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(b) Find the equation of the sphere, which contains the circle
x^2 + y^2 + z^2 = 9, 3x + 3y + 3z = 5 and passes through the origin.
THAT IS CONTAINS CIRCLE
x^2 + y^2 + z^2 - 9=0=P SAY ,AND 3x + 3y + 3z - 5 =0=L SAY
ANY SPHERE THROUGH THE ABOVE CIRCLE IS GIVEN BY
P+KL=0...WHERE K IS A CONSTANT TO BE FOUND..WE ARE GIVEN IT PASSES THROUGH ORIGIN.HENCE
0^2+0^2+0^2-9+K(0+0+0-5)=0
-9-5K=0...........5K=-9
K=-9/5
HENCE EQN OF SPHERE IS
5(X^2+Y^2+Z^2-9)-9(3X+3Y+3Z-5)=0
5X^2+5Y^2+5Z^2-45-27X-27Y-27Z+45=0
5X^2+5Y^2+5Z^2-27X-27Y-27Z=0

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Find two positive real numbers..X AND Y SAY that differ by 1 and have a product of 1.
X-Y=1..................................I
XY=1
(X+Y)^2=(X-Y)^2+4XY=1^2+4*1=5
X+Y=SQRT(5).................................II
EQN.I+EQN.II GIVES
2X=1+SQRT(5)
X=(1+SQRT(5))/2=(2.236+1)/2=1.618
EQN.II-EQN.I..GIVES
2Y=SQRT(5)-1
Y=(SQRT(5)-1)/2=(2.236-1)/2=0.618

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f(x)=4x-1=Y SAY
g(x)=3-5x^2
find the compostion g of f(x)=G(Y)=3-5Y^2=3-5(4X-1)^2=3-5(16X^2-8X+1)
=3-80X^2+40X-5=-80X^2+40X-2

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SEE THE FOLLOWING EXAMPLES AND COME BACK IF STILL IN DOUBT
When using the quadratic formula to solve a quadratic equation ax2 + bx + c = 0, the discriminant is b2 - 4ac. This discriminant can be positive, zero, or negative. (When the discriminate is negative, then we have the square root of a negative number. This is called an imaginary number, sqrt(-1) = i. )
Explain what the value of the discriminant means to the graph of y = ax2 + bx + c. Hint: Chose values of a, b and c to create a particular discriminant. Then, graph the corresponding equation.
CASE 1....DISCRIMINANT (D SAY) IS POSITIVE.....
EX..LET Y = X^2-5X+6=0..
D=5^2-4*1*6=25-24=1...
HENCE ROOTS ARE
REAL,THAT IS THE GRAPH CUTS THE X AXIS AT 2 REAL POINTS
DISTINCT ...2 AND 3
AND THE FUNCTION Y COULD BE POSITIVE OR NEGATIVE WITH A MAXIMUM OR MINIMUM
SEE GRAPH BELOW

CASE 2.....D=0
EX....LET...Y=X^2-2X+1=0
D=2^2-4*1*1=4-4=0
HENCE ROOTS ARE
REAL.THAT IS THE GRAPH CUTS THE X AXIS AT 1 REAL POINT.
EQUAL...1 AND 1
AND THE FUNCTION Y IS ALWAYS NON NEGATIVE OR NON POSITIVE DEPENDING ON THE SIGN OF COEFFICIENT OF X^2 BEING POSITIVE OR NEGATIVE , WITH A MINIMUM OR MAXIMUM VALUE OF ZERO.
SEE GRAPH BELOW.

CASE 3.......D IS NEGATIVE
EX....LET Y = X^2+X+1=0
D=1^2-4*1*1=1-4=-3
HENCE ROOTS ARE
IMAGINARY.THAT IS THE GRAPH DOES NOT CUT THE X AXIS.
DISTINCT.....(-1+iSQRT(3))/2....AND (-1-iSQRT(3))/2
AND THE FUNCTION Y IS ALWAYS POSITIVE SINCE COEFFICIENT OF X^2 IS POSITIVE.
SEE THE GRAPH BELOW...


EX....LET Y = -X^2+X-1=0
D=1^2-4*(-1)*(-1)=1-4=-3
HENCE ROOTS ARE
IMAGINARY.THAT IS THE GRAPH DOES NOT CUT THE X AXIS.
DISTINCT.....(1-iSQRT(3))/2....AND (1+iSQRT(3))/2
AND THE FUNCTION Y IS ALWAYS NEGATIVE SINCE COEFFICIENT OF X^2 IS NEGATIVE.
SEE THE GRAPH BELOW...

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find f^-1(x) if LET Y=f(x) =2x-1
2X=Y+1
X=(Y+1)/2
F^-1(X)=(F(X)+1)/2

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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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TIP:
1.IF X^2 AND Y^2 HAVE EQUAL COEFFICIENTS AND THERE IS NO XY TERM ,IT
COULD BE A CIRCLE.
1.) X2 + y2 + 4x – 6y = 3
COMPLETE SQUARE.
(X^2+2*2X+2^2)-2^2+(Y^2+2*3Y+3^2)-3^2=3
(X+2)^2+(Y+3)^2=3+9+4=16=4^2
THIS IS THE EQN OF A CIRCLE WHOSE STD.FORM IS
(X-H)^2+(Y-K)^2=R^2,WHERE (H,K) IS THE CENTRE OF THE CIRCLE AND R IS
THE RADIUS.HENCE
(-2,-3) IS THE CENTRE AND 4 IS THE RADIUS.

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TIP:
1.IF X^2 AND Y^2 HAVE EQUAL COEFFICIENTS AND THERE IS NO XY TERM ,IT
COULD BE A CIRCLE.
1.) X2 + y2 + 4x – 6y = 3
COMPLETE SQUARE.
(X^2+2*2X+2^2)-2^2+(Y^2+2*3Y+3^2)-3^2=3
(X+2)^2+(Y+3)^2=3+9+4=16=4^2
THIS IS THE EQN OF A CIRCLE WHOSE STD.FORM IS
(X-H)^2+(Y-K)^2=R^2,WHERE (H,K) IS THE CENTRE OF THE CIRCLE AND R IS
THE RADIUS.HENCE
(-2,-3) IS THE CENTRE AND 4 IS THE RADIUS.

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TIP:
2.IF X^2 OR Y^2 IS ONLY PRESENT ,IT COULD BE A PARABOLA.
2.) x2 – 4x – 8y = 12
COMPLETE SQUARE
(X^2-2*2X+2^2)-2^2=12+8Y
(X-2)^2=8Y+16=8(Y+2)
THIS IS THE EQN.OF A PARABOLA .STD EQN. IS
(X-H)^2=4A(Y-K),WHERE
(H,K) IS THE VERTEX....(2,-2) HERE.
4A=LATUS RECTUM =8 HERE...A=2
FOCUS IS (H+A,K).....(2+2,-2)=(4,-2)..HERE.
DIRECTRIX IS X-H+A=0...
X-2+2=0...OR....X=0..
AXIS IS Y-K =0..Y+2=0

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TIP
IF X^2 AND Y^2 HAVE DIFFERENT COEFFICIENTS WITH OPPOSITE SIGNS THEN IT
COULD BE HYPERBOLA
3.) 9x2 – 96y = 16y2 + 18x + 279
COMPLETE SQUARE...
{(3X)^2-2*3X*3+3^2}-3^2-{(4Y)^2+2*4Y*12+12^2}-12^2=279
(3X+3)^2-(4Y+12)^2=279+9+144=432
9(X+1)^2-16(Y+3)^2=432......NOW DIVIDE THROUGH OUT WITH 432 TO GET 1 ON THE RHS.
(X+1)^2/(432/9) -(Y+3)^2/(432/16)=1
(X+1)^2/48 - (Y+3)^2/27 =1
THIS THE EQN. OF A HYPERBOLA.STD.EQN.IS.
(X-H)^2/A^2 - (Y-K)^2/B^2=1..WHERE
(H,K) IS CENTRE.....(-1,-3) HERE
TRANSVERSE AXIS IS Y=K...Y=-3
LENGTH OF TRANSVERSE AXIS=2A..
...=2SQRT(48)
CONJUGATE AXIS IS X=H.......X=-1
LENGTH OF CONJUGATE AXIS = 2B
=2SQRT(27)
ECCENTRICITY=E=SQRT{(A^2+B^2)/A^2}
=SQRT{(48+27)/48}=SQRT(75/48)
A*E=SQRT(48)*SQRT(75/48)=SQRT(75)
FOCI ARE (H+-AE,K)......(-1+SQRT(75),-3)
AND ......(-1-SQRT(75),-3)
A/E=SQRT(48)/SQRT(75/48)=48/SQRT(75)
DIRECTRIX ARE X=H+-A/E....
X=-1+48/SQRT(75)...AND
X=-1-48/SQRT(75)
ASYMPTOTES ARE GIVEN BY
(X-H)^2/A^2 = (Y-K)^2/B^2
OR
(X-H)/A=+(Y-K)/B AND............(X+1)/SQRT(48) =(Y+3)/SQRT(27)
(X-H)/A=-(Y-K)/B.............(X+1)/SQRT(48) = -(Y+3)/SQRT(27)
GRAPH IS GIVEN BELOW..

--------------------------------------------------------

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TIP:
IF X^2 AND Y^2 HAVE DIFFERENT COEFFICIENTS OF SAME SIGN ,THEN IT COULD
BE ELLIPSE.
4.) 4(x-1)2 = 4-y2
I HOPE IT IS
4(X-1)^2+Y^2=4...DIVIDE WITH 4
(X-1)^2/1^2+Y^2/2^2=1
THIS THE EQN.OF AN ELLIPSE.STD.EQN. IS
(X-H)^2/A^2 +(Y-K)^2/B^2=1
WHERE
(H,K) IS CENTRTE....(1,0).HERE
LENGTH OF MAJOR AXIS IS 2B...2*2=4
MINOR AXIS IS ALONG Y=K....Y=0
LENGTH OF MINOR AXIS IS 2A...2*1=2
ECCENTRICITY=E=SQRT{(B^2-A^2)/A^2}
=SQRT(4-1)/1=SQRT(3)
B*E=2SQRT(3)
B/E=2/SQRT(3)
FOCI ARE (H,K+BE) AND (H,K-BE)
(1,2SQRT(3)) AND (1,-2SQRT(3))
DIRECTRIX ARE.Y=K+B/E AND K-B/E
Y=2/SQRT(3)..AND...Y=-2/SQRT(3)
GRAPH IS GIVEN BELOW...

-----------------------------------------------------------------

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Why is it true that any two points satisfying a linear equation will give you the same graph for the line represented by the equation?
Please Help...WELL THE ANSWER IS IN YOUR QUESTION ITSELF!!LET ME ELABORATE...
FIRSTLY WE SHOULD UNDERSTAND WHAT IS A GRAPH,WHAT IS ITS EQUATION AND VICEVERSA..
AND WHAT IS MEANT BY SATISFYING AN EQUATION.
1.WE SAY A POINT SATISFIES AN EQUATION WHEN ITS COORDINATES ARE SUBSTITUTED IN THE EQUATION MAKES ITS RHS EQUAL ITS LHS.
2.WE SAY A POINT DOES NOT SATISFY AN EQUATION WHEN ITS COORDINATES ARE SUBSTITUTED IN THE EQUATION DOES NOT MAKE ITS RHS EQUAL ITS LHS.
NEXT....
1.WE SAY A GRAPH IS REPRESENTED BY AN EQUATION OR THE GIVEN EQN.IS FOR A PARTICULAR GRAPH,WHEN THE POINTS OBTAINED AS ABOVE FROM THE GIVEN EQN.,ARE PLOTTED AND JOINED
BY A LINE ,THEY FORM THE GRAPH UNDER CONSIDERATION
THESE APPEAR TO BE ELEMENTARY ,BUT THEY STIPULATE 4 IMPORTANT CONSIDERATION AT THE SAME TIME WHICH WILL ANSWER YOUR QUESTION...THEY ARE
GIVEN AS MENTIONED ABOVE AN EQUATION AND ITS GRAPH.....THEN....
1.IF A POINT SATISFIES THAT EQN. THEN IT LIES ON THE GRAPH.
2.IF A POINT DOES NOT SATISFY THAT EQN. THEN IT DOES NOT LIE ON THE GRAPH.
3.IF A POINT IS ON THE GRAPH,THEN IT SATISFIES THE EQN.
4.IF A POINT IS NOT ON THE GRAPH,THEN IT DOES NOT SATISFY THE EQN.
WITH THE ABOVE UNDERSTANDING LET US ANSWER YOUR DOUBT.
THE EQN. IS A LINEAR EQN.YOU WILL SOON LEARN THAT GRAPH OF A LINEAR EQN.IS A STRAIGHT LINE.
SO ONCE IT IS GIVEN THAT THE EQN.IS LINEAR IN X AND Y IT AUTOMATICALLY IMPLIES THAT THE GRAPH IS A STRAIGHT LINE IN PLANE GEOMETRY.
NOW WE KNOW THAT WE NEED ONLY 2 POINTS TO FIX A STRAIGHT LINE.
HENCE ONCE THE LINE IS FIXED BY THE 2 GIVEN POINTS SATISFYING THE GIVEN EQN.ANY OTHER POINT ON THE LINE SHOULD SATISFY THE EQN. AS ELABORATED ABOVE IN THE 4 STIPULATIONS.THAT IS THEY GIVE THE SAME STRAIGHT LINE WHATEVER 2 POINTS YOU TAKE.HENCE ANY 2 POINTS SATISFYING A LINEAR EQN. WILL GIVE YOU THE SAME GRAPH
FOR THE LINE REPRESENTED BY THE EQN. SINCE
THE GRAPH IS A STRAIGHT LINE FOR LINEAR EQN. AND ANY 2 POINTS ON THE LINE UNIQUELY DETERMINE THE LINE.

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The sum of a geometric series is 57232. The common ratio is 2 and the last term is 28672. What is the first term?
FORMULA FOR N TH TERM IS ....
TN=A*R^(N-1)..HENCE
28672=A*2^(N-1).......................I
FORMULA FOR SUM IS
SN=A*{R^N - 1 }/(R-1)...HENCE
57232 = A*(2^N-1)/(2-1)=A*(2^N-1)....................II
EQN.II/EQN.I...
57232/28672=A*(2^N -1)/{A*2^(N-1)}=(2^N -1)/2^(N-1)
57232/28672=2^N/2^(N-1) - 1/2^(N-1)=2-1/2^(N-1)
1/2^(N-1)=2-57232/28672=(57344-57232)/28672=112/28672
2^(N-1)=28672/112=256=2^7
N-1=7….OR……N=8…SUBSTITUTING IN EQN.I…
28672=A*2^(8-1)=256A…..OR…..A = 28672/256=112

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GOOD PROBLEM .YOU CAN USE IT TO
GOOD EFFECT TO PUT ACROSS A PROPOSAL TO YOUR FRIEND ...
YOU PROMISS HIM A 100,000..POUNDS EVERY DAY FOR 30
DAYS.LET HIM GIVE YOU IN RETURN JUST A PENNY ON THE
FIRST DAY,2 PENNIES ON THE SECOND DAY ,4 PENNIES ON
THE THIRD DAY,......ETC FOR 30 DAYS ..AS IN YOUR
PROBLEM... I PRESUME HE WILL
LAP ON THE PROPOSAL AND I CAN ASSURE YOU THAT YOU WONT
REGRET THE PROPOSAL!!!!
OK NOW LET US GET BACK TO YOUR PROBLEM AS WELL AS MY
SUGGESTION ,WHICH IS JUST A PART OF IT .I AM GIVING
BELOW THE MONEY TO BE PAID ON THIS BASIS FOR 30 AND 64
DAYS ,ROUNDED OFF TO POUNDS OR MILLION POUNDS AT THE
LATER DAYS.
DAY.....MONEY TO BE PAID........MONEY TO BE PAID BY
YOUR FRIEND TO YOU IN
............TO YOUR FRIEND.....
CENTS...POUNDS...MILLION POUNDS
................POUNDS
1 100000 1 0 0
2 100000 2 0 0
3 100000 4 0 0
4 100000 8 0 0
5 100000 16 0 0
6 100000 32 0 0
7 100000 64 1 0
8 100000 128 1 0
9 100000 256 3 0
10 100000 512 5 0
11 100000 1024 10 0
12 100000 2048 20 0
13 100000 4096 41 0
14 100000 8192 82 0
15 100000 16384 164 0
16 100000 32768 328 0
17 100000 65536 655 0
18 100000 131072 1311 0
19 100000 262144 2621 0
20 100000 524288 5243 0
21 100000 1048576 10486 0
22 100000 2097152 20972 0
23 100000 4194304 41943 0
24 100000 8388608 83886 0
25 100000 16777216 167772 0
26 100000 33554432 335544 0
27 100000 67108864 671089 1
28 100000 134217728 1342177 1
29 100000 268435456 2684355 3
30 100000 536870912 5368709 5
31 100000 1073741824 10737418 11
32 100000 2147483648 21474836 21
33 100000 4294967296 42949673 43
34 100000 8589934592 85899346 86
35 100000 17179869184 171798692 172
36 100000 34359738368 343597384 344
37 100000 68719476736 687194767 687
38 100000 1.37439E+11 1374389535 1374
39 100000 2.74878E+11 2748779069 2749
40 100000 5.49756E+11 5497558139 5498
41 100000 1.09951E+12 10995116278 10995
42 100000 2.19902E+12 21990232556 21990
43 100000 4.39805E+12 43980465111 43980
44 100000 8.79609E+12 87960930222 87961
45 100000 1.75922E+13 175921860444 175922
46 100000 3.51844E+13 351843720888 351844
47 100000 7.03687E+13 703687441777 703687
48 100000 1.40737E+14 1407374883553 1407375
49 100000 2.81475E+14 2814749767107 2814750
50 100000 5.6295E+14 5629499534213 5629500
51 100000 1.1259E+15 11258999068426 11258999
52 100000 2.2518E+15 22517998136853 22517998
53 100000 4.5036E+15 45035996273705 45035996
54 100000 9.0072E+15 90071992547410 90071993
55 100000 1.80144E+16 180143985094820 180143985
56 100000 3.60288E+16 360287970189640 360287970
57 100000 7.20576E+16 720575940379279 720575940
58 100000 1.44115E+17 1441151880758560 1441151881
59 100000 2.8823E+17 2882303761517120 2882303762
60 100000 5.76461E+17 5764607523034230 5764607523
61 100000 1.15292E+18 11529215046068500 11529215046
62 100000 2.30584E+18 23058430092136900 23058430092
63 100000 4.61169E+18 46116860184273900 46116860184
64 100000 9.22337E+18 92233720368547800 92233720369
30
DAY..3000000.........1073741823......10737418.....................11
TOTAL
64
DAY..6400000.........1.84467E+19.....184467440737095000......184467440737
TOTAL
YOUR GAIN IN 30 DAYS = 8 MILLION POUNDS..GOT IT BUT
ONLY BE CAREFULL THAT YOUR FRIEND WONT RUN OUT OF YOU
ON THE 25 TH. DAY.!!!
YOUR GAIN IN 30 DAYS = 184467440731 MILLION POUNDS
NOW COMING TO THE MATHS PART OF THIS ,THIS SEQUENCE
WHERE EACH NUMBER BEARS A CONSTANT RATIO TO ITS
PREDECESSOR IS CALLED GEOMETRIC PROGRESSION..HERE YOU
FIND EACH NUMBER IS OBTAINED FROM THE PREVIOUS ONE BY
MULTIPLYING WITH 2 , CALLED COMMON RATIO.
THE LAST NUMBER ON NTH. DAY AND SUM OF SUCH SERIES OF
NUMBERS UPTO THE N TH.DAY IS GIVEN BY THE FOLLOWING
FORMULAE...
N TH. NUMBER = FIRST NUMBER *(C0MM0N RATIO
)^(N-1)=A*(R)^(N-1)=(2)^(N-1) IN THIS CASE.
SUM UP TO N TH.NUMBER =A*{((R)^N -
1))/(R-1)}={(2)^N-1}IN THIS CASE

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When using the quadratic formula to solve a quadratic equation ax2 + bx + c = 0, the discriminant is b2 - 4ac. This discriminant can be positive, zero, or negative. (When the discriminate is negative, then we have the square root of a negative number. This is called an imaginary number, sqrt(-1) = i. )
Explain what the value of the discriminant means to the graph of y = ax2 + bx + c. Hint: Chose values of a, b and c to create a particular discriminant. Then, graph the corresponding equation.
CASE 1....DISCRIMINANT (D SAY) IS POSITIVE.....
EX..LET Y = X^2-5X+6=0..
D=5^2-4*1*6=25-24=1...
HENCE ROOTS ARE
REAL,THAT IS THE GRAPH CUTS THE X AXIS AT 2 REAL POINTS
DISTINCT ...2 AND 3
AND THE FUNCTION Y COULD BE POSITIVE OR NEGATIVE WITH A MAXIMUM OR MINIMUM
SEE GRAPH BELOW

CASE 2.....D=0
EX....LET...Y=X^2-2X+1=0
D=2^2-4*1*1=4-4=0
HENCE ROOTS ARE
REAL.THAT IS THE GRAPH CUTS THE X AXIS AT 1 REAL POINT.
EQUAL...1 AND 1
AND THE FUNCTION Y IS ALWAYS NON NEGATIVE OR NON POSITIVE DEPENDING ON THE SIGN OF COEFFICIENT OF X^2 BEING POSITIVE OR NEGATIVE , WITH A MINIMUM OR MAXIMUM VALUE OF ZERO.
SEE GRAPH BELOW.

CASE 3.......D IS NEGATIVE
EX....LET Y = X^2+X+1=0
D=1^2-4*1*1=1-4=-3
HENCE ROOTS ARE
IMAGINARY.THAT IS THE GRAPH DOES NOT CUT THE X AXIS.
DISTINCT.....(-1+iSQRT(3))/2....AND (-1-iSQRT(3))/2
AND THE FUNCTION Y IS ALWAYS POSITIVE SINCE COEFFICIENT OF X^2 IS POSITIVE.
SEE THE GRAPH BELOW...


EX....LET Y = -X^2+X-1=0
D=1^2-4*(-1)*(-1)=1-4=-3
HENCE ROOTS ARE
IMAGINARY.THAT IS THE GRAPH DOES NOT CUT THE X AXIS.
DISTINCT.....(1-iSQRT(3))/2....AND (1+iSQRT(3))/2
AND THE FUNCTION Y IS ALWAYS NEGATIVE SINCE COEFFICIENT OF X^2 IS NEGATIVE.
SEE THE GRAPH BELOW...

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Solve the determinant
3 x
1 1 = 2
that is
3*1-1*x=2
3-x=2
x=3-2=1
x=1

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a) What is the remainder left after dividing
1! + 2! + 3! +…………………+ 100! by 7?
GROUP AS FOLLOWS
=(1!+3!)+(2!+4!)+(5!+6!)+(7!+8!+9!+..........+100!)
=(7)+(26)+5!(1+6)+7!(1+8+8.9+............+8.9.10......100)...WE FIND THAT EXCEPT 26 THE SECOND TERM, ALL THE OTHER TERMS ARE DIVISIBLE BY 7.HENCE THE REMAINDER WE GET ON DIVIDING WITH 7 IS THE REMAINDER WE GET WHEN WE DO 26/7...THAT IS ...5 IS THE REMAINDER .
b) If n >=4 then find whether 2.n is less than 4.n or not.
CHECK THE QUESTION 2.N MEAns 2*n ..if n=5 say 2*5=10.....it is definitely less than 4*n=4*5=20
what is there to prove in that?

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The sides of the triangle are 6,7,and x. What is the largest value that could be the area of such a triangle?
HOPE YOU KNOW ELEMENTARY TRIGNOMMETRY.THEN IT IS VERY EASY.OTHERWISE ,PLEASE INFORM AND I SHALL SHOW YOUB ANOTHER METHOD
AREA OF TRIANGLE =(1/2)*BASE*ALTITUDE=0.5*b*h=0.5*b*c*SIN(A)=0.5*6*7*SIN(A)
=21SIN(A).....WE KNOW THAT MAXIMUM VALUE OF SIN(A)=1 WHEN A=90 DEGREES
HENCE MAXIMUM AREA =21*1=21 WHEN IT IS RIGHT ANGLED TRIANGLE WITH LEGS EQUAL TO 6 AND 7.

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STD.EQN. OF A CIRCLE IS
(X-H)^2+(Y-K)^2=R^2...
WHERE (H,K)IS THE CENTRE AND R IS THE RADIUS.USING THIS FORMULA...
(1) find the center of the circle whose equation is
x^2+ y^2-3x+8y-8=0
{X^2-2*(3/2)X+(3/2)^2}+{Y^2+2*4*Y+4^2}-(3/2)^2-4^2-8=0
{X-(3/2)}^2+{Y+4}^2=24+9/4=105/4..HENCE CENTRE OF CIRCLE IS {(3/2),-4}
(2) write the equation of the circle with center (-2,1) and which passes through (3,4).
EQN IS (X+2)^2+(Y-1)^2=R^2...IT IS` PASSING THROUGH (3,4)..SO...
(3+2)^2+(4-1)^2=R^2
25+9=34=R^2.HENCE EQN.OF CIRCLE IS
(X+2)^2+(Y-1)^2=34

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Step by step
Original- ax^2 + bx + c =0
1. subtract c from each side...OK...WE GET AX^2+BX=-C
2. Divide each side by a......OK...WE GET..X^2+BX/A=-C/A
3. Add he square of half the coefficient of x to each side..OK..WE GET
X^2+BX/A+(B/2A)^2=(B/2A)^2-(C/A)
4.write the left side as a perfect square...OK...WE GET ..
{X+(B/2A)}^2=B^2/4A^2 - C/A
5.use a common denominator to express the right side as a single fraction...OK...
{X+(B/2A)}^2=(B^2-4AC)/4A^2
6. find the square root of eac side....OK......WE GET
X+(B/2A)=+ OR - {SQRT(B^2-4AC)}/2A
7. solve for x by subtracting the same term form each side...OK....
X=-(B/2A)+OR-{SQRT(B^2-4AC)}/2A
8. use a common denominator to express the right side as a single fraction
X=[-B+OR-{SQRT(B^2-4AC)}]/(2A)
THAT IS IT GOT IT?..OK...
once all the steps are done you should end up with the quadratice formula but i cant get past step three and i need each step. PLEASE HELp

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A tower is 275 feet high. From it's top, the angle of depression to a rock on the ground is 25degrees. Find the distance from the base of the building of the the tower to the rock.
LET BT BE THE TOWER WITH B ON THE GROUND AND T THE TOP.
BT=275
LET THE ROCK ON GROUND BE AT R.
TRIANGLE BTR IS RIGHT ANGLED AT B.
ANGLE TBR=90
ANGLE BRT=25
TAN(25) =BT/BR=275/BR
BR =275/TAN(25)=590

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Find the equation of the line joining the points
(─5, 2, 3) and (5, ─ 2, 3).
THE FORMULA IS
(X-X1)/(X2-X1)=(Y-Y1)/(Y2-Y1)=(Z-Z1)/(Z2-Z1)
(X+5)/(5+5)=(Y-2)/(-2-2)=(Z-3)/(3-3)
(X+5)/10=-(Y-2)/4=(Z-3)/0
OR
Z=3 AND
4(X+5)=-10(Y-2)..OR
2(X+5)+5(Y-2)=0...OR
2X+5Y=0

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Could you help me understand standard deviation? Here is my problem:
Consider the following data set: 28, 34, 41, 19, 17, 23.
a) Compute the mean and standard deviation of the data set
b) Suppose each value is mulitplied by two and then 5 is added to each of the values after each is multiplied by two. Compute the mean and standard deviation of this new data set.
Thanks for your help!!!
......... DEVIATION SQUARE OF
S.NO..... X WRT.MEAN DEVIATION
1.......... 28 1 .............1
2....... 34 7 ............49
3....... 41 14 ...........196
4....... 19 -8 ............64
5....... 17 -10 ............100
6....... 23 -4 .............16
---------------------------------------------------
TOTAL... 162 0 .............426
MEAN=162/6=... 27...... MEAN.OF SQUARES =426/6=71
........OF DEVIATION
R.M.S.DEVN.=SQRT(71)= 8.43
OR STD.DEVN.

DEVIATION SQUARE OF
S.NO. X WRT.MEAN DEVIATION
1 61 34...... 1156
2 73 46......... 2116
3 87 60...... 3600
4 43 16....... 256
5 39 12...... 144
6 51 24...... 576
TOTAL 354 192..... 7848
MEAN=354/6= 59...... MEAN.OF SQUARES
................OF DEVNS.
.........=7848/6=1308

R.M.S.DEVN. = SQRT(1308)=36.16628264
OR STD.DEVN.

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The angles of a polygon add up to 2700 degrees. How many sides does the polygon have?
SUM OF ANGLES =(2N-4)*90 DEGRES=2700
2N-4=2700/90=30
2N=30+4=34
N=34/2=17

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what is the sum of the angles of a polygon with 22 sides?
SUM OF ANGLES IN A POLYGON OF N SIDES IS GIVEN BY
S=(2N-4)*90..DEGREES
FOR N=22
S=(2*22-4)*90=18*90=1620 DEGREES

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given: points N (-1,-5), O( 0,0),P (3,2) AND Q(8,1)SHOW NPQT IS AN ISOSCELES TRAPEZOID
SLOPE OF OP =(Y2-Y1)/(X2-X1)=(2-0)/(3-0)=2/3
SLOPE OF NQ =(1+5)/(8+1)=6/9=2/3
HENCE OP IS PARALLEL TO NQ
FURTHER
ON^2=(0+1)^2+(0+5)^2=1+25=26
PQ^2=(8-3)^2+(1-2)^2=5^2+1^2=25+1=26
HENCE ON =PQ
HENCE OPQN IS AN ISOCELLES TRAPEZIUM.

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given: points N (-1,-5), O( 0,0),P (3,2) AND Q(8,1)SHOW NPQT IS AN ISOSCELES TRAPEZOID
SLOPE OF OP =(Y2-Y1)/(X2-X1)=(2-0)/(3-0)=2/3
SLOPE OF NQ =(1+5)/(8+1)=6/9=2/3
HENCE OP IS PARALLEL TO NQ
FURTHER
ON^2=(0+1)^2+(0+5)^2=1+25=26
PQ^2=(8-3)^2+(1-2)^2=5^2+1^2=25+1=26
HENCE ON =PQ
HENCE OPQN IS AN ISOCELLES TRAPEZIUM.

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How do mathematicians get the formula: lal = 1/4c (absolute value of 'a' equals one-fourth times 'c')? This formula is used to determine the equation of a parabola. Please explain how mathematicians got this formula. Thank you for your time.
Julia
I TAKE IT THAT YOU ARE REFERRING TO EQN. OF PARABOLA..IN STD. FORM
Y^2=4AX...........WITH AN EQN.IN THE FORM .......Y^2=CX...OR...SAY...Y^=X..OR SO...WELL

PARABOLA IS THE LOCUS OF A POINT (P SAY ..WITH COOORDINATES X AND Y )WHICH MOVES SUCH THAT ITS DISTANCE FROM A POINT CALLED FOCUS (F SAY) = ITS DISTANCE FROM A LINE CALLED DIRECTRIX.
IN THE STANDARD FORM , A PERPENDICULAR FROM FOCUS ( F G SAY)TO DIRECTRIX IS TAKEN AS X AXIS.
MID POINT ON THIS IS TAKEN AS THE ORIGIN O (0,0).AND A LINE PARALLEL TO DIRECTRIX THROUGH O AS THE Y AXIS.
LET OF = A ....SINCE OF=OG BY CONSTRUCTION,AS O IS THE MIDPOINT OF FG, WE HAVE OF=OG=A.
HENCE AS PER THE DEFINITION OF PARABOLA ,O IS A POINT SUCH THAT ITS DISTANCE FROM FOCUS F =OF IS AME AS ITS DISTANCE FROM DIRECTRIX =OG.
HENCE O LIES ON THE PARABOLA.IT IS CALLED VERTEX.
HENCE COORDINATES OF F..FOCUS ARE (A,0)
EQN. OF DIRECTRIX IS X=-A..SINCE IT IS PARALLEL TO Y AXIS AND IS AT A CONSTANT DISTANCE OF -A FROM IT.
NOW IF P IS ANY POINT ON THE PARABOLA , ITS DISTANCE FROM F IS GIVEN BY

D^2=(X-A)^2+(Y-0)^2
THIS IS EQUAL TO ITS DISTANCE FROM DIRECTRIX = X+A
HENCE WE HAVE
(X+A)^2=(X-A)^2+Y^2
Y^2=X^2+A^2+2AX-(X^2+A^2-2AX)=4AX
HENCE Y^2 =4AX IS THE EQN.OF PARABOLA IN ITS STANDARD FORM
NOW IN A GENERAL PROBLEM IF IT IS GIVEN THAT
Y^2=8X ..0R..IN GENERAL....CX...THEN IT MEANS
Y^2=4*2X...OR...A=2.....IN GENERAL .....Y^2=4*(C/4)*(X)...HENCE COMPARING WE GET
A=C/4
HOPE YOU UNDERSTOOD.

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Write the first five terms of the follownig sequence.
a sub n = n + 1/ n IS AN INDEPENDENT PROBLEM..FOR THIS
A1=1+1/1=2
A2=2+1/2=5/2
A3=3+1/3=10/3
A4=4+1/4=17/4
A5=5+1/5=26/5
with the condition: a sub n= a sub n-1 + 2, where a sub n= 1....THIS IS A SEPERATE PROBLEM....
A1=1
A2=A1+2=1+2=3
A3=A2+2=3+2=5
A4=A3+2=5+2=7
A5=A4+2=7+2=9

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135x2 - 222x + 91
135X^2-105X-117X+91=0
15X(9X-7)-13(9X-7)=0
(15x- 13)(9x-7)=0
15X-13=0......THAT IS .....15X=13....X=13/15
OR
9X-7=0........THAT IS ......9X=7.......X=7/9

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apply the quadratic formula to find the roots of the given function, and then graph the function.
f(x) = x^2 + 4
D=B^2-4AC=0-16....
NO REAL ROOTS....2 COMPLEX CONJUGATE ROOTS