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Calculate the NPV(Net Present Value) for a project costing $250,000 which will then provide $25,000 annually for 25 years. The discount rate is 5%. What would the NPV be if the inflows were $20.000 annually for only 20 years?
TABULATE AS FOLLOWS IN XL WORK SHEET WHICH IS SELF EXPLANATORY.IF YOU GO TO FUNCTIONS-FINANCIALS AT THE END AND GIVE THE DISCOUNT RATE OF 0.05 AND NET INFLOW COLUMN AS THE DATA,YOU WILL GET NPV AT THE END.THE FORMULA IN XL IS.
=NPV(0.05,D256:D281)WHERE 0.05 IS THE DISCOUNT RATE GIVEN AS 5% PER YEAR AND D256:D281 IS THE COLUMN WHERE NETINFLOW IS DEPICTED.IF YOU WANT TO KNOW MATHMATICAL FORMULA FOR THIS PLEASE COME BACK AND WE SHALL EXPLAIN

YEAR INFLOW OUTFLOW NETINFLOW YEAR INFLOW OUTFLOW NETINFLOW
1 250000 -250000... 1 250000 -250000
2 25000 25000... 2 20000 20000
3 25000 25000... 3 20000 20000
4 25000 25000... 4 20000 20000
5 25000 25000... 5 20000 20000
6 25000 25000... 6 20000 20000
7 25000 25000... 7 20000 20000
8 25000 25000... 8 20000 20000
9 25000 25000... 9 20000 20000
10 25000 25000... 10 20000 20000
11 25000 25000... 11 20000 20000
12 25000 25000... 12 20000 20000
13 25000 25000... 13 20000 20000
14 25000 25000... 14 20000 20000
15 25000 25000... 15 20000 20000
16 25000 25000... 16 20000 20000
17 25000 25000... 17 20000 20000
18 25000 25000... 18 20000 20000
19 25000 25000... 19 20000 20000
20 25000 25000... 20 20000 20000
21 25000 25000... 21 20000 20000
22 25000 25000
23 25000 25000
24 25000 25000
25 25000 25000
26 25000 25000
NPV $97,474.87 NPV-NEGATIVE ($719.80)

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A Regualr polygon of n sides inscribed in a circle of radius r has 3r squared. What is the value of n?
YOU MEAN AREA =3R^2...ASSUMING SO...
THERE WILL BE N TRIAGLES WHEN N VERTICES ARE JOINED TO THE CENTRE OF THE CIRCUM CIRCLE.
SO EACH RIANGLE HAS AREA OF 3R^2/N
NOW AREA OF EACH TRIANGLE IS GIVEN BY 0.5*R^2*SIN(X)...WHERE X IS THE ANGLE SUBTENDED BY ONE SIDE OF REGULAR POLYGON AT CENTRE OF CIRCUMCIRCLE.
HENCE 0.5*R^2*SIN(X)=3R^2/N
SIN(X)=3/(N*0.5)= 6/N
NOW IF THERE ARE N SIDES THEN EACH SIDE SUBTENDS 360/N ANGLE AT CENTRE .HENCE
X=2PI/N...OR....N=2PI/X
SIN(X)=6*X/2PI=3X/PI
WE FIND THAT AT X=30...OR...PI/6,WE HAVE SIN(X)=SIN(30)=SIN(PI/6)=0.5
AND 3X/PI=(3*PI)/(6*PI)=0.5
HENCE X=PI/6 IS THE SOLUTION
HENCE N=2PI/X=(2PI)/(PI/6)=2*6=12
HENCE THE REGULR POLYGON IS 12 SIDED

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A box contains 2 fifty paise coins,5 twenty five paise coins and a certain fixed number(N<=2) of ten and five paise
coins.If five coins are taken out of the box at random,then the probability that the total value of these 5 coins is
less than one rupee and fifty paise is which of the following:-
SINCE IT IS EASIER TO FIND THE NUMBER OF WAYS BY WHICH THE SUM OF 5 COINS CAN EQUAL OR EXCEED RS.1.50,LET US FIND IT OUT AND SUBTRACT THE RESULTANT PROBABILITY FROM ONE TO GET THE ANSWER
WE HAVE IN ALL 2+5+N=N+7 COINS
5 CAN BE TAKEN IN ...(N+7)C5 WAYS
CASE 1.....1*50+4*25=1.50....IN..(2C1)(5C4)=2*5=10
CASE 2.....2*50+3*25=1.75....IN..(2C2)(5C3)=1*10=10
CASE 3.....2*50+2*25+1*5 OR 10=1.55 OR 1.60...IN...(2C2)(5C2)(NC1)=1*10*N=10N
TOTAL NUMBER .....................=10+10+10N=20+10N
PROBABILITY OF GETTING 1.50 OR MORE =(20+10N)/{(N+7)C5}
PROBABILITY OF GETTING LESS THAN 1.50 =1- [(20+10N)/{(N+7)C5}]

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.|3-5i|
ABSOLUTE VALUE OR MOD OF A COMPLEX NUMBER X+iY IS GIVEN BY SQUAREROOT OF (X^2+Y^2) AND IS SYMBOLISED AS
|X+iY|=SQRT(X^2+Y^2)...HENCE WE GET
|3=5i|=SQRT.(3^2+5^2)=SQRT(9+25)=SQRT(34)

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I DONT THINK THE PROBLEM IS CORRECT
IN Z2
F(T)=T^2+T+1...IS A SECOND DEGREE POLYNOMIAL...IT HAS NO ZERO IN REAL FIELD OF Z2, AND IS NOT REDUCIBLE.PLEASE CHECK BACK WITH YOUR BOOK/QUESTION SOURCE OR CLASS TEACHER

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1) The denominator(DR) of a fraction is three more than the numerator(NR). If the numerator is doubled and the denominator is increased by nine, the resulting fraction is equal to ½. What is the original fraction?
LET NR =N
3 MORE =N+3=DR...SO FRACTION IS N/(N+3)
NR IS DOUBLED...WE GET 2N
DR INCREASED BY 9...WE GET DR+9=N+3+9=N+12
NEW FRACTION=2N/(N+12)=1/2
2*2N=N+12
4N-N=12.....3N=12.....N=4
HENCE ORIGINAL FRACTION =N/(N+3)=4/7
2) The denominator of a fraction is three more than numerator. If the numerator is tripled and the denominator is increased by 17, the resulting fraction is ½. What is the original fraction?
PROCEDING IN SAME WAY AS ABOVE
ORIGINAL FRACTION =N/(N+3)...AND
3N/(N+3+17)=1/2
2*3N=N+20
6N-N=20
N=4
ORIGINAL FRACTION =4/7

3) Bob, Celia, Sam and Daniel contributed money to buy their boss a retirement gift. Sam and Daniel each gave the same amount of money. Celia gave $12 more than Daniel. Bob gave half as much as Celia gave. If the four workers gave a total of $81, how much did Sam give?
LSET SAM GIVE =S
DANIEL GAVE =S
12 MORE =S+12=CELIA GAVE
HALF THIS =(S+12)/2=BOB GAVE
TOTAL GIVEN =S+S+S+12+(S+12)/2=81
{2(3S+12)+S+12}=2*81=162
6S+24+S+12=162
7S=162-36=126
S=126/7=18
HENCE SAM GAVE 18 $

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SEE THE FOLLOWING AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
Rate-of-work-word-problems/30847: Please help me with this problem. I think it needs to be set up as a linear equation. Also, it would help if you could explain it step by step for me. Thanks so much! ** An empty swimming pool can be filled in 10 hours. When full, the pool can be drained in 16 hours. How long will it take to fill the pool if the drain is left open?
1 solutions
Answer 17587 by venugopalramana(1167) About Me on 2006-03-20 21:52:19 (Show Source):
Please help me with this one. ** An empty swimming pool can be filled in 10 hours.
SO IN 1 HOUR WE CAN FILL 1/10 POOL
When full, the pool can be drained in 16 hours.
SO IN 1 HOUR 1/16 POOL WILL BE DRAINED
How long will it take to fill the pool if the drain is left open?
SO IN 1 HOUR 1/10 - 1/16 = (8-5)/80 = 3/80 POOL WILL BE FILLED
SO IT TAKES 80/3 = 26.67 HRS TO FILL THE POOL
Miscellaneous_Word_Problems/30717: Please help me with this one. ** An empty swimming pool can be filled in 10 hours. When full, the pool can be drained in 16 hours. How long will it take to fill the pool if the drain is left open?
1 solutions
Answer 17442 by venugopalramana(1167) About Me on 2006-03-19 21:28:52 (Show Source):
Please help me with this one. ** An empty swimming pool can be filled in 10 hours.
SO IN 1 HOUR WE CAN FILL 1/10 POOL
When full, the pool can be drained in 16 hours.
SO IN 1 HOUR 1/16 POOL WILL BE DRAINED
How long will it take to fill the pool if the drain is left open?
SO IN 1 HOUR 1/10 - 1/16 = (8-5)/80 = 3/80 POOL WILL BE FILLED
SO IT TAKES 80/3 = 26.67 HRS TO FILL THE POOL
Mixture_Word_Problems/30424: Dennis can eat a pie in 6 min. David can eat a pie in 12 min. Robert can eat a pie in 14.639145073927682623334797247841 min. If they share the same pie, how lomg will it take the three guys to scarf down the whole thing. (Hint: First figure out how many pies each guy can eat in an hour)
1 solutions
Answer 17065 by venugopalramana(1167) About Me on 2006-03-15 22:41:46 (Show Source):
Dennis can eat a pie in 6 min. David can eat a pie in 12 min. Robert can eat a pie in 14.639145073927682623334797247841 min. If they share the same pie, how lomg will it take the three guys to scarf down the whole thing. (Hint: First figure out how many pies each guy can eat in an hour)
DENIS CAN EAT IN 1 MINUTE....1/6 =0.166666667 PIE
DAVID CAN EAT IN 1 MINUTE....1/12 =0.083333333 PIE
ROBERT CAN EAT IN 1 MINUTE...1/14.63914507=0.06831 PIE
HENCE THE 3 TOGETHER CAN EAT IB 1 MINUTE=0.31831 PIE
HENCE THEY TAKE ...1/0.31831 = 3.14159153 MINUTES TO EAT THE PIE.
Rate-of-work-word-problems/29587: need help plz...heres the question "There are three friends named Allan, Bobby and Charlie. The three friends want to know their individual rate in finishing a job. Allan and Bobby can finish the job in 42 days, Bobby and Charlie can finish the job in 31 days, and Allan and Charlie can finish the job in 20 days. Solve the rate of each individual."
1 solutions
Answer 16715 by venugopalramana(1167) About Me on 2006-03-12 11:04:12 (Show Source):
need help plz...heres the question "There are three friends named Allan, Bobby and Charlie. The three friends want to know their individual rate in finishing a job. Allan and Bobby can finish the job in 42 days, Bobby and Charlie can finish the job in 31 days, and Allan and Charlie can finish the job in 20 days. Solve the rate of each individual."
LET ALLAN TAKE A DAYS TO DO THE JOB ALONE
HENCE ALLAN ALONE CAN DO IN 1 DAY 1/A JOB
LET BOBBY TAKE B DAYS TO DO THE JOB ALONE
HENCE BOBBY ALONE CAN DO IN 1 DAY 1/B JOB
LET CHARLIE TAKE C DAYS TO DO THE JOB ALONE
HENCE CHARLIE ALONE CAN DO IN 1 DAY 1/C JOB
FROM ABOVE WE GET ...
ALLAN AND BOBBY CAN DO IN 1 DAY ..1/A +1/B =(A+B)/AB JOB..
SO DAYS THEY TAKE TO COMPLETE THE JOB =AB/(A+B)=42..OR..1/A+1/B=1/42..........I BOBBY AND CHARLIE CAN DO IN 1 DAY ..1/B +1/C =(B+C)/BC JOB..
SO DAYS THEY TAKE TO COMPLETE THE JOB =BC/(B+C)=31..OR..1/B+1/C=1/31.........II
ALLAN AND CHARLIE CAN DO IN 1 DAY ..1/A +1/C =(A+C)/AC JOB..
SO DAYS THEY TAKE TO COMPLETE THE JOB =AC/(A+C)=20..OR..1/A+1/C=1/20........III
EQNI+EQNII+EQNIII GIVES...
2{(1/A)+(1/B)+(1/C)}=(1/42)+(1/31)+(1/20)
(1/A)+(1/B)+(1/C)=(1/2)*{(1/42)+(1/31)+(1/20)}.........IV
EQN.IV-EQN.I GIVES
1/C= (1/2)*{(1/42)+(1/31)+(1/20)}-1/42....OR.....C=34.2 DAYS
EQN.IV-EQN.II GIVES.......
1/A=(1/2)*{(1/42)+(1/31)+(1/20)}- 1/31...OR......A=48.1 DAYS
EQN.IV-EQN.III GIVES....
1/B=(1/2)*{(1/42)+(1/31)+(1/20)}- 1/20....OR.....B=329.6 DAYS

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Mixture_Word_Problems/30725: How many gallons of pure alcohol solution should be added to 63 gal of 32% alcohol solution to make a 37% solution? Thank you.
1 solutions
Answer 17458 by venugopalramana(1167) About Me on 2006-03-19 22:24:14 (Show Source):
PLEASE SEE THE FOLLOWING EXAMPLE AND TRY.IF STILL IN DIFFICULTY,PLEASE COME BACK
PLEASE HELP ASAP: Ziggy's famous yogurt blends regular yogurt that is 3% fat with its no fat yogurt to obtain low fat yogurt that is 1% fat. How many pounds of regular and how many pounds of non-fat yogurt should be mixed to obtain 60 pounds of lowfat yogurt.
PLEASE HELP ASAP. thank you
THESE ARE MATERIAL BALANCE PROBLEMS.THE PRINCIPLE IS TO APPLY
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS..
THIS PRINCIPLE CAN BE APPLIED TO TOTAL MIXTURE AS A WHOLE AS WELL AS INDIVIDUAL COMPONENTS OF THE MIXTURE.LET US SEE THE APPLICATION USING YOUR PROBLEM.
HERE THE MIXTURE COMPRISES 2 INPUTS-REGULAR YOGURT (RY) & NO FAT YOGURT (NFY)
AND ONE OUT PUT-LOW FAT YOGURT (LFY).THE COMPONENT OF IMPORTANCE IN THE MIXTURE IS FAT CONTENT.SO WE TAKE 2 BALANCES HERE ..ONE FOR THE TOTAL MIXTURE AND ANOTHER FOR COMPONENT OF FAT IN THE MIXTURE.
I..TOTAL BALANCE...
INPUTS
1.QTY.OF.RY=X POUNDS
2.QTY OF NFY=Y POUNDS
OUT PUT
1.QTY.OF LFY=60 POUNDS
SO APPLYING
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS.....WE GET
X+Y=60.............................I
II..COMPONENT BALANCE..HERE IT IS FAT .
INPUTS
1.QTY.OF FAT IN RY=X*3/100=3X/100 POUNDS
2.QTY OF FAT IN NFY=Y*0/100=0 POUNDS
OUT PUT
1.QTY.OF FAT IN LFY=60*1/100=60/100 POUNDS
SO APPLYING
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS.....WE GET
3X/100 + 0=60/100.............................II
3X=60
X=20 POUNDS. OF REGULAR YOGURT
Y=60-20=40 POUNDS OF NO FAT Yogurt
Mixture_Word_Problems/30587: The amount (by weight) of gold, silver and lead in three alloys of these metals are in ratios:
4:3:2 - Alloy 1
3:5:1 - Alloy 2
2:2:5 - Alloy 3
It is desired to make a fourth alloy containing equal amounts of gold, silver and lead. How many grams each alloy should be used for every 10 grams of the new alloy?
1 solutions
Answer 17262 by venugopalramana(1167) About Me on 2006-03-18 04:34:35 (Show Source):
The amount (by weight) of gold, silver and lead in three alloys of these metals are in ratios:
4:3:2 - Alloy 1
3:5:1 - Alloy 2
2:2:5 - Alloy 3
It is desired to make a fourth alloy containing equal amounts of gold, silver and lead. How many grams each alloy should be used for every 10 grams of the new alloy?
LET X GMS OF ALLOY1 ,Y GMS OF ALLOY2 AND 10-X-Y GMS OF ALLOY3 BE USED TO GET
X+Y+10-X-Y=10 GMS OF ALLOY 4
SO.................GOLD..............SIVER............LEAD IN THE MIX IS GIVEN BY
X GMS A1...........4X/9..............3X/9..........2X/9
Y GMS A2...........3Y/9..............5Y/9...........Y/9....
10-X-Y GMS A3....(20-2X-2Y)/9 ......(20-2X-2Y)/9...(50-5X-5Y)/9
-------------------------------------------------------------------------------
10 GMS A4.......(20+2X+Y)/9........(20+X+3Y)/9.....(50-3X-4Y)/9
THESE ARE ALL EQUAL...HENCE
20+2X+Y = 20+X+3Y...OR......................X-2Y=0..............I
20+2X+Y = 50-3X-4Y...OR...5X+5Y=30....OR....X+Y=6......II
EQN.II - EQN I...GIVES
X+Y-X+2Y=6......OR 3Y=6.....Y=2
SO X=6-Y=6-2=4
Z=10-4-2=4...
HENCE 4 GMS OF A1,2 GMS OF A2 AND 4 GMS OF A3 ARE TO BE ADDED TO GET 10 GMS OF A4.
Money_Word_Problems/29402: The nut store sells walnuts for 4.00 a pound and cashews for 7.00 a pound how many pounds of cashews should be mixed with 10 pounds of walnuts to obtain a mixture that sells for 5.50 a pound?
1 solutions
Answer 16262 by venugopalramana(1088) About Me on 2006-03-07 05:47:32 (Show Source):
SEE THE FOLLOWING EXAMPLE WHICH WILL HELP YOU TO GET AN INSIGHT INTO THESE PROBLEMS AND THE METHOD OF APPROACH TO SOLVE THEM
--------------------------------------------------------------------------
How much pure alcohol must a nurse add to 8cc of a 70% solution to strengthen it to a 90% solution? Just need to know if this is right and if not what is right.
.90(8)+ 0(x) = .70(x+8)....NO ...FIRST WRITE DOWN WHAT IS X?I THINK X IS QTY.OF
PURE ALCOHOL TO BE ADDED TO 8 CC OF 70% ALCOHOL TO GET X+8 CC OF 90% SOLUTION.SO THE EQN. SHOULD BE X*100/100+8*70/100=(X+8)*90/100
7.2 = .70x+5.6.....X+5.6=0.9(X+8)
1.6 = 70x.......X-0.9X=7.2-5.6=1.6
2.29c = x.......0.1X=1.6
..................OR X=1.6/0.1=16 CC.
SEE THE FOLLOWING EXAMPLE WHICH WILL HELP YOU TO GET AN INSIGHT INTO THESE PROBLEMS AND THE METHOD OF APPROACH TO SOLVE THEM
------------------------------------------------------------------------------
answer 15572 by venugopalramana(1008) About Me on 2006-02-26 07:15:31 (Show Source):
PLEASE HELP ASAP: Ziggy's famous yogurt blends regular yogurt that is 3% fat with its no fat yogurt to obtain low fat yogurt that is 1% fat. How many pounds of regular and how many pounds of non-fat yogurt should be mixed to obtain 60 pounds of lowfat yogurt.
PLEASE HELP ASAP. thank you
THESE ARE MATERIAL BALANCE PROBLEMS.THE PRINCIPLE IS TO APPLY
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS..
THIS PRINCIPLE CAN BE APPLIED TO TOTAL MIXTURE AS A WHOLE AS WELL AS INDIVIDUAL COMPONENTS OF THE MIXTURE.LET US SEE THE APPLICATION USING YOUR PROBLEM.
HERE THE MIXTURE COMPRISES 2 INPUTS-REGULAR YOGURT (RY) & NO FAT YOGURT (NFY)
AND ONE OUT PUT-LOW FAT YOGURT (LFY).THE COMPONENT OF IMPORTANCE IN THE MIXTURE IS FAT CONTENT.SO WE TAKE 2 BALANCES HERE ..ONE FOR THE TOTAL MIXTURE AND ANOTHER FOR COMPONENT OF FAT IN THE MIXTURE.
I..TOTAL BALANCE...
INPUTS
1.QTY.OF.RY=X POUNDS
2.QTY OF NFY=Y POUNDS
OUT PUT
1.QTY.OF LFY=60 POUNDS
SO APPLYING
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS.....WE GET
X+Y=60.............................I
II..COMPONENT BALANCE..HERE IT IS FAT .
INPUTS
1.QTY.OF FAT IN RY=X*3/100=3X/100 POUNDS
2.QTY OF FAT IN NFY=Y*0/100=0 POUNDS
OUT PUT
1.QTY.OF FAT IN LFY=60*1/100=60/100 POUNDS
SO APPLYING
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS.....WE GET
3X/100 + 0=60/100.............................II
3X=60
X=20 POUNDS. OF REGULAR YOGURT
Y=60-20=40 POUNDS OF NO FAT YOGURT.
logarithm/29283: If 800 mL of a juice drink is 15% grape juice, how much grape juice should be added to make a drink that is 20% grape juice?
1 solutions
Answer 16163 by venugopalramana(1088) About Me on 2006-03-06 11:02:24 (Show Source):
SEE THE FOLLOWING EXAMPLE WHICH WILL HELP YOU TO GET AN INSIGHT INTO THESE PROBLEMS AND THE METHOD OF APPROACH TO SOLVE THEM
--------------------------------------------------------------------------
How much pure alcohol must a nurse add to 8cc of a 70% solution to strengthen it to a 90% solution? Just need to know if this is right and if not what is right.
.90(8)+ 0(x) = .70(x+8)....NO ...FIRST WRITE DOWN WHAT IS X?I THINK X IS QTY.OF
PURE ALCOHOL TO BE ADDED TO 8 CC OF 70% ALCOHOL TO GET X+8 CC OF 90% SOLUTION.SO THE EQN. SHOULD BE X*100/100+8*70/100=(X+8)*90/100
7.2 = .70x+5.6.....X+5.6=0.9(X+8)
1.6 = 70x.......X-0.9X=7.2-5.6=1.6
2.29c = x.......0.1X=1.6
..................OR X=1.6/0.1=16 CC.
SEE THE FOLLOWING EXAMPLE WHICH WILL HELP YOU TO GET AN INSIGHT INTO THESE PROBLEMS AND THE METHOD OF APPROACH TO SOLVE THEM
------------------------------------------------------------------------------
answer 15572 by venugopalramana(1008) About Me on 2006-02-26 07:15:31 (Show Source):
PLEASE HELP ASAP: Ziggy's famous yogurt blends regular yogurt that is 3% fat with its no fat yogurt to obtain low fat yogurt that is 1% fat. How many pounds of regular and how many pounds of non-fat yogurt should be mixed to obtain 60 pounds of lowfat yogurt.
PLEASE HELP ASAP. thank you
THESE ARE MATERIAL BALANCE PROBLEMS.THE PRINCIPLE IS TO APPLY
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS..
THIS PRINCIPLE CAN BE APPLIED TO TOTAL MIXTURE AS A WHOLE AS WELL AS INDIVIDUAL COMPONENTS OF THE MIXTURE.LET US SEE THE APPLICATION USING YOUR PROBLEM.
HERE THE MIXTURE COMPRISES 2 INPUTS-REGULAR YOGURT (RY) & NO FAT YOGURT (NFY)
AND ONE OUT PUT-LOW FAT YOGURT (LFY).THE COMPONENT OF IMPORTANCE IN THE MIXTURE IS FAT CONTENT.SO WE TAKE 2 BALANCES HERE ..ONE FOR THE TOTAL MIXTURE AND ANOTHER FOR COMPONENT OF FAT IN THE MIXTURE.
I..TOTAL BALANCE...
INPUTS
1.QTY.OF.RY=X POUNDS
2.QTY OF NFY=Y POUNDS
OUT PUT
1.QTY.OF LFY=60 POUNDS
SO APPLYING
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS.....WE GET
X+Y=60.............................I
II..COMPONENT BALANCE..HERE IT IS FAT .
INPUTS
1.QTY.OF FAT IN RY=X*3/100=3X/100 POUNDS
2.QTY OF FAT IN NFY=Y*0/100=0 POUNDS
OUT PUT
1.QTY.OF FAT IN LFY=60*1/100=60/100 POUNDS
SO APPLYING
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS.....WE GET
3X/100 + 0=60/100.............................II
3X=60
X=20 POUNDS. OF REGULAR YOGURT
Y=60-20=40 POUNDS OF NO FAT YOGURT.

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Find a polynomial equation of lowest degree with integer coefficients and has 0, 3, and plus/minus sq rt of 2 as zeros. Please, anyone help.
ROOTS ARE ...0,3,SQRT(2),-SQRT(2)...SO EQN IS
(X-0)(X-3)(X-SQRT2)(X+SQRT2)=0
(X^2-3X)(X^2-2)=0
X^4-3X^3-2X^2+6X=0

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y = (x - 2)^2
VERTEX IS WHERE Y BECOMES MAXIMUM OR MINIMUM...WE FIND THAT IT BECOMES MINIMUM HERE OF ZERO WHEN X=2...HENCE VERTEX IS X=2 AND Y=0...OR...(2,0)

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If we apply the quadratic formula and find that the value of b^2 - 4ac equals zero, what can we conclude about the solutions?
A) The equation has no real number solutions.
B) The equation has exactly one irrational solution.
C) The equation has two different rational solutions
D) The equation has exactly one rational solution.


X=-B/2A......
The equation has exactly one rational solution.

D IS THE ANSWER..

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Graphs/24460: find the slope,m,and the y intercept,b,of the line equation 2y=28
1 solutions
Answer 13033 by venugopalramana(1088) About Me on 2006-01-15 10:53:09 (Show Source):
SEE THE FOLLOWING WHICH IS SIMILAR TO YOUR PROBLEM AND DO
GIVEN:
· There is a line (L1) that passes through the points
(8,-3) and (3,3/4).
· There is another line (L2) with slope m=2/3 that
intersects L1 at the point (-4,6).
· A third line (L3) is parallel to L2 that passes
through the (7,-13 1/2).
· Yet another line (L4) is perpendicular to L3, and
passes through the point (1/2,5 2/3).
· The fifth line (L5) has the equation
2/5y-6/10x=24/5.
Using whatever method, find the following:
2. The point of intersection of L1 and L3
3. The point of intersection of L1 and L4
4. The point of intersection of L1 and L5
5. The point of intersection of L2 and L3
6. The point of intersection of L2 and L4
7. The point of intersection of L2 and L5
8. The point of intersection of L3 and L4
9. The point of intersection of L3 and L5
10. The point of intersection of L4 and L5
PLEASE NOTE THE FOLLOWING FORMULAE FOR EQUATION OF A
STRAIGHT LINE:
slope(m)and intercept(c) form...y=mx+c
point (x1,y1) and slope (m) form...y-y1=m(x-x1)
two point (x1,y1)and(x2,y2)form.....................
y-y1=((y2-y1)/(x2-x1))*(x-x1)
standard linear form..ax+by+c=0..here by transforming
we get by=-ax-c..or y=(-a/b)x+(-c/b)..comparing with
slope intercept form we get ...slope = -a/b and
intercept = -c/b
*****************************************************
line (L1) that passes through the points (8,-3) and
(3,3/4).
eqn.of L1..y-(-3)=((3/4+3)/(3-8))*(x-8)
y+3=((15/4(-5)))(x-8)=(-3/4)(x-8)
or multiplying with 4 throughout
4y+12=-3x+24
3x+4y+12-24=0
3x+4y-12=0.........L1
There is another line (L2) with slope m=2/3 that
intersects L1 at the point (-4,6).
This means (-4,6)lies on both L1 and L2.(you can check
the eqn.of L1 we got by substituting this point in
equation of L1 and see whether it is satisfied).So
eqn.of L2...
y-6=(2/3)(x+4)..multiplying with 3 throughout..
3y-18=2x+8
-2x+3y-26=0.........L2
A third line (L3) is parallel to L2 that passes
through the (7,-13 1/2).
lines are parallel mean their slopes are same . so we
keep coefficients of x and y same for both parallel
lines and change the constant term only..
eqn.of L2 from above is ...-2x+3y-26=0.........L2
hence L3,its parallel will be ...-2x+3y+k=0..now it
passes through (7,-13 1/2)=(7,-13.5)......substituting
in L3..we get k
-2*7+3*(-13.5)+k=0...or k=14+40.5=54.5..hence eqn.of
L3 is........-2x+3y+54.5=0......................L3
Yet another line (L4) is perpendicular to L3, and
passes through the point (1/2,5 2/3).
lines are perpendicular when the product of their
slopes is equal to -1..so ,we interchange coefficients
of x and y from the first line and insert a negative
sign to one of them and then change the constant term.
L3 is........-2x+3y+54.5=0......................L3
hence L4,its perpendicular will be ..3x+2y+p=0...L4
this passes through (1/2,5 2/3)=(1/2,17/3).hence..
3*1/2+2*17/3+p=0..or ..p= -77/6.so eqn.of L4 is
3x+2y-77/6=0..............L4
The fifth line (L5) has the equation 2/5y-6/10x=24/5.
now to find a point of intersection means to find a
point say P(x,y) which lies on both the lines ..that
is, it satisfies both the equations..so we have to
simply solve the 2 equations of the 2 lines for x and
y to get their point of intersection.For example to
find the point of intersection of L1 and L3 we have to
solve for x and y the 2 equations....
3x+4y-12=0.........L1....(1) and
-2x+3y+40.5=0......L3.....(2)
I TRUST YOU CAN CONTINUE FROM HERE TO GET THE
ANSWERS.If you have any doubts or get into any
difficulty ,please ask me.

You can put this solution on YOUR website!
Graphs/24460: find the slope,m,and the y intercept,b,of the line equation 2y=28
1 solutions
Answer 13033 by venugopalramana(1088) About Me on 2006-01-15 10:53:09 (Show Source):
SEE THE FOLLOWING WHICH IS SIMILAR TO YOUR PROBLEM AND DO
GIVEN:
· There is a line (L1) that passes through the points
(8,-3) and (3,3/4).
· There is another line (L2) with slope m=2/3 that
intersects L1 at the point (-4,6).
· A third line (L3) is parallel to L2 that passes
through the (7,-13 1/2).
· Yet another line (L4) is perpendicular to L3, and
passes through the point (1/2,5 2/3).
· The fifth line (L5) has the equation
2/5y-6/10x=24/5.
Using whatever method, find the following:
2. The point of intersection of L1 and L3
3. The point of intersection of L1 and L4
4. The point of intersection of L1 and L5
5. The point of intersection of L2 and L3
6. The point of intersection of L2 and L4
7. The point of intersection of L2 and L5
8. The point of intersection of L3 and L4
9. The point of intersection of L3 and L5
10. The point of intersection of L4 and L5
PLEASE NOTE THE FOLLOWING FORMULAE FOR EQUATION OF A
STRAIGHT LINE:
slope(m)and intercept(c) form...y=mx+c
point (x1,y1) and slope (m) form...y-y1=m(x-x1)
two point (x1,y1)and(x2,y2)form.....................
y-y1=((y2-y1)/(x2-x1))*(x-x1)
standard linear form..ax+by+c=0..here by transforming
we get by=-ax-c..or y=(-a/b)x+(-c/b)..comparing with
slope intercept form we get ...slope = -a/b and
intercept = -c/b
*****************************************************
line (L1) that passes through the points (8,-3) and
(3,3/4).
eqn.of L1..y-(-3)=((3/4+3)/(3-8))*(x-8)
y+3=((15/4(-5)))(x-8)=(-3/4)(x-8)
or multiplying with 4 throughout
4y+12=-3x+24
3x+4y+12-24=0
3x+4y-12=0.........L1
There is another line (L2) with slope m=2/3 that
intersects L1 at the point (-4,6).
This means (-4,6)lies on both L1 and L2.(you can check
the eqn.of L1 we got by substituting this point in
equation of L1 and see whether it is satisfied).So
eqn.of L2...
y-6=(2/3)(x+4)..multiplying with 3 throughout..
3y-18=2x+8
-2x+3y-26=0.........L2
A third line (L3) is parallel to L2 that passes
through the (7,-13 1/2).
lines are parallel mean their slopes are same . so we
keep coefficients of x and y same for both parallel
lines and change the constant term only..
eqn.of L2 from above is ...-2x+3y-26=0.........L2
hence L3,its parallel will be ...-2x+3y+k=0..now it
passes through (7,-13 1/2)=(7,-13.5)......substituting
in L3..we get k
-2*7+3*(-13.5)+k=0...or k=14+40.5=54.5..hence eqn.of
L3 is........-2x+3y+54.5=0......................L3
Yet another line (L4) is perpendicular to L3, and
passes through the point (1/2,5 2/3).
lines are perpendicular when the product of their
slopes is equal to -1..so ,we interchange coefficients
of x and y from the first line and insert a negative
sign to one of them and then change the constant term.
L3 is........-2x+3y+54.5=0......................L3
hence L4,its perpendicular will be ..3x+2y+p=0...L4
this passes through (1/2,5 2/3)=(1/2,17/3).hence..
3*1/2+2*17/3+p=0..or ..p= -77/6.so eqn.of L4 is
3x+2y-77/6=0..............L4
The fifth line (L5) has the equation 2/5y-6/10x=24/5.
now to find a point of intersection means to find a
point say P(x,y) which lies on both the lines ..that
is, it satisfies both the equations..so we have to
simply solve the 2 equations of the 2 lines for x and
y to get their point of intersection.For example to
find the point of intersection of L1 and L3 we have to
solve for x and y the 2 equations....
3x+4y-12=0.........L1....(1) and
-2x+3y+40.5=0......L3.....(2)
I TRUST YOU CAN CONTINUE FROM HERE TO GET THE
ANSWERS.If you have any doubts or get into any
difficulty ,please ask me.

You can put this solution on YOUR website!
A regular Pentagon ABCDE is incribed in circle O. Chords EC and DB intersect at F (not diameter). Chord DB is Extended to G (outside circle), and Tangent GA is drawn.
find measure of angle BDE
find measure of angle BFC
find measure of angle AGD
HOPE YOU HAVE A FIGURE OF THE ABOVE..ASSUMING THAT I AM GIVING THE SOLUTION.
SUM OF ALL INTERIOR ANGLES IN A PENTAGON =(2*5-4)*90 =540
SINCE IT IS REGULAR PENTAGON EACH ANGLE =540/5=108
AS PER GIVEN DATA,A,B,E,D LIE ON A CIRCLE...SO ABED IS A CYCLIC QUADRILATERAL..HENCE SUM OF OPPOSITE ANGLES=180=ANGLE EAB+ANGLE BDE
BUT ANGLE EAB=108 AS PROVED ABOVE
-----------------------------------------
HENCE ANGLE BDE=180-108=72
-------------------------------------------
SIMILARLY FROM CYCLIC QUADRILATERAL ECBA,WE GET ANGLE ECB=72
NOW IN A REGULAR PENTAGON EACH SIDE SPANS 1/5 OF PERIMETER OF CIRCUM CIRCLE WITH O AS CENTRE.HENCE EACH SIDE SUBTENDS AN ANGLE OF 360/5=72 AT CENTRE AND HALF OF THAT THAT IS 36 AT THE CIRCUMFERENCE..
HENCE ANGLE ADB=ANGLE DBC =36
NOW IN TRIANGLE CFB...WE HAVE...
ANGLE FCB =ANGLE ECB =72..PROVED ABOVE.
ANGLE FBC=ANGLE DBC=36......PROVED ABOVE
--------------------------------------------
HENCE ANGLE BFC=180-72-36=72
--------------------------------
ABDE IS A CYCLIC QUADRILATERAL...AB IS A CHORD...AG IS A TANGENT TO CIRCUM CIRCLE..HENCE ANGLE BAG = ANGLE IN THE OPPOSITE SEGMENT =ANGLE ADB=36..PROVED ABOVE
HENCE ANGLE EAG=ANGLE EAB + ANGLE BAG=108+36=144
IN QUADRILATERALDEAG...WE HAVE
ANGLE AGD=360-ANGLE EDB-ANGLE EAG-ANGLE DEA
=360-72-144-108=36
--------------------------
HENCE ANGLE AGD=36
-------------------------------

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SEE THE FOLLOWING AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK.
---------------------------------------------------------------
Linear_Algebra/30362: Question: Find the equation of the ellipse whose center is (5,-3) that has a vertex at 13,-3) and a minor axis of lenght 10.
POssible Answers:
(A) (x-5)^2/64 + (y+3)^2/25 = 1
(B) (x+5)^2/64 + (y-3)^2/25 = 1
(C) x^2/64 + y^2/25 = 1
(D) none of these
1 solutions
Answer 17014 by venugopalramana(1167) About Me on 2006-03-15 11:21:03 (Show Source):
SEE THE FOLLOWING AND TRY..IF STILL IN DIFFICULTY PLEASE COME BACK...
OK I WORKED IT OUT FOR YOU NOW
I TOLD YOU EQN IS
(X-H)^2/A^2 + (Y-K)^2/B^2....
WHERE H,K IS CENTRE...SO H=5 AND K=-3 AS CENTRE IS GIVEN AS (5,-3)....NOW VERTEX IS (13,-3)...IT LIES ON ELLIPSE..SO IT SATISFIES THE EQN
(13-5)^2/A^2 +(-3+3)^2/B^2 =1
HENCE A^2=64...OR A=8
MINOR AXIS =10=2B...HENCE B=5..SO EQN.S
(X-H)^2/64 + (Y+3)^2/25 =1
THAT IS A IS CORRECT.

Can you help me write an equation for an ellipse with a major axis with endpoints of (0,8), and (0,-8) with foci of (0,5) and (0,-5)?
1 solutions
--------------------------------------------------------------------------------
Answer 16810 by venugopalramana(1120) on 2006-03-13 11:19:12 (Show Source):
Can you help me write an equation for an ellipse with a major axis with endpoints of (0,8), and (0,-8) with foci of (0,5) and (0,-5)?
THIS SHOWS THAT X AXIS IS THE MAJOR AXIS
STANDARD EQN.OF ELLIPSE IS
(X-H)^2/A^2 +(Y-K)^2/B^2=1
CENTRE IS (H,K)..AS PER THE PROBLEM H=K=0 AS CENTRE OF ELLIPSE IS AT (0,0)..SINCE major axis with endpoints ARE (0,8), and (0,-8)
WHERE MAJOR AXIS =2A=8+8=16...SO A=8..SINCE major axis with endpoints ARE (0,8), and (0,-8)
FOCI ARE GIVEN BY
AE,0 AND -AE,0...SO AE =5...SO E=5/A=5/8
BUT E=SQRT{(A^2-B^2)/A^2}=5/8...SQUARING
25/64=(A^2-B^2)/A^2=1-B^2/A^2
B^2/64=1-25/64=49/64
B^2=49
B=7
HENCE EQN. OF ELLIPSE IS
X^2/64 + Y^2/49 = 1

Quadratic-relations-and-conic-sections/30009: Can you help me write an equation for an ellipse with a major axis with endpoints of (0,8), and (0,-8) with foci of (0,5) and (0,-5)?

1 solutions
Answer 16810 by venugopalramana(1167) About Me on 2006-03-13 11:19:12 (Show Source):
Can you help me write an equation for an ellipse with a major axis with endpoints of (0,8), and (0,-8) with foci of (0,5) and (0,-5)?
THIS SHOWS THAT X AXIS IS THE MAJOR AXIS
STANDARD EQN.OF ELLIPSE IS
(X-H)^2/A^2 +(Y-K)^2/B^2=1
CENTRE IS (H,K)..AS PER THE PROBLEM H=K=0 AS CENTRE OF ELLIPSE IS AT (0,0)..SINCE major axis with endpoints ARE (0,8), and (0,-8)
WHERE MAJOR AXIS =2A=8+8=16...SO A=8..SINCE major axis with endpoints ARE (0,8), and (0,-8)
FOCI ARE GIVEN BY
AE,0 AND -AE,0...SO AE =5...SO E=5/A=5/8
BUT E=SQRT{(A^2-B^2)/A^2}=5/8...SQUARING
25/64=(A^2-B^2)/A^2=1-B^2/A^2
B^2/64=1-25/64=49/64
B^2=49
B=7
HENCE EQN. OF ELLIPSE IS
X^2/64 + Y^2/49 = 1

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SEE THE FOLLOWING EXAMPLES AND TRY..IF STILL IN DOUBT PLEASE COME BACK
I need help with usings Cramer's Rule. The question is: Solve the following system of equations using Cramer's Rule.
2x + 4y - 3z = 2
2x - 2y + 3z = 3
3x - 4y + 5z = 4
I understand that I have to first have to set up the determinants. Which are
D = 2 4 -3 Answer Column is 2
2 -2 3 3
3 -4 5 4
D of x = 2 4 -3
3 -2 3
4 -4 5
D of y = 2 2 -3
2 3 3
3 4 5
D of Z = 2 4 2
2 -2 3
3 -4 4
Now my problem is how do I evaluate the determinants? I think that I understand how to set it up. I'm just not sure what to do after that. Any help you can give me would be greatly appreciated. I'm not necessarily looking for the entire question, just an explanation of what to do next. I've looked over the web for hours and none of the Algebra websites explains how to do this step. Thanks for the help and have a great night.
Jonna
1 solutions
Answer 19794 by venugopalramana(1400) About Me on 2006-04-12 11:21:15 (Show Source):
IT IS REALLY HEARTENING TO SEE YOUR SINCERE ENDEAVOUR TO LEARN AND DO INSTEAD OF GETTING THE READY MADE SOLUTIONS.THIS APPROACH WILL TAKE YOU A LONG WAY UP THE LADDER IN FUTURE.KEEP IT UP.WE ARE ALWAYS THERE TO HELP SUCH DESERVING STUDENTS.
-----------------------------------------------------------
SEE BELOW MY COMMENTS AND SUGGESTIONS AND THE EXAMPLE AT THE END...........
I need help with usings Cramer's Rule. The question is: Solve the following system of equations using Cramer's Rule.
2x + 4y - 3z = 2
2x - 2y + 3z = 3
3x - 4y + 5z = 4
I understand that I have to first have to set up the determinants. Which are
D = 2 4 -3 Answer Column is 2
2 -2 3 3
3 -4 5 4
------------------------------------------------------------------
OK…BUT LET US INTRODUCE SOME NOMENCLATURE TO MAKE IT EASY TO REMEMBER.
FIRST MAKE D= DETERMINANT OF COEFFICIENTS.THAT IS WRITE COEFFICIENTS OF X,Y AND Z IN THE SAME ORDER IN ALL 3 EQNS.AS SHOWN BELOW. WE PUT THEM APART TO WRITE COLUMN NUMBERS.
COLUMN I(X)....COL.II(Y)...COL.III(Z)
2,................4,.........-3..............I ROW
2,...............-2,..........3...............II ROW
3,...............-4,..........5..................III ROW
OK ...NOW WE SHALL FIND ITS VALUE...WE DO IT IN A STEP-WISE PROCESS AS GIVEN BELOW..THERE ARE 3 ROWS AND 3 COLUMNS.WE CAN FIND ITS VALUE USING ANY ONE ROW OR COLUMN.LET US USE FIRST COLUMN TO EXPAND IN THIS EXAMPLE.FIRST COLUMN HAS 3 ELEMENTS...2,2,AND 3.LET US CALL THEM E11,E21 AND E31 TO SHOW THAT THEY ARE FROM ROW1,COL.1;ROW2,COL.1;AND ROW3,COL.1.THE FORMULA FOR VALUE OF DETERMINANT IS
D=(E11)*(S11)*(M11)+(E21)*(S21)*(M21)+(E31)*(S31)*(M31)……………………………………..I
THAT IS WE MULTIPLY EACH ELEMENT WITH ITS CO-FACTOR AND ADD ALL THE RESULTS.COFACTOR OF AN ELEMENT IS PRODUCT OF 2 NUMBERS...ONE IS FOR SIGN(S11,S21,S31) AND ANOTHER FOR VALUE(M11,M21,M31).
THE NUMBER FOR SIGN IS (-1)^(ROW NUMBER+COLUMN NUMBER)OF THE ELEMENT.HERE THE FIRST ELEMENT 2 IS FROM ROW1 AND COL.1.SO ITS SIGN NUMBER IS (-1)^(1+1)=(-1)^2=+1=S11
THE NUMBER FOR VALUE IS CALLED THE MINOR OF THE ELEMENT.IT IS OBTAINED BY DELETING THE ROW AND COLUMN CONTAINING THE ELEMENT.OUR ELEMENT IS 2 FROM ROW1 AND COLUMN 1, AS GIVEN ABOVE.SO REMOVE THE ROW1 AND COLUMN 1.SO WE GET A 2 BY 2 DETERMINANT.IT IS
-2,3
-4,5
LET US CALL THIS EQUAL TO M11 TO INDICATE IT IS MINOR OF ELEMENT 11 (E11)THAT IS ROW 1 AND COLUMN 1.
SO THE FIRST ELEMENT GIVES US ITS CONTRIBUTION AS...(2)*(+1)(M11)...WE SHALL FIND THE VALUE OF M11 IN THE SAME WAY AS ABOVE.BUT LET US SKIP THIS FOR A MOMENT AND COMPLETE OUR JOB.
NOW TAKING THE NEXT NUMBER 2 AGAIN WHICH IS FROM ROW 2 AND COL.1.SO ITS
SIGN =S21=(-1)^(2+1)=(-1)^3=-1
MINOR IS = M21 GIVEN BY
4,-3
-4,5
SO CONTRIBUTION OF E21 IS
(2)(-1)(M21)
SIMILARLY CONTRIBUTION OF E31 IS
(3)*{(-1)^(3+1)}*(M31)=(3)*(1)*(M31) WHERE M31 IS THE 2 BY 2 DETERMINANT
4,-3
-2,3
SO NOW WE GOT THE 3 BY 3 DETERMINANT CONVERTED TO 2 BY 2 DETERMINANT.THE SAME METHOD WE CAN USE TO EVALUATE ANY ORDER DETERMINANT BY REDUCING IT TO A LOWER ORDER DETERMINANT IN EACH STEP.4 BY 4 TO 3 BY 3 TO 2 BY 2 WHICH IS THE END AS WE SHOW BELOW.
LET US FINISH THIS NOW BY ONE EXAMPLE OF 2 BY 2 DETERMINANT SAY M11
-2,3
-4,5
M11=BY THE SAME METHOD ….=(-2)(1)(5)+(-4)(-1)(3)=-10+12=2
SO BY THIS WAY WE CAN FIND M21 AND M31 AND HENCE FIND D BY THE ABOVE FORMULA-I...VIZ.. D=(E11)*(S11)*(M11)+(E21)*(S21)*(M21)+(E31)*(S31)*(M31)……………………………………..I
NEXT MAKE DX…..BY REPLACING COEFFICIENTS OF X IN D WITH CONSTANT TERMS…SO DX IS
COLUMN I(CONSTANTS)....COL.II(Y)...COL.III(Z)
2,.………………………...............4,.........-3..............I ROW
3,………………………...............-2,..........3...............II ROW
4,...………………………............-4,..........5..................III ROW
-------------------------
D of x = 2 4 -3
3 -2 3
4 -4 5
SO YOURS IS OK TOO
------------------
SIMILARLY DY IS
COLUMN I(X)....COL.II(CONSTANT)...COL.III(Z)
2,…………………….............…2,.........-3..............I ROW
2,...…………………….........…3,..........3...............II ROW
3,........…………………….......4,..........5..................III ROW --------------------------------
D of y = 2 2 -3
2 3 3
3 4 5
SO YOURS IS OK
----------------------
SIMILARLY DZ IS
COLUMN I(X)....COL.II(Y)...COL.III(CONSTANT)
2,...........………….....4,......….2..............I ROW
2,.........…………......-2,..........3...............II ROW
3,......……………......-4,.......…4..................III ROW
-----------------------
D of Z = 2 4 2
2 -2 3
3 -4 4
SO YOURS IS OK TOO
------------------------------
Now my problem is how do I evaluate the determinants? I think that I understand how to set it up. I'm just not sure what to do after that. Any help you can give me would be greatly appreciated. I'm not necessarily looking for the entire question, just an explanation of what to do next. I've looked over the web for hours and none of the Algebra websites explains how to do this step. Thanks for the help and have a great night.
Jonna
HOPE NOW YOUR PROBLEM IS SOLVED AS FAR AS EVALUATING THE DETERMINANTS IS COCERNED.LET US FINISH BY GIVING FINAL FORMULAE TO FIND X,Y AND Z
X=DX/D……..;…..Y=DY/D……………..; AND………………….Z=DZ/D………………………………….II
----------------------------------------------------------------------------------
NOW SEE THE FOLLOWING ADDITIONAL PROBLEMS WORKED EARLIER.
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Matrices-and-determiminant/17820: Please help I have no idea how to do this problem: Use Cramer's Rule to solve each system.
1. 2x+y=4
3x-y=6


2. 2x+3y+ z= 5
x+y-2z= -2
-3x +z=-7
1 solutions
Answer 8586 by venugopalramana(1088) About Me on 2005-10-31 03:43:14 (Show Source):
2x+y=4
3x-y=6
make a deteminant with coefficients of x (2,3)and y(1,-1) in the 2 eqns.call it C.(Actually for a determinant as you know ,the numbers are contained in vertical bars at either end like |xx|,but in the following the bars are omitted due to difficulty in depiction.you may assume the bars are present)
C=matrix(2,2,2,1,3,-1)=2*(-1)-(1*3)=-5
..now use the constants (4,6)to replace coefficients of x(2,3) in the above determinant C...call it CX..
CX=matrix(2,2,4,1,6,-1)=4*(-1)-1*6=-4-6=-10
..now use the constants (4,6)to replace coefficients of y(1,-1) in the above determinant C...call it CY..
CY=matrix(2,2,2,4,3,6)=2*6-3*4=12=12=0
..now cramers rule says that
(x/CX)=(y/CY)=(1/C)..so we get
x/(-10)=y/0=1/-5
x=-10/-5=10/5=2
y=0/-5=0
************************************
so using the above method you can do the next problem ..here due to presence of 3 variables you will get 3rd.order determinants...4 in all...namely C,CX,CY and CZ,the last formula also extends to include z ,
(x/CX)=(y/CY)=(z/CZ)=(1/C)..
but the procedure is same ..
2x+3y+ z= 5
x+y-2z= -2
-3x +z=-7 ...
...just to give you the idea
C=matrix(3,3,2,3,1,1,1,-2,-3,0,1)..and
CZ=matrix(3,3,2,3,5,1,1,-2,-3,0,7)..etc..hope you can work out the rest
-------------------------------------------------------------

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For each of the following equations either prove that it is correct (by using the rules of logarithms and exponents) or else show that it is not correct (by finding numerical values for the variables that make the values of the two sides of the equation different).
log (x/y) = log x/ log y
LHS....LOG(10/1)=LOG(10)=1...BUT RHS... LOG(10)/LOG(10)=1/0.....NOT CORRECT
log x - log y = log (x/y)
OK..AS PER FORMULA..
log (2x) = 2 log x
LHS.....LOG(2*1)=LOG(2).....RHS....2*LOG(1)=2*0=0....NOT CORRECT
2 log x = log (x^2)
CORRECT AS PER FORMULA LOG(X)^N=N*LOG(X)
log {(x+1) / (x+3)} = log (x+1) - log (x+3)...I SUPPOSE THIS IS THE PROBLEM.
CORRECT AS PER FORMULA...LOG(A/B)=LOG(A)-LOG(B)
log {(x *square root (x^2 + 1)} = log x + 1/2 (x^2 +1)
WRONG...LHS....LOG{1*SQRT(1^2+1)}=LOG{SQRT(2)}
RHS......LOG(1)+(1/2)*(1+1)=0+1=1...WHICH IS NOT CORRECT.
log (x^2 +1) = 2 log x + log 1
WRONG...LHS.....LOG(1^2+1)=LOG(2)
RHS.....2*LOG(1)+LOG(1)=0+0=0.......WHICH IS NOT CORRECT...

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If a+b+c=0,then 3a*x*x+2b*x+c=0 has
A.atleast one root in [-1,0]
B.atleast one root in [0,1]
C.atleast one root in [-1,1]
D.atleast one root in [0,2]
CONSIDER F(X)=AX^3+BX^2+CX...
F(0)=0
F(1)=A+B+C=0...... GIVEN
HENCE F(X) HASS 2 ZEROS AT EITHER END OF THE INTERVAL [0,1]
DIFFERENTIATING F(X)... W.R.T.X,WE GET
DF/DX=3AX^2+2BX+C....AS PER RTO;;ES THEOREM THIS MUST HAVE A ZERO IN THE INTERVAL [0,1]................SO...............B..IS CORRECT

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XY=Z....................I
Z=85000....................II
Y=0.09.............................III
SUBSTITUTING II AND III IN I
X*0.09=85000
X=85000/9=9444.44

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An elevator containing 5 people can stop at any one of 7 floors. What is the probability that no 2 people get off on the same floor? (Assume each floor is equally like for each person)
LET US DEVELOP ALOGIC LIKE THIS
THE FIRST PERSON CAN GET DOWN AT ANY ONE OF THE 7 FLOORS..IN....7 WAYS.
THE SECOND PERSON CAN GET DOWN AT ANY FLOOR OTHER THAN THE ABOVE THAT IS IN 6 WAYS
LIKE THAT IF WE CONTINUE WE GET THE ANSWER AS
7*6*5*4*3=2520 WAYS

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I have a right triangle with one known angle in addition to the right angle and the length of the side (b) between these two angles. The length of side (b) is five. How do I find the length of side (a)? Side (c) is also unknown.
LET THE TRIANGLE BE ABC
b=AC=5
ANGLE BAC=90
ANGLE ACB IS KNOWN= C SAY
YOU REFER TABLES OR A CALCULATOR TO FIND COS(C) AND TAN(C) TO DETERMINE SIDES BC=a AND c=AB USING THE FOLLOWING EQUATIONS
COS(C)=AC/BC=5/a....OR.......a=5/COS(C)...KNOWING COS(C)FROM CALCULATOR / TABLE YOU CAN FIND a...IF ANGLE C=60 SAY....THEN COS(C)=COS(60)=0.5....a=5/0.5=10
TAN(C)=AB/AC=c/5....OR.......c=5*TAN(C)...KNOWING TAN(C)FROM CALCULATOR / TABLE YOU CAN FIND c...IF ANGLE C=60 SAY....THEN TAN(C)=TAN(60)=SQRT(3)=1.732....c=5*1.732=8.66

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log(10x)^2=2+2logx This one is false right? Thank you for helping me.
LHS=2LOG(10X)=2{LOG(10)+LOG(X)}=2+2LOG(X)

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ln(1/x)+lnx=0
LHS=LN(X*1/X)=LN(1)=0

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log(2x)=2logx I believe this one is true. Am I right? Thank you.
NO THE FORMULA IS LOG(A*B)=LOG(A)+LOG(B)
SO......LOG(2X)=LOG(2)+LOG(X)

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WHEN WE CUT X LONG PIECES ON ALL 4 SIDES THE CARD BOARD WILL GET REDUCED BY
X+X=2X...ALONG LENGTH AND...X+X=2X.....ALONG WIDTH
SO OPEN BOX LENGTH = 8-2X AND WIDTH = 6-2X..AND HEIGHT =X ...SO VOLUME V IS GIVEN BY LEMGTH*WIDTH*HEIGHT
V=(8-2X)(6-2X)X...DOMAIN OF V IS GIVEN BY THE FACT THAT LENGTH OR WIDTH CAN NOT BE NEGATIVE...CRITICAL VALUE BEING WIDTH WE GET ....
8-2X>0...AND 6-2X>0...SO X <3
RANGE.....MAXIMUM VALUE........
V=X(8-2X)(6-2X)=X{48-16X-12X+4X^2)=4X^3-28X^2+48X...IF YOU KNOW CALCULUS
DV/DX=12X^2-56X+48=0..OR...3X^2-14X+12=0....
X=(14+SQRT.(52))/6...OR......(7+SQRT.(13))/3...OR....(7-SQRT.13)/3
X=3.54..OR...1.13.
D2V/DX2=6X-14=- VE AT X=1.13...SO MAXIMUM VOLUME IS OBTAINED AT X=1.13'
YOU CAN SEE IT BY PLOTTING THE GRAPH.

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The volume of a cylinder (think about the volume of a can) is given by V = pr2h where r is the radius of the cylinder and h is the height of the cylinder. Suppose the volume of the can is 100 cubic centimeters.
a)Write h as a function of r.
Answer
100=PI*R^2*H
H=100/(PI*R^2)
b)What is the measurement of the height if the radius of the cylinder is 2 centimeters?
Answer
H=100/(PI*2^2)=25/PI=7.96 CM.
Show work in this space
c)Graph this function.
Show graph here

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SCALE IS 1:3000,000 MEANS 1 MM IN PICTURE IS ACTUALLY 3000000 MM IN REALITY...OR...3000 METRES....OR.....3 KM.IN ACTUAL
SO IF WE HAVE 25 MM IN PICTURE IT IS EQUAL TO 25*3=75 KM.

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KEEP YOUR OBJECTIVE CLEAR.....
Find the asymptotes of the hyperbola. Then sketch its graph.
35.
I have spent well over an hour attempting to solve this problem, and have had others try to help me with no succes. There is the answer in the back of the book, but i still want to know how to solve the problem. I hope that you will be able to help me, and tell me where I have gone wrong.
SEE MY COMMENTS BELOW AND THE EXAMPLE AT THE END.
Tried methods
1) multiply by 36 to get rid of 36 in the denominator, and then multiply the other side of the equation by 36, I then added the section over to the other side of the equation. Then I divided both sides by 25
YOU ARE MISSING THE KEY POINT...SEE BELOW
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I know the Standard Equation for a hyperbola is for horizontal and for vertical. After that I am somewhat lost...
THIS IS THE KEY..TRY TO GET THE EQN.IN THIS FORM
SEND 25 TO THE DENOMINATOR.THAT IS ALL...THAT IS
X^2/(36/25) - Y^2/(121)=1...SO A^2=36/25...OR...A=6/5 AND B^2=121...OR...B=11
OK BUT WE WANT ASYMPTOTES..SO...
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SEE THE FOLLOWING EXAMPLES TO KNOW THE METHOD.AND FINALLY ANSWER FOR YOUR PROBLEM AT THE END.
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Quadratic-relations-and-conic-sections/28184: What is the vertices, foci, and slope of the asymptotes for the hyperbola whose equation is, y^2/16 - x^2/25 =25?
SEE THE FOLLOWING AND YOU SHOULD BE ABLE TO SOLVE YOUR PROBLEM BY YOUR SELF.....
THE ANSWERS FOR YOUR CASE...H=0...K=0..A=4...B=5....EQN IS OF THE TYPE
(Y-K)^2/B^2-(X-H)^2/A^2=1....
SO VERTICES ARE...(H,(K-B)) AND (H,(K+B)) ...(0,-5) AND (0,5)
FOCI ARE {H,(K-BE)} AND {H,(K+BE)}...WHERE E IS
ECCENTRICITY =SQRT{(A^2+B^2)/B^2}=SQRT((16+25)/25)=SQRT(41/25)
SO FOCI ARE =(0,-5SQRT(41/25) AND (0,5SQRT(41/25)...
OR....(0,-SQRT(41)) AND (0,SQRT(41)
ASYMPTOTES ARE GIVEN BY
Y^2/16-X^2/25=K
25Y^2-16X^2-400K=0
(5Y+4X+A)(5Y-4X+B)=0
SLOPES OF ASYMPTOTES ARE
-4/5 AND 4/5
THE GRAPHS LOOK LIKE THIS
graph( 600, 600, -10, 10, -10, 10, 4*(1+x^2/25)^0.5,-4*(1+x^2/25)^0.5,4*x/5,-4*x/5 )
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What is the vertices, foci, and slope of the asymptote for the hyperbola whose equation is, y^2 - 4x^2 - 2y - 16x + 1 = 0?
(Y^2-2*Y*1+1^2)-{(2X)^2+2*(2X)*4+4^2}-1^2+4^2+1=0
(Y-1)^2-(2X+4)^2=-16
4(X+2)^2-(Y-1)^2=16
{4(X+2)^2}/16-{(Y-1)^2}/16=1
(X+2)^2/2^2-(Y-1)^2/4^2=1...
COMPARING WITH STANDARD EQN.
(X-H)^2/A^2-(Y-K)^2/B^2=1....WE HAVE
VERTICES ARE {(H-A),K} AND {(H+A),K}=(-2-2,1) AND (-2+2,1)=(-4,1) AND (0,1)
FOCI ARE {(H-AE),K} AND {(H+AE),K}...WHERE E IS
ECCENTRICITY =SQRT{(A^2+B^2)/A^2}=SQRT((4+16)/4)=SQRT(5)
SO FOCI ARE =(-2-2SQRT(5),1) AND (-2+2SQRT(5),1)
SLOPE OF ASYMPTOTE IS GIVEN BY DIFFERENTIATION.HAVE YOU BEEN TAUGHT?PLEASE INFORM.I SHALL COME BACK ON HEARING FROM YOU.
or you can take this proposition as proved formula
the pair of asymptotes for a conic is given by the same equation as the conic except for the constant term which has to be found using the condition for the equation to represent a pair of straight lines.
HENCE EQN OF ASYMPTOTES IS GIVEN BY
y^2 - 4x^2 - 2y - 16x + K=0 , WHERE K IS DETERMINED using the condition for the equation to represent a pair of straight lines.
SINCE WE ARE TO FIND ONLY SLOPES ,WE NEED NOT DETERMINE THE CONSTANT BUT ASSUME THAT THIS EQN REPRESENTS A PAIR OF STRAIGHT LINES.SO
y^2 - 4x^2 - 2y - 16x + K=0 = (Y+2X+A)(Y-2X+B)
HENCE SLOPES ARE +2 AND -2
graph( 600, 600, -10, 10, -10, 10, 1-2*((x+2)^2-4)^0.5,1+2*((x+2)^2-4)^0.5,2x+5,-2x-3 )
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UNDERSTOOD THE METHOD?WE HAVE TO WRITE THE QUADRATIC IN TO 2 LINEAR FACTORS
THE EQN.IS....
X^2/(36/25) - Y^2/(121)=1 SO EQN.OF ASYMPTOTES IS
X^2/(36/25) - Y^2/(121)=K...BUT K=0 SINCE THE ASYMPTOTES PASS THROUGH THE CENTRE OF HYPERBOLA WHICH IS ORIGIN IN THIS CASE.
SO WE GET
[{X/(6/5)} + {Y/11}]=0
HENCE THE 2 ASYMPTOTES ARE...
{X/(6/5)} + {Y/11}=0 AND {X/(6/5)}-{Y/11}=0
MULTIPLYING BY LCM...66...
55X+6Y=0......AND......55X-6Y=0
OK.....IF STILL IN DOUBT PLEASE COME BACK

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If it has been determined that the probability of an earthquake occurring on a certain day in a certain area is 0.03125, what are the odds AGAINST an earthquake?
ODDS IN FAVOUR =0.03125
ODDS AGAINST =1-0.03125=0.96875
ODDS AGAINST/ODDS IN FAVOUR =0.96875/0.03125=96875/3125=19375/625=3875/125
=775/25=155/5=31/1

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SEE THE FOLLOWING EXAMPLES TO KNOW
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Linear_Algebra/30362: Question: Find the equation of the ellipse whose center is (5,-3) that has a vertex at 13,-3) and a minor axis of lenght 10.
POssible Answers:
(A) (x-5)^2/64 + (y+3)^2/25 = 1
(B) (x+5)^2/64 + (y-3)^2/25 = 1
(C) x^2/64 + y^2/25 = 1
(D) none of these
1 solutions
Answer 17014 by venugopalramana(1167) About Me on 2006-03-15 11:21:03 (Show Source):
SEE THE FOLLOWING AND TRY..IF STILL IN DIFFICULTY PLEASE COME BACK...
OK I WORKED IT OUT FOR YOU NOW
I TOLD YOU EQN IS
(X-H)^2/A^2 + (Y-K)^2/B^2....
WHERE H,K IS CENTRE...SO H=5 AND K=-3 AS CENTRE IS GIVEN AS (5,-3)....NOW VERTEX IS (13,-3)...IT LIES ON ELLIPSE..SO IT SATISFIES THE EQN
(13-5)^2/A^2 +(-3+3)^2/B^2 =1
HENCE A^2=64...OR A=8
MINOR AXIS =10=2B...HENCE B=5..SO EQN.S
(X-H)^2/64 + (Y+3)^2/25 =1
THAT IS A IS CORRECT.

Can you help me write an equation for an ellipse with a major axis with endpoints of (0,8), and (0,-8) with foci of (0,5) and (0,-5)?
1 solutions
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Answer 16810 by venugopalramana(1120) on 2006-03-13 11:19:12 (Show Source):
Can you help me write an equation for an ellipse with a major axis with endpoints of (0,8), and (0,-8) with foci of (0,5) and (0,-5)?
THIS SHOWS THAT X AXIS IS THE MAJOR AXIS
STANDARD EQN.OF ELLIPSE IS
(X-H)^2/A^2 +(Y-K)^2/B^2=1
CENTRE IS (H,K)..AS PER THE PROBLEM H=K=0 AS CENTRE OF ELLIPSE IS AT (0,0)..SINCE major axis with endpoints ARE (0,8), and (0,-8)
WHERE MAJOR AXIS =2A=8+8=16...SO A=8..SINCE major axis with endpoints ARE (0,8), and (0,-8)
FOCI ARE GIVEN BY
AE,0 AND -AE,0...SO AE =5...SO E=5/A=5/8
BUT E=SQRT{(A^2-B^2)/A^2}=5/8...SQUARING
25/64=(A^2-B^2)/A^2=1-B^2/A^2
B^2/64=1-25/64=49/64
B^2=49
B=7
HENCE EQN. OF ELLIPSE IS
X^2/64 + Y^2/49 = 1

Quadratic-relations-and-conic-sections/30009: Can you help me write an equation for an ellipse with a major axis with endpoints of (0,8), and (0,-8) with foci of (0,5) and (0,-5)?

1 solutions
Answer 16810 by venugopalramana(1167) About Me on 2006-03-13 11:19:12 (Show Source):
Can you help me write an equation for an ellipse with a major axis with endpoints of (0,8), and (0,-8) with foci of (0,5) and (0,-5)?
THIS SHOWS THAT X AXIS IS THE MAJOR AXIS
STANDARD EQN.OF ELLIPSE IS
(X-H)^2/A^2 +(Y-K)^2/B^2=1
CENTRE IS (H,K)..AS PER THE PROBLEM H=K=0 AS CENTRE OF ELLIPSE IS AT (0,0)..SINCE major axis with endpoints ARE (0,8), and (0,-8)
WHERE MAJOR AXIS =2A=8+8=16...SO A=8..SINCE major axis with endpoints ARE (0,8), and (0,-8)
FOCI ARE GIVEN BY
AE,0 AND -AE,0...SO AE =5...SO E=5/A=5/8
BUT E=SQRT{(A^2-B^2)/A^2}=5/8...SQUARING
25/64=(A^2-B^2)/A^2=1-B^2/A^2
B^2/64=1-25/64=49/64
B^2=49
B=7
HENCE EQN. OF ELLIPSE IS
X^2/64 + Y^2/49 = 1

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Show that b^3+b+1=a^3 has no positive integer solutions.
LET US TEST IT UNDER 2 STEPS...
1..POSITIVE ROOT EXISTS...AND THEN
2..IT SHOULD BE AN INTEGER.
F(B)= b^3+b+1-a^3=0..
USING THE RULE OF SIGNS ,FOR THIS TO HAVE A POSITIVE ROOT 1-A^3 HAS TO BE NEGATIVE,SINCE B IS POSITIVE.
THAT IS A>1..LET US ASSUME SO..THEN WE SHALL HAVE
BASIS .....A>1...
F(B)= b^3+b-(A^3-1)=0...WHERE A^3-1 IS POSITIVE.
NOW FOR THE ROOT TO BE RATIONAL ,THE ROOT HAS TO BE FACTOR OF A^3-1...SINCE COEFFICIENT OF B IS 1,WE SHALL HAVE AN INTEGER ROOT IF WE GET A RATIONAL ROOT.
FACTORS OF A^3-1 ARE (A-1) AND A^2+A+1...WHICH ARE NO MORE REDUCIBLE.TESTING THEM
I..FIRSTLY FOR A-1..WE GET...
F(A-1)=(A-1)^3+A-1-A^3+1=A^3-1-3A^2+3A+A-1-A^3+1=-3A^2+4A-1 SHOULD EQUAL ZERO
=-3A^2+3A+A-1=3A(1-A)-1(1-A)=0
(3A-1)(1-A)=0
CASE 1..............3A-1=0...OR....A=1/3 IS NOT POSSIBLE AS A>1 AS PER OUR BASIS FOR THE ROOT TO BE POSITIVE
CASE 2....1-A=0.....OR...A=1....IS NOT POSSIBLE AS A>1 AS PER OUR BASIS FOR THE ROOT TO BE POSITIVE.
SO A-1 IS NOT A ROOT..
II.....SECONDLY FOR A^2+A+1...WE GET...
F(A^2+A+1)=(A^2+A+1)^3+A^2+A+1-A^3+1 SHOULD EQUAL ZERO OR A VALUE OF A>1
WE FIND THAT THERE IS ONLY ONE NEGATIVE NUMBER -A^3,WHICH GETS CANCELLED OUT BY +A^3 IN THE EXPANSION OF (A^+A+1)^3....LEAVING ALL OTHERS AS SUM OF POSITIVE TERMS WHICH CAN NEVER EQUAL ZERO.HENCE A^2+A+1 IS ALSO NOT A FACTOR.
HENCE F(B) HAS NO RATIONAL ROOT FOR A>1
HENCE F(B) HAS NO POSITIVE INTEGRAL ROOT

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SEE THE FOLLOWING AND TRY
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127t^5 - 43t^2 + 7=F(T)
FIND FACTORS FOR CONSTANT TERM ...THEY ARE 1,-1,7 AND -7......CALL THIS N
FIND FACTORS FOR HIGHEST DEGREE TERM..THEY ARE 1,-1,127,-127....CALL THIS...D
RATIONAL ROOTS COULD THEN BE ONLY FROM N/D...THAT IS
1/1=1......1/-1=-1........1/127......1/-127
7/1=7......7/-1=-7........7/127......7/-127
THAT IS ALL......... SO TEST EACH OF THEM IN THE GIVEN EQN. TO FIND IF IT GIVES A ZERO..THEN IT IS REDUCIBLE OTHERWISE NOT
WE FIND THAT F(T) IS NOT ZERO FOR ANY OF THESE VALUES..SO IT IS IRREDUCIBLE OVER Q(RATIONAL)