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# Recent problems solved by 'venugopalramana'

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 Square-cubic-other-roots/36365: (b) Find the equation of the sphere, which contains the circle x^2 + y^2 + z^2 = 9, 3x + 3y + 3z = 5 and passes through the origin.1 solutions Answer 22506 by venugopalramana(3286)   on 2006-05-07 09:16:51 (Show Source): You can put this solution on YOUR website! (b) Find the equation of the sphere, which contains the circle x^2 + y^2 + z^2 = 9, 3x + 3y + 3z = 5 and passes through the origin. THAT IS CONTAINS CIRCLE x^2 + y^2 + z^2 - 9=0=P SAY ,AND 3x + 3y + 3z - 5 =0=L SAY ANY SPHERE THROUGH THE ABOVE CIRCLE IS GIVEN BY P+KL=0...WHERE K IS A CONSTANT TO BE FOUND..WE ARE GIVEN IT PASSES THROUGH ORIGIN.HENCE 0^2+0^2+0^2-9+K(0+0+0-5)=0 -9-5K=0...........5K=-9 K=-9/5 HENCE EQN OF SPHERE IS 5(X^2+Y^2+Z^2-9)-9(3X+3Y+3Z-5)=0 5X^2+5Y^2+5Z^2-45-27X-27Y-27Z+45=0 5X^2+5Y^2+5Z^2-27X-27Y-27Z=0
 Quadratic_Equations/36378: instruction: find exact & approximate solutions to the problem. 106. One on One. Find two positive real numbers that differ by 1 and have a product of 1.1 solutions Answer 22505 by venugopalramana(3286)   on 2006-05-07 09:01:57 (Show Source): You can put this solution on YOUR website!Find two positive real numbers..X AND Y SAY that differ by 1 and have a product of 1. X-Y=1..................................I XY=1 (X+Y)^2=(X-Y)^2+4XY=1^2+4*1=5 X+Y=SQRT(5).................................II EQN.I+EQN.II GIVES 2X=1+SQRT(5) X=(1+SQRT(5))/2=(2.236+1)/2=1.618 EQN.II-EQN.I..GIVES 2Y=SQRT(5)-1 Y=(SQRT(5)-1)/2=(2.236-1)/2=0.618
 Functions/36481: This question is from textbook f(x)=4x-1 g(x)=3-5x^2 find the compostion g of f(x)1 solutions Answer 22504 by venugopalramana(3286)   on 2006-05-07 08:54:26 (Show Source): You can put this solution on YOUR website!f(x)=4x-1=Y SAY g(x)=3-5x^2 find the compostion g of f(x)=G(Y)=3-5Y^2=3-5(4X-1)^2=3-5(16X^2-8X+1) =3-80X^2+40X-5=-80X^2+40X-2
 Quadratic_Equations/36483: Use the quadratic formula to solve a quadratic equation ax^2 + bx + c = 0, the discriminant is b^2-4ac. This discriminant can be positive, zero, or negative. Explain what the value of the disciminant means to the graph of y= ax^2 + bx + c. Use these values of a=2, b=4,and c=5. Then, graph the corresponding equation. How do I know what the discriminant is? If I use the quadratic equation, does this show how to graph the equation? I think the vertex is -1,4 but how do I know how to graph the rest of the parabola? I really need some help with this one. Thanks 1 solutions Answer 22502 by venugopalramana(3286)   on 2006-05-07 08:49:50 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND COME BACK IF STILL IN DOUBT When using the quadratic formula to solve a quadratic equation ax2 + bx + c = 0, the discriminant is b2 - 4ac. This discriminant can be positive, zero, or negative. (When the discriminate is negative, then we have the square root of a negative number. This is called an imaginary number, sqrt(-1) = i. ) Explain what the value of the discriminant means to the graph of y = ax2 + bx + c. Hint: Chose values of a, b and c to create a particular discriminant. Then, graph the corresponding equation. CASE 1....DISCRIMINANT (D SAY) IS POSITIVE..... EX..LET Y = X^2-5X+6=0.. D=5^2-4*1*6=25-24=1... HENCE ROOTS ARE REAL,THAT IS THE GRAPH CUTS THE X AXIS AT 2 REAL POINTS DISTINCT ...2 AND 3 AND THE FUNCTION Y COULD BE POSITIVE OR NEGATIVE WITH A MAXIMUM OR MINIMUM SEE GRAPH BELOW CASE 2.....D=0 EX....LET...Y=X^2-2X+1=0 D=2^2-4*1*1=4-4=0 HENCE ROOTS ARE REAL.THAT IS THE GRAPH CUTS THE X AXIS AT 1 REAL POINT. EQUAL...1 AND 1 AND THE FUNCTION Y IS ALWAYS NON NEGATIVE OR NON POSITIVE DEPENDING ON THE SIGN OF COEFFICIENT OF X^2 BEING POSITIVE OR NEGATIVE , WITH A MINIMUM OR MAXIMUM VALUE OF ZERO. SEE GRAPH BELOW. CASE 3.......D IS NEGATIVE EX....LET Y = X^2+X+1=0 D=1^2-4*1*1=1-4=-3 HENCE ROOTS ARE IMAGINARY.THAT IS THE GRAPH DOES NOT CUT THE X AXIS. DISTINCT.....(-1+iSQRT(3))/2....AND (-1-iSQRT(3))/2 AND THE FUNCTION Y IS ALWAYS POSITIVE SINCE COEFFICIENT OF X^2 IS POSITIVE. SEE THE GRAPH BELOW... EX....LET Y = -X^2+X-1=0 D=1^2-4*(-1)*(-1)=1-4=-3 HENCE ROOTS ARE IMAGINARY.THAT IS THE GRAPH DOES NOT CUT THE X AXIS. DISTINCT.....(1-iSQRT(3))/2....AND (1+iSQRT(3))/2 AND THE FUNCTION Y IS ALWAYS NEGATIVE SINCE COEFFICIENT OF X^2 IS NEGATIVE. SEE THE GRAPH BELOW...
 Functions/36517: find f^-1(x) if f(x) =2x-11 solutions Answer 22501 by venugopalramana(3286)   on 2006-05-07 08:45:26 (Show Source): You can put this solution on YOUR website! find f^-1(x) if LET Y=f(x) =2x-1 2X=Y+1 X=(Y+1)/2 F^-1(X)=(F(X)+1)/2
 Quadratic-relations-and-conic-sections/36546: Determine what type of figure is given, then graph. Show all work and label all key points (asymptotes, foci, vertices, the directrix, the center) where applicable. X2 + y2 + 4x  6y = 31 solutions Answer 22497 by venugopalramana(3286)   on 2006-05-07 08:37:49 (Show Source): You can put this solution on YOUR website!TIP: 1.IF X^2 AND Y^2 HAVE EQUAL COEFFICIENTS AND THERE IS NO XY TERM ,IT COULD BE A CIRCLE. 1.) X2 + y2 + 4x  6y = 3 COMPLETE SQUARE. (X^2+2*2X+2^2)-2^2+(Y^2+2*3Y+3^2)-3^2=3 (X+2)^2+(Y+3)^2=3+9+4=16=4^2 THIS IS THE EQN OF A CIRCLE WHOSE STD.FORM IS (X-H)^2+(Y-K)^2=R^2,WHERE (H,K) IS THE CENTRE OF THE CIRCLE AND R IS THE RADIUS.HENCE (-2,-3) IS THE CENTRE AND 4 IS THE RADIUS.
 Quadratic-relations-and-conic-sections/36547: Determine what type of figure is given, then graph. Show all work and label all key points (asymptotes, foci, vertices, the directrix, the center) where applicable. X^2 + y^2 + 4x  6y = 31 solutions Answer 22496 by venugopalramana(3286)   on 2006-05-07 08:35:59 (Show Source): You can put this solution on YOUR website!TIP: 1.IF X^2 AND Y^2 HAVE EQUAL COEFFICIENTS AND THERE IS NO XY TERM ,IT COULD BE A CIRCLE. 1.) X2 + y2 + 4x  6y = 3 COMPLETE SQUARE. (X^2+2*2X+2^2)-2^2+(Y^2+2*3Y+3^2)-3^2=3 (X+2)^2+(Y+3)^2=3+9+4=16=4^2 THIS IS THE EQN OF A CIRCLE WHOSE STD.FORM IS (X-H)^2+(Y-K)^2=R^2,WHERE (H,K) IS THE CENTRE OF THE CIRCLE AND R IS THE RADIUS.HENCE (-2,-3) IS THE CENTRE AND 4 IS THE RADIUS.
 Quadratic-relations-and-conic-sections/36549: I was wondering if anyone could help me determine what type of figure is given, then graph. Show all work and label all key points (asymptotes, foci, vertices, the directrix, the center) where applicable. x^2  4x  8y = 121 solutions Answer 22495 by venugopalramana(3286)   on 2006-05-07 08:31:50 (Show Source): You can put this solution on YOUR website!TIP: 2.IF X^2 OR Y^2 IS ONLY PRESENT ,IT COULD BE A PARABOLA. 2.) x2  4x  8y = 12 COMPLETE SQUARE (X^2-2*2X+2^2)-2^2=12+8Y (X-2)^2=8Y+16=8(Y+2) THIS IS THE EQN.OF A PARABOLA .STD EQN. IS (X-H)^2=4A(Y-K),WHERE (H,K) IS THE VERTEX....(2,-2) HERE. 4A=LATUS RECTUM =8 HERE...A=2 FOCUS IS (H+A,K).....(2+2,-2)=(4,-2)..HERE. DIRECTRIX IS X-H+A=0... X-2+2=0...OR....X=0.. AXIS IS Y-K =0..Y+2=0
 Quadratic-relations-and-conic-sections/36550: I was wondering if anyone could help me determine what type of figure is given, then graph. Show all work and label all key points (asymptotes, foci, vertices, the directrix, the center) where applicable. 9x^2  96y = 16y^2 + 18x + 2791 solutions Answer 22493 by venugopalramana(3286)   on 2006-05-07 08:27:57 (Show Source): You can put this solution on YOUR website!TIP IF X^2 AND Y^2 HAVE DIFFERENT COEFFICIENTS WITH OPPOSITE SIGNS THEN IT COULD BE HYPERBOLA 3.) 9x2  96y = 16y2 + 18x + 279 COMPLETE SQUARE... {(3X)^2-2*3X*3+3^2}-3^2-{(4Y)^2+2*4Y*12+12^2}-12^2=279 (3X+3)^2-(4Y+12)^2=279+9+144=432 9(X+1)^2-16(Y+3)^2=432......NOW DIVIDE THROUGH OUT WITH 432 TO GET 1 ON THE RHS. (X+1)^2/(432/9) -(Y+3)^2/(432/16)=1 (X+1)^2/48 - (Y+3)^2/27 =1 THIS THE EQN. OF A HYPERBOLA.STD.EQN.IS. (X-H)^2/A^2 - (Y-K)^2/B^2=1..WHERE (H,K) IS CENTRE.....(-1,-3) HERE TRANSVERSE AXIS IS Y=K...Y=-3 LENGTH OF TRANSVERSE AXIS=2A.. ...=2SQRT(48) CONJUGATE AXIS IS X=H.......X=-1 LENGTH OF CONJUGATE AXIS = 2B =2SQRT(27) ECCENTRICITY=E=SQRT{(A^2+B^2)/A^2} =SQRT{(48+27)/48}=SQRT(75/48) A*E=SQRT(48)*SQRT(75/48)=SQRT(75) FOCI ARE (H+-AE,K)......(-1+SQRT(75),-3) AND ......(-1-SQRT(75),-3) A/E=SQRT(48)/SQRT(75/48)=48/SQRT(75) DIRECTRIX ARE X=H+-A/E.... X=-1+48/SQRT(75)...AND X=-1-48/SQRT(75) ASYMPTOTES ARE GIVEN BY (X-H)^2/A^2 = (Y-K)^2/B^2 OR (X-H)/A=+(Y-K)/B AND............(X+1)/SQRT(48) =(Y+3)/SQRT(27) (X-H)/A=-(Y-K)/B.............(X+1)/SQRT(48) = -(Y+3)/SQRT(27) GRAPH IS GIVEN BELOW.. --------------------------------------------------------
 Quadratic-relations-and-conic-sections/36551: I was wondering if anyone could help me determine what type of figure is given, then graph. Show all work and label all key points (asymptotes, foci, vertices, the directrix, the center) where applicable. 4(x-1)2 = 4-y^21 solutions Answer 22486 by venugopalramana(3286)   on 2006-05-07 06:33:13 (Show Source): You can put this solution on YOUR website!TIP: IF X^2 AND Y^2 HAVE DIFFERENT COEFFICIENTS OF SAME SIGN ,THEN IT COULD BE ELLIPSE. 4.) 4(x-1)2 = 4-y2 I HOPE IT IS 4(X-1)^2+Y^2=4...DIVIDE WITH 4 (X-1)^2/1^2+Y^2/2^2=1 THIS THE EQN.OF AN ELLIPSE.STD.EQN. IS (X-H)^2/A^2 +(Y-K)^2/B^2=1 WHERE (H,K) IS CENTRTE....(1,0).HERE LENGTH OF MAJOR AXIS IS 2B...2*2=4 MINOR AXIS IS ALONG Y=K....Y=0 LENGTH OF MINOR AXIS IS 2A...2*1=2 ECCENTRICITY=E=SQRT{(B^2-A^2)/A^2} =SQRT(4-1)/1=SQRT(3) B*E=2SQRT(3) B/E=2/SQRT(3) FOCI ARE (H,K+BE) AND (H,K-BE) (1,2SQRT(3)) AND (1,-2SQRT(3)) DIRECTRIX ARE.Y=K+B/E AND K-B/E Y=2/SQRT(3)..AND...Y=-2/SQRT(3) GRAPH IS GIVEN BELOW... -----------------------------------------------------------------
 Sequences-and-series/36552: The sum of a geometric series is 57232. The common ratio is 2 and the last term is 28672. What is the first term?1 solutions Answer 22483 by venugopalramana(3286)   on 2006-05-07 01:58:32 (Show Source): You can put this solution on YOUR website!The sum of a geometric series is 57232. The common ratio is 2 and the last term is 28672. What is the first term? FORMULA FOR N TH TERM IS .... TN=A*R^(N-1)..HENCE 28672=A*2^(N-1).......................I FORMULA FOR SUM IS SN=A*{R^N - 1 }/(R-1)...HENCE 57232 = A*(2^N-1)/(2-1)=A*(2^N-1)....................II EQN.II/EQN.I... 57232/28672=A*(2^N -1)/{A*2^(N-1)}=(2^N -1)/2^(N-1) 57232/28672=2^N/2^(N-1) - 1/2^(N-1)=2-1/2^(N-1) 1/2^(N-1)=2-57232/28672=(57344-57232)/28672=112/28672 2^(N-1)=28672/112=256=2^7 N-1=7.ORN=8SUBSTITUTING IN EQN.I 28672=A*2^(8-1)=256A..OR..A = 28672/256=112
 Quadratic_Equations/36575: When using the quadratic formula to solve a quadratic equation ax2 + bx + c = 0, the discriminant is b2 - 4ac. This discriminant can be positive, zero, or negative. (When the discriminate is negative, then we have the square root of a negative number. This is called an imaginary number, sqrt(-1) = i. ) Explain what the value of the discriminant means to the graph of y = ax2 + bx + c. Hint: Chose values of a, b and c to create a particular discriminant. Then, graph the corresponding equation. 1 solutions Answer 22478 by venugopalramana(3286)   on 2006-05-06 23:33:16 (Show Source): You can put this solution on YOUR website!When using the quadratic formula to solve a quadratic equation ax2 + bx + c = 0, the discriminant is b2 - 4ac. This discriminant can be positive, zero, or negative. (When the discriminate is negative, then we have the square root of a negative number. This is called an imaginary number, sqrt(-1) = i. ) Explain what the value of the discriminant means to the graph of y = ax2 + bx + c. Hint: Chose values of a, b and c to create a particular discriminant. Then, graph the corresponding equation. CASE 1....DISCRIMINANT (D SAY) IS POSITIVE..... EX..LET Y = X^2-5X+6=0.. D=5^2-4*1*6=25-24=1... HENCE ROOTS ARE REAL,THAT IS THE GRAPH CUTS THE X AXIS AT 2 REAL POINTS DISTINCT ...2 AND 3 AND THE FUNCTION Y COULD BE POSITIVE OR NEGATIVE WITH A MAXIMUM OR MINIMUM SEE GRAPH BELOW CASE 2.....D=0 EX....LET...Y=X^2-2X+1=0 D=2^2-4*1*1=4-4=0 HENCE ROOTS ARE REAL.THAT IS THE GRAPH CUTS THE X AXIS AT 1 REAL POINT. EQUAL...1 AND 1 AND THE FUNCTION Y IS ALWAYS NON NEGATIVE OR NON POSITIVE DEPENDING ON THE SIGN OF COEFFICIENT OF X^2 BEING POSITIVE OR NEGATIVE , WITH A MINIMUM OR MAXIMUM VALUE OF ZERO. SEE GRAPH BELOW. CASE 3.......D IS NEGATIVE EX....LET Y = X^2+X+1=0 D=1^2-4*1*1=1-4=-3 HENCE ROOTS ARE IMAGINARY.THAT IS THE GRAPH DOES NOT CUT THE X AXIS. DISTINCT.....(-1+iSQRT(3))/2....AND (-1-iSQRT(3))/2 AND THE FUNCTION Y IS ALWAYS POSITIVE SINCE COEFFICIENT OF X^2 IS POSITIVE. SEE THE GRAPH BELOW... EX....LET Y = -X^2+X-1=0 D=1^2-4*(-1)*(-1)=1-4=-3 HENCE ROOTS ARE IMAGINARY.THAT IS THE GRAPH DOES NOT CUT THE X AXIS. DISTINCT.....(1-iSQRT(3))/2....AND (1+iSQRT(3))/2 AND THE FUNCTION Y IS ALWAYS NEGATIVE SINCE COEFFICIENT OF X^2 IS NEGATIVE. SEE THE GRAPH BELOW...
 Matrices-and-determiminant/36582: Solve the determinant 3 x 1 1 = 2 (the 3 x and 1 1 are inside determinant signs)1 solutions Answer 22475 by venugopalramana(3286)   on 2006-05-06 22:37:34 (Show Source): You can put this solution on YOUR website!Solve the determinant 3 x 1 1 = 2 that is 3*1-1*x=2 3-x=2 x=3-2=1 x=1
 Divisibility_and_Prime_Numbers/36307: a) What is the remainder left after dividing 1! + 2! + 3! ++ 100! by 7? b) If n >=4 then find whether 2.n is less than 4.n or not. 1 solutions Answer 22474 by venugopalramana(3286)   on 2006-05-06 22:33:18 (Show Source): You can put this solution on YOUR website!a) What is the remainder left after dividing 1! + 2! + 3! ++ 100! by 7? GROUP AS FOLLOWS =(1!+3!)+(2!+4!)+(5!+6!)+(7!+8!+9!+..........+100!) =(7)+(26)+5!(1+6)+7!(1+8+8.9+............+8.9.10......100)...WE FIND THAT EXCEPT 26 THE SECOND TERM, ALL THE OTHER TERMS ARE DIVISIBLE BY 7.HENCE THE REMAINDER WE GET ON DIVIDING WITH 7 IS THE REMAINDER WE GET WHEN WE DO 26/7...THAT IS ...5 IS THE REMAINDER . b) If n >=4 then find whether 2.n is less than 4.n or not. CHECK THE QUESTION 2.N MEAns 2*n ..if n=5 say 2*5=10.....it is definitely less than 4*n=4*5=20 what is there to prove in that?
 Miscellaneous_Word_Problems/36379: The sides of the triangle are 6,7,and x. What is the largest value that could be the area of such a triangle?1 solutions Answer 22473 by venugopalramana(3286)   on 2006-05-06 22:14:21 (Show Source): You can put this solution on YOUR website!The sides of the triangle are 6,7,and x. What is the largest value that could be the area of such a triangle? HOPE YOU KNOW ELEMENTARY TRIGNOMMETRY.THEN IT IS VERY EASY.OTHERWISE ,PLEASE INFORM AND I SHALL SHOW YOUB ANOTHER METHOD AREA OF TRIANGLE =(1/2)*BASE*ALTITUDE=0.5*b*h=0.5*b*c*SIN(A)=0.5*6*7*SIN(A) =21SIN(A).....WE KNOW THAT MAXIMUM VALUE OF SIN(A)=1 WHEN A=90 DEGREES HENCE MAXIMUM AREA =21*1=21 WHEN IT IS RIGHT ANGLED TRIANGLE WITH LEGS EQUAL TO 6 AND 7.
 Circles/36510: (1) find the center of the circle whose equation is x^2+ y^2-3x+8y-8=0 (2) write the equation of the circle with center (-2,1) and which passes through (3,4).1 solutions Answer 22426 by venugopalramana(3286)   on 2006-05-06 12:34:24 (Show Source): You can put this solution on YOUR website!STD.EQN. OF A CIRCLE IS (X-H)^2+(Y-K)^2=R^2... WHERE (H,K)IS THE CENTRE AND R IS THE RADIUS.USING THIS FORMULA... (1) find the center of the circle whose equation is x^2+ y^2-3x+8y-8=0 {X^2-2*(3/2)X+(3/2)^2}+{Y^2+2*4*Y+4^2}-(3/2)^2-4^2-8=0 {X-(3/2)}^2+{Y+4}^2=24+9/4=105/4..HENCE CENTRE OF CIRCLE IS {(3/2),-4} (2) write the equation of the circle with center (-2,1) and which passes through (3,4). EQN IS (X+2)^2+(Y-1)^2=R^2...IT IS` PASSING THROUGH (3,4)..SO... (3+2)^2+(4-1)^2=R^2 25+9=34=R^2.HENCE EQN.OF CIRCLE IS (X+2)^2+(Y-1)^2=34