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 Linear-equations/34427: For the systems of linear equations in question 1 - Determine how many solutions exist - Use either elimination or substitution to find the solutions (if any) - Graph the two lines, labeling the x-intercepts, y-intercepts, and points of intersection 1. (8 points) y = 2x + 3 and y = -x - 41 solutions Answer 20828 by venugopalramana(3286)   on 2006-04-22 08:48:16 (Show Source): You can put this solution on YOUR website!For the systems of linear equations in this question - Determine how many solutions exist - Use either elimination or substitution to find the solutions (if any) - Graph the two lines, labeling the x-intercepts, y-intercepts, and points of intersection y = 2x + 3 and y = -x - 4 -X-4=2X+3 2X+X=-4-3=-7 3X=-7 X=-7/3 Y=-(-7/3)-4=7/3-4=(7-12)/3=-5/3 HENCE THERE IS ONE UNIQUE SOLUTION.
 Linear-equations/34428: For the systems of linear equations in this question - Determine how many solutions exist - Use either elimination or substitution to find the solutions (if any) - Graph the two lines, labeling the x-intercepts, y-intercepts, and points of intersection 2x + 3y = 8 and 3x + 2y = 71 solutions Answer 20826 by venugopalramana(3286)   on 2006-04-22 08:47:15 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLE AND TRY ------------------------------------------------------------- For the systems of linear equations in this question - Determine how many solutions exist - Use either elimination or substitution to find the solutions (if any) - Graph the two lines, labeling the x-intercepts, y-intercepts, and points of intersection y = 2x + 3 and y = -x - 4 -X-4=2X+3 2X+X=-4-3=-7 3X=-7 X=-7/3 Y=-(-7/3)-4=7/3-4=(7-12)/3=-5/3 HENCE THERE IS ONE UNIQUE SOLUTION.
 Linear-equations/34429: For the systems of linear equations in this question - Determine how many solutions exist - Use either elimination or substitution to find the solutions (if any) - Graph the two lines, labeling the x-intercepts, y-intercepts, and points of intersection y = 2x + 3 and y = -x - 41 solutions Answer 20825 by venugopalramana(3286)   on 2006-04-22 08:45:44 (Show Source): You can put this solution on YOUR website!For the systems of linear equations in this question - Determine how many solutions exist - Use either elimination or substitution to find the solutions (if any) - Graph the two lines, labeling the x-intercepts, y-intercepts, and points of intersection y = 2x + 3 and y = -x - 4 -X-4=2X+3 2X+X=-4-3=-7 3X=-7 X=-7/3 Y=-(-7/3)-4=7/3-4=(7-12)/3=-5/3 HENCE THERE IS ONE UNIQUE SOLUTION.
 Linear-equations/34430: For the systems of linear equations in this question - Determine how many solutions exist - Use either elimination or substitution to find the solutions (if any) - Graph the two lines, labeling the x-intercepts, y-intercepts, and points of intersection x + y = 3 and y = x + 31 solutions Answer 20824 by venugopalramana(3286)   on 2006-04-22 08:40:43 (Show Source): You can put this solution on YOUR website!For the systems of linear equations in this question - Determine how many solutions exist - Use either elimination or substitution to find the solutions (if any) - Graph the two lines, labeling the x-intercepts, y-intercepts, and points of intersection x + y = 3 and y = x + 3 X+X+3=3 2X=0 X=0 Y=3...HENCE THERE IS ONLY ONE UNIQUE SOLUTION. (
 Polynomials-and-rational-expressions/34464: Factor each polynomial completely, given that the binomial following it is a factor of the polynomial. 1. x^3-4x^2-3x-10,x-5 2. w^3+5w^2-w=5 3. t^2+1=13/6t1 solutions Answer 20823 by venugopalramana(3286)   on 2006-04-22 08:34:00 (Show Source): You can put this solution on YOUR website!Factor each polynomial completely, given that the binomial following it is a factor of the polynomial. 1. x^3-4x^2-3x-10,x-5 5|1...........-4..........-3..........-10 .|0............5...........5...........10 ---------------------------------------------------------- ..1............1...........2............0 SO QUOTIENT IS X^2+X+2...THIS NO MORE REDUCIBLE. HENCE ANSWER IS (X-5)(X^2+X+2) 2. w^3+5w^2-w=5....PLEASE TYPE PROPERLY...WHERE IS THE BINOMIAL FACTOR? 3. t^2+1=13/6t....PLEASE TYPE PROPERLY...WHERE IS THE BINOMIAL FACTOR?
 Polynomials-and-rational-expressions/34467: Factor the polynomial completely, given that the binomial following it is a factor of the polynomial. w^3+5w^2-w=51 solutions Answer 20822 by venugopalramana(3286)   on 2006-04-22 08:29:49 (Show Source): You can put this solution on YOUR website!factor the polynomial completely, given that the binomial followin it is a factor of the polynomial. w^3+5w^2-w=5 PLEASE TYPE PROPERLY..WHERE IS THE BINOMIAL FACTOR FOLLOWING THE POLYNOMIAL?
 Polynomials-and-rational-expressions/34468: factor the polynomial completely, given that the binomial followin it is a factor of the polynomial. w^3+5w^2-w=51 solutions Answer 20820 by venugopalramana(3286)   on 2006-04-22 08:29:11 (Show Source): You can put this solution on YOUR website!factor the polynomial completely, given that the binomial followin it is a factor of the polynomial. w^3+5w^2-w=5 PLEASE TYPE PROPERLY..WHERE IS THE BINOMIAL FACTOR FOLLOWING THE POLYNOMIAL?
 Polynomials-and-rational-expressions/34469: factor the polynomial completely, given that the binomial followin it is a factor of the polynomial. w^3+5w^2-w=51 solutions Answer 20819 by venugopalramana(3286)   on 2006-04-22 08:28:27 (Show Source): You can put this solution on YOUR website!factor the polynomial completely, given that the binomial followin it is a factor of the polynomial. w^3+5w^2-w=5 PLEASE TYPE PROPERLY..WHERE IS THE BINOMIAL FACTOR FOLLOWING THE POLYNOMIAL?
 logarithm/34526: So here is the question I've been mulling over.. 10^3log(x)= 1/9 Log is base 10.. Thanks a million for any help!1 solutions Answer 20818 by venugopalramana(3286)   on 2006-04-22 08:25:34 (Show Source): You can put this solution on YOUR website!10^3log(x)= 1/9HOPE YOU TYPED THIS PROPERLY..ANY WAY AS TYPED IT SAYS TAKING LOGS 3LOG(X)*LOG(10)=LOG(1/9) 3LOG(X)=LOG(1/9) LOG(X^3)=LOG(1/9) X^3=1/9 X=CUBEROOT OF (1/9)
 logarithm/34528: Solve: 8^(4x-3)= 1/16, find the exact answer.1 solutions Answer 20816 by venugopalramana(3286)   on 2006-04-22 08:20:13 (Show Source): You can put this solution on YOUR website!8^(4x-3)= 1/16, (2^3)^(4X-3)=1/(2^4)=2^(-4) 2^(12X-9)=2^(-4) 12X-9=-4 12X=-4+9=-5 X=-5/12 find the exact answer.
 Systems-of-equations/34505: How do you solve this problem (0,4),(2,8) & (3,1) to get the quadratic equation by solving systems of Three Linear Equations to find quadratic equations?1 solutions Answer 20811 by venugopalramana(3286)   on 2006-04-22 05:04:56 (Show Source): You can put this solution on YOUR website!THE QUESTION IS NOT COMPLETE.PLEASE CHECK AND RETYPE
 Length-and-distance/34510: A triangle with sides of 5, 12 and 13 has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles.1 solutions Answer 20810 by venugopalramana(3286)   on 2006-04-22 05:02:21 (Show Source): You can put this solution on YOUR website!WE FIND THAT 5^2+12^2=25+144=169=13^2...SO IT IS RIGHT ANGLED TRIANGLE..IF LET OA=5,OB=12,AB=13...THEN ANGLE AOB =90....LET O BE ORIGIN.SO A IS (5,0) AND B IS (0,12).SO CIRCUM CENTRE S IS THE MID POINT OF AB THE HYPOTENUSE. HENCE S IS {5/2,12/2)=(5/2,6) INCENTRE I IS GIVEN BY X COORDINATE....(5*0+12*5+13*0)/(5+12+13)=60/30=2 Y COORDINATE....(5*12+12*0+13*0)/(5+12+13)=60/30=2 HENCE S IS (2,2) SI =SQRT.{(5/2-2)^2+(6-2)^2}=SQRT(1/4+16)=SQRT(16.25)
 Circles/34511: A triangle with sides of 5, 12 and 13 has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles. 1 solutions Answer 20809 by venugopalramana(3286)   on 2006-04-22 04:59:31 (Show Source): You can put this solution on YOUR website!WE FIND THAT 5^2+12^2=25+144=169=13^2...SO IT IS RIGHT ANGLED TRIANGLE..IF LET OA=5,OB=12,AB=13...THEN ANGLE AOB =90....LET O BE ORIGIN.SO A IS (5,0) AND B IS (0,12).SO CIRCUM CENTRE S IS THE MID POINT OF AB THE HYPOTENUSE. HENCE S IS {5/2,12/2)=(5/2,6) INCENTRE I IS GIVEN BY X COORDINATE....(5*0+12*5+13*0)/(5+12+13)=60/30=2 Y COORDINATE....(5*12+12*0+13*0)/(5+12+13)=60/30=2 HENCE S IS (2,2) SI =SQRT.{(5/2-2)^2+(6-2)^2}=SQRT(1/4+16)=SQRT(16.25)
 Vectors/34516: A straight line L and a plane P in R3 are defined by the equations 2x − 3y + z = 1 and (x, y, z) = (1,−1, 2) + t (1, 2, 3), t any real number. (a) Say briefly but clearly how you know the line L does not pass through the point (−1,−5,−3). (b) Say briefly but clearly how you know that the plane P does not pass through the origin. (c) Write the equation for the line perpendicular to the plane P and passing through the origin. (d) Write the equation for the plane perpendicular to L and passing through the point (0, 1, 3). (e) Write the equation for the xz-plane.1 solutions Answer 20808 by venugopalramana(3286)   on 2006-04-22 02:57:39 (Show Source): You can put this solution on YOUR website!SEE ANSWERS BELOW ------------------------------ A straight line L and a plane P in R3 are defined by the equations EQN.OF PLANE P........ 2x − 3y + z = 1 and EQN.OF LINE L.....(x, y, z) = (1,−1, 2) + t (1, 2, 3), t any real number. (a) Say briefly but clearly how you know the line L does not pass through the point (−1,−5,−3)....THIS DOES NOT SATISFY EQN.L.FOR ONE VALUE OF T AS SHOWN BELOW -1=1+T...OR....T=-2... -5=-1+2T...OR....T=-2..OK -3=2+3T.....OR....T=-5/3....NOT TALLYING (b) Say briefly but clearly how you know that the plane P does not pass through the origin....ORIGIN IS (0,0,0)...THIS DOES NOT SATISFY EQN.P...WE GET 0=1 (c) Write the equation for the line perpendicular to the plane P and passing through the origin....LINE PERPENDICULAR TO PLANE IS PARALLEL TO PLANE'S NORMAL WHICH HAS DRS OF 2,-3,1...LINE PASSES THROUGH (0,0,0)..SO EQN OF LINE IS X/2 = Y/-3 = Z/1 (d) Write the equation for the plane perpendicular to L and passing through the point (0, 1, 3)..........DRS OF LINE ARE 1,2,3...THE REQD.PLANE IS PERPENDICULAR TO THE LINE.HENCE ITS NORMAL IS PARALLEL TO LINE.HENCE DRS OF NORMAL ARE SAME AS DRS OF LINE..THAT IS 1,2,3...HENCE EQN.OF PLANE IS X+2Y+3Z=K...IT PASSES THROUGH (0,1,3)....SO.....0+1*2+3*3=K=11 EQN. OF REQD. PLANE IS ..................X+2Y+3Z=11 (e) Write the equation for the xz-plane. Y=0
 Radicals/34412: This question is from textbook College Algebra The linear function is an arithmetic sequence. The exponential function is related to the geometric sequence. Give at least two real-life examples of a sequences or series. One example should be arithmetic, and the second should be geometric. Explain how these examples would affect you personally. Can someone please help me with this question(s). Maybe the example could apply to me on a job. I need help, please? 1 solutions Answer 20752 by venugopalramana(3286)   on 2006-04-21 12:26:30 (Show Source): You can put this solution on YOUR website!REAL LIFE EXAMPLES... ARITHMATIC / PROGRESSION/SERIES....A.P. SUPPOSE YOU ARE SAVING YOUR MONEY IN A BANK BY SYSTEMATIC PLAN OF DEPOSITING 100 \$ A MONTH...THEN STARTING WITH A 100\$ ACCOUNT YOUR MONEY IN CREDIT WITHOUT INTEREST ,WOULD BE AN EXAMPLE OF A.P...IT WILL BE..... TAKING MONTH AS AN INDEX... 100,200,300,400...... GEOMETRIC PROGRESSION/SERIES.....G.P. IN A SIMILAR WAY SUPPOSE YOU DEPOSITED 1000 \$ IN A BANK FOR INTEREST OF 4% PER YEAR,AND THE INTEREST IS CALCULATED AT THE END EVERY YEAR AND ADDED TO THE PRINCIPAL,THEN THE AMOUNT GROWN AT THE END OF YEAR IS AN EXAMPLE OF G.P.TAKING YEAR AS N INDEX....THE AMOUNT AT THE END OF SUCCESSIVE YEARS IS ..... 1000,1000*1.04,1000*1.04^2,1000*1.04^3....ETC...
 Rational-functions/34471: Hi my name is Eddie and I am a 11th grade student. I am have trouble with Graphing Radical Functions. I have a online text book that shows the problems http://www.phsuccessnet.com/altcontent/products/0-13-037881-X/Ch07/07-08/PH_Alg2_ch07-08_Ex.pdf I need help on problems 2,6,10, and 18. Any help would be helpful. Thank you.1 solutions Answer 20743 by venugopalramana(3286)   on 2006-04-21 10:52:31 (Show Source): You can put this solution on YOUR website!THE SITE YOU MENTIONED IS RESTRICTED AND I AM NOT ABLE TO ACCESS THE PROBLEMS .EITHER YOU HAVE TO REPRODUCE THEM OR ENABLE ME ACCESS TO YOUR STUDENT REGISTRATION BY SOME MEANS.PLEASE SEE WHAT CAN BE DONE. VENUGOPALRAMANA
 Miscellaneous_Word_Problems/31898: Five men took turns weighing themselves on a scale. However, they weighed each other two at a time, ten times total , so that every possible combination of two men was weighed. The scale had the following readouts for the pairs: 236 pounds, 244 pounds, 228 pounds, 250 pounds, 258 pounds, 230 pounds, 246 pounds, 238 pounds, 242 pounds, and 252 pounds. How heavy was the heaviest man?1 solutions Answer 20709 by venugopalramana(3286)   on 2006-04-21 06:34:25 (Show Source): You can put this solution on YOUR website!Five men took turns weighing themselves on a scale. However, they weighed each other two at a time, ten times total , so that every possible combination of two men was weighed. The scale had the following readouts for the pairs: 236 pounds, 244 pounds, 228 pounds, 250 pounds, 258 pounds, 230 pounds, 246 pounds, 238 pounds, 242 pounds, and 252 pounds. How heavy was the heaviest man? LET THE WEIGHTS 5 MEN IN ORDER OF DECREASING WEIGHT BE A,B,C,D,E WHEN THEY WEIGH IN PAIRS WE GET 5C2=5*4/2=10 DIFFERENT PAIRS OR 20 MEN HENCE THE TOTAL WEIGHT IS FOUR TIMES THE WEIGHT OF THE 5 MEN'S TOTAL WEIGHT 4(A+B+C+D+E)=236+244+228+250+258+230+246+238+242+252=2424 A+B+C+D+E=606..............I NOW HEAVIEST IS A+B=258.........II NEXT HEAVY IS A+C=252...........III LIGHTEST =D+E=228.................IV EQN.I-EQN.IV...GIVES.. A+B+C+D+E-D-E=606-228=374 A+B+C=374.....................V EQN.II+EQN.III GIVES A+B+A+C=258+252=510 2A+B+C=510.................VI EQN.VI-EQ.EQN.V...GIVES... 2A+B+C-A-B-C=510-374=136 A=136 SO HEAVIEST MAN'S WEIGHT =136
 Quadratic_Equations/33912: x^3-3x^2+3x-9=0 I got x=3; x=3i; x=-3i...how do I plug answers back in to check solution?1 solutions Answer 20706 by venugopalramana(3286)   on 2006-04-21 05:56:05 (Show Source): You can put this solution on YOUR website!LET Y = x^3-3x^2+3x-9=0 I got x=3; x=3i; x=-3i...how do I plug answers back in to check solution? HOW DID YOU GET THOSE ANSWERS?...ANY WAY ,YOU HAVE TO JUST SUBSTITUTE THOSE VALUES FOR X AND CHECK 1...X=3...Y=3^3-3*3^2+3*3-9=27-27+9-9=0...OK.. 2...X=3i..Y=(3i)^3-3*(3i)^2+3*(3i)-9=27*(i^2)(i)-27(i)^2+9i-9=-27i+27+9i-9=-18i+18... THIS IS NOT ZERO HENCE THIS IS NOT A ROOT 3...X=-3i..Y=(-3i)^3-3*(-3i)^2+3(-3i)-9=-27(i^2)(i)-27(i^2)-9i-9=27i+27-9i-9=18i+18... THIS IS NOT ZERO ..HENCE THIS IS NOT A ROOT.. THE CORRECT ANSWERS ARE X=3 X=i*SQRT(3) X=-i*SQRT(3) YOU CAN CHECK BY SUBSTITUTION AS SHOWN ABOVE
 Functions/34421: find the domain of the sqrt (3-x)/x1 solutions Answer 20705 by venugopalramana(3286)   on 2006-04-21 05:34:31 (Show Source): You can put this solution on YOUR website!LET Y =SQRT{(3-X)/X}THIS IS THE PROBLEM OR IS IT {SQRT(3-X)}/X..PLEASE PUT BRACKETS APPROPRIATELY..OK LET US TAKE Y={SQRT(3-X)}/X SO (3-X) CAN NOT BE NEGATIVE..THAT IS 3>=X...OR....X<=3 FURTHER WE HAVE X IN DENOMINATOR ..HENCE X SHOULD NOT BE ZERO. SO DOMAIN OF THE FUNCTION IS X<=3 BUT NOT EQUAL TO ZERO.