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# Recent problems solved by 'venugopalramana'

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 Linear_Algebra/30525: please help! I'm getting stuck in the middle of this problem. I know to separate the radicals square both sides, then isolate radical but I am not coming up with the correct answer. Any help is appreciated! Thanks Jane P. sqrt(2x+5) + sqrt(x+6) = 91 solutions Answer 17157 by venugopalramana(3286)   on 2006-03-16 23:04:18 (Show Source): You can put this solution on YOUR website!sqrt(2x+5) + sqrt(x+6) = 9 sqrt(2x+5)= 9 - sqrt(x+6) ...square both sides 2X+5=81+X+6-18*SQRT.(X+6) X-82 = -18*SQRT.(X+6)...SQUARING AGAIN X^2+6724-164X =324(X+6)=324X+1944 X^2- 488X+4780 = 0 X^2-10X - 478X+4780=0 X(X-10)- 478(X - 10)=0 (X-10)(X-478)=0...HENCE X-10 = 0 ....OR....X=10 WHICH IS CORRECT SINCE sqrt(2x+5) + sqrt(x+6) =sqrt(2*10+5) + sqrt(10+6) = 5 + 4 = 9 X-478 = 0 OR ..X=478 IS AN EXTRANEOUS ANSWER OBTAINED DUE TO REPEATED SQUARING ...WE SHOULD IGNORE THAT
 Linear-systems/30467: Please help me work this problem. I have the answer but don't know how to work it. Please explain in very plain English terms! Point A (-4,1) is in the standard (x,y) coordinate plane. What must be the coordinates of Point B so that the line x = 2 is the perpendicular bisector of AB ? (note: there is a line over AB) Thank you for your help...Karen1 solutions Answer 17114 by venugopalramana(3286)   on 2006-03-16 11:03:08 (Show Source): You can put this solution on YOUR website!Point A (-4,1) is in the standard (x,y) coordinate plane. What must be the coordinates of Point B so that the line x = 2 is the perpendicular bisector of AB ? (note: there is a line over AB) Thank you for your help...Karen LINE X=2 IS PARALLEL TO Y AXIS SO ITS PERPENDICULAR WOULD BE PARALLEL TO X AXIS HENCE ITS EQUATION SHALL BE Y=K......THAT IS EQN.OF AB SHALL BE Y=CONSTANT FURTHER X=2 BISECTS AB.HENCE MID POINT OF AB SHALL BE ON X=2 THAT IS X COORDINATE OF MIDPOINT SHALL BE EQUAL TO 2.IF WE TAKE POINT B AS (P,Q) THEN X COORDINATE OF MID POINT OF AB SHALL BE = (-4+P)/2=2 -4+P=4 P=8 WE SPROVED EARLIER THAT EQN OF AB SHALL BE IN THE FORM Y=CONSTANT..OR ITS SLOPE IS ZERO SINCE Y=CONSTANT MEANS ...Y=0*X+CONSTANT SLOPE OF AB =(Q-1)/(P-(-4))=0...SO Q-1=0...SO Q=1... HENCE POINT B IS (8,1) YOU WILL SEE ITS EQN. IS Y=1..TO UNDERSTAND HOW THEY LOOK LIKE SEE THE GRAPH BELOW
 Linear_Algebra/30459: (8) Question: Choose the nonlinear systesm below that have (x,y)=(4,3) as on of their solutions. (A) x^2+y^2=25 x+y=7 (B) y=x^2+2x-21 x-y+1 (C) x^2-y^2=7 2x+7=0 (D) both a and b (E) both b and c Please be specific as to which letter answer is correct.1 solutions Answer 17108 by venugopalramana(3286)   on 2006-03-16 10:27:45 (Show Source): You can put this solution on YOUR website!(8) Question: Choose the nonlinear systesm below that have (x,y)=(4,3) as on of their solutions. SUBSTITUTE X=4 AND Y=3 AND CHECK...... (A) x^2+y^2=25...........4^2+3^2=16+9=25..OK x+y=7....4+3=7...OK.. (B) y=x^2+2x-21.......3=4^2+2*4-21=16+8-21=3...OK x-y+1..DO YOU MEAN X-Y=1...IF SO ....4-3=1....OK (C) x^2-y^2=7...........4^2-3^2=16-9=7...OK 2x+7=0.....2*4+7=8+7=15....NOT CORRECT..... (D) both a and b..........OK................... (E) both b and c ..........NOT OK............ Please be specific as to which letter answer is correct. SUBJECT TO ABOVE CORRECTION D IS CORRECT
 Linear_Algebra/30436: On the vector space M2,2 of 2*2 real matrices, show that defining = tr(A(B^T))1 solutions Answer 17067 by venugopalramana(3286)   on 2006-03-15 22:51:56 (Show Source): You can put this solution on YOUR website!On the vector space M2,2 of 2*2 real matrices, show that.....WHAT?SHOW WHAT?..PLEASE WRITE QUESTION PROPERLY AND YOU WILL GET THE ANSWER...PLEASE ACKNOWLEDGE THIS TO ALGEBRA.COM GIVING THE PROPER QUESTION defining = tr(A(B^T))
 Polynomials-and-rational-expressions/30435: Please explain how to factor the following problem by grouping to solve the equation. x^3+3x^2-4x-12=01 solutions Answer 17066 by venugopalramana(3286)   on 2006-03-15 22:49:18 (Show Source): You can put this solution on YOUR website!Please explain how to factor the following problem by grouping to solve the equation. LET F(X)= x^3+3x^2-4x-12=0 TRY PUTTING X=0,1,-1,2,-2..ETC..TO GET A ZERO WE FIND AT X=2..F(2)=2^3+3*2^2-4*2-12=8+12-8-12=0...SO X-2 IS A FACTOR.DIVIDE F(X) WITH X-2 NOW 2...|1..........3..........-4.........-12 ....|0..........2..........10..........12 .................................................. .....1..........5...........6...........0 HENCE X^2+5X+6 IS THE QUOTIENT =X^2+2X+3X+6=X(X+2)+3(X+2)=(X+2)(X+3) HENCE F(X)=(X-2)(X+2)(X+3)
 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(3286)   on 2006-03-15 22:41:46 (Show Source): You can put this solution on YOUR website!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.
 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 these1 solutions Answer 17014 by venugopalramana(3286)   on 2006-03-15 11:21:03 (Show Source): You can put this solution on YOUR website!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
 Polynomials-and-rational-expressions/30341: Which of the following is a factor of the polynomial x^2 - x - 20? A) x-5 B) x-4 C) x+2 D) x+5 E) x+10 1 solutions Answer 17009 by venugopalramana(3286)   on 2006-03-15 10:53:26 (Show Source): You can put this solution on YOUR website!Which of the following is a factor of the polynomial x^2 - x - 20? THERE ARE 2 METHODS 1 FACTORISE AND CHECK..F(X)=X^2-X-20=X^2-5X+4X-20=X(X-5)+4(X-5)=(X-5)(X+4) SO...A...X-5 IS THE ANSWER. 2...PUT X=5,4,-2..ETC..AS PER THE ANSWERS AND CHECK WHEN FUNCTION F(X) BECOMES ZERO..WE SEE THAT F(5)=0..SO X-5 IS A FACTOR. A) x-5 B) x-4 C) x+2 D) x+5 E) x+10
 Complex_Numbers/30282: (6+8i)- (-7+i) I come up with 13+ 9i but I am not sure, can you help?1 solutions Answer 16974 by venugopalramana(3286)   on 2006-03-14 23:06:12 (Show Source): You can put this solution on YOUR website!(6+8i)- (-7+i) I come up with 13+ 9i but I am not sure, can you help? GOOD JOB!!YOU DID CORRECTLY FOR REAL NUMBERS BUT FOR COMPLEX NUMBER IT IS 8i-i=7i...SO THE ANSWER IS 13+7i
 Travel_Word_Problems/30292: Please answer this question for me: A missle is fired with an initial velocity of 1600 feet per second. When will it be 40,000 feet above the starting point? The answer I got was: 25 seconds because I did 40000 ------ = 25 I think its wrong though 16001 solutions Answer 16973 by venugopalramana(3286)   on 2006-03-14 23:02:42 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING AND DO IN FPS SYSTEM ....G=32 FT/SEC^2..SO USE -32 INSTEAD OF -9.8 IN THE FOLLOWING -------------------------------------------------------------- PLEASE HELP ME A rocket is fired with an initial velocity of 800 meters per second. When will it be 32000 meters above the starting point? Than answer I got was : 40 second because I divided... I think its wrong thought AS THE ROCKET GOES UP AGAINST EARTHS GRAVITATIONAL FORCE , ITS VELOCITY REDUCES,GIVEN BY ACCELERATION DUE TO GRAVITY =-9.8 M/SEC^2..THE FORMULA TO BE USED IS S = UT + (G/2)T^2....WHERE U = INIYIAL VELOCITY = 800 M/SEC T = TIME ELAPSED IN SECS. G=ACCELERATION DUE TO GRAVITY = - 9.8 M/SEC^2 FOR UPWARD TRAVEL...SO..WE GET 32000 = 800T -(9.8/2)T^2=800T-4.9 T^2 4.9T^2-800T+32000 =0 T=70.1 WHILE GOING UP AND T=93.2 WHILE FALLING DOWN.
 Travel_Word_Problems/30301: PLEASE HELP ME A rocket is fired with an initial velocity of 800 meters per second. When will it be 32000 meters above the starting point? Than answer I got was : 40 second because I divided... I think its wrong thought 1 solutions Answer 16965 by venugopalramana(3286)   on 2006-03-14 22:06:23 (Show Source): You can put this solution on YOUR website!PLEASE HELP ME A rocket is fired with an initial velocity of 800 meters per second. When will it be 32000 meters above the starting point? Than answer I got was : 40 second because I divided... I think its wrong thought AS THE ROCKET GOES UP AGAINST EARTHS GRAVITATIONAL FORCE , ITS VELOCITY REDUCES,GIVEN BY ACCELERATION DUE TO GRAVITY =-9.8 M/SEC^2..THE FORMULA TO BE USED IS S = UT + (G/2)T^2....WHERE U = INIYIAL VELOCITY = 800 M/SEC T = TIME ELAPSED IN SECS. G=ACCELERATION DUE TO GRAVITY = - 9.8 M/SEC^2 FOR UPWARD TRAVEL...SO..WE GET 32000 = 800T -(9.8/2)T^2=800T-4.9 T^2 4.9T^2-800T+32000 =0 T=70.1 WHILE GOING UP AND T=93.2 WHILE FALLING DOWN.
 Quadratic_Equations/30208: when a question asks you to show something algebraically, DO NOT substitute numbers for the variables and give me an example. On the other hand, when a question asks you to show something numerically or to provide an example, you may substitute numbers for the variables. Show all work. Let p(x) = x4 - 6x3 + 11x2 - 6x. Using the fact that p(1) = 0, completely factor p (i.e., express p as a product of linear factors).1 solutions Answer 16909 by venugopalramana(3286)   on 2006-03-14 11:20:03 (Show Source): You can put this solution on YOUR website!Let p(x) = x4 - 6x3 + 11x2 - 6x. Using the fact that p(1) = 0, completely factor p (i.e., express p as a product of linear factors SO X-1 IS A FACTOR.. p(x) = x4 - 6x3 + 11x2 - 6x. =X(X^3-6X^2+11X-6)=X(X-1)(AX^2+BX+C)...COMPARING COEFFICIENTS OF LIKE TERMS COEFFICIENT OF X^4.......GIVES US....A=1....... COEFFICIENT OF X GIVES US............C=6...... COEFFICIENT OF X^2 GIVES US ....C-B=11...SINCE C IS 6 , WE GET B=-5.. SO WE GET p(x) = x4 - 6x3 + 11x2 - 6x. =X(X^3-6X^2+11X-6)=X(X-1)(AX^2+BX+C)=X(X-1)(X^2-5X+6)=X(X-1)(X^2-3X-2X+6)= X(X-1){X(X-3)-2(X-3)}=X(X-1)(X-3)(X-2)
 Polynomials-and-rational-expressions/30217: Explain why x to the 4th+2xsquared+4 has no real root while every polynomial function of degree 3 has at least 1 real root1 solutions Answer 16908 by venugopalramana(3286)   on 2006-03-14 11:06:59 (Show Source): You can put this solution on YOUR website!Explain why x to the 4th+2xsquared+4 has no real root X^4+2X^2+4={(X^2)^2+2(X^2)(1)+(1)^2}-(1)^2+4 = (X^2+1)^2+3...OR...(X^2+1)^2=-3 WE FIND THT THE SQUARE OF A NUMBER X^+1 IS NEGATIVE WHICH HAS NO REAL SOLUTION.(X^2+1)=i*SQRT.3 HENCE X^2+1= i*SQRT.3 X^2=-1+i*SQRT.3 = A COMPLEX NUMBER..HENCE ALL FOUR ROOTS OF X ARE COMPLEX/IMAGINARY NUMBERS.THAT IS THERE ARE NO REAL SOLUTIONS FOR X. IT IS POSSIBLE HERE SINCE THE TWO PAIRS OF COMPLEX ROOTS ARE CONJUGATE WITH THE RESULT THAT THE FUNCTIONS OF THEIR SUM/PRODUCT IS REAL.SINCE THE COEFFICIENTS OF VARIABLE IN THE POLYNOMIAL ARE FUNCTIONS OF SUM/PRODUCT OF THE ROOTS,IT IS NECESSARY THAT THE ROOTS SHOULD BE REAL OR CONJUGATE COMPLEX NUMBERS FOR THIS TO HAPPEN IN CASE OF A POLYNOMIAL WITH REAL COEFFICIENTS.IN THIS CASE WE HAVE IF A,A',B,B' ARE ROOTS (A AND A' ARE CONJUGATE ...B AND B' ARE CONJUGATE).. THEN A+A'+B+B'=0....POSSIBLE SINCE A+A' IS REAL AND B+B' IS REAL AA'+AB+AB'+A'B+A'B'+BB'=2...POSSIBLE SINCE AA' AND BB' ARE REAL AND AB+AB'+A'B+A'B'=A(B+B')+A'(B+B')=(A+A')(B+B')=REAL*REAL=REAL .....ETC... while every polynomial function of degree 3 has at least 1 real root IN CASE OF 3 RD. DEGREE POLYNOMIAL ,WE HAVE 3 ROOTS..SO IF THEY ARE COMPLEX THEN AS WE SHOWED ABOVE FOR POLYNOMIAL WITH REAL COEFFICIENTS ..COMPLEX ROOTS OCCUR IN PAIRS OF CONJUGATE NUMBERS ONLY.SO IN 3 ROOTS 2 COULD BE COMPLEX CONJUGATES ,BUT THE 3 RD. HAS TO BE REAL IF THE COEFFICIENTS ARE REAL.
 Matrices-and-determiminant/30198: when a question asks you to show something algebraically, DO NOT substitute numbers for the variables and give me an example. On the other hand, when a question asks you to show something numerically or to provide an example, you may substitute numbers for the variables. Show all work A square matrix is lower triangular if every entry above the diagonal is 0. a. Give an example of a 3x3 lower triangular matrix. b. Using your example from part (a), show that the determinant of a lower triangular matrix is the product of the entries on the diagonal. c. Show algebraically that the determinant of a 2x2 lower triangular matrix will always be the product of the entries on the diagonal. 1 solutions Answer 16903 by venugopalramana(3286)   on 2006-03-14 10:34:46 (Show Source): You can put this solution on YOUR website!a. Give an example of a 3x3 lower triangular matrix. (A)= b. Using your example from part (a), show that the determinant of a lower triangular matrix is the product of the entries on the diagonal. |A|=1(3*2-0*1)-0+0 =6 PRODUCT OF DIAGONAL ELEMENTS =1*3*2=6 =|A| c. Show algebraically that the determinant of a 2x2 lower triangular matrix will always be the product of the entries on the diagonal. LET (B) = |B|= a*c-0*b=ac PRODUCT OF DIAGONAL ELEMENTS =a*c = ac =|B|
 logarithm/30061: Show that log[base 2]9 x log[base 3]8 = log[base 2]8 x log[base 3]9. Thank you.1 solutions Answer 16811 by venugopalramana(3286)   on 2006-03-13 11:23:58 (Show Source): You can put this solution on YOUR website!log[base 2]9 x log[base 3]8 = log[base 2]8 x log[base 3]9. USING THE FORMULA LOG (A) TO BASE B =LOG(A)/LOG(B)....TO ANY COMMON BASE WE GET log[base 2]9 x log[base 3]8 = log[base 2]8 x log[base 3]9. LHS={LOG(9)/LOG(2)}*{LOG(8)/LOG(3)}=LOG(9)*LOG(8)/{LOG(2)*LOG(3)} RHS={LOG(8)/LOG(2)}*{LOG(9)/LOG(3)}=LOG(9)*LOG(8)/{LOG(2)*LOG(3)} =LHS
 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(3286)   on 2006-03-13 11:19:12 (Show Source): You can put this solution on YOUR website!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
 Graphs/30060: i need help with writing equasions in slope- intercept form and point slope equasion of a line and Slope intercept of a line1 solutions Answer 16809 by venugopalramana(3286)   on 2006-03-13 11:06:04 (Show Source): You can put this solution on YOUR website!1. Write in slope intercept the equation of theline passing through the two points.Show that the line is perpindicular to the given line. (-2,-2), (1,3); y= 3x-1 EQN.OF LINE JOINING (X1,Y1) AND (X2,Y2) IS GIVEN BY Y-Y1=(X-X1)*(Y2-Y1)/(X2-X1), WHERE (Y2-Y1)/(X2-X1) IS THE SLOPE OF THE LINE AND Y1-X1*(Y2-Y1)/(X2-X1)IS THE INTERCEPT. HENCE EQN.OF LINE IS Y+2=(X+2)*(3+2)/(1+2)=X(5/3)+2*5/3 Y=X(5/3)+10/3-2=X(5/3)+4/3.....HENCE SLOPE IS 5/3 AND INTERCEPT IS 4/3 SLOPE OF GIVEN LINE Y=3X-1 IS 3 ,,,FOR 2 LINES TO BE PERPENDICULAR,THE PRODUCT OF THEIR SLOPES SHOULD BE -1.HERE THE PRODUCT IS 3*5/3=5..HENCE THEY ARE NOT PERPENDICULAR TO EACH OTHER..CHECK BACK YOUR NUMBERS...COPY THE PROBLEM PROPERLY.AS GIVEN THE LINES ARE NOT PERPENDICULAR. 2. write in slope-intercept form the equation of the line passing through the given point and perpindicular to the given line. (-4,-7), y=-4x-7 SLOPE OF Y=-4X-7 IS -4 AS GIVEN ABOVE FOR 2 LINES TO BE PERPENDICULAR,THE PRODUCT OF THEIR SLOPES SHOULD BE -1.HENCE THE SLOPE OF THE REQUIRED LINE IS -1/-4=1/4 EQN.OF REQD. LINE IS Y+7=(1/4)(X+4)=X/4 + 1 Y= X/4 - 6
 Matrices-and-determiminant/30044: The value of square-root(2/pi) * Integral(e to the power -square(t)/2 ), where "t" is a variable from 4 to infinity is equal to what? 1 solutions Answer 16807 by venugopalramana(3286)   on 2006-03-13 10:57:06 (Show Source): You can put this solution on YOUR website!The value of square-root(2/pi) * Integral(e to the power -square(t)/2 ), where "t" is a variable from 4 to infinity is equal to what? THIS IS CALLED PROBABILITY INTEGRAL FOR WHICH YOU HAVE TO REFER STANDARD TABLES THE TABLE IS MADE FOR P(T)=(1/SQRT(2PI))*INTEGRAL{0 TO T OF (E^(-T^2/2) DT)} IT IS THE AREA UNDER THE CURVE.PRACTICALLY,THE ENTIRE CURVE LIES BETWEEN T=- 3AND +3...99.73% IN FACT..SO IF YOU WANT AREA BEYOND T=4 IT IS ALMOST ZERO..THE ANSWER IS NEARLY ZERO.
 Equations/30056: I need to determine the following for these two problems :HOW MANY X-INTERCEPTS THE PARABOLA HAS, and WHETER ITS VERTEX LIES ABOVE OR BELOW OR ON THE X-AXIS. 1.problem y=x^2-5x+6 2.problem y=-x^2+2x-11 solutions Answer 16805 by venugopalramana(3286)   on 2006-03-13 10:41:52 (Show Source): You can put this solution on YOUR website!I need to determine the following for these two problems :HOW MANY X-INTERCEPTS THE PARABOLA HAS, and WHETER ITS VERTEX LIES ABOVE OR BELOW OR ON THE X-AXIS. 1.problem y=x^2-5x+6 PUT Y=0 AND SOLVE FOR X TO GET X INTERCEPTS. X^2-5X+6=0=X^2-2X-3X+6=0=X(X-2)-3(X-2)=0=(X-2)(X-3)=0....X=2 AND 3... SO THE X INTERCEPTS ARE AT X = 2 AND X = 3 Y=X^2-5X+6={X^2-2*X*5/2+(5/2)^2}-(5/2)^2+6 =(X-2.5)^2 - 0.25 SO THE VERTEX IS AT X=2.5 AND Y=-0.25...THAT IS BELOW X AXIS 2.problem y=-x^2+2x-1 DOING THE SAME WAY WE GET Y=-(X-1)^2=0 AND HENCE X INTERCEPTS ARE X=1 AND VERTEX IS AT X=1 AND Y=0 SO THE VERTEX IS ON THE X AXIS.
 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(3286)   on 2006-03-12 11:04:12 (Show Source): You can put this solution on YOUR website!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
 Quadratic_Equations/29894: a. if sin=-7/8,then the value of 1/cotA in the interval 3pi/2 b.What is the exact value of sin 5pi/6 c.sketch the graph of y=cos 2x and y=-0.5 over the domain -pi<-x<-pi.determine the exact values of x where these 2 graphs intersect. d. draw a graph that represents a sine function in the form y=A sin B {x+C}+D or a cosine function in the form y=AcosB(x+C}+D. Write an equation of the graph in both forms 1 solutions Answer 16699 by venugopalramana(3286)   on 2006-03-12 04:07:26 (Show Source): You can put this solution on YOUR website!a. if sin=-7/8,then the value of 1/cotA in the interval 3pi/2 <2<2pi SIN A = -7/8 COS A =SQRT(1-(SIN A)^2)=SQRT(1-49/64)=SQRT(15/64)=SQRT(15)/8 1/COT A = TAN A=SIN A/COS A=-(7/8)/(SQRT(15)/8)=-7/SQRT(15)...IN THE GIVEN RANGE TAN A IS NEGATIVE. b.What is the exact value of sin 5pi/6 SIN 5PI/6=SIN (PI-PI/6)=SIN (PI/6)=1/2 c.sketch the graph of y=cos 2x and y=-0.5 over the domain -pi<-x<-pi.determine td,he exact values of x where these 2 graphs intersect. THE 2 GRAPHS INTERSECT AT X=PI-PI/3=2PI/3 AND PI+PI/3=4PI/3
 Quadratic_Equations/29895: a. if sin=-7/8,then the value of 1/cotA in the interval 3pi/2 b.What is the exact value of sin 5pi/6 c.sketch the graph of y=cos 2x and y=-0.5 over the domain -pi<-x<-pi.determine td,he exact values of x where these 2 graphs intersect.1 solutions Answer 16675 by venugopalramana(3286)   on 2006-03-11 23:06:41 (Show Source): You can put this solution on YOUR website!a. if sin=-7/8,then the value of 1/cotA in the interval 3pi/2 <2<2pi SIN A = -7/8 COS A =SQRT(1-(SIN A)^2)=SQRT(1-49/64)=SQRT(15/64)=SQRT(15)/8 1/COT A = TAN A=SIN A/COS A=-(7/8)/(SQRT(15)/8)=-7/SQRT(15)...IN THE GIVEN RANGE TAN A IS NEGATIVE. b.What is the exact value of sin 5pi/6 SIN 5PI/6=SIN (PI-PI/6)=SIN (PI/6)=1/2 c.sketch the graph of y=cos 2x and y=-0.5 over the domain -pi<-x<-pi.determine td,he exact values of x where these 2 graphs intersect. THE 2 GRAPHS INTERSECT AT X=PI-PI/3=2PI/3 AND PI+PI/3=4PI/3
 Coordinate-system/29860: I am working with parabolas. For this problem I need to Complete the square Give the Vertex Give the Axis Give the x-intercepts Give the y-intercepts Give a point symmetric to the y-intercept Draw the graph The problem is y=x^2-2x-3. Can anyone help me solve this. I am working on it and would like to have something to check my answer with. 1 solutions Answer 16624 by venugopalramana(3286)   on 2006-03-11 07:58:25 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLE WHICH IS ALMOST SAME AS YOUR PROBLEM AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK. Y=X^2-2X-3=X^2-2*X*1+1^2-1-3=(X-1)^2-1-3=(X-1)^2-4=0 X INTERCEPT IS OBTAINED BY PUTTING Y=0....WE GET X^2-2X-3=0=X^2-3X+X-3=X(X-3)+1(X-3)=(X-3)(X+1)=0...SO..X=3 OR -1... Y INTERCEPT IS GOT BY PUTTING X =0...WE GET Y=0-0-3=-3 SEE THE GRAPH BELOW ...... THE AXIS IS X=1 AS YOU CAN SEE FROM THE GRAPH ONE POINT SYMMETRIC TO Y INTERCEPT??NO...SYMMETRIC TO AXIS IT IS......(-1,3) ****************************************************************************** I have to put this equation y=x^2-2x-15 into this form y=a(x-h)^2+k. MAKE A PERFECT SQUARE USING X^2 AND X TERMS.ADD AND SUBTRACT THE REQUIRED CONSTANT FOR THE PURPOSE. Y=(X-1)^2-1-15=(X-1)^2-16...COMPARING WITH THE ABOVE y=a(x-h)^2+k. WE GET A=1 AND K=-16 I have to find the line of symmetry. X-1=0 IS THE LINE OF SYMMETRY SINCE ON EITHER SIDE OF X=1,WE GET SYMMETRIC/SAME VALUES FOR Y..AT X=1+2=3..Y IS -12 AND AT X=1-2=-1 ALSO WE GET Y=-12 (h,k)=vertex I think you use complete the square technique. YA ..THE VERTEX AS YOU SHOULD KNOW NOW IS AT X=1 AND AT X=1 ,Y=-16 .SO (1,-16) IS THE VERTEX THE GRAPH WILL LOOK LIKE THIS Please help. thanks
 Trigonometry-basics/29855: prove each of the following identities algebraically a. cosA+cosA tan squareA=secA b.cosA(secA-cscA)=1-cotA c.Prove the identity sin squareA divided 1-cosA=secA+1 divided by secA d.Verify sin square x+cos square x=1 for x=5pi divided by 3.Graph y=sin square x+cos square x and describe the result.Also explain the result. Thank you so much . This is my first attempt at pure math so i am very bad at it.1 solutions Answer 16623 by venugopalramana(3286)   on 2006-03-11 07:27:42 (Show Source): You can put this solution on YOUR website!prove each of the following identities algebraically a. cosA+cosA tan squareA=secA LHS=COS A (1+TANSQUARE A)=COS A * SECSQ.A=COS A/COSSQ.A =1/COS A = SEC A =RHS. b.cosA(secA-cscA)=1-cotA LHS=COS A*1/COS A-COS A/SIN A = 1-COT A = RHS c.Prove the identity sin squareA divided 1-cosA=secA+1 divided by secA LHS = (1-COSSQ.A)/(1-COS A)=1+COS A RHS={(1/COS A) + 1 }/(1/COS A)=(1+COS A)COS A/COS A=1+COS A=LHS d.Verify sin square x+cos square x=1 for x=5pi divided by 3. SINSQ.5PI + COSSQ. 5PI = SINSQ.(2*2PI+PI) + COSSQ.(2*2PI+PI) =SINSQ.PI + COSSQ. PI = (0)^2 + (-1)^2 =0+1=1 Graph y=sin square x+cos square x and describe the result. Also explain the result...WE GET A LINE PARALLEL TO XAXIS AT A DISTANCE OF PLUS ONE UNIT FROM IT.THAT IS Y=1..WHATEVER MAY BE THE VALUE OF X .
 Trigonometry-basics/29856: if 9cos square A+ 3 cos A = 0where 90degrees<0<180degrees what is the value of A to the nearest 10th.1 solutions Answer 16622 by venugopalramana(3286)   on 2006-03-11 06:34:55 (Show Source): You can put this solution on YOUR website!9cos square A+ 3 cos A = 0 =3COS A(3COS A+1)=0 SO .....COS A = 0 ...HENCE A=90 ..IN 90degrees 3COS A +1 =0...OR...COS A =-1/3 A=180-70.5=109.5 DEGREES
 Trigonometry-basics/29858: What is the exact value of sin square pi/ 6 - 2sin pi/ 6 cos pi/ 6 =cos square (-pi/6)? I got 2-Sqrt3/2 1 solutions Answer 16621 by venugopalramana(3286)   on 2006-03-11 06:22:43 (Show Source): You can put this solution on YOUR website!What is the exact value of sin square pi/ 6 - 2sin pi/ 6 cos pi/ 6 ...............=cos square (-pi/6)?...WHAT IS THIS EQUAL TO ?? =(1/2)^2-2*(1/2)*(SQRT 3)/2=1/4 - SQRT(3)/2={2-SQRT(3)}/2 I got 2-Sqrt3/2 CORRECT .BUT PUT BRACKET ..OTHERWISE IT MAY BE TAKEN AS 2 -(SQRTRT 3)/2 RHS = cos square (-pi/6)?...= {SQRT(3)/2}^2 = 3/4
 Trigonometry-basics/29857: another form of cos(30 degrees-A)+sin(60degrees+A)= I got sqrt of 3 divided by 2 sin A + cos A1 solutions Answer 16620 by venugopalramana(3286)   on 2006-03-11 05:20:04 (Show Source): You can put this solution on YOUR website!another form of cos(30 degrees-A)+sin(60degrees+A)= = COS 30 COS A +SIN 30 SIN A + SIN 60 COS A + COS 60 SIN A ={(SQRT 3)/2}COS A +(1/2) SIN A + {(SQRT 3)/2}COS A +(1/2) SIN A =(SQRT 3)COS A + SIN A IS THE ANSWER... I got sqrt of 3 divided by 2 sin A + cos A.....NO IT IS NOT CORRECT...
 Complex_Numbers/29843: how do you find the nth term of a negative complex number? for example the fourth roots of -41 solutions Answer 16595 by venugopalramana(3286)   on 2006-03-10 23:07:03 (Show Source): You can put this solution on YOUR website!how do you find the nth term of a negative complex number? I HOPE YOU MEAN N th. ROOT NOT AND NOT N th. TERM..OF A COMPLEX NUMBER ..CAN BE POSITIVE OR NEGATIVE... LET THE COMPLEX NUMBER BE Z = A+Bi IN GENERAL WHICH CAN ALWAYS BE WRITTEN AS Z = R*(COS X + i SIN X ) WHERE R = SQRT(A^2+B^2) AND COS X = A/(SQRT(A^2+B^2) AND SIN X = B/(SQRT(A^2+B^2) SINCE N th.ROOT OF R IS EASY TO FIND (R^1/N),R BEING POSITIVE, WE NEED TO ONLY FIND Nth.ROOT OF SAY Z = COS X +i SIN X Z^1/N = (COS X +i SIN X)^1/N...AS PER DEMOVIERS THEOREM THIS IS EQUAL TO = COS (X/N) +i SIN (X/N)..SO NOW WE USE THE GENERAL FORMULA FOR TRIGNOMETRIC EQNS ..THAT IS ..IF COS X = P SAY THEN X = COS ^-1(P)..ITS GENERAL SOLUTION IS X=2K*(PI)+COS^-1(P)...GIVE K=1,2,3 ETC.. TO GET ALL THE DIFFERENT ROOTS. for example the fourth roots of -4 Z = -4 =4*(-1)...SINCE 4 TH. ROOT OF 4 CAN BE EASILY FOUND OUT .LET US FIND FOURTH ROOTS OF -1.WE CAN MULTIPLY THAT ANSWER WITH 4TH ROOT OF +4 TO GET THE TOTAL ANSWER. Z = COS (0) +i SIN (0).. = COS (2K*PI+0)+iSIN (2K*PI+0)=COS (2K*PI)+i SIN (2K*PI) 4TH.ROOTS OF THIS ARE AS PER ABOVE FORMULA... Z^1/4 =COS (2K*PI)/4+i SIN (2K*PI)/4...PUTTING K=1,2,3 AND 4 TO GET THE 4 ROOTS WE GET... COS (2*1*PI)/4+i SIN (2*1*PI)/4..., COS (2*2*PI)/4+i SIN (2*2*PI)/4..., COS (2*3*PI)/4+i SIN (2*3*PI)/4..., COS (2*4*PI)/4+i SIN (2*4*PI)/4...AS THE 4 ROOTS.