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# Recent problems solved by 'venugopalramana'

venugopalramana answered: 3288 problems
 Numeric_Fractions/26458: Hi i'm trying to solve for X, a/bx+c/d=e. I have reduced if to a/bx=ed-c/d but can't see how to get to the answer. which i belive is x=da/b(ed-c). Please help Thank Darryl 1 solutions Answer 14288 by venugopalramana(3286)   on 2006-02-03 00:08:20 (Show Source): You can put this solution on YOUR website! Hi i'm trying to solve for X, a/bx+c/d=e. USE BRACKETS TO SHOW THE THINGS PROPERLY AND TO AVOID MISTAKES. (a/b)x+(c/d)=e. I have reduced if to a/bx=ed-c/d NO...IT IS (a/b)x=e-(c/d) = (ED-C)/D SO NOW WE GET X={(ED-C)}/(A/B) =B(ED - C )/A......IS THE ANSWER.HOWEVER IF YOUR PROBLEM IS REALLY (a/bx)+(c/d)=e....THEN (A/BX)= E - C/D = (ED-C)/D A/B = X*(ED - C)/D X = (A/B)/{(ED-C)/D} X = AD/B(ED - C)WHICH IS THE ANSWER YOU ARE SHOWING BELOW...SO NOW YOU UNDERSTAND THE IMPORTANCE OF PUTTING PROPER BRACKETS. but can't see how to get to the answer. which i belive is x=da/b(ed-c).
 Distributive-associative-commutative-properties/26383: Calculate gcd(x,y) and express it as an integral linear combination of x and y. a) x=3113, y=1331 The gcd(3113,1331) is 451, but i don't know how to express as a integral combination of x and y. 1 solutions Answer 14287 by venugopalramana(3286)   on 2006-02-02 23:58:16 (Show Source): You can put this solution on YOUR website! x=3113, y=1331 The gcd(3113,1331) is 451, but i don't know how to express as a integral combination of x and y. GCD IS NOT 451 IT IS 11SEE BELOW. 1331 3113 2 .. .. .. .. .. .. 451 1331 2 .. .. .. .. .. .. 429 451 1 .. .. .. .. .. .. 22 429 19 .. .. .. .. .. .. 11 22 2 .. .. .. .. .. .. 0 .. GCD=11 WE KNOW THAT IF N IS DIVIDED BY D TO GIVE QUOTIENT OF Q AND REMAINDER OF R THEN N=Q*D+R.WE USE THIS TO WRITE GCD AS LINEAR COMBINATION OF THE 2 GIVEN NUMBERS. THAT IS TO WRITE.GCD = X * N1 + Y * N2 FROM THE ABOVE DIVISIONS WE DID TO FIND GCD,WE GET 3113=1331*2+451OR..451=3113 - 1331*2..I 1331=451*2+429OR.429 = 1331 - 451*2.....II 451=429*1+22OR..22 = 451 - 429*1.....III 429=22*19+11OR..11 = 429 - 22*19...IV 22=11*2+0.V HENCE GCD =11 WE NOW SUBSTITUTE BACK WARDS FROM EQN.IV TO EQN.I,REPLACING THE REMAINDERS IN EACH EQN.SUCCESSIVELY 11=429 - 22*19.................IV = 429 - (451 - 429*1)*19 = 429 - 451*19+429*19= 429*20 - 451*19 =(1331 - 451*2)*20 - (3113 - 1331*2)*19 = 1331*20 - 451*40 -3113*19 + 1331*38 =1331*58 - 3113*19 - 451*40 =1331*58 - 3113*19 - (3113 - 1331*2)*40 = 1331*138 - 3113*59 HENCE GCD = 11 = 1331*138 - 3113*59 WHICH YOU CAN EASILY VERIFY.HOPE THE METHOD IS CLEAR.IT IS A BIT LONG PROCEDURE.
 Exponential-and-logarithmic-functions/26254: How do you solve this problem? The cube root of (9x + 19) = x + 1 I know that it would be simplified to 9x + 19 = (x + 1)^3, but how do you cube a binomial? 1 solutions Answer 14206 by venugopalramana(3286)   on 2006-02-02 11:06:18 (Show Source): You can put this solution on YOUR website! The cube root of (9x + 19) = x + 1 I know that it would be simplified to 9x + 19 = (x + 1)^3, but how do you cube a binomial? IF YOU DO NOT KNOW OR NOT TAUGHT THE FORMLA FOR CUBING A BINOMIAL...YOU CAN USE A SMALL TRICK..YOU DONOT KNOW (X+1)^3...SO PUT X+1=Y...THEN IT IS SIMPLY Y^3 ON THE R.H.S. ON THE L.H.S..............9X+19=9(Y-1)+19=9Y-9+19=9Y+10...SO Y^3=9Y+10...YOU WILL GET A VALUE FOR Y BETWEEN 3 AND 4...BUT I THINK YOUR PROBLEM SHOULD READ,IF YOU ARE IN AN ELEMENTARY STAGE IN THE TOPIC, The cube root of (9x + 9) = x + 1 WHEN YOU WILL GET Y^3=9Y Y^3-9Y=0 Y(Y^2-9)=0 Y=0.....0R.....Y^2=9.....THAT IS Y=3......OR....-3
 absolute-value/26302: Find values for m and b in the following system so that the solution to the system is (-3,4). 5x + 7y= b mx + y = 22 I really need help. Thank you, I appreciate it. 1 solutions Answer 14204 by venugopalramana(3286)   on 2006-02-02 10:54:26 (Show Source): You can put this solution on YOUR website! Find values for m and b in the following system so that the solution to the system is (-3,4). 5x + 7y= b mx + y = 22 SUBSTITUTE X=-3 AND Y=4 IN THE 2 EQNS. AND FIND B AND M 5*-3+7*4=B=-15+28=13 M*-3+4=22 -3M=22-4=18 M=-18/3=-6
 Systems-of-equations/26316: The perimeter of a rectangle is 20 inches. The perimeter of the inscribed triangle is 20 inches. Find the dimensions of the rectangle. 1 solutions Answer 14203 by venugopalramana(3286)   on 2006-02-02 10:50:51 (Show Source): You can put this solution on YOUR website! CAN YOU PLEASE CHECK BACK ,WHETHER TRIANGLE IS INSCRIBED IN RECTANGLE OR RECTANGLE IS INSCRIBED IN A TRIANGLE?
 Equations/26349: Can you please help me with this problem. We've figured out one answer but there is another answer. So how do you get two answers. Please help us. Rhonda and Tiffany x^2 + 9x + 18 = 0 1 solutions Answer 14202 by venugopalramana(3286)   on 2006-02-02 10:44:29 (Show Source): You can put this solution on YOUR website! x^2 + 9x + 18 = 0 =X^2+3X+6X+18=0 =X(X+3)+6(X+3)=0 =(X+3)(X+6)=0 HENCE X+3=0...THAT IS....X=-3.....OR X+6=0....THAT IS......X=-6...
 Linear-systems/26347: My son is having problems with this type of equation. He has to find out if the linear equation is consistent or inconsistent.Dependant or independant. Example: x+y=4 5x+y=8 How do you graph this equation. How do you work 5x+y=8 to know where to draw the second line of this equation.What would the graph look like .Thanks 1 solutions Answer 14201 by venugopalramana(3286)   on 2006-02-02 10:41:20 (Show Source): You can put this solution on YOUR website! SEE THE FOLLOWING TO GET AN IDEA OF THE TOPIC.IF STILL IN DIFFICULTY PLEASE CONTACT ME AND I SHALL HELP FURTHER AS REGARDS GRAPHING GIVE DIFFERENT VALUES TO X AND FIND Y FROM THE GIVEN RELATION .SEE BELOW I EQN....X+Y=4 X.............0.........1...........2...........3..............4........ETC... Y=4-X.........4.........3...........2...........1..............0.........ETC II EQN.....5X+Y=8 X...........0...........1............2.............3......ETC......... Y=8-5X......8...........3............-2............-7.....ETC........... THE GRAPHS WILL LOOK LIKE THIS. THE 2 LINES INTERSECT AT X=1 AND Y=3 WHICH IS THE UNIQUE SOLUTION TO THESE EQNS.THAT IS THEY ARE CONSISTENT AND INDEPENDENT.IF THEY ARE DEPENDENT YOU WILL GET ONE IDENTICAL LINE FOR BOTH EQNS.THAT IS THEY HAVE INFINITE SOLUTIONS ON THAT LINE. IF THEY ARE INCONSISTENT YOU WILL GET 2 PARALLEL LINES WHICH DO NOT INTERSECT AND HENCE THERE IS NO SOLUTION -------------------------------------------------------------------------------- What is a coinciding equation? 1 solutions -------------------------------------------------------------------------------- Answer 13848 by venugopalramana(791) on 2006-01-28 11:16:56 (Show Source): consider these 2 eqns x+y=2...........i 2x+2y=4.........ii if the first eqn is given to us ,the second one can be derived by us by multiplying the first eqn. with 2 on either side. so the second eqn.does not give us any additional information ,but is a derivative or dependent eqn. of eqn.i.read the following for additional related information.in case of more than 2 or for that matter any number of equations , if one or more of them could be sinilarly derived by a suitable combination of other eqns.,then we say they or dependent eqns. --------------------------------------------------------------------------- When using gaussian elimination, can an inconsistant solution have a general solution? i worked out the matrix and came to a spot where 1=-3 and 1=10.4 (inconsistant 1 can't equal 2 different solutions); would there still be a way to formulate a general solution such as x=(2-4x,2-3x,x,0) = (2,2,0,0) + x(-4,-3,1,0) I hope this is enough information to clearly explain my question. thanks I THINK YOU ARE CONFUSED BETWEEN ...INCONSISTENT,AND DEPENDENT EQUATIONS. AS THE NAME IMPLIES AN INCONSISTENT SET OF EQUATIONS HAS NO SOLUTION AS THEY ARE INCONSISTENT.FOR EXAMPLE X+Y=2 2X+2Y=3....HENCE THERE IS NO QUESTION OF GETTING A GENERAL SOLUTION FOR THAT BY ANY METHOD GAUSSIAN ELIMINATION OR ANY.. CONSISTENT AND DEPENDENT EQNS.HAVE INFINITE SOLUTIONS GIVEN BY A GENERAL FORMULA AS YOU HAVE MENTIONED..FOR EX. X+Y=1 2X+2Y=2....THE GERAL FORMULA IS Y=1-X...OR (X,1-X)IS A SOLUTION SET..LIKE (1,0),(2,-1) ETC.... FINALLY CONSISTENT AND INDEPENDENT EQNS.HAVE UNIQUE SOLUTION..FOR EXAMPLE X+Y=2 X-Y=0...HAVE ONE UNIQUE SOLUTION X=1 AND Y=1 NOW YOU CAN USE GAUSSIAN ELIMINATION TO THE 3 EXAMPLES ABOVE AND SEE WHAT YOU GET.IF YOU STILL AVE DIFFICULTY COME BACK
 Quadratic_Equations/26264: I am hung up on a word problem. Can anyone help? It is: A rectangle with length 2x-1 inches has an area of 2x^2 + 5x - 3 square inches. Write a binomial that represents its width. Thank you for any help you can give me. 1 solutions Answer 14189 by venugopalramana(3286)   on 2006-02-02 06:33:58 (Show Source): You can put this solution on YOUR website! A rectangle with length 2x-1 inches has an area of 2x^2 + 5x - 3 square inches. Write a binomial that represents its width. Thank you for any help you can give me. AREA=LENGTH*WIDTH (2X-1)*WIDTH=2X^2+5X-3=2X^2+6X-X-3=2X(X+3)-1(X+3)=(2X-1)(X+3) WIDTH=(2X-1)(X+3)/(2X-1)=X+3
 Miscellaneous_Word_Problems/26267: An object drops from a cliff that is 150m high. The distance, d, in meters that the object has dropped at t seconds is modelled by d(t)=4.9t^2. a. Find the average rate of change of distance with respect to time from 2 s to 5 s. b. Find the rate at which the object hits the ground to the nearest tenth. THANK YOU. 1 solutions Answer 14187 by venugopalramana(3286)   on 2006-02-02 05:40:23 (Show Source): You can put this solution on YOUR website! LET ME USE SYMOL S INSTEAD OF D FOR DISTANCE S=4.9T^2 RATE OF CHANGE OF DISTANCE (MORE APPROPRIATE WORD IS RATE OF CHANGE OF POSITION OR RATE OF DISPLACEMENT IN PHYSICS) IS VELOCITY.HOPE YOU ARE TAUGHT DIFFERENTIATION.. V=DS/DT=4.9*2*T=9.8T AT T= 2 SEC.,V=9.8*2=19.6 M/SEC. AT T=5 SEC....V=4.9*5=24.5 M/SEC. AVERAGE VELOCITY DURING THIS PERIOD=(19.6+24.5)/2=22.05 M/SEC. S=4.9T^2=150 FOR THE OBJECT TO HIT THE GROUND T=5.53 SEC AT T= 5.53 SEC. V= 4.9*5.53=27.11 M/SEC.
 logarithm/26275: Can you please help me graph a log: y=log base 2 of x. Thanks 1 solutions Answer 14186 by venugopalramana(3286)   on 2006-02-02 05:37:02 (Show Source): You can put this solution on YOUR website! y=log base 2 of x. THAT IS 2^Y=X...WE USE THIS RELATION TO MAKE A TABLE OX AND Y VALUES AS THIS IS EASIER AND KNOWN TO CLACULATE..FURTHER WE GIVE HERE Y VAUE FIRST AND CALCULATE X AS THAT IS EASIER. X=2^Y.................1.......2.......4.......8.........16.....ETC...... Y=log base 2 of x. ...0.......1.......2.......3.........4.....ETC... NOW YOU CAN PLOT IT
 Numbers_Word_Problems/26277: An object drops from a cliff that is 150m high. The distance, d, in meters that the object has dropped at t seconds is modelled by d(t)=4.9t^2. a. Find the average rate of change of distance with respect to time from 2 s to 5 s. b. Find the rate at which the object hits the ground to the nearest tenth. THANK YOU. 1 solutions Answer 14185 by venugopalramana(3286)   on 2006-02-02 05:11:56 (Show Source): You can put this solution on YOUR website! LET ME USE SYMOL S INSTEAD OF D FOR DISTANCE S=4.9T^2 RATE OF CHANGE OF DISTANCE (MORE APPROPRIATE WORD IS RATE OF CHANGE OF POSITION OR RATE OF DISPLACEMENT IN PHYSICS) IS VELOCITY.HOPE YOU ARE TAUGHT DIFFERENTIATION.. V=DS/DT=4.9*2*T=9.8T AT T= 2 SEC.,V=9.8*2=19.6 M/SEC. AT T=5 SEC....V=4.9*5=24.5 M/SEC. AVERAGE VELOCITY DURING THIS PERIOD=(19.6+24.5)/2=22.05 M/SEC. S=4.9T^2=150 FOR THE OBJECT TO HIT THE GROUND T=5.53 SEC AT T= 5.53 SEC. V= 4.9*5.53=27.11 M/SEC.
 Distributive-associative-commutative-properties/26115: Show for integers a,b and k that gcd(a,b)=gcd(a,b+ka). 1 solutions Answer 14181 by venugopalramana(3286)   on 2006-02-02 04:30:21 (Show Source): You can put this solution on YOUR website! Show for intergers a,b and k that gcd(a,b)=gcd(a,b+ka). LET GCD OF A AND B BE G HENCE G|A AND G|B....THAT IS A=GA1.....AND B=GB1.....WHERE A1 AND B1 ARE INTEGERS RELATIVELY PRIME TO EACH OTHER..THAT IS GCD OF A1 AND B1 IS 1. NOW WE HAVE B+KA=GB1+KGA1=G(B1+KA1)...SINCE B1,A1 AND K ARE INTEGERS,THIS MEANS G|(B+KA)...SO G|A AND G|(B+KA)...SO G IS A COMMON DIVISOR OF A AND B+KA NOW TO SHOW THAT IT IS THE GREATEST INTEGER OR A1 AND B1+KA1 ARE RELATIVELY PRIME TO EACH OTHER.WE KNOW ALREADY A1 AND B1 ARE PRIME TO EACH OTHER.HENCE ONLY IF K IS A MULTIPLE OF B THEN ONLY B1+KA1 CAN HAVE A COMMON FACTOR OF GB1 OR ITS MULTIPLE WHICH IS DEFINITELY MORE THAN G.HENCE WE HAVE A=GA1 AND IF K=B1 OR NB1 THEN B+KA COULD BE =GB1(1+A1)...OR.....GB1(1+NA1) BUT GB1 CANNOT DIVIDE A SINCE A1=A/G IS ALREADY PRIME TO B1.HENCE THE RESULT
 Distributive-associative-commutative-properties/26326: For integers a and b show that a|b also equals -a|b and a|-b and -a|-b 1 solutions Answer 14179 by venugopalramana(3286)   on 2006-02-02 04:24:52 (Show Source): You can put this solution on YOUR website! For integers a and b show that a|b also equals -a|b and a|-b and -a|-b A|B HENCE B=M*A WHERE M IS AN INTEGER. T.P.T. A|-B...SAY... WE HAVE FROM B=M*A -B=-M*A...WHERE -M IS ALSO AN INTEGER HENCE A|-B....NOTE THE DEFINITION OF PERFECT DIVISION.IT SAYS THAT M HAS TO BE AN INTEGER ONLY.POSITIVE AND NEGATIVE NUMBERS ARE BOTH INTEGERS.SO IT DOES NOT MAKE ANY DIFFERENCE IF M IS PLUS OR MINUS.
 Triangles/26341: Show that for two right angles, if the hypotenuse and leg of one are congruent to the hypotenuse and the leg of the other, then the two triangles are congruent. 1 solutions Answer 14173 by venugopalramana(3286)   on 2006-02-02 03:42:52 (Show Source): You can put this solution on YOUR website! Show that for two right angles, if the hypotenuse and leg of one are congruent to the hypotenuse and the leg of the other, then the two triangles are congruent. YOU CAN USE PYTHOGARUS THOREM TO PROVE THAT SECOND LEGS OF THE 2 TRIANGLES ARE EQUAL.AND HENCE ALL 3 SIDES OF ONE TRIANGLE ARE EQUAL TO THE CORRESPONDING SIDES OF THE SECOND TRIANGLE HENCE BY SSS THEOREM THEY ARE CONGRUENT. NOTE THAT SAS IS ALSO HOLDING NOW AS WE HAVE PROVED THAT THE 2 LEGSF RIGHTANGLE IN ONE TRIANGLE ARE EQUAL TO 2 LEGS OF RIGHT ANGLE IN ANOTHER. IF ABC AND DEF ARE THE 2 TRIANGLES WITH ANGLE ABC =ANGLE DEF =90 AC = HYPOTENUSE = DF AND ONE LEG AB = DE ,WE HAVE BY PYTHOGARUS THEOREM BC^2=AC^2-AB^2=DF^2-DE^2=EF^2 BC=EF SO USING SSS THE 2 TRIANGLES ARE CONGRUENT
 Permutations/26342: how many 6 number combinations will be made from 1 2 3 4 5 6 7 8 9, where 1 2 and 3 must appear on the 6-number combination? 1 solutions Answer 14172 by venugopalramana(3286)   on 2006-02-02 03:24:09 (Show Source): You can put this solution on YOUR website! 1,2 AND 3 MUST APPEAR IN THE COMBINATION. SO WE NEED TO SELECT ANOTHER 3 NUMBERS ONLY OUT OF REMAINING 6 NUMBERS WHICH CAN BE DONE IN 6C3 WAYS 6C3=6*5*4/(3*2*1)=20
 Triangles/26340: The triangle A (lower left hand corner of the triangle) B (lower right hand corner) C (top of triangle) =ABC, is an equilateral triangle where each side of the triangle is 6. What is the height of h if h is the dotted vertical line down the center? I am having a hard time trying to copy and paste the triangle with my question. Hopefully, you can imagine what I am trying to say. The picture looks like an equilateral triangle w/ a dotted vertical line down the middle. The three sides of the triangle is 6 and the three points of the triangle is ABC. Thanks, Kristin 1 solutions Answer 14171 by venugopalramana(3286)   on 2006-02-02 03:20:51 (Show Source): You can put this solution on YOUR website! LET ABC BE THE TRIANGLE. AND LET AD BE THE HEIGHT SHOWN BY A PERPENDICULAR FRO A TO BC.D IS ON BC. IN TRIANGLES ABD AND ACD WE HAVE AB=AC=6 AD=AD ANGLE ADB = 90 = ANGLE ADC HENCE THE 2 TRIANGLES ARE IDENTICAL HENCE BD = DC BUT BC=BD+DC=6 ...SO BD=DC=3 IN RIGHT ANGLED TRIANGLE ABD WE HAVE ,ANGLE ADB =90 DEGREES HYPOTENUSE = AB =6 BD =3 SO AD^2+BD^2=AB^2 AD^2+3^2=6^2 AD^2=36-9=27 AD=SQ.RT.27 = 3*SQRT3
 Polynomials-and-rational-expressions/26337: I have to Evaluate this problem (g-6h)(g+6h) I got =(g)^2-6h^2 = g^2-36h is this correct? 1 solutions Answer 14170 by venugopalramana(3286)   on 2006-02-02 01:25:15 (Show Source): You can put this solution on YOUR website! I have to Evaluate this problem (g-6h)(g+6h) I got =(g)^2-6h^2 = g^2-36h is this correct? SEE BELOW MY COMMENTS I have to Evaluate this problem (g-6h)(g+6h) PUT BRACKETS TO AVOID MISTAKES I got =(g)^2-(6h)^2 SO YOU GET = g^2-36h^2 is this correct?ALMOST...YOU ARE GOOD...YOU CAN IMPROVE FURTHER..DEVELOP GOOD PRACTICES LIKE PUTTING APPROPRIATE BRACKETS...LITTLE MORE OF THEM IS BETTER THAN LESS AS YOUR PRESENT SMALL MISTAKE SHOWS...
 Travel_Word_Problems/26328: I was wondering if you could help me with this problem. I need it as soon as possible. Thank you! A train averaged 120 km/h for the first two thirds of a trip and 100 km/h for the whole trip. Find its average speed for the last third of the trip. I think the answer is 75km/h but i don't know how to set it up. Thank you. 1 solutions Answer 14169 by venugopalramana(3286)   on 2006-02-02 00:40:53 (Show Source): You can put this solution on YOUR website! A train averaged 120 km/h for the first two thirds of a trip and 100 km/h for the whole trip. Find its average speed for the last third of the trip. I think the answer is 75km/h but i don't know how to set it up. Thank you. LET THE TRIP BE FOR SAY X KM. 2/3 RD. OF TRIP =X*2/3=2X/3 KM.....IT TRAVELLED AT 120 KMPH SO TIME TAKEN = 2X/(3*120)=X/180 HRS. WHOLE TRIP AVERAGE SPEED =100 KMPH ..SO TIME TAKEN FOR WHOLE TRIP = X/100 HRS. SO TIME TAKEN FOR 1/3 RD. OF TRIP=X/100 - X/180 =X(180-100)/18000=X/225 HRS. DISTANCE TRAVELLED =X*1/3=X/3 SO AVERAGE SPEED FOR THIS PART = (X/3)/(X/225)=225/3 =75 KMPH
 Travel_Word_Problems/26329: Hi could you please solve this problem for me. I don't understand how to set it up. Helped by a strong jet stream, a Los-Angeles-to-Boston plane flew 10% faster than usual and made the 4400km trip in 30min less time then usual. At what speed does the plane usually fly? What i came up with: x=old speed x-30=new speed? 1 solutions Answer 14168 by venugopalramana(3286)   on 2006-02-02 00:28:50 (Show Source): You can put this solution on YOUR website! Hi could you please solve this problem for me. I don't understand how to set it up. Helped by a strong jet stream, a Los-Angeles-to-Boston plane flew 10% faster than usual and made the 4400km trip in 30min less time then usual. At what speed does the plane usually fly? What i came up with: x=old speed...OK x-30=new speed?..NO 30 MINUTES LESS IS TIME TAKEN.NOT SPEED.HE SAID SPEED IS 10% FASTER,SO IF OLD SPEED IS X NEW SPEED IS FASTER BY =X*10/100=0.1X SO NEW SPEED =X+0.1X=1.1X SO NOW TIME TAKEN AT NEW SPEED =DISTANCE/SPEED=4400/(1.1X) TIME TAKEN AT OLD SPEED =4400/X DIFFERENCE = 30 MTS=30/60 HRS=0.5 HRS. SO WE GET 4400/X - 4400//1.1X = 0.5 4400{1/X - 1/1.1X)=0.5 4400(1.1-1)/1.1X=0.5 4400(0.1/1.1X)=0.5 4400*1/11X=0.5 11X*0.5=4400 X=4400/5.5=800 KMPH
 Geometric_formulas/26332: how many conversation will there be at a party of 50 people. (each person only talk to each other once.) 1 solutions Answer 14167 by venugopalramana(3286)   on 2006-02-02 00:16:16 (Show Source): You can put this solution on YOUR website! THE ANSWER IS 50C2..HOPE YOU KNOW COMBINATIONS. =50*49/2=1225
 Distributive-associative-commutative-properties/26202: Show for intergers a,b and k that gcd(a,b)=gcd(a,b+ka). PLEASE NEED HELP WITH THIS QUESTION 1 solutions Answer 14107 by venugopalramana(3286)   on 2006-02-01 11:30:56 (Show Source): You can put this solution on YOUR website! Show for intergers a,b and k that gcd(a,b)=gcd(a,b+ka). LET GCD OF A AND B BE G HENCE G|A AND G|B....THAT IS A=GA1.....AND B=GB1.....WHERE A1 AND B1 ARE INTEGERS RELATIVELY PRIME TO EACH OTHER..THAT IS GCD OF A1 AND B1 IS 1. NOW WE HAVE B+KA=GB1+KGA1=G(B1+KA1)...SINCE B1,A1 AND K ARE INTEGERS,THIS MEANS G|(B+KA)...SO G|A AND G|(B+KA)...SO G IS A COMMON DIVISOR OF A AND B+KA NOW TO SHOW THAT IT IS THE GREATEST INTEGER OR A1 AND B1+KA1 ARE RELATIVELY PRIME TO EACH OTHER.WE KNOW ALREADY A1 AND B1 ARE PRIME TO EACH OTHER.HENCE ONLY IF K IS A MULTIPLE OF B THEN ONLY B1+KA1 CAN HAVE A COMMON FACTOR OF GB1 OR ITS MULTIPLE WHICH IS DEFINITELY MORE THAN G.HENCE WE HAVE A=GA1 AND IF K=B1 OR NB1 THEN B+KA COULD BE =GB1(1+A1)...OR.....GB1(1+NA1) BUT GB1 CANNOT DIVIDE A SINCE A1=A/G IS ALREADY PRIME TO B1.HENCE THE RESULT
 Volume/26217: 1) An open-top box is to be constructed from a 6 by 8 foot rectangular cardboard by cutting out equal squares at each corner and the folding up the flaps. Let x denote the length of each side of the square to be cut out. a) Find the function V that represents the volume of the box in terms of x. Answer b) Graph this function. Show Graph here c) Using the graph, what is the value of x that will produce the maximum volume? Answer 1 solutions Answer 14103 by venugopalramana(3286)   on 2006-02-01 10:57:07 (Show Source): You can put this solution on YOUR website! An open-top box is to be connected from a 6 by 8 foot rectangular cardboard by cutting out equal squares at each corner and then folding up the flaps. Let x denote the length of each side of the square to be cut out. Find the function V that represents the volume of the box in terms of x. This is what I got: 6x+8x+x=14x² x=14 LENGTH OF BOARD=6 FT....WIDTH OF BOARD =8 FT. IF X LONG SQUARE PIECES ARE CUT AT 4 CORNERS WE SHALL HAVE THE BOX DIMENSIONS AS HEIGHT =X.....LENGTH = 6-2X.....WIDTH =8-2X VOLUME = V = L*W*H =X(6-2X)(8-2X)=X(48-28X+4X^2)=48X-28X^2+4X^3 THE GRAPH WILL LOOK LIKE THIS FROM THE GRAPH YOU CAN SEE V HAS A PEAK AT X=ABOUT 3.5 FEET
 Graphs/26216: I don't understand the wording or what formula I need to plug this problem into. Find the equations of the circles that are tangent to the x axis and have a radius of length five units. In each case, the abscissa of the center is -3.(There is more than one circle that satisfies these conditions) 1 solutions Answer 14102 by venugopalramana(3286)   on 2006-02-01 10:51:20 (Show Source): You can put this solution on YOUR website! Find the equations of the circles that are tangent to the x axis and have a radius of length five units. In each case, the abscissa of the center is -3.(There is more than one circle that satisfies these conditions) WHEN THE CIRCLE IS TANGENT TO X AXIS ,IT MEANS ITS CENTRE IS AT A DISTANCE EQUAL TO RADIUS OF THE CIRCLE FROM THE X AXIS .THAT IS Y COORDINATE OF CENTRE OF CIRCLE = 5 OR -5 ABSCISSA OF CENTRE IS GIVEN AS -3. HENCE CENTRE IS EITHER (-3,5).....OR (-3,-5)..ITS RADIUS =5 HENCE EQN OF 2 POSSIBLE CIRCLES ARE (X+3)^2+(Y-5)^2 = 5^2=25...................I OR (X+3)^2+(Y+5)^2=25........................II
 Geometric_formulas/26224: 4) CLASSIC PROBLEM - A traveling salesman (selling shoes) stops at a farm in the Midwest. Before he could knock on the door, he noticed an old truck on fire. He rushed over and pulled a young lady out of the flaming truck. Farmer Brown came out and gratefully thanked the traveling salesman for saving his daughters life. Mr. Brown insisted on giving the man an award for his heroism. So, the salesman said, If you insist, I do not want much. Get your checkerboard and place one penny on the first square. Then place two pennies on the next square. Then place four pennies on the third square. Continue this until all 64 squares are covered with pennies. As hed been saving pennies for over 25 years, Mr. Brown did not consider this much of an award, but soon realized he made a miscalculation on the amount of money involved. a) How much money would Mr. Brown have to put on the 32nd square? b) How much would the traveling salesman receive if the checkerboard only had 32 squares? Calculate the amount of money necessary to fill the whole checkerboard (64 squares). How much money would the farmer need to give the salesman? Thank you so, much for your help. 1 solutions Answer 14101 by venugopalramana(3286)   on 2006-02-01 10:42:32 (Show Source): You can put this solution on YOUR website! ok, this is one that has baffled not only myself but my parents and boyfriend as well. This is a problem that no one seems to be able to figure out. Do you think you could help me? GOOD PROBLEM YOUNG GIRL!NOW THAT YOU SAY THAT YOUR BOY FRIEND IS ALSO FOXED BY THE PROBLEM ,YOU CAN USE IT TO GOOD EFFECT TO PUT ACROSS A PROPOSAL TO HIM ... YOU PROMISS HIM A 100,000..POUNDS EVERY DAY FOR 30 DAYS.LET HIM GIVE YOU IN RETURN JUST A PENNY ON THE FIRST DAY,2 PENNIES ON THE SECOND DAY ,4 PENNIES ON THE THIRD DAY,......ETC FOR 30 DAYS ..AS IN YOUR PROBLEM...BASED ON YOUR FEED BACK , I PRESUME HE WILL LAP ON THE PROPOSAL AND I CAN ASSURE YOU THAT YOU WONT REGRET THE PROPOSAL!!!! OK NOW LET US GET BACK TO YOUR PROBLEM AS WELL AS MY SUGGESTION ,WHICH IS JUST A PART OF IT .I AM GIVING BELOW THE MONEY TO BE PAID ON THIS BASIS FOR 30 AND 64 DAYS ,ROUNDED OFF TO POUNDS OR MILLION POUNDS AT THE LATER DAYS. DAY.....MONEY TO BE PAID........MONEY TO BE PAID BY YOUR FRIEND TO YOU IN ............TO YOUR FRIEND..... CENTS...POUNDS...MILLION POUNDS ................POUNDS 1 100000 1 0 0 2 100000 2 0 0 3 100000 4 0 0 4 100000 8 0 0 5 100000 16 0 0 6 100000 32 0 0 7 100000 64 1 0 8 100000 128 1 0 9 100000 256 3 0 10 100000 512 5 0 11 100000 1024 10 0 12 100000 2048 20 0 13 100000 4096 41 0 14 100000 8192 82 0 15 100000 16384 164 0 16 100000 32768 328 0 17 100000 65536 655 0 18 100000 131072 1311 0 19 100000 262144 2621 0 20 100000 524288 5243 0 21 100000 1048576 10486 0 22 100000 2097152 20972 0 23 100000 4194304 41943 0 24 100000 8388608 83886 0 25 100000 16777216 167772 0 26 100000 33554432 335544 0 27 100000 67108864 671089 1 28 100000 134217728 1342177 1 29 100000 268435456 2684355 3 30 100000 536870912 5368709 5 31 100000 1073741824 10737418 11 32 100000 2147483648 21474836 21 33 100000 4294967296 42949673 43 34 100000 8589934592 85899346 86 35 100000 17179869184 171798692 172 36 100000 34359738368 343597384 344 37 100000 68719476736 687194767 687 38 100000 1.37439E+11 1374389535 1374 39 100000 2.74878E+11 2748779069 2749 40 100000 5.49756E+11 5497558139 5498 41 100000 1.09951E+12 10995116278 10995 42 100000 2.19902E+12 21990232556 21990 43 100000 4.39805E+12 43980465111 43980 44 100000 8.79609E+12 87960930222 87961 45 100000 1.75922E+13 175921860444 175922 46 100000 3.51844E+13 351843720888 351844 47 100000 7.03687E+13 703687441777 703687 48 100000 1.40737E+14 1407374883553 1407375 49 100000 2.81475E+14 2814749767107 2814750 50 100000 5.6295E+14 5629499534213 5629500 51 100000 1.1259E+15 11258999068426 11258999 52 100000 2.2518E+15 22517998136853 22517998 53 100000 4.5036E+15 45035996273705 45035996 54 100000 9.0072E+15 90071992547410 90071993 55 100000 1.80144E+16 180143985094820 180143985 56 100000 3.60288E+16 360287970189640 360287970 57 100000 7.20576E+16 720575940379279 720575940 58 100000 1.44115E+17 1441151880758560 1441151881 59 100000 2.8823E+17 2882303761517120 2882303762 60 100000 5.76461E+17 5764607523034230 5764607523 61 100000 1.15292E+18 11529215046068500 11529215046 62 100000 2.30584E+18 23058430092136900 23058430092 63 100000 4.61169E+18 46116860184273900 46116860184 64 100000 9.22337E+18 92233720368547800 92233720369 30 DAY..3000000.........1073741823......10737418.....................11 TOTAL 64 DAY..6400000.........1.84467E+19.....184467440737095000......184467440737 TOTAL YOUR GAIN IN 30 DAYS = 8 MILLION POUNDS..GOT IT BUT ONLY BE CAREFULL THAT YOUR FRIEND WONT RUN OUT OF YOU ON THE 25 TH. DAY.!!! YOUR GAIN IN 30 DAYS = 184467440731 MILLION POUNDS NOW COMING TO THE MATHS PART OF THIS ,THIS SEQUENCE WHERE EACH NUMBER BEARS A CONSTANT RATIO TO ITS PREDECESSOR IS CALLED GEOMETRIC PROGRESSION..HERE YOU FIND EACH NUMBER IS OBTAINED FROM THE PREVIOUS ONE BY MULTIPLYING WITH 2 , CALLED COMMON RATIO. THE LAST NUMBER ON NTH. DAY AND SUM OF SUCH SERIES OF NUMBERS UPTO THE N TH.DAY IS GIVEN BY THE FOLLOWING FORMULAE... N TH. NUMBER = FIRST NUMBER *(C0MM0N RATIO )^(N-1)=A*(R)^(N-1)=(2)^(N-1) IN THIS CASE. SUM UP TO N TH.NUMBER =A*{((R)^N - 1))/(R-1)}={(2)^N-1}IN THIS CASE
 Complex_Numbers/26058: ok, i just DONT understand imaginary numbers in problems! i really need help! heres a problem: 7i^3 + 8i^12 +9i^17 -5i^16 = ? PLEASE HELP! i dont know what to do with the numbers in front...!!! 1 solutions Answer 14033 by venugopalramana(3286)   on 2006-01-31 11:27:06 (Show Source): You can put this solution on YOUR website! 7i^3 + 8i^12 +9i^17 -5i^16 = PLEASE NOTE THE FOLLOWING I=I I^2=-1 I^3=I^2*I=-I I^4=(I^2)^2=(-1)(-1)=1 HENCE IN GENERAL (I)^EVEN POWER = -1 OR +1..IT IS +1 IF THE POWER IS DIVISIBLE BY 4 OTHER WISE -1 (I)^ODD POWER = I OR -I ....IT IS +I IF THE POWER LESS ONE IS DIVISIBLE BY 4 OTHER WISE -I... SO NOW YOU CAN DO THE PROBLEM I THINK ..LET US SEE 7i^3 + 8i^12 +9i^17 -5i^16 = -7I+8+9I-5=3-2I...OK...CLEAR?
 Equations/26102: f(x)=4+x/(1+2x)(1-x)^2 a) Express F(x) in partial fractions b) Given that |x|<0.5, expand f(x) in ascending powers of x, up to and including the term x^3 1 solutions Answer 14031 by venugopalramana(3286)   on 2006-01-31 11:12:49 (Show Source): You can put this solution on YOUR website! a) Express F(x) in partial fractions f(x)=(4+x)/(1+2x)(1-x)^2 = A/(1+2X) + B/(1-X) + C /(1-X)^2 PUT 2X+1=0 OR X=-1/2=-0.5 TO GET A A=(4-0.5)/(1+0.5)^2=3.5/2.25=14/9 AND PUT 1-X =0 OR X=1 TO GET C C=(4+1)/(1+2*1)=5/3....HENCE.... f(x)=(4+x)/(1+2x)(1-x)^2 = A/(1+2X) + B/(1-X) + C /(1-X)^2 =14/9*(1+2X) + B / (1-X) + 5 /3*(1-X)^2 NOW PUT X=0 TO GET B 4+0=14/9 +B +5/3....OR ....B=7/9 HENCE WE GET f(x)=(4+x)/(1+2x)(1-x)^2 = A/(1+2X) + B/(1-X) + C /(1-X)^2 =14/9*(1+2X) + 7 /9*(1-X) + 5 /3*(1-X)^2 b) Given that |x|<0.5, expand f(x) in ascending powers of x, up to and including the term x^3 F(X)=14/9*(1+2X) + 7 /9*(1-X) + 5 /3*(1-X)^2 =(14/9)(1+2X)^(-1)+(7/9)*(1-X)^(-1)+(5/3)*(1-X)^(-2) =(14/9)(1-2X+4X^2-8X^3)+(7/9)(1+X+X^2+X^3)+(5/3)(1+2X+3X^2+4X^3) REST YOU CAN SIMPLIFY IF NEEDED OR IT CAN BE LEFT AS SUCH..NOTE THAT THE EXPANSIONS WE DID UP TO X^3 AS DESIRED WILL HOLD GOOD ONLY IF |2X|<1...OR...|X|<0.5..HENCE THE NECESSITY FOR THE DATA THAT |X|<0.5 IN THE PROBLEM
 Geometric_formulas/26074: 1) Use the geometric series of numbers 1, 2, 4, 8,to find the following: a) What is r, the ratio between 2 consecutive terms? b) Using the formula for the nth term of a geometric series, what is the 24th term? c) Using the formula for the sum of a geometric series, what is the sum of the first 10 terms? pls help 1 solutions Answer 14006 by venugopalramana(3286)   on 2006-01-31 05:56:57 (Show Source): You can put this solution on YOUR website! ) Use the geometric series of numbers 1, 2, 4, 8,to find the following: a) What is r, the ratio between 2 consecutive terms? b) Using the formula for the nth term of a geometric series, what is the 24th term? c) Using the formula for the sum of a geometric series, what is the sum of the first 10 terms? SUM IS GIVEN BY SN=A(R^N-1)/(R-1) S10=1*(2^10-1)/(2-1)=2^10-1 SEE THE FOLLOWING EXAMPLE AND SOLVE Use the geometric series of numbers 1, 1/2, 1/4, 1/8,to find the following: a) What is r, the ratio between 2 consecutive terms? R=(1/2)/1=1/2...IT IS SAME FOR ANY 2 CONSECUTIVE TERMS...SAY (1/8)/(1/4)=1/2...ETC...... b) Using the formula for the nth term of a geometric series, what is 10th term? TN=A*(R)^(N-1)=1*(1/2)^(N-1) T10=(1/2)^(10-1)=(1/2)^9 c) Using the formula for the nth term of a geometric series, what is 12th term? T12=(1/2)^11 d) What observation can make about these sums? In particular, what number does it appear that the sum will always be smaller than? THE NUMBERS TEND TO ZERO AS N INCREASES TO LARGE NUMBERS AND TOWARDS INFINITY. SUM OF A G.P. TO INFINITE TERMS WITH R<1 IS GIVEN BY A/(1-R)=1/(1-0.5)=2
 Geometric_formulas/26075: 10) Use the geometric series of numbers 1, 1/2, 1/4, 1/8,to find the following: a) What is r, the ratio between 2 consecutive terms? b) Using the formula for the nth term of a geometric series, what is 10th term? c) Using the formula for the nth term of a geometric series, what is 12th term? d) What observation can make about these sums? In particular, what number does it appear that the sum will always be smaller than? 1 solutions Answer 14005 by venugopalramana(3286)   on 2006-01-31 05:54:54 (Show Source): You can put this solution on YOUR website! Use the geometric series of numbers 1, 1/2, 1/4, 1/8,to find the following: a) What is r, the ratio between 2 consecutive terms? R=(1/2)/1=1/2...IT IS SAME FOR ANY 2 CONSECUTIVE TERMS...SAY (1/8)/(1/4)=1/2...ETC...... b) Using the formula for the nth term of a geometric series, what is 10th term? TN=A*(R)^(N-1)=1*(1/2)^(N-1) T10=(1/2)^(10-1)=(1/2)^9 c) Using the formula for the nth term of a geometric series, what is 12th term? T12=(1/2)^11 d) What observation can make about these sums? In particular, what number does it appear that the sum will always be smaller than? THE NUMBERS TEND TO ZERO AS N INCREASES TO LARGE NUMBERS AND TOWARDS INFINITY. SUM OF A G.P. TO INFINITE TERMS WITH R<1 IS GIVEN BY A/(1-R)=1/(1-0.5)=2
 Quadratic_Equations/25934: Please help me find the vertex and intercepts for this quadratic function, and help me to sketch its graph. y = x^2 + 4x 1 solutions Answer 13949 by venugopalramana(3286)   on 2006-01-30 11:26:27 (Show Source): You can put this solution on YOUR website! PLEASE SEE THE FOLLOWING EXAMPLE AND DO ACCORDINGLY.IF YOU STILL HAVE DIFFICULTY COME BACK AND WE WILL HELP YOU. ------------------------------------------------------------------------------ Please help me find the vertex and intercepts for this quadratic function, and help me to sketch its graph. y = x^2 - 6x Y=(X-3)^2-9...THIS HAS VERTEX AT X=3 AND ITS VALUE IS Y=(3-3)^2-9=-9 GRAPH IS OBTAINED BY GIVING DIFFERENT VALUES TO X AND INDING THE CORRESPONDING VALUES OF Y AND PLOTTING THEM TO A SUITABLE SCALE X=........0........1.........2...........3..........ETC..... Y=X^2-6X..0........-5........-8..........-9.........ETC.... THE GRAPH WILL LOOK LIKE THIS
 Quadratic_Equations/25937: Please help me find the vertex and intercepts for this quadratic function, and sketch its graph. y = -2x^2 + 8x 1 solutions Answer 13948 by venugopalramana(3286)   on 2006-01-30 11:25:43 (Show Source): You can put this solution on YOUR website! PLEASE SEE THE FOLLOWING EXAMPLE AND DO ACCORDINGLY.IF YOU STILL HAVE DIFFICULTY COME BACK AND WE WILL HELP YOU. ------------------------------------------------------------------------------ Please help me find the vertex and intercepts for this quadratic function, and help me to sketch its graph. y = x^2 - 6x Y=(X-3)^2-9...THIS HAS VERTEX AT X=3 AND ITS VALUE IS Y=(3-3)^2-9=-9 GRAPH IS OBTAINED BY GIVING DIFFERENT VALUES TO X AND INDING THE CORRESPONDING VALUES OF Y AND PLOTTING THEM TO A SUITABLE SCALE X=........0........1.........2...........3..........ETC..... Y=X^2-6X..0........-5........-8..........-9.........ETC.... THE GRAPH WILL LOOK LIKE THIS
 Quadratic_Equations/25938: Please help me find the vertex and intercepts for this quadratic function, and sketch its graph. y = -3x^2 + 6x 1 solutions Answer 13947 by venugopalramana(3286)   on 2006-01-30 11:25:00 (Show Source): You can put this solution on YOUR website! PLEASE SEE THE FOLLOWING EXAMPLE AND DO ACCORDINGLY.IF YOU STILL HAVE DIFFICULTY COME BACK AND WE WILL HELP YOU. ------------------------------------------------------------------------------ Please help me find the vertex and intercepts for this quadratic function, and help me to sketch its graph. y = x^2 - 6x Y=(X-3)^2-9...THIS HAS VERTEX AT X=3 AND ITS VALUE IS Y=(3-3)^2-9=-9 GRAPH IS OBTAINED BY GIVING DIFFERENT VALUES TO X AND INDING THE CORRESPONDING VALUES OF Y AND PLOTTING THEM TO A SUITABLE SCALE X=........0........1.........2...........3..........ETC..... Y=X^2-6X..0........-5........-8..........-9.........ETC.... THE GRAPH WILL LOOK LIKE THIS