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# Recent problems solved by 'venugopalramana'

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 Trigonometry-basics/33174: Using DeMoivre's Theorem, find three distinct cube roots of 1251 solutions Answer 19632 by venugopalramana(3286)   on 2006-04-11 00:38:01 (Show Source): You can put this solution on YOUR website!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
 expressions/33209: A three digit number multiplied by a three digit number is equal to the product of 67,392. What are the two sets of three digits that can give you 67,392 when multiplied? Or what equstion can be usd to solve problem? My son has this problem for homework and is unable to derive the answer. Can you help us? Thanks so much for your assistance. Dee1 solutions Answer 19631 by venugopalramana(3286)   on 2006-04-11 00:23:56 (Show Source): You can put this solution on YOUR website!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
 Angles/33241: Hi. Can you please answer this question for me? Thanks so much: 1 solutions Answer 19630 by venugopalramana(3286)   on 2006-04-11 00:02:22 (Show Source): You can put this solution on YOUR website!they are called complementary angles
 Exponential-and-logarithmic-functions/33246: This question is from textbook College Algebra ln(x+1)=ln(3x+1) - lnx solve exactly1 solutions Answer 19629 by venugopalramana(3286)   on 2006-04-10 23:57:48 (Show Source): You can put this solution on YOUR website!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
 Functions/33121: In my math class we are doing Functions right now. The specific three we are using are Linear, Quadratic, and Exponential. Linear is y=mx+b, Quadratic is y=ax^2(+b) and Exponential is y=a*b^x. Each problem gives ordered pairs, or a table. I'll show you a table, and my answer, which I do not believe to be correct, since when I plug in some numbers from the table, doesnt not equal out. EX: X Y -2 1.5 0 .5 2 -.5 4 -1.5 y=-x +.5 Now, that is how my math teacher taught us to solve it; find the slope, y intercept when x=0, etc, and when you plug numbers in from the table, my answer does not work. Obviously, since you are very intelligent, you will be able to see my flaw. Thank you.1 solutions Answer 19562 by venugopalramana(3286)   on 2006-04-10 12:31:22 (Show Source): You can put this solution on YOUR website!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
 Trigonometry-basics/33123: find the exact value, leaving the answer in a proper form of sec 540 degrees.1 solutions Answer 19560 by venugopalramana(3286)   on 2006-04-10 12:22:24 (Show Source): You can put this solution on YOUR website!SEC(540)=SEC(360+180)=SEC(180)=SEC(180-0)=-SEC(0)=-1
 Trigonometry-basics/33125: Simplify the following as much as possible leaved the answer in exact form: sin 225 degrees ctn 300 degrees + sec 150 degrees cos 315 degrees 1 solutions Answer 19559 by venugopalramana(3286)   on 2006-04-10 12:20:05 (Show Source): You can put this solution on YOUR website!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
 Numbers_Word_Problems/33138: cannot seem to get this!!! from my teacher's own problem solving packet. it reads... The larger of two consecutive even integers is six less than twice the smaller. Find the numbers. thank you tons!!!1 solutions Answer 19555 by venugopalramana(3286)   on 2006-04-10 12:04:10 (Show Source): You can put this solution on YOUR website!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.
 Geometry_proofs/32977: Given: :Line segment AB is congruent to line segment AE. Given: Angle C is congruent to angle F Given: Line segment BD is congruent to line segment ED Prove: Line segment AC is congruent to line segment AF1 solutions Answer 19550 by venugopalramana(3286)   on 2006-04-10 11:23:49 (Show Source): You can put this solution on YOUR website!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
 Trigonometry-basics/32481: Prove the following identity: cos 4x = cos^4x - 6cos^2xsin^2s + sin^4x1 solutions Answer 19373 by venugopalramana(3286)   on 2006-04-08 12:34:32 (Show Source): You can put this solution on YOUR website!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)
 Trigonometry-basics/32484: Develop the formula for the tangent of the sum of angles in terms of the tangent of the separate angles alpha and beta. (I know i have to use tan(alpha + beta) = sin(a + B) ---------- cos(a + B) 1 solutions Answer 19372 by venugopalramana(3286)   on 2006-04-08 12:29:32 (Show Source): You can put this solution on YOUR website!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)}
 Equations/32500: How does finding a solution to 8=2x-6 help find the coordinates of a point on the line of the equasion y=2x-6?1 solutions Answer 19369 by venugopalramana(3286)   on 2006-04-08 12:15:12 (Show Source): You can put this solution on YOUR website!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.
 Distributive-associative-commutative-properties/32844: (a) Prove that the set S of rational numbers (in lowest term) with odd denominators is a subring of Q. (b) Let I be the set of elements of S with even numerators. Prove that I is an ideal in S. (c) Show that S/I consists of exactly two elements. Thank You1 solutions Answer 19368 by venugopalramana(3286)   on 2006-04-08 12:00:37 (Show Source): You can put this solution on YOUR website! 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)
 Radicals/32745: 2) Is sqrt x^2 = x an identity (true for all values of x)? Explain answer please?1 solutions Answer 19367 by venugopalramana(3286)   on 2006-04-08 11:39:41 (Show Source): You can put this solution on YOUR website!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
 Equations/32734: Isolate for "t": m=16t^2+vt+7 thanks, i cant seem to figure it out... are you supposed to divide it out?1 solutions Answer 19366 by venugopalramana(3286)   on 2006-04-08 11:31:34 (Show Source): You can put this solution on YOUR website!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
 Trigonometry-basics/32892: For my college Pre-calc. class I need to prove this: tan^3x-1/tanx-1 = sec^2x+tan x1 solutions Answer 19365 by venugopalramana(3286)   on 2006-04-08 11:20:35 (Show Source): You can put this solution on YOUR website!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)
 logarithm/32907: The solution of the equation log(log x)=1 is x=10^10. Is this one true or false. I believe that it is false but want to make sure before I turn it in that it is correct.1 solutions Answer 19364 by venugopalramana(3286)   on 2006-04-08 11:08:41 (Show Source): You can put this solution on YOUR website!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
 Miscellaneous_Word_Problems/32723: 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?1 solutions Answer 19335 by venugopalramana(3286)   on 2006-04-08 01:46:53 (Show Source): You can put this solution on YOUR website!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
 Miscellaneous_Word_Problems/32755: Dealing with contradiction: Show that for any positive integer a and any prime p, if p is divides a, then p does not divide a+1.1 solutions Answer 19334 by venugopalramana(3286)   on 2006-04-08 00:49:10 (Show Source): You can put this solution on YOUR website!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.
 Miscellaneous_Word_Problems/32880: Show that the polynomial p(x)= x^4 + 2x^2 + 2x + 2 cannot be expressed as the product of two quadratic polynomials of form x^2 + ax + b and x^2 + cx +d where a,b,c, and d are any integers.1 solutions Answer 19332 by venugopalramana(3286)   on 2006-04-08 00:45:20 (Show Source): You can put this solution on YOUR website!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.
 Triangles/32801: This question is from textbook Addison Wesley Geometry Chapter is tittled "Right Triangle Relationships" 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? 1 solutions Answer 19263 by venugopalramana(3286)   on 2006-04-07 12:52:44 (Show Source): You can put this solution on YOUR website!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
 Miscellaneous_Word_Problems/32757: Dealing with contradictions: If a, b, and c are nonzero real numbers such that a>0 and the equation ax^2+bx+c has no real solutions, then show that ax^2+bx+c<0 cannot be true for all values of x.1 solutions Answer 19261 by venugopalramana(3286)   on 2006-04-07 12:38:29 (Show Source): You can put this solution on YOUR website!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
 Linear_Algebra/30774: The Question Reads: List all of the potential rational zeros of the polynomial function. Do not attempt to find the zeros. f(x) =2x^3+x^2+3x+8 The Answers to choose are: (A) +2, +4, +8, +1/2, +7/2 (B) +2, +4, +8, +16, +7/2 (C) 0, +1, +2, +4, +1/2 (D) +1, +2, +4, +8, +1/2 Can you please help me! Thank you!1 solutions Answer 19182 by venugopalramana(3286)   on 2006-04-06 12:30:48 (Show Source): You can put this solution on YOUR website!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
 Surface-area/32729: 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?1 solutions Answer 19180 by venugopalramana(3286)   on 2006-04-06 12:18:25 (Show Source): You can put this solution on YOUR website!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.
 Trigonometry-basics/32644: show the steps to simplify the first expression to the second expression: I'll let theta be equal to x. csc^2x - cot^2x/1+tan^2x , cos^2x thanks so much to whomever can help me with this! 1 solutions Answer 19179 by venugopalramana(3286)   on 2006-04-06 12:12:00 (Show Source): You can put this solution on YOUR website!(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
 Miscellaneous_Word_Problems/32697: 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?1 solutions Answer 19178 by venugopalramana(3286)   on 2006-04-06 12:08:40 (Show Source): You can put this solution on YOUR website!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
 Quadratic-relations-and-conic-sections/32653: Can someone explain to me how to graph (x+2)²/9+(y-3)²/4=1 and other ellipses and how to find the equation of an ellipse from its graph?1 solutions Answer 19177 by venugopalramana(3286)   on 2006-04-06 11:49:53 (Show Source): You can put this solution on YOUR website!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
 Quadratic-relations-and-conic-sections/32650: Find the equation of a hyperbola with vertices at (0,±5) and asymtotes y=±2x.1 solutions Answer 19149 by venugopalramana(3286)   on 2006-04-06 09:01:24 (Show Source): You can put this solution on YOUR website!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