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H(s)= 2s^3-7s^2-43s-42 try s=0,1,2,-1,-2 etc..
we find that at s=2 ....H=0...HENCE S+2 IS A FACTOR
DIVIDING WITH S+2 WE GET
-2...|...2......-7.........-43.........-42
.....|...0......-4..........22..........42
..............................................
.........2......-11.........-21..........0
HENCE WE GET 2S^2-11S-21
=2S^2-14S+3S-21=2S(S-7)+3(S-7)=(2S+3)(S-7)..HENCE
H(S)=(S+2)(2S+3)(S-7)

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SEE THE FOLLOWING EXAMPLE AND TRY.IF YOU HAVE DIFFICULTY ,I SHALL WORK OUT FOR YOU.KEY IS TO FIND HOW MUCH WILL BE FILLED OR DRAINED IN 1 HOUR BY EACH PIPE AND THEN ESTIMATR FOR THE COMBINATION IN 1 HOUR AND THEN TAKE RECIPROCAL TO FIND THE HOURS NEEDED TO FILL OR EMPTY THE FULL TANK OR WHATEVER IS NEEDED
Mixture_Word_Problems/18599: Need help with this, The inlet pipe of an oil tank can fill the tank in 1 hour, 30 minutes. The outlet pipe can empty the tank in 1 hour. How long does it take to empty the tank if both pipes are open?1 solutions Answer 9051 by venugopalramana(345) on 2005-11-08 02:05:51 (Show Source): The inlet pipe of an oil tank can fill the tank in 1 hour, 30 minutes. The outlet pipe can empty the tank in 1 hour. How long does it take to empty the tank if both pipes are open?inlet pipe fills 1 tank in 1hr 30 mts =1.5=3/2 hrs.or inlet pipe fills ...1/(3/2)=2/3 tank in 1 hr.out let pipe empties 1 tank in 1 hr....so assuming the tank is full at the start water to be emptied =1 tank and in 1 hr inlet fills 2/3 tank and out let empties 1 tank ..so net out flow is 1-2/3=(3-2)/3=1/3 tank in 1 hour.so 1/3 tank emptying needs .....1 hrthen 1 tank emptying needs ......? hrs...= 1*1/(1/3)=3 hrs.

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SEE THIS WHICH IS SIMILAR TO YOUR PROBLEM AND SOLVE YOUR PROBLEMA boat travels at the speed of 20 Miles per hour in still water. It travels 48 miles up stream, and then returns to the starting point in a total of five hours. The speed of the current is ____ miles per hour.distance =speed * time..or...time=distance/speedStill water speed of boat =20 mphlet the stream velocity be =x mphso its speed upstream =20-xits speed down stream =20+xdistance travelled either way =48 milestime taken to travel upstream=48/(20-x)time taken to travel down stream =48/(20+x)total time of travel =5 hours..hence[48/(20-x)]+[48/(20+x)]=548[(20+x+20-x)/((20-x)(20+x))]=548*40=5(20^2-x^2)=2000-5x^25x^2=2000-1920=8=x^2=80/5=16x=4 mph1 solutions Answer 9323 by venugopalramana(345) on 2005-11-12 06:55:46 (Show Source):

Travel_Word_Problems/19151: OK...Brent rides a moped at 24 miles per hour. His wife Jan rides in the opposite direction at 8 miles per hour. How much time will pass when they are 24 miles apart? I know the answer is 3/4 hour, but I'm having a hard time getting that using the old rate x time = distance method. Can you bail me out? 1 solutions Answer 9226 by venugopalramana(345) on 2005-11-10 04:38:26 (Show Source): Brent rides a moped at 24 miles per hour. His wife Jan rides in the opposite direction at 8 miles per hour. How much time will pass when they are 24 miles apart? I know the answer is 3/4 hour, but I'm having a hard time getting that using the old rate x time = distance method. Can you bail me out? I'm having a hard time getting that using the................................ old rate x time = distance method. Can you bail me out? your formula is ok but you have to understand relative speed cocept and use it ..it is as followsBrent rides a moped at 24 miles per hour...so in one hour he goes 24 miles in one directionHis wife Jan rides in the opposite direction at 8 miles per hour...so in one hour she goes 8 miles in opposite direction.so initially if they were together ,after 1 hour their distance of seperation is 24+8=32 miles as they were going in opposite directions.(just for your idea if they were going in same direction their distance of seperation would have been 24-8=16 miles)so now we define relative speed as sum of the 2 speeds if they are in opposite directions (and difference between 2 speeds ifthey are in same direction)so relative speed in this case =24+8=32 mph now for distance again you have to consider 2 aspects ..one...what is the distance between them at start it is 0 here as they started at the same place.....next...we have to use the distance between them at the end ..here it is 24 miles ...so the additional distance of seperation between them from begining to end is 24-0=24(had the start been not 0 distance ,you will account for that here) now for time same way as above for distance..both started at same time ...so begining time =0end time = not known =t say...so difference in time =t-0=tnow you can use your formula with 3 new terms..relative speed ,difference in distance,difference in time,inplace of speed ,distance and time....so we get rate x time = distance....gives32*t=24t=24/32=3/4 hour =3*60/4 =45 mts....hope it is clear...

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SEE THE FOLLOWING EXAMPLE AND TRY.IF YOU HAVE DIFFICULTY ,I SHALL WORK OUT FOR YOU.KEY IS TO FIND HOW MUCH WILL BE FILLED OR DRAINED IN 1 HOUR BY EACH PIPE AND THEN ESTIMATR FOR THE COMBINATION IN 1 HOUR AND THEN TAKE RECIPROCAL TO FIND THE HOURS NEEDED TO FILL OR EMPTY THE FULL TANK OR WHATEVER IS NEEDED
Mixture_Word_Problems/18599: Need help with this, The inlet pipe of an oil tank can fill the tank in 1 hour, 30 minutes. The outlet pipe can empty the tank in 1 hour. How long does it take to empty the tank if both pipes are open?1 solutions Answer 9051 by venugopalramana(345) on 2005-11-08 02:05:51 (Show Source): The inlet pipe of an oil tank can fill the tank in 1 hour, 30 minutes. The outlet pipe can empty the tank in 1 hour. How long does it take to empty the tank if both pipes are open?inlet pipe fills 1 tank in 1hr 30 mts =1.5=3/2 hrs.or inlet pipe fills ...1/(3/2)=2/3 tank in 1 hr.out let pipe empties 1 tank in 1 hr....so assuming the tank is full at the start water to be emptied =1 tank and in 1 hr inlet fills 2/3 tank and out let empties 1 tank ..so net out flow is 1-2/3=(3-2)/3=1/3 tank in 1 hour.so 1/3 tank emptying needs .....1 hrthen 1 tank emptying needs ......? hrs...= 1*1/(1/3)=3 hrs.

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I believe this would be the best place to ask...Okay, this is a relation (set of ordered pairs) that I have to write an equation for. The set is :
SEE MY COMMENTS BELOW
(-3,10)(-2,5)(-1,2)(0,1)(1,2)(2,5)(3,10)…OK
x : -3, -2, -1, 0, 1, 2, 3………GOOD
y : 10, 5, 2, 1, 2, 5, 10 ………GOOD
I'm supposed to divide the range differences by the domain differences :THIS IS HINT GIVEN TO HELP YOU..OK…
range difference : 5, 3 1…THIS IS NOT CORRECT.IT IS A GOOD PRACTICE TO FOLLOW AN ORDER OR CONVENTION.SUBTRACT PREVIOUS ENTRY FROM SUCCEEDING ENTRY ALWAYS.SO THE RANGE DIFFERENCE IS –5,-3,-1,1,3,5.
domain difference : 1 ..CONSTANT ALWAYS..GOOD.HERE YOU FOLLOWED THE CONVNTION I MENTIONED ABOVE
But, if I divide 5/1, 3/1 and 1/1, I end up with 5, 3 and 1, as any number divided by 1 is that number…OK ..BUT YOU GET –5,-3,-1,1,3,5 AS WE CORRECTED ABOVE.
So where do I go from here and how do I get there???…LOOK AT THE OBJECTIVE..WE HAVE TO FIND A PATTERN..A HINT IS GIVEN..WE USED THE HINT AND GOT THE NUMBERS AS –5,-3,-1,1,3,5…THERE IS A PATTERN HERE..THEY STARTED WITH –5 AND INCREASED BY A CONSTANT DIFFERENCE OF 2 ..THAT GIVES US THE CLUE TO DERIVE THE RELATION
LET US CALL THE FIRST NUMBER AS X AS YOU NAMED IT .SECOND NUMBER IS Y.
SO WE GOT (X1,Y1),(X2,Y2)…ETC
AND NOW YOU HAVE TO FIND A RELATION BETWEEN X AND Y BY ANALYSING THEM SEPERATELY OR TOGETHER BY COMBINING THEM
LET ME SHOW YOU BOTH METHODS.
FIRSTLY LET US ANALYSE SEPERATELY.WE IND THAT BOTH ARE LINKED TO N THE NUMBER OF THE TERM.SO LET US TRY TO RELATE EACH WITH N WHICH WILL THEN GIVE US WHAT IS CALLED AS PARAMETRIC RELATION.THAT IS X IS RELATED TO N AND Y IS RELATED TO N .SO INDIRECTLY X AND Y ARE RELATED THROUGH A PARAMETER N.
LET US TAKE UP X FIRST.WE FOUND
X2-X1=1
X3-X2=1
X4-X3=1
X5-X4=1
Etc…..
X(N+1)-X(N)=1

LET US ADD THEM ALL TOGETHER..WE FIND THAT X2,X3,X4…X(N) CANCEL OUT LEAVING ONLY X(N+1)-X1 ON THE L.H.S…ON THE R.H.S. WE GET 1+1+1…N TIMES =1*N=N
HENCE WE GET X(N+1)-X1=N
BUT X1=-3…SO ….X(N+1)=N+X1=N-3..WE CAN CHECK BCK BY PUTTING N=0,1,2,3,…..ETC…WE GET X(0+1)=X1= 0-3,….X(1+1)=X2=1-3=-2…ETC..OK.
NOW LET US SEE Y ,WE FOUND
Y2-Y1=-5
Y3-Y2=-3
Y4-Y3=-1
Y5-Y4=1
Y6-Y5=3
Y7-Y6=5
ETC…..
Y(N+1)-YN= -5 +(N-1)*2= -5+2N-2=2N-7…….LET ME EXPLIN THIS TO YOU.WE FIND THE NUMBERS ON THE R.H.S ARE –5,-3,-1 ETC….DIFFERING BY A CONSTANT VALUE 2..THIS IS CALLED ARITHMATIC PROGRESSION(A.P)……. 2 ND TERM I OBTAINED BY ADDING 2 O THE FIRT TERM .THIRD TERM IS GOT BY ADDING 2*2 TO THE FIRST TERM.FOURTH IS OBTAINED BY ADDING 2*3 TO THE FIRST TERM……N TH TERM IS OBTAINED BY ADDING (N-1)*2 TO THE FIRST TERM
THE FORMULAE ARE …NTH TERM =TN=A+(N-1)*D WHERE A IS THE FIRST TERM AND D IS THE CONSTANT DIFFERENCE.
SUM OF N TERMS OF AN A.P. IS GIVEN BY
SN=(N/2){2A+(N-1)*D}…WHICH WE SHALL USE BELOW
NOW AS WE DID EARLIER LET US ADD ALL THE ABOVE TO GET
Y(N+1)-Y1={-5+(-3)+(-1)+1+3………..+(2N-7)}=AS PER THE FORMULA ABOVE
Y(N+1)-Y1=(N/2){2*(-5)+(N-1)*2}=(N/2){-10+2N-2}=(N/2)(2N-12)
=(N/2)2(N-6)=N(N-6)=N^2-6N….BUT Y1=10 SO WE GET
Y(N+1)-Y1=Y(N+1)-10=N^2-6N….OR…….
Y(N+1)=N^2-6N+10……LET US CHECK BACK BY PUTTING N+0,1,2,ETC….
Y(0+1)=Y1=0-0+10=10…OK
Y(1+1)=Y2=1-6+10=5…..OK
Y(2+1)=Y3=2^2-6*2+10=2……OK
Y(3+1)=Y4=3^2-6*3+10=1…….OK
Y(4+1)=Y5=4^2-6*4+10=2………OK
Y(5+1)=Y6=5^2-6*5+10=5…….OK
Y(6+1)=Y7=6^2-6*6+10=10…OK…
SO NOW WE GOT …Y(N+1)=N^2-6N+10
THIS TOGETHER WITH X(N+1)=N-3….GIVES US THE RELATION BETWEEN X AND Y IN TERMS OF ACOMMON PARAMETER N.
NOW IF WE WANT X AND Y DIRECTLY RELATED WE HAVE TO ELIMINATE N ..WE NOTE THAT X(N+1)=N-3…OR N=X(N+1)+3…BY PUTTING THIS IN EQUATION FOR Y(N+1)=N^2-6N+10,WE GET A DIRET RELATION BETWEEN X AND Y …
Y(N+1)={X(N+1)+3}^2-6{X(N+1)+3}+10
IF YOU ARE MORE COMFORTABLE WITH XN AND YN YOU CAN WRITE THE ABOVE AS
YN=(XN+3)^2-6(XN+3)+10......OR IN CONVENTIONAL FORM WE CAN WRITE THIS AS
Y=(X+3)^2-6(X+3)+10=X^2+9+6X-6X-18+10=X^2+1...OR
Y=X^2+1.....YOU CAN CHECK ALL THE ORDERED PAIRS AND THEY SATISFY THIS EQUATION

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I was doing a test in algebra and I wanted to know how to slove this, it takes 1/2 a gallon of paint to fill 530 square feet of a wall and 3/4 of a gallon to fill up 735 sqaure feet of a wall it also says write an equation to find out how many gallons you need to fill up a certain amount of space in a wall let g be gallons, s for square feet.
THE DATA IS SUPERFLUOUS AND ERRONEUS.IF IT TAKES 1/2 GALLON TO PAINT 530 SQ.FT.,IT SHOULD MAKE 3/4 GALLON SUFFICIENT TO PAINT 795 SQ.FT.SO LET US TAKE ONLY THE FIRST PART
LET THE SPACE TO BE FILLED =S SQ.FT.
LET THE PAINT REQD. = G GALLONS
BUT 1/2 GALLON CAN PAINT 530 SQ.FT
SO 1 GALLON CAN PAINT 530*2=1060 SQ.FT
SO G GALLONS CAN PAINT 1060*G SQ.FT = S SQ.FT. WE CALLED SO THE EQN.IS
S=1060G .................
CHECK BACK YOUR QUESTION AND WRITE BACK PROPERLY IF THERE IS A MISTAKE FOR US TO UPDATE THE ANSWER.

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LET THE NUMBER BE X
3 MORE THAN IT =X+3...THIS IS DIVIDED BY..
3 LESS THAN THE ORIGINAL NUMBER = X-3...HENCE QUOTIENT IS (X+3)/(X-3)

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WHAT DO YOU WANT FACTORISATION?PLEAE WRITE QUESTION CORRECTLY
x^2-16 =(X)^2-(4)^2=(X+4)(X-4)...USING FORMULA A^2-B^2=(A+B)(A-B)
x^2-6x+8 =X^2-4X-2X+8=X(X-4)-2(X-4)=(X-4)(X-2)
these are two different Questions
x^2-10x+25=X^2-2*X*5+5^2=(X-5)^2...USING FORMULA A^2-2*A*B+B^2=(A-B)^2
x-5=X-5 ONLY ...IF YOU WANT DIVISION THEN
x^2-10x+25.....X^2-2*X*5+5^2...............(X-5)^2......X-5
--------- = -------------...............=--------- = ---- IS THE ANSWER
x-5.................X-5......................X-5..........1

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LET TIME TAKEN BY REENA ALONE TO DO 1 JOB =X HRS
BILLY ALONE TAKES 2 HRS. MORE =X+2
SO REENA CAN DO ALONE IN 1 HOUR =1/X JOB
BILLY CAN DO ALONE IN 1 HOUR =1/(X+2) JOB
TOGETHER THEY CAN DO IN 1 HR.=(1/X)+(1/(X+2)) JOB
IN 2 HRS TOGETHER THEY CAN DO =2*{(1/X)+(1/(X+2))} JOB
NOW REENA LEFT AND BILL ONLY WORKED FOR 1 HOUR = BILLY ALONE DOES 1/(X+2) JOB IN 1 HOUR
SO TOTAL JOB DONE =2*{(1/X)+(1/(X+2))}+(1/(X+2)) JOB..BUT THIS COMPLETED THE 1 JOB .SO
2*{(1/X)+(1/(X+2))}+(1/(X+2)) =1
2/(X)+2/(X+2)+1/(X+2)=1..L.C.M IS (X)(X+2)
{2(X+2)+2(X)+1(X)}/(X)(X+2)=1
MULTIPLYING BY LCM (X)(X+2) BOTH SIDES WE GET
2X+4+2X+X=(X)(X+2)=X^2+2X
5X+4=X^2+2X
X^2+2X-5X-4=0
X^2-3X-4=0
X^2-4X+X-4=0
X(X-4)+(X-4)=0
(X+1)(X-4)=0
X-4=0
X=4
SO REENA CAN DO THE JOB IN 4 HRS AND BILL CAN DO IN X+2=4+2=6 HRS THE JOB.

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SEE THE FOLLOWING PROBLEM WHICH IS SIMILAR TO YOURS AND SOLVE YOUR PROBLEM.IN CASE OF DIFFICULTY COME BACK
log(x+1) + log(x-5) = 4
USING THE FORMULA LOG A TO BASE X = LOG A/LOG X TO A COMMON BASE OR STANDARD BASE.HENCE WE HAVE
{LOG (X+1)/LOG 2} + {LOG (X-5)/LOG 2}=4
LOG(X+1)+LOG(X-5)=4*LOG 2=LOG 2^4=LOG 16 (SINCE LOG X^N=N*LOG X)
LOG(X+1)(X-5)=LOG 16 (SINCE LOG X + LOG Y =LOG(X*Y))
TAKING ANTILOGS
(X+1)(X-5)=16
X^2-5X+X-5=16
X^2-4X-5-16=0
X^2-7X+3X-21=0
X(X-7)+3(X-7)=0
(X-7)(X+3)=0
HENCE X=7 OR X=-3 AS X=-3 LEADS TO LOG OF NEGATIVE NUMBERS WHICH DO NOT EXIT , THE ANSWER IS X=7

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log(x+1) + log(x-5) = 4
USING THE FORMULA LOG A TO BASE X = LOG A/LOG X TO A COMMON BASE OR STANDARD BASE.HENCE WE HAVE
{LOG (X+1)/LOG 2} + {LOG (X-5)/LOG 2}=4
LOG(X+1)+LOG(X-5)=4*LOG 2=LOG 2^4=LOG 16 (SINCE LOG X^N=N*LOG X)
LOG(X+1)(X-5)=LOG 16 (SINCE LOG X + LOG Y =LOG(X*Y))
TAKING ANTILOGS
(X+1)(X-5)=16
X^2-5X+X-5=16
X^2-4X-5-16=0
X^2-7X+3X-21=0
X(X-7)+3(X-7)=0
(X-7)(X+3)=0
HENCE X=7 OR X=-3 AS X=-3 LEADS TO LOG OF NEGATIVE NUMBERS WHICH DO NOT EXIT , THE ANSWER IS X=7

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PUT Z+2=0...SO Z=-2
-2....|4.....0......0......1.......-4......9
......|0....-8.....16......-32.....62....-116
---------------------------------------------------
......|4....-8.....16......-31.....-58....-107
HENCE QUOTIENT IS 4Z^4-8Z^3+16Z^2-31Z-58 AND REMAINDER IS -107 WHEN (4z^5+z^2-4z+9) IS DIVIDED BY (z+2)
SEE BELOW FOR EXPLANATION IN SIMILAR PROBLEM
Divide x^3 - 1 by x - 1, x^4 - 1 by x - 1, and x^5 - 1 by x - 1. What is the quotient when x^9 - 1 is divided by x - 1?
1 solutions
Answer 8067 by venugopalramana(454) About Me on 2005-10-22 01:41:44 (Show Source):
there is a procedure for short division called horner's method which can be used here. i shall show you 2 examples ..rest you can do in a similar manner..
first write the given expression in decreasing order of powers listing missing terms also as zeros...
For example x^3-1 = 1*x^3+0*x^2+0*x-1..........(A)
now check if we have the divisor in the form of x+a or x-a or not .we have here x-1 which is in the form x-a...then let x-1 =0 which gives us x=1...(B)
now do the division as follows
write 1 0 0 -1......these are coefficients of powers of x obtained above under (A)from the problem
put 1 as divisor as obtained under (B) above
do as follows
divisor...1|.. 1..0..0..-1......row 1
...........|.. 0..1..1..1.......row 2
----------------------------------------------
...........|...1..1..1..0 ......row 3
explanation: row 1 has divisor on the left as 1 followed by a seperation bracket..then put coefficients of powers of x obtained above under (A)from the problem in different columns
row 2 ..start with a zero under the coefficient in row 1 in 1st.column..now add the row 1 and row 2 numbers in 1st. column (which are there vertically one below the other )and put the sum under the same column in row 3...here 1+0=1..now multiply this number in row 3 with divisor and put it in row 2 in second column...now add numbers in row 1 and row 2 in column 2 put the sum in row 3 under same column no.2....repeat the procedure till the end
row 3 represents the quotient and remainder...number in last column in row 3 is remainder here it is zero.the other numbers read from right to left give the coefficients of increasing powers of x ...that is the answer on division is 1x^2+1x+1..remainder =0
to explain i am giving another example below...x^2-5x+6 devide by x-3
put x-3 =0 so x=3 is the divisor...
3] 1 -5 6
..... 0 3 -6
--------------
..... 1 -2 0 answer is x-2 and remainder is 0

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PERIMETER =2(L+B)=30
L+B=30/2=15
L=-B+15
GIVE DIFFERENT VALUES TO B ,FIND CORRESPONDING VALUES OF L FROM THE ABOVE FORMULA,TABULATE AS BELOW AND PLOT THEM TO GET THE GRAPH AS SHOWN BELOW
B....1....4.....8....12.....
L....14...11....7.....3....

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SEE THE FOLLOWING WHICH IS SIMILAR AND DO ACCORDINGLY.IN CASE OF DIFFICULTY PLEASE WRITE BACK

Please! If you could be so kind.
Question: If Sally can paint a house in 4 hours, and John can paint the same house in 6 hour, how long will it take for both of them to paint the house together?
The answer is suppose to be 2 hours and 24 minutes, but I can't seem to get the solution to it. Thank You.
1 solutions
Answer 10642 by venugopalramana(454) About Me on 2005-12-08 08:48:07 (Show Source):
I THINK I ANSWERED THIS QUESTION GIVING EXAMPLES.IF YOU HAVE NOT UNDERSTOOD IT,LET ME SOLVE..THE KEY IN THESE PROBLEMS IS TO FIND WORK DONE PER UNIT TIME FIRST FROM GIVEN DATA,COMBINE THEM AS REQUIRED AND THEN FIND THE TIME REQUIRED TO DO THE FULL JOB(THAT IS REVERSE THE FIRST STEP)WE HAVE....
SALLY CAN PAINT 1 HOUSE IN 4 HRS..SO IN 1 HR SHE CAN DO 1/4 HOUSE
JOHN CAN PAINT 1 HOUSE IN 6 HRS...SO IN 1 HR HE CAN PAINT 1/6 HOUSE...
SO TOGETHER THEY CAN PAINT 1/4 +1/6 HOUSE IN 1 HR =(3+2)/12 HOUSE IN 1 HR.=5/12 HOUSE IN 1 HOUR....
SO FULL HOUSE THEY CAN PAINT IN 1/(5/12)=12/5=2.4 HRS.=2HRS AND 0.4*60=24 MTS.OR 2HRS 24 MTS. TIME.

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LET Y=25^(LOG X TO BASE 5)
USE FORMULA LOG X TO BASE 5 = LOG X/LOG 5 TO COMMON BASE
Y=25^(LOG X/LOG 5)
TAKING LOGS BOTH SIDES
LOG Y ={(LOG X)/(LOG 5)}*LOG 25
=LOG X*LOG 5^2/LOG 5=LOG X*(2*LOG 5)/LOG 5(SINCE LOG X^N=N*LOG X)
=2LOG X=LOG X^2
HENCE LOG Y = LOG X^2
TAKING ANTILOGS
Y=X^2
HENCE
Y=25^(LOG X TO BASE 5)=X^2

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SEE THE FOLLOWING EXAMPLE..IT IS SIMILAR TO YOUR PROBLEM....COEFFICIENT MATRIX AUGMENTED BY CONSTANTS IS CALLED AUGMENTED MATRIX ....IF YOU HAVEW DOUBTS COME BACK
COEFFICIENT MATRIX...........AUGMENTED MATRIX
[ 7 -4 -2]................. | 9 ]
[-5 8 5 ]...................|-8 ]...............
[-5 5 8 ]...................|-2 ]

Hi! I'm in homeschooling and I've been having trouble with matrices. I asked a question the other day about them, and I'm still having trouble with another problem on my lesson. I'd appreciate the help!
Here's the matrix:
[ 7 -4 -2 | 9 ]
[-5 8 5 |-8 ]
[-5 5 8 |-2 ]
Thank you!
Caitlyn
1 solutions
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Answer 10184 by venugopalramana(450) on 2005-12-01 06:05:36 (Show Source):
SEE THE FOLLOWING EXAMPLE SOLVED FOR YOU ONLY WHICH IS ESSENTIALLY SAME BUT FOR THE NUMBERS , FOLLOW STEP BY STEP AND DO BY YOUR SELF,THERE BY YOU WILL LEARN HOW TO DO AND BECOME AN EXPERT BY YOUR SELF!!!!IF YOU DO NOT UNDERSTAND ASK WHERE IT IS NOT CLEAR...BUT IT IS NOT GOOD PRACTICE TO GET ALL YOUR HOME WORKS DONE PARTICULARLY WHEN THEY ARE IDENTICAL PROBLEMS...THIS SITE ALSO MENTIONS THAT ASPECT IN ITS MOTTOS...BY FOLLOWING SUCH PRACTICES YOU WILL E THE LOSER AS YOU ARE NOT TRYING TO LEARN!!!!!!!!!
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Linear_Algebra/20750: Hi, I'm in homeschooling and I'm having trouble with matrices. I was wondering how to solve the problem where you have to find the x,y, and z values in the matrix:
[7 -7 5 | 9]
[9 5 -7 | -17]
[6 1 -7 | -2]
I'd appreciate the help. Thank you!
Caitlyn Reese
1 solutions
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Answer 9969 by venugopalramana(390) on 2005-11-28 07:14:11 (Show Source):
the 4 column heads represent x,y,z and constant term in the matrix of system of eqns.
then each row gives us one eqn.like say row 1 gives us that 7x-7y+5z=9..etc…
hence if we can make the matrix to become
1 0 0 ?
0 1 0 ??
0 0 1 ???
then from the explantion given above it means 1x=?.1y=?? And 1z=???
so we try to transform the matrix in to that form..by the following steps.
in fact using the above explanation,you can see that what we do at each step is just
divide each eqn. with a constant/add/subtract etc which does not change the basic
eqn.for ex. dividing row 1 by 7 means change the given eqn.7x-7y+5z=9 to x-y+5z/7=9/7
legend:- or1 means old row 1..nr1 means new row 1…r1 means the existing row 1 please note that no changes are made in rows other than those mentioned at each step.
start with given matrix …
7...... -7..... 5...... 9
9...... 5...... -7..... -17
6...... 1...... -7..... -2
step 1…we want to make 1st.row 1st.column as 1….so….nr1=or1/7...
1...... -1..... (5/7).. (9/7)
9...... 5...... -7..... -17
6...... 1...... -7..... -2
step 2..we want to make 2nd/3rd.rows,col.1 as 0...so...nr2=or2-9*r1........nr3=or3-6*r1
1...... -1..... (5/7).. (9/7)
0... 14..... (-7-9*5/7).... (-17-9*9/7)
0...... 7...... (-7-6*5/7)..... (-2-6*9/7)
step 3…we want to make 2nd.row.2nd.col.as 1..so..nr2=or2/14
1...... -1..... 5/7.... 9/7
0...... 1...... (-94/7)/14..... (-200/7)/14
0...... 7...... (-7-6*5/7)..... (-2-6*9/7)
step 4..we want to make 3rd.row.2nd.col.as 0…so….nr3=or3-7*r2
1...... -1..... (5/7).. (9/7)
0...... 1...... (-94/7)/14..... (-200/7)/14
0 0 (-79/7)-7*(-94/98) (68/7)-7*(-200/98)
step 5….we want to make 3rd.row.3rd.col.as 1…so….nr3=or3/(-32/7)
1...... -1..... (5/7).. (9/7)
0...... 1...... (-94/98)... (-200/98)
0...... 0..... 1...... -1
step 6…we want to make 1st/2nd.row 3rd.col.as 0..so..nr1=or1-5*r3/7...nr2=or2+94*r3/98
1...... -1..... 0..... 2
0...... 1...... 0...... -3
0...... 0...... 1...... -1
step7….we want to make 1st.row 2nd.col.as 0..so….nr1=or1+r2
1...... 0...... 0...... -1
0...... 1...... 0...... -3
0...... 0...... 1...... -1
so x=-1.....y=-3.....and z=-1...you can check back
YOU CAN SEE THE FOLLOWING ADDITIONAL MATERIAL FOR REFERENCE
How do I perform the next required row operation on the following matrix and provide only the next table:
x y z
1 28 14 245
0 3 7 42
0 7 7 -38
1 solutions
Answer 9892 by venugopalramana(370) About Me on 2005-11-25 08:01:32 (Show Source):
trust you want to solve the equations for x,y and z and you are at this stage now....assuming that .....our objective is to finally get the matrix if possible into the following form ....(i am using ....to seperate the numbers with suitable gaps..your typing is giving raise to uneven gaps bringing a little lack of clarity)
1.....0.....0.....x
0.....1.....0.....y
0.....0.....1.....z
now we have
1......28.....14.....245
0.......3......7......42
0.......7......7.....-38
new row2=old row2/3.......to get 1 as required in row2.so we get...
1......28.....14.....245
0......3/3....7/3....42/3
0.......7......7.....-38
new row3=oldrow3-7*row2 to get 0 as required in row3
1......28.....14...........245
0.......1.....7/3...........14
0......7-7*1..7-7*7/3......-38-7*14
new row3 = old row3/(-28/3)..to get 1 as required in row3
1......28.....14...................245
0.......1.....7/3...................14
0.......0....(-28/3)/(-28/3).....(-136)/(-28/3)
this gives us finally in the following form
1......28.....14............245
0.......1.....7/3...........14
0.......0......1............102/7
now we go back in the same way to get 0 in row2 and row3
new row2=old row2-row3*7/3...and new row1=old row1-row3*14...so we get
1......28......14-1*14.......245-(102/7)*14
0.......1.......7/3-(7/3)*1...14-(102/7)*(7/3)
0.......0.........1.............102/7
the above on simplification gives us
1.......28.......0..........41
0........1.......0..........-20
0........0.......1..........102/7
now finally we take new row1=old row1-28*row2
1.......28-28*1......0.......41-(-28*20)
0........1.......0...........-20
0........0.......1...........102/7
so the final answer is
1......0.......0.......601
0......1.......0.......-20
0......0.......1.......102/7
which tells us that
1*x+0*y+0*z=x=601
0*x+1*y+0*z=y=-20
0*x+0*y+1*z=z=102/7
note that each and every transformation we did above can be interpreted as given in the last statement given above...this i hope will give you the insight of the process at every step.you can also substitute these values of x,y and z in each and every matrix above to see that they satify all the equations given by the different matrices..in general each mtrix can be taken as a set of simltanous equations in x,y and z...they can be written as follows..take column 1 is for x,column 2 is for y and column 3 is for z.so the first matrix you gave
1......28.....14.....245
0.......3......7......42
0.......7......7.....-38
tells us that
1*x+28*y+14*z=245....etc...

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2 3
5 2
6 1
7 0
WHY DOUBT FOLLOW SAME PROCEDURE
(2*3+5*2+6*1+7*0)/(2+5+6+7)=19/20

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SEE MY COMMENTS IN CAPITAL LETTERS
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4. When a pair of dice is rolled, the total will range from 2 (1,1) to 12 (6,6)....OK
It is a fact that some numbers will occur more frequently than others as the dice are rolled over and over. ....YA
Why will some numbers come up more frequently than others? BECAUSE OF 2 DICE AND TOTAL OF TWO NUMBERS SHOWING UP ON 2 DICE BEING TAKEN.SEE THE TABLE BELOW.IT WONT HAPPEN WITH 1 DICE THROW.ALL NUMBERS FROM 1 TO 6 IN THAT CASE ARE EQUALLY LIKELY.
DICE 1 DICE 2 TOTAL DICE 1 DICE 2 TOTAL total FREQ. %
1 1 2 4 1 5 1 0 2.8
1 2 3 4 2 6 2 1 5.6
1 3 4 4 3 7 3 2 8.3
1 4 5 4 4 8 4 3 11.1
1 5 6 4 5 9 5 4 13.9
1 6 7 4 6 10 6 5 16.7
2 1 3 5 1 6 7 6 19.4
2 2 4 5 2 7 8 5 22.2
2 3 5 5 3 8 9 4 25.0
2 4 6 5 4 9 10 3 27.8
2 5 7 5 5 10 11 2 30.6
2 6 8 5 6 11 12 1 33.3
3 1 4 6 1 7
3 2 5 6 2 8
3 3 6 6 3 9
3 4 7 6 4 10
3 5 8 6 5 11
3 6 9 6 6 12
Each die has six sides numbered from 1 to 6. How many possible ways can a number be rolled? In other words, we can roll (2,3) or (3,2) or (6,1) and so on. What are the total (x,y) outcomes that can occur? SEE TABLE ABOVE
How might you then estimate the percentage of the time a particular number will come up if the dice are rolled over and over? SEE TABLE ABOVE FOR PERCENT
Once these percentages have been calculated, how might the mean value of the all the numbers thrown be determined? BY TAKING WEIGHTED AVERAGE OF THE VALUE MULTIPLIED BY FREQUENCY..THAT IS
(1*0+2*1+3*2+4*3+5*4+6*5+7*6+8*5+9*4+10*3+11*2+12*1)/36=7

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LET THE 2 A.M.S BE X,Y...HENCE -2,X,Y,58 ARE IN A.P.WE FIND THAT 58 IS THE FOURTH TERM OR N=4 , A=-2 AND D=?...USING TN=A+(N-1)D,WE GET....
T4=A+(4-1)D
58=-2+3D
3D=58+2=60
D=20
SO X=-2+1D=-2+20=18
Y=-2+2D=-2+2*20=38
SO THE 2 A.MS ARE 18 AND 38

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216x^3+1...use the formula a^3+b^3=(a+b)(a^2-ab+b^2)
216x^3+1=(6x)^3+1^3=(6x+1){(6x)^2-(6x)*1+1^2}
=(6x+1)(36x^2-6x+1)

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YOU ARE PROCEEDING ON RIGHT LINES GO AHEAD...
OFCOURSE THERE IS A FUNDAMENTAL ASSUMPTION MADE TO START WITH THAT IS ELIMINATION IS BY KNOCK-OUT...NOT ON LEAGUE BASIS IN WHICH CASE EACH PLAYER WILL PLAY WITH ALL OTHERS TO MAKE A LONG SERIES OF
3003 MATCHES WHICH YOU WILL KNOW IN HIGHER CLASS..
.NOW FOR THE PRESENT TAKE KNOCK-OUT BASIS AS YOU HAVE TAKEN GIVING A BYE TO ODD PLAYER AND INDUCTING HIM INTO THE NEXT ROUND.IT WILL RESULT IN SOME UNEVEN MATCHES PLAYED BY DIFFERENT PLAYERS BUT THAT GOES BY LUCK OF THE DRAW...
SO WE HAVE
I ROUND =78/2=39 MATCHES
II ROUND =38/2=19 MATCHES WITH 1 PLAYER GIVEN BYE BY LOTS
III ROUND =(19+1 BYE)/2=10....LIKE THAT YOU CAN PROCEED...BUT YOU ARE RIGHT IN INFERING IT ENTAILS UNEVENPLAY OFFS...SO ATLEAST AFTER SOME STAGE A LEAGUE BASIS WOULD BE GOOD SAY FOR THE LAST 5 WHEN WE NEED 10 MATCHES FOR 5 PEOPLE PLAYING LEAGUE

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SEE THE FOLLOWING EXAMPLE FOR DIFFERENT SIZES 9,7 AND 3..... AND DO AS PER THAT
FORMULA FOR AREA OF A TRAPEZIUM IS
A=(1/2)*{SUM OF PARALLEL SIDES}*HEIGHT..........HERE WE HAVE
A=(1/2)(9+7)*3=24 SQ.CM.

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FORMULA FOR AREA OF A TRAPEZIUM IS
A=(1/2)*{SUM OF PARALLEL SIDES}*HEIGHT..........HERE WE HAVE
A=(1/2)(9+7)*3=24 SQ.CM.

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7480000000......THIS WITH SO MANY ZEROES IS NOT CONVENIENT TO WRITE OR READ.SO IN SCIENTIFICATION IT IS WRITTEN AS A UNIT QUANTITY WITH DECIMAL AND EXPONENT..THAT IS WE WRITE
7480000000=7.48*10^9...AGAIN FOR CONVENIENCE WE WRITE 10^9 AS E 9
SO WE GET
7480000000=7.48 E 9

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0.000000543 THIS WITH SO MANY ZEROES IS NOT CONVENIENT TO WRITE OR READ.SO IN SCIENTIFICATION IT IS WRITTEN AS A UNIT QUANTITY WITH DECIMAL AND EXPONENT..THAT IS WE WRITE
0.000000543=5.43*10^(-7)...AGAIN FOR CONVENIENCE WE WRITE 10^(-7) AS E-7
SO WE GET
0.000000543=5.43 E -7

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Determine if the ordered pair (0,2) is a solution of -3x + y = -2 ....IT MEANS YOU SUBSTITUTE 0 FOR X AND 2 FOR Y IN THE GIVEN EQN.AND VERIFY WHETHER IT IS SATISFIED OR NOT...THAT IS IN THIS CASE FIND L.H.S.AND R.H.S. SEPERATELY AND CHECK WHETHER THEY COME OUT SAME OR NOT
L.H.S=-3X+Y=-3*0+2=0+2=2
R.H.S.=-2...
SO L.H.S IS NOT EQUAL TO R.H.S.
HENCE (0,2)IS NOT A SOLUTION OF THE GIVEN EQN.

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simplify,
16x^7y^2 =
____________
2^x2
i came up with 8x^5y^2......VERY GOOD CORRECT...BUT YOU BETTER TYPE THIS AS
simplify,
16(x^7)(y^2) =
____________
2(x^2)
i came up with 8(x^5)(y^2)TO MAKE IT AMPLY CLEAR

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I THINK I ANSWERED THIS QUESTION GIVING EXAMPLES.IF YOU HAVE NOT UNDERSTOOD IT,LET ME SOLVE..THE KEY IN THESE PROBLEMS IS TO FIND WORK DONE PER UNIT TIME FIRST FROM GIVEN DATA,COMBINE THEM AS REQUIRED AND THEN FIND THE TIME REQUIRED TO DO THE FULL JOB(THAT IS REVERSE THE FIRST STEP)WE HAVE....
SALLY CAN PAINT 1 HOUSE IN 4 HRS..SO IN 1 HR SHE CAN DO 1/4 HOUSE
JOHN CAN PAINT 1 HOUSE IN 6 HRS...SO IN 1 HR HE CAN PAINT 1/6 HOUSE...
SO TOGETHER THEY CAN PAINT 1/4 +1/6 HOUSE IN 1 HR =(3+2)/12 HOUSE IN 1 HR.=5/12 HOUSE IN 1 HOUR....
SO FULL HOUSE THEY CAN PAINT IN 1/(5/12)=12/5=2.4 HRS.=2HRS AND 0.4*60=24 MTS.OR 2HRS 24 MTS. TIME.

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sqrt(2x-1)=x-2i SQUARING BOTH SIDES,WE GET
2X-1=X^+4I^2-4IX
X^2-X(2+4I)-3=0
USING QUADRATIC FORMULA

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16t^2 -180+464?..TAKE OUT COMMON FACTOR IF ANY FIRST..HERE 4 IS COMMON...SO..
=4(4T^2-45T+116)...NOW TAKE THE PRODUCT OF T^2 AND CONSTANT TERMS=4*116=464..TRY TO FIND 2 FACTORS FOR THIS PRODUCT WHOSE SUM IS THE COEFFICIENT OF X TERM =-45...WE HAVE (-16)*(-29)=+464 AND -16-29=-45...SO NOW REWRITE THE GIVEN EXPRESSION BREAKING MIDDLE TERM OF -45T AS -16T-29T AND TAKE OUT COMMON FACTORS..
=4(4T^2-16T-29T+116)
=4{4T(T-4)-29(T-4)}
=4(T-4)(4T-29)

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h(x)=-2x^2+6x-3

so far I have done is -2(x^2-3x)-3....OK
I'm not sure if I am suppose to find the perfect square ....YES DO IT ...YOU HAVE TO USE THE GIVEN X^2 AND X TERMS TO MAKE A PERFECT SQUARE
=-2(X^2-2*1.5*X+1.5^2-1.5^2)-3
=-2{X-1.5)^2-1.5^2}-3
=-2(X-1.5)^2+2*1.5*1.5-3
=-2(X-1.5)^2+1.5=Y SAY...Y=1.5-2(X-1.5)^2
NOW ,WE REASON OUT THAT (X-1.5)^2 BEING PERFECT SQUARE IT IS ALWAYS POSITIVE. SO ITS MIIMUM VALUE IS ZERO WHEN X=1.5
NOW WHEN THIS IS ZERO,Y WILL BECOME MAXIMUM SINCE WE ARE SUBTRACTING THE MINIMUM POSSIBLE VALUE FROM 1.5....YO Y HAS A MAXIMUM VALUE OR VERTEX AT X=1.5