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# Recent problems solved by 'venugopalramana'

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 Quadratic_Equations/38622: find the condition that the root of the equation ax^2+bx+c=0 may be in the ratio m:n1 solutions Answer 24164 by venugopalramana(3286)   on 2006-05-20 04:55:32 (Show Source): You can put this solution on YOUR website!AX^2+BX+C=0 ROOTS ARE IN RATIO M:N...LET THE ROOTS BE MY AND NY SUM OF ROOTS =MY+NY=Y(M+N)=-B/A Y=-B/A(M+N) PRODUCT OF ROOTS=MNY^2=C/A MN*{-B/A(M+N)}^2=C/A MNB^2=CA^2(M+N)^2/A=CA(M+N)^2
 Permutations/38620: How many different license plate numbers can be made using 2 letters followed by 4 digits selected from digits 0 through 9 if letters and digits may be repeated?1 solutions Answer 24163 by venugopalramana(3286)   on 2006-05-20 04:49:20 (Show Source): You can put this solution on YOUR website!I PLACE...2 LETTERS.....2 WAYS II PLACE..2 LETTERS.....2 WAYS III PLACE.10 NUMBERS....10 WAYS IV PLACE..10.NUMBERS....10 WAYS V PLACE ..10.NUMBERS....10 WAYS VI PLACE..10 NUMBERS....10 WAYS ------------------------------------ TOTAL POSSIBILITIES = 2*2*10*10*10*10=40000
 Trigonometry-basics/38147: I need help using special right triangles to find the exact values of sines and cosines for each angle. For example: find the values of sine and cosine for the angle -135 degrees.1 solutions Answer 24065 by venugopalramana(3286)   on 2006-05-19 12:28:01 (Show Source): You can put this solution on YOUR website!I need help using special right triangles to find the exact values of sines and cosines for each angle. For example: find the values of sine and cosine for the angle -135 degrees. FIRST LEARN THE TRIGNOMETRIC RATIOS FOR STANDARD TRIANGLES. THEN LEARN WHICH QUADRANT THE ANGLE IS USE ALL SILVER TEA CUPS CONVENTION ,TO KNOW THE SIGN OF THE TRIGNOMETRIC RATIO. IF ANGLE IS IN I QUADRANT ...THEN...ALL ARE +VE ...............II.......................SINE ONLY IS +VE ...............III......................TANGENT ONLY IS +VE ................IV.......................COSINE ONLY IS +VE. NOW WRITE GIVEN ANGLE AS 0+X..OR...0-X...OR....180+ OR - X ....OR....360+ OR -X ETC.... YOUR EXAMPLE OF -135 SINE (0-135)=-SIN(135)..........0-X..MEANS IV Q...SINE IS -VE. -1*SIN(135)=-SIN(180-45)=-SIN(45).....180-X MEANS II Q. SINE IS POSITIVE. -1*SIN45=-1/SQRT(2) GET BY HEART THE STANDARD TABLES FOR 0,30,45,60,90. SIMILARLY COS (0-135)=COS(135)=COS(180-45)=-COS(45)=-1/SQRT2 0
 Geometry_proofs/38152: Question: Prove: If a diagonal of a parallelogram bisects an angle of the parallelogram, the parallelogram is a rhombus. (State your plan and give a proof. Given: ABCD is a parallelogram with <1~= <2 To Prove: ABCD is a rhombus So I can see that angles 1 and 2 are equal. In theorm 5-13 says the diagonals are pependicular but there is only one and then therom 5-14 says that each diagnoal... bisects two angles of the rhombus. Am I making this too difficult or do I just need to state 5-14 Thanks1 solutions Answer 24063 by venugopalramana(3286)   on 2006-05-19 12:16:51 (Show Source): You can put this solution on YOUR website!THIS IS NOT CORRECT.THERE IS SOME THING WRONG IN YOUR HYPOTHESIS.DIAGONALS OF A PARALLELOGRAM DEFINITELY BISECT THEIR ANGLES ,WHETHER IT IS A RHOMBUS OR NOT.THE DIAGONALS NEED TO BE PERPENDICULAR TO EACH OTHER FOR IT TO BE A RHOMBUS
 Expressions-with-variables/38540: Hello! I've tried to approach this from a substiution standpoint but I'm at a loss : Find the number of solutions for the system (the following equations are all together in one bracket) {f + g + h = 4 f = 5 - h - g g + h = 7 - f If you could help me on how I should approach this problem I would be very grateful!! Thanks!!!!!1 solutions Answer 24062 by venugopalramana(3286)   on 2006-05-19 12:10:24 (Show Source): You can put this solution on YOUR website!Hello! I've tried to approach this from a substiution standpoint but I'm at a loss : Find the number of solutions for the system (the following equations are all together in one bracket) {f + g + h = 4.........................................I f = 5 - h - g.....OR.....F+G+H=5.....................II g + h = 7 - f.....OR.....F+G+H=7 If you could help me on how I should approach this problem I would be very grateful!! Thanks!!!!! THESE 3 ARE IDENTICAL ON LHS BUT HAVE DIFFERENT VALUES ON RHS.HENCE THEY ARTE INCONSISTENT AND NO SOLUTION EXISTS WHICH CAN SATISFY ALL 3 EQNS.
 Quadratic-relations-and-conic-sections/38557: This question is from textbook The question is write an equation for the perpendicular bisector of the line segment joining the two points? the points are (0,0)(-8,-10) how would you solve this1 solutions Answer 24061 by venugopalramana(3286)   on 2006-05-19 12:06:57 (Show Source): You can put this solution on YOUR website!the question is write an equation for the perpendicular bisector of the line segment joining the two points? the pionts are A-(0,0).............AND B-(-8,-10) how would you solve this MID POINT OF AB ...SAY....C...IS (0-8)/2,(0-10)/2=(-4,-5) SLOPE OF AB =(0+10)/(0+8)=10/8=5/4 SLOPE OF PERPENDICULAR TO AB IS -1/(5/4)=-4/5 PERPENDICULAR BISECTOR IS PERPENDICULAR TO AB AND PASSES THROUGH ITS MID POINT C.HENCE ITS EQN. IS Y+5=(-4/5)(X+4) 5Y+25=-4X-16 4X+5Y+41=0
 Trigonometry-basics/38524: Find the exact value of sec(225 degrees)1 solutions Answer 24029 by venugopalramana(3286)   on 2006-05-19 00:07:36 (Show Source): You can put this solution on YOUR website!SEC(225)=SEC(180+45)=-SEC(45)=-SQRT(2)=-1.414
 expressions/38171: Find the sum of the fcllowing geometric series. 8 + 12 +..... + 60.751 solutions Answer 23877 by venugopalramana(3286)   on 2006-05-18 12:26:34 (Show Source): You can put this solution on YOUR website!Find the sum of the fcllowing geometric series. 8 + 12 +..... + 60.75 A=8 R=12/8=1.5 TN=AR^(N-1)=8*(3/2)^(N-1)=60.75 N=6 BY INSPECTION SINCE 8*3*3*3*3*3/(2*2*2*2*2)=60.75 SN=S6=A*{R^N - 1)/(R-1)=(60.75*1.5-8)/0.5=166.25
 Trigonometry-basics/38174: A geometric sequenee has first term -4 and a constant ratio of 5. a) Write the first six terms of this sequence. _____,______,_____,_______,_______,_______ b) Find the sum of the first eight terms of this sequence1 solutions Answer 23874 by venugopalramana(3286)   on 2006-05-18 12:19:18 (Show Source): You can put this solution on YOUR website!A geometric sequenee has first term -4 and a constant ratio of 5. a) Write the first six terms of this sequence. A=-4...R=5 T1=A=-4 T2=AR=-4*5=-20 T3=A*R^2=-4*5^2=-100 T4=A*R^3=-500 T5=-2500 T6=--12500 _____,______,_____,_______,_______,_______ b) Find the sum of the first eight terms of this sequence S8=A*{R^8-1)}/(R-1)=-4*(5^8-1)/(5-1)=-(5^8-1)
 Length-and-distance/38264: i wasn't sure where to go with this one but here it is: Point A is equidistant from P and Q. The midpoint, M, of line PQ is 12 units form A and 7 1/2 units from the midpoint of line AQ. Find the length of line PQ. (you have to draw your own picture, which im not very good at, i keep getting weird square roots for PQ, please help!)1 solutions Answer 23867 by venugopalramana(3286)   on 2006-05-18 11:53:49 (Show Source): You can put this solution on YOUR website!i wasn't sure where to go with this one but here it is: Point A is equidistant from P and Q. The midpoint, M, of line PQ is 12 units form A and 7 1/2 units from the midpoint(SAY N ) of line AQ THAT IS AN =7.5. Find the length of line PQ. (you have to draw your own picture, which im not very good at, i keep getting weird square roots for PQ, please help!) IN TRIANGLE APQ,WE HAVE M IS MID POINT OF PQ ...GIVEN N IS MID POINT OF AQ...GIVEN. HENCE MN=AP/2...OR...AP =2*MN=2*7.5=15 IN TRIANGLES APM AND APQ PM=MQ...GIVEN AM=AM AP=AQ...GIVEN HENCE THEY ARE CONGRUENT AND HENCE ANGLE AMP=ANGLE AMQ =180/2=90 HENCE IN THIS RIGHT TRIANGLE AP^2=AM^2+MP^2 MP^2=AP^2-AM^2=15^2-12^2=225-144=81 MP=9 HENCE PQ =2*MP=2*9=18
 Equations/38365: x+y=10,x*y=20 find the value of x,y1 solutions Answer 23866 by venugopalramana(3286)   on 2006-05-18 11:37:53 (Show Source): You can put this solution on YOUR website!x+y=10...................................i x*y=20 find the value of x,y (x-y)^2=(x+y)^2-4y=10^2-4*20=100-80=20 x-y=sqrt(20)=2sqrt(5)....................................ii i+ii 2x=10+2sqrt(5) x=5+sqrt(5) i-ii 2y=10-2sqrt(5) y=5-sqrt(5)
 Complex_Numbers/38350: Express the following as a complex number in the form a + ib: The complex conjugate of i4n-3, with n Î Z Please show you working out, and thank you in advance for your help1 solutions Answer 23841 by venugopalramana(3286)   on 2006-05-18 09:05:16 (Show Source): You can put this solution on YOUR website!I^(4N-3)=I^4N/I^3={(I^2)^2N}/{(I^2)(I)}={(-1)^2}^N/{(-1)(I)=(1)^N/(-I)=-1/I =(-1)(I)/{(I)(I)}=(-1)(I)/(-1)=I=0+1*I
 Graphs/38042: Are the following lines parallel perpendicular or neither L1 with equation x-5y=10 L2 with equation 5x+y=5 a. Parallel B. Perpendicular C. Neither1 solutions Answer 23540 by venugopalramana(3286)   on 2006-05-16 09:35:26 (Show Source): You can put this solution on YOUR website!yes perpendicular..b is ok. product of slopes =-5*1/5=-1
 Probability-and-statistics/38037: Each question on a five-question multiple choice quiz has answer choices labeled A, B, C, and D. How many different ways can a student answer the five questions?1 solutions Answer 23539 by venugopalramana(3286)   on 2006-05-16 09:32:54 (Show Source): You can put this solution on YOUR website!FORV EACH QUESTION HE CAN ANSWER IN 4 WAYS..THERTE ARE 5 QUESTIONS SO THERE ARE 4*4*4*4*4=1024WAYS
 Coordinate-system/37610: Find the two remaining vertices of square ABCD that has A=(-2,1), B=(6,-5) and a vertex in quadrant III.1 solutions Answer 23337 by venugopalramana(3286)   on 2006-05-15 09:23:20 (Show Source): You can put this solution on YOUR website!Find the two remaining vertices of square ABCD that has A=(-2,1), B=(6,-5) and a vertex in quadrant III. AB=SQRT((6+2)^2+(5+1)^2)=10 EQN. OF AB IS (Y-1)=[(1+5)/(-6-2)](X+2)} 4y-4=-3x-6 3X+4Y=-2 ITS DISTANCE FROM ORIGIN IS 2/5 CD IS PARALLEL TO AB AND IS AT 10 DIUSTANCEB FROM AB OR 10+2/5 =52/5 DISTANCE FROM ORIGIN. HENCE ITS EQN,IS 3X+4Y=-52................................I..OR...3H+4K=-52 AS C H,K LIES ON IT. SLOPE OF AB =-3/4 SLOPE OF BC =4/3= LET C BE H,K SLOPE OF BC = (K+5)/(H-6)=4/3 3K+15=4H-24 4H-3K=39...........................II SOLVING I AND II H=0...K=-13.SO C IS (0,-13) SIMILARLY FOR D WE CAN FIND SLOPE OF AD =4/3= LET D BE H,K SLOPE OF AD = (K-1)/(H+2)=4/3 3K-3=4H+8 4H-3K=-11...........................III SOLVING I AND III H=-8 AND K=-7
 Circles/37625: if a Quad is inscribed in a circle, how do i find the measures of the angles of the quad? ex: left: 26y on top 21y on bottom Right: 3x on top 2x on bottom thanks.1 solutions Answer 23254 by venugopalramana(3286)   on 2006-05-13 23:25:27 (Show Source): You can put this solution on YOUR website!IN A CYCLIC QUADRILATERAL SUM OF OPPOSITE ANGLES IS 180 HENCE left: 26y on top 21y on bottom Right: 3x on top 2x on bottom thanks. 26Y+2X=180......X+13Y=90...............I 3X+21Y=180......X+7Y=60..............II EQN.I-EQN.II 13Y-7Y=90-60=30 6Y=30...Y=5 X+7*5=60 X=60-35=25 HENCE THE ANGLES ARE 26*5=130 21*5=105 3*25=75 2*25=50
 Polynomials-and-rational-expressions/37619: The problem is : . We have to factor each expression as a trinomial. I found the factors of three : 1 & 3 , but it says in the book that one of the factor pairs has to equal "b" which is supposed to be 5. This doesn't make sense to me. AJ1 solutions Answer 23217 by venugopalramana(3286)   on 2006-05-13 12:32:46 (Show Source): You can put this solution on YOUR website!The problem is : . We have to factor each expression as a trinomial. I found the factors of three : 1 & 3 , but it says in the book that one of the factor pairs has to equal "b" which is supposed to be 5. This doesn't make sense to me. LET ME EXPLAIN YOU IN DETAIL 1.TAKE COEFFICIENT OF A^2 =2 HERE 2. TAKE CONSTANT TERM = 3 HERE. 3.FIND THEIR PRODUCT =2*3=6 4.KEEP IN MIND ALL FACTORS FOR THE ABOVE PRODUCT ... 1*6,-1*-6,2*3,-2*-3. 5.CHOOSE THAT PAIR WHOSE SUM IS EQUAL TO COEFFICIENT OF A =+5 HERE 6.2*3=6 SATISFY THIS REQUIREMENT SINCE 2+3=5 7.NOW SPLIT THE X TERM AS ABOVE ,TAKE COMMON FACTORS AND FACTORISE AS SHOWN BELOW 2A^2+2A+3A+3=2A(A+1)+3(A+1)=(A+1)(2A+3) GOT IT??
 Miscellaneous_Word_Problems/37605: The volume of an ideal gas varies directly with the temperature T and inversely with the Pressure P. A cylinder contains 15 liters of oxygen at a temperature of 240 K and a pressure of 80 atmospheres. If a piston is lowered into the cylinder decreasing the volume occupied by the gas to 10 liters and the temperature remains constant, what is the pressure?1 solutions Answer 23216 by venugopalramana(3286)   on 2006-05-13 11:58:29 (Show Source): You can put this solution on YOUR website!The volume - V of an ideal gas varies directly with the temperature T and inversely with the Pressure P. V=K*T/P...WHERE K IS A CONSTANT A cylinder contains 15 liters of oxygen at a temperature of 240 K and a pressure of 80 atmospheres. 15=K*240/80=3K K=15/3=5 LITRE.ATM./DEG.K V=5T/P If a piston is lowered into the cylinder decreasing the volume occupied by the gas to 10 liters and the temperature remains constant, what is the pressure? 10=5*240/P P=5*240/10=120 ATM.
 Quadratic_Equations/37613: I am to solve this problem finding all real or imaginary solutions. I believe I solved it, but I question whether it is correct. The question is: 6x=19x+25/x+1 My work: 6x(x+1)=6x^2+6x Then I put it into the equation form of -6x^2-25x+25=0 I used the b^2-4(a)(c) formula which =625-4(-6)(25) The whole thing turned out to be 625-600=25 Because the outcome is positive, does it not have to have two solutions? I am unsure what the second one is? Thank you1 solutions Answer 23214 by venugopalramana(3286)   on 2006-05-13 11:54:21 (Show Source): You can put this solution on YOUR website!I am to solve this problem finding all real or imaginary solutions. I believe I solved it, but I question whether it is correct. The question is: 6x=19x+25/x+1 My work: 6x(x+1)=6x^2+6x...OK Then I put it into the equation form of -6x^2-25x+25=0 NO..YOU MISSED 19X 6X^2+6X=19X+25 6X^2-13X-25=0 ARE THE 2 SOLUTIONS. I used the b^2-4(a)(c) formula which =625-4(-6)(25) The whole thing turned out to be 625-600=25 Because the outcome is positive, does it not have to have two solutions? I am unsure what the second one is? Thank you
 Linear-equations/37611: Find the slope of the sides of rectangle ABCD having vertices A=(1,4) and B=(-5,3).1 solutions Answer 23213 by venugopalramana(3286)   on 2006-05-13 11:46:41 (Show Source): You can put this solution on YOUR website!Find the slope of the sides of rectangle ABCD having vertices A=(1,4)=(X1,Y1) and B=(-5,3)=(X2,Y2) SLOPE OF AB =(Y2-Y1)/(X2-X1)=(3-4)/(-5-1)=-1/-6=1/6 IN A RECTANGLE OPPOSITE SIDES ARE PARALLEL AND ADJACENT SIDES ARE PERPENDICULAR.HENCE SLOPE OF CD=1/6 SLOPE BC=SLOPE OF AD =-1/(1/6)=-6
 Quadratic_Equations/37614: 400 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). She wants to maximize the area of her patio (area of a rectangle is length times width). What should the dimensions of the patio be, and show how the maximum area of the patio is calculated from the algebraic equation. Answer: 1 solutions Answer 23212 by venugopalramana(3286)   on 2006-05-13 11:42:10 (Show Source): You can put this solution on YOUR website!Amanda has 400 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). She wants to maximize the area of her patio (area of a rectangle is length times width). What should the dimensions of the patio be, and show how the maximum area of the patio is calculated from the algebraic equation. Answer: IF L AND B ARE DIMENSIONS WE HAVE PERIMETER=2(L+B)=400.....OR.....L+B=200..OR......B=200-L.................I AREA=A =LB=L(200-L)=200L-L^2=-{L^2-200L}=-{(L^2)-2(L)(100)+100^2-100^2} A=10000-(L-100)^2 (L-100)^2 BEING PERFECT SQUARE,ITS MINIMUM VALUE IS ZERO. HENCE AREA IS MAXIMUM WHEN L-100 IS ZERO,OR WHEN L=100 AND THEN THE MAXIMUM AREA WOULD BE A-MAX.=10000-0=10000 DIMENSIONS ARE 100*100