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# Recent problems solved by 'venugopalramana'

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 Numbers_Word_Problems/26807: Show that the sum of an odd integer and an even integer is always odd.1 solutions Answer 14613 by venugopalramana(3286)   on 2006-02-08 10:40:25 (Show Source): You can put this solution on YOUR website!ANY INTEGER IS DENOTED BY N IT WILL BE EVEN IF WE MULTIPLY BY 2...THAT IS 2N IS EVEN INTEGER. IT WILL BE ODD IF WE ADD 1 TO IT..THAT IS 2N+1 IS ODD INTEGER. SUM OF ODD AND EVEN INTEGERS IS (2N+1)+(2M)=2N+2M+1=2(N+M)+1=2P+1 WHERE P IS AN INTEGER. HENCE BY THE ABOVE LOGIC THE SUM IS AN ODD INTEGER.
 Numbers_Word_Problems/26809: show that the product of two even integers is always even.1 solutions Answer 14612 by venugopalramana(3286)   on 2006-02-08 10:37:56 (Show Source): You can put this solution on YOUR website!ANY INTEGER IS DENOTED BY N IT WILL BE EVEN IF WE MULTIPLY BY 2...THAT IS 2N IS EVEN INTEGER. PRODUCT OF 2 EVEN INTEGERS IS (2N)(2M)=4MN=2(2MN)=2P WHERE P IS AN INTEGER. HENCE BY THE ABOVE LOGIC THE PRODUCT IS AN EVEN INTEGER.
 Numbers_Word_Problems/26810: show that the product of an odd integer and an even integer is always even.1 solutions Answer 14611 by venugopalramana(3286)   on 2006-02-08 10:35:55 (Show Source): You can put this solution on YOUR website!ANY INTEGER IS DENOTED BY N IT WILL BE EVEN IF WE MULTIPLY BY 2...THAT IS 2N IS EVEN INTEGER. IT WILL BE ODD IF WE ADD 1 TO IT..THAT IS 2N+1 IS ODD INTEGER. PRODUCT OF 1 ODD AND 1 EVEN INTEGER IS (2N)(2M+1)=4MN+2N=2(2MN+N)=2P WHERE P IS AN INTEGER. HENCE BY THE ABOVE LOGIC THE PRODUCT IS AN EVEN INTEGER.
 Numbers_Word_Problems/26812: show that the product of two odd integers is always odd.1 solutions Answer 14610 by venugopalramana(3286)   on 2006-02-08 10:33:08 (Show Source): You can put this solution on YOUR website!ANY INTEGER IS DENOTED BY N IT WILL BE EVEN IF WE MULTIPLY BY 2...THAT IS 2N IS EVEN INTEGER. IT WILL BE ODD IF WE ADD 1 TO IT..THAT IS 2N+1 IS ODD INTEGER. PRODUCT OF 2 ODD INTEGERS IS (2N+1)(2M+1)=4MN+2N+2M+1=2(2MN+M+N)+1=2P+1 WHERE PIS AN INTEGER. HENCE BY THE ABOVE LOGIC THE PRODUCT IS AN ODD INTEGER.
 Linear-equations/26847: Help! trying to solve a problem given g(x)=x-3 solve if g[g(6)]1 solutions Answer 14609 by venugopalramana(3286)   on 2006-02-08 10:26:03 (Show Source): You can put this solution on YOUR website!g(x)=x-3 solve if g[g(6)] G(6)=6-3=3...(TO FIND G(A)..PUT X=A IN THE GIVEN EQN. FOR G(X) SO NOW G(G(6))=G(3)=3-3=0
 Graphs/26849: Find the slope and y-intercept of the line represented by each of the following equations. 2x - 3y = 61 solutions Answer 14608 by venugopalramana(3286)   on 2006-02-08 10:23:06 (Show Source): You can put this solution on YOUR website!2x-3y=6 3y=2x-6 y=2x/3-2 slope=2/3 y intercept point is (0,2) see the following example and do the same way.. ------------------------------------------------- Slope: ; -3/4 y-intercept: (0, 8) eqn.of a line with slope..m ..and y intercept..c...is given by y=mx+c...hence..eqn.of reqd.line is y=(-3x/4)+8..or....4y=-3x+32 determine points and graph as follows x....................0.................4.....................8.....etc y=(-3x/4)+8..........8.................5.....................2...etc..
 Graphs/26850: Write the equation of the line with given slope and y-intercept. Then graph each line using the slope and y-intercept. Slope: -2; y-intercept: (0, 4) and Slope: 5; y-intercept: (0, -2)1 solutions Answer 14607 by venugopalramana(3286)   on 2006-02-08 10:21:09 (Show Source): You can put this solution on YOUR website!see the following example and do the same way.. ------------------------------------------------- Slope: ; -3/4 y-intercept: (0, 8) eqn.of a line with slope..m ..and y intercept..c...is given by y=mx+c...hence..eqn.of reqd.line is y=(-3x/4)+8..or....4y=-3x+32 determine points and graph as follows x....................0.................4.....................8.....etc y=(-3x/4)+8..........8.................5.....................2...etc..
 Graphs/26851: Write the equation of the line with given slope and y-intercept. Then graph each line using the slope and y-intercept. Slope: ; -3/4 y-intercept: (0, 8) 1 solutions Answer 14606 by venugopalramana(3286)   on 2006-02-08 10:18:56 (Show Source): You can put this solution on YOUR website!Slope: ; -3/4 y-intercept: (0, 8) eqn.of a line with slope..m ..and y intercept..c...is given by y=mx+c...hence..eqn.of reqd.line is y=(-3x/4)+8..or....4y=-3x+32 determine points and graph as follows x....................0.................4.....................8.....etc y=(-3x/4)+8..........8.................5.....................2...etc..
 Linear_Algebra/26765: Let S be the basis {1,t,t^2} of P2. Set P1= 2+t^2 P2= 1-t+t^2 P3= 3-t+t^2 in P2. Show that P1,P2,P3 isa basis of P2 and find the coordinates of 1,t,t^2 in this new basis. 1 solutions Answer 14605 by venugopalramana(3286)   on 2006-02-08 08:24:25 (Show Source): You can put this solution on YOUR website!Let S be the basis {1,t,t^2} of P2.BETTER CALL THIS Q2 TO AVOID CONFUSION WITH P2 BELOW Set P1= 2+t^2 P2= 1-t+t^2 P3= 3-t+t^2 in P2. Show that P1,P2,P3 isa basis of Q2 and find the coordinates of 1,t,t^2 in this new basis. TO SHOW THAT P1,P2,P3 IS A BASIS,WE NEED TO PROVE THAT THEY ARE INDEPENDENT. THAT IS IF AP1+BP2+CP3=0...THEN A=B=C=0 AP1+BP2+CP3=A(2+T^2)+B(1-T+T^2)+C(3-T+T^2)=0 T^2(A+B+C)-T(B+C)+(2A+B+3C)=0.....................I..........HENCE A+B+C=0 ............................II B+C=0...............................III 2A+B+3C=0...........................IV EQN.II - EQN.III..GIVES.....A=0......V...SUBSTITUTING IN IV,WE GET B+3C=0...............................VI EQN.VI-EQN.III...GIVES....2C=0...OR...C=0..WHICH ON SUBSTITUTING IN EQN.I GIVES B==0...SO A=B=C=0...HENCE P1,P2,P3 ARE INDEPENDENT.HENCE THEY FORM A BASIS. NOW TO WRITE Q2 IN THE EW BASIS WE EQUATE EQN.I TO Q2 AND FIND A,B,C. T^2(A+B+C)-T(B+C)+(2A+B+3C)=Q2=1+T+T^2...HENCE A+B+C=1.....................VII B+C=-1......................VIII 2A+B+3C=1...................IX PROCEEDING AS BEFORE WE GET A=2,B=0,C=-1..HENCE Q2=1+T+T^2=2(2+T^2)-(3-T+T^2)
 Rational-functions/26803: find a third-degree polynomial equation with rational coefficients that has the roots -5 and 4+i1 solutions Answer 14602 by venugopalramana(3286)   on 2006-02-08 01:03:22 (Show Source): You can put this solution on YOUR website!find a third-degree polynomial equation with rational coefficients that has the roots -5 and 4+i IF THE COEFFICIENTS OF POLYNOMIAL ARE REAL ,THEN COMPLEX ROOTS CAN OCCUR ONLY IN CONJUGATES.SO IF 4+I IS A ROOT THEN 4-I IS ALSO A ROOT.SO,WE HAVE ALL THE 3 ROOTS FOR THE 3 RD. DEGREE POLYNOMIAL...HENCE THE POLYNOMIAL IS GIVEN BY... (X+5)(X-(4+I))((X-(4-I))=0 (X+5){(X-4)^2-I^2}=0 (X+5)(X^2+16-8X+1)=0 =(X+5)(X^2-8X+17)=0 =X^3-8X^2+17X+5X^2-40X+85=0 =X^3-3X^2-23X+85=0
 logarithm/26823: For what integral values of n is the expression 1^n + 2^n + 3^n + 4^n divisible by 5? I'd really appreciate any offered help. Thanks.1 solutions Answer 14601 by venugopalramana(3286)   on 2006-02-08 00:52:57 (Show Source): You can put this solution on YOUR website!For what integral values of n is the expression 1^n + 2^n + 3^n + 4^n divisible by 5? WE CAN EASILY SEE THAT FOR N=1,2,3 ETC..IT IS DIVISIBLE BY 5..FOR EX...FOR N=1,WE HAVE 1^1+2^1+3^1+4^1=1+2+3+4=10 WHICH IS DIVISIBLE BY 5..SIMILARLY FOR N=2,3 ETC.. IF YOU WANT A PROOF OF THE THEOREM USE INDUCTION METHOD ...
 Proportions/26824: A bag contains 6 red marbles, 9 blue marbles, and 5 green marbles. You withdrew one marble,replace it, and then withdraw another marble. what is the probability that you do NOT pick two green marbles?1 solutions Answer 14600 by venugopalramana(3286)   on 2006-02-08 00:34:03 (Show Source): You can put this solution on YOUR website!A bag contains 6 red marbles, 9 blue marbles, and 5 green marbles. You withdrew one marble,replace it, and then withdraw another marble. what is the probability that you do NOT pick two green marbles? in such problems it is easier to find probability of an event happening and subtract it from total probability of 1 to get probability of event not happening. SO LET US CONSIDER PROBABILITY OF DRAWING 2 GREEN MARBLES. THERE ARE 6 RED+9 BLUE+5 GREEN MARBLES=20 MARBLES. PROBABILITY OF GETTING A GREEN MARBLE ON FIRST ATTEMPT=5/20=1/4 PROBABILITY OF GETTING A GREEN MARBLE ON SECOND ATTEMPT= =PROBABILITY OF GETTING A GREEN MARBLE ON FIRST ATTEMPT AS THE MARBLE IS REPLACED =5/20=1/4 HENCE PROBABILITY OF DRAWING 2 GREEN MARBLES = (1/4)8(1/4)=1/16 HENCE PROBABILITY OF NOT DRAWING 2 GREEN MARBLES =1-1/16=15/16
 Distributive-associative-commutative-properties/26683: Verify that thesetof orthogonal nxn matrices form a subgroup of the general linear group GLn(R).1 solutions Answer 14599 by venugopalramana(3286)   on 2006-02-08 00:19:41 (Show Source): You can put this solution on YOUR website!ORTHOGONAL MATRICES ARE SQUARE MATRICES SUCH THAT PRODUCT OF AN ORTHOGONAL MATRIX WITH ITS TRANSPOSE IS AN IDENTITY MATRIX.THAT IS IF A*A'=A'*A=I,THEN A IS AN ORTHOGONAL MATRIX. TO PROVE THAT THEY FORM A SUB GROUP UNDER THE GENERAL LINEAR GROUP OF MATRICES,WE NEED TO SHOW THAT I.ORTHOGONAL MATRICES ARE A SUB SET OF THE GENERAL GROUP OF MATRICES. II.UNDER THE SAME COMPOSITION AS IN THE GENERAL GROUP,THE SUB GROUP ITSELF IS A GROUP.HERE WE CONSIDER THE COMPOSITION OF MULTIPLICATION.THIS MEANS THAT WE HAVE TO SHOW WITHIN THE SUBGROUP FOR MULTIPLICATION...... 1.CLOSOURE 2.ASSOCIATIVITY 3.EXISTENCE OF IDENTITY 4.EXISTENCE OF INVERSE NOW I CRITERIA IS OBVIOUS AS THE SET OF ORTHOGONAL MATRICES (CALLED S) AS DEFINED ARE INDEED A SUB SET OF GENERAL LINEAR GROUP OF MATRICES(CALLED G). II....MULTIPLICATION IS INDEED WELL DEFINED AND EXISTS FOR ORTHOGONAL MATRICES AS IT IS FOR THE GENERAL GROUP OF MATRICES. 1 AND 2.ASSOCIATIVITY HOLDS FOR MATRIX MULTIPLCATION IN GENERAL SO WE SHALL PROVE CLOSOURE HERE. LET A AND B BE ORTHOGONAL MATRICES IN THE SUB SET OF ORTHOGONAL MATRICES S .SINCE A' AND B' ARE ALSO ORTHOGONAL MATRICES,THEY ARE ALSO ELEMENTS OF S. SO A*A'=A'*A=I...AND...B*B'=B'*B=I... NOW.. IF WE SHOW THAT A*B IS ALSO AN ORTHOGONAL MATIX THEN IT WILL BE AN ELEMENT OF S WHICH PROVES CLOSOURE...... TST (A*B)*(A*B)'=I WE KNOW FOR MATRIX MULTIPLICATION THAT (A*B)'=(B'*A')...HENCE (A*B)*(A*B)'=(A*B)*(B'*A')=A*(B*B')*A'=A*I*A'=A*A'=AA'=I HENCE A*B IS ALSO AN ORTHOGONAL MATRIX.HENCE IT IS AN ELEMENT OF S. 3. IDENTITY ELEMENT IN S IS THE UNIT MATRIX WHICH IS SAME AS IN G.SINCE,WE HAVE A*I=I*A=A...WHICH FOLLOWS OBVIOUSLY FROM MULTIPLICATIVE PROPERTY. 4.WE HAVE A*A'=A'*A=I.....IF A IS AN ORTHOGONAL MATRIX ITS INVERSE A' IS ALSO AN ORTHOGONAL MATRIX. HENCE IT IS AN ELEMENT OF S. THUS S IS A SUB GROUP OF G.
 Circles/26653: The bisector of a vertical angle A of a triangle ABC meets the circumcircle of triangle ABC at D. Show that D is the mid point of arc BDC. Prove with a diagram.1 solutions Answer 14554 by venugopalramana(3286)   on 2006-02-07 11:19:11 (Show Source): You can put this solution on YOUR website!DRAW A CIRCLE.INSCRIBE A TRIANGLE ABC IN IT.DRAW AD BISECTING ANGLE BAC AND MEETING THE CIRCLE AT D.SO IN THE 2 SEGMENTS OF THE CIRCLE..NAMELY ARC BD AND ARC DC WE HAVE,BOTH ARCS SUBTENDING EQUAL ANGLES AT A .SINCE ANGLE BAD=ANGLE DAC. SO THE ARCS SHALL BE EQUAL IN LENGTH.SO ARC BD = ARC DC SO D IS THE MID POINT OF ARC BDC
 Quadratic-relations-and-conic-sections/26655: Find the foci of the ellipse whose equation is (y^2/16) + (x^2/12) = 11 solutions Answer 14553 by venugopalramana(3286)   on 2006-02-07 11:12:10 (Show Source): You can put this solution on YOUR website! Find the foci of the ellipse whose equation is (y^2/16)+(x^2/12)=1. > COMPARING WITH STANDARD EQN. OF ELLIPSE > (X^2/A^2)+(Y^2/B^2)+1...WE GET > FOCI=(AE,0) AND (-AE,0)..WHERE E=SQRT.{(A^2-B^2)/A^2} > WE HAVE A=4 AND E=SQRT.{(16-12)/16}=1/2..SO FOCI ARE..(2,0),(-2,0)
 Probability-and-statistics/26658: Find the probability of drawing an ace or a face card from a standard 52 card deck.1 solutions Answer 14551 by venugopalramana(3286)   on 2006-02-07 10:49:49 (Show Source): You can put this solution on YOUR website!THERE ARE 4 ACES PLUS 12 FACE CARDS =16 IN ALL IN A PACK OF 52 CARDS SO PROBABILITY OF DRAWING ONE OF THEM IS =16/52=4/13
 Probability-and-statistics/26659: Identify the number of ways seven books can be arranged on a shelf.1 solutions Answer 14549 by venugopalramana(3286)   on 2006-02-07 10:44:27 (Show Source): You can put this solution on YOUR website!THIS IS GIVEN BY 7P7=7!=7*6*5*4*3*2*1=5040
 Equations/26667: Solve the equation Inx=2. Round the answer to four decimal places. Please help!1 solutions Answer 14548 by venugopalramana(3286)   on 2006-02-07 10:41:21 (Show Source): You can put this solution on YOUR website! Solve the equation 1nx=2. Round the answer to four decimal places. > LN X=2 > X=E^2=7.389056..FROM CALCULATOR..
 Equations/26666: Solve the equation 25^(x+1)=125^x Help!1 solutions Answer 14547 by venugopalramana(3286)   on 2006-02-07 10:38:41 (Show Source): You can put this solution on YOUR website!LET 5^X=Y > 25^X=(5^2)^X=5^2X=(5^X)^2=Y^2 > 125^X=(5^3)^X=5^3X=(5^X)^3=Y^3 WE HAVE....25^(X+1)=125^X... (25^X)(25)=125^X...OR... 25Y^2=Y^3 Y^2(Y-25)=0...OR...Y=25.. 5^X=25...SO..X=2
 Coordinate-system/26668: Find the midpoint and the distance between the points (6,-8) and (14,7). Thanks for your help!!!1 solutions Answer 14535 by venugopalramana(3286)   on 2006-02-07 05:53:46 (Show Source): You can put this solution on YOUR website!Find the midpoint and the distance between the points (6,-8) and (14,7). DISTANCE BETWEEN (X1,Y1) AND (X2,Y2) IS GIVEN BY SQRT.{(X2-X1)^2+(Y2-Y1)^2} AND THEIR MID POINT IS GIVEN BY {(X1+X2)/2,(Y1+Y2)/2} SO WE GET DISTANCE =SQRT.{(14-6)^2+(7+8)^2}=SQRT(64+225)=SQRT.(289) MIDPOINT IS {(6+14)/2,(-8+7)/2}=(10,-1/2)
 Probability-and-statistics/26669: Find the probability of drawing a red card and then a black card from a standard 52 card deck. (without replacement)1 solutions Answer 14534 by venugopalramana(3286)   on 2006-02-07 05:47:09 (Show Source): You can put this solution on YOUR website!THERE ARE 26 RED CARDS OUT OF 52. HENCE PROBABILITY OF DRAWING RED CARD FIRST=26/52=1/2 NOW THERE ARE 26 BLACK CARDS AND 51 TOTAL CARDS PROBABILITY OF DRAWING BLACK CARD IS =26/51 SO PROBABILITY OF DRAWING FIRST RED AND THEN BLACK CARD =(1/2)(26/51) THIS ENDS YOUR PROBLEM. ***************************************************************************** BUT IF PROBILITY OF ONE RED AND ONE BLACK CARD IS NEEDED THEN..... SIMILARLY PROBABILITY OF DRAWING FIRST BLACK AND THEN RED CARD =(1/2)(26/51) HENCE TOTAL PROBABILITY OF DRAWING ONE BLACK AND ONE RED CARD =SUM OF THE ABOVE 2 PROBABILITIES.=26/51
 Distributive-associative-commutative-properties/26682: Let a not equal 0, b and c be integers with a and b relatively prime. Show that if a|b*c then a|c. how do i give equations for relatively prime and divisible and then substitute. 1 solutions Answer 14522 by venugopalramana(3286)   on 2006-02-07 05:16:20 (Show Source): You can put this solution on YOUR website!Let a not equal 0, b and c be integers with a and b relatively prime. Show that if a|b*c A|BC...HENCE.... BC=A*M......I.......WHERE M IS AN INTEGER AND B ARE RELATIVELY PRIME...THAT IS THEY HAVE NO COMMON FACTORS OR THEIR GCD IS 1. HENCE GCD =1 =AX+BY...WHERE X AND Y ARE INTEGERS..MULTIPLYING BY C THROUGHOUT,WE GET.. C=CAX+CBY...SUBSTITUTING EQN.I... C=CAX+AMY=A(CX+MY)=A*K SAY.....WHERE K=CX+MY... BUT C,X,M,Y ARE ALL INTEGERS...HENCE K IS AN INTEGER. C=A*K....OR...A DIVIDES C....A|C then a|c. how do i give equations for relatively prime and divisible and then substitute. USE GCD OF A AND B =XA+YB...TO CONVERT INTO EQNS.HOPE YOU UNDERSTOOD.
 Distributive-associative-commutative-properties/26684: show that the binary operation is a group and determine if it is Abelian H = {(a,b)belong to R^(2): a can not equal 0} with product (a,b)(c,d) = (ac,ad + b)1 solutions Answer 14520 by venugopalramana(3286)   on 2006-02-07 03:41:54 (Show Source): You can put this solution on YOUR website!(a,b)*(c,d) = (ac,ad + b) TO SHOW THIS OPERATION * IS A GROUP,WE NEED TO PROVE... 1..CLOSOURE... AC IS ALSO REAL NUMBER SINCE A AND C ARE REAL...AD IS ALSO REAL AND SO IS AD+B.SO (AC,AD+B)IS AN ELEMENT OF R^2 2..ASSOCIATIVE.... TPT...{(A,B)*(C,D)}*(E,F)=(A,B)*{(C,D)}*(E,F)}.. LHS=(AC,AD+B)(E,F)=(ACE,ACF+AD+B) RHS=(A,B)(CE,CF+D)=(ACE,ACF+AD+B)=LHS 3...EXISTENCE OF IDENTITY....LET IT BE (I1,I2) WE SHOULD HAVE (A,B)*(I1,I2)=(A,B) AND (I1,I2)*(A,B)=(A,B) AI1=A...SO...I1=1 AI2+B=B...SO....I2=0....SO...(1,0) IS THE IDENTITY ELEMENT..IT IS AN ELEMENT OF R^2...OK.... LET US CHECK... (1,0)*(A,B)=(1A,1B+0)=(A,B)...OK... 4...EXISTENCE OF INVERSE.... WE SHOULD FIND (X,Y)SO THAT (X,Y)(A,B)=(1,0)=(A,B)*(X,Y) XA=1....OR..X=1/A..... XB+Y=0.....Y=-XB=-B/A.....SO INVERSE IS (1/A,-B/A)..SINCE...A IS NOT ZERO ,WE HAVE THE INVERSE AS AN ELEMENT OF R^2. LET US CHECK... (A,B)(1/A,-B/A)=(A*1/A,((A*-B)/A)+B)=(1,0) ....OK.... HENCE * AS A BINARY OPERATION IS A GROUP. WE FIND THAT (a,b)*(c,d) = (ac,ad + b)..WHERE AS (C,D)*(A,B) = (CA,CB+D) WHICH ARE NOT EQUAL. HENCE THIS IS NOT ABELIAN
 Functions/26686: For what value of 'c' is 'ax - by = c' a direct variation? 1 solutions Answer 14519 by venugopalramana(3286)   on 2006-02-07 00:52:11 (Show Source): You can put this solution on YOUR website!For what value of 'c' is 'ax - by= c' a direct variation? 1 solutions -------------------------------------------------------------------------------- Answer 14515 by venugopalramana(833) on 2006-02-07 00:05:37 (Show Source): WE SAY X VARIES DIRECTLY AS Y IF X=K*Y,WHERE K IS A CONSTANT..HERE WE HAVE AX-BY=C..OR...AX=BY+C...OR...X=(BY+C)/A = (B/A)Y+(C/A)..BUT FOR DIRECT PROPORTION X=K*Y...HENCE C/A=0...OR...C=0
 Distributive-associative-commutative-properties/26701: Let r=12*95*1750 and h=38*25^(2)*16^(3). Give all common divisors, gcd, and lcm of a and b1 solutions Answer 14518 by venugopalramana(3286)   on 2006-02-07 00:48:18 (Show Source): You can put this solution on YOUR website!Let r=12*95*1750...WRITE AS PRODUCT OF PRIMES... R=(2*2*3)*(5*19)*(2*5*5*5*7)=(2^3)(3)(5^4)(7)(19) and h=38*25^(2)*16^(3). H=(2*19)((5*5)^2)((2*2*2*2)^3)=(2^13)(5^4) HENCE GCD =(2^3)(5^4) LCM=(2^13)(3)(5^4)(7)(19) COMMON DIVISORS ARE ALL FACTORS OBTAINED FROM GCD =(2^3)(5^4)...THAT IS (2^0)(5),(2^0)(5^2),(2^0)(5^3),(2^0)(5^4) (2^1)(5^0),(2^1)(5^1).....ETC........ Give all common divisors, gcd, and lcm of a and b
 Distributive-associative-commutative-properties/26646: Let (Fn)=(1,1,2,3,5,8,13,21,34,55,...) be the fibonacci sequence defined by F1=F2=1, Fn=F(n-1)+F(n-2) if n>2. Show that it holds for n which is greator and equal to 1. Fn < 2^(n) 1 solutions Answer 14517 by venugopalramana(3286)   on 2006-02-07 00:31:31 (Show Source): You can put this solution on YOUR website!LET US COMPARE THE GIVEN SERIES WITH A GEOMETRIC SERIES WHOSE WE CAN DO... THE G.P IS ...1,2,4,8,16,.......WITH C.R. OF 2 AND A=1. FIRSTLY WE FIND THAT THE GIVEN SERIES IS ALL POSITIVE NUMBERS AND INCREASING SERIES AFTER F2,SINCE EVERY TERM IS THE SUM OF PREVIOUS 2 TERMS WHICH ARE POSITIVE...JENCE..F2 NOW WE HAVE.. F1=1............................................G1=1.........F1=G1 F2=1............................................G2=2G1.......F2 F3=F1+F2 F4=F2+F3 .................................................................... ..................................................................... FN=F(N-2)+F(N-1) ------------------------------------------------------------------------ ADDING ALL THE ABOVE WE GET...(F1+F2+F3+F4+...+FN)<(G1+G2+G3+.....+GN) BUT WE KNOW THAT G1+G2+G3+.....+GN=1+2+4+8+.........2^(N-1) A=1...R=2...SO...SUM IS ..A*(R^N-1)/(R-1)=1*(2^N-1)/(2-1)=2^N-1 HENCE (F1+F2+F3+F4+...+FN)<(G1+G2+G3+.....+GN)=2^N-1<2^N...THEN OBVIOUSLY FN<2^N
 Numbers_Word_Problems/26723: Its from differential calculus. LEt y= 5/³√(2x-5)^2 So far I got upto: d/dx=5(2x^2-10x+25)^(2/3) However, I can not continue on from there.1 solutions Answer 14516 by venugopalramana(3286)   on 2006-02-07 00:07:53 (Show Source): You can put this solution on YOUR website!Its from differential calculus. LEt y= 5/³√(2x-5)^2 So far I got upto: d/dx=5(2x^2-10x+25)^(2/3) However, I can not continue on from there. the answer at the back on my book is: -20 ---------- 3(2x-5)^(5/3) Your Answer: see my working and comments on your work----------- -------------------------------------------------------- Its from differential calculus. LEt y= 5/³√(2x-5)^2 ..USE X^(-N)=1/(X^N)AND N th. ROOT OF X =X^(1/N) Y=5*(2X-5)^(-2/3)... DY/DX=5*(2X-5)^(-2/3-1)*2..USING DY/DX=DY/DU*DU/DX..HERE U=2X-5 AND DERIVATIVE OF X^N = N*X^(N-1) DY/DX=(-20/3)*(2X-5)^(-5/3) (-20)/3*(2X-5)^(5/3) ------------------------------------------------------ So far I got upto: d/dx=5(2x^2-10x+25)^(2/3)..YOU SQUARED 2X-5 ..THEN WHY AGAIN POWER OF 2/3..IT SHOULD BE ONLY 1/3..FURTHER IT SHOULD BE -1/3 AND NOT + However, I can not continue on from there. the answer at the back on my book is: -20 ---------- 3(2x-5)^(5/3)
 Linear-equations/26730: For what value of 'c' is 'ax - by= c' a direct variation?1 solutions Answer 14515 by venugopalramana(3286)   on 2006-02-07 00:05:37 (Show Source): You can put this solution on YOUR website!WE SAY X VARIES DIRECTLY AS Y IF X=K*Y,WHERE K IS A CONSTANT..HERE WE HAVE AX-BY=C..OR...AX=BY+C...OR...X=(BY+C)/A = (B/A)Y+(C/A)..BUT FOR DIRECT PROPORTION X=K*Y...HENCE C/A=0...OR...C=0
 Numbers_Word_Problems/26735: Its from differential calculus. LEt y= 5/³√(2x-5)^2 So far I got upto: d/dx=5(2x^2-10x+25)^(2/3) However, I can not continue on from there. 1 solutions Answer 14514 by venugopalramana(3286)   on 2006-02-07 00:01:09 (Show Source): You can put this solution on YOUR website!Its from differential calculus. LEt y= 5/³√(2x-5)^2 So far I got upto: d/dx=5(2x^2-10x+25)^(2/3) However, I can not continue on from there. the answer at the back on my book is: -20 ---------- 3(2x-5)^(5/3) Your Answer: see my working and comments on your work----------- -------------------------------------------------------- Its from differential calculus. LEt y= 5/³√(2x-5)^2 ..USE X^(-N)=1/(X^N)AND N th. ROOT OF X =X^(1/N) Y=5*(2X-5)^(-2/3)... DY/DX=5*(2X-5)^(-2/3-1)*2..USING DY/DX=DY/DU*DU/DX..HERE U=2X-5 AND DERIVATIVE OF X^N = N*X^(N-1) DY/DX=(-20/3)*(2X-5)^(-5/3) (-20)/3*(2X-5)^(5/3) ------------------------------------------------------ So far I got upto: d/dx=5(2x^2-10x+25)^(2/3)..YOU SQUARED 2X-5 ..THEN WHY AGAIN POWER OF 2/3..IT SHOULD BE ONLY 1/3..FURTHER IT SHOULD BE -1/3 AND NOT + However, I can not continue on from there. the answer at the back on my book is: -20 ---------- 3(2x-5)^(5/3)
 Numbers_Word_Problems/26736: Its from differential calculus. LEt y= 5/³√(2x-5)^2 So far I got upto: d/dx=5(2x^2-10x+25)^(2/3) However, I can not continue on from there. the answer at the back on my book is: -20 ---------- 3(2x-5)^(5/3)1 solutions Answer 14513 by venugopalramana(3286)   on 2006-02-06 23:48:53 (Show Source): You can put this solution on YOUR website!see my working and comments on your work----------- -------------------------------------------------------- Its from differential calculus. LEt y= 5/³√(2x-5)^2 ..USE X^(-N)=1/(X^N)AND N th. ROOT OF X =X^(1/N) Y=5*(2X-5)^(-2/3)... DY/DX=5*(2X-5)^(-2/3-1)*2..USING DY/DX=DY/DU*DU/DX..HERE U=2X-5 AND DERIVATIVE OF X^N = N*X^(N-1) DY/DX=(-20/3)*(2X-5)^(-5/3) (-20)/3*(2X-5)^(5/3) ------------------------------------------------------ So far I got upto: d/dx=5(2x^2-10x+25)^(2/3)..YOU SQUARED 2X-5 ..THEN WHY AGAIN POWER OF 2/3..IT SHOULD BE ONLY 1/3..FURTHER IT SHOULD BE -1/3 AND NOT + However, I can not continue on from there. the answer at the back on my book is: -20 ---------- 3(2x-5)^(5/3) You may edit the question. Maybe convert formulae to the same formula notation as in your solutions.
 logarithm/26651: Solve the equation log^9(3x+14)-log^95=log^92x.1 solutions Answer 14498 by venugopalramana(3286)   on 2006-02-06 11:17:23 (Show Source): You can put this solution on YOUR website!log^9(3x+14)-log^95=log^92x. i think you mean all logs are to base 9...in that case since all are to a common base we can drop the base. log(3x+1)-log 5=log(2x)....or.....log(3x+1)-log 5-log(2x)=0 log{(3x+1)/(5*2x)=0..taking antilogs,we get (3x+1)/10x=1 3x+1=10x...or...10x-3x=1... 7x=1 x=1/7