 1 4 1 5 1 2 1 3 1 4 1 5 1 6 1 3 1 4 1 5 1 6 1 7 1 4 1 5 1 6 1 7 1 8 1 5 1 6 1 7 1 8 1 9
3.34 Construct the 5 5 identity matrix.
SOLUTION
IdentityMatrix[5]//MatrixForm 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1
3.35 Construct a 5 5 matrix having the first five primes as diagonal entries and 0s elsewhere.
SOLUTION
diag = Table[Prime[k],{k, 1, 5}] {2, 3, 5, 7, 11} DiagonalMatrix[diag] //MatrixForm 2 0 0 0 0 0 3 0 0 0 0 0 5 0 0 0 0 0 7 0 0 0 0 0 11
3.36 Construct a table having three columns. The first column lists the consecutive integers 1 through 10 and the second and third columns are their squares and cubes. Label the three columns integers, squares, and cubes.
SOLUTION
list = Table[{k, k2, k3},{k, 1, 10}]; TableForm[list, TableHeadings {None, {"integers","squares", "cubes"}}, TableAlignments Right]
Lists
integers 1 2 3 4 5 6 7 8 9 10
squares 1 4 9 16 25 36 49 64 81 100
cubes 1 8 27 64 125 216 343 512 729 1000
3.37 If c represents the temperature in degrees Celsius, its corresponding Fahrenheit temperature is f = 9 c + 32 . Construct a labeled table showing, horizontally, the Fahrenheit equivalents of Celsius 5 temperatures from 1 to 10 in increments of 1 .
SOLUTION
f = 9 c + 32 5 list = Table[{c, PaddedForm[ N[f], {3, 1}]}, {c, 1, 10}] TableForm[list, TableDirections Row, TableHeadings {None, {"Celsius", "Fahrenheit"}}, TableAlignments Center] Celsius 1 2 35.6 3 37.4 4 39.2 5 41.0 6 42.8 7 44.6 8 9 10 50.0
Fahrenheit 33.8
46.4 48.2
3.38 Construct a table showing the radian equivalents of angles from 0 to 30 in increments of 5 .
SOLUTION
Degree is a Mathematica constant (see 2).
list = Table[{deg, N[deg Degree]}, {deg, 0, 30, 5}]; TableForm[list, TableDirections Row, TableHeadings {None, {"Degrees","Radians"}}, TableAlignments Center] Degrees 0 5 10 15 20 25 30
Radians 0
0.174533 0.261799 0.349066 0.436332
3.39 If p dollars is invested for t years in a bank account paying an annual interest rate of r compounded k n times a year, the amount of money after k periods is p 1 + r dollars. If $1,000 is invested in an n account paying 6% compounded quarterly, make a table showing how much money has accumulated during a three-year period.
SOLUTION
p = 1000; r = .06; n = 4; t = 3; a[k_]= p (1 + r / n)k ; list = Table[{k, a[k]}, {k, 1, n * t}] TableForm[list, TableHeadings {None, {"period", "amount"}}]
Lists
period 1 2 3 4 5 6 7 8 9 10 11 12
amount 1015. 1030.22 1045.68 1061.36 1077.28 1093.44 1109.84 1126.49 1143.39 1160.54 1177.95 1195.62
3.40 If p dollars is invested in a bank account paying a rate of r compounded n times a year, the amount of money after t years is p 1 + r dollars. If interest is compounded continuously, the amount after t years is pert. If n $1,000 is invested in an account paying 6% annually, make a table showing how much money is in the account at the end of each year for 10 years if interest is compounded quarterly, monthly, daily, and continuously.
SOLUTION
p = 1000; r = .06; a = p(1 + r / 4)4 t ; b = p(1 + r / 12)12 t ; c = p(1 + r / 365)365 t ; d = p Exp[r t];
tt = PaddedForm[t, 2]; aa = PaddedForm[a, {7, 2}]; bb = PaddedForm[b, {7, 2}]; cc = PaddedForm[c, {7, 2}]; dd = PaddedForm[d, {7, 2}];
list = Table[{tt, aa, bb, cc, dd},{t, 1, 10}]; TableForm[list, TableHeadings {None, {"year"," quarterly"," year 1 2 3 4 5 6 7 8 9 10 quarterly 1061.36 1126.49 1195.62 1268.99 1346.86 1429.50 1517.22 1610.32 1709.14 1814.02 monthly"," daily","continuously"}}] continuously 1061.84 1127.50 1197.22 1271.25 1349.86 1433.33 1521.96 1616.07 1716.01 1822.12
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