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sqrt(2x+5) + sqrt(x+6) = 9
sqrt(2x+5)= 9 - sqrt(x+6) ...square both sides
2X+5=81+X+6-18*SQRT.(X+6)
X-82 = -18*SQRT.(X+6)...SQUARING AGAIN
X^2+6724-164X =324(X+6)=324X+1944
X^2- 488X+4780 = 0
X^2-10X - 478X+4780=0
X(X-10)- 478(X - 10)=0
(X-10)(X-478)=0...HENCE X-10 = 0 ....OR....X=10 WHICH IS CORRECT SINCE
sqrt(2x+5) + sqrt(x+6) =sqrt(2*10+5) + sqrt(10+6) = 5 + 4 = 9
X-478 = 0 OR ..X=478 IS AN EXTRANEOUS ANSWER OBTAINED DUE TO REPEATED SQUARING ...WE SHOULD IGNORE THAT

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Point A (-4,1) is in the standard (x,y) coordinate plane. What must be the coordinates of Point B so that the line x = 2 is the perpendicular bisector of AB ? (note: there is a line over AB) Thank you for your help...Karen
LINE X=2 IS PARALLEL TO Y AXIS SO ITS PERPENDICULAR WOULD BE PARALLEL TO X AXIS
HENCE ITS EQUATION SHALL BE Y=K......THAT IS EQN.OF AB SHALL BE Y=CONSTANT
FURTHER X=2 BISECTS AB.HENCE MID POINT OF AB SHALL BE ON X=2
THAT IS X COORDINATE OF MIDPOINT SHALL BE EQUAL TO 2.IF WE TAKE POINT B AS (P,Q)
THEN X COORDINATE OF MID POINT OF AB SHALL BE = (-4+P)/2=2
-4+P=4
P=8
WE SPROVED EARLIER THAT EQN OF AB SHALL BE IN THE FORM Y=CONSTANT..OR ITS SLOPE IS ZERO SINCE Y=CONSTANT MEANS ...Y=0*X+CONSTANT
SLOPE OF AB =(Q-1)/(P-(-4))=0...SO Q-1=0...SO Q=1...
HENCE POINT B IS (8,1)
YOU WILL SEE ITS EQN. IS Y=1..TO UNDERSTAND HOW THEY LOOK LIKE SEE THE GRAPH BELOW

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(8) Question: Choose the nonlinear systesm below that have (x,y)=(4,3) as on of their solutions.
SUBSTITUTE X=4 AND Y=3 AND CHECK......
(A) x^2+y^2=25...........4^2+3^2=16+9=25..OK
x+y=7....4+3=7...OK..
(B) y=x^2+2x-21.......3=4^2+2*4-21=16+8-21=3...OK
x-y+1..DO YOU MEAN X-Y=1...IF SO ....4-3=1....OK
(C) x^2-y^2=7...........4^2-3^2=16-9=7...OK
2x+7=0.....2*4+7=8+7=15....NOT CORRECT.....
(D) both a and b..........OK...................
(E) both b and c ..........NOT OK............
Please be specific as to which letter answer is correct.
SUBJECT TO ABOVE CORRECTION D IS CORRECT

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Please explain how to factor the following problem by grouping to solve the equation.
LET F(X)= x^3+3x^2-4x-12=0
TRY PUTTING X=0,1,-1,2,-2..ETC..TO GET A ZERO
WE FIND AT X=2..F(2)=2^3+3*2^2-4*2-12=8+12-8-12=0...SO X-2 IS A FACTOR.DIVIDE F(X) WITH X-2 NOW
2...|1..........3..........-4.........-12
....|0..........2..........10..........12
..................................................
.....1..........5...........6...........0
HENCE X^2+5X+6 IS THE QUOTIENT
=X^2+2X+3X+6=X(X+2)+3(X+2)=(X+2)(X+3)
HENCE F(X)=(X-2)(X+2)(X+3)

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Dennis can eat a pie in 6 min. David can eat a pie in 12 min. Robert can eat a pie in 14.639145073927682623334797247841 min. If they share the same pie, how lomg will it take the three guys to scarf down the whole thing. (Hint: First figure out how many pies each guy can eat in an hour)
DENIS CAN EAT IN 1 MINUTE....1/6 =0.166666667 PIE
DAVID CAN EAT IN 1 MINUTE....1/12 =0.083333333 PIE
ROBERT CAN EAT IN 1 MINUTE...1/14.63914507=0.06831 PIE
HENCE THE 3 TOGETHER CAN EAT IB 1 MINUTE=0.31831 PIE
HENCE THEY TAKE ...1/0.31831 = 3.14159153 MINUTES TO EAT THE PIE.

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SEE THE FOLLOWING AND TRY..IF STILL IN DIFFICULTY PLEASE COME BACK...
OK I WORKED IT OUT FOR YOU NOW
I TOLD YOU EQN IS
(X-H)^2/A^2 + (Y-K)^2/B^2....
WHERE H,K IS CENTRE...SO H=5 AND K=-3 AS CENTRE IS GIVEN AS (5,-3)....NOW VERTEX IS (13,-3)...IT LIES ON ELLIPSE..SO IT SATISFIES THE EQN
(13-5)^2/A^2 +(-3+3)^2/B^2 =1
HENCE A^2=64...OR A=8
MINOR AXIS =10=2B...HENCE B=5..SO EQN.S
(X-H)^2/64 + (Y+3)^2/25 =1
THAT IS A IS CORRECT.

Can you help me write an equation for an ellipse with a major axis with endpoints of (0,8), and (0,-8) with foci of (0,5) and (0,-5)?
1 solutions
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Answer 16810 by venugopalramana(1120) on 2006-03-13 11:19:12 (Show Source):
Can you help me write an equation for an ellipse with a major axis with endpoints of (0,8), and (0,-8) with foci of (0,5) and (0,-5)?
THIS SHOWS THAT X AXIS IS THE MAJOR AXIS
STANDARD EQN.OF ELLIPSE IS
(X-H)^2/A^2 +(Y-K)^2/B^2=1
CENTRE IS (H,K)..AS PER THE PROBLEM H=K=0 AS CENTRE OF ELLIPSE IS AT (0,0)..SINCE major axis with endpoints ARE (0,8), and (0,-8)
WHERE MAJOR AXIS =2A=8+8=16...SO A=8..SINCE major axis with endpoints ARE (0,8), and (0,-8)
FOCI ARE GIVEN BY
AE,0 AND -AE,0...SO AE =5...SO E=5/A=5/8
BUT E=SQRT{(A^2-B^2)/A^2}=5/8...SQUARING
25/64=(A^2-B^2)/A^2=1-B^2/A^2
B^2/64=1-25/64=49/64
B^2=49
B=7
HENCE EQN. OF ELLIPSE IS
X^2/64 + Y^2/49 = 1

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Which of the following is a factor of the polynomial x^2 - x - 20?
THERE ARE 2 METHODS
1 FACTORISE AND CHECK..F(X)=X^2-X-20=X^2-5X+4X-20=X(X-5)+4(X-5)=(X-5)(X+4)
SO...A...X-5 IS THE ANSWER.
2...PUT X=5,4,-2..ETC..AS PER THE ANSWERS AND CHECK WHEN FUNCTION F(X) BECOMES ZERO..WE SEE THAT F(5)=0..SO X-5 IS A FACTOR.
A) x-5
B) x-4
C) x+2
D) x+5
E) x+10

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(6+8i)- (-7+i)
I come up with 13+ 9i but I am not sure, can you help?
GOOD JOB!!YOU DID CORRECTLY FOR REAL NUMBERS BUT FOR COMPLEX NUMBER IT IS
8i-i=7i...SO THE ANSWER IS 13+7i

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SEE THE FOLLOWING AND DO
IN FPS SYSTEM ....G=32 FT/SEC^2..SO USE -32 INSTEAD OF -9.8 IN THE FOLLOWING
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PLEASE HELP ME A rocket is fired with an initial velocity of 800 meters per second. When will it be 32000 meters above the starting point?
Than answer I got was : 40 second because I divided... I think its wrong thought
AS THE ROCKET GOES UP AGAINST EARTHS GRAVITATIONAL FORCE , ITS VELOCITY REDUCES,GIVEN BY ACCELERATION DUE TO GRAVITY =-9.8 M/SEC^2..THE FORMULA TO BE USED IS
S = UT + (G/2)T^2....WHERE
U = INIYIAL VELOCITY = 800 M/SEC
T = TIME ELAPSED IN SECS.
G=ACCELERATION DUE TO GRAVITY = - 9.8 M/SEC^2 FOR UPWARD TRAVEL...SO..WE GET
32000 = 800T -(9.8/2)T^2=800T-4.9 T^2
4.9T^2-800T+32000 =0

T=70.1 WHILE GOING UP AND
T=93.2 WHILE FALLING DOWN.

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PLEASE HELP ME A rocket is fired with an initial velocity of 800 meters per second. When will it be 32000 meters above the starting point?
Than answer I got was : 40 second because I divided... I think its wrong thought
AS THE ROCKET GOES UP AGAINST EARTHS GRAVITATIONAL FORCE , ITS VELOCITY REDUCES,GIVEN BY ACCELERATION DUE TO GRAVITY =-9.8 M/SEC^2..THE FORMULA TO BE USED IS
S = UT + (G/2)T^2....WHERE
U = INIYIAL VELOCITY = 800 M/SEC
T = TIME ELAPSED IN SECS.
G=ACCELERATION DUE TO GRAVITY = - 9.8 M/SEC^2 FOR UPWARD TRAVEL...SO..WE GET
32000 = 800T -(9.8/2)T^2=800T-4.9 T^2
4.9T^2-800T+32000 =0

T=70.1 WHILE GOING UP AND
T=93.2 WHILE FALLING DOWN.

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I have been asked to simplify an expression containing complex numbers and I am not sure how to do it. Please help me!
-3i/5 + 4i Translation: -3i divided by 5 + 4i Thanks a lot. I am not doing
that well in Algebra Two right now.
COMBINE ALL TERMS WITH i ,TAKING i AS A COMMON FACTOR.
COMBINE TERMS NOT CONTAINING i IN THE USUAL MANNER.
THAT IS THE WAY TO GO ABOUT..
HERE WE HAVE ONLY TERMS CONTAINING i..SO...WE GET...
i(-3/5 +4)=i{(-3+4*5)/5}= i(-3+20)/5=i17/20......OR.......17i/20...AS IT IS USUALLY WRITTEN.

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Let p(x) = x4 - 6x3 + 11x2 - 6x. Using the fact that p(1) = 0, completely factor p (i.e., express p as a product of linear factors
SO X-1 IS A FACTOR..
p(x) = x4 - 6x3 + 11x2 - 6x.
=X(X^3-6X^2+11X-6)=X(X-1)(AX^2+BX+C)...COMPARING COEFFICIENTS OF LIKE TERMS
COEFFICIENT OF X^4.......GIVES US....A=1.......
COEFFICIENT OF X GIVES US............C=6......
COEFFICIENT OF X^2 GIVES US ....C-B=11...SINCE C IS 6 , WE GET B=-5..
SO WE GET
p(x) = x4 - 6x3 + 11x2 - 6x.
=X(X^3-6X^2+11X-6)=X(X-1)(AX^2+BX+C)=X(X-1)(X^2-5X+6)=X(X-1)(X^2-3X-2X+6)=
X(X-1){X(X-3)-2(X-3)}=X(X-1)(X-3)(X-2)

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Explain why x to the 4th+2xsquared+4 has no real root
X^4+2X^2+4={(X^2)^2+2(X^2)(1)+(1)^2}-(1)^2+4 = (X^2+1)^2+3...OR...(X^2+1)^2=-3
WE FIND THT THE SQUARE OF A NUMBER X^+1 IS NEGATIVE WHICH HAS NO REAL SOLUTION.(X^2+1)=i*SQRT.3
HENCE
X^2+1= i*SQRT.3
X^2=-1+i*SQRT.3 = A COMPLEX NUMBER..HENCE ALL FOUR ROOTS OF X ARE COMPLEX/IMAGINARY NUMBERS.THAT IS THERE ARE NO REAL SOLUTIONS FOR X.
IT IS POSSIBLE HERE SINCE THE TWO PAIRS OF COMPLEX ROOTS ARE CONJUGATE WITH THE RESULT THAT THE FUNCTIONS OF THEIR SUM/PRODUCT IS REAL.SINCE THE COEFFICIENTS OF VARIABLE IN THE POLYNOMIAL ARE FUNCTIONS OF SUM/PRODUCT OF THE ROOTS,IT IS NECESSARY THAT THE ROOTS SHOULD BE REAL OR CONJUGATE COMPLEX NUMBERS FOR THIS TO HAPPEN IN CASE OF A POLYNOMIAL WITH REAL COEFFICIENTS.IN THIS CASE WE HAVE
IF A,A',B,B' ARE ROOTS (A AND A' ARE CONJUGATE ...B AND B' ARE CONJUGATE).. THEN
A+A'+B+B'=0....POSSIBLE SINCE A+A' IS REAL AND B+B' IS REAL
AA'+AB+AB'+A'B+A'B'+BB'=2...POSSIBLE SINCE AA' AND BB' ARE REAL AND
AB+AB'+A'B+A'B'=A(B+B')+A'(B+B')=(A+A')(B+B')=REAL*REAL=REAL
.....ETC...
while every polynomial function of degree 3 has at least 1 real root
IN CASE OF 3 RD. DEGREE POLYNOMIAL ,WE HAVE 3 ROOTS..SO IF THEY ARE COMPLEX THEN AS WE SHOWED ABOVE FOR POLYNOMIAL WITH REAL COEFFICIENTS ..COMPLEX ROOTS OCCUR IN PAIRS OF CONJUGATE NUMBERS ONLY.SO IN 3 ROOTS 2 COULD BE COMPLEX CONJUGATES ,BUT THE 3 RD. HAS TO BE REAL IF THE COEFFICIENTS ARE REAL.

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a. Give an example of a 3x3 lower triangular matrix.
(A)=
b. Using your example from part (a), show that the determinant of a lower triangular matrix is the product of the entries on the diagonal.
|A|=1(3*2-0*1)-0+0 =6
PRODUCT OF DIAGONAL ELEMENTS =1*3*2=6 =|A|
c. Show algebraically that the determinant of a 2x2 lower triangular matrix will always be the product of the entries on the diagonal.
LET (B) =
|B|= a*c-0*b=ac
PRODUCT OF DIAGONAL ELEMENTS =a*c = ac =|B|

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log[base 2]9 x log[base 3]8 = log[base 2]8 x log[base 3]9.
USING THE FORMULA
LOG (A) TO BASE B =LOG(A)/LOG(B)....TO ANY COMMON BASE WE GET
log[base 2]9 x log[base 3]8 = log[base 2]8 x log[base 3]9.
LHS={LOG(9)/LOG(2)}*{LOG(8)/LOG(3)}=LOG(9)*LOG(8)/{LOG(2)*LOG(3)}
RHS={LOG(8)/LOG(2)}*{LOG(9)/LOG(3)}=LOG(9)*LOG(8)/{LOG(2)*LOG(3)}
=LHS

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Can you help me write an equation for an ellipse with a major axis with endpoints of (0,8), and (0,-8) with foci of (0,5) and (0,-5)?
THIS SHOWS THAT X AXIS IS THE MAJOR AXIS
STANDARD EQN.OF ELLIPSE IS
(X-H)^2/A^2 +(Y-K)^2/B^2=1
CENTRE IS (H,K)..AS PER THE PROBLEM H=K=0 AS CENTRE OF ELLIPSE IS AT (0,0)..SINCE major axis with endpoints ARE (0,8), and (0,-8)
WHERE MAJOR AXIS =2A=8+8=16...SO A=8..SINCE major axis with endpoints ARE (0,8), and (0,-8)
FOCI ARE GIVEN BY
AE,0 AND -AE,0...SO AE =5...SO E=5/A=5/8
BUT E=SQRT{(A^2-B^2)/A^2}=5/8...SQUARING
25/64=(A^2-B^2)/A^2=1-B^2/A^2
B^2/64=1-25/64=49/64
B^2=49
B=7
HENCE EQN. OF ELLIPSE IS
X^2/64 + Y^2/49 = 1

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1. Write in slope intercept the equation of theline passing through the two points.Show that the line is perpindicular to the given line.
(-2,-2), (1,3); y= 3x-1
EQN.OF LINE JOINING (X1,Y1) AND (X2,Y2) IS GIVEN BY
Y-Y1=(X-X1)*(Y2-Y1)/(X2-X1), WHERE (Y2-Y1)/(X2-X1) IS THE SLOPE OF THE LINE AND Y1-X1*(Y2-Y1)/(X2-X1)IS THE INTERCEPT.
HENCE EQN.OF LINE IS
Y+2=(X+2)*(3+2)/(1+2)=X(5/3)+2*5/3
Y=X(5/3)+10/3-2=X(5/3)+4/3.....HENCE SLOPE IS 5/3 AND INTERCEPT IS 4/3
SLOPE OF GIVEN LINE Y=3X-1 IS 3 ,,,FOR 2 LINES TO BE PERPENDICULAR,THE PRODUCT OF THEIR SLOPES SHOULD BE -1.HERE THE PRODUCT IS 3*5/3=5..HENCE THEY ARE NOT PERPENDICULAR TO EACH OTHER..CHECK BACK YOUR NUMBERS...COPY THE PROBLEM PROPERLY.AS GIVEN THE LINES ARE NOT PERPENDICULAR.
2. write in slope-intercept form the equation of the line passing through the given point and perpindicular to the given line.
(-4,-7), y=-4x-7
SLOPE OF Y=-4X-7 IS -4
AS GIVEN ABOVE FOR 2 LINES TO BE PERPENDICULAR,THE PRODUCT OF THEIR SLOPES SHOULD BE -1.HENCE THE SLOPE OF THE REQUIRED LINE IS -1/-4=1/4
EQN.OF REQD. LINE IS
Y+7=(1/4)(X+4)=X/4 + 1
Y= X/4 - 6

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The value of square-root(2/pi) * Integral(e to the power -square(t)/2 ),
where "t" is a variable from 4 to infinity is equal to what?
THIS IS CALLED PROBABILITY INTEGRAL FOR WHICH YOU HAVE TO REFER STANDARD TABLES
THE TABLE IS MADE FOR
P(T)=(1/SQRT(2PI))*INTEGRAL{0 TO T OF (E^(-T^2/2) DT)}
IT IS THE AREA UNDER THE CURVE.PRACTICALLY,THE ENTIRE CURVE LIES BETWEEN
T=- 3AND +3...99.73% IN FACT..SO IF YOU WANT AREA BEYOND T=4 IT IS ALMOST ZERO..THE ANSWER IS NEARLY ZERO.

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I need to determine the following for these two problems :HOW MANY X-INTERCEPTS THE PARABOLA HAS, and WHETER ITS VERTEX LIES ABOVE OR BELOW OR ON THE X-AXIS.
1.problem
y=x^2-5x+6
PUT Y=0 AND SOLVE FOR X TO GET X INTERCEPTS.
X^2-5X+6=0=X^2-2X-3X+6=0=X(X-2)-3(X-2)=0=(X-2)(X-3)=0....X=2 AND 3...
SO THE X INTERCEPTS ARE AT X = 2 AND X = 3
Y=X^2-5X+6={X^2-2*X*5/2+(5/2)^2}-(5/2)^2+6 =(X-2.5)^2 - 0.25
SO THE VERTEX IS AT X=2.5 AND Y=-0.25...THAT IS BELOW X AXIS
2.problem
y=-x^2+2x-1
DOING THE SAME WAY WE GET
Y=-(X-1)^2=0 AND HENCE
X INTERCEPTS ARE X=1
AND VERTEX IS AT X=1 AND Y=0 SO THE VERTEX IS ON THE X AXIS.

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need help plz...heres the question "There are three friends named Allan, Bobby and Charlie. The three friends want to know their individual rate in finishing a job. Allan and Bobby can finish the job in 42 days, Bobby and Charlie can finish the job in 31 days, and Allan and Charlie can finish the job in 20 days. Solve the rate of each individual."
LET ALLAN TAKE A DAYS TO DO THE JOB ALONE
HENCE ALLAN ALONE CAN DO IN 1 DAY 1/A JOB
LET BOBBY TAKE B DAYS TO DO THE JOB ALONE
HENCE BOBBY ALONE CAN DO IN 1 DAY 1/B JOB
LET CHARLIE TAKE C DAYS TO DO THE JOB ALONE
HENCE CHARLIE ALONE CAN DO IN 1 DAY 1/C JOB
FROM ABOVE WE GET ...
ALLAN AND BOBBY CAN DO IN 1 DAY ..1/A +1/B =(A+B)/AB JOB..
SO DAYS THEY TAKE TO COMPLETE THE JOB =AB/(A+B)=42..OR..1/A+1/B=1/42..........I BOBBY AND CHARLIE CAN DO IN 1 DAY ..1/B +1/C =(B+C)/BC JOB..
SO DAYS THEY TAKE TO COMPLETE THE JOB =BC/(B+C)=31..OR..1/B+1/C=1/31.........II
ALLAN AND CHARLIE CAN DO IN 1 DAY ..1/A +1/C =(A+C)/AC JOB..
SO DAYS THEY TAKE TO COMPLETE THE JOB =AC/(A+C)=20..OR..1/A+1/C=1/20........III
EQNI+EQNII+EQNIII GIVES...
2{(1/A)+(1/B)+(1/C)}=(1/42)+(1/31)+(1/20)
(1/A)+(1/B)+(1/C)=(1/2)*{(1/42)+(1/31)+(1/20)}.........IV
EQN.IV-EQN.I GIVES
1/C= (1/2)*{(1/42)+(1/31)+(1/20)}-1/42....OR.....C=34.2 DAYS
EQN.IV-EQN.II GIVES.......
1/A=(1/2)*{(1/42)+(1/31)+(1/20)}- 1/31...OR......A=48.1 DAYS
EQN.IV-EQN.III GIVES....
1/B=(1/2)*{(1/42)+(1/31)+(1/20)}- 1/20....OR.....B=329.6 DAYS

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SEE THE FOLLOWING EXAMPLE WHICH IS ALMOST SAME AS YOUR PROBLEM AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK.
Y=X^2-2X-3=X^2-2*X*1+1^2-1-3=(X-1)^2-1-3=(X-1)^2-4=0
X INTERCEPT IS OBTAINED BY PUTTING Y=0....WE GET
X^2-2X-3=0=X^2-3X+X-3=X(X-3)+1(X-3)=(X-3)(X+1)=0...SO..X=3 OR -1...
Y INTERCEPT IS GOT BY PUTTING X =0...WE GET
Y=0-0-3=-3
SEE THE GRAPH BELOW ......

THE AXIS IS X=1 AS YOU CAN SEE FROM THE GRAPH
ONE POINT SYMMETRIC TO Y INTERCEPT??NO...SYMMETRIC TO AXIS IT IS......(-1,3)
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I have to put this equation y=x^2-2x-15 into this form y=a(x-h)^2+k.
MAKE A PERFECT SQUARE USING X^2 AND X TERMS.ADD AND SUBTRACT THE REQUIRED CONSTANT FOR THE PURPOSE.
Y=(X-1)^2-1-15=(X-1)^2-16...COMPARING WITH THE ABOVE
y=a(x-h)^2+k.
WE GET A=1 AND K=-16
I have to find the line of symmetry.
X-1=0 IS THE LINE OF SYMMETRY SINCE ON EITHER SIDE OF X=1,WE GET SYMMETRIC/SAME VALUES FOR Y..AT X=1+2=3..Y IS -12 AND AT X=1-2=-1 ALSO WE GET Y=-12
(h,k)=vertex I think you use complete the square technique.
YA ..THE VERTEX AS YOU SHOULD KNOW NOW IS AT X=1 AND AT X=1 ,Y=-16 .SO (1,-16) IS THE VERTEX
THE GRAPH WILL LOOK LIKE THIS

Please help. thanks

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prove each of the following identities algebraically a. cosA+cosA tan squareA=secA
LHS=COS A (1+TANSQUARE A)=COS A * SECSQ.A=COS A/COSSQ.A =1/COS A = SEC A =RHS.
b.cosA(secA-cscA)=1-cotA
LHS=COS A*1/COS A-COS A/SIN A = 1-COT A = RHS
c.Prove the identity sin squareA divided 1-cosA=secA+1 divided by secA
LHS = (1-COSSQ.A)/(1-COS A)=1+COS A
RHS={(1/COS A) + 1 }/(1/COS A)=(1+COS A)COS A/COS A=1+COS A=LHS
d.Verify sin square x+cos square x=1 for x=5pi divided by 3.
SINSQ.5PI + COSSQ. 5PI = SINSQ.(2*2PI+PI) + COSSQ.(2*2PI+PI)
=SINSQ.PI + COSSQ. PI = (0)^2 + (-1)^2 =0+1=1
Graph
y=sin square x+cos square x and describe the result.

Also explain the result...WE GET A LINE PARALLEL TO XAXIS AT A DISTANCE OF PLUS ONE UNIT FROM IT.THAT IS Y=1..WHATEVER MAY BE THE VALUE OF X .

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9cos square A+ 3 cos A = 0
=3COS A(3COS A+1)=0
SO .....COS A = 0 ...HENCE A=90 ..IN 90degrees 3COS A +1 =0...OR...COS A =-1/3
A=180-70.5=109.5 DEGREES

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What is the exact value of
sin square pi/ 6 - 2sin pi/ 6 cos pi/ 6
...............=cos square (-pi/6)?...WHAT IS THIS EQUAL TO ??
=(1/2)^2-2*(1/2)*(SQRT 3)/2=1/4 - SQRT(3)/2={2-SQRT(3)}/2
I got 2-Sqrt3/2 CORRECT .BUT PUT BRACKET ..OTHERWISE IT MAY BE TAKEN AS
2 -(SQRTRT 3)/2
RHS = cos square (-pi/6)?...= {SQRT(3)/2}^2 = 3/4

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another form of cos(30 degrees-A)+sin(60degrees+A)=
= COS 30 COS A +SIN 30 SIN A + SIN 60 COS A + COS 60 SIN A
={(SQRT 3)/2}COS A +(1/2) SIN A + {(SQRT 3)/2}COS A +(1/2) SIN A
=(SQRT 3)COS A + SIN A IS THE ANSWER...
I got sqrt of 3 divided by 2 sin A + cos A.....NO IT IS NOT CORRECT...

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how do you find the nth term of a negative complex number?
I HOPE YOU MEAN N th. ROOT NOT AND NOT N th. TERM..OF A COMPLEX NUMBER ..CAN BE POSITIVE OR NEGATIVE...
LET THE COMPLEX NUMBER BE Z = A+Bi IN GENERAL WHICH CAN ALWAYS BE WRITTEN AS
Z = R*(COS X + i SIN X ) WHERE R = SQRT(A^2+B^2) AND COS X = A/(SQRT(A^2+B^2)
AND SIN X = B/(SQRT(A^2+B^2)
SINCE N th.ROOT OF R IS EASY TO FIND (R^1/N),R BEING POSITIVE, WE NEED TO ONLY FIND Nth.ROOT OF SAY Z = COS X +i SIN X
Z^1/N = (COS X +i SIN X)^1/N...AS PER DEMOVIERS THEOREM THIS IS EQUAL TO
= COS (X/N) +i SIN (X/N)..SO NOW WE USE THE GENERAL FORMULA FOR TRIGNOMETRIC EQNS ..THAT IS ..IF COS X = P SAY THEN X = COS ^-1(P)..ITS GENERAL SOLUTION IS
X=2K*(PI)+COS^-1(P)...GIVE K=1,2,3 ETC.. TO GET ALL THE DIFFERENT ROOTS.
for example the fourth roots of -4
Z = -4 =4*(-1)...SINCE 4 TH. ROOT OF 4 CAN BE EASILY FOUND OUT .LET US FIND FOURTH ROOTS OF -1.WE CAN MULTIPLY THAT ANSWER WITH 4TH ROOT OF +4 TO GET THE TOTAL ANSWER.
Z = COS (0) +i SIN (0)..
= COS (2K*PI+0)+iSIN (2K*PI+0)=COS (2K*PI)+i SIN (2K*PI)
4TH.ROOTS OF THIS ARE AS PER ABOVE FORMULA...
Z^1/4 =COS (2K*PI)/4+i SIN (2K*PI)/4...PUTTING K=1,2,3 AND 4 TO GET THE 4 ROOTS WE GET...
COS (2*1*PI)/4+i SIN (2*1*PI)/4...,
COS (2*2*PI)/4+i SIN (2*2*PI)/4...,
COS (2*3*PI)/4+i SIN (2*3*PI)/4...,
COS (2*4*PI)/4+i SIN (2*4*PI)/4...AS THE 4 ROOTS.

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Help please :) I really don't want the answer....please just help me w/ the proper steps to solve.
A child in an airport is able to cover meters in minutes running at a steady speed down a moving sidewalk in the direction of the sidewalk's motion. Running at the same speed in the direction opposite to the sidewalk's movement, the child is able to cover meters in minutes. What is the child's running speed on a still sidewalk, and what is the speed of the moving sidewalk?
VERY GOOD TO SEE YOUR KEEN DESIRE TO LEEARN..IN THESE TYPES OF TIME AND DISTANCE PROBLEMS,WE USE NORMAL FORMULAE WITH SLIGHT MODIFICATION
STANDARD FORMULA FOR TIME AND DISTANCE IS
DISTANCE TRAVELLED = D = SPEED * TIME =S*T
REVISED FORMULA WE USE IN GENERAL CASES IS
DISTANCE OF SEPERATION FROM INITIAL STAGE TO FINAL STAGE=D2-D1=D
TIME = TIME AT END - TIME AT THE BEGINING =T2-T1=T
RELATIVE SPEED = SUM OF SPEEDS WHEN THE SPEEDS OF 2 MOVING OBJECTS AID EACH OTHER ....... = S1+S2=S
RELATIVE SPEED = DIFFERENCE IN SPEEDS WHEN THE SPEEDS OF 2 MOVING OBJECTS OPPOSE EACH OTHER ....... = S1-S2=S
LET US EXPAND A BIT ON THIS RELATIVE SPEED..SUPPOSE YOU WALK AT 3 MPH IN A MOVING TRAIN IN THE SAME DIRECTION AS TRAIN IS MOVING,THEN WITH RESPECT TO THE TRAIN YOU ARE ONLY WALKING AT 3 MPH, BUT W.R.T. A STATIONARY GROUND ,YOUR RELATIVE SPEED WILL BE 3 MPH+SPEED OF TRAIN SAY 50 MPH =53 MPH.
SIMILARLY WHEN YOU ARE WALKING IN THE OPPOSITE DIRECTION TO THAT OF TRAIN ,YOUR RELATIVE SPEED W.R.T.TO THE GROUND =50-3=47 MPH.
SAME THING APPLIES WHEN YOU ARE SWIMMING IBN A CURRENT OF WATER OR WALKING ON A MOVING SIDE WALK OR A PLANE GOES AGAINST THE WIND OR PROWIND ETC...
IN YOUR PRESENT PROBLEM YOU CAN TAKE
LET SPEED OF CHILD ON STILL SIDE WALK =S1 AND
SPEED OF MOVING SIDE WALK =S2
SO WHEN THEY ARE GOING IN THE SAME DIRECTION RELATIVE SPEED W.R.T GROUND =S1+S2
& WHEN THEY ARE GOING IN OPPOSITE DIRECTIONS RELATIVE SPEED W.R.T GROUND =S1-S2
HENCE TIME TAKEN WILL BE D/(S1+S2) IN ONE CASE AND D/(S1-S2) IN THE OPPOSITE CASE.HOPE YOU CAN SOLVE YOUR PROBLEM NOW.

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i am currentley revising for my gcse and i am stuck on algebra. the school have given me a a word dotted paper and told me to work out the area by drawing the shapes.There is a word on the paper i do not understand the word is trapezium. what does that word mean can you help me please thank you.
SEE THE FOLLOWING EXAMPLES TO KNOW THE TRAPEZIUM..IT IS A QUADRILATERAL WITH 2 OPPOSITE SIDES PARALLEL
IF YOU WANT MORE OR IN DOUBT COME BACK.
see the following example and try
SEE THE TRAPEZOID ....ABCD below
AREA OF TRAPEZOID IS GIVEN BY
AREA=(1/2)*ALTITUDE*(SUM OF PARALLEL SIDES)=(1/2)*DP*(AB+CD)
WHERE ALTITUDE IS THE PERPENDICULAR DISTANCE BETWEEN PARALLEL SIDES =DP
PARALLEL SIDES ARE AB AND CD.
WE HAVE AB=40
BC=AD=20
ANGLE DAP=63
DAP IS A RIGHT ANGLED TRIANGLE...DP/DA=SIN(63)
DP=DA*SIN(63)=20*0.89=17.8
SO ALTITUDE=17.8
AP/DA=COS(63)
AP=DA*COS(63)=20*0.4545=9.1
SO CD =AB-2*AP SINCE TRAPEZOID IS ISOCELLES..
CD=40-2*9.1=21.8
AREA=(1/2)17.8(40+21.8)=550 SQUARE UNITS.