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Linearsystems/26347: My son is having problems with this type of equation. He has to find out if the linear equation is consistent or inconsistent.Dependant or independant. Example: x+y=4 5x+y=8 How do you graph this equation. How do you work 5x+y=8 to know where to draw the second line of this equation.What would the graph look like .Thanks
1 solutions
Answer 14201 by venugopalramana(3286) on 20060202 10:41:20 (Show Source):
You can put this solution on YOUR website! SEE THE FOLLOWING TO GET AN IDEA OF THE TOPIC.IF STILL IN DIFFICULTY PLEASE CONTACT ME AND I SHALL HELP FURTHER
AS REGARDS GRAPHING GIVE DIFFERENT VALUES TO X AND FIND Y FROM THE GIVEN RELATION .SEE BELOW
I EQN....X+Y=4
X.............0.........1...........2...........3..............4........ETC...
Y=4X.........4.........3...........2...........1..............0.........ETC
II EQN.....5X+Y=8
X...........0...........1............2.............3......ETC.........
Y=85X......8...........3............2............7.....ETC...........
THE GRAPHS WILL LOOK LIKE THIS.
THE 2 LINES INTERSECT AT X=1 AND Y=3 WHICH IS THE UNIQUE SOLUTION TO THESE EQNS.THAT IS THEY ARE CONSISTENT AND INDEPENDENT.IF THEY ARE DEPENDENT YOU WILL GET ONE IDENTICAL LINE FOR BOTH EQNS.THAT IS THEY HAVE INFINITE SOLUTIONS ON THAT LINE. IF THEY ARE INCONSISTENT YOU WILL GET 2 PARALLEL LINES WHICH DO NOT INTERSECT AND HENCE THERE IS NO SOLUTION

What is a coinciding equation?
1 solutions

Answer 13848 by venugopalramana(791) on 20060128 11:16:56 (Show Source):
consider these 2 eqns
x+y=2...........i
2x+2y=4.........ii
if the first eqn is given to us ,the second one can be derived by us by multiplying the first eqn. with 2 on either side. so the second eqn.does not give us any additional information ,but is a derivative or dependent eqn. of eqn.i.read the following for additional related information.in case of more than 2 or for that matter any number of equations , if one or more of them could be sinilarly derived by a suitable combination of other eqns.,then we say they or dependent eqns.

When using gaussian elimination, can an inconsistant solution have a general solution? i worked out the matrix and came to a spot where 1=3 and 1=10.4 (inconsistant 1 can't equal 2 different solutions); would there still be a way to formulate a general solution such as x=(24x,23x,x,0) = (2,2,0,0) + x(4,3,1,0)
I hope this is enough information to clearly explain my question. thanks
I THINK YOU ARE CONFUSED BETWEEN ...INCONSISTENT,AND DEPENDENT EQUATIONS.
AS THE NAME IMPLIES AN INCONSISTENT SET OF EQUATIONS HAS NO SOLUTION AS THEY ARE INCONSISTENT.FOR EXAMPLE
X+Y=2
2X+2Y=3....HENCE THERE IS NO QUESTION OF GETTING A GENERAL SOLUTION FOR THAT BY ANY METHOD GAUSSIAN ELIMINATION OR ANY..
CONSISTENT AND DEPENDENT EQNS.HAVE INFINITE SOLUTIONS GIVEN BY A GENERAL FORMULA AS YOU HAVE MENTIONED..FOR EX.
X+Y=1
2X+2Y=2....THE GERAL FORMULA IS Y=1X...OR (X,1X)IS A SOLUTION SET..LIKE (1,0),(2,1) ETC....
FINALLY CONSISTENT AND INDEPENDENT EQNS.HAVE UNIQUE SOLUTION..FOR EXAMPLE
X+Y=2
XY=0...HAVE ONE UNIQUE SOLUTION X=1 AND Y=1
NOW YOU CAN USE GAUSSIAN ELIMINATION TO THE 3 EXAMPLES ABOVE AND SEE WHAT YOU GET.IF YOU STILL AVE DIFFICULTY COME BACK

Geometric_formulas/26224: 4) CLASSIC PROBLEM  A traveling salesman (selling shoes) stops at a farm in the Midwest. Before he could knock on the door, he noticed an old truck on fire. He rushed over and pulled a young lady out of the flaming truck. Farmer Brown came out and gratefully thanked the traveling salesman for saving his daughters life. Mr. Brown insisted on giving the man an award for his heroism.
So, the salesman said, If you insist, I do not want much. Get your checkerboard and place one penny on the first square. Then place two pennies on the next square. Then place four pennies on the third square. Continue this until all 64 squares are covered with pennies. As hed been saving pennies for over 25 years, Mr. Brown did not consider this much of an award, but soon realized he made a miscalculation on the amount of money involved.
a) How much money would Mr. Brown have to put on the 32nd square?
b) How much would the traveling salesman receive if the checkerboard only had 32 squares?
Calculate the amount of money necessary to fill the whole checkerboard (64 squares). How much money would the farmer need to give the salesman?
Thank you so, much for your help.
1 solutions
Answer 14101 by venugopalramana(3286) on 20060201 10:42:32 (Show Source):
You can put this solution on YOUR website! ok, this is one that has baffled not only myself but my parents and boyfriend as well. This is a problem that no one seems to be able to figure out. Do you think you could help me?
GOOD PROBLEM YOUNG GIRL!NOW THAT YOU SAY THAT YOUR BOY FRIEND IS ALSO FOXED BY THE PROBLEM ,YOU CAN USE IT TO GOOD EFFECT TO PUT ACROSS A PROPOSAL TO HIM ...
YOU PROMISS HIM A 100,000..POUNDS EVERY DAY FOR 30 DAYS.LET HIM GIVE YOU IN RETURN JUST A PENNY ON THE FIRST DAY,2 PENNIES ON THE SECOND DAY ,4 PENNIES ON THE THIRD DAY,......ETC FOR 30 DAYS ..AS IN YOUR PROBLEM...BASED ON YOUR FEED BACK , I PRESUME HE WILL LAP ON THE PROPOSAL AND I CAN ASSURE YOU THAT YOU WONT REGRET THE PROPOSAL!!!!
OK NOW LET US GET BACK TO YOUR PROBLEM AS WELL AS MY SUGGESTION ,WHICH IS JUST A PART OF IT .I AM GIVING BELOW THE MONEY TO BE PAID ON THIS BASIS FOR 30 AND 64 DAYS ,ROUNDED OFF TO POUNDS OR MILLION POUNDS AT THE LATER DAYS.
DAY.....MONEY TO BE PAID........MONEY TO BE PAID BY YOUR FRIEND TO YOU IN
............TO YOUR FRIEND..... CENTS...POUNDS...MILLION POUNDS
................POUNDS
1 100000 1 0 0
2 100000 2 0 0
3 100000 4 0 0
4 100000 8 0 0
5 100000 16 0 0
6 100000 32 0 0
7 100000 64 1 0
8 100000 128 1 0
9 100000 256 3 0
10 100000 512 5 0
11 100000 1024 10 0
12 100000 2048 20 0
13 100000 4096 41 0
14 100000 8192 82 0
15 100000 16384 164 0
16 100000 32768 328 0
17 100000 65536 655 0
18 100000 131072 1311 0
19 100000 262144 2621 0
20 100000 524288 5243 0
21 100000 1048576 10486 0
22 100000 2097152 20972 0
23 100000 4194304 41943 0
24 100000 8388608 83886 0
25 100000 16777216 167772 0
26 100000 33554432 335544 0
27 100000 67108864 671089 1
28 100000 134217728 1342177 1
29 100000 268435456 2684355 3
30 100000 536870912 5368709 5
31 100000 1073741824 10737418 11
32 100000 2147483648 21474836 21
33 100000 4294967296 42949673 43
34 100000 8589934592 85899346 86
35 100000 17179869184 171798692 172
36 100000 34359738368 343597384 344
37 100000 68719476736 687194767 687
38 100000 1.37439E+11 1374389535 1374
39 100000 2.74878E+11 2748779069 2749
40 100000 5.49756E+11 5497558139 5498
41 100000 1.09951E+12 10995116278 10995
42 100000 2.19902E+12 21990232556 21990
43 100000 4.39805E+12 43980465111 43980
44 100000 8.79609E+12 87960930222 87961
45 100000 1.75922E+13 175921860444 175922
46 100000 3.51844E+13 351843720888 351844
47 100000 7.03687E+13 703687441777 703687
48 100000 1.40737E+14 1407374883553 1407375
49 100000 2.81475E+14 2814749767107 2814750
50 100000 5.6295E+14 5629499534213 5629500
51 100000 1.1259E+15 11258999068426 11258999
52 100000 2.2518E+15 22517998136853 22517998
53 100000 4.5036E+15 45035996273705 45035996
54 100000 9.0072E+15 90071992547410 90071993
55 100000 1.80144E+16 180143985094820 180143985
56 100000 3.60288E+16 360287970189640 360287970
57 100000 7.20576E+16 720575940379279 720575940
58 100000 1.44115E+17 1441151880758560 1441151881
59 100000 2.8823E+17 2882303761517120 2882303762
60 100000 5.76461E+17 5764607523034230 5764607523
61 100000 1.15292E+18 11529215046068500 11529215046
62 100000 2.30584E+18 23058430092136900 23058430092
63 100000 4.61169E+18 46116860184273900 46116860184
64 100000 9.22337E+18 92233720368547800 92233720369
30 DAY..3000000.........1073741823......10737418.....................11
TOTAL
64 DAY..6400000.........1.84467E+19.....184467440737095000......184467440737
TOTAL
YOUR GAIN IN 30 DAYS = 8 MILLION POUNDS..GOT IT BUT ONLY BE CAREFULL THAT YOUR FRIEND WONT RUN OUT OF YOU ON THE 25 TH. DAY.!!!
YOUR GAIN IN 30 DAYS = 184467440731 MILLION POUNDS
NOW COMING TO THE MATHS PART OF THIS ,THIS SEQUENCE WHERE EACH NUMBER BEARS A CONSTANT RATIO TO ITS PREDECESSOR IS CALLED GEOMETRIC PROGRESSION..HERE YOU FIND EACH NUMBER IS OBTAINED FROM THE PREVIOUS ONE BY MULTIPLYING WITH 2 , CALLED COMMON RATIO.
THE LAST NUMBER ON NTH. DAY AND SUM OF SUCH SERIES OF NUMBERS UPTO THE N TH.DAY IS GIVEN BY THE FOLLOWING FORMULAE...
N TH. NUMBER = FIRST NUMBER *(C0MM0N RATIO )^(N1)=A*(R)^(N1)=(2)^(N1) IN THIS CASE.
SUM UP TO N TH.NUMBER =A*{((R)^N  1))/(R1)}={(2)^N1}IN THIS CASE

Geometric_formulas/26074: 1) Use the geometric series of numbers 1, 2, 4, 8,
to find the following:
a) What is r, the ratio between 2 consecutive terms?
b) Using the formula for the nth term of a geometric series, what is the 24th term?
c) Using the formula for the sum of a geometric series, what is the sum of the first 10 terms?
pls help
1 solutions
Answer 14006 by venugopalramana(3286) on 20060131 05:56:57 (Show Source):
You can put this solution on YOUR website! ) Use the geometric series of numbers 1, 2, 4, 8,
to find the following:
a) What is r, the ratio between 2 consecutive terms?
b) Using the formula for the nth term of a geometric series, what is the 24th term?
c) Using the formula for the sum of a geometric series, what is the sum of the first 10 terms? SUM IS GIVEN BY
SN=A(R^N1)/(R1)
S10=1*(2^101)/(21)=2^101
SEE THE FOLLOWING EXAMPLE AND SOLVE
Use the geometric series of numbers 1, 1/2, 1/4, 1/8,
to find the following:
a) What is r, the ratio between 2 consecutive terms?
R=(1/2)/1=1/2...IT IS SAME FOR ANY 2 CONSECUTIVE TERMS...SAY
(1/8)/(1/4)=1/2...ETC......
b) Using the formula for the nth term of a geometric series, what is 10th term?
TN=A*(R)^(N1)=1*(1/2)^(N1)
T10=(1/2)^(101)=(1/2)^9
c) Using the formula for the nth term of a geometric series, what is 12th term?
T12=(1/2)^11
d) What observation can make about these sums? In particular, what number does it appear that the sum will always be smaller than?
THE NUMBERS TEND TO ZERO AS N INCREASES TO LARGE NUMBERS AND TOWARDS INFINITY.
SUM OF A G.P. TO INFINITE TERMS WITH R<1 IS GIVEN BY
A/(1R)=1/(10.5)=2

Geometric_formulas/26075: 10) Use the geometric series of numbers 1, 1/2, 1/4, 1/8,
to find the following:
a) What is r, the ratio between 2 consecutive terms?
b) Using the formula for the nth term of a geometric series, what is 10th term?
c) Using the formula for the nth term of a geometric series, what is 12th term?
d) What observation can make about these sums? In particular, what number does it appear that the sum will always be smaller than?
1 solutions
Answer 14005 by venugopalramana(3286) on 20060131 05:54:54 (Show Source):
You can put this solution on YOUR website! Use the geometric series of numbers 1, 1/2, 1/4, 1/8,
to find the following:
a) What is r, the ratio between 2 consecutive terms?
R=(1/2)/1=1/2...IT IS SAME FOR ANY 2 CONSECUTIVE TERMS...SAY
(1/8)/(1/4)=1/2...ETC......
b) Using the formula for the nth term of a geometric series, what is 10th term?
TN=A*(R)^(N1)=1*(1/2)^(N1)
T10=(1/2)^(101)=(1/2)^9
c) Using the formula for the nth term of a geometric series, what is 12th term?
T12=(1/2)^11
d) What observation can make about these sums? In particular, what number does it appear that the sum will always be smaller than?
THE NUMBERS TEND TO ZERO AS N INCREASES TO LARGE NUMBERS AND TOWARDS INFINITY.
SUM OF A G.P. TO INFINITE TERMS WITH R<1 IS GIVEN BY
A/(1R)=1/(10.5)=2

