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LET N=125=Z^3
SINCE 125^1/3=5...WE CAN WRITE Z^3=(5^3)*1=(5^3){COS(2KPI)+iSIN(2KPI)}
SO Z =5*{COS(2KPI)+iSIN(2KPI)}^(1/3)
=5*{COS(2KPI/3)+iSIN(2KPI/3)}BY DEMOVIERS THEOREM...PUTTING K=1,2 AND 3 WE GET THE 3 CUBE ROOTS
5*{COS(2*1PI/3)+iSIN(2*1PI/3)}=5(-0.5+i*0.5*SQRT(3))=2.5(-1+i*SQRT(3))
5*{COS(2*2PI/3)+iSIN(2*2PI/3)}=5(-0.5-i*0.5SQRT(3))= -2.5(1+i*SQRT(3)
5*{COS2*3PI)+iSIN(2*3PI/3)}=5

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one method is to find all prime factors one by one
2|67392
-----------
2|33696
-------------
2|16848
------------
2|8424
-------------
2|4212
-------------
2|2106
------------
3|1053
--------------
3|351
-----------
3|117
----------------
3|39
---------------
..13
hence 67392=2^6*3^4*13...now take 2 combinations so that they make 2 threedigit numbers
like 2^5*13=32*13=416 and
2*3^4=2*81=162

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they are called complementary angles

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ln(x+1)=ln(3x+1) - lnx
ln(x+1)=ln{(3x+1)/x}...taking antilogs
x+1=(3x+1)/x
x(x+1)=3x+1
x^2+x-3x-1=0
x^2-2x-1=0..using quadratic formula


x=1+sqrt.(2)....or......1-sqrt.(2)..but since logof negative number does not exist...1-sqrt.(2) cannot be a solution.
so...x=1+sqrt.2=2.4142

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IF YOU KNOW EQNS...STRAIGHT FORWARD APPROACH IS SOLVE FOR THE UNKNOWNS ASSUMING AN EQN.HERE YOU ARE TRYING LINEAR..OK...
FOR -2,1.5......WE GET
1.5=-2A+B.....FOR THE NEXT PAIR OF 0,0.5
0.5=2*0+B...OR...B=0.5
SO 1.5=-2A+0.5....OR......-2A=1.5-0.5=1....OR...A=-1/2=-0.5
SO EQN.IS
Y=-0.5X+0.5...NOW TEST THE OTHER PAIRS TO SEE IF THIS IS CORRECT....
-0.5=-0.5*2+0.5=-1+0.5=-0.5...OK
-1.5=-0.5*4+0.5=-2+0.5=-1.5...OK
SO THE EQN.IS LINEAR AND IT IS
Y=-0.5X+0.5

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SEC(540)=SEC(360+180)=SEC(180)=SEC(180-0)=-SEC(0)=-1

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sin 225 degrees ctn 300 degrees + sec 150 degrees cos 315 degrees
={SIN(180+45)}*{COT(360-60)}+{SEC(180-30)}*{COS(360-45)}
=-SIN(45)*{-COT(60)}+{-SEC(30)}*COS(45)
=(1/SQRT.2)((1/SQRT.3)-(2/SQRT.3)(1/SQRT.2)
=(1-2)/SQRT(3*2)=-1/SQRT.6=-SQRT.(6)/6

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I AM ABLE TO FIND ONE SUCH QUESTION I ANSWERED EARLIER.SEE THAT AND BY THE SIDE CORRESPONDING ANSWER TO YOUR QUESTION
Volume/30504: an open box is to be constructed from a piece of cardboard 15 inches by 25 inches by cutting squares of length x from each corner and folding up the sides. Express the volume of the box as a function of x. what is the domain v?
1 solutions .........IN YOUR CASE THE DIMENSIONS ARE 6' AND 8'



Answer 17192 by venugopalramana(1167) About Me on 2006-03-17 06:08:55 (Show Source):
an open box is to be constructed from a piece of cardboard 15 inches by 25 inches by cutting squares of length x from each corner and folding up the sides. Express the volume of the box as a function of x. what is the domain v?
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 = 25-2X ..(IN YOUR CASE 8-2X) AND WIDTH = 15-2X..(IN YOUR CASE 6-2X)..AND HEIGHT =X ...SO VOLUME V IS GIVEN BY LEMGTH*WIDTH*HEIGHT
V=(25-2X)(15-2X)X...(IN YOUR CASE (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 ....
15-2X>0...OR....15>2X...OR....7.5>X....OR X<7.5...(IN YOUR CASE 8-2X>0...AND 6-2X>0...SO X <3)
RANGE.....MAXIMUM VALUE....IN YOUR CASE....
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 larger of two consecutive even integers is six less than twice the smaller. Find the numbers.
thank you tons!!!
LET SMALLER EVEN NUMBER BE =2N...
TWICE THIS =2*2N=4N....6 LESS =4N-6...THIS IS EQUAL TO NEXT EVEN NUMBER =2N+2
4N-6=2N+2
4N-2N=2+6=8
2N=8
N=4....SO THE 2 NUMBERS ARE 2*4=8 AND 10.

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pleasae give a sketch or describe the figure in full.what is c,e,f,d etc..are they all lines triangles..or what.then only we can help you

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COS(4X)=COS(2(2X))=COS^2(2X)-SIN^2(2X)
={COS^2(X)-SIN^2(X)}^2-{2SIN(X)COS(X)}^2
=COS^4(X)+SIN^4(X)-2COS^2(X)SIN^2(X)-4SIN^2(X)COS^2(X)
=COS^4(X)-6COS^2(X)SIN^2(X)+SIN^4(X)

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TAN(A+B)=SIN(A+B)/COS(A+B)
={SIN(A)COS(B)+COS(A)SIN(B)}/{COS(A)COS(B)-SIN(A)SIN(B)}...DIVIDING THROUGHOUT BY COS(A)COS(B) WE GET THIS
=[{SIN(A)COS(B)/COS(A)COS(B)}+{COS(A)SIN(B)/COS(A)COS(B)}]/[{COS(A)COS(B)/COS(A)COS(B)}-{SIN(A)SIN(B)/COS(A)COS(B)]
={TAN(A)+TAN(B)}/{1-TAN(A)TAN(B)}

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USE FIRST a^2=b^2+c^2-2bcCOS(A)..
COS(A)=(b^2+c^2-a^2)/(2bc)
TO FIND ANGLE A AND THEN FOLLOW THE EXAMPLE GIVEN BELOW

SEE THE FOLLOWING EXAMPLE WHICH IS SAME AS YOUR QUESTION AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
please help me with this long question. i PROMISE it really is only one question in my book, but its super long. please believe me that its only one question...
Given triangle ABC with C=48 degrees, a=12 and b=8:
a. Solve for c, rounding your answer first to the nearest tenth and then to the nearest hundredth.
b. Use the law of sines with your first answer for c to find A
c. use the law of sines with your second answer for c to find A
d. Use the law of cosines to find angle A using your first answer for c
e. use the law of cosines to find angle A using your second answer for c
f. Find angle B using the two measures found in part a.
g. is angle A acute or abtuse and why?
im so sorry this is so long, its seriously how it is in the book... its a C excersise and they're usually the hard ones, and i dont get it at all!
1 solutions
Answer 19063 by venugopalramana(1346) About Me on 2006-04-05 12:11:24 (Show Source):
Given triangle ABC with C=48 degrees, a=12 and b=8:
a. Solve for c, rounding your answer first to the nearest tenth and then to the nearest hundredth.
C^2=a^2+b^2-2ab*COS(C),SUBSTITUTING THE ABOVE VALUES WE GET
c=8.9.....
c=8.92
b. Use the law of sines with your first answer for c to find A
a/SINA(A) = c/SIN(C)......
WITH c=8.9......NO ANSWER...AS VALUE EXCEEDS RANGE OF SIN(A)
c. use the law of sines with your second answer for c to find A
WITH c=8.92......A=88.71 DEG.
d. Use the law of cosines to find angle A using your first answer for c
COS(A)=(b^2+c^2-a^2)/2bc......A=90.3 DEG.WITH c=8.9
e. use the law of cosines to find angle A using your second answer for c
COS(A)=(b^2+c^2-a^2)/2bc......A=90.3 DEG.WITH c=8.92
f. Find angle B using the two measures found in part a.
b/SIN(B)=c/SIN(C)......B=41.9..WITH c=8.9...AND B=41.8...WITH c=8.92
g. is angle A acute or abtuse and why?
FOR ACTUAL VALUE OF c WE GET A=89.8...SO IT IS ACUTE.
im so sorry this is so long, its seriously how it is in the book... its a C excersise and they're usually the hard ones, and i dont get it at all!

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SEE THE FOLLOWING EXAMPLE WHICH IS SAME AS YOUR QUESTION AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
please help me with this long question. i PROMISE it really is only one question in my book, but its super long. please believe me that its only one question...
Given triangle ABC with C=48 degrees, a=12 and b=8:
a. Solve for c, rounding your answer first to the nearest tenth and then to the nearest hundredth.
b. Use the law of sines with your first answer for c to find A
c. use the law of sines with your second answer for c to find A
d. Use the law of cosines to find angle A using your first answer for c
e. use the law of cosines to find angle A using your second answer for c
f. Find angle B using the two measures found in part a.
g. is angle A acute or abtuse and why?
im so sorry this is so long, its seriously how it is in the book... its a C excersise and they're usually the hard ones, and i dont get it at all!
1 solutions
Answer 19063 by venugopalramana(1346) About Me on 2006-04-05 12:11:24 (Show Source):
Given triangle ABC with C=48 degrees, a=12 and b=8:
a. Solve for c, rounding your answer first to the nearest tenth and then to the nearest hundredth.
C^2=a^2+b^2-2ab*COS(C),SUBSTITUTING THE ABOVE VALUES WE GET
c=8.9.....
c=8.92
b. Use the law of sines with your first answer for c to find A
a/SINA(A) = c/SIN(C)......
WITH c=8.9......NO ANSWER...AS VALUE EXCEEDS RANGE OF SIN(A)
c. use the law of sines with your second answer for c to find A
WITH c=8.92......A=88.71 DEG.
d. Use the law of cosines to find angle A using your first answer for c
COS(A)=(b^2+c^2-a^2)/2bc......A=90.3 DEG.WITH c=8.9
e. use the law of cosines to find angle A using your second answer for c
COS(A)=(b^2+c^2-a^2)/2bc......A=90.3 DEG.WITH c=8.92
f. Find angle B using the two measures found in part a.
b/SIN(B)=c/SIN(C)......B=41.9..WITH c=8.9...AND B=41.8...WITH c=8.92
g. is angle A acute or abtuse and why?
FOR ACTUAL VALUE OF c WE GET A=89.8...SO IT IS ACUTE.
im so sorry this is so long, its seriously how it is in the book... its a C excersise and they're usually the hard ones, and i dont get it at all!

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8=2X-6...COMPARE THIS WITH Y=2X-6...WE CAN SEE THAT Y IS REPLACED BY 8..
FIRST LET US SOVE 8=2X-6...WE GET 2X=8+6=14....OR...X=14/2=7
SO WE CONCLUDE THAT X=7 AND Y=8 SATISFIES THE EQN.Y=2X-6
NOW SATISFYING AN EQN.MEANS 4 THINGS
1. ANY POINT P WITH COORDINATES (H,K) SATISFIES THE EQN. THEN IT WILL LIE ON THE GRAPH OF THAT EQN.
2. ANY POINT P WITH COORDINATES (H,K)WHICH DOES NOT SATISFY THE EQN. THEN IT WILL NOT LIE ON THE GRAPH OF THAT EQN.
3.IF A POINT P WITH COORDINATES (H,K) LIES ON THE GRAPH OF THE GIVEN EQN.THEN IT WILL SATISFY THE EQN.OF THE GRAPH.
4.IF A POINT P WITH COORDINATES (H,K) DOES NOT LIE ON THE GRAPH OF THE GIVEN EQN.THEN IT WILL NOT SATISFY THE EQN.OF THE GRAPH.
HENCE YOU CAN SEE NOW THAT FINDING A SET OF VALUES OF X AND Y SATISFYING THE EQN. GIVES US THE COODINATES OF THE POINT LYING ON THAT EQN.'S GRAPH.WHICH IS A STRAIGHT LINE IN THIS CASE.

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S IS THE SET OF RATIONAL NUMBERS WITH D.R. BEING ODD NUMBER ..THAT
IS ELEMENTS ARE OF THE TYPE P/(2Q-1)
TST S IS A SUB RING OF Q.
FOR THIS WE NEED TO SHOW
1.IF A AND B ARE ELEMENTS OF S THEN A-B IS ALSO AN ELEMENT OF S.
LET A=P1/(2Q1-1)...AND....B=P2/(2Q2-1)
A-B={P1(2Q2-1)-P2(2Q1-1)}/(2Q1-1)(2Q2-1)...WE FIND THAT D.R IS PRODUCT
OF 2 ODD NUMBERS AND HENCE IS ODD.SO THE RESULTANT VALUE OF A-B IS
STILL IN THE FORM OF SOME P/(2Q-1)...SO THIS BELONGS TO S
2.IF A AND B ARE ELEMENTS OF S THEN AB IS ALSO AN ELEMENT OF S
WITH SAME A AND B AS ABOVE WE HAVE
AB=(P1P2)/(2Q1-1)(2Q2-1)...WHICH IS AGAIN OF THE TYPE SOME P/(2Q-1)
AND HENCE AN ELEMENT OF S
THESE 2 CRITERIA BEING NECESSARY AND SUFFICIENT FOR S TO BE A SUB RING
OF Q WE CONCLUDE THE RESULT.
----------------------------------------------------------------------------------------------------------------------
I IS SET OF EVEN N.R....SO THE ELEMENTS ARE 2P/(2Q-1)..TYPE.
TO SHOW THAT THIS IS AN IDEAL OF S.
FOR THIS WE HAVE TO PROVE THAT
1.IF A AND B ARE ELEMENTS OF I ,THEN A-B IS AN ELEMENT OF I.
LET A =2P1/(2Q-1)....B=2P2/(2Q2-1)....
A-B = 2[P1(2Q2-1)-P2(2Q1-1)]/(2Q1-1)(2Q2-1).....WE FIND THAT N.R IS A
MULTIPLE OF 2 AND HENCE AN EVEN NUMBER.DENOMINATOR IS PRODUCT OF 2 ODD
NUMBERS AND HENCE AN ODD NUMBER.
HENCE A-B IS EQUAL TO SOME 2P/(2Q-1)...THAT IS IT BELONGS TO I
2.IF A IS AN ELEMENT OF I AND R IS AN ELEMENT OF S THEN A*R AND R*A
ARE ELEMENTS OF I.
LET R=R1/(2Q3-1)....WE HAVE
A*R=R*A=[(R1)(2P1)]/(2Q1-1)(2Q3-1)...WHICH IS AGAIN OF THE TYPE SOME
2P/(2Q-1)..FOR THE SAME REASONS GIVEN ABOVE.
THESE 2 CRITERIA BEING NECESSARY AND SUFFICIENT CRITERIA FOR I TO BE
AN IDEAL OF S ,WE CONCLUDE THE RESULT.
BY DEFINITION S/I IS THE SET OF COSETS OF I IN S
I/S={S+A:A IS AN ELEMENT OF S}
NOW WE HAVE I AS THE SET OF ELEMENTS OF THE TYPE...EVEN /ODD NUMBER
WHERE AS S IS THE SET OF ELEMENTS OF THE TYPE,....EVEN/ODD AND ODD/ODD
ELEMENTS...SO WHEN YOU MAKE A COSET , YOU WILL HAVE ONLY 2 DISTINCT
CLASSES ALWAYS NAMELY ODD AND EVEN IN NR.SINCE S/I IS THE SET OF
DISTINCT COSETS ,WE HAVE ONLY 2 DISTINCT CLASSES OR ELEMENTS IN THIS
QUOTIENT RING OR RESIDUAL CLASS AS IT IS CALLED.
HENCE THERE ARE ONLY 2 DISTINCT ELEMENTS IN S/I,ONE CLASS OF ELEMENTS OS 2P/(2Q-1)AND ANOTHER CLASS OF ELEMENTS OF TYPE (2P-1)/(2Q-1)

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Is sqrt x^2 = x an identity (true for all values of x)? Explain answer please?
IN MATHS AS A RULE/COVENTION WE TAKE SQRT(A) AS THE POSITIVE VALUE ONLY.THAT IS SQRT(4) IS TAKEN AS +2 ONLY.,THOUGH -2 IS ALSO POSSIBLE,UNLESS OTHERWISE INDICATED.
CASE 1....X IS POSITIVE
SO WE GET LHS=SQRT.(X^2)=+X
RHS =+ X...
SO WE HAVE X=X WHICH IS AN IDENTITY AS IT IS TRUE FOR ANY VALUE OF X.
CASE 2......X IS NEGATIVE
SO WE GET LHS=SQRT.{(-X)^2}=SQRT.(X^2)=+X
RHS = -X...
SO WE HAVE X=-X WHICH IS NOT AN IDENTITY AS IT IS TRUE FOR ONLY X=0

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Isolate for "t": m=16t^2+vt+7
thanks, i cant seem to figure it out... are you supposed to divide it out?
YOU CAN DO THIS BY COMPLETING THE SQUARE TECHNIQUE..UTILISING THE T^2 AND T TERMS.
M=[(4T)^2+2(4T)(V/8)+(V/8)^2]-(V/8)^2+7
={4T+(V/8)}^2+7-V^2/64
{4T+(V/8)}^2=M+V^2/64-7=(64M+V^2-7*64)/64
{4T+(V/8)}=(64M+V^2-448)^0.5/8
4T=(64M+V^2-448)^0.5/8 -V/8={(64M+V^2-448)^0.5-V}/8
T={(64M+V^2-448)^0.5-V}/32

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tan^3x-1/tanx-1 = sec^2x+tan x..USING....A^3-B^3=(A-B)(A^2+AB+B^2)
LHS=[{TAN(X)-1}{TAN^2(X)+TAN(X)+1}]/[(TAN(X)-1]=SEC^2(X)+TAN(X)=RHS...SINCE 1+TAN^2(X)=SEC^2(X)

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LOG(LOG(X))=1
PUT LOG(X)=Y
SO...LOG(Y)=1
AS PER DEFINITION OF LOG.....IF B^P=N...THEN....LOG(N)TO BASE B IS P
SO....HERE AS BASE IS 10..WE HAVE
10^1=Y...SO Y=10
NOW Y = LOG(X)=10
SO 10^10=X

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Suppose 25 people are arranged in a circle. Three of them are selected at a random. what is the probability that none of those selected were next to each other?
KEY TO SOLVING THESE PROBLEMS IS TO DEVELOP A LOGICAL PROCEDURE WHICH
1.IS AMENABLE TO CALCULATION WITH KNOWN/STANDARD FORMULAE OR FROM FUNDAMENTALS
2.SATISFIES ALL POSITIVE REQUIREMENTS OF THE PROBLEM
3.EXCLUDES ALL NEGATIVE REQUIREMENTS OR EXCEPTIONS IN THE PROBLEM.
4.NO DUPLICTION OR OVERLAP OF THE DONE THINGS OCCUR AT ANY STAGE IN THE SOLUTION
THESE PROBLEMS ARE MANY TIMES TRICKY OR DECEPTIVE IN THAT THEY GIVE US A FALSE SENSE OF CORRECTNESS AND SOME FLAW IN OUR LOGIC IS NOT SHOWN UP TILL A LATER DATE WHEN WE REALSE IT OUR SELVES OR SOME ONE ELSE POINTS OUT.
HENCE THE BEST WAY TO LEARN THESE THINGS IS TO HAVE AN OPEN MIND IN OUR APPROACH,DISCUSS THE LOGIC WITH OTHERS ACCEPTING THEIR SUGGESTIONS`ON ANY OF OUR FLAWS AND TO CONTINUOUSLY KEEP IMPROVING BY CORRECTING OUR MISTAKES.
IN ONE WORD SAY "WE WELCOME CRITICISM"..THE SAME THING APPLIES TO ME TOO!FEEL FREE TO POINT OUT IF YOU FIND ANY LOGIC NOT CLEAR OR DOUBT FULL.WE OWE YOU A THOUSAND THANKS FOR EACH FLAW POINTED OUT IN OUR LOGIC.
SO NOW ON TO THE PROBLEM....
1.LET US NAME THE PEOPLE FOR CONVENIENCE AS P1,P2.......P25 IN THEIR ORDER OF ORIGINAL ARRANGEMENT IN A CIRCLE.
2.TOTAL NUMBER OF COMBINATIONS OF 3 PEOPLE FROM THESE 25 ARE 25C3=25!/(3!*22!)
3.WE HAVE TO FIND THE NUMBER OF SELECTIONS WHERE NONE OF THOSE SELECTED WERE NEXT TO EACH OTHER.
4.NOW LET US SAY WE PICKED ONE PERSON FIRST..THIS CAN BE DONE IN 25 WAYS.
5.LET US SAY THE PERSON SELECTED IS P1
6.THEN THE NEXT PERSON CAN BE FROM PEOPLE OTHER THAN P2 AND P24,AS THEY ARE NEIGHBORS TO P1.SO WE HAVE 25-P1-P2-P25=22 PEOPLE TO CHOSE FROM FOR THE SECOND PERSON IN 22 WAYS.
7.LET US SPLIT THESE 22 POSSIBILITIES INTO 3 CASES AS FOLLOWS OR REASON YOU WOULD KNOW SOON.
8.CASE 1....P3 IS SELECTED....IN 1 WAY..
9.CASE 2....P24 IS SELECTED...IN 1 WAY..
10.CASE 3....ANY OTHER PERSON IS SELECTED...IN 20 WAYS.
11.NOW IN CASE 1. THE THIRD PERSON BY THE SAME LOGIC (NO.6) AS ABOVE CAN BE FROM PEOPLE OTHER THAN P4,P2,P1,P25,P3....THAT IS 25-5=20 WAYS.
12.SIMILARLY IN CASE 2. THE THIRD PERSON BY THE SAME LOGIC (NO.6) AS ABOVE CAN BE FROM PEOPLE OTHER THAN P23,P25,P1,P2,P24....THAT IS 25-5=20 WAYS.
13.BUT IN THE THIRD CASE THEY CAN BE ANY ONE FROM 25-P1-P2-P25-PSECOND PICK -2 PERSONS ON EITHER SIDE OF SECOND PICK=19 WAYS.
14. (I THINK NOW IT IS CLEAR WHY WE MADE THE SPLIT INTO 3 CASES)
15.SO TOTAL NUMBER OF ARRANGEMENTS ARE
25*{1*20+1*20+20*19)=25*21*20....BUT THESE ARE ARRANGEMENTS...SO NUMBER OF COMBINATIONS ARE 25*21*20/3!
PROBABILITY THAT NONE OF THOSE SELECTED WERE NEXT TO EACH OTHER ORIGINALL=
={25*21*20/3!}/(25!/(3!*22!)
=25*21*20*3!/25*24*23*3!
=(21*20)/(24*23)=
=0.76

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P|A...SO A=P*M....WHERE M IS AN INTEGER
FOR P TO DIVIDE A+1...
P|PM+1...IMPLIES
(PM+1)/P IS AN INTEGER...THAT IS M+(1/P)IS AN INTEGER...THAT IS 1/P IS
AN INTEGER...THIS IS NOT POSSIBLE AS 1 HAS NO DIVIDERS EXCEPT IT SELF
AND P BEING A PRIME NUMBER DIFFERENT FROM 1 CANNOT DIVIDE 1.

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F(X)=X^+2X^2+2X+2.....IF THIS HAS TO BE PRODUCT OF 2 QUADRATICS THEN
THEY HAVE TO BE
X^2+AX+1 & X^2-AX+2...OR......
X^2+AX-1 & X^2-AX-2.....AS CONSTANT TERM =2 .ITS INTEGRAL FACTORS ARE
1 &2 OR -1 & -2.
FURTHER COEFFICIENT OF X^3=0 .,SO IF THERE IS AX IN ONE FACTOR,THERE
SHOULD BE -AX IN ANOTHER FACTOR.
THIS LEAVES US....... BY EQUATING COEFFICIENTS OF X AND X^2 TERMS...
(X^2+AX+1)(X^2-AX+2)=X^4-AX^3+2X^2+AX^3-(A^2)(x^2)+2AX+X^2-AX+2
=X^4+X^2(3-A^2)+AX+2=X^4+2X^2+2X+2...SO
3-A^2=2....AND.....A=2.....WHICH ARE NOT CONSISTENT.HENCE THERE IS NO
SOLUTION....SIMILARLY...WE CAN PROVE THIS FOR
(X^2+AX-1)(X^2-AX-2)=X^4-AX^3-2X^2+AX^3-(A^2)(x^2)-2AX-X^2+AX+2
=X^4-X^2(3+A^2)-AX+2=X^4+2X^2+2X+2...SO
3+A^2=2....AND.....A= -2.....WHICH ARE NOT CONSISTENT.HENCE THERE IS
NO SOLUTION....
HENCE F(X) IS NOT REDUCIBLE OR CAN NOT BE FACTORISED INTO 2 QUADRATIC
EXPRESSIONS.

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PLEASE GIVE COMPLETE NOMENCLATURE LIKE WHAT IS THE TRIANGLE NAME ETC..
Where the altitude is MQ and the hypotenuse is NP.......... The question is this: If MN = 8, QN = 6, What is PN? I have tried to help my 9th grader, but Geometry was not my strong subject when I was in high school. He has a quiz over this tomorrow. Any chance anyone is out there who can help with the answer and explanation yet tonight so he can look at it and understand it in the morning before he goes to school?
OK ASSUMING MNP IS THE TRIANGLE,ANGLE NMP=90...AND SINCE MQ IS THE ALTITUDE,ANGLE NQM =90
IN TRIANGLES MNQ AND PNM...
ANGLE MQN=90=ANGLE NMP
ANGLE MNQ=ANGLE MNP....
SO THE 2 TRIANLES ARE SIMILAR..HENCE
MN/PN=NQ/NM
8/PN=6/8
6*PN=8*8=64
PN=64/6=32/3=10.667

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AX^2+BX+C=0...HAS NO REAL SOLUTION...SO DISCRIMINANT = B^2-4AC ..IS
NEGATIVE....OR....4AC-B^2 IS POSITIVE
LET F(X)=AX^2+BX+C=A{X^2+BX/A+C/A}=A[{X^2+2(X)(B/2A)+(B/2A)^2}-(B/2A)^2+(C/A)]
=A[{X+(B/A)}^2}+{(C/A)-B^2/4A^2}]=A[{X+(B/A)}^2+{(4AC-B^2)/4A^2}]
SO F(x)=A..SINCE {X+(B/A)}^2 IS
ALWAYS POSITIVE BEING A PERFECT SQUAREAND {(4AC-B^2)} IS POSITIVE AS
SHOWN ABOVE AND 4A^2 IS ALSO POSITIVE BEING A PERFECT SQUARE...
HENCE F(X) IS ONLY DEPENDENT ON SIGN OF A...
F(X) IS POSITIVE IF A IS POSITIVE ..AND
F(X) IS NEGATIVE IF A IS NEGATIVE..
SO IF WE ARE GIVEN THAT A >0..OR POSITIVE THEN
F(X)=AX^2 + BX + C CAN NOT BE NEGATIVE

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f(x) =2x^3+x^2+3x+8
THE METHOD IS TO LIST ALL FACTORS OF COEFFICIENT OF HIGHEST DEGREE TERM(SAY Q1,Q2...ETC...) AND THE CONSTANT TERM...(SAY P1,P2...ETC)...THEN ALL POSSIBLE RATIONAL ZEROS OF THE POLYNOMIAL ARE ALL POSSIBLE RATIOS OF P'S TO Q'S...HERE
COEFFICIENT OF HIGHEST DEGREE TERM X^3 IS=2..FACTORS ARE 1,-1,2,-2..THESE ARE Q'S CONSTANT TERM = 8....FACTORS ARE 1,-1,2,-2,4,-4,8,-8.....THESE ARE P'S
SO ALL POSSIBLE ZEROS OF THE P[OLYNOMIAL ARE....
1/1,-1/1,2/1,-2/1,4/1,-4/1,8/1,-8/1
1/-1,-1/-1,2/-1,-2/-1,4/-1,-4/-1,8/-1,-8/-1
1/2,-1/2,2/2,-2/2,4/2,-4/2,8/2,-8/2
1/-2,-1/-2,2/-2,-2/-2,4/-2,-4/-2,8/-2,-8/-2

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The midpoints of square ABCD are connected to form square EFGH. If EF=10units,
what is the area of squareABCD? Could u solve this please?
JOIN AC...A DIAGONAL OF THE SQUARE.
IN TRIANGLE ABC,WE HAVE E IS THE MIDPOINT OF AB AND F IS THE MIDPOINT OF BC.
SO EF IS EQUAL TO HALF OF AC
EF=10
AC=2*10=20....SIMILARLY IN THE SQUARE THE OTHER DIAGONAL BD=20
AREA OF A SQUARE =PRODUCT OF DIAGONALS/2=20*20/2=200 SQUARE UNITS.

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(csc^2x - cot^2x)/(1+tan^2x)
=1/SEC^2 X...SINCE COSEC^2 X - COT^2 X=1..AND 1+TAN^2 X = SEC^2 X
=COS^2 X

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The temperature of an ideal gas was held at 400 degrees C. The initial pressure and volume were 750 torr and 50 ml respectively. What would the final pressure be if the volume were increased to 1500 ml?
AT CONSTANT TEMPERATURE,FOR AN IDEAL GAS PRODUCT OF TOTAL PRESSURE AND VOLUME IS CONSTANT....
P1=750 TORR.....V1=50 ML.
P2=?............V2=1500 ML
P1V1=P2V2
P2=P1V1/V2=750*50/1500=25 TORR

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EQN.OF ELLIPSE IS ….(X+2)^2/9 +(Y-3)^2/4 = 1………….I
STD.EQN.OF ELLIPSE IS
(X-H)^2/A^2 + (Y-K)^2/B^2=1…………………..II………….COMPARING WITH ABOVE EQN.I
WHERE H,K IS CENTRE...SO H=-2 AND K=3 HENCE CENTRE IS (-2,3)..
A^2=9….OR……A=-3…. AND…….A =+3
B^2=4….OR……B=-2….AND….B=+2
MINOR AXIS =2B=4…
MAJOR AXIS =2A=6

NOW TO PLOT THE ELLIPSE YOU NEED
1.CENTRE COORDINATES…HERE THEY ARE (-2,3)
2.LENGTHS OF AXES….HERE…..6 AND 4.
3.EQNS.OF AXES…..MINOR AXIS IS…….X=H…OR…X=-2….OR…X+2=0
MAJOR AXIS IS …….Y=K…..OR……Y=3….OR……..Y-3=0
HAVING FOUND THE REQUIRED DATA,TAKE A SUITABLE SCALE…SAY 1 CM=1 UNIT OR SO AND PLOT
1.MARK ORIGIN…O,X AXIS AND Y AXIS
2.PLOT CENTRE (-2,3)…CALL IT C
3.DRAW MAJOR AXIS Y-3=0…THIS IS A LINE THROUGH CENTRE PARALLEL TO X AXIS
4.DRAW MINOR AXIS X+2=0….THIS IS A LINE THROUGH THE CENTRE PARALLEL TO Y AXIS.
5.MARK MAJOR AXIS LENGTH OF 6 ALONG THE MAJOR AXIS PLOTTED ABOVE….TAKING 3 UNITS ON EITHER SIDE OF CENTRE.CALL IT AA'
6.MARK MINOR AXIS LENGTH OF 4 ALONG THE MINOR AXIS PLOTTED ABOVE…TAKING 2 UNITS ON EITHER SIDE OF CENTRE.CALL IT BB'
7.ELLIPSE PASSES THROUGH A,A',B AND B'.JOIN THEM SMOOTHLY BY A SMOOTH ELLIPTICAL CURVE.
8.IF YOU NEED YOU CAN PLOT ADDITIONAL POINTS BY GIVING ARBITRARY VALUES FOR X AND FINDING CORRESPONDING VALUE OF Y

THE SAME PROCEDURE YOU CAN REVERSE TO FIND THE EQN.OF AN ELLIPSE FROM THE GRAPH

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ASYMPTOTES ARE
Y=2X…OR….Y-2X=0……I………………..AND
Y=-2X…OR….Y+2X=0……II…...SO THEIR COMBINED EQN.IS GIVEN BY EQN.I*EQN.II=0
(Y-2X)(Y+2X)=0=Y^2-4X^2=0………EQN.3.
IT IS KNOWN THAT THE EQN.OF HYPERBOLA DIFFERS FROM THE COMBINED EQN.OF ASYMPTOTES ONLY BY A CONSTANT.
SO…EQN.OF HYPERBOLA IS ….
Y^2-4X^2=C……..III…………...WHERE C IS A ONSTANT TO BE FOUND…PUTTING THIS EQN.IN STD.FORM WE GET
Y^2/C -4X^2/C =1….OR…..
(Y-0)^2/C - (x-0)^2/(C/4) =1…………………………………………IV
COMPARING WITH STANDARD EQN.OF HYPERBOLA…
(Y-K)^2/B^2-(X-H)^2/A^2=1.………………………………………V
WHERE VERTICES ARE...(H,(K-B)) AND (H,(K+B)) …GIVEN HERE AS (0,-5) AND (0,5)
WE GET H=0
K+B=5
K-B=-5
ADDING THE ABOVE 2 EQNS…..2K=0…OR…K=0
SO…B=5
COMPARING EQN.IV,WITH STD.EQN.V,WE GET….
H=0…..K=0…….C=B^2……..C/4=A^2…….BUT B=5 WE FOUND .SO
C=5^2=25…….A^2=25/4
HENCE EQN.OF HYPERBOLA IS AS PER EQN.III IS
Y^2-4X^2=25