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# Recent problems solved by 'venugopalramana'

venugopalramana answered: 3288 problems
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 Rational-functions/20983: 3(x+2)(x-3)(x+9)=01 solutions Answer 12178 by venugopalramana(3286)   on 2006-01-03 10:52:36 (Show Source): You can put this solution on YOUR website!3(x+2)(x-3)(x+9)=0 IF A PRODUCT IS ZERO THEN ONE OF THE FACTORS SHOULD BE ZERO. HENCE WE GET X+2=0...OR X-3=0....OR X+9=0...THAT IS X=-2 OR 3 OR -9 ..SO THEY ARE THE POSSIBLE SOLUTIONS FOR THIS EQUATION.
 Quadratic_Equations/20884: Write a matrix equation equivalent to the system of equations. 9x + 9y = -9 5x - 2y = 6 Cramer's rule. 6x + 4y = -4 y = -3x - 71 solutions Answer 12176 by venugopalramana(3286)   on 2006-01-03 10:24:38 (Show Source): You can put this solution on YOUR website!Write a matrix equation equivalent to the system of equations. 9x + 9y = -9 5x - 2y = 6 we write the matrix equation as (A)*(X)=(C)..where A is the (2,2)matrix of coefficients namely,9,9,5 and -2 here.X is the (2,1) matrix of unknowns x and y and C is the constants (2,1)matrix on the right side of the eqn NAMELY -9 AND 6.you will find that from the rule for equality of matrices,the above matrix eqn.in effect means the same as that of the given equations. ()*()=() Cramer's rule. 6x + 4y = -4 y = -3x - 7 GIVING BELOW EXAMPE OF CRAMERS RULE. My question is.... I was wondering how to do the cramer rule on a 3x3. I have found a bunch of examples and stuff, but I want to know how in the world do you find the determinants of the D, Dx, Dy.and Dz. If you could just tell me how, that would be great. 1 solutions Answer 9496 by venugopalramana(585) About Me on 2005-11-15 10:49:54 (Show Source): SEE THE FOLLOWING AND COME BACK IF YOU HAVE DIFFICULTY.HERE C,CX,CY,CZ REFER TO YOUR D,DX,DY,DZ...JUST A DIFFERENCE IN NOMENCLATURE.I SHOWED IN DETAIL A 2X2 DETERMINANT AND THEN IN BRIEF A 3X3 DETERMINANT 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
 Mixture_Word_Problems/20900: A certain quantity of gas, when compressed to a volume of 0.82 m^3 has a pressure of 76Pa. Find the pressure of the gas when it is compressed to 0.55. Assume a constant temperature.1 solutions Answer 12167 by venugopalramana(3286)   on 2006-01-03 05:43:10 (Show Source): You can put this solution on YOUR website!A certain quantity of gas, when compressed to a volume of 0.82 m^3 has a pressure of 76Pa. Find the pressure of the gas when it is compressed to 0.55. Assume a constant temperature. AS PER BOYLES LAW at constant temperature PRODUCT OF PRESSURE AND VOLUME REMAINS CONSTANT ...SO..... 76*0.82=P*0.55 P=76*0.82/0.55 = 113.3 Pa
 Human-and-algebraic-language/23378: A group of men can do a job in as many as thre are men in the group. If the number of men are increased by 5, the no. of days to do the job will be decreased by 4. Find the number of men in the group...i really need your help!!!1 solutions Answer 12166 by venugopalramana(3286)   on 2006-01-03 05:09:30 (Show Source): You can put this solution on YOUR website!A group of men can do a job in as many as there are men in the group. If the number of men is increased by 5, the number of days to do the job will be decreased by 4. Find the number of men in the group. Thanks! 1 solutions Answer 11344 by venugopalramana(581) About Me on 2005-12-19 11:17:56 (Show Source): LET X PEOPLE DO THE JOB IN X DAYS....SO IN ONE DAY X PEOPLE CAN DO 1/X JOB...OR..1 PERSON CAN DO IN ONE DAY 1/(X)^2 JOB 5 MORTE MEANS X+5 PEOPLE........X+5 CAN DO IN ONE DAY (X+5)/(X)^2 JOB ...HENCE HEY REQUIRE (X)^2/(X+5) DAYS TO DO 1 JOB..BUT THIS IS GIVEN AS EQUAL TO 4 LESS DAYS OR X-4 DAYS. (X)^2/(X+5)=X-4 X^2=(X+5)(X-4)=X^2+5X-4X-20 X^2-X^2+X-20=0 X-20=0 X=20...
 Mixture_Word_Problems/23380: Please give me some examples of Mixtures Problems. I really need it now. 1 solutions Answer 12165 by venugopalramana(3286)   on 2006-01-03 05:08:08 (Show Source): You can put this solution on YOUR website! A chemist needs 10 liters of a 50% salt solution. All she has available is a 20% salt solution and a 70% salt solution. How much of each of the two solutions should she mix to obtain her desired solution? 1 solutions Answer 11589 by venugopalramana(581) About Me on 2005-12-23 11:03:10 (Show Source): A chemist needs 10 liters of a 50% salt solution. All she has available is a 20% salt solution and a 70% salt solution. How much of each of the two solutions should she mix to obtain her desired solution? LET X LITRES OF 20% SOLUTION BE REQUIRED...SINCE TOTAL IS 10 LITRES ,WE HAVE TO ADD 10-X LITRES OF 70% SOLUTION..SO TAKING A SALT BALANCE SALT IN X LITRES OF 20 % SOLUTION......=X*20/100=0.2X SALT IN 10-X LITRES OF 10 % SOLUTION.. =(10-X)*70/100=7-0.7X SALT IN 10 LITRES OF FINAL 50 % SOLUTION =10*50/100=5 HENCE 0.2X+7-0.7X=5 0.7X-0.2X=7-5=2 0.5X=2 X=2/0.5=20/5=4 LITRES OF 20 % SWOLUTION AND 10-4=6 LITRES OF 70% ARE NEEDED TO BE MIXED TO GET 10 LITRES OF 50% SOLUTION -------------------------------------------------------------------------- SEE THIS EXAMPLE WHICH IS SIMILAR TO YOUR PROBLEM AND TRY.IF YOU STILL HAVE DIFFICULTY PLEASE COME BACK In a chemistry class, 7 liters of a 4% silver iodide solution must be mixed with a 10% solution to get a 6% solution. How many liters of a 10% solution are needed? A. 4.5L LET X LITRES OF 10% SOLUTION BE NEEDED.IT HAS X*10/100 LITRES OF SILVER IODIDE. B. 2.5L 7 LITRES OF 4% SILVER IODIDE SOLUTION HAS 7*4/100=28/100 LITRES OF SILVER IODIDE C. 7.0L TOTAL SOLUTION =X+7 LITRES…….IT HAS TOTAL 10X/100+28/100=(10X+28)/100 LITRES OF SILVER IODIDE D. 3.5L FINAL CONCENTRATION =6 %…SO 6/100=(10X+28)/(X+7)*100..OR…6=(10X+28)/(X+7)...CROSS MULTIPLYING.. 6(X+7)=(10X+28)…OR…6X+42=10X+28..OR….10X-6X=42-28=14…OR….4X=28..OR..X=7 LITRES.C IS THE ANSWER **********************************************************************
 Polynomials-and-rational-expressions/20942: what is the seventh term of the expansion (2a-3b) to the ninth power?1 solutions Answer 12164 by venugopalramana(3286)   on 2006-01-03 02:17:36 (Show Source): You can put this solution on YOUR website!7 TH TERM OF (2A-3B)^9 AS PER BINOMIAL THEOREM R+1 ST. TERM Tr+1 OF EXPANSION (X+A)^N IS GIVEN BY Tr+1=nCr*{X^(N-R)}*(A^R) ..WE WANT 7TH.TERM HENCE R=6...SO T7=9C6*{(2A)^(9-6)}*(3B)^6 =9C6*8*A^3*3^6*B^6 =(9!/6!*3!)*8*A^3*3^6*B^6
 Linear-systems/23376: I have two questions I just don't get them. Solve the system of linear equations by graphing: x+2y=5 x-y=2 that is the first one the second one is: 4x-y=6 3x+y=1 these are very hard for me please help1 solutions Answer 12163 by venugopalramana(3286)   on 2006-01-03 02:05:55 (Show Source): You can put this solution on YOUR website!SECOND PROBLEM IS SHOWN BELOW. ******************************************************************************** SOLVE BY GRAPHING MEANS DRAW THE GRAPHS AND FIND THE 2 GRAPHS OF 2 EQUATIONS CUT EACH OTHER .THAT IS THE SOLUTION.LET ME SHOW YOU THE SECOND PROBLEM. 4X-Y=6..OR...4X-6=Y..... 3X+Y=1...OR....Y=1-3X......... LET X=...........................0......1......2........3.........4......ETC.. Y=4X-6..IN FIRST EQN........-6......-2.....2........6.........10.......ETC Y=1-3X..IN SECOND EQN.......1.......-2.....-5.......-8........-11......ETC THE GRAPHS WILL LOOK LIKE THIS WE FIND THAT THE TWO GRAPHS ARE CUTTING AT X=1 AND Y=-2.HENCE THAT IS THE SOLUTION..YOU CAN CHECK BACK EQN.I.....4X-Y=4*1-(-2)=4+ 2 = 6 ..OK EQN.II.....3X+Y=3*1+(-2)=3-2=1........OK. ******************************************************************************* Solve the system of linear equations by graphing: x+2y=5.. OR 2Y=5-X..OR Y=(5-X)/2 x-y=2... OR Y=X-2 that is the first one the second one is: 4x-y=6 3x+y=1 SOLVE BY GRAPHING MEANS DRAW THE GRAPHS AND FIND THE 2 GRAPHS OF 2 EQUATIONS CUT EACH OTHER .THAT IS THE SOLUTION.LET ME SHOW YOU THE FIRST PROBLEM. THEN YOU CAN DO SECOND PROBLEM THE SAME WAY LET X=.........................0......1......2........3.........4......5....ETC. Y=(5-X)/2..IN FIRST EQN........5/2....2......3/2......1.........1/2....0...ETC Y=(X-2).....IN SECOND EQN.......-2....-1......0........1.........2......3...ETC THE GRAPHS WILL LOOK LIKE THIS WE FIND THAT THE TWO GRAPHS ARE CUTTING AT X=3 AND Y=1.HENCE THAT IS THE SOLUTION..YOU CAN CHECK BACK EQN.I.....X+2Y=3+2*1=3+2=5..OK EQN.II.....X-Y=3-1=2........OK.
 Length-and-distance/23351: Two friends are swimming laps back and forth across the length of a pool. They swim with constant speeds and essentially instantaneous turns at the ends of the pool. If at the beginning they leave from opposite ends of the pool and meet for the first time 21 meters from one end and continue on their ways to their own ends of the pool, trun around and meet the second tome 8 meters from the opppsite end (from which they first met), how long is the pool?1 solutions Answer 12162 by venugopalramana(3286)   on 2006-01-03 01:51:17 (Show Source): You can put this solution on YOUR website!Two friends are swimming laps back and forth across the length of a pool. They swim with constant speeds and essentially instantaneous turns at the ends of the pool. If at the beginning they leave from opposite ends of the pool and meet for the first time 21 meters from one end and continue on their ways to their own ends of the pool, trun around and meet the second tome 8 meters from the opppsite end (from which they first met), how long is the pool? LET THE LENGTH OF POOL = L......LET THE TWO FRIENDS BE F1 AND F2. LET THE SPEEDS OF F1 AND F2 BE X AND Y AND LET THEM START FROM ENDS.1 & 2 RESPECTIVELY I SWIM..... DISTANCE BETWEEN THE TWO =L...SINCE THEY LEAVE FROM OPPOSITE ENDS. RELATIVE SPEED =X+Y..SINCE THEY ARE SWIMMING IN OPPOSITE DIRECTIONS. TIME TAKEN TO MEET IN I SWIM =L/(X+Y) DISTANCE COVERED BY F1 DURING THIS TIME = (L*X)/(X+Y)= 21 METRES FROM END 1 SAY. DISTANCE COVERED BY F2 DURING THIS TIME = (L*Y)/(X+Y)= L-21 METRES FROM END 2. II SWIM........THERE IS AN AMBIGUITY IN WORDING OF THE PROBLEM HERE QUOTE ......meet for the first time 21 meters from one end and continue on their ways to their own ends of the pool(THEIR OWN ENDS MEAN THEY GO BACK TO THEIR ORIGINAL STARTING ENDS???OR CONTINUE ON THEIR WAYS MEANS THEY GO TO OPPOSITE ENDS FROM WHERE THEY STARTED) , trun around and meet the second tome 8 meters from the opppsite end UNQUOTE...CAPITAL LETTERS INDICATE THE AMBIGUITY IN WORDING..LET US WORK OUT THE PROBLEM ON THE FOLLOWING ASSUMPTION. ...THEY GO BACK TO THEIR OWN ENDS THAT IS TO REACH THE SAME ENDS AS THEY STARTED FROM.... DISTANCE TRAVELLED BY F1 TWO TO MEET A SECOND TIME =21 M.TO GO BACK TO END 1 + L-8 M. = 21+L-8=L+13 DISTANCE TRAVELLED BY F2 TWO TO MEET A SECOND TIME = L-21 M.TO GO TO END 2 + 8 M.= L-21+8=L-13. ANALYSIS....... SINCE THEY WERE TOGETHER AT THE SAME TIME / PLACE WHEN THEY FIRST MET IN I SWIM,THE DURATION FOR WHICH THE TWO SWAM TO MEET A SECOND TIME IS SAME SINCE THEY WERE MEETING AT THE SAME TIME THERE DURING II SWIM. HENCE .. (L+13)/X=(L-13)/Y...OR....Y/X=(L-13)/(L+13)......EQN.III BUT WE GOT FROM I SWIM THAT (L*X)/(X+Y)=21...AND........EQN.I (L*Y)/(X+Y)=L-21............EQN.II EQN.II/EQN.I...GIVES Y/X=(L-21)/L..........EQN.IV..USING EQN.III...WE GET (L-21)/L=(L-13)/(L+13)...CROSS MULTIPLYING (L-21)(L+13)=L(L-13) L^2+13L-21L-21*13=L^2-13L L^2+13L-21L-L^2+13L=21*13 5L=273 L=273/5=54.6 M ...THIS APPEARS TO BE MORE REASONABLE ASSUMPTION.
 Number-Line/20938: I need to help my son , when no books are sent home to complete school work. He has to answer the question on the definitions of number lines . Like coordinate plane, coordinate Axes, gragh of a ordered pair, funtion, origin, bar gragh, vertical line test, funtion notation, ordered pair ect. He has questions where these words muct complete the sentence. Well Im stuck. Help. Thanks Mom1 solutions Answer 12126 by venugopalramana(3286)   on 2006-01-02 11:38:25 (Show Source): You can put this solution on YOUR website!PLEASE GIVE THE QUESTIONS WE SHALL DEFINITELY HELP YOU
 Equations/20937: Please help with this equation: 4/3(WPB)=XY, solve for X I multiplied both sides by 3 to get 4WPB=3XY and then divided both sides by 3Y to get 4WPB/3Y = X. Does this sound right? Thank You!1 solutions Answer 12125 by venugopalramana(3286)   on 2006-01-02 11:36:46 (Show Source): You can put this solution on YOUR website!4/3(WPB)=XY, solve for X I multiplied both sides by 3 to get 4WPB=3XY and then divided both sides by 3Y to get 4WPB/3Y = X.... Does this sound right? Thank You PERFECT ..YOU DID A GREAT JOB ..BELIEVE IN YOUR SELF ..HAVE CONFIDENCE..YOU WILL COME OUT ON TOP
 Numbers_Word_Problems/20931: The sum of the squares of two positive consecutive odd integers is 394. Find the integer. 1 solutions Answer 12124 by venugopalramana(3286)   on 2006-01-02 11:31:32 (Show Source): You can put this solution on YOUR website!let any number be x ...then 2x will be always even....if we subtract 1 from it it will be always odd so 2x-1 is one odd number ..its next odd number will be 2x-1+2=2x+1 so sum of their squares (2x-1)^2+(2x+1)^2=394 2[(2x)^2+(1)^2]=394...using formula (a+b)^2+(a-b)^2=2[(a)^2+(b)^2] (4x^2+1)=394/2=197 4x^2=197-1=196 x^2=196/4=49 x=7...hence the 2 numbers are 2*7-1=14-1=13 and 15...you can easily verify 13^2=169 15^2=225 sum is 394
 Quadratic_Equations/20906: I have been asked to solve the equations simultaneously, and i believe one of them to be a quadratic equation, and so am stuck - please can you help me? x-y=3 2 x -10y-6=0 P.s. If you could show me how to achieve the answer, possibly it would help me understand it? Thankyou so much.1 solutions Answer 12123 by venugopalramana(3286)   on 2006-01-02 11:23:49 (Show Source): You can put this solution on YOUR website!x-y=3 x=y+3...or...y=x-3.... x^2 -10y-6=0 ,substituting y=x-3 from above eqn. in this we get , x^2-10(x-3)-6=0 x^2-10x+30-6=0 x^2-10x+24=0 find 2 factors for 24 which add up to -10.we find that -6*-4=+24 and -6-4=-10...so x^2-6x-4x+24=0 x(x-6)-4(x-6)=0 (x-6)(x-4)=0 x-6=0 ....or...x-4=0 x=6...or...x=4
 Expressions-with-variables/20929: if (a+b)/a = 6 and (b+c)/c = 9, compute the numerical value of a/c1 solutions Answer 12122 by venugopalramana(3286)   on 2006-01-02 11:17:26 (Show Source): You can put this solution on YOUR website!(a+b)/a = 6 ... (a/a)+(b/a)=6... 1+b/a=6 b/a=6-1=5..................I. and (b+c)/c = 9 (b/c)+(/c)=9 b/c=9-1=8......................II dividing II with I,we get (b/c)/(b/a)=8/5 (b*a)/(b*c)=8/5 a/c=8/5
 Functions/20925: Write a function rule for the table of values; x=0,1,2,-3 y=1,3,5,-51 solutions Answer 12102 by venugopalramana(3286)   on 2006-01-02 04:30:27 (Show Source): You can put this solution on YOUR website!SINCE THERE IS NO HINT ON THE NATURE OF RELATION,WE HAVE TO GO BY TRIAL AND ERROR LET US TRY LINEAR RELATION..THAT IS Y=AX+B AT X=0,Y=1....SO A*0+B=1...OR B=1 SO WE HAVE Y=AX+1 AT X=1,Y=3....SO A*1+1=3...ORA=3-1=2 SO WE HAVE Y=2X+1 NOW LET US TET FOR OTHER CASES WHETHER IT HOLDS AT X=2,Y=5....2*2+1=5.....OK AT X=-3,Y=-5.......2*-3+1=-6+1=-5.....OK HENCE THE RELATION IS Y=2X+1
 Linear_Algebra/20889: Find the range of the function: g(x) = x/3 + 5 if the domain is {-6, -3, 0, 3, 6} I can't even find any examples of this in my textbook (although it's on my homework), and I don't know where to even begin to find an answer. 1 solutions Answer 12071 by venugopalramana(3286)   on 2006-01-01 11:20:24 (Show Source): You can put this solution on YOUR website!IT IS NOT SO DIFFICULT AS YOU IMAGINE..IT IS VERY SIMPLE IF YOU KNOW WHAT IS REQUIRED TO BE DONE.. Find the range of the function: g(x) = x/3 + 5 if the domain is {-6, -3, 0, 3, 6} I can't even find any examples of this in my textbook (although it's on my homework), and I don't know where to even begin to find an answer. DOMAIN IS THE VALUES X TAKES .RANGE IS THE CORRESPONDING VALUES OF THE FUNCTION TAKES..HERE IT IS G(X)=X/3+5..FOR EX. WHEN X=-6 , WE GET -6/3+5=-2+5=3 WHEN X=-3,WE GET -3/3+5=4 WHEN X=0,WE GET 0+5=5 WHEN X = 3 ,WE GET 3/3+5=6 WHEN X=6 , WE GET 6/3+5=7 SO THE RANGE IS (3,4,5,6,7)
 Functions/20904: Four equations are shown. Which lines are parallel? Which lines are perpendicular? a. y=-1/8x+2 b. y=8x+4 c. y=-8x+2 d. 8x+y=4 Fill in the blanks with the correct leters. Lines______ and ______ are parallel. Lines______ and ______ are perpendicular.1 solutions Answer 12066 by venugopalramana(3286)   on 2006-01-01 08:28:26 (Show Source): You can put this solution on YOUR website!LINES A AND B ARE PERPENDICULAR. LINES C AND D ARE PARALLEL. SEE BELOW FOR EXPLANATION -------------------------------------------------------------------------- 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) What is the point of interectstion of line L1 and L2 1 solutions Answer 8530 by venugopalramana(345) About Me on 2005-10-29 23:45:54 (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. venugopal
 Geometry_Word_Problems/20895: How do i find the gradient of a line parallel to y=3x? hope you can help1 solutions Answer 12065 by venugopalramana(3286)   on 2006-01-01 08:23:26 (Show Source): You can put this solution on YOUR website!GRADIENT OR SLOPE OF A LINE IS GIVEN BY M WHERE THE STANDARD EQUATION OF LINE IS PUT IN THE FORM Y=M*X+C...HERE WE HAVE Y=3X..HENCE M=3. PARALLEL LINES ARE THOSE HAVING SAME GRADIENT OR SLOPE .HENCE GRADIENT OF A PARALLEL LINE IS ALSO 3
 Miscellaneous_Word_Problems/20896: A horizontal bar of negligible weight has an 82.6 lb weight hanging from the left end and a 43.0 lb weight hanging from the right end. The bar is seen to balance 56.2 in. from the right end. Find the length of the bar.1 solutions Answer 12064 by venugopalramana(3286)   on 2006-01-01 08:20:25 (Show Source): You can put this solution on YOUR website!A horizontal bar of negligible weight has an 82.6 lb weight hanging from the left end and a 43.0 lb weight hanging from the right end. The bar is seen to balance 56.2 in. from the right end. Find the length of the bar. LET AB BE THE BAR AND ITS LENGTH BE L.WE HAVE 82.6 POUNDS AT A AND 43 POUNDS AT B.IT IS BALANCING AT C SAY AND WE ARE GIVEN THAT CB=56.2..HENCE TAKING MOMENTS ABOUT C WE GET 82.6*AC=43*CB 82.6(AB-CB)=43*CB 82.6(L-56.2)=43*56.2 82.6L-82.6*56.2=43*56.2 82.6L=43*56.2+82.6*56.2=56.2(43+82.6)=56.2*125.6 L=56.2*125.6/82.6=85.4566 INCHES
 Equations/20899: What values of x, if any are not permitted? x-3/x^2-16 Thank You1 solutions Answer 12063 by venugopalramana(3286)   on 2006-01-01 08:11:36 (Show Source): You can put this solution on YOUR website!x-3/x^2-16 IN MATHS CERTAIN OPERATIONS LIKE DIVISION BY ZERO ,LOG OF ZERO OR A NEGATIVE NUMBER ;OR SQUARE ROOT OF A NEGATIVE NUMBER ETC...ARE NOT PERMITTED IN MATHS .HENCE WE SHOULD SEE THAT X DOES NOT TAKE SUCH VALUES WHICH WILL RESULT IN ADOPTING SUCH PROHIBITED PRACTICES. HERE IN DENOMINATOR,WE HAVE X^2-16..HENCE IT SHOULD NOT BE ZERO THAT IS X^2-16=0..OR X^2=16..OR...X=+4 AND -4 ...THERE IS NO OTHER PROHIBITED AREA .HENCE X SHOULD NOT EQUAL +4 AND -4.
 Money_Word_Problems/20905: Hi, I looked at like problems but did not see one for 3 types of coins... A coin collecton contains 29 coins made up of pennies, nickles, and quarters. The number of quarters is 8 less than the number of pennies. Thr total face value of the coins is \$1.77 How many of each denomination are there?1 solutions Answer 12061 by venugopalramana(3286)   on 2006-01-01 08:04:34 (Show Source): You can put this solution on YOUR website!Hi, I looked at like problems but did not see one for 3 types of coins... --------------------------------------------------------------------- HERE IS AN EXAMPLE OF 3 COIN COLLECTION DESIRED BY YOU.TRY TO WORK OUT YOUR PROBLEM AND COME BACK IF NEEDED --------------------------------------------------------------------------- a coins collection consists of 4, 2, and 3 cent coins. the number of 2 cent coin is one less than twice the number of 4 cent coins. the number 3 cent coins is five more than the number of 2 cent coins. the total value of all coins is \$3.00 find the number of 2 cent coins in the collection. 1 solutions -------------------------------------------------------------------------------- Answer 11647 by venugopalramana(563) on 2005-12-26 11:22:14 (Show Source): a coins collection consists of 4, 2, and 3 cent coins. the number of 2 cent coin is one less than twice the number of 4 cent coins. the number 3 cent coins is five more than the number of 2 cent coins. the total value of all coins is \$3.00 find the number of 2 cent coins in the collection. LET THE NUMBER OF 4 CENT COINS =X TWICE THIS =2X ONE LESS THAN THIS =2X-1 ..THIS IS EQUAL TO NUMBER OF 2 CENT COINS 5 MORE THAN THIS =2X-1+5=2X+4..THIS IS EQUAL TO NUMBER OF 3 CENT COINS SO TOTAL VALUE OF COINS =4X+2(2X-1)+3(2X+4)=4X+4X-2+6X+12 =14X+10...THIS EQUAL TO \$3 OR 300 CENTS HENCE 14X+10=300 14X=300-10=290 X=290/14..THERE IS A MISTAKE IN NUMBERS SINCE X IS COMING A FRACTION LET US TAKE \$2.90 INSTEAD OF \$3 TO ILLUSTRATE THE ANSWER. WE GET 14X+10=290 CENTS 14X=290-10=280 X=280/14=20 HENCE THERE ARE 2O COINS OF 4 CENTS ...................OF VALUE =20*4=80 CENTS NUMBER OF 2 CENT COINS =2X-1=2*20-1=40-1=39............OF VALUE =39*2=78 CENTS NUMBER OF 3 CENT COINS = 2X+4=2*20+4=40+4=44...........OF VALUE=44*3=132 CENTS ...................................................................------ TOTAL VALUE .........................................................290 CENTS
 Radicals/20909: How do you rationalize denominators and reduce the fraction completely? For example - 9 square root 5 over square root 3.1 solutions Answer 12054 by venugopalramana(3286)   on 2006-01-01 07:21:23 (Show Source): You can put this solution on YOUR website!-9squareroot5/squareroot3 square root of 3 can be rationalised by multiplying with square root of 3 .it is true for any number which requires rationalisation.so multiply n.r and d.r with that number that is all .here we get -(9squareroot5)*(squareroot3)/[(squareroot3)*(squareroot3)=-9square root(5*3)/3 =-(9squareroot15)/3=-3squareroot15
 Complex_Numbers/20910: Imagnary numbers: 1 solutions Answer 12052 by venugopalramana(3286)   on 2006-01-01 07:15:08 (Show Source): You can put this solution on YOUR website!(1-3i)/(2+i)=...we rationalise the denominator by multiplying with its conjugate...conjugate of 2+i is 2-i...hence multiplying n.r and d.r with 2-i, (1-3i)(2-i)/(2+i)(2-i)=[1*2-1*i-3i*2+(-3i)(-i)]/[2^2-i^2]=[2-7i-3]/(4+1) =(-1-7i)/5=-(1+7i)/5....using (a+b)(a-b)=a^2-b^2 and i^2=-1
 Complex_Numbers/20912: The distance betwwen the points (2,3)and (1,1) is1 solutions Answer 12050 by venugopalramana(3286)   on 2006-01-01 07:04:43 (Show Source): You can put this solution on YOUR website!FORMULA FOR DISTANCE BETWEEN 2 POINTS (X1,Y1) AND (X2,Y2) = [(X2-X1)^2+(Y2-Y1)^2]^0.5...HENCE DISTANCE BETWEEN (1,1) AND (2,3)=[(2-1)^2+(3-1)^2]^0.5=(1^2+2^2)^0.5=(5)^0.5=SQUARE ROOT OF 5
 Graphs/20923: If you know the vertex of the graph of a quadratic function, can you specify the range of the function? If so, what is the range? If not what additional information do you need?1 solutions Answer 12049 by venugopalramana(3286)   on 2006-01-01 06:59:44 (Show Source): You can put this solution on YOUR website!If you know the vertex of the graph of a quadratic function, can you specify the range of the function? If so, what is the range? If not what additional information do you need? LET US SEE THIS FIRST VISUALLY BY PLOTTING SOME GRAPHS FOR 1....Y=(X-2)^2+4 2....Y=4-(X-2)^2 3....Y=(X-2)^2-4 SO YOU FIND THAT WHEN QUADRATIC EQUATION GIVES US A VERTEX ,AS THE NAME IMPLIES ,WE ARE GETTING A PEAK VALUE OR TROUGH VALUE FOR THE FUNCTION Y AT THAT POINT .DEPENDING ON WHETHER IT IS PEAK (MAXIMUM)OR TROUGH (MINIMUM) ,ONE BOUNDARY FOR Y IS FIXED ,THE OTHER BOUNDARY BEING PLUS INFINITY OR MINUS INFINITY AGAIN DEPENDING ON EHETHER WE GOT A PEAK (MAXIMUM)OR TROUGH (MINIMUM). SO IN THIS WAY THE RANGE OF Y GETS FIXED.HENCE WE NEED THE VERTEX ,MORE PRECISELY THE Y COORDINATE OF THE VERTEX AND ITS NATURE..... PEAK (MAXIMUM)OR TROUGH (MINIMUM)...TO DEFINE THE RANGE OF THE FUNCTION. YOU CAN SEE IT ALGEBRAICALLY ALSO AS A PERFECT SQUARE IS ALWAYS POSITIVE OR ZERO....ITS MINIMUM VALUE CAN BE ZERO.....HENCE IT WILL CONTRIBUTE TO MAXIMIXE OR MINIMISE THE FUNCTION (VALUE OF Y)AT THAT POINT DEPENDING ON THE SIGN INFRONT OF THE PERFECT SQUARE BEING NEGATIVE OR POSITIVE RESPECTIVELY.IN EXAMPLES 1 AND 3 IT IS POSITIVE +(X-2)^2. HENCE VERTEX OR Y VALUE BECOMES MINIMUM THERE SINCE YOU GET THE LOWEST VALUE BY ADDING THE LEAST VALUE OF ZERO. IN EXAMPLE 2 IT IS NEGATIVE -(X-2)^2.HENCE VERTEX OR Y VALUE IS MAXIMUM THERE SINCE YOU GET THE HIGHEST VALUE BY SUBTRACTING THE LEAST VALUE OF ZERO.
 Geometric_formulas/23282: the transformation with rule T(x,y)=(x+2,y+6)is translation a.graph (7,3)and T(7,3) B.FIND THE SLOPE OF THE LINE THROUGH (7,3)and its image1 solutions Answer 12044 by venugopalramana(3286)   on 2006-01-01 06:18:00 (Show Source): You can put this solution on YOUR website!SEE MY WORKING BELOW T(x,y)=(x+2,y+6)is translation a.graph (7,3)and T(7,3)....SINCE T(x,y)=(x+2,y+6),WE GET T(7,3)=(7+2,3+6)=(9,9) B.FIND THE SLOPE OF THE LINE THROUGH (7,3)and its image...FROM ABOVE WE GOT ITS IMAGE THAT IS T(7,3) AS EQUAL TO (9,9)... HENCE SLOPE OF LINE CONNECTING (7,3) AND (9,9) IS =(9-3)/(9-7)=6/2=3
 Travel_Word_Problems/23284: Please help! Like all planets Jupiter has an elliptical orbit, with the centre of the sun located at a focus. Write an equation of the ellipse that models Jupiter's orbit around the sun. Assume that the centre of the sun in on x-axis. ------------------jupiter __740_____sun___810______ --------> assuming this is a ellipse and Jupiter is around the orbit of the sun.1 solutions Answer 12043 by venugopalramana(3286)   on 2006-01-01 06:12:22 (Show Source): You can put this solution on YOUR website!standard eqn. of an ellipse with foci on x axis,centre as origin and directrices parallel to y axis is (X^2/A^2)+(Y^2/B^2)=1,where A and B are semi major/minor axes foci are given by (A*E,0),(-A*E,0) eccentricity =E =[(A^2-B^2)/A^2]^0.5 we are given the sun is at one focus and the 2 numbers you have given as 740 and 810 are NOT ELABORATED AS TO WHAT THESE NUMBERS ARE. I back calculated the major / minor axes as per the answer given by you and came out with following interpretation of the numbers given by you. 740 is the minimum distance of jupiter from sun 810 is the maximum distance of jupiter from sun so total distance =major axis ,since sun is at one of the foci and is on x axis or major axis major axis =2*A=740+810=1550..or A=1550/2= 775 as given above foci are given by (A*E,0) and (-A*E,0),with centre taken as origin. since sun is at focus and its coordinate is (775-740,0) and (775-810,0) or (35,0),(-35,0) we have A*E=35..that is.... 775*E=35...or....E=35/775=7/155 further we have E=[(A^2-B^2)/A^2]^0.5 so B=[A^2-(E^2)(A^2)]^0.5 =A*(1-E^2)^0.5=774.2093 hence eqn. Of ellipse is (X^2/775^2)+(Y^2/774.2093^2)=1 (X^2/600625)+(Y^2/599400)=1
 logarithm/20958: Solve using properties of logarithms: log2 (x+1)- log2 x = log2 5 note- the numbers directly following "log" are suppossed to be the spaces. I do not know how to type them to be small and at the bottom1 solutions Answer 12011 by venugopalramana(3286)   on 2005-12-31 09:08:47 (Show Source): You can put this solution on YOUR website!Solve using properties of logarithms: log2 (x+1)- log2 x = log2 5 note- the numbers directly following "log" are suppossed to be the spaces. I do not know how to type them to be small and at the bottom YOU MEAN ALL LOGS ARE TO BASE 2....IF ALL LOGS ARE TO SAME BASE ITDOES NOT MATTER IF WE OMIT WRITING THAT BASE...SINCE LOG X TO BASE TO LOG 2 = LOG X/LOG 2...LIKE THAT EVERY TERM HAS LOG 2 IN THE DENOMINATOR ON BOTH SIDES WHICH WE CAN CANCEL OUT. SO WE GET LOG(X+1)-LOG X=LOG 5 USING LOG A - LOG B = LOG (A/B) ..WE GET... LOG(X+1)/X = LOG 5 ...TAKING ANTILOGS ,WE GET .. (X+1)/X=5 5X=X+1 5X-X=1 4X=1 X=1/4