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AX^2+BX+C=0
ROOTS ARE IN RATIO M:N...LET THE ROOTS BE MY AND NY
SUM OF ROOTS =MY+NY=Y(M+N)=-B/A
Y=-B/A(M+N)
PRODUCT OF ROOTS=MNY^2=C/A
MN*{-B/A(M+N)}^2=C/A
MNB^2=CA^2(M+N)^2/A=CA(M+N)^2

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I PLACE...2 LETTERS.....2 WAYS
II PLACE..2 LETTERS.....2 WAYS
III PLACE.10 NUMBERS....10 WAYS
IV PLACE..10.NUMBERS....10 WAYS
V PLACE ..10.NUMBERS....10 WAYS
VI PLACE..10 NUMBERS....10 WAYS
------------------------------------
TOTAL POSSIBILITIES = 2*2*10*10*10*10=40000

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I need help using special right triangles to find the exact values of sines and cosines for each angle. For example: find the values of sine and cosine for the angle -135 degrees.
FIRST LEARN THE TRIGNOMETRIC RATIOS FOR STANDARD TRIANGLES.
THEN LEARN WHICH QUADRANT THE ANGLE IS
USE ALL SILVER TEA CUPS CONVENTION ,TO KNOW THE SIGN OF THE TRIGNOMETRIC RATIO.
IF ANGLE IS IN I QUADRANT ...THEN...ALL ARE +VE
...............II.......................SINE ONLY IS +VE
...............III......................TANGENT ONLY IS +VE
................IV.......................COSINE ONLY IS +VE.
NOW WRITE GIVEN ANGLE AS 0+X..OR...0-X...OR....180+ OR - X ....OR....360+ OR -X ETC....
YOUR EXAMPLE OF -135
SINE (0-135)=-SIN(135)..........0-X..MEANS IV Q...SINE IS -VE.
-1*SIN(135)=-SIN(180-45)=-SIN(45).....180-X MEANS II Q. SINE IS POSITIVE.
-1*SIN45=-1/SQRT(2)
GET BY HEART THE STANDARD TABLES FOR 0,30,45,60,90.
SIMILARLY COS (0-135)=COS(135)=COS(180-45)=-COS(45)=-1/SQRT2









0

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THIS IS NOT CORRECT.THERE IS SOME THING WRONG IN YOUR HYPOTHESIS.DIAGONALS OF A PARALLELOGRAM DEFINITELY BISECT THEIR ANGLES ,WHETHER IT IS A RHOMBUS OR NOT.THE DIAGONALS NEED TO BE PERPENDICULAR TO EACH OTHER FOR IT TO BE A RHOMBUS

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Hello! I've tried to approach this from a substiution standpoint but I'm at a loss :
Find the number of solutions for the system
(the following equations are all together in one bracket)
{f + g + h = 4.........................................I
f = 5 - h - g.....OR.....F+G+H=5.....................II
g + h = 7 - f.....OR.....F+G+H=7
If you could help me on how I should approach this problem I would be very grateful!! Thanks!!!!!
THESE 3 ARE IDENTICAL ON LHS BUT HAVE DIFFERENT VALUES ON RHS.HENCE THEY ARTE INCONSISTENT AND NO SOLUTION EXISTS WHICH CAN SATISFY ALL 3 EQNS.

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the question is write an equation for the perpendicular bisector of the line segment joining the two points?
the pionts are A-(0,0).............AND B-(-8,-10)
how would you solve this
MID POINT OF AB ...SAY....C...IS (0-8)/2,(0-10)/2=(-4,-5)
SLOPE OF AB =(0+10)/(0+8)=10/8=5/4
SLOPE OF PERPENDICULAR TO AB IS -1/(5/4)=-4/5
PERPENDICULAR BISECTOR IS PERPENDICULAR TO AB AND PASSES THROUGH ITS MID POINT C.HENCE ITS EQN. IS
Y+5=(-4/5)(X+4)
5Y+25=-4X-16
4X+5Y+41=0

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SEC(225)=SEC(180+45)=-SEC(45)=-SQRT(2)=-1.414

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Find the sum of the fcllowing geometric series.
8 + 12 +..... + 60.75
A=8
R=12/8=1.5
TN=AR^(N-1)=8*(3/2)^(N-1)=60.75
N=6 BY INSPECTION SINCE 8*3*3*3*3*3/(2*2*2*2*2)=60.75
SN=S6=A*{R^N - 1)/(R-1)=(60.75*1.5-8)/0.5=166.25

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A geometric sequenee has first term -4 and a constant ratio of 5.
a) Write the first six terms of this sequence.
A=-4...R=5
T1=A=-4
T2=AR=-4*5=-20
T3=A*R^2=-4*5^2=-100
T4=A*R^3=-500
T5=-2500
T6=--12500
_____,______,_____,_______,_______,_______
b) Find the sum of the first eight terms of this sequence

S8=A*{R^8-1)}/(R-1)=-4*(5^8-1)/(5-1)=-(5^8-1)

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i wasn't sure where to go with this one but here it is: Point A is equidistant from P and Q. The midpoint, M, of line PQ is 12 units form A and 7 1/2 units from the midpoint(SAY N ) of line AQ THAT IS AN =7.5. Find the length of line PQ. (you have to draw your own picture, which im not very good at, i keep getting weird square roots for PQ, please help!)
IN TRIANGLE APQ,WE HAVE
M IS MID POINT OF PQ ...GIVEN
N IS MID POINT OF AQ...GIVEN.
HENCE
MN=AP/2...OR...AP =2*MN=2*7.5=15
IN TRIANGLES APM AND APQ
PM=MQ...GIVEN
AM=AM
AP=AQ...GIVEN
HENCE THEY ARE CONGRUENT AND HENCE ANGLE AMP=ANGLE AMQ =180/2=90
HENCE IN THIS RIGHT TRIANGLE
AP^2=AM^2+MP^2
MP^2=AP^2-AM^2=15^2-12^2=225-144=81
MP=9
HENCE PQ =2*MP=2*9=18

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x+y=10...................................i
x*y=20 find the value of x,y
(x-y)^2=(x+y)^2-4y=10^2-4*20=100-80=20
x-y=sqrt(20)=2sqrt(5)....................................ii
i+ii
2x=10+2sqrt(5)
x=5+sqrt(5)
i-ii
2y=10-2sqrt(5)
y=5-sqrt(5)

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I^(4N-3)=I^4N/I^3={(I^2)^2N}/{(I^2)(I)}={(-1)^2}^N/{(-1)(I)=(1)^N/(-I)=-1/I
=(-1)(I)/{(I)(I)}=(-1)(I)/(-1)=I=0+1*I

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yes perpendicular..b is ok.
product of slopes =-5*1/5=-1

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FORV EACH QUESTION HE CAN ANSWER IN 4 WAYS..THERTE ARE 5 QUESTIONS
SO THERE ARE 4*4*4*4*4=1024WAYS

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Find the two remaining vertices of square ABCD that has A=(-2,1), B=(6,-5) and a vertex in quadrant III.
AB=SQRT((6+2)^2+(5+1)^2)=10
EQN. OF AB IS
(Y-1)=[(1+5)/(-6-2)](X+2)}
4y-4=-3x-6
3X+4Y=-2
ITS DISTANCE FROM ORIGIN IS 2/5
CD IS PARALLEL TO AB AND IS AT 10 DIUSTANCEB FROM AB OR 10+2/5 =52/5 DISTANCE FROM ORIGIN.
HENCE ITS EQN,IS
3X+4Y=-52................................I..OR...3H+4K=-52 AS C H,K LIES ON IT.
SLOPE OF AB =-3/4
SLOPE OF BC =4/3=
LET C BE H,K
SLOPE OF BC = (K+5)/(H-6)=4/3
3K+15=4H-24
4H-3K=39...........................II
SOLVING I AND II
H=0...K=-13.SO C IS (0,-13)
SIMILARLY FOR D WE CAN FIND
SLOPE OF AD =4/3=
LET D BE H,K
SLOPE OF AD = (K-1)/(H+2)=4/3
3K-3=4H+8
4H-3K=-11...........................III
SOLVING I AND III
H=-8 AND K=-7

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IN A CYCLIC QUADRILATERAL SUM OF OPPOSITE ANGLES IS 180
HENCE
left: 26y on top 21y on bottom
Right: 3x on top 2x on bottom
thanks.
26Y+2X=180......X+13Y=90...............I
3X+21Y=180......X+7Y=60..............II
EQN.I-EQN.II
13Y-7Y=90-60=30
6Y=30...Y=5
X+7*5=60
X=60-35=25
HENCE THE ANGLES ARE 26*5=130
21*5=105
3*25=75
2*25=50

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The problem is : . We have to factor each expression as a trinomial. I found the factors of three : 1 & 3 , but it says in the book that one of the factor pairs has to equal "b" which is supposed to be 5. This doesn't make sense to me.
LET ME EXPLAIN YOU IN DETAIL
1.TAKE COEFFICIENT OF A^2 =2 HERE
2. TAKE CONSTANT TERM = 3 HERE.
3.FIND THEIR PRODUCT =2*3=6
4.KEEP IN MIND ALL FACTORS FOR THE ABOVE PRODUCT ...
1*6,-1*-6,2*3,-2*-3.
5.CHOOSE THAT PAIR WHOSE SUM IS EQUAL TO COEFFICIENT OF A =+5 HERE
6.2*3=6 SATISFY THIS REQUIREMENT SINCE 2+3=5
7.NOW SPLIT THE X TERM AS ABOVE ,TAKE COMMON FACTORS AND FACTORISE AS SHOWN BELOW
2A^2+2A+3A+3=2A(A+1)+3(A+1)=(A+1)(2A+3)
GOT IT??

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The volume - V of an ideal gas varies directly with the temperature T and inversely with the Pressure P.
V=K*T/P...WHERE K IS A CONSTANT
A cylinder contains 15 liters of oxygen at a temperature of 240 K and a pressure of 80 atmospheres.
15=K*240/80=3K
K=15/3=5 LITRE.ATM./DEG.K
V=5T/P
If a piston is lowered into the cylinder decreasing the volume occupied by the gas to 10 liters and the temperature remains constant, what is the pressure?
10=5*240/P
P=5*240/10=120 ATM.

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I am to solve this problem finding all real or imaginary solutions. I believe I solved it, but I question whether it is correct.
The question is: 6x=19x+25/x+1
My work:
6x(x+1)=6x^2+6x...OK
Then I put it into the equation form of -6x^2-25x+25=0
NO..YOU MISSED 19X
6X^2+6X=19X+25
6X^2-13X-25=0




ARE THE 2 SOLUTIONS.
I used the b^2-4(a)(c) formula which =625-4(-6)(25)
The whole thing turned out to be 625-600=25
Because the outcome is positive, does it not have to have two solutions? I am unsure what the second one is?
Thank you

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Find the slope of the sides of rectangle ABCD having vertices A=(1,4)=(X1,Y1) and
B=(-5,3)=(X2,Y2)
SLOPE OF AB =(Y2-Y1)/(X2-X1)=(3-4)/(-5-1)=-1/-6=1/6
IN A RECTANGLE OPPOSITE SIDES ARE PARALLEL AND ADJACENT SIDES ARE PERPENDICULAR.HENCE
SLOPE OF CD=1/6
SLOPE BC=SLOPE OF AD =-1/(1/6)=-6

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Amanda has 400 feet of lumber to frame a
rectangular patio (the perimeter of a rectangle is 2
times length plus 2 times width). She wants to
maximize the area of her patio (area of a rectangle is
length times width). What should the dimensions of the
patio be, and show how the maximum area of the patio
is calculated from the algebraic equation.
Answer:
IF L AND B ARE DIMENSIONS WE HAVE
PERIMETER=2(L+B)=400.....OR.....L+B=200..OR......B=200-L.................I
AREA=A
=LB=L(200-L)=200L-L^2=-{L^2-200L}=-{(L^2)-2(L)(100)+100^2-100^2}
A=10000-(L-100)^2
(L-100)^2 BEING PERFECT SQUARE,ITS MINIMUM VALUE IS
ZERO.
HENCE AREA IS MAXIMUM WHEN L-100 IS ZERO,OR WHEN L=100
AND THEN THE MAXIMUM AREA WOULD BE
A-MAX.=10000-0=10000
DIMENSIONS ARE 100*100

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Suppose you throw a baseball straight up at a
velocity of 32 feet per second. A function can be
created by expressing distance above the ground, s, as
a function of time, t. This function is s = -16t2 +
v0t + s0
� 16 represents �g, the gravitational
pull due to
gravity (measured in feet per second 2).
� v0 is the initial velocity (how hard do you
throw
the object, measured in feet per second).
� s0 is the initial distance above ground (in
feet).
If you are standing on the ground, then s0 = 0.
a) What is the function that describes this problem?
Answer:
THIS IS WHAT IS GIVEN ABOVE
S=-16*T^2+V0*T+S0
.....................................I
WITH THE ELABORATIONS GIVEN ABOVE.
b) The ball will be how high above the ground after 1
second?
Answer:
Show work in this space.
WE ARE GIVEN
V0=32 FPS
S0=0...BALL THROWN FROM GROUND AND DISTANCE MEASURED
FROM GROUND
T=1 SEC...
S=?
HENCE SUBSTITUTING IN EQN.I,WE GET
S=-16*1^2+32*1+0=-16+32=16 FT.
c) How long will it take to hit the ground?
Answer:
Show work in this space.
WE ARE GIVEN HERE
V0=32
S0=0
S=0
T=?
HENCE SUBSTITUTING IN EQN.I,WE GET
0=-16T^2+32T+0
16T(-T+2)=0
T=0...OR....-T+2=0....OR....T=2 SEC.
SINCE T=0 REPRESENTS THE INTIAL POSITION
T=2 SEC.IS THE ANSWER WHEN IT HITS THE GROUND AGAIN
AFTER GOING UP AND FALLING DOWN.
d) What is the maximum height of the ball?
Answer:
Show work in this space.
HERE WE ARE GIVEN
V0=32
S0=0....AND THERE ARE 2 ASPECTS TO TAKE NOTE OF.
1.THE BALL GOES UP FIRST SLOWING DOWN AS IT GOES UP
DUE TO EARTHS GRAVITATION...NOTE -16 GRAVITATIONAL
ACCELERATIONED MENTIONED IN THE PROBLEM. IT GOES UP
TILL ITS VELOCITY BECOMES ZERO FROM THE INITIAL
VELOCITY OF THROW OF 32 FPS.AND THEN FALLS DOWN
REGAINING THE SAME VELOCITY WHEN IT HITS THE GROUND AS
PER PHYSICS LAWS,NEGLECTING AIR DRAG.
2.IT IS ALSO PROVED IN PHYSICS IN SUCH CASES,THAT
i)THE TIME OF ASCENT = THE TIME OF DESCENT
ii)THE DISTANCE TRAVELLED UPWARD=THE DISTANCE TRVELLED
DOWN WARD
iii)THE VELOCITY WITH WHICH IT IS THROWN UP= THE
VELOCITY WITH WHICH IT HITS THE GROUND.
HENCE USING i)PRINCIPLE ,SINCE TOTAL TIME OF TRAVEL AS
PER C) AS WE ALCULATED IS 2 SEC.,TIME OF ASCENT =1
SEC.HENCE MAXIMUM HEIGHT REACHED IS DISTANCE TRAVELLED
IN 1 SEC=16 FT.AS SHOWN IN B).
---------
ANOTHER METHOD IS TO FIND WHEN THE VELOCITY WILL
BECOME ZERO AS IT GOES UP AND FIND THE DISTANCE
TRAVELLED IN THAT TIME.
THE FORMULA FOR THAT IS OBTAINED BY DIFFERENTIATING
THE GIVEN EQN.II
AS FOLLOWS
DS/DT=VELOCITY=V=-32*T+V0...WHERE V IS THE FINAL
VELOCITY AFTER T SECS.
SINCE AT MAXMUM HEIGHT FINAL VELOCITY =0 (AS THEN ONLY
IT FALLS BACK TOWARDS GROUND.)
HENCE 0=-32T+32
32T=32
T=1...AS WE USED EARLIER NOW WE FIND S=16 FT.AS
BEFORE.
HOPE YOU UNDERSTOOD.

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Suppose you throw a baseball straight up at a
velocity of 32 feet per second. A function can be
created by expressing distance above the ground, s, as
a function of time, t. This function is s = -16t2 +
v0t + s0
� 16 represents �g, the gravitational pull due to
gravity (measured in feet per second 2).
� v0 is the initial velocity (how hard do you throw
the object, measured in feet per second).
� s0 is the initial distance above ground (in feet).
If you are standing on the ground, then s0 = 0.
a) What is the function that describes this problem?
Answer:
THIS IS WHAT IS GIVEN ABOVE
S=-16*T^2+V0*T+S0
.....................................I
WITH THE ELABORATIONS GIVEN ABOVE.
b) The ball will be how high above the ground after 1
second?
Answer:
Show work in this space.
WE ARE GIVEN
V0=32 FPS
S0=0...BALL THROWN FROM GROUND AND DISTANCE MEASURED
FROM GROUND
T=1 SEC...
S=?
HENCE SUBSTITUTING IN EQN.I,WE GET
S=-16*1^2+32*1+0=-16+32=16 FT.
c) How long will it take to hit the ground?
Answer:
Show work in this space.
WE ARE GIVEN HERE
V0=32
S0=0
S=0
T=?
HENCE SUBSTITUTING IN EQN.I,WE GET
0=-16T^2+32T+0
16T(-T+2)=0
T=0...OR....-T+2=0....OR....T=2 SEC.
SINCE T=0 REPRESENTS THE INTIAL POSITION
T=2 SEC.IS THE ANSWER WHEN IT HITS THE GROUND AGAIN
AFTER GOING UP AND FALLING DOWN.
d) What is the maximum height of the ball?
Answer:
Show work in this space.
HERE WE ARE GIVEN
V0=32
S0=0....AND THERE ARE 2 ASPECTS TO TAKE NOTE OF.
1.THE BALL GOES UP FIRST SLOWING DOWN AS IT GOES UP
DUE TO EARTHS GRAVITATION...NOTE -16 GRAVITATIONAL
ACCELERATIONED MENTIONED IN THE PROBLEM. IT GOES UP
TILL ITS VELOCITY BECOMES ZERO FROM THE INITIAL
VELOCITY OF THROW OF 32 FPS.AND THEN FALLS DOWN
REGAINING THE SAME VELOCITY WHEN IT HITS THE GROUND AS
PER PHYSICS LAWS,NEGLECTING AIR DRAG.
2.IT IS ALSO PROVED IN PHYSICS IN SUCH CASES,THAT
i)THE TIME OF ASCENT = THE TIME OF DESCENT
ii)THE DISTANCE TRAVELLED UPWARD=THE DISTANCE TRVELLED
DOWN WARD
iii)THE VELOCITY WITH WHICH IT IS THROWN UP= THE
VELOCITY WITH WHICH IT HITS THE GROUND.
HENCE USING i)PRINCIPLE ,SINCE TOTAL TIME OF TRAVEL AS
PER C) AS WE ALCULATED IS 2 SEC.,TIME OF ASCENT =1
SEC.HENCE MAXIMUM HEIGHT REACHED IS DISTANCE TRAVELLED
IN 1 SEC=16 FT.AS SHOWN IN B).
---------
ANOTHER METHOD IS TO FIND WHEN THE VELOCITY WILL
BECOME ZERO AS IT GOES UP AND FIND THE DISTANCE
TRAVELLED IN THAT TIME.
THE FORMULA FOR THAT IS OBTAINED BY DIFFERENTIATING
THE GIVEN EQN.II
AS FOLLOWS
DS/DT=VELOCITY=V=-32*T+V0...WHERE V IS THE FINAL
VELOCITY AFTER T SECS.
SINCE AT MAXMUM HEIGHT FINAL VELOCITY =0 (AS THEN ONLY
IT FALLS BACK TOWARDS GROUND.)
HENCE 0=-32T+32
32T=32
T=1...AS WE USED EARLIER NOW WE FIND S=16 FT.AS
BEFORE.
HOPE YOU UNDERSTOOD.

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factor by grouping with complete solution
1. x² - y² - x –y
=(X+Y)(X-Y)-1(X+Y)=(X+Y)(X-Y-1)
2. x³ - y³ - x² + 2xy – y²
(X-Y)(X^2+XY+Y^2)-(X-Y)^2
=(X-Y)(X^2+XY+Y^2-X+Y)

3. x² + xy – 2y² - x³ + y³
{X^2+2XY-XY-2Y^2}-(X^3-Y^3)
={X(X+2Y)-Y(X+2Y)}-(X-Y)(X^2+XY+Y^2)
(X-Y)(X+2Y)-(X-Y)(X^2+XY+Y^2)
(X-Y)(X+2Y-X^2-XY-Y^2)

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Find the distance between the points: (4, 1) and (-2, 5)
Thank you in advance!
DISTANCE BETWEEN 2 POINTS (X1,Y1) AND (X2,Y2) IS GIVEN BY
D=SQRT{(X2-X1)^2+(Y2-Y1)^2}=SQRT{(4+2)^2+(1-5)^2}=SQRT(36+16)=SQRT(52)=2SQRT(13)

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Write the first four terms of the sequence an=64(1/4)^n, for n=0, 1, 2, 3,
How do I finish problem?
PUT N=0,1,2,3 IN THE ABOVE..WE GET
FOR N=0................A0=64(1/4)^0=64
FOR N=1..........A1=64(1/4)^1=64/4=16
FOR N=2..........A2=64(1/4)^2=64/16=4
FOR N=3..........A3=64(1/4)^3=64/64=1

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Find the general term aVn of a sequence whose first four terms are giving: 5,9,13,17,...for n=1,2,3,4,... Please help me with this problem.
WE FIND
9-5=13-9=17-13=4...SO THIS IS A.P
A=5 AND D=4
GENERAL TERM
TN=A+(N-1)D=5+(N-1)4=5+4N-4=4N+1

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PLEASE CONFIRM WHETHER AND V ARE DIFFERENT VARIABLES OR NOT?IF DIFFERENT WE NEED ADDITIONAL DATA
Find aV8 if aV1 =-21,
IF AV1 IS ONE VARIABLE...THEN AV2=(AV1)^-3=(-21)^-3=1/(-21)^3..ETC
aVn=aVn-1^-3; for n=2,3,4,... I have no idea how to do any of these problems! This is going to be so very hard.

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PLEASE CONFIRM WHETHER AND V ARE DIFFERENT VARIABLES OR NOT?IF DIFFERENT WE NEED ADDITIONAL DATA
Find aV8 if aV1 =-21,
IF AV1 IS ONE VARIABLE...THEN AV2=(AV1)^-3=(-21)^-3=1/(-21)^3..ETC
aVn=aVn-1^-3; for n=2,3,4,... I have no idea how to do any of these problems! This is going to be so very hard.

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Find aV6 for aVn=8(1/2)^n, n=1,2,3,4,... Please help me to solve this problem.
IT APPEARS A=8 AND VN=(1/2)^N...IF SO WE HAVE FOR
N=1.......AV1=8(1/2)=4
N=2.......AV2=8(1/2)^2=8/4=2
N=3.......AV3=8(1/2)^3=8/8=1
N=4.......AV4=8(1/2)^4=8/16=1/2.....ETC

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3X^2-2X-8=0
3X^2-6X+4X-8=0
3X(X-2)-4(X-2)=0
(X-2)(3X-4)=0
X=2....OR......3X=4...OR...X=4/3