a word is represented by only one set of numbers as given in anyone of the alternatives are represents by two classes of alphabets as shown in two matrices given below the column and rows of matrix i are numbered from 0 to 4 and that of matrix ii are numbered from 5 to 9 a letter from these matrices can be represented first by its row and next by its column e lsquohrsquo can be represented by 14 20 42 etc etc and lsquodrsquo can be represented by 59 65 86 etc identify the set for the world railmatrix ndash inbsp012340rheli1ielrh2hlrie3erihl4lihermatrix ndash inbsp567895bsnad6dnbsa7abdns8sdabn9nasdbe b l e l e l e e a e e e e 1 0 4 ii 2 5 9 i e a l u b u le h 14 20 42 aa d 59 65 86 aa i aa rail b le ndash inbsp012340rheli1ielrh2hlrie3erihl4lihernbsp ndash iinbsp567895bsnad6dnbsa7abdns8sdabn9nasdb
31, 87, 32, 21
finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). we have, 6 objects {b}, {h}, {a}, {r}, {a}, {t} and there are 2 a's.so, the words can be formed out of the letters of the word ‘bharat’ taking 3 at a time can be done in 2 ways:case-1: when all the letters are distinct.we have, 5 distinct letters, out of which taking three at a time, the number of words that can be formed = 5p3= 60case-2: when 2 a’s are selected.so, we have, 2a’s and 1 letter is to selected out of the 4 distinct letters, which can be done in = 4p1= 4 ways.now, the 3 letters can be arranged among themselves, but there are 2 a’s, so the number of ways in which arrangement can be done isso, in this case, total number of words that can be formed = 12.the number of arrangements of the letters of the word bharat taking 3 at a time is = (60 + 12)= 72. direction: a word and number arrangement machine when given an input line of word and numbers rearranges them following a particular rule in each step. the following is an illustration of input and rearrangement. input: politics 61 policy 116 partner 93 role 45 clear 27 38 point step i: role politics policy 116 partner 93 45 clear 27 38 point 61 step ii: role clear politics policy partner 93 45 27 38 point 116 61 step iii: role clear partner politics policy 93 45 38 point 27 116 61 step iv: role clear partner politics policy 93 38 point 45 27 116 61 step v: role clear partner politics point policy 93 38 45 27 116 61 step v, is the last step of the above arrangement as the intended arrangement is obtained. as per the rule followed in the given steps find out the appropriate steps for the given input. input: energy 54 78 place 23 issue 211 145 flat rent taxi 17in step iii, what is the difference between fifth element from left and second element from right? dear student
red is coded as 2076
which means
r=20
e=7
d=6
now lets us number the alphabets as follows
a=1
b=2
c= 3
d= 4
e= 5




r=18


y=25
z=26

now according to the question
r=20=18+2
e=7=5+2
d=6=4+2

therefore we can say that
g = 7+2 =9
a=1+2=3


gag will be coded as 939


regards
â€‹â€‹â€‹â€‹â€‹â€‹â€‹ hello, 

this code checks to see if the diagonal elements of a given matrix a (assuming n x n) are larger in magnitude than the sum of the magnitude of the non-diagonal elements in its row.

line by line explanation:

the first line loops through all the rows of a.

 for i = 1:n

the second line is defining j = [1 2 3 ... n], so a vector containing 1..n,
 
 j = 1:n;

the third line is removing the diagonal element, aka, the ith row and this removes the ith column, so the i,i th element,
 
 j(i) = [];

the fourth line takes the absolute value of all the elements in the ith row of a, excluding the diagonal element: a(i,j) = a(ith row, elements 1,2,...,i-1,i+1,...,n) since we removed the ith element from j previously.
 
 b = abs(a(i,j));

the fifth line computes the absolute value of the diagonal element minus the sum of the absolute value of all the other elements in the row. if check is positive, we know the magnitude of the diagonal element was greater than the sum of all the magnitudes of the other elements in its row.
 
 check(i) = abs(a(i,i)) - sum(b);

the sixth line checks if it's less than zero, in which case the diagonal elements magnitude is not greater than or equal to the sum of the magnitudes of the other elements.
 
 if check(i) < 0

if this is the case, the matrix is not strictly diagonally dominant, which the seventh line prints.
 
 fprintf('the matrix is not strictly diagonally dominant at row %2i

',i)

the eighth and ninth lines end the cycle.
 
 end
 end

sorry if this was too much detail,

hope this helps! >> get(groot,'defaultaxescolororder')
 ans =
 0 0.4470 0.7410
 0.8500 0.3250 0.0980
 0.9290 0.6940 0.1250
 0.4940 0.1840 0.5560
 0.4660 0.6740 0.1880
 0.3010 0.7450 0.9330
 0.6350 0.0780 0.1840
 >> 

the letter monikers for colors are the pure colors, 'b' --> [0 0 1], etc., ... we know that- - - - - (i)taking 3 as common, we get3; solving only bracket term first,here a = 2a, b = 3b; putting values in eq.i}applying }}ans: - 3}note: must memorize cubes upto 12 here is a list of the symbols by their names: ampersand & apostrophe or single quote ' asterisk * at @ back quote ` back slash \ braces or curly braces {} brackets [] carat ^ close brace } close bracket ] close parenthesis ) colon : comma , dollar $ equal = exclamation point ! greater than > less than < minus or hyphen - open brace { open bracket [ open parenthesis ( parenthesis () percent % pipe or bar | plus + octothorpe # quote " semi-colon ; slash or forward slash / tilde ~ underscore \_ 10.1 sequences polynomials: easy to evaluate, integrate, diﬀerentiate general functions: -a sequence is a list of #s -sequences are written as: -many series correspond to functions formally, the sequence converges to l if for every positive number , there corresponds an integer n such that for all we have to show that: a = b
given: p (a) = p (b)
let x be any element of set a,p(a) is the power set of set a, hence it contains all the subsets of set a. thus set a is contained in set p(a).

now let c be an element of set b
∴ 
now, we have:
c ⊂ b
∴ 
∴ a ⊂ b

similarly, we have:
b ⊂ a
now if a ⊂ b and b ⊂ a
∴ a = b