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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).

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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.

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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

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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

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CAN YOU PLEASE CHECK BACK ,WHETHER TRIANGLE IS INSCRIBED IN RECTANGLE OR RECTANGLE IS INSCRIBED IN A TRIANGLE?

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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...

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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
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What is a coinciding equation?
1 solutions
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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.
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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

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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

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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.

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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

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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.

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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

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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.

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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

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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

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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

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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...

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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

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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

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THE ANSWER IS 50C2..HOPE YOU KNOW COMBINATIONS.
=50*49/2=1225

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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

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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

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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

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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

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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?

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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

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) 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

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

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.
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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

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

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