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Exponential and Logarithmic Functions.
Compound Interest. If $500 is deposited in an account paying 8.5% annual interest, compounded semiannually, how long will it take for ethe account to increase to $800?
A=P(1+R/100)^N
800=500*(1+8.5/100)^N
(1.085)^N=800/500=1.6
NLOG(1.085)=LOG(1.6)
N=LOG(1.6)/LOG(1.085)=0.2041/0.0354=5.76 YRS...ROUNDED ...6 YEARS

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I am required to solve using the quadratic formula, if there are no real roots, then that can be an answer too. Thank you.
3=7x-4x(2) the two in parentheses is a power.
4X^2-7X+3=0...IF AX^2+BX+C=0...THEN

A=4...B=-7...C=3


x=(7+1)/6=8/8=1.....and
x=(7-1)/8=6/8=3/4

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The amount A in account after t years of an initial principle P invested at an annual rate r compounded continuiusly is given by A = Pe^rt where r is expressed as a decimal. What is the amount in the account if $500 is invested for 10 years at the rate of 5% compounded continuously? My answer is $750.00, thanks for checking my answer.
P=500 $
R=5%=5/100=0.05
T=10 YRS.
A=500*E^10*0.05=500*E^0.5=824.36

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Find the number
let x = log5(1/5).
hence 5^x=1/5=5^-1
hence x=-1
my answer is -1 but not sure...PERFECT 100% OK. thanks for checking.

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solve log a (8x+5)=log a (4x+29)
8x+5=4x+29
8x-4x=29-5=24
4x=24
x=6
My answer is 6, is that right. thanks for checking

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Given that log a (x)=3.58 and log a (y)=4.79. find loga(y/x).
log(y/x)=log(y)-log(x)=4.79-3.58=1.21
My answer is 8.37, is this right. ...no you did log(xy)which is log(x)+log(y)=8.37
thanks for looking at answer.

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DIVIDE X^3-64
=(x)^3-(4)^3=(x-4)(x^2+4x+16)...hence this divided by x-4
gives

x^2+4x+16

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Determine the average rate of change of the function between the given values of the variable.
f(t) = Sqrt(t); t = a, t= a+h
I came out with
f(a+h)= sqrt(a+h)
f(a)= Sqrt (a)
f(a+h)-f(a)=sqrt(a+h)-sqrt(a)
avg rate of change={f(a+h)-f(a)}/(a+h-a)={sqrt(a+h)-sqrt(a)}/h

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ok here is a problem that is driving me nuts i can't understand what to do. the question says " for the given functions f and g, find the indicated composition." and here is the problem f(x)=x^2+2x+3
g(x)=x^2+2x+2
f of g(-5)=f{g(-5)}
g(-5)=(-5)^2+2(-5)+2=25-10+2=17
f{g(-5)}=f(17)=17^2+2*17+3=289+34+3=326

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A linear function is given. (a) Find the average rate of change of the function between x = a and x = a + h. (b). Show that the average rate of change is the same as the slope of the line.
g(x) = -4x + 2
g(a+h)=-4(a+h)+2=-4a-4h+2
g(a)=-4a+2
g(a+h)-g(a)=-4h
avg.rate of change = {g(a+h)-g(a)}/(a+h-a)=-4h/h=-4
slope of the given line =-4
I came out with 4 for both answers....no -4 is the answer
Thank you so much!

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Find an exponential function of the form f(x)=Y= ba^x+c with y intercept 2,
THAT FOR X=0...Y=2
Y=2=B*A^0+C=B+C...............................I
horizontal asymptote=-2,
THAT IS AS X TENDS TO + OR - INFINITY Y TENDS TO -2
A^X WILL TEND TO ZERO AS X TENDS TO +INFINITY IF A<1 AND
A^X WILL TEND TO ZERO AS X TENDS TO -INFINITY IF A>1..HENCE WE TAKE THAT A^X WILL TEND TO ZERO AS X TENDS TO + OR - INFINITY TO GET
Y=-2=B*0+C...............................II
C=-2
FROM EQN.I....B-2=2..
B=4
that passes through the point P(1,4). My answer is f(x)= 2(1.5^x)-2. Thanks for the help.
Y=4=4*A^1-2..................III
4A=4+2=6
A=6/4=1.5
ANSWER IS F(X)=Y=4*1.5^X-2...GRAPH IS GIVEN BELOW

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SEE THE FOLLOWING AND TRY
-----------
SEE
\solve each of the follwing systems by substitution.
5x-2y=-5............I
y-5x=3......................II...Y=3+5X..........PUT IN EQN.I
5X-2(3+5X)=-5
5X-6-10X=-5
-5X=6-5=1
X=-1/5
Y=3+5X=3+5*(-1/5)=3-1=2
THESE ARE CONSISTENT INDEPENDENT EQNS.AND HENCE HAVE A UNIQUE SOLUTUIN NAMELY...
=-1/5 AND Y=2THE FOLLOWING AND TRY
Equations/44050: solve each of the following systems by graphing
3x-6y=9
x-2y=3
1 solutions
Answer 28988 by venugopalramana(2194) About Me on 2006-06-28 00:45:24 (Show Source):
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SEE THE FOLLOWING AND TRY
solve each of the following systems by sustitution.
8x-4y=16....................I
y=2x-4....................II....PUT THIS IN EQN.I
8X-4(2X-4)=16
8X-8X+16=16
16=16.....OK...
HENCE THE EQNS.ARE DEPENDENT ANE THERE ARE INFINITE SETS OF SOLUTIONS..ANY SET WITH
X=ANY VALUE......SAY'0'
AND Y=2X-4=2*0-4=-4 IS A SOLUTION SET.
YOU CAN GIVE ANY VALUE TO X,FIND CORRESPONDING Y AND THAT IS THE SOLUTION SET.
Equations/44051: Solve each of the following systems by addition. If a unique solution does not exist, state whether the system is inconsistent or dependent.
2x+3y=1
5x+3y=16
1 solutions
Answer 28987 by venugopalramana(2194) About Me on 2006-06-28 00:44:02 (Show Source):
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SEE THE FOLLOWING AND TRY
-------------------------------
solve each of the follwing systems by substitution.
5x-2y=-5............I
y-5x=3......................II...Y=3+5X..........PUT IN EQN.I
5X-2(3+5X)=-5
5X-6-10X=-5
-5X=6-5=1
X=-1/5
Y=3+5X=3+5*(-1/5)=3-1=2
THESE ARE CONSISTENT INDEPENDENT EQNS.AND HENCE HAVE A UNIQUE SOLUTUIN NAMELY...
=-1/5 AND Y=2
Equations/44052: slove each of the following systems by addition. If a unique solution does not exist, state whether the system is inconsistent or dependent.
x+5y=10
-2x-10y=-20
1 solutions
Answer 28986 by venugopalramana(2194) About Me on 2006-06-28 00:42:35 (Show Source):
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SEE THE FOLLOWING AND TRY
-----------------------------
solve each of the following systems by sustitution.
8x-4y=16....................I
y=2x-4....................II....PUT THIS IN EQN.I
8X-4(2X-4)=16
8X-8X+16=16
16=16.....OK...
HENCE THE EQNS.ARE DEPENDENT ANE THERE ARE INFINITE SETS OF SOLUTIONS..ANY SET WITH
X=ANY VALUE......SAY'0'
AND Y=2X-4=2*0-4=-4 IS A SOLUTION SET.
YOU CAN GIVE ANY VALUE TO X,FIND CORRESPONDING Y AND THAT IS THE SOLUTION SET.
Equations/44055: solve each of the follwing systems by substitution.
5x-2y=-5
y-5x=3
1 solutions
Answer 28985 by venugopalramana(2194) About Me on 2006-06-28 00:39:28 (Show Source):
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solve each of the follwing systems by substitution.
5x-2y=-5............I
y-5x=3......................II...Y=3+5X..........PUT IN EQN.I
5X-2(3+5X)=-5
5X-6-10X=-5
-5X=6-5=1
X=-1/5
Y=3+5X=3+5*(-1/5)=3-1=2
THESE ARE CONSISTENT INDEPENDENT EQNS.AND HENCE HAVE A UNIQUE SOLUTUIN NAMELY...
=-1/5 AND Y=2
Equations/44056: solve each of the following systems by sustitution.
8x-4y=16
y=2x-4
1 solutions
Answer 28984 by venugopalramana(2194) About Me on 2006-06-28 00:35:11 (Show Source):
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solve each of the following systems by sustitution.
8x-4y=16....................I
y=2x-4....................II....PUT THIS IN EQN.I
8X-4(2X-4)=16
8X-8X+16=16
16=16.....OK...
HENCE THE EQNS.ARE DEPENDENT ANE THERE ARE INFINITE SETS OF SOLUTIONS..ANY SET WITH
X=ANY VALUE......SAY'0'
AND Y=2X-4=2*0-4=-4 IS A SOLUTION SET.
YOU CAN GIVE ANY VALUE TO X,FIND CORRESPONDING Y AND THAT IS THE SOLUTION SET.
Equations/44057: solve each of the following systmes by substitution.
4x-12y=5
-x+3y=-1
1 solutions
Answer 28983 by venugopalramana(2194) About Me on 2006-06-28 00:31:42 (Show Source):
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solve each of the following systmes by substitution.
4x-12y=5................I
-x+3y=-1...............II.....X=3Y+1..PUT THIS IN EQN.I
4(3Y+1)-12Y=5
12Y+4-12Y=5
4=5...NOT POSSIBLE
HENCE EQNS. ARE INCONSISTENT AND THERE IS NO SOLUTION
Equations/44059: solve each of the following systems by using either addition or substitution. If a unique solution does not exist, state whether the system is dependent or inconsistent.
10x+2y=7
y=-5x+3
1 solutions
Answer 28982 by venugopalramana(2194) About Me on 2006-06-28 00:22:33 (Show Source):
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SEE THE FOLLOWING AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
------------------------------------------------------------------------------
IS the given system of equations consistent
MEANS WHETHER THERE IS ANY SET OF VALUES FOR X AND Y
WHICH SATISFIES BOTH THE GIVEN EQUATIONS.
and dependent,MEANS WHETHER ONE EQUATION CAN BE GOT
FROM THE OTHER BY SUITABLE AND PERMISSIBLE
TRANSFORMATION
consistent and independent,MEANS THAT ONE EQUATION
CANNOT BE OBTAINED FROM THE OTHER EQUATION BY BY ANY
SUITABLE AND PERMISSIBLE TRANSFORMATION .
or inconsistent.MEANS THERE ARE NO SET OF VALUES FOR X
AND Y WHICH SATISFY BOTH THE EQUATIONS.
3x-y=7.....I.......
6x-2Y=9..........II
what does these terms mean?NOW I THINK YOU KNOW WHAT
THE TERMS MEAN.THERE ARE TESTS FOR FINDING THESE
SITUATIONS.WE SHALL RESTRICT THE ANALYSIS TO YOUR
PROBLEM
WE FIND THAT ON L.H.S OF THE 2 EQNS.WE HAVE 3X-Y AND
6X-2Y WHICH ARE DIRECT MULTIPLES BY 2 .THAT IS WE GET
LHS OF EQN.2 BY MULTIPLYING THE EQN 1 WITH 2
WE GET EQN.I *2 ....6X-2Y=14
WHERE AS EQN.II SAYS 6X-2Y=9..SO IF WE TRY TO SATISFY
EQN.I ,EQN II WILL NOT BE SATISFIED AND IF WE TRY TO
SATISFY EQN II ,EQN.I WILL NOT BE SATISFIED.SO THERE
ARE NO VALUES OF X AND Y WHICH SATISFY BOTH
EQUATIONS.SO WE SAY THE GIVEN EQNS. ARE INCONSISTENT.
--------------------------------------------------------------------------------
Solve the following system by addition. If a unique
solution does not exist, state
whether the system is inconsistent or dependent.
x + 5y =
10...................................................I
-2x – 10y = -20.................................II
-2*EQN.I GIVES
-2X-10Y=-20....WHICH IS SAME AS EQN.II...
HENCE THEY ARE DEPENDENT EQUATIONS.
SO INEFFECT WE HAVE ONLY ONE EQN.TO SOLVE FOR 2
UNKOWNS.
HENCE WE SHALL HAVE INFINTE SOLUTIONS LIKE IN THIS
CASE
X=0....Y=2
X=10...Y=0...ETC..
IMPORTANT NOTE.
THEY HAVE INFINITE SOLUTIONS BUT NOT ANY AND EVERY
VALUE OF X AND Y.FOR EXAMPLE IF X=0 ,THEN Y HAS TO
EQUAL 2 ONLY NOT ANY OTHER VALUE.SO IT HAS INFINITE
SETS OF X AND Y PAIRS FOR ITS SOLUTION
--------------------------------------------------------------------------------------------

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SEE THE FOLLOWING AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
-----------------------
x - y + z = 2
-x + y + z = 4
-x + z = 2
augmented matrix is
1.....................-1.......................1.............................2
-1.....................1.......................1.............................4
-1.....................0.......................1.............................2
nr2=r2+r1 and nr3=r3+r1
1..................-1.................1.............2
1-1=0............1-1=0............1+1=2..........2+4=6
1-1=0...........-1+0=-1...........1+1=2..........2+2=4
nr2=r2/2....nr3=r3-r2
1....................-1...............1..............2
0/2=0.............0/2=0..............2/2=1........6/2=3
0-0=0............-1-0=-1..............2-2=0.......4-6=-2
nr1=r1-r3-r2.......nr2=2r2........nr3=-r3
1-0-0=1............-1+1-0=0.........1-0-1=0.......2+2-3=1
0.....................0...................1............3
0.....................1...................0.............2
exchange r2 and r3
1....0........0.......1
0....1........0.......2
0....0........1.......3
hence x=1......y=2.....z=3

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A trough, in the form of an open rectangular box, is 1.85 m long, 45 cm wide and 28 cm deep externally. If the trough is made of wood 2.5 cm thick, find, in cubic centimetres, the volume of wood used.
EXTERNAL DIMENSIONS
L=1.85 M =185 CM.
B=45 CM
H=28 CM
OPEN AT TOP.
EXTERNAL VOLUME OF BOX =LBH=185*45*28=233100 CC
INTERNAL DIMENSIONS
WOOD THICKNESS=2.5 CM
L'=185-2*2.5=180 CM
B'=45-2*2.5=40 CM
H'=28-1*2.5=25.5 CM...SINCE THE BOX IS OPEN AT TOP
INTERNAL VOLUME OF BOX =L'*B'*H'=180*40*25.5=183600 CC
HENCE VOLUME OF WOOD =233100-183600=49500 CC

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Im not sure what topic it would fall under. But here is my questions
the variable x is 4 less than the variable y?
THINK THIS WAY..WHICH IS LESS OR MORE..X IS LESS..Y IS MORE..SO TO GET SMALLER NUMBER X FROM A BIGGER NUMBER Y YOU NEED TO SUBTRACT 4 FROM BIGGER NUMBER.HENCE ANSWER IS X=Y-4
answer I think it is, x - 4 = y...NO...Y-4=X
next question
if the variable t is multiplied by 5, the result is equal to the variable r.
answer I think it is, t * 5 = r...OK..GOOD
next question
if 3 is subtracted from the product of the variables x and y the result is equal to 4 time the variable z
GO STEP BY STEP
PRODUCT OF X AND Y =XY
3 IS SUBTRACTE..FROM THE PRODUCT ..SO WE GET
XY-3
4 TIMES Z =4Z
BOTH ARE SAME..SO
XY-3=4Z
this one I really couldn't understand
next question
the quotient of the variable x with the variable y is 2 more than o.7 of the variable z.
this is another one I don't understand.

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A tin which is 12 cm long, 9 cm wide and 4 cm deep holds 120 g of tea. If 1 kg of the same tea is packed into a tin which has a 12 cm square base, how tall will the tin have to be ?
VOLUME OF TIN =L*B*H=12*9*4=432 CC
IT HOLDS 120 GM.OF TEA
TO HOLD 1 KG=1000 GMS..WE NEED
1000*432/120 =36000 CC
BASE AREA OF NEW TIN =A=12*12=144 SQ.CM.
HENCE HT NEEDED =V/A=36000/144=250 CM
Please help me to solve this. Thanks.

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OK..THE PROBLEM,WE FACE IN ANSWERING IS THAT WE ARE NOT AWARE OF YOUR COURSE OF STUDY AND BACKGROUND.THE SOLUTION I GAVE IS BY GAUSS JORDAN ELIMINATION METHOD USED IN MATRICES AS WANTED BY SOME STUDENTS.IF YOU ARE NOT TAUGHT THAT ,WE SHALL DO BY NORMAL ELIMINATION.
equations with 3 linear variables
6x-4y+5z=31.......................................I
5x+2y+2z=13...........................II
x+y+z=2.......................III
EQN.II - 2*EQN.III
5X+2Y+2Z-2X-2Y-2Z=13-2*2=9
3X=9
X=9/3=3
EQN.I+4*EQN.III
6X-4Y+5Z+4X+4Y+4Z=31+4*2=39
10X+9Z=39...BUT...X=3..SO
10*3+9Z=39
9Z=39-30=9
Z=9/9=1
PUTTING IN EQN.III
3+Y+1=2
Y=2-4=-2
HOPE YOU UNDERSTOOD.
------------------------------------------------------------------------
SEE THE FOLLOWING AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
YOU MAY MULTIPLY EQN.II WITH 2 AND ADD TO EQN.I TO ELIMINATE Y
x - y + z = 2
-x + y + z = 4
-x + z = 2
augmented matrix is
1.....................-1.......................1.............................2
-1.....................1.......................1.............................4
-1.....................0.......................1.............................2
nr2=r2+r1 and nr3=r3+r1
1..................-1.................1.............2
1-1=0............1-1=0............1+1=2..........2+4=6
1-1=0...........-1+0=-1...........1+1=2..........2+2=4
nr2=r2/2....nr3=r3-r2
1....................-1...............1..............2
0/2=0.............0/2=0..............2/2=1........6/2=3
0-0=0............-1-0=-1..............2-2=0.......4-6=-2
nr1=r1-r3-r2.......nr2=2r2........nr3=-r3
1-0-0=1............-1+1-0=0.........1-0-1=0.......2+2-3=1
0.....................0...................1............3
0.....................1...................0.............2
exchange r2 and r3
1....0........0.......1
0....1........0.......2
0....0........1.......3
he
nce x=1......y=2.....z=3

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SEE THE FOLLOWING AND TRY.IF TILL IN DIFFICULTY PLEASE COME BACK
YOU CAN DIVIDE EQN.I WITH 2 AND ADD TO EQN.III TO ELIMINATE X
-------------------------------------------------------------------
x - y + z = 2
-x + y + z = 4
-x + z = 2
augmented matrix is
1.....................-1.......................1.............................2
-1.....................1.......................1.............................4
-1.....................0.......................1.............................2
nr2=r2+r1 and nr3=r3+r1
1..................-1.................1.............2
1-1=0............1-1=0............1+1=2..........2+4=6
1-1=0...........-1+0=-1...........1+1=2..........2+2=4
nr2=r2/2....nr3=r3-r2
1....................-1...............1..............2
0/2=0.............0/2=0..............2/2=1........6/2=3
0-0=0............-1-0=-1..............2-2=0.......4-6=-2
nr1=r1-r3-r2.......nr2=2r2........nr3=-r3
1-0-0=1............-1+1-0=0.........1-0-1=0.......2+2-3=1
0.....................0...................1............3
0.....................1...................0.............2
exchange r2 and r3
1....0........0.......1
0....1........0.......2
0....0........1.......3
hence x=1......y=2.....z=3

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SEE THE FOLLOWING AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
x - y + z = 2
-x + y + z = 4
-x + z = 2
augmented matrix is
1.....................-1.......................1.............................2
-1.....................1.......................1.............................4
-1.....................0.......................1.............................2
nr2=r2+r1 and nr3=r3+r1
1..................-1.................1.............2
1-1=0............1-1=0............1+1=2..........2+4=6
1-1=0...........-1+0=-1...........1+1=2..........2+2=4
nr2=r2/2....nr3=r3-r2
1....................-1...............1..............2
0/2=0.............0/2=0..............2/2=1........6/2=3
0-0=0............-1-0=-1..............2-2=0.......4-6=-2
nr1=r1-r3-r2.......nr2=2r2........nr3=-r3
1-0-0=1............-1+1-0=0.........1-0-1=0.......2+2-3=1
0.....................0...................1............3
0.....................1...................0.............2
exchange r2 and r3
1....0........0.......1
0....1........0.......2
0....0........1.......3
hence x=1......y=2.....z=3

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SEE THE FOLLOWING EXAMPLE AND TRY.COME BACK IF STILL IN DIFFICULTY
--------------------------------------------------------------------------------
x - y + z = 2
-x + y + z = 4
-x + z = 2
augmented matrix is
1.....................-1.......................1.............................2
-1.....................1.......................1.............................4
-1.....................0.......................1.............................2
nr2=r2+r1 and nr3=r3+r1
1..................-1.................1.............2
1-1=0............1-1=0............1+1=2..........2+4=6
1-1=0...........-1+0=-1...........1+1=2..........2+2=4
nr2=r2/2....nr3=r3-r2
1....................-1...............1..............2
0/2=0.............0/2=0..............2/2=1........6/2=3
0-0=0............-1-0=-1..............2-2=0.......4-6=-2
nr1=r1-r3-r2.......nr2=2r2........nr3=-r3
1-0-0=1............-1+1-0=0.........1-0-1=0.......2+2-3=1
0.....................0...................1............3
0.....................1...................0.............2
exchange r2 and r3
1....0........0.......1
0....1........0.......2
0....0........1.......3
hence x=1......y=2.....z=3

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x - y + z = 2
-x + y + z = 4
-x + z = 2
augmented matrix is
1.....................-1.......................1.............................2
-1.....................1.......................1.............................4
-1.....................0.......................1.............................2
nr2=r2+r1 and nr3=r3+r1
1..................-1.................1.............2
1-1=0............1-1=0............1+1=2..........2+4=6
1-1=0...........-1+0=-1...........1+1=2..........2+2=4
nr2=r2/2....nr3=r3-r2
1....................-1...............1..............2
0/2=0.............0/2=0..............2/2=1........6/2=3
0-0=0............-1-0=-1..............2-2=0.......4-6=-2
nr1=r1-r3-r2.......nr2=2r2........nr3=-r3
1-0-0=1............-1+1-0=0.........1-0-1=0.......2+2-3=1
0.....................0...................1............3
0.....................1...................0.............2
exchange r2 and r3
1....0........0.......1
0....1........0.......2
0....0........1.......3
hence x=1......y=2.....z=3

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or sets H and K, we define the inersection H intersect k by
H intersect k = {x|x in H and x in k}
show that if H <= G and k <= G, then H intersect K <= g. ( remember: <= denotes "is a sub group of," not "is a subgroup of.")
CASE 1....H=G AND K=G....
THEN IT IMPLIES THAT IF H1 IS ANY ELEMENT OF H THEN IT IS ALSO AN ELEMENT OF G AND HENCE K.
SO H1 WILL BE AN ELEMENT OF H INTERSECTION K
SO H I.S. K IS A SUBSET OF G.................I
SIMILARLY ,WE CAN PROVE THAT IF G1 IS ANY ELEMENT OF G THEN IT IMPLIES IT IS ALSO AN ELEMENT OF H ,K AND HENCE H I.S. K
SO G IS A SUB SET OF H I.S. K.............II
FROM I AND II H I.S. K=G
CASE 2 .....H < G......AND K < G
HENCE IF H1 IS AN ELEMENT OF H IT IMPLIES IT IS AN ELEMENT OF G.
BUT IT MAY OR MAY NOT BE AN ELEMENT OF K
IF IT IS THEN H1 IS AN ELEMENT OF H I.S.K...HENCE H I.S.K IS A SUB SET OF G
IF IT IS NOT THEN IT IS NOT AN ELEMENT OF H I.S.K..SO OK.
NOW THE OTHER WAY ROUND..
SINCE H THERE WILL BE AT LEAST ONE ELEMENT OF G SAY G1 WHICH IS NOT AN ELEMENT OF H.
HENCE G1 WILL NOT BE AN ELEMENT OF H I.S. K
HENCE H I.S.K IS A SUBSET OF G
SIMILARLY WE CAN ARGUE FOR K....TO SHOW H I.S.K IS A SUBSET OF G

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elimination method
2x/3 + 3y/4 =11/12..........I
x/3 + 7y/18 =1/2..........II
wouldn't 12 be the common denominator in the first equation
YES SO MULTIPLY WITH 12...EQN.I
8X+9Y=11...............III
SIMILARLY MULTIPLY EQN.II WITH 18
6X+7Y=9..............IV
7*EQNIII-9*EQN.IV
56X+63Y-72X-63Y=77-81
-16X=-4
X=1/4...PUT IN EQN.III
8*1/4+9Y=11
9Y=11-2=9
Y=1

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Write the equation of the line that passes through point A..(X1,Y1).. (-2,3) with a slope of ...M=-4.
EQN IS
Y-Y1=M(X-X1)
Y-3=-4(X+2)
Y+4X=3-8=-5
Y+4X+5=0

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2x-4y=7......................I
4x+2y=9.................II
EQN.I+EQN.II*2
2X-4Y+8X+4Y=7+18=25
10X=25
X=25/10=2.5
4*2.5+2Y=9
2Y=9-10=-1
Y=-1/2=-0.5

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Solve the system by graphing.
3x+y=6
3x-y=0
X.................0..............1...........................5........................
Y=6-3X..........=6-3**0=3.....=6-3*1=3....................=6-3*5=-9.................
Y=3X............=3*0=0........=3*1=3......................=3*5=15

WE SEE THAT THE 2 LINES INTERSECT AT X=1 AND Y=3 WHICH IS THE ANSWER

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Fact: 1 a divided by c = a
A/C=A....
CASE 1.A=0
CASE 2.IF A IS NOT ZERO THEN C=A/A=1
Fact: 2 a plus b = a
A+B=A.......B=0
Fact: 3 abc = 1
CASE 1.
ABC=0*0*C=0..WHATEVER MAY BE C
HENCE ABC CANNOT EQUAL 1
CASE 2.
IF A IS NOT EQUAL TO ZERO THEN
ABC=A*0*1=0...WHATEVER MAY BE A
HENCE ABC CANNOT EQUAL 1
If fact 1 and 2 are correct why is fact 3 wrong?
What are the value of each of the variables in fact 1 and 2?

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IF a+b=1,a>0,b>0,then show 1/a+1/b>=4
1/A + 1/B = (A+B)/AB
TST (A+B)/AB>=4
SINCE A>0 AND B>0 ,AB>0..SO WE CAN MULTIPLY BOTH SIDES WITH AB
TST A+B>=4AB
TST A+B-4AB>=0
A+B=1
B=1-A
A+B-4AB=1-4A(1-A)=1-4A+4A^2=(2A-1)^2..THIS BEING A PERFECT SQUARE IS ALWAYS >=0...
PROVED

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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 expressed in dollars would Mr. Brown have to put on the 32nd square?
Answer:
$21474836.48

Show work in this space
an = .01*2n-1
a32 = .01*232-1
a32 = .01*231
a32 = .01 * 2147483648
a32 = 21474836.48

b) How much money expressed in dollars would the traveling salesman receive in total if the checkerboard only had 32 squares?
Answer:

$42949673.00


Show work in this space

Sum an = .01 * (2n - 1) / ( 2 - 1 )
Sum a32 = .01 * (232 - 1) / ( 2 - 1 )
Sum a32 = 42949673.00
c) Calculate the amount of money necessary to fill the whole checkerboard (64 squares). How money expressed in dollars would the farmer need to give the salesman?
Answer:

$184467000000000000.00

Show work in this space

Sum an = .01 * (2n - 1) / ( 2 - 1 )
Sum a64 = .01 * (264 - 1) / ( 2 - 1 )
Sum a64 = 184467000000000000.00

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THE NUMBERS GIVEN IN THIS PROBLEM ALL LEAD TO FRACTIONS INVOLVING HRS,MTS AND SECS.ARE YOU SURE YOU GOT THE NUMBERS RIGHT?IS DISTANCE BETWEN P AND Q =34 KM OR 35 KM?
Two towns P and Q are 35 km apart. Peter starts cycling from P towards Q at 12 p.m. at 20 km/h until he is 16 km from P, when he changes speed so that he arrives at Q at 2 p.m. John leaves Q at 12:30 p.m. and cycles to P at a constant speed 26 km/h, find
a)the time when Peter and John meet.
b)the distance that they will meet
PETER TRAVELS STARTING FROM P AT 12=00 PM GOING AT 20 KMPH FOR 16 KM
TIME TAKEN =16/20=0.8 HRS=0.8*60=48 MTS….THAT IS AT 12 =48
BALANCE DISTANCE TO Q =35-16=19 KM.
TIME TAKEN TO REACH =2=00 PM THAT IS 14=00 HRS
TRAVEL TIME IS 14=00 - 12 =48 =1 HR12 MTS. = 1 +12/60 HRS =6/5 HRS.
PETERS SPEED DURING THIS TRAVEL =19/(6/5)=5*19/6=95/6 KMPH
JOHN LEAVES Q AT 12=30 PM.
TILL 12=48 HE TRAVELS FOR 12=48 -12=00=18 MTS=18/60 HRS =0.3 HRS.
DISTANCE TRAVELLED BY JOHN DURING THIS PERIOD AT 26 KMPH=26*0.3=7.8 KM
HENCE DISTANCE OF SEPERATION BETWEEN PETER AND JOHN AT 12 = 48 IS (35-16)-7.8=11.2 KM
THEIR RELATIVE SPEED WHILE TRAVELLING IN OPPOSITE DIRECTIONS =95/6 +26=(95+156)/6=251/6 KMPH
TIME TAKEN TO MEET =11.2/(251/6)=67.2/251 HRS = 16.06 MTS
HENCE PETER AND JOHN MEET AT 12=48 + 16.06 MTS..THAT IS AT 13 HRS.04 MTS.36 SEC
THE DISTANCE AT WHICH THEY MEET =16.06*95/6=4.24 KM THAT IS AT 16+4.24=18.24 KM FROM P.

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(x+8)/5 - (x-8)/7 = 2..MULTIPLY THROUGH OUT WITH 7*5=35
7(X+8)-5(X-8)=2*35=70
7X+56-5X+40=70
2X=70-40-56=-26
X=-26/2=-13