You can put this solution on YOUR website!
3(x+2)(x-3)(x+9)=0
IF A PRODUCT IS ZERO THEN ONE OF THE FACTORS SHOULD BE ZERO.
HENCE WE GET
X+2=0...OR
X-3=0....OR
X+9=0...THAT IS
X=-2 OR 3 OR -9 ..SO THEY ARE THE POSSIBLE SOLUTIONS FOR THIS EQUATION.

You can put this solution on YOUR website!
SEE THE FOLLOWING EXAMPLE ND TRY.IN CASE OF DIFFICULTY PLEASE COME BACK
Hi, I'm in homeschooling and I'm having trouble with matrices. I was wondering how to solve the problem where you have to find the x,y, and z values in the matrix:
[7 -7 5 | 9]
[9 5 -7 | -17]
[6 1 -7 | -2]
I'd appreciate the help. Thank you!
Caitlyn Reese
1 solutions
Answer 9969 by venugopalramana(585) About Me on 2005-11-28 07:14:11 (Show Source):
the 4 column heads represent x,y,z and constant term in the matrix of system of eqns.
then each row gives us one eqn.like say row 1 gives us that 7x-7y+5z=9..etc…
hence if we can make the matrix to become
1 0 0 ?
0 1 0 ??
0 0 1 ???
then from the explantion given above it means 1x=?.1y=?? And 1z=???
so we try to transform the matrix in to that form..by the following steps.
in fact using the above explanation,you can see that what we do at each step is just
divide each eqn. with a constant/add/subtract etc which does not change the basic
eqn.for ex. dividing row 1 by 7 means change the given eqn.7x-7y+5z=9 to x-y+5z/7=9/7
legend:- or1 means old row 1..nr1 means new row 1…r1 means the existing row 1 please note that no changes are made in rows other than those mentioned at each step.
start with given matrix …

7...... -7..... 5...... 9
9...... 5...... -7..... -17
6...... 1...... -7..... -2
step 1…we want to make 1st.row 1st.column as 1….so….nr1=or1/7...
1...... -1..... (5/7).. (9/7)
9...... 5...... -7..... -17
6...... 1...... -7..... -2
step 2..we want to make 2nd/3rd.rows,col.1 as 0...so...nr2=or2-9*r1........nr3=or3-6*r1
1...... -1..... (5/7).. (9/7)
0... 14..... (-7-9*5/7).... (-17-9*9/7)
0...... 7...... (-7-6*5/7)..... (-2-6*9/7)
step 3…we want to make 2nd.row.2nd.col.as 1..so..nr2=or2/14
1...... -1..... 5/7.... 9/7
0...... 1...... (-94/7)/14..... (-200/7)/14
0...... 7...... (-7-6*5/7)..... (-2-6*9/7)
step 4..we want to make 3rd.row.2nd.col.as 0…so….nr3=or3-7*r2
1...... -1..... (5/7).. (9/7)
0...... 1...... (-94/7)/14..... (-200/7)/14
0 0 (-79/7)-7*(-94/98) (68/7)-7*(-200/98)
step 5….we want to make 3rd.row.3rd.col.as 1…so….nr3=or3/(-32/7)
1...... -1..... (5/7).. (9/7)
0...... 1...... (-94/98)... (-200/98)
0...... 0..... 1...... -1
step 6…we want to make 1st/2nd.row 3rd.col.as 0..so..nr1=or1-5*r3/7...nr2=or2+94*r3/98
1...... -1..... 0..... 2
0...... 1...... 0...... -3
0...... 0...... 1...... -1
step7….we want to make 1st.row 2nd.col.as 0..so….nr1=or1+r2
1...... 0...... 0...... -1
0...... 1...... 0...... -3
0...... 0...... 1...... -1
so x=-1.....y=-3.....and z=-1...you can check back
YOU CAN SEE THE FOLLOWING ADDITIONAL MATERIAL FOR REFERENCE
How do I perform the next required row operation on the following matrix and provide only the next table:
x y z
1 28 14 245
0 3 7 42
0 7 7 -38
1 solutions
Answer 9892 by venugopalramana(370) About Me on 2005-11-25 08:01:32 (Show Source):
trust you want to solve the equations for x,y and z and you are at this stage now....assuming that .....our objective is to finally get the matrix if possible into the following form ....(i am using ....to seperate the numbers with suitable gaps..your typing is giving raise to uneven gaps bringing a little lack of clarity)
1.....0.....0.....x
0.....1.....0.....y
0.....0.....1.....z
now we have
1......28.....14.....245
0.......3......7......42
0.......7......7.....-38
new row2=old row2/3.......to get 1 as required in row2.so we get...
1......28.....14.....245
0......3/3....7/3....42/3
0.......7......7.....-38
new row3=oldrow3-7*row2 to get 0 as required in row3
1......28.....14...........245
0.......1.....7/3...........14
0......7-7*1..7-7*7/3......-38-7*14
new row3 = old row3/(-28/3)..to get 1 as required in row3
1......28.....14...................245
0.......1.....7/3...................14
0.......0....(-28/3)/(-28/3).....(-136)/(-28/3)
this gives us finally in the following form
1......28.....14............245
0.......1.....7/3...........14
0.......0......1............102/7
now we go back in the same way to get 0 in row2 and row3
new row2=old row2-row3*7/3...and new row1=old row1-row3*14...so we get
1......28......14-1*14.......245-(102/7)*14
0.......1.......7/3-(7/3)*1...14-(102/7)*(7/3)
0.......0.........1.............102/7
the above on simplification gives us
1.......28.......0..........41
0........1.......0..........-20
0........0.......1..........102/7
now finally we take new row1=old row1-28*row2
1.......28-28*1......0.......41-(-28*20)
0........1.......0...........-20
0........0.......1...........102/7
so the final answer is
1......0.......0.......601
0......1.......0.......-20
0......0.......1.......102/7
which tells us that
1*x+0*y+0*z=x=601
0*x+1*y+0*z=y=-20
0*x+0*y+1*z=z=102/7
note that each and every transformation we did above can be interpreted as given in the last statement given above...this i hope will give you the insight of the process at every step.you can also substitute these values of x,y and z in each and every matrix above to see that they satify all the equations given by the different matrices..in general each mtrix can be taken as a set of simltanous equations in x,y and z...they can be written as follows..take column 1 is for x,column 2 is for y and column 3 is for z.so the first matrix you gave
1......28.....14.....245
0.......3......7......42
0.......7......7.....-38
tells us that
1*x+28*y+14*z=245....etc...

You can put this solution on YOUR website!
Write a matrix equation equivalent to the system of equations.
9x + 9y = -9
5x - 2y = 6
we write the matrix equation as (A)*(X)=(C)..where A is the (2,2)matrix of coefficients namely,9,9,5 and -2 here.X is the (2,1) matrix of unknowns x and y and C is the constants (2,1)matrix on the right side of the eqn NAMELY -9 AND 6.you will find that from the rule for equality of matrices,the above matrix eqn.in effect means the same as that of the given equations.
()*()=()
Cramer's rule.
6x + 4y = -4
y = -3x - 7
GIVING BELOW EXAMPE OF CRAMERS RULE.
My question is.... I was wondering how to do the cramer rule on a 3x3. I have found a bunch of examples and stuff, but I want to know how in the world do you find the determinants of the D, Dx, Dy.and Dz. If you could just tell me how, that would be great.
1 solutions
Answer 9496 by venugopalramana(585) About Me on 2005-11-15 10:49:54 (Show Source):
SEE THE FOLLOWING AND COME BACK IF YOU HAVE DIFFICULTY.HERE C,CX,CY,CZ REFER TO YOUR D,DX,DY,DZ...JUST A DIFFERENCE IN NOMENCLATURE.I SHOWED IN DETAIL A 2X2 DETERMINANT AND THEN IN BRIEF A 3X3 DETERMINANT
2x+y=4
3x-y=6
make a deteminant with coefficients of x (2,3)and y(1,-1) in the 2 eqns.call it C.(Actually for a determinant as you know ,the numbers are contained in vertical bars at either end like |xx|,but in the following the bars are omitted due to difficulty in depiction.you may assume the bars are present)
C=matrix(2,2,2,1,3,-1)=2*(-1)-(1*3)=-5
..now use the constants (4,6)to replace coefficients of x(2,3) in the above determinant C...call it CX..
CX=matrix(2,2,4,1,6,-1)=4*(-1)-1*6=-4-6=-10
..now use the constants (4,6)to replace coefficients of y(1,-1) in the above determinant C...call it CY..
CY=matrix(2,2,2,4,3,6)=2*6-3*4=12=12=0
..now cramers rule says that
(x/CX)=(y/CY)=(1/C)..so we get
x/(-10)=y/0=1/-5
x=-10/-5=10/5=2
y=0/-5=0
************************************
so using the above method you can do the next problem ..here due to presence of 3 variables you will get 3rd.order determinants...4 in all...namely C,CX,CY and CZ,the last formula also extends to include z ,
(x/CX)=(y/CY)=(z/CZ)=(1/C)..
but the procedure is same ..
2x+3y+ z= 5
x+y-2z= -2
-3x +z=-7 ...
...just to give you the idea
C=matrix(3,3,2,3,1,1,1,-2,-3,0,1)..and
CZ=matrix(3,3,2,3,5,1,1,-2,-3,0,7)..etc..hope you can work out the rest

You can put this solution on YOUR website!
my daughter brought home her homework of sample space:
1 cube rolled twice (36 outcomes)
2 cubes rolled together (21 outcomes)...NO NOT CORRECT .IT IS ALSO 36 OUTCOMES
16
25 = 9 How do you go about retrieving the answer
34
43
52
YOUR QUESTION IS INCOMPLETE,IT IS NOT MENTIONED WHAT IS TO BE FOUND..SO PLEASE COMPLETE THE QUESTION SO THAT WE CAN HELP YOU.
ANY WAY LET ME GIVE SOME IDEAS ON THIS
WHEN YOU A ROLL ONE DICE TWICE ,OR ROLL 2 DICE TOGETHER ,THE POSSIBLE OUT COMES OR THE SAMPLE SPACE AS IT IS CALLED IS SAME .IT IS 36 POSSIBILITIES..FOR UNDERSTANDING CALL ONE DICE OR THE FIRST THROUGH OF SAME DICE AS A AND CALL SECOND DICE OR THE SECOND THROUGH OF SAME DICE AS B.
SO NOW DICE/THROUGH A CAN SHOW UP 6 POSSIBILITIES..1,2,3,4,5,6
SIMILARLY DICE/THROUGH B CAN ALSO SHOW UP 6 POSSIBILITIES..1,2,3,4,5,6..SO WE GET 6*6=36 POSSIBILITIES LIKE (1,1),(1,2).........(1,6)
(2,1),(2,2)........(2,6)
......................
........................
(6,1),(6,2).....(6,6)...THESE ARE THE TOTAL POSSIBLE OUTCOMES WHICH WE CALL OUR UNIVERSE OR SAMPLE SPACE.....NOW SUPPOSE IT IS DESIRED THAT YOU WIN IF YOU GET A 7 AS THE SUM OF THE 2 THROWS OR DICE...SO WE HAVE 7 POSSIBILITIES FOR THIS AS FOLLOWS
1+6
2+5
3+4
4+3
5+2
6+1
WE SAY 7 IS THE POSSIBLE SUCCESS COUNT THEN PROBABILITY OF SUCCESS THAT IS GETTING A 7 IS 7/36
SO NOW PLEASE COMEOUT WITH WHAT YOU WANT EXACTLY...

You can put this solution on YOUR website!
A certain quantity of gas, when compressed to a volume of 0.82 m^3 has a pressure of 76Pa. Find the pressure of the gas when it is compressed to 0.55. Assume a constant temperature.
AS PER BOYLES LAW at constant temperature PRODUCT OF PRESSURE AND VOLUME REMAINS CONSTANT ...SO.....
76*0.82=P*0.55
P=76*0.82/0.55 = 113.3 Pa

You can put this solution on YOUR website!
A group of men can do a job in as many as there are men in the group. If the number of men is increased by 5, the number of days to do the job will be decreased by 4. Find the number of men in the group. Thanks!
1 solutions
Answer 11344 by venugopalramana(581) About Me on 2005-12-19 11:17:56 (Show Source):
LET X PEOPLE DO THE JOB IN X DAYS....SO IN ONE DAY X PEOPLE CAN DO 1/X JOB...OR..1 PERSON CAN DO IN ONE DAY 1/(X)^2 JOB
5 MORTE MEANS X+5 PEOPLE........X+5 CAN DO IN ONE DAY (X+5)/(X)^2 JOB ...HENCE HEY REQUIRE (X)^2/(X+5) DAYS TO DO 1 JOB..BUT THIS IS GIVEN AS EQUAL TO 4 LESS DAYS OR X-4 DAYS.
(X)^2/(X+5)=X-4
X^2=(X+5)(X-4)=X^2+5X-4X-20
X^2-X^2+X-20=0
X-20=0
X=20...

You can put this solution on YOUR website!
A chemist needs 10 liters of a 50% salt solution. All she has available is a 20% salt solution and a 70% salt solution. How much of each of the two solutions should she mix to obtain her desired solution?
1 solutions
Answer 11589 by venugopalramana(581) About Me on 2005-12-23 11:03:10 (Show Source):
A chemist needs 10 liters of a 50% salt solution. All she has available is a 20% salt solution and a 70% salt solution. How much of each of the two solutions should she mix to obtain her desired solution?
LET X LITRES OF 20% SOLUTION BE REQUIRED...SINCE TOTAL IS 10 LITRES ,WE HAVE TO ADD 10-X LITRES OF 70% SOLUTION..SO TAKING A SALT BALANCE
SALT IN X LITRES OF 20 % SOLUTION......=X*20/100=0.2X
SALT IN 10-X LITRES OF 10 % SOLUTION.. =(10-X)*70/100=7-0.7X
SALT IN 10 LITRES OF FINAL 50 % SOLUTION =10*50/100=5
HENCE
0.2X+7-0.7X=5
0.7X-0.2X=7-5=2
0.5X=2
X=2/0.5=20/5=4 LITRES OF 20 % SWOLUTION AND 10-4=6 LITRES OF 70% ARE NEEDED TO BE MIXED TO GET 10 LITRES OF 50% SOLUTION
--------------------------------------------------------------------------
SEE THIS EXAMPLE WHICH IS SIMILAR TO YOUR PROBLEM AND TRY.IF YOU STILL HAVE DIFFICULTY PLEASE COME BACK
In a chemistry class, 7 liters of a 4% silver iodide solution must be mixed with a 10% solution to get a 6% solution. How many liters of a 10% solution are needed?
A. 4.5L LET X LITRES OF 10% SOLUTION BE NEEDED.IT HAS X*10/100 LITRES OF SILVER IODIDE.
B. 2.5L 7 LITRES OF 4% SILVER IODIDE SOLUTION HAS 7*4/100=28/100 LITRES OF SILVER IODIDE
C. 7.0L TOTAL SOLUTION =X+7 LITRES…….IT HAS TOTAL 10X/100+28/100=(10X+28)/100 LITRES OF SILVER IODIDE
D. 3.5L FINAL CONCENTRATION =6 %…SO 6/100=(10X+28)/(X+7)*100..OR…6=(10X+28)/(X+7)...CROSS MULTIPLYING..
6(X+7)=(10X+28)…OR…6X+42=10X+28..OR….10X-6X=42-28=14…OR….4X=28..OR..X=7 LITRES.C IS THE ANSWER
**********************************************************************

You can put this solution on YOUR website!
7 TH TERM OF (2A-3B)^9
AS PER BINOMIAL THEOREM R+1 ST. TERM Tr+1 OF EXPANSION (X+A)^N IS GIVEN BY
Tr+1=nCr*{X^(N-R)}*(A^R) ..WE WANT 7TH.TERM HENCE R=6...SO
T7=9C6*{(2A)^(9-6)}*(3B)^6
=9C6*8*A^3*3^6*B^6
=(9!/6!*3!)*8*A^3*3^6*B^6

You can put this solution on YOUR website!
SECOND PROBLEM IS SHOWN BELOW.
********************************************************************************
SOLVE BY GRAPHING MEANS DRAW THE GRAPHS AND FIND THE 2 GRAPHS OF 2 EQUATIONS CUT EACH OTHER .THAT IS THE SOLUTION.LET ME SHOW YOU THE SECOND PROBLEM.
4X-Y=6..OR...4X-6=Y.....
3X+Y=1...OR....Y=1-3X.........
LET X=...........................0......1......2........3.........4......ETC..
Y=4X-6..IN FIRST EQN........-6......-2.....2........6.........10.......ETC
Y=1-3X..IN SECOND EQN.......1.......-2.....-5.......-8........-11......ETC
THE GRAPHS WILL LOOK LIKE THIS

WE FIND THAT THE TWO GRAPHS ARE CUTTING AT X=1 AND Y=-2.HENCE THAT IS THE SOLUTION..YOU CAN CHECK BACK
EQN.I.....4X-Y=4*1-(-2)=4+ 2 = 6 ..OK
EQN.II.....3X+Y=3*1+(-2)=3-2=1........OK.
*******************************************************************************
Solve the system of linear equations by graphing: x+2y=5..
OR 2Y=5-X..OR Y=(5-X)/2
x-y=2...
OR Y=X-2
that is the first one
the second one is: 4x-y=6
3x+y=1
SOLVE BY GRAPHING MEANS DRAW THE GRAPHS AND FIND THE 2 GRAPHS OF 2 EQUATIONS CUT EACH OTHER .THAT IS THE SOLUTION.LET ME SHOW YOU THE FIRST PROBLEM. THEN YOU CAN DO SECOND PROBLEM THE SAME WAY
LET X=.........................0......1......2........3.........4......5....ETC.
Y=(5-X)/2..IN FIRST EQN........5/2....2......3/2......1.........1/2....0...ETC
Y=(X-2).....IN SECOND EQN.......-2....-1......0........1.........2......3...ETC
THE GRAPHS WILL LOOK LIKE THIS

WE FIND THAT THE TWO GRAPHS ARE CUTTING AT X=3 AND Y=1.HENCE THAT IS THE SOLUTION..YOU CAN CHECK BACK
EQN.I.....X+2Y=3+2*1=3+2=5..OK
EQN.II.....X-Y=3-1=2........OK.

You can put this solution on YOUR website!
Two friends are swimming laps back and forth across the length of a pool. They swim with constant speeds and essentially instantaneous turns at the ends of the pool. If at the beginning they leave from opposite ends of the pool and meet for the first time 21 meters from one end and continue on their ways to their own ends of the pool, trun around and meet the second tome 8 meters from the opppsite end (from which they first met), how long is the pool?
LET THE LENGTH OF POOL = L......LET THE TWO FRIENDS BE F1 AND F2.
LET THE SPEEDS OF F1 AND F2 BE X AND Y AND LET THEM START FROM ENDS.1 & 2 RESPECTIVELY
I SWIM.....
DISTANCE BETWEEN THE TWO =L...SINCE THEY LEAVE FROM OPPOSITE ENDS.
RELATIVE SPEED =X+Y..SINCE THEY ARE SWIMMING IN OPPOSITE DIRECTIONS.
TIME TAKEN TO MEET IN I SWIM =L/(X+Y)
DISTANCE COVERED BY F1 DURING THIS TIME = (L*X)/(X+Y)= 21 METRES FROM END 1 SAY.
DISTANCE COVERED BY F2 DURING THIS TIME = (L*Y)/(X+Y)= L-21 METRES FROM END 2.
II SWIM........THERE IS AN AMBIGUITY IN WORDING OF THE PROBLEM HERE
QUOTE
......meet for the first time 21 meters from one end and continue on their ways to their own ends of the pool(THEIR OWN ENDS MEAN THEY GO BACK TO THEIR ORIGINAL STARTING ENDS???OR CONTINUE ON THEIR WAYS MEANS THEY GO TO OPPOSITE ENDS FROM WHERE THEY STARTED) , trun around and meet the second tome 8 meters from the opppsite end
UNQUOTE...CAPITAL LETTERS INDICATE THE AMBIGUITY IN WORDING..LET US WORK OUT THE PROBLEM ON THE FOLLOWING ASSUMPTION.
...THEY GO BACK TO THEIR OWN ENDS THAT IS TO REACH THE SAME ENDS AS THEY STARTED FROM....
DISTANCE TRAVELLED BY F1 TWO TO MEET A SECOND TIME
=21 M.TO GO BACK TO END 1 + L-8 M. = 21+L-8=L+13
DISTANCE TRAVELLED BY F2 TWO TO MEET A SECOND TIME
= L-21 M.TO GO TO END 2 + 8 M.= L-21+8=L-13.
ANALYSIS.......
SINCE THEY WERE TOGETHER AT THE SAME TIME / PLACE WHEN THEY FIRST MET IN I SWIM,THE DURATION FOR WHICH THE TWO SWAM TO MEET A SECOND TIME IS SAME SINCE THEY WERE MEETING AT THE SAME TIME THERE DURING II SWIM.
HENCE ..
(L+13)/X=(L-13)/Y...OR....Y/X=(L-13)/(L+13)......EQN.III
BUT WE GOT FROM I SWIM THAT
(L*X)/(X+Y)=21...AND........EQN.I
(L*Y)/(X+Y)=L-21............EQN.II
EQN.II/EQN.I...GIVES
Y/X=(L-21)/L..........EQN.IV..USING EQN.III...WE GET
(L-21)/L=(L-13)/(L+13)...CROSS MULTIPLYING
(L-21)(L+13)=L(L-13)
L^2+13L-21L-21*13=L^2-13L
L^2+13L-21L-L^2+13L=21*13
5L=273
L=273/5=54.6 M ...THIS APPEARS TO BE MORE REASONABLE ASSUMPTION.

You can put this solution on YOUR website!
PLEASE GIVE THE QUESTIONS WE SHALL DEFINITELY HELP YOU

You can put this solution on YOUR website!
4/3(WPB)=XY, solve for X
I multiplied both sides by 3 to get 4WPB=3XY and then divided both sides by 3Y to get 4WPB/3Y = X....
Does this sound right? Thank You
PERFECT ..YOU DID A GREAT JOB ..BELIEVE IN YOUR SELF ..HAVE CONFIDENCE..YOU WILL COME OUT ON TOP

You can put this solution on YOUR website!
let any number be x ...then 2x will be always even....if we subtract 1 from it it will be always odd
so 2x-1 is one odd number ..its next odd number will be 2x-1+2=2x+1
so sum of their squares
(2x-1)^2+(2x+1)^2=394
2[(2x)^2+(1)^2]=394...using formula (a+b)^2+(a-b)^2=2[(a)^2+(b)^2]
(4x^2+1)=394/2=197
4x^2=197-1=196
x^2=196/4=49
x=7...hence the 2 numbers are 2*7-1=14-1=13 and 15...you can easily verify
13^2=169
15^2=225
sum is 394

You can put this solution on YOUR website!
x-y=3
x=y+3...or...y=x-3....
x^2 -10y-6=0 ,substituting y=x-3 from above eqn. in this we get ,
x^2-10(x-3)-6=0
x^2-10x+30-6=0
x^2-10x+24=0
find 2 factors for 24 which add up to -10.we find that -6*-4=+24 and -6-4=-10...so
x^2-6x-4x+24=0
x(x-6)-4(x-6)=0
(x-6)(x-4)=0
x-6=0 ....or...x-4=0
x=6...or...x=4

You can put this solution on YOUR website!
(a+b)/a = 6 ...
(a/a)+(b/a)=6...
1+b/a=6
b/a=6-1=5..................I.
and (b+c)/c = 9
(b/c)+(/c)=9
b/c=9-1=8......................II
dividing II with I,we get
(b/c)/(b/a)=8/5
(b*a)/(b*c)=8/5
a/c=8/5

You can put this solution on YOUR website!
Four equations are shown. Which lines are parallel? Which lines are perpendicular?
a. y=-1/8x+2
b. y=8x+4
c. y=-8x+2
d. 8x+y=4
Fill in the blanks with the correct leters.
Lines______ and ______ are parallel.
Lines______ and ______ are perpendicular.
1 solutions
--------------------------------------------------------------------------------
Answer 12066 by venugopalramana(571) on 2006-01-01 08:28:26 (Show Source):
LINES A AND B ARE PERPENDICULAR.
LINES C AND D ARE PARALLEL.
SEE BELOW FOR EXPLANATION
--------------------------------------------------------------------------
There is a line (L1) that passes through the points (8,-3) and (3,3/4)
There is another line (L2)with slope M=2/3 that intersects L1 at the point
(-4,6)
What is the point of interectstion of line L1 and L2
1 solutions
Answer 8530 by venugopalramana(345) About Me on 2005-10-29 23:45:54 (Show Source):
SEE THE FOLLOWING WHICH IS SIMILAR TO YOUR PROBLEM AND DO
GIVEN:
· There is a line (L1) that passes through the points
(8,-3) and (3,3/4).
· There is another line (L2) with slope m=2/3 that
intersects L1 at the point (-4,6).
· A third line (L3) is parallel to L2 that passes
through the (7,-13 1/2).
· Yet another line (L4) is perpendicular to L3, and
passes through the point (1/2,5 2/3).
· The fifth line (L5) has the equation
2/5y-6/10x=24/5.
Using whatever method, find the following:
2. The point of intersection of L1 and L3
3. The point of intersection of L1 and L4
4. The point of intersection of L1 and L5
5. The point of intersection of L2 and L3
6. The point of intersection of L2 and L4
7. The point of intersection of L2 and L5
8. The point of intersection of L3 and L4
9. The point of intersection of L3 and L5
10. The point of intersection of L4 and L5
PLEASE NOTE THE FOLLOWING FORMULAE FOR EQUATION OF A
STRAIGHT LINE:
slope(m)and intercept(c) form...y=mx+c
point (x1,y1) and slope (m) form...y-y1=m(x-x1)
two point (x1,y1)and(x2,y2)form.....................
y-y1=((y2-y1)/(x2-x1))*(x-x1)
standard linear form..ax+by+c=0..here by transforming
we get by=-ax-c..or y=(-a/b)x+(-c/b)..comparing with
slope intercept form we get ...slope = -a/b and
intercept = -c/b
*****************************************************
line (L1) that passes through the points (8,-3) and
(3,3/4).
eqn.of L1..y-(-3)=((3/4+3)/(3-8))*(x-8)
y+3=((15/4(-5)))(x-8)=(-3/4)(x-8)
or multiplying with 4 throughout
4y+12=-3x+24
3x+4y+12-24=0
3x+4y-12=0.........L1
There is another line (L2) with slope m=2/3 that
intersects L1 at the point (-4,6).
This means (-4,6)lies on both L1 and L2.(you can check
the eqn.of L1 we got by substituting this point in
equation of L1 and see whether it is satisfied).So
eqn.of L2...
y-6=(2/3)(x+4)..multiplying with 3 throughout..
3y-18=2x+8
-2x+3y-26=0.........L2
A third line (L3) is parallel to L2 that passes
through the (7,-13 1/2).
lines are parallel mean their slopes are same . so we
keep coefficients of x and y same for both parallel
lines and change the constant term only..
eqn.of L2 from above is ...-2x+3y-26=0.........L2
hence L3,its parallel will be ...-2x+3y+k=0..now it
passes through (7,-13 1/2)=(7,-13.5)......substituting
in L3..we get k
-2*7+3*(-13.5)+k=0...or k=14+40.5=54.5..hence eqn.of
L3 is........-2x+3y+54.5=0......................L3
Yet another line (L4) is perpendicular to L3, and
passes through the point (1/2,5 2/3).
lines are perpendicular when the product of their
slopes is equal to -1..so ,we interchange coefficients
of x and y from the first line and insert a negative
sign to one of them and then change the constant term.
L3 is........-2x+3y+54.5=0......................L3
hence L4,its perpendicular will be ..3x+2y+p=0...L4
this passes through (1/2,5 2/3)=(1/2,17/3).hence..
3*1/2+2*17/3+p=0..or ..p= -77/6.so eqn.of L4 is
3x+2y-77/6=0..............L4
The fifth line (L5) has the equation 2/5y-6/10x=24/5.
now to find a point of intersection means to find a
point say P(x,y) which lies on both the lines ..that
is, it satisfies both the equations..so we have to
simply solve the 2 equations of the 2 lines for x and
y to get their point of intersection.For example to
find the point of intersection of L1 and L3 we have to
solve for x and y the 2 equations....
3x+4y-12=0.........L1....(1) and
-2x+3y+40.5=0......L3.....(2)
I TRUST YOU CAN CONTINUE FROM HERE TO GET THE
ANSWERS.If you have any doubts or get into any
difficulty ,please ask me.
venugopal



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

Geometry_Word_Problems/20895: How do i find the gradient of a line parallel to y=3x? hope you can help
1 solutions
--------------------------------------------------------------------------------
Answer 12065 by venugopalramana(571) on 2006-01-01 08:23:26 (Show Source):
GRADIENT OR SLOPE OF A LINE IS GIVEN BY M WHERE THE STANDARD EQUATION OF LINE IS PUT IN THE FORM Y=M*X+C...HERE WE HAVE Y=3X..HENCE M=3.
PARALLEL LINES ARE THOSE HAVING SAME GRADIENT OR SLOPE .HENCE GRADIENT OF A PARALLEL LINE IS ALSO 3

You can put this solution on YOUR website!
Four equations are shown. Which lines are parallel? Which lines are perpendicular?
a. y=-1/8x+2
b. y=8x+4
c. y=-8x+2
d. 8x+y=4
Fill in the blanks with the correct leters.
Lines______ and ______ are parallel.
Lines______ and ______ are perpendicular.
1 solutions
--------------------------------------------------------------------------------
Answer 12066 by venugopalramana(571) on 2006-01-01 08:28:26 (Show Source):
LINES A AND B ARE PERPENDICULAR.
LINES C AND D ARE PARALLEL.
SEE BELOW FOR EXPLANATION
--------------------------------------------------------------------------
There is a line (L1) that passes through the points (8,-3) and (3,3/4)
There is another line (L2)with slope M=2/3 that intersects L1 at the point
(-4,6)
What is the point of interectstion of line L1 and L2
1 solutions
Answer 8530 by venugopalramana(345) About Me on 2005-10-29 23:45:54 (Show Source):
SEE THE FOLLOWING WHICH IS SIMILAR TO YOUR PROBLEM AND DO
GIVEN:
· There is a line (L1) that passes through the points
(8,-3) and (3,3/4).
· There is another line (L2) with slope m=2/3 that
intersects L1 at the point (-4,6).
· A third line (L3) is parallel to L2 that passes
through the (7,-13 1/2).
· Yet another line (L4) is perpendicular to L3, and
passes through the point (1/2,5 2/3).
· The fifth line (L5) has the equation
2/5y-6/10x=24/5.
Using whatever method, find the following:
2. The point of intersection of L1 and L3
3. The point of intersection of L1 and L4
4. The point of intersection of L1 and L5
5. The point of intersection of L2 and L3
6. The point of intersection of L2 and L4
7. The point of intersection of L2 and L5
8. The point of intersection of L3 and L4
9. The point of intersection of L3 and L5
10. The point of intersection of L4 and L5
PLEASE NOTE THE FOLLOWING FORMULAE FOR EQUATION OF A
STRAIGHT LINE:
slope(m)and intercept(c) form...y=mx+c
point (x1,y1) and slope (m) form...y-y1=m(x-x1)
two point (x1,y1)and(x2,y2)form.....................
y-y1=((y2-y1)/(x2-x1))*(x-x1)
standard linear form..ax+by+c=0..here by transforming
we get by=-ax-c..or y=(-a/b)x+(-c/b)..comparing with
slope intercept form we get ...slope = -a/b and
intercept = -c/b
*****************************************************
line (L1) that passes through the points (8,-3) and
(3,3/4).
eqn.of L1..y-(-3)=((3/4+3)/(3-8))*(x-8)
y+3=((15/4(-5)))(x-8)=(-3/4)(x-8)
or multiplying with 4 throughout
4y+12=-3x+24
3x+4y+12-24=0
3x+4y-12=0.........L1
There is another line (L2) with slope m=2/3 that
intersects L1 at the point (-4,6).
This means (-4,6)lies on both L1 and L2.(you can check
the eqn.of L1 we got by substituting this point in
equation of L1 and see whether it is satisfied).So
eqn.of L2...
y-6=(2/3)(x+4)..multiplying with 3 throughout..
3y-18=2x+8
-2x+3y-26=0.........L2
A third line (L3) is parallel to L2 that passes
through the (7,-13 1/2).
lines are parallel mean their slopes are same . so we
keep coefficients of x and y same for both parallel
lines and change the constant term only..
eqn.of L2 from above is ...-2x+3y-26=0.........L2
hence L3,its parallel will be ...-2x+3y+k=0..now it
passes through (7,-13 1/2)=(7,-13.5)......substituting
in L3..we get k
-2*7+3*(-13.5)+k=0...or k=14+40.5=54.5..hence eqn.of
L3 is........-2x+3y+54.5=0......................L3
Yet another line (L4) is perpendicular to L3, and
passes through the point (1/2,5 2/3).
lines are perpendicular when the product of their
slopes is equal to -1..so ,we interchange coefficients
of x and y from the first line and insert a negative
sign to one of them and then change the constant term.
L3 is........-2x+3y+54.5=0......................L3
hence L4,its perpendicular will be ..3x+2y+p=0...L4
this passes through (1/2,5 2/3)=(1/2,17/3).hence..
3*1/2+2*17/3+p=0..or ..p= -77/6.so eqn.of L4 is
3x+2y-77/6=0..............L4
The fifth line (L5) has the equation 2/5y-6/10x=24/5.
now to find a point of intersection means to find a
point say P(x,y) which lies on both the lines ..that
is, it satisfies both the equations..so we have to
simply solve the 2 equations of the 2 lines for x and
y to get their point of intersection.For example to
find the point of intersection of L1 and L3 we have to
solve for x and y the 2 equations....
3x+4y-12=0.........L1....(1) and
-2x+3y+40.5=0......L3.....(2)
I TRUST YOU CAN CONTINUE FROM HERE TO GET THE
ANSWERS.If you have any doubts or get into any
difficulty ,please ask me.
venugopal



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

Geometry_Word_Problems/20895: How do i find the gradient of a line parallel to y=3x? hope you can help
1 solutions
--------------------------------------------------------------------------------
Answer 12065 by venugopalramana(571) on 2006-01-01 08:23:26 (Show Source):
GRADIENT OR SLOPE OF A LINE IS GIVEN BY M WHERE THE STANDARD EQUATION OF LINE IS PUT IN THE FORM Y=M*X+C...HERE WE HAVE Y=3X..HENCE M=3.
PARALLEL LINES ARE THOSE HAVING SAME GRADIENT OR SLOPE .HENCE GRADIENT OF A PARALLEL LINE IS ALSO 3

You can put this solution on YOUR website!
SINCE THERE IS NO HINT ON THE NATURE OF RELATION,WE HAVE TO GO BY TRIAL AND ERROR
LET US TRY LINEAR RELATION..THAT IS Y=AX+B
AT X=0,Y=1....SO A*0+B=1...OR B=1
SO WE HAVE Y=AX+1
AT X=1,Y=3....SO A*1+1=3...ORA=3-1=2
SO WE HAVE Y=2X+1
NOW LET US TET FOR OTHER CASES WHETHER IT HOLDS
AT X=2,Y=5....2*2+1=5.....OK
AT X=-3,Y=-5.......2*-3+1=-6+1=-5.....OK
HENCE THE RELATION IS Y=2X+1

You can put this solution on YOUR website!
IT IS NOT SO DIFFICULT AS YOU IMAGINE..IT IS VERY SIMPLE IF YOU KNOW WHAT IS REQUIRED TO BE DONE..
Find the range of the function: g(x) = x/3 + 5 if the domain is {-6, -3, 0, 3, 6}
I can't even find any examples of this in my textbook (although it's on my homework), and I don't know where to even begin to find an answer.
DOMAIN IS THE VALUES X TAKES .RANGE IS THE CORRESPONDING VALUES OF THE FUNCTION TAKES..HERE IT IS G(X)=X/3+5..FOR EX. WHEN X=-6 , WE GET -6/3+5=-2+5=3
WHEN X=-3,WE GET -3/3+5=4
WHEN X=0,WE GET 0+5=5
WHEN X = 3 ,WE GET 3/3+5=6
WHEN X=6 , WE GET 6/3+5=7
SO THE RANGE IS (3,4,5,6,7)

You can put this solution on YOUR website!
LINES A AND B ARE PERPENDICULAR.
LINES C AND D ARE PARALLEL.
SEE BELOW FOR EXPLANATION
--------------------------------------------------------------------------
There is a line (L1) that passes through the points (8,-3) and (3,3/4)
There is another line (L2)with slope M=2/3 that intersects L1 at the point
(-4,6)
What is the point of interectstion of line L1 and L2
1 solutions
Answer 8530 by venugopalramana(345) About Me on 2005-10-29 23:45:54 (Show Source):
SEE THE FOLLOWING WHICH IS SIMILAR TO YOUR PROBLEM AND DO
GIVEN:
· There is a line (L1) that passes through the points
(8,-3) and (3,3/4).
· There is another line (L2) with slope m=2/3 that
intersects L1 at the point (-4,6).
· A third line (L3) is parallel to L2 that passes
through the (7,-13 1/2).
· Yet another line (L4) is perpendicular to L3, and
passes through the point (1/2,5 2/3).
· The fifth line (L5) has the equation
2/5y-6/10x=24/5.
Using whatever method, find the following:
2. The point of intersection of L1 and L3
3. The point of intersection of L1 and L4
4. The point of intersection of L1 and L5
5. The point of intersection of L2 and L3
6. The point of intersection of L2 and L4
7. The point of intersection of L2 and L5
8. The point of intersection of L3 and L4
9. The point of intersection of L3 and L5
10. The point of intersection of L4 and L5
PLEASE NOTE THE FOLLOWING FORMULAE FOR EQUATION OF A
STRAIGHT LINE:
slope(m)and intercept(c) form...y=mx+c
point (x1,y1) and slope (m) form...y-y1=m(x-x1)
two point (x1,y1)and(x2,y2)form.....................
y-y1=((y2-y1)/(x2-x1))*(x-x1)
standard linear form..ax+by+c=0..here by transforming
we get by=-ax-c..or y=(-a/b)x+(-c/b)..comparing with
slope intercept form we get ...slope = -a/b and
intercept = -c/b
*****************************************************
line (L1) that passes through the points (8,-3) and
(3,3/4).
eqn.of L1..y-(-3)=((3/4+3)/(3-8))*(x-8)
y+3=((15/4(-5)))(x-8)=(-3/4)(x-8)
or multiplying with 4 throughout
4y+12=-3x+24
3x+4y+12-24=0
3x+4y-12=0.........L1
There is another line (L2) with slope m=2/3 that
intersects L1 at the point (-4,6).
This means (-4,6)lies on both L1 and L2.(you can check
the eqn.of L1 we got by substituting this point in
equation of L1 and see whether it is satisfied).So
eqn.of L2...
y-6=(2/3)(x+4)..multiplying with 3 throughout..
3y-18=2x+8
-2x+3y-26=0.........L2
A third line (L3) is parallel to L2 that passes
through the (7,-13 1/2).
lines are parallel mean their slopes are same . so we
keep coefficients of x and y same for both parallel
lines and change the constant term only..
eqn.of L2 from above is ...-2x+3y-26=0.........L2
hence L3,its parallel will be ...-2x+3y+k=0..now it
passes through (7,-13 1/2)=(7,-13.5)......substituting
in L3..we get k
-2*7+3*(-13.5)+k=0...or k=14+40.5=54.5..hence eqn.of
L3 is........-2x+3y+54.5=0......................L3
Yet another line (L4) is perpendicular to L3, and
passes through the point (1/2,5 2/3).
lines are perpendicular when the product of their
slopes is equal to -1..so ,we interchange coefficients
of x and y from the first line and insert a negative
sign to one of them and then change the constant term.
L3 is........-2x+3y+54.5=0......................L3
hence L4,its perpendicular will be ..3x+2y+p=0...L4
this passes through (1/2,5 2/3)=(1/2,17/3).hence..
3*1/2+2*17/3+p=0..or ..p= -77/6.so eqn.of L4 is
3x+2y-77/6=0..............L4
The fifth line (L5) has the equation 2/5y-6/10x=24/5.
now to find a point of intersection means to find a
point say P(x,y) which lies on both the lines ..that
is, it satisfies both the equations..so we have to
simply solve the 2 equations of the 2 lines for x and
y to get their point of intersection.For example to
find the point of intersection of L1 and L3 we have to
solve for x and y the 2 equations....
3x+4y-12=0.........L1....(1) and
-2x+3y+40.5=0......L3.....(2)
I TRUST YOU CAN CONTINUE FROM HERE TO GET THE
ANSWERS.If you have any doubts or get into any
difficulty ,please ask me.
venugopal

You can put this solution on YOUR website!
GRADIENT OR SLOPE OF A LINE IS GIVEN BY M WHERE THE STANDARD EQUATION OF LINE IS PUT IN THE FORM Y=M*X+C...HERE WE HAVE Y=3X..HENCE M=3.
PARALLEL LINES ARE THOSE HAVING SAME GRADIENT OR SLOPE .HENCE GRADIENT OF A PARALLEL LINE IS ALSO 3

You can put this solution on YOUR website!
A horizontal bar of negligible weight has an 82.6 lb weight hanging from the left end and a 43.0 lb weight hanging from the right end. The bar is seen to balance 56.2 in. from the right end. Find the length of the bar.
LET AB BE THE BAR AND ITS LENGTH BE L.WE HAVE 82.6 POUNDS AT A AND 43 POUNDS AT B.IT IS BALANCING AT C SAY AND WE ARE GIVEN THAT CB=56.2..HENCE TAKING MOMENTS ABOUT C WE GET
82.6*AC=43*CB
82.6(AB-CB)=43*CB
82.6(L-56.2)=43*56.2
82.6L-82.6*56.2=43*56.2
82.6L=43*56.2+82.6*56.2=56.2(43+82.6)=56.2*125.6
L=56.2*125.6/82.6=85.4566 INCHES

You can put this solution on YOUR website!
x-3/x^2-16
IN MATHS CERTAIN OPERATIONS LIKE DIVISION BY ZERO ,LOG OF ZERO OR A NEGATIVE NUMBER ;OR SQUARE ROOT OF A NEGATIVE NUMBER ETC...ARE NOT PERMITTED IN MATHS .HENCE WE SHOULD SEE THAT X DOES NOT TAKE SUCH VALUES WHICH WILL RESULT IN ADOPTING SUCH PROHIBITED PRACTICES.
HERE IN DENOMINATOR,WE HAVE X^2-16..HENCE IT SHOULD NOT BE ZERO
THAT IS X^2-16=0..OR X^2=16..OR...X=+4 AND -4 ...THERE IS NO OTHER PROHIBITED AREA .HENCE X SHOULD NOT EQUAL +4 AND -4.

You can put this solution on YOUR website!
Hi, I looked at like problems but did not see one for 3 types of coins...
---------------------------------------------------------------------
HERE IS AN EXAMPLE OF 3 COIN COLLECTION DESIRED BY YOU.TRY TO WORK OUT YOUR PROBLEM AND COME BACK IF NEEDED
---------------------------------------------------------------------------
a coins collection consists of 4, 2, and 3 cent coins. the number of 2 cent coin is one less than twice the number of 4 cent coins. the number 3 cent coins is five more than the number of 2 cent coins. the total value of all coins is $3.00 find the number of 2 cent coins in the collection.
1 solutions
--------------------------------------------------------------------------------
Answer 11647 by venugopalramana(563) on 2005-12-26 11:22:14 (Show Source):
a coins collection consists of 4, 2, and 3 cent coins. the number of 2 cent coin is one less than twice the number of 4 cent coins. the number 3 cent coins is five more than the number of 2 cent coins. the total value of all coins is $3.00 find the number of 2 cent coins in the collection.
LET THE NUMBER OF 4 CENT COINS =X
TWICE THIS =2X
ONE LESS THAN THIS =2X-1 ..THIS IS EQUAL TO NUMBER OF 2 CENT COINS
5 MORE THAN THIS =2X-1+5=2X+4..THIS IS EQUAL TO NUMBER OF 3 CENT COINS
SO TOTAL VALUE OF COINS =4X+2(2X-1)+3(2X+4)=4X+4X-2+6X+12
=14X+10...THIS EQUAL TO $3 OR 300 CENTS
HENCE 14X+10=300
14X=300-10=290
X=290/14..THERE IS A MISTAKE IN NUMBERS SINCE X IS COMING A FRACTION
LET US TAKE $2.90 INSTEAD OF $3 TO ILLUSTRATE THE ANSWER.
WE GET 14X+10=290 CENTS
14X=290-10=280
X=280/14=20
HENCE THERE ARE 2O COINS OF 4 CENTS ...................OF VALUE =20*4=80 CENTS
NUMBER OF 2 CENT COINS =2X-1=2*20-1=40-1=39............OF VALUE =39*2=78 CENTS
NUMBER OF 3 CENT COINS = 2X+4=2*20+4=40+4=44...........OF VALUE=44*3=132 CENTS
...................................................................------
TOTAL VALUE .........................................................290 CENTS

You can put this solution on YOUR website!
-9squareroot5/squareroot3
square root of 3 can be rationalised by multiplying with square root of 3 .it is true for any number which requires rationalisation.so multiply n.r and d.r with that number that is all .here we get
-(9squareroot5)*(squareroot3)/[(squareroot3)*(squareroot3)=-9square root(5*3)/3
=-(9squareroot15)/3=-3squareroot15

You can put this solution on YOUR website!
(1-3i)/(2+i)=...we rationalise the denominator by multiplying with its conjugate...conjugate of 2+i is 2-i...hence multiplying n.r and d.r with 2-i,
(1-3i)(2-i)/(2+i)(2-i)=[1*2-1*i-3i*2+(-3i)(-i)]/[2^2-i^2]=[2-7i-3]/(4+1)
=(-1-7i)/5=-(1+7i)/5....using (a+b)(a-b)=a^2-b^2 and i^2=-1

You can put this solution on YOUR website!
FORMULA FOR DISTANCE BETWEEN 2 POINTS (X1,Y1) AND (X2,Y2)
= [(X2-X1)^2+(Y2-Y1)^2]^0.5...HENCE
DISTANCE BETWEEN (1,1) AND (2,3)=[(2-1)^2+(3-1)^2]^0.5=(1^2+2^2)^0.5=(5)^0.5=SQUARE ROOT OF 5

You can put this solution on YOUR website!
If you know the vertex of the graph of a quadratic function, can you specify the range of the function? If so, what is the range? If not what additional information do you need?
LET US SEE THIS FIRST VISUALLY BY PLOTTING SOME GRAPHS FOR
1....Y=(X-2)^2+4
2....Y=4-(X-2)^2
3....Y=(X-2)^2-4



SO YOU FIND THAT WHEN QUADRATIC EQUATION GIVES US A VERTEX ,AS THE NAME IMPLIES ,WE ARE GETTING A PEAK VALUE OR TROUGH VALUE FOR THE FUNCTION Y AT THAT POINT .DEPENDING ON WHETHER IT IS PEAK (MAXIMUM)OR TROUGH (MINIMUM) ,ONE BOUNDARY FOR Y IS FIXED ,THE OTHER BOUNDARY BEING PLUS INFINITY OR MINUS INFINITY AGAIN DEPENDING ON EHETHER WE GOT A PEAK (MAXIMUM)OR TROUGH (MINIMUM).
SO IN THIS WAY THE RANGE OF Y GETS FIXED.HENCE WE NEED THE VERTEX ,MORE PRECISELY THE Y COORDINATE OF THE VERTEX AND ITS NATURE..... PEAK (MAXIMUM)OR TROUGH (MINIMUM)...TO DEFINE THE RANGE OF THE FUNCTION.
YOU CAN SEE IT ALGEBRAICALLY ALSO AS A PERFECT SQUARE IS ALWAYS POSITIVE OR ZERO....ITS MINIMUM VALUE CAN BE ZERO.....HENCE IT WILL CONTRIBUTE TO MAXIMIXE OR MINIMISE THE FUNCTION (VALUE OF Y)AT THAT POINT DEPENDING ON THE SIGN INFRONT OF THE PERFECT SQUARE BEING NEGATIVE OR POSITIVE RESPECTIVELY.IN EXAMPLES 1 AND 3 IT IS POSITIVE +(X-2)^2. HENCE VERTEX OR Y VALUE BECOMES MINIMUM THERE SINCE YOU GET THE LOWEST VALUE BY ADDING THE LEAST VALUE OF ZERO. IN EXAMPLE 2 IT IS NEGATIVE -(X-2)^2.HENCE VERTEX OR Y VALUE IS MAXIMUM THERE SINCE YOU GET THE HIGHEST VALUE BY SUBTRACTING THE LEAST VALUE OF ZERO.

You can put this solution on YOUR website!
SEE MY WORKING BELOW
T(x,y)=(x+2,y+6)is translation
a.graph (7,3)and T(7,3)....SINCE T(x,y)=(x+2,y+6),WE GET T(7,3)=(7+2,3+6)=(9,9)
B.FIND THE SLOPE OF THE LINE THROUGH (7,3)and its image...FROM ABOVE WE GOT ITS IMAGE THAT IS T(7,3) AS EQUAL TO (9,9)...
HENCE SLOPE OF LINE CONNECTING (7,3) AND (9,9) IS
=(9-3)/(9-7)=6/2=3

You can put this solution on YOUR website!
standard eqn. of an ellipse with foci on x axis,centre as origin and directrices parallel to y axis is
(X^2/A^2)+(Y^2/B^2)=1,where A and B are semi major/minor axes
foci are given by (A*E,0),(-A*E,0)
eccentricity =E =[(A^2-B^2)/A^2]^0.5
we are given the sun is at one focus and the 2 numbers you have given as 740 and 810 are NOT ELABORATED AS TO WHAT THESE NUMBERS ARE.
I back calculated the major / minor axes as per the answer given by you and came out
with following interpretation of the numbers given by you.
740 is the minimum distance of jupiter from sun
810 is the maximum distance of jupiter from sun
so total distance =major axis ,since sun is at one of the foci and is on x axis or major axis
major axis =2*A=740+810=1550..or A=1550/2= 775
as given above foci are given by (A*E,0) and (-A*E,0),with centre taken as origin.
since sun is at focus and its coordinate is (775-740,0) and (775-810,0) or (35,0),(-35,0) we have
A*E=35..that is.... 775*E=35...or....E=35/775=7/155
further we have E=[(A^2-B^2)/A^2]^0.5
so B=[A^2-(E^2)(A^2)]^0.5 =A*(1-E^2)^0.5=774.2093
hence eqn. Of ellipse is
(X^2/775^2)+(Y^2/774.2093^2)=1
(X^2/600625)+(Y^2/599400)=1

You can put this solution on YOUR website!
Solve using properties of logarithms:
log2 (x+1)- log2 x = log2 5
note- the numbers directly following "log" are suppossed to be the spaces. I do not know how to type them to be small and at the bottom
YOU MEAN ALL LOGS ARE TO BASE 2....IF ALL LOGS ARE TO SAME BASE ITDOES NOT MATTER IF WE OMIT WRITING THAT BASE...SINCE LOG X TO BASE TO LOG 2 = LOG X/LOG 2...LIKE THAT EVERY TERM HAS LOG 2 IN THE DENOMINATOR ON BOTH SIDES WHICH WE CAN CANCEL OUT. SO WE GET
LOG(X+1)-LOG X=LOG 5
USING LOG A - LOG B = LOG (A/B) ..WE GET...
LOG(X+1)/X = LOG 5 ...TAKING ANTILOGS ,WE GET ..
(X+1)/X=5
5X=X+1
5X-X=1
4X=1
X=1/4