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For the systems of linear equations in this question
- Determine how many solutions exist
- Use either elimination or substitution to find the solutions (if any)
- Graph the two lines, labeling the x-intercepts, y-intercepts, and points of intersection
y = 2x + 3 and y = -x - 4
-X-4=2X+3
2X+X=-4-3=-7
3X=-7
X=-7/3
Y=-(-7/3)-4=7/3-4=(7-12)/3=-5/3
HENCE THERE IS ONE UNIQUE SOLUTION.

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SEE THE FOLLOWING EXAMPLE AND TRY
-------------------------------------------------------------
For the systems of linear equations in this question
- Determine how many solutions exist
- Use either elimination or substitution to find the solutions (if any)
- Graph the two lines, labeling the x-intercepts, y-intercepts, and points of intersection
y = 2x + 3 and y = -x - 4
-X-4=2X+3
2X+X=-4-3=-7
3X=-7
X=-7/3
Y=-(-7/3)-4=7/3-4=(7-12)/3=-5/3
HENCE THERE IS ONE UNIQUE SOLUTION.

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For the systems of linear equations in this question
- Determine how many solutions exist
- Use either elimination or substitution to find the solutions (if any)
- Graph the two lines, labeling the x-intercepts, y-intercepts, and points of intersection
y = 2x + 3 and y = -x - 4
-X-4=2X+3
2X+X=-4-3=-7
3X=-7
X=-7/3
Y=-(-7/3)-4=7/3-4=(7-12)/3=-5/3
HENCE THERE IS ONE UNIQUE SOLUTION.

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For the systems of linear equations in this question
- Determine how many solutions exist
- Use either elimination or substitution to find the solutions (if any)
- Graph the two lines, labeling the x-intercepts, y-intercepts, and points of intersection
x + y = 3 and y = x + 3
X+X+3=3
2X=0
X=0
Y=3...HENCE THERE IS ONLY ONE UNIQUE SOLUTION.
(

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Factor each polynomial completely, given that the binomial following it is a factor of the polynomial.
1. x^3-4x^2-3x-10,x-5
5|1...........-4..........-3..........-10
.|0............5...........5...........10
----------------------------------------------------------
..1............1...........2............0
SO QUOTIENT IS X^2+X+2...THIS NO MORE REDUCIBLE.
HENCE ANSWER IS (X-5)(X^2+X+2)
2. w^3+5w^2-w=5....PLEASE TYPE PROPERLY...WHERE IS THE BINOMIAL FACTOR?
3. t^2+1=13/6t....PLEASE TYPE PROPERLY...WHERE IS THE BINOMIAL FACTOR?

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factor the polynomial completely, given that the binomial followin it is a factor of the polynomial.
w^3+5w^2-w=5
PLEASE TYPE PROPERLY..WHERE IS THE BINOMIAL FACTOR FOLLOWING THE POLYNOMIAL?

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factor the polynomial completely, given that the binomial followin it is a factor of the polynomial.
w^3+5w^2-w=5
PLEASE TYPE PROPERLY..WHERE IS THE BINOMIAL FACTOR FOLLOWING THE POLYNOMIAL?

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factor the polynomial completely, given that the binomial followin it is a factor of the polynomial.
w^3+5w^2-w=5
PLEASE TYPE PROPERLY..WHERE IS THE BINOMIAL FACTOR FOLLOWING THE POLYNOMIAL?

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10^3log(x)= 1/9HOPE YOU TYPED THIS PROPERLY..ANY WAY AS TYPED IT SAYS
TAKING LOGS
3LOG(X)*LOG(10)=LOG(1/9)
3LOG(X)=LOG(1/9)
LOG(X^3)=LOG(1/9)
X^3=1/9
X=CUBEROOT OF (1/9)

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8^(4x-3)= 1/16,
(2^3)^(4X-3)=1/(2^4)=2^(-4)
2^(12X-9)=2^(-4)
12X-9=-4
12X=-4+9=-5
X=-5/12
find the exact answer.

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THE QUESTION IS NOT COMPLETE.PLEASE CHECK AND RETYPE

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WE FIND THAT 5^2+12^2=25+144=169=13^2...SO IT IS RIGHT ANGLED TRIANGLE..IF
LET OA=5,OB=12,AB=13...THEN ANGLE AOB =90....LET O BE ORIGIN.SO A IS (5,0) AND B IS (0,12).SO CIRCUM CENTRE S IS THE MID POINT OF AB THE HYPOTENUSE.
HENCE S IS {5/2,12/2)=(5/2,6)
INCENTRE I IS GIVEN BY
X COORDINATE....(5*0+12*5+13*0)/(5+12+13)=60/30=2
Y COORDINATE....(5*12+12*0+13*0)/(5+12+13)=60/30=2
HENCE S IS (2,2)
SI =SQRT.{(5/2-2)^2+(6-2)^2}=SQRT(1/4+16)=SQRT(16.25)

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WE FIND THAT 5^2+12^2=25+144=169=13^2...SO IT IS RIGHT ANGLED TRIANGLE..IF
LET OA=5,OB=12,AB=13...THEN ANGLE AOB =90....LET O BE ORIGIN.SO A IS (5,0) AND B IS (0,12).SO CIRCUM CENTRE S IS THE MID POINT OF AB THE HYPOTENUSE.
HENCE S IS {5/2,12/2)=(5/2,6)
INCENTRE I IS GIVEN BY
X COORDINATE....(5*0+12*5+13*0)/(5+12+13)=60/30=2
Y COORDINATE....(5*12+12*0+13*0)/(5+12+13)=60/30=2
HENCE S IS (2,2)
SI =SQRT.{(5/2-2)^2+(6-2)^2}=SQRT(1/4+16)=SQRT(16.25)

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SEE ANSWERS BELOW
------------------------------
A straight line L and a plane P in R3 are defined by the equations
EQN.OF PLANE P........ 2x − 3y + z = 1
and EQN.OF LINE L.....(x, y, z) = (1,−1, 2) + t (1, 2, 3), t any real number.
(a) Say briefly but clearly how you know the line L does not pass through the point
(−1,−5,−3)....THIS DOES NOT SATISFY EQN.L.FOR ONE VALUE OF T AS SHOWN BELOW
-1=1+T...OR....T=-2...
-5=-1+2T...OR....T=-2..OK
-3=2+3T.....OR....T=-5/3....NOT TALLYING
(b) Say briefly but clearly how you know that the plane P does not pass through the origin....ORIGIN IS (0,0,0)...THIS DOES NOT SATISFY EQN.P...WE GET 0=1
(c) Write the equation for the line perpendicular to the plane P and passing through the
origin....LINE PERPENDICULAR TO PLANE IS PARALLEL TO PLANE'S NORMAL WHICH HAS DRS OF 2,-3,1...LINE PASSES THROUGH (0,0,0)..SO EQN OF LINE IS
X/2 = Y/-3 = Z/1
(d) Write the equation for the plane perpendicular to L and passing through the point
(0, 1, 3)..........DRS OF LINE ARE 1,2,3...THE REQD.PLANE IS PERPENDICULAR TO THE LINE.HENCE ITS NORMAL IS PARALLEL TO LINE.HENCE DRS OF NORMAL ARE SAME AS DRS OF LINE..THAT IS 1,2,3...HENCE EQN.OF PLANE IS X+2Y+3Z=K...IT PASSES THROUGH
(0,1,3)....SO.....0+1*2+3*3=K=11
EQN. OF REQD. PLANE IS ..................X+2Y+3Z=11
(e) Write the equation for the xz-plane.
Y=0

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REAL LIFE EXAMPLES...
ARITHMATIC / PROGRESSION/SERIES....A.P.
SUPPOSE YOU ARE SAVING YOUR MONEY IN A BANK BY SYSTEMATIC PLAN OF
DEPOSITING 100 $ A MONTH...THEN STARTING WITH A 100$ ACCOUNT YOUR
MONEY IN CREDIT WITHOUT INTEREST ,WOULD BE AN EXAMPLE OF A.P...IT
WILL BE..... TAKING MONTH AS AN INDEX...
100,200,300,400......
GEOMETRIC PROGRESSION/SERIES.....G.P.
IN A SIMILAR WAY SUPPOSE YOU DEPOSITED 1000 $ IN A BANK FOR INTEREST
OF 4% PER YEAR,AND THE INTEREST IS CALCULATED AT THE END EVERY YEAR
AND ADDED TO THE PRINCIPAL,THEN THE AMOUNT GROWN AT THE END OF YEAR IS
AN EXAMPLE OF G.P.TAKING YEAR AS N INDEX....THE AMOUNT AT THE END OF
SUCCESSIVE YEARS IS .....
1000,1000*1.04,1000*1.04^2,1000*1.04^3....ETC...

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THE SITE YOU MENTIONED IS RESTRICTED AND I AM NOT ABLE TO ACCESS THE PROBLEMS .EITHER YOU HAVE TO REPRODUCE THEM OR ENABLE ME ACCESS TO YOUR STUDENT REGISTRATION BY SOME MEANS.PLEASE SEE WHAT CAN BE DONE.
VENUGOPALRAMANA

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SEE MY COMMENTS BELOW..AND THE ADDDITIONAL EXAMPLE GIVEN BELOW
Can someone help? and check my answer to see if im doing this right?
For the function y = x^2 - 4x - 5, perform the following tasks:
(1)Put the function in the form y = a(x - h)^2 + k.
I got the answer: y = (x-2)^2 - 9......................OK
What is the line of symmetry?
I got the answer: x = 2...OK..VERY GOOD
Graph the function using the equation in part (1).

Explain why it is not necessary to plot points to graph when using y = a (x – h)^2 + k.
YOU CAN DRAW LINE OF SYMMETRY.,PLOT VERTEX AND SKETCH THE GRAPH ON EITHER SIDE OF LINE OF SYMMETRY BY SMOOTH LINES NOTING,THE X INTERCEPTS OF -1 AND 5 AND THAT Y TENDS TO INFINITY AS X TENDS TO INFINITY.
how does this graph compare to the graph of y = x^2?

---------------------------------------------------------------------------
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 ......
graph( 500, 500, -10, 20, -20, 20,x^2-2x-3 )
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)
******************************************************************************
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
graph( 500, 500, -10, 20, -20, 20, x^2-2x-15 )
Please help. thanks

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Five men took turns weighing themselves on a scale. However, they weighed each other two at a time, ten times total , so that every possible combination of two men was weighed. The scale had the following readouts for the pairs: 236 pounds, 244 pounds, 228 pounds, 250 pounds, 258 pounds, 230 pounds, 246 pounds, 238 pounds, 242 pounds, and 252 pounds.
How heavy was the heaviest man?
LET THE WEIGHTS 5 MEN IN ORDER OF DECREASING WEIGHT BE A,B,C,D,E
WHEN THEY WEIGH IN PAIRS WE GET 5C2=5*4/2=10 DIFFERENT PAIRS OR 20 MEN
HENCE THE TOTAL WEIGHT IS FOUR TIMES THE WEIGHT OF THE 5 MEN'S TOTAL WEIGHT
4(A+B+C+D+E)=236+244+228+250+258+230+246+238+242+252=2424
A+B+C+D+E=606..............I
NOW HEAVIEST IS A+B=258.........II
NEXT HEAVY IS A+C=252...........III
LIGHTEST =D+E=228.................IV
EQN.I-EQN.IV...GIVES..
A+B+C+D+E-D-E=606-228=374
A+B+C=374.....................V
EQN.II+EQN.III GIVES
A+B+A+C=258+252=510
2A+B+C=510.................VI
EQN.VI-EQ.EQN.V...GIVES...
2A+B+C-A-B-C=510-374=136
A=136
SO HEAVIEST MAN'S WEIGHT =136

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LET Y = x^3-3x^2+3x-9=0
I got x=3; x=3i; x=-3i...how do I plug answers back in to check solution?
HOW DID YOU GET THOSE ANSWERS?...ANY WAY ,YOU HAVE TO JUST SUBSTITUTE THOSE VALUES FOR X AND CHECK
1...X=3...Y=3^3-3*3^2+3*3-9=27-27+9-9=0...OK..
2...X=3i..Y=(3i)^3-3*(3i)^2+3*(3i)-9=27*(i^2)(i)-27(i)^2+9i-9=-27i+27+9i-9=-18i+18...
THIS IS NOT ZERO HENCE THIS IS NOT A ROOT
3...X=-3i..Y=(-3i)^3-3*(-3i)^2+3(-3i)-9=-27(i^2)(i)-27(i^2)-9i-9=27i+27-9i-9=18i+18...
THIS IS NOT ZERO ..HENCE THIS IS NOT A ROOT..
THE CORRECT ANSWERS ARE
X=3
X=i*SQRT(3)
X=-i*SQRT(3)
YOU CAN CHECK BY SUBSTITUTION AS SHOWN ABOVE

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LET Y =SQRT{(3-X)/X}THIS IS THE PROBLEM OR IS IT {SQRT(3-X)}/X..PLEASE PUT BRACKETS APPROPRIATELY..OK LET US TAKE
Y={SQRT(3-X)}/X
SO (3-X) CAN NOT BE NEGATIVE..THAT IS 3>=X...OR....X<=3
FURTHER WE HAVE X IN DENOMINATOR ..HENCE X SHOULD NOT BE ZERO.
SO DOMAIN OF THE FUNCTION IS X<=3 BUT NOT EQUAL TO ZERO.

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SEE MY COMMENTS BELOW
can someone please check my answers, and explain if and where i went right, thank you so much..
solve equations for the variable x
4x=y
my answer 4x/4=y/4...OK...CANCEL 4/4 ON LHS YOU GET X ON LHS..ON RHS YOU GOT Y/4
x=4/y...HOW?...YOU GOT ABOVE X=Y/4
xyz=t
my answer xyz/xy=t/xy .....NO ...XYZ/YZ=T/YZ...R...X=T/YZ
x=xyz/t
4xz=p
my answer 4xz/4=p/4..OK...XZ=P/4....XZ/Z=P/4Z....X=P/4Z
x=4xz/p
y=xtz
my answer y/y=xtz/y x=xtz/y...NO...YOU HAVE TO SEND EVERY THING OTHER THAN X TO THE OTHER SIDE...SO ...XTZ/TZ=Y/TZ....OR...X=Y/TZ
xyz/t
my answer xyz/xy=t/xy x=xyz/t
3.5tpx=r...........................SAME WAY AS ABOVE X=R/3.5TP
my answer 3.5/tpx=r/3.5 x=tpx/r
YOU ARE MAKING SAME ERROR EVERY TIME.STUDY THE ABOVE EXAMPLES AND TRY TO CORRECT YOUR SELF.IF STILL IN DOUBT PLEASE COME BACK

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Find two numbers, given that the second is 3 times the first. If you multiply the first number by the second number and the second number by 4, the sum of the products is 420.
LET FIRST NUMBER =X
SEOND NUMBER =3 TIMES FIRST =3*X
MULTIPLY THE FIRST NUMBER BY SECOND WE GET =X*3X=3X^2
MULTIPLY SECOND NUMBER BY 4 WE GET =4*3*X=12X
SUM OF 2 PRODUCTS =3X^2+12X=420...DIVIDING WITH 3 WE GET
X^2+4X=140
X^2+4X-140=0
X^2+14X-10X-140=0
X(X+14)-10(X+14)=0
(X-10)(X+14)=0
X-10=0 ....OR....X=10
SO THE 2 NUMBER ARE 10 AND 30....
OTHER POSSIBILITY IS X+14=0...OR X=-14...THE SECOND NUMBER IS -42

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SEE MY WORKING BELOW...
I AM USING * SYMBOL FOR INTERSECTION...X*Y MEANS X INTERSECTION Y.
AND # SYMBOL FOR UNION .....X#Y MEANS X UNION Y.
AND | MEANS COMPLEMENT OR NEGATION.....X| MEANS NOT X.
In the Junior Class the Followin statistics are true...
*116 students take algebra II...=A SAY
*99 students take Chemistry.....=C SAY
*49 take Spanish................=S SAY
*15 students that take algebra II also take Chemistry...= A AND C = A*C
*20 students that take algebra II also take Spanish.....= A AND S = A*S
*22 students that take Chemistry also take Spanish......= C AND S = C*S
*12 students take Chemistry algebra II and Spanish......= C AND A AND S = C*A*S
*31 students take none of th three classes..............= NEITHER C NOR A NOR S = (C|)#(A|)#(S|)
How many students are in the class?
I know the answer is 250... i solved it using a venn diagram and the teacher said it was right...
NOW LOOK AT YOUR VENN DIAGRAM AND ANALYSE....YOU WILL FIND...
A INCLUDES FOUR DISJOINT OR SEPERATE PARTS.. A*C ONLY,A*S ONLY, A ONLY AND FINALLY A AND C AND S
SIMILARLY
C INCLUDES FOUR DISJOINT OR SEPERATE PARTS.. C*A ONLY,C*S ONLY,C ONLY AND FINALLY C AND A AND S
SIMILARLY
S INCLUDES FOUR DISJOINT OR SEPERATE PARTS.. S*A ONLY,S*C ONLY,S ONLY AND
FINALLY S AND A AND C.
HENCE IF WE TOTAL A,C AND S WE GET A#C#S=TOTAL NUMBER OF STUDENTS TAKING ANY ONE OR MORE OF THESE 3 TOPICS BUT INCUDING THE DUPLICATED COUNTS OF THOSE TAKING MORE THAN ONE TOPIC...FURTHER WE FIND WITHIN THOSE ARE THERE TAKING ONLY ONE TOPIC,ONLY 2 TOPICS AND ALL 3 TOPICS..THAT IS MATHEMATICALLY SPEAKING
A INCUDES A*C AND A*C INCLUDES A*C*S...SIMILAR LOGIC HOLDS FOR ALL OTHER PARAMETERS...NOW WITH THE HELP OF VENN DIAGRAM ,WE CAN REMOVE THIS DUPLICATION..METHODICLLY AS FOLLOWS
TOTAL NUMBER OF STUDENTS TAKING ANY ONE OF THE 3 TOPICS(T)
T=(A#C#S)-{A*C+A*S+C*S-(A*C*S)}=A#C#S-A*C-A*S-C*S+A*C*S
TO GET THE TOTAL NUMBER OF STUDENTS IN THE CLASS (G.T)WE SHOULD ADD TO THE ABOVE THE NUMBER OF STUDENTS TAKING NONE OF THE ABOVE 3 TOPICS..THAT IS
G.T=A#C#S-A*C-A*S-C*S+A*C*S+A|#C|#S|
=116+99+49-15-20-22+12+31=250
THAT IS YOUR MATHEMATICAL FORMULA,,
HOPE YOU UNDERSTOOD AND YOU WILL GET YOUR BONUS SCORE...DONT FORGET TO REPLACE THE ABOVE SYMBOLS FOR INTERSECTION(*),UNION(#) AND NEGATION (|)WITH PROPER MATHS
SYMBOLS

but he said that there is a formulaic way to solve it and is giving MAJOR bonus points if i figure it out... so PLEASE i nned it solved out mathatmatically (meaning a formula)

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(t^3-t^2+t-1)divided by (t+1)
DO IT BY SHORT DIVISION METHOD...PUT T+1=0...SO T=-1
USE THIS TO DIVIDE....WRITE COEFFICIENTS IN ORDER
-1|.....1......-1........1..........-1
........0......-1........2..........-3
----------------------------------------------
........1......-2........3..........-4
SO ANSWER IS
T^2-2T+3.....AND REMAINDER IS -4.

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SEE THE FOLLOWING AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
This is a systems of equations problem and I need to solve for x,y,z. I do not how to get any of the variables or what steps to take to get the answers.
The problem:
2x-5y+ ___=-22
__+y+3z=10
x+___+8z=15
The underscores represents blanks or O
If this had to be an augmented matrix, how would I do that and get those variables?
1 solutions
Answer 20202 by venugopalramana(1462) About Me on 2006-04-16 22:01:33 (Show Source):
2x-5y+ ___=-22...........................I
__+y+3z=10.....................II
x+___+8z=15...........................III
2*EQN.III-EQN.I
2X+16Z-2X+5Y=30+22=52
5Y+16Z=52...........................IV
EQN.IV-5*EQN.II
5Y+16Z-5Y-15Z=52-50=2
Z=2....SUBSTITUTING IN EQN.III
X+8*2=15
X=15-16=-1
SUBSTITUTING FOR Z IN EQN.II
Y+3*2=10
Y=10-6=4
USING MATRIX METHOD...AUGMENTED MATRIX IS
2 -5 0 -22 1 0 0 ?
0 1 3 10 0 1 0 ?
1 0 8 15 0 0 1 ?
NR1=R1/2
1 -2.5 0 -11
0 1 3 10
1 0 8 15
NR3=R3-R1
1 -2.5 0 -11
0 1 3 10
0 2.5 8 26
NR3=R3-2.5*R2
1 -2.5 0 -11
0 1 3 10
0 0 0.5 1
NR3=R3/0.5
1 -2.5 0 -11
0 1 3 10
0 0 1 2
NR2=R2-3R3
1 -2.5 0 -11
0 1 0 4
0 0 1 2
NR1=R1+2.5R2
1 0 0 -1
0 1 0 4
0 0 1 2
HENCE
X=-1
Y=4
Z=2

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SEE THE FOLLOWING EXAMPLES TO KNOW ABOUT ELLIPSE..ITS EQN SHALL BE
(X-H)^2/A^2 + (Y-K)^2/B^2=1
WE HAVE SUCH POSSIBILITY IN A AND B ONLY..THERE IS A TYPING MISTAKE IN A CHECK UP

Linear_Algebra/30362: Question: Find the equation of the ellipse whose center is (5,-3) that has a vertex at 13,-3) and a minor axis of lenght 10.
POssible Answers:
(A) (x-5)^2/64 + (y+3)^2/25 = 1
(B) (x+5)^2/64 + (y-3)^2/25 = 1
(C) x^2/64 + y^2/25 = 1
(D) none of these
1 solutions
Answer 17014 by venugopalramana(1167) About Me on 2006-03-15 11:21:03 (Show Source):
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
--------------------------------------------------------------------------------
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
Quadratic-relations-and-conic-sections/30009: 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
Answer 16810 by venugopalramana(1167) About Me 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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SEE THE FOLLOWING EXAMPLES TO KNOW ABOUT ELLIPSE..ITS EQN SHALL BE
(X-H)^2/A^2 + (Y-K)^2/B^2=1
WE HAVE SUCH POSSIBILITY IN A AND B ONLY..THERE IS A TYPING MISTAKE IN A CHECK UP


Linear_Algebra/30362: Question: Find the equation of the ellipse whose center is (5,-3) that has a vertex at 13,-3) and a minor axis of lenght 10.
POssible Answers:
(A) (x-5)^2/64 + (y+3)^2/25 = 1
(B) (x+5)^2/64 + (y-3)^2/25 = 1
(C) x^2/64 + y^2/25 = 1
(D) none of these
1 solutions
Answer 17014 by venugopalramana(1167) About Me on 2006-03-15 11:21:03 (Show Source):
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
Quadratic-relations-and-conic-sections/30009: 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
Answer 16810 by venugopalramana(1167) About Me 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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A box contains 2 fifty paise coins,5 twenty five paise coins and a certain fixed number(N<=2) of ten and five paise
coins.If five coins are taken out of the box at random,then the probability that the total value of these 5 coins is
less than one rupee and fifty paise is which of the following:-
SINCE IT IS EASIER TO FIND THE NUMBER OF WAYS BY WHICH THE SUM OF 5 COINS CAN EQUAL OR EXCEED RS.1.50,LET US FIND IT OUT AND SUBTRACT THE RESULTANT PROBABILITY FROM ONE TO GET THE ANSWER
WE HAVE IN ALL 2+5+N=N+7 COINS
5 CAN BE TAKEN IN ...(N+7)C5 WAYS
CASE 1.....1*50+4*25=1.50....IN..(2C1)(5C4)=2*5=10
CASE 2.....2*50+3*25=1.75....IN..(2C2)(5C3)=1*10=10
CASE 3.....2*50+2*25+1*5 OR 10=1.55 OR 1.60...IN...(2C2)(5C2)(NC1)=1*10*N=10N
TOTAL NUMBER .....................=10+10+10N=20+10N
PROBABILITY OF GETTING 1.50 OR MORE =(20+10N)/{(N+7)C5}
PROBABILITY OF GETTING LESS THAN 1.50 =1- [(20+10N)/{(N+7)C5}]

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Question: Evaluate: I SUPPOSE YOU MEAN.....PUT BRACKETS TO AVOID CONFUSION....1n((1)/(5sqrt e))=LN(1)-LN(5SQRT(E))=0-LN(5)-LN(E^0.5)
=-LN(5)-0.5LN(E)=-LN(5)-0.5...CHECK THE PROBLEM FOR PROPER TYPING
I THINK THE QUESTION IS LN(1/FIFTH ROOT OF E)
IT IS NOT SQUARE ROOT
IF NOTHING IS PUT IN V GROOVE OF THE SYMBOL IT IS SQUARE ROOT
IF 3 IS PUT IN V GROOVE OF THE SYMBOL IT IS CUBE ROOT
IF 4 IS PUT IN V GROOVE OF THE SYMBOL IT IS FOURTH ROOT
IF 5 IS PUT IN V GROOVE OF THE SYMBOL IT IS FIFTH ROOT...ETC....
I HOPE YOU KNOW 5 TH ROOT
IF 2*2*2*2*2=2^5=32
WE SAY 5TH.ROOT OF 32 =32^(1/5)= 32^0.2= 2...SO ASSUMING THAT WE GET
LN(1/FIFTH ROOT OF E)=LN(1/E^(1/5))=-LN(E^(1/5))=-(1/5)LN(E)=-1/5 IS THE ANSWER.
HENCE B IS THE ANSWER

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SEE THE FOLLOWING AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
This is a systems of equations problem and I need to solve for x,y,z. I do not how to get any of the variables or what steps to take to get the answers.
The problem:
2x-5y+ ___=-22
__+y+3z=10
x+___+8z=15
The underscores represents blanks or O
If this had to be an augmented matrix, how would I do that and get those variables?
1 solutions
Answer 20202 by venugopalramana(1462) About Me on 2006-04-16 22:01:33 (Show Source):
2x-5y+ ___=-22...........................I
__+y+3z=10.....................II
x+___+8z=15...........................III
2*EQN.III-EQN.I
2X+16Z-2X+5Y=30+22=52
5Y+16Z=52...........................IV
EQN.IV-5*EQN.II
5Y+16Z-5Y-15Z=52-50=2
Z=2....SUBSTITUTING IN EQN.III
X+8*2=15
X=15-16=-1
SUBSTITUTING FOR Z IN EQN.II
Y+3*2=10
Y=10-6=4
USING MATRIX METHOD...AUGMENTED MATRIX IS
2 -5 0 -22 1 0 0 ?
0 1 3 10 0 1 0 ?
1 0 8 15 0 0 1 ?
NR1=R1/2
1 -2.5 0 -11
0 1 3 10
1 0 8 15
NR3=R3-R1
1 -2.5 0 -11
0 1 3 10
0 2.5 8 26
NR3=R3-2.5*R2
1 -2.5 0 -11
0 1 3 10
0 0 0.5 1
NR3=R3/0.5
1 -2.5 0 -11
0 1 3 10
0 0 1 2
NR2=R2-3R3
1 -2.5 0 -11
0 1 0 4
0 0 1 2
NR1=R1+2.5R2
1 0 0 -1
0 1 0 4
0 0 1 2
HENCE
X=-1
Y=4
Z=2

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Let P (n) represent the statment 2+6+10+...+(4n-2)=2n^2
In the proof that P (n) is true for all integers n, n>1, what term must be added to both sides of P (k) to show P (k+1) follows form p(k)?
(A) 4k+2
(B) P(k+1)
(C) 4k+6
(D) 4k-2
P(K)=(4K-2)...SO WE HAVE TO ADD THE NEXT TERM THAT IS P(K+1) TO BOTH SIDES TO PROVE THE EQUALITY BY INDUCTION METHOD.
SO P(K+1)=4(K+1)-2=4K+4-2=4K+2 IS THE NUMBER WE SHOULD ADD TO BOTH SIDES ....A IS THE ANSWER.