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YOU CAN USE IT TO
GOOD EFFECT TO PUT ACROSS A PROPOSAL TO A FRIEND ...
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... 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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PLEASE BE CLEAR IN DESCRIPTION OR GIVE A SKETCH.WHERE IS D?THEN ONLY WE CAN HLP YOU

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THEY ARE ALL OF SAME MODEL .I WILL SHOW YOU ONE .YOU TRY OTHERS.COME BACK IF IN DIFFICULTY.
(1)A=[-3 2] B= [0 2] (find AB and BA if possible)
[2 -2] [-2 4]
FIRST CHECK COMPATABILITY...THAT IS FIND THE ORDER OF A (M,N) = (2,2) MATRIX HERE.
AND ORDER OF B (P,Q)= (2,2)..N SHOULD EQUAL P .THEN WE SAY THEY ARE COMPATABLE AND THEY CAN BE MULTIPLIED.OTHERWISE NOT.THE ANSWER WILL BE C (M,Q) MATRIX THEN .HERE WE HAVE N=2=P...SO THEY ARE COMPATABLE AND CAN BE MULTIPLIED.THE NSWER WILL BE M=2 AND Q=2...THAT IS C(2,2)MATRIX
NEXT HOW DO WE FIND C =....
E11,E12
E21,E22
E11 IS OBTAINED BY MULTIPLYING EACH ELEMENT OF ROW 1 OF A WITH EACH ELEMENT OF COLUMN 1 OF B AND ADDING THEM UP.THAT IS
E11=-3*0+2*-2=-4
E12 IS OBTAINED BY MULTIPLYING EACH ELEMENT OF ROW 1 OF A WITH EACH ELEMENT OF COLUMN 2 OF B AND ADDING THEM UP.THAT IS
E12=-3*2+2*4=2
E21 IS OBTAINED BY MULTIPLYING EACH ELEMENT OF ROW 2 OF A WITH EACH ELEMENT OF COLUMN 1 OF B AND ADDING THEM UP.THAT IS
E21=2*0+-2*-2=4
E22 IS OBTAINED BY MULTIPLYING EACH ELEMENT OF ROW 2 OF A WITH EACH ELEMENT OF COLUMN 2 OF B AND ADDING THEM UP.THAT IS
E22=2*2+-2*4=-4
HENCE THE ANSWER IS C=
-4,2
4,-4

(2)A=[-2 3] B=[ 3 ] (find AB if possible)
[1 -2] [-2 ]
[0 0]
(3)A=[2 -2 4] B=[2 1 -3 0]
[1 0 -1] [0 -2 1 -2] (find AB if possible)
[2 1 3] [1 -1 0 2]

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Prove: If an isosceles triangle has an altitude from the vertex to the base, then the altitude bisects the vertex angle.
Given: Triangle ABC is isoceles;
AC=CB
Segment CD is the altitude to base Segment AB...
WHERE D IS ON AB.
HENCE ANGLE CDA=90=ANGLE CDB
To Prove: Segment CD bisects angle C
THAT IS ANGLE DCA=ANGLE DCB
Plan: ??.
WE SHALL TRY TO PROVE BOTH TRIANGLES CDA AND CDB TO BE CONGRUENT
Proof:??
IN TRIANGLES CDA AND CDB WE HAVE
AC=CB........GIVEN
CD=CD......SAME COMMON SIDE
ANGLE CDA=90=ANGLE CDB........GIVEN
HENCE THE 2 TRIANGLES ARE CONGRUENT BY THEOREM ON CONGUENCE OF RIGHT TRIANGLES
HENCE ANGLE DCA=ANGLE DCB

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cos theta = 5/12 and tan theta is negative, find the exact value for ctn theta.
USING ALL SILVER TEA CUPS,SINCE COS IS +VE AND TAN IS -VE,THETA IS IN IV Q.WHERE CTAN IS ALSO -VE.
HENCE COMPLETING TRIANGLE
12^2=5^2+X^2
X=SQRT(144-25)=SQRT(119)
HENCE COTAN(THETA)= 5/-SQRT(119)=-(5SQRT(119)/119)

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We are working on combinations and permutaions which involve factorials most of the time.
How many 4 letter combinations can be made from the letters f,g,h,i,j & k. No letters can repeat themselves and the letter do not have to form a word. How do you solve this?
COUNT TOTAL NUMBER OF LETTERS=6
TO MAKE 4 LETTER COMBINATIONS YOU SAID THAT IS NOT PROPER I THINK..THAT MEANS A TEAM OF 4 PLAYERS FROM 6 PERSONS.THAT MEANS ABCD..OR...BCDA...OR...CDBA..ETC...ARE ALL SAME TEAM.THE ARRANGEMENT IS NOT IMPORATANT.THEN ONLY IT IS CALLED COMBINATION.THEN THE ANSWER IS 6C4..THE FOMULA FOR NCR = N!/{(R!)*(N-R)!}
SO WE GET 6!/2!*(4-2)!=6!/2!*4!=6*5*4*3*2*1/2*1*4*3*2*1=15...
BUT I THINK YOU WANT ARRANGEMENTS .THEN ITS ANSWER IS 6P4..THE FORMULA FOR
NPR=N!/(N-R)!
6P4=6!/(6-4)!=6!/2!=360

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cos theta = 5/12 and tan theta is negative, find the exact valcue for sin theta.
USING ALL SILVER TEA CUPS,SINCE COS IS +VE AND TAN IS -VE,THETA IS IN IV Q.WHERE SINE IS ALSO -VE.
HENCE COMPLETING TRIANGLE
12^2=5^2+X^2
X=SQRT(144-25)=SQRT(119)
HENCE SIN(THETA)= -SQRT(119)/12

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I have constructed a pentagon with a base of 529.2 ft.=AB SAY The other sides follow clockwise in this order: 57.6 ft=BC, 248 ft=CD, 323.4 ft=DE, and 48.5 ft=EA. How do I find the measure of a perpendicular drawn from..NO ..IT IS TO the base AB to..NO IT IS FROM the opposite vertex D?
Base = 529.2 ft
Vertex formed by sides measuring 248 ft and 323.4 ft.
USING ABOVE NOMENCLATURE.....
TAKE A COMPASS WITH SOME CONVENIENT RADIUS BIG ENOUGH SO THAT WHEN YOU DRAW AN ARC WITH THAT RADIUS KEEPING THE CENTRE AT D,THE ARC CUTS AB OR EXTENDED AB ON EITHER SIDE AT 2 POINTS ...SAY...P AND Q.
NOW DRAW 2 ARCS WITH SAME RADIUS ,KEEPING THE CENTRE AT P AND Q..ON TO THE OTHER SIDE OF AB .LET THE 2 ARCS CUT AT R.JOIN DR.LET DR CUT AB AT Z.NOW DZ IS THE PERPENDICULAR FROM VERTEX...D TO BASE....AB.

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X=28.................D IS THE ANSWER

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VOLUME OF A PRISM =BASE AREA *HEIGHT
IF BASE IS RECTANGLE THEN
VOLUME = L*B*H

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VOLUME OF PRISM =BASE AREA *HEIGHT=TAKING BASE AS EQUILATERAL TRIANGLE
VOLUME =A*A*H*SQRT(3)/4

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A= [1 1]
[2 1]....
METHOD...WRITE IT IN NORMAL FORM.......1,0
.......................................0,1
STEP.1
USE....R21(-2)...THAT IS NEW ROW 2=OLD ROW 2 + (OLD ROW 1)*(-2)...WE GET
1,1
0,-1
STEP.2
USE R12(1).....THAT IS NEW ROW 1=OLD ROW 1 + OLD ROW 2 *1...WE GET..
1,0
0,-1
STEP.3.
USE C2(-1)....THATIS NEW COLUMN 2 =OLD COLUMN 2*(-1)
1,0
0,1
HENCE WE GOT
I = R12(1)*R21(-2)*A*C2(-1)
HENCE A = {R21(-2)}^(-1)*{R12(1)}^(-1)}*I*{C2(-1)}^(-1)
A = {R21(-2)}^(-1)*{R12(1)}^(-1)}*{C2(-1)}^(-1)
A = {R21(2)}*{R12(-1)}*{C2(-1)}
NOW WE KNOW THAT
I =
1,0
0,1
R21(2) =
1,0
2,1
R12(-1) =
1,-1
0,1
C2(-1) =
1,0
0,-1
HENCE
A = |1,0|*|1,-1|*|1,.0|
..……|2,1|*|0,.1|*|0,-1|
THUS A IS EXPRESSED AS PRODUCT OF ELEMENTARY MATRICES.YOU CAN MULTIPLY AND CHECK
HOPE YOU ARE TAUGHT THE PROCEDURE AND YOU KNOW IT.OTHERWISE PLEASE COME BACK AND WE SHALL EXPLAIN

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I need as quick as possible for HW: FInd the volume of a regular octahedron with sides equalling fifty square root of three cubic units? I don't know what to do please help me
FORMULA IS
VOLUME =SQRT(2)*S^3/3={SQRT(2)/3}*{52*SQRT(3)}^3=344418

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In a right triangle one side is 8 ft and the other side is 6 ft. What are the measures of angle A and B if C is 90 degrees? What is the hypotenuse?
HYPOTENUSE IS THE SIDE OPPOSITE RIGHT ANGLE OR 90 DEGREE ANGLE WHICH IS C HERE.SO AB IS THE HYPOTENUSE.AS PER PYTHOGARUS THEOREM
AC^2=AB^2+BC^2=8^2+6^2=64+36=100
AC=10

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sum of interior angles of any polygon OF N SIDES=(2*N-4)*90 DEGREES

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sum of interior angles of any polygon WITH N SIDES =(2*N-4)*90 DEGREES
EX....N=4...QUADRILATERAL.......(2*4-4)*90=4*90=360

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is triangle DEF, d=5 e=6 f=3 then COS E=????
i have the right set up i believe but i cant get the right answer helpp!!!
FORMULA IS
e^2=d^2+f^2+2df*COS(E)
6^2=5^2+3^2+2*5*3*COS(E)
COS(E)=(36-25-9)/(2*5*3)=1/6

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1. graph x-3= -I/8 (y=2)^2. Write the coordinates of the vertex and
> the
> > > focus and the equation of the directrix.
X-3=-(1/8)(Y-2)^2
(Y-2)^2=-8(X-3)
COMPARING WITH STD.EQN.OF PARABOLA
(Y-K)^2=4A(X-H)^2...WE GET
VERTEX=(H,K)=(3,2)
A=-2
FOCUS =(H+A,K)=(3-2,2)=(1,2)
DIRECTIX IS ... X-H+A=0...THAT IS
X-3-2=0...OR...X=5
GRAPH IS GIVEN BELOW

> > > 2.Find all solution to each system of equations algerbaiclly.
> > > x^2 + y^2 =25
SINCE SUM OF TWO SQUARES = 25 ,AND SQUARES ARE ALWYS POSITIVE ,WE GET THE SOLUTION SETS AS X^2<=25 AND Y^2<=25....THAT IS
1.0<=|X|<=5 AND 0<=|Y|<=5
> > > 3x^2 + 2y^2 = 59...YOU CAN DO THE SAME WAY AS ABOVE
3X^2<=59 AND 2Y^2<=59..ETC.......
> > > 3. Write the standard equations of the conic section with the given
> > > characteristics
> > > hyperbola with foci at (5,-7) and (14,0) and vertices at (12,0) and
> > > (4,0)
STANDARD EQN.OF HYPERBOLA IS
(X-H)^2/A^2 -(Y-K)^2/B^2 =1...WHERE
CENTRE IS (H,K)...WE ARE GIVEN VERTICES ARE AT (12,0) AND (4,0)...SO CENTRE IS AT
(12+4)/0,(0+0)/2=(8,0)
HENCE H=8....K=0
TRANSVERSE AXIS IS Y=K=0
LENGTH OF TRANVERSE AXIS =2A=16
A=8
CONJUGATE AXIS IS X=H=8
PLEASE CHECK DATA....CENTRE WHICH IS MID POINT OF FOCI AND ALSO MIDPOINT OF VERTICES ARE NOT TALLYING

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Assume that the only area formula you know is the formula for the area of a triangle. Explain how to derive the formula for the area of the trapezoid
LET ABCD BE THE TRAPEZOID WITH AB PARALLEL TO CD.SEE DIAGRAM BELOW.
FROM C AND D DRAW CF AND DE PERPENDICULARS TO AB MEETING AB AT F AND E
JOIN EC
NOW AREA OF TRAPEZIUM =AREA OF TRIANGLE ADE+AREA OF TRIANGLE DEC+AREA OF TRIANGLE EFC+TRIANGLE CFB....ALL ARE RIGHT ANGLED TRIANGLES HENCE AREA =BASE*HT/2..HT.FOR ALL TRIANGLES =H=DE=CF=DISTANCE BETWEN PARALLEL SIDES OF TRAPEZIUM
AREA=(1/2)*H{AE+DC+EF+FB)=(1/2)*H(AE+EF+FB+DC)=(1/2)*H(AB+DC)
AREA OF TRAPEZIUM =(1/2)(SUM OF PARALLEL SIDES )(PERPENDICULAR DISTANCE BETWEEN THE 2 PARALLEL SIDES)

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If each dimension of a box is quadrupled, how are the surface area and the volume affected?
SINCE AREA IS PROPORTIONAL TO SQUARE OF THE SIDE...AREA WILL INCREASE BY 4^2=16 TIMES
SINCE VOLUME IS PROPORTIONAL TO CUBE OF THE SIDE...VOLUME WILL INCREASE BY 4^3=64 TIMES

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A building casts a 52-foot shadow at the same time taylor casts a 4-foot shadow. Taylor and her shadow form a triangle that is similiar to the triangle formed by the building and it's shadow. If Taylor is 5 feet tall, what is the height h of the building?
SINCE IT IS GIVEN THAT TAYLOR AND THE BULDING AND THEIR SHADOWS FORM SIMILAR TRIANGLES...THEIR SIDES WILL BE IN PROPORTION.SO
SHADOW OF TAYLOR/HER HEIGHT=SHADOW OF BUILDING /ITS HT.
4/5=52/H
H=52*5/4=65'

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A right triangle has legs of lengths 7m and 9m.
HYPOTENUSE ^2=7^2+9^2=49+81=130
Which is the best estimate of the length of the hypotenuse? 10m,,11m,12m,13mSQUARES ARE....100,121,144,169...SO 11 IS NEAREST

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square root x+7 equals x+1 ..SQUARE BOTH SIDES
X+7=(X+1)^2=X^2+2X+1
X^2+2X+1-X-7=0
X^2+X-6=0
X^2+3X-2X-6=0
X(X+3)-2(X+3)=0
(X-2)(X+3)=0
X-2=0...THAT IS...X=2....OR......X+3=0...THAT IS X=-3

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I need this immediately for my H.W : The total surfac area of a cone is 616 sq cm.Its slant height i.e. length is equal to three times the radius of the base.Find its radius and slant height.
Please please please send the answer fast...
TSA=PI*R^2+PI*R*L=616
LET R BE THE RADIUS.SO SLANT HT.=L=3R
PI(R^2+R*3R)=616
PI*4R^2=616
R^2=616/(4PI)=616*7=/(4*22)=7*7
R=7
SLANT HT.=3*7=21

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F(N)=N^3-N=N(N^2-1)=N(N+1)(N-1)=(N-1)N(N+1)=PRODUCT OF 3 CONSECUTIVE INTEGERS.
SO ONE OF THEM IS DEFINITELY DIVISIBLE BY 2.
FURTHER WHEN WE DIVIDE A NUMBER WITH 3 WE CAN GET A REMAINDER OF 1 OR 2 OR 0 THAT IS DIVISIBLE BY 3.SO SINCE THIS IS A PRODUCT OF 3 CONSECUTIVE INTEGERS,ONE OF THEM HAS TO BE DIVISIBLE BY 3.
SO WE GET THAT THE PRODUCT IS DIVISIBLE BY 2 AS WELL AS 3.HENCE THE PRODUCT IS DIVISIBLE BY 6

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Lisa is flying her kite with 62 feet of string attached to the ground. She measures 5feet of string from the ground to a point 4 feet high. How high is her kite?
5 FEET LENGTH TO 4 FEET HEIGHT..THIS RATIO IS SAME FOR THE FULL LENGTH OF STRING
HENCE HEIGHT OF KITE = 62*4/5=248/5=49.6 FEET.

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Demonstrate that 2+6i is a solution to the equation x^2-40x+40=0
SUBSTITUTING FOR X ..
(2+6i)^2-40(2+6i)+40
=4+36i^2+2*2*6i-80-240i+40
=4-36+24i-80-240i+40=
=NO...IT IS NOT A SOLUTION AS IT IS NOT ZERO

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find the zeros (i.e. x-intercepts) of the following functions:
1) f(x)=-48x^2+3x^4
3X^2(X^2-16)=0
X^2=0...THAT IS X=0...OR
X^2-16=0...THAT IS X^2=16...THAT IS X=4 OR -4
2) f(x)=-4x^3-15x^2+18x
X(-4X^2-15X+18)=0
X=0...OR
4X^2+15X-18=0....USE QUADRATIC FORMULA`TO GET THE ROOTS

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find the third, fourth and fifth terms.
t sub(1) = 6; t sub(n) = t sub(n-1) + 4
T1=6
TN=T(N-1)+4
PUT N=1,2,3,4,5,6....
N=1............T1=6...GIVEN
N=2....T2=T1+4=6+4=10
N=3....T3=T2+4=10+4=14
N=4....T4=T3+4=14+4=18
N=5....T5=T4+4=18+4=22

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there are 3 times as many pears as oranges. If a group of children receives 5 organges each, there will be no oranges left over. If the same group of children receives 8 pears each, there will be 21 pears left over. How many children and organges are there?
LET ORANGES=X
PEARS ARE 3 TIMES =3X
LET CHILDREN=Y
IF WE GIVE 5 ORANGES PER CHILD WE NEED 5Y ORANGES.SO THIS =X
X=5Y........................I
IF WE GIVE 8 PEARS PER CHILD WE NEED 8Y PERARS...THERE ARE STILL 21 PEARS LEFT OUT AFTER THIS THAT IS
8Y+21=3X....................II
SUBSTITUTING FOR X FROM EQN.I..
8Y+21=3*5Y
15Y-8Y=21
7Y=21
Y=3
X==5Y=5*3=15..
SO CHILDREN =3
ORANGES=15
PEARS=3*15=45

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GOOD TO SEE YOUR ATTEMPT..SEE MY COMMENTS BELOW
-------------------------------------------------------------------------------
A girl bought some pencils, erasers, and paper clips at the stationary store. The pencils cost 10 cents each, the erasers cost 5 cents each, and the clips cost 2 for 1 cent. If she bought 100 items altogether at a total cost of $1, how many of each item did she buy?
I set up the variables as such:
X=the number of pencils
GOOD
Y=the number of erasers
GOOD
Z=the number of paper clips
GOOD
THen set up the problem
X+Y+Z= 100
VERY GOOD....CALL IT EQN.I
.10X + .05Y +.01 Z/2 = 1.00
EXCELLENT..YOU SHOWED GOOD MATURITY IN CONVERTING ALL O SAME CURRENCY THAT IS DOLLARS!!!KEEP IT UP!!BUT IN GENERAL WE ARE MORE COMFORTABLE WITH INTEGERS THAN DECIMALS OR FRACTIONS.SO BETTER WRITE IT AS
10X+5Y+Z/2=100....MULTIPLY BY 2 THROUGH OUT TO GET RID OF Z/2..
20X+10Y+Z=200...........CALL THIS EQN.II..............
I tried solving for X, but it still left two variables.
VERY CORRECT..THERE ARE 3 UNKNOWNS AND ONLY 2 EQNS..SO YOU CANNOT FIND UNIQUE SOLUTION UNLESS....????...LET US SEE...THERE IS A HIDDEN CONDITION..OBVIOUS BUT NOT MENTIONED...KNOW WHAT??IT IS CALLED "ONION PEEL" APPROACH.YOU HAVE TO PEEL THE ONION TO GET THE REAL STUFF!HERE YOU HAVE TO UNVEIL THE HIDDEN FACT........ IT IS THAT THE ANSWERS FOR X,Y AND Z SHOULD BE INTEGERS AND THEY CANNOT BE NEGATIVE!!!WE CANNOT BUY HALF PENCIL OR -2 ERASERS???IS IT NOT??LET US SEE WHETHER IT HELPS
LET US DO EQN.II-EQN.I....WE GET
20X+10Y+Z-X-Y-Z=200-100=100......
19X+9Y=100................EQN.III
NOW LET US PATIENTLY TRY OUR ARITHMATIC..PUTTING X=1,2,3,4,5..THAT IS ALL AS IF X IS MORE THAN 5 ,Y WILL BECOME NEGATIVE.
X=1....19*1+9Y=100....OR....9Y=100-19=81...OR...X=9..EUREKA! GOT IT!!BUT LET US SEE IF THERE ARE ANY OTHER ANSWERS.
X=2......9Y=100-38=62....NOT POSSIBLE
X=3......9Y=100-57=43....NOT POSSIBLE
X=4......9Y=100-76=24....NOT POSSIBLE
X=5......9Y=100-95=5.....NOT POSSIBLE
NO SO THE ONLY SOLUTION IS X=1,Y=9 AND Z=100-1-9=90...GOT IT??
So I tried to substitue for X then solve for Y. Needless to say I came up with bunk. My sister, nephew and I spent 3 hours on this problem and can't seem to solve it.
DONT DESPAIR THAT FAST...THE MORAL OF THE STORY IS THAT SOME TIMES WE MISS
THE OBVIOUS INFRONT OF US(IS IT BECAUSE OF PROXIMITY?OR IT IS TAKEN FOR GRANTED?)AND SEARCH FOR ILLUSIONS!!YOU HAVE DONE EXTREMELY WELL TO COME THIS FAR AND YOU ARE SURE TO SUCCEED WITH THIS APPROACH AND DETERMINATION.KEEP IT UP!
My homework is due next week and as it is the last 3 weeks of school I am trying to get it out of the way. ANy help you can give would be greatly appreciated.
Sincerely Jennifer Smith