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A plane is observed approaching your home and you assume the its speed is 550 km/hr. ......OK.....
The angle of elevation of the plane is 16 degrees at one time......OK
and 57 degrees .....OK
one minute later....OK
Approximate the altitude of the plane....OK
Let r be the approximate altitude of the plane.....OK
I AM NOT ABLE TO FOLLOW THE LATER PART AS YOU HAVE NOT GIVEN A SKETCH..BUT IT LOOKS AS IF YOU ASSUMED A CIRCULAR FIGURE OR SO....OK ..LET ME TRY...LET YOUR POSITION ON GROUND BE A.LET THE PLANE BE AT P FIRST AND Q LATER.SO PQ = DISTANCE TRAVELLED BY PLANE IN 1 MINUTE=550/60=55/6 KM.FROM P AND Q DROP PERPENDICULARS PB AND QC TO GROUND LEVEL WHERE YOU ARE.SO PB=QC=ALTITUDE OF PLANE = R ASSUMED.
APB IS A RIGHT ANGLED TRIANGLE WITH ANGLE PAB=16...HENCE TAN 16 =R/AB....
AB=R/TAN(16)
SIMILARLY AC=R/TAN(57)
BC=AB-AC=R/TAN(16)-R/(TAN(57)=R(1/TAN16-1/TAN57)..BUT...BC=PQ=55/6..SO WE HAVE
55/6=R(1/TAN16-1/TAN57)...FROM TABLES YOU CAN FIND TAN 16 AND TAN 57 AND USING THOSE VALUES YOU CAN FIND R FROM THE ABOVE EQUATION.

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A plane is observed approaching your home and you assume the its speed is 550 km/hr. ......OK.....
The angle of elevation of the plane is 16 degrees at one time......OK
and 57 degrees .....OK
one minute later....OK
Approximate the altitude of the plane....OK
Let r be the approximate altitude of the plane.....OK
I AM NOT ABLE TO FOLLOW THE LATER PART AS YOU HAVE NOT GIVEN A SKETCH..BUT IT LOOKS AS IF YOU ASSUMED A CIRCULAR FIGURE OR SO....OK ..LET ME TRY...LET YOUR POSITION ON GROUND BE A.LET THE PLANE BE AT P FIRST AND Q LATER.SO PQ = DISTANCE TRAVELLED BY PLANE IN 1 MINUTE=550/60=55/6 KM.FROM P AND Q DROP PERPENDICULARS PB AND QC TO GROUND LEVEL WHERE YOU ARE.SO PB=QC=ALTITUDE OF PLANE = R ASSUMED.
APB IS A RIGHT ANGLED TRIANGLE WITH ANGLE PAB=16...HENCE TAN 16 =R/AB....
AB=R/TAN(16)
SIMILARLY AC=R/TAN(57)
BC=AB-AC=R/TAN(16)-R/(TAN(57)=R(1/TAN16-1/TAN57)..BUT...BC=PQ=55/6..SO WE HAVE
55/6=R(1/TAN16-1/TAN57)...FROM TABLES YOU CAN FIND TAN 16 AND TAN 57 AND USING THOSE VALUES YOU CAN FIND R FROM THE ABOVE EQN.
41 degrees = 9.17 km
360 degrees = 80.52 km
80.52=2pier
r=80.52/2pie
r=12.82 km

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Midpoint Coordinates:
L(-2,1)
M(2,3)
N(3,-2)
LET ABC BE THE TRIANGLE WITH COORDINATES OF VERTICES AS A(X1,Y1),B(X2,Y2),C(X3,Y3).LET L,M,N BE THE MIDPOINTS OF AB,BC,CA RESPECTIVELY.AS PER COORDINATE GEOMETRY MIDPOINT'S COORDINATES ARE GIVEN BY THE ARITHMATIC MEAN OF COORINATES OF THE END POINTS..SO..
L={(X1+X2)/2,(Y1+Y2)/2}
HENCE (X1+X2)/2=-2...OR X1+X2=-2*2=-4
X1+X2=-4......(1)
SIMILARLY WE GET FROM M AND N
X2+X3=4.........(2)
X3+X1=6...........(3)
ADDING ALL 3 EQNS.1,2,3,WE GET
X1+X2+X2+X3+X3+X1=2(X1+X2+X3)=-4+4+6=6
X1+X2+X3=3.......(4)
NOW SUBTRACTING EQNS..1,2,AND 3 FROM THIS ,WE GET
X3=7
X2=-1
X1=-3
SIMILARLY YOU CAN FIND Y1,Y2,Y3 WHICH WILL COMPLETE YOUR REQUIREMENT..

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4(7-i)-5(2-6i)
HOPE YOU KNOW COMPLEX NUMBERS...FOR ELEMENTARY UNDERSTANDING IT SUFFICES TO KNOW THAT WE DESIGNATE SQUARE ROOT OF -1 AS i WHICH IS AN IMAGINARY NUMBER..WHERE AS 1,2,3 ETC..ARE ALL REAL NUMBERS..WE CANT MIX THEM LIKE WE CANT ADD 2 APPLES TO 3 GRAPES AND SAY THAT THERE ARE 5 APPLES OR 5 GRAPES..WE CAN ONLY SAY THAT THERE ARE 2 APPLES AND 3 GRAPES..SO WE DEAL WITH REAL NUMBERS AND IMAGINARY NUMBERS SEPERATELY AND CALL THEM TOGETHER AS COMPLEX NUMBERS
4(7-i)-5(2-6i)=4*7-4i-(5*2-5*6i)=28-4i-10+30i=(28-10)+i(30-4)=18+26i

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THERE IS NOTHING WRONG WITH YOUR ANSWER..THE PROBLEM GIVEN IS LIKE THAT.. IT IS AN IDENTITY...WHEN LEFT HAND SIDE OF AN EQUATION IS SAME AS THAT ON THE RIGHT HAND SIDE IT IS CALLED AN IDENTITY..IT IS TRUE FOR ANY VALUE OF THE VARIABLE INVOLVED.SO YOU CAN SAY THAT Y = 0,1,-1,1/2....ETC..ANY AND EVERY VALUE. SOME MORE EXAMPLES OF IDENTITIES ARE
(X+Y)^2=X^2+Y^2+2*X*Y WHICH IS TRUE FOR ANY VALUE OF X AND Y.

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i think you mean logs are to base 2..taking that and using the formula,
log x to base y = log x/log y
so we get
log(2x+8)/log 2+log (2x^2+21x+61)/log 2=-3
multiplying through out with log 2
log(2x+8)+log(2x^2+21x+61)=-3log 2=log 2^(-3)==log(1/2^3)=log(1/8)
log(2x+8)(2x^2+21x+61)=log(1/8)
(2x+8)(2x^2+21x+61)=(1/8)
i hope now you can continue from here
to simplify and factorise to get the answers

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.............................................................................
> Five years ago, $10000 was invested at 6%/a
> compounded semi-annually. Today
> the investment rates have risen to 7%/a compounded
> annually. If the original
> investment and accumulated interest is rolled into
> the new investment
> conditions, how much will it be worth in five years.
>
> Use A=P(1+i)^n..HERE
A=AMOUNT=PRINCIPAL+INTEREST..P=PRINCIPAL
YOU USED I FOR RATE PER TERM PER DOLLAR.. WE USE R FOR
RATE AND I FOR INTEREST..N=NUMBER OF TERMS...
> I know for 6% P=10000,CORRECT i = 0.03 VERY GOOD..IT
IS RATE PER TERM HERE TERM IS SEMIANNUAL SO INTERET
RATE IS HALF OF 6% THAT IS 3% THAT IS = 0.03 PER
DOLLAR..but I don't know
> what n =? AS I SAID N = NUMBER OF TERMS FOR 5 YEARS
=10 HALF YEARS....SO N=10
> I know for 7% P=10000,NO...IT IS ACUMULATED INTEREST
AND PRINCIPAL..SO YOU SHOULD USE A OBTAINED FROM ABOVE
CALCULATION AS P HERE. i = 0.07 VERY GOOD..But I don't
know
> what n =?..N=5 IN THIS CASE AS NUMBER OF TERMS IN 5
YEARS IS 5.
> NOW FIND A USING THE ABOVE VALUES.

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i think you mean logs to base 8 ...assuming that and using the formula
log x to base y = log x /log y...where all logs are taken to any common base.
so we have
log(n-3)/log 8 + log(n+4)/log 8 =1....multipling through out with log 8,we get
log (n-3)+log (n+4)=log 8
log(n-3)(n+4)=log 8
hence (n-3)(n+4)=8
n^2-3n+4n-12-8=0
n^2+n-20=0
n^2+5n-4n-20=0
n(n+5)-4(n+5)=0
(n+5)(n-4)=0
hence n=-5 or n=4

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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...

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Answer 9943 by venugopalramana(368) About Me on 2005-11-27 11:27:42 (Show Source):
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Can anyone help with this problem?
The tens digit of a two-digit number is 7 less than twice the units digit. If the digits are reversed, the new number is 13 more than twice the original number. Find the original number.
I started with
x = units............OK
y = tens .....SO THE NUMBER IS YX.....AND ...[2x-7=y]......GOOD..NOW YOU BETTER WRITE THE VALUE OF THIS NUMBER
VALUE OF NUMBER = 10Y+X=10(2X-7)+X=20X-70+X=21X-70
NOW THE DIGITS ARE REVERSED...SO THE NUMBER IS XY...THE VALUE OF THIS NUMBER IS
10X+Y =10X+2X-7=12X-7...NOW WE ARE GIVEN THAT
2*(21X-70)+13=12X-7.
42X-140+13=12X-7
42X-12X=140-13-7=120
30X=120
X=4
SO Y = 2X-7=2*4-7=1
SO THE ORIGINAL NUMBER IS YX = 14....WHICH YOU CAN EASILY VERIFY..NOW YOU DID EXTREMELY WELL I SHOULD SAY BUT MADE A SMALL AITHMATIC ERROR AS SHOWN BELOW
2(10y + x )+ 13 = 10x + y..............EXCELLENT
but end up with
y = 8/19x - 13/19...VERY GOOD
2x-7 = 8/19x + 13/9..............HOW YOU MADE A SIMPLE COPYING MISTAKE..IT IS
2X-7=(8/19)X-13/19..SO
(30/19)X=120/19
X=(120/19)*(19/30)=4..GOT IT ....I AM REALLY IMPRESSED BY YOUR WORKING IT IS VERY GOOD AND YOU GOT THE ANSWER ALMOST EXCEPT FOR THAT SIMPLE MITAKE...CONCENTRATE A BIT MORE AND YOU WILL SUCCEED..GOD LUCK
but ended up with
30/19x = 76/9.....
I know the answer is 14
but I am so lost!l Please help!
Thanks,
Sandy
0 solutions

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Can anyone help with this problem?
The tens digit of a two-digit number is 7 less than twice the units digit. If the digits are reversed, the new number is 13 more than twice the original number. Find the original number.
I started with
x = units............OK
y = tens .....SO THE NUMBER IS YX.....AND ...[2x-7=y]......GOOD..NOW YOU BETTER WRITE THE VALUE OF THIS NUMBER
VALUE OF NUMBER = 10Y+X=10(2X-7)+X=20X-70+X=21X-70
NOW THE DIGITS ARE REVERSED...SO THE NUMBER IS XY...THE VALUE OF THIS NUMBER IS
10X+Y =10X+2X-7=12X-7...NOW WE ARE GIVEN THAT
2*(21X-70)+13=12X-7.
42X-140+13=12X-7
42X-12X=140-13-7=120
30X=120
X=4
SO Y = 2X-7=2*4-7=1
SO THE ORIGINAL NUMBER IS YX = 14....WHICH YOU CAN EASILY VERIFY..NOW YOU DID EXTREMELY WELL I SHOULD SAY BUT MADE A SMALL AITHMATIC ERROR AS SHOWN BELOW
2(10y + x )+ 13 = 10x + y..............EXCELLENT
but end up with
y = 8/19x - 13/19...VERY GOOD
2x-7 = 8/19x + 13/9..............HOW YOU MADE A SIMPLE COPYING MISTAKE..IT IS
2X-7=(8/19)X-13/19..SO
(30/19)X=120/19
X=(120/19)*(19/30)=4..GOT IT ....I AM REALLY IMPRESSED BY YOUR WORKING IT IS VERY GOOD AND YOU GOT THE ANSWER ALMOST EXCEPT FOR THAT SIMPLE MITAKE...CONCENTRATE A BIT MORE AND YOU WILL SUCCEED..GOD LUCK
but ended up with
30/19x = 76/9.....
I know the answer is 14
but I am so lost!l Please help!
Thanks,
Sandy
0 solutions

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you mean log 5 to base 10=0.6990 etc..since all logs are to base 10 we may take it for granted and drop writing that base..so we have
log 5 = 0.6990
log 7 =0.8451
to find log 0.05
log 0.05=log (5/100)=log 5 - log 100 = 0.6990 - 2 = -1.3010...
since log 100 to base 10 =2...note that 10^2 = 100 ....
hence by definition log 100 to base 10 =2..

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2(3)^x=5^(x-1)
trust you know logarithms...take logs on both sides ...use the formulae..
log x*y = log x + log y
log x^n= n*log x
taking logs we get
log 2+ log 3^x=(x-1)*log 5
log2+x log 3=(x-1) log 5=x log 5 - log 5
x(log 5-log 3)=log 2+ log 5=log 2*5 = log 10 =1
x log(5/3)=1
x = 1/log(5/3)

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logb^2 = .25596, logb^3 = .40568, logb^5 = .59432
this is not properly written..i think by the above you meant log 2 to base b =0.25596..etc....then the above has some meaning..otherwise they are not correct and represent contradictory things
logb^(5/12)^3=?...as given above this should mean that we want log (5/12)^3 to base b.
use the following formulae in logs which are true for any base..here wew can take b as base
log x*y=log x+log y
log x/y=log x-log y
log x^n=n*log x
log x to base b=log x /log b
LET US TAKE ALL LOGS TO BASE b AS WE HAVE THEM TO BASE b IN THE PROBLEM.
log(5/12)^3=3*log(5/12)
=3*(log 5-log 12)=3*(log 5-log 2*2*3)
=3*(log 5-log 2-log 2-log 3)=3*(0.59432-0.25596-0.25596-0.40568)
3*(-0.3238)=-0.96984
YOU GAVE THE ANSWER AS 0.096984...IN WHICH CASE THE QUESTION SHOULD BE log(5/12)^(-0.3) to base b =????.PLEASE CHECK BACK THE QUESTION WRITE IT DOWN PROPERLY AS AS GIVEN IN THE TEXT AND COME BACK

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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...

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production cost...all figures in $....
basis...number of candles made = x
start up cost =200
making cost per candle =1.5
cost of making x candles =1.5*x
total production cost =200+1.5*x
sales revenue....
sale price per candle=4
sales revenue for x candles =4*x
now to break even............. sales revenue =production cost
so.........4*x=200+1.5*x
4*x-1.5*x=200
2.5*x=200
x=200/2.5=80
hence 80 candles are to be made to break even..

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Basis...1 week...let sales be $ x
plan A....
3% commission on sales =3*x/100
weekly salary = $ 450
total emoluments in $ =450+3*x/100
plan B.......
7% commission on sales =7*x/100
total emoluments in $=7*x/100

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the problem is not very explicit..what are the 3 numbers ?..let us take them as 2,3,4.so let us write 8 different expressions using these 3 numbers using each number once only in each expression which give different answers...
1.....2+3+4=9
2.....2+3-4=1
3........2-3-4=-5
4.........2*3*4=24
5........2*(3+4)=14
6...........2*(3-4)=-2
7..........2*(4-3)=2
8.......3*4/2=6

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there are 12 numbers..say x1,x2,x3,.....x11,x12 when arranged in increasing order.
mean is 7.so sum of numbers by /total no.of numbers=7=(x1+x2+....+x11+x12)/12
hence sum of numbers =7*12=84=(x1+x2+....+x11+x12)
median is 6..that is the middle term is 6..but total number of terms is 12 so there would be 2 middle terms...namely x6 and x7...by convention in such cases their average is taken as median..so (x6+x7)/2=6...or....x6+x7=2*6=12.
mode is 8 ...that is the frequency of occurence of mode or number 8 should be maximum...
now there are no unique answers for this... we are just to give any set of 12 numbers satisfying the above 3 conditions.
say one set of numbers could be then ....
let us put median as 6 ..by taking both x6 and x7 as 6 each...
now 8 should occur maximm numer of times ...so let x8=x9=x10=x11=8..here 8 occurs 4 times.
let us take x12=16 say...
total of numbers we have taken so far is 6+6+8*4+16=60
now take x1 to x5 so that thy add up to 84-60=24...since total should be 84 to get mean 7 as given above...we only have to see that no number should occur more than 4 times as mod of 8 ,we took with frequency of 4 which should be the highest...
so let us take x1=4,x2=4,x3=5,x4=5,x5=6..which satisfies all the above conditions
so the set of numbers in increasing order are 4,4,5,5,6,6,6,8,8,8,8,16....
their mean is (4+4+5+5+6+6+6+8+8+8+8+16)/12=84/12=7
median is 6 ,since middle most number /numers are 6.
mode is 8 as 8 occurs 4 times ,the maximum frequency..

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do you mean that y axis is plotted on log scale and x axis is plotted on normal scale..or both x and y axis are plotted to log scale ,the folowing example will help to understand
let us take an example..let the graph be a straight line while plotted on log scale for y axis and normal scale for x axis set of data could be
x.....1....10.....100......1000......10000....etc
log x.....0.....1.......2........3..........4....etc...
y.....10....1.......1/10.....1/100.......1/1000...etc...
log y..1....0........-1.......-2............-3......
while taking a scale of 1 cm as 1 unit you will find that in log graph ,if there is 1 at 0 cm,the next point after 1 cm will show 10 , and the next point after another cm will show 100 ,the next point after another 1 cm will show 1000 , and the next point after another cm will show 10000...etc..that is ineffect they are log 1 at zero cm,log 10 at 1 cm ,log 100 at 2 cm ...etc...
hence if we get a straight line on such a plot ...that means really ,assuming both axes are plotted to lg scale,log x is linearly related to log y...or
log y =m*log x+log c...where m is the slope of the line and log c is the y intercept of the line ..drawing comparison from linear scale plots wher we call the eaqn.as y=mx+c
so on a log plot instead of y=mx+c we have log y = m log x +log c
that is using log formula...log x*y=log x +log y,and a log x = log (x^a)we get
log y =log (x^m)*c
y=(x^m)*c...is the real eqn.without logs ..so the key is take the log given graph ...see its pattern and make the eqn in normal manner and finally replace x with log x and y with log y to get the real relation ship..simplify if neede as above to get eqn. without logs.
hope it is clear if you have any difficulty please come back ...

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SEE THE FOLLOWING EXAMPLE TO KNOW


Is the square root of 9 equal to -3i?
1 solutions
--------------------------------------------------------------------------------
Answer 9707 by venugopalramana(356) on 2005-11-19 00:41:47 (Show Source):
NO NO >>>
-3i*-3i=+9*(i*i)=9*(-1)=-9
square root of a negative number only gives raise to complex numbers involving i..square root of a positive number is always real and does not involve complex or imaginary number i
so square root of +9 could be +3 or -3
where as square root of -9 is +3i or -3i...i being defined with the property i*i=-1...which as you can see from above is an imaginary number

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yes....you are correct ..there is no limitation on x for the function e^(x+1)to be meaning full or legal mathematically
so set of real numbers is the domain

You can put this solution on YOUR website!
NO NO >>>
-3i*-3i=+9*(i*i)=9*(-1)=-9
square root of a negative number only gives raise to complex numbers involving i..square root of a positrive number is always real and does not involve complex or imaginary number i
so square root of +9 could be +3 or -3

You can put this solution on YOUR website!
SEE THE FOLLOWING EXAMPLE AND DO THE SAME WAY>>>
2x-3y=5
4x-6y=10
the second eqn. is obtainedby multiplying with 2 the first eqn.
so they are dependent eqns..so in effect we have only one eqn.with 2 variables which give raise to infinte solutions
let us take 2x-3y=5
put x=0,1,2,3,etc..and get corresponding values of y
x..0......1......2...3...4...5..
y..-5/3...-1...-1/3.....etc..infinite solutions
.

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let 5^(log base5 15)=x...so taking logs to a common base 10 say,we get
[using the formula log x to base y =log x / log y ]
{log base5 (15)}log 5 =logx
(log15/log5)log 5 = logx
log 15 = log x
so x=15

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the second eqn. is obtainedby multiplying with 2 the first eqn.
so they are dependent eqns..so in effect we have only one eqn.with 2 variables which give raise to infinte solutions
let us take 2x-3y=5
put x=0,1,2,3,etc..and get corresponding values of y
x..0......1......2...3...4...5..
y..-5/3...-1...-1/3.....etc..infinite solutions
.

You can put this solution on YOUR website!
SEE THE FOLLWING SOLVED EARLIER AND YOU CAN ANSWER YOUR QUESTION BY YOUR SELF!!!!HORIZONTL LINE MEANS SLOPE=0....VERTICAL LINE MEANS SLOPE=ININITY..that is COEFFIIENT OF Y WILL BE ZERO OR THERE WILL BE NO Y TERM
write an equation for the line, in point slope form, that passes through the points (-5,7) and (-4,-3)
1 solutions
Answer 9227 by venugopalramana(345) About Me on 2005-11-10 05:01:36 (Show Source):
write an equation for the line, in point slope form, that passes through the points (-5,7) and (-4,-3)
formula for eqn.of line connecting 2 points x1,y1 and x2,y2 is given by
y-y1=(x-x1)*{(y2-y1)/(x2-x1)}....and slope of the line =(y2-y1)/(x2-x1)
so eqn.of line is y-(-3)=(x-(-4))*{(-3-(7))/(-4-(-5))
y+3=(x+4)*(-10/1)=-10(x+4)=-10x-40
y=-10x-40-3
y=-10x-43
-------------------------------------------------------------------------------
i.. i need help with writing and solving an equation in standard form. for instance.. y- intercept=8 and the slope is 3. does that mean the equation in standard form is 3x+y=-8?? I also need help with.. if the slope = 3/4 and y-intercept= -2 what's the equation in standard form and how can you get 3/4 to not be a fraction or negative? same with slope= 3/5 and passes through 0, -6.. i think that means the y-int. is -6 but how do you graph slopes that are fractions and write them in an equation. sorry it's such a long question but please help me.. thanks!!
1 solutions
Answer 8968 by venugopalramana(345) About Me on 2005-11-07 01:15:23 (Show Source):
hi.. i need help with writing and solving an equation in standard form. for instance.. y- intercept=8 and the slope is 3. does that mean the equation in standard form is 3x+y=-8?? I also need help with.. if the slope = 3/4 and y-intercept= -2 what's the equation in standard form and how can you get 3/4 to not be a fraction or negative? same with slope= 3/5 and passes through 0, -6.. i think that means the y-int. is -6 but how do you graph slopes that are fractions and write them in an equation. sorry it's such a long question but please help me.. thanks!!
The slope intercept form of eqn.of a st.line is y=mx+c..where m is the slope and c is the y intercept.so your answer should be
..y=3x+8 and not y+3x=8
now slope=3/4.. we are given that y intercept is –2.. so the eqn.is..
..y=(3/4)x-2….to remove the fraction multiply both sides of the eqn.with 4 we get
4y=(3*4/4)x-2*4…or …4y=3x-8…if you so desire you put all of them on one side as
4y-3x+8=0
now slope is 3/5 and passes through 0, -6.. i think that means the y-int. is -6 YES CORRECT YOU ARE ..so proceed as above to get the eqn.and remove fraction in slope as is convenient.the resulting eqn.you should be able to graph easily by giving different values for x say 0,1,2,3..etc and finding corresponding values of y and plotting.
************
peter wanted to draw a line parallel to the line of a window with equation y=x+2.if the line is supposed to pass through point (2,3).what is the equation of the line desired by peter.
1 solutions
Answer 8914 by venugopalramana(345) About Me on 2005-11-06 00:23:15 (Show Source):
y=x+2....line parallel to this is given by y=x+k..it is passing through (2,3)
so substituting...3=2+k..or k=3-2=1
hence the eqn.of the parallel line is y=x+1
-------------------------------------------------------------------------
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!
SEE THE FOLLWING SOLVED EARLIER AND YOU CAN ANSWER YOUR QUESTION BY YOUR SELF!!!!
write an equation for the line, in point slope form, that passes through the points (-5,7) and (-4,-3)
1 solutions
Answer 9227 by venugopalramana(345) About Me on 2005-11-10 05:01:36 (Show Source):
write an equation for the line, in point slope form, that passes through the points (-5,7) and (-4,-3)
formula for eqn.of line connecting 2 points x1,y1 and x2,y2 is given by
y-y1=(x-x1)*{(y2-y1)/(x2-x1)}....and slope of the line =(y2-y1)/(x2-x1)
so eqn.of line is y-(-3)=(x-(-4))*{(-3-(7))/(-4-(-5))
y+3=(x+4)*(-10/1)=-10(x+4)=-10x-40
y=-10x-40-3
y=-10x-43
-------------------------------------------------------------------------------
i.. i need help with writing and solving an equation in standard form. for instance.. y- intercept=8 and the slope is 3. does that mean the equation in standard form is 3x+y=-8?? I also need help with.. if the slope = 3/4 and y-intercept= -2 what's the equation in standard form and how can you get 3/4 to not be a fraction or negative? same with slope= 3/5 and passes through 0, -6.. i think that means the y-int. is -6 but how do you graph slopes that are fractions and write them in an equation. sorry it's such a long question but please help me.. thanks!!
1 solutions
Answer 8968 by venugopalramana(345) About Me on 2005-11-07 01:15:23 (Show Source):
hi.. i need help with writing and solving an equation in standard form. for instance.. y- intercept=8 and the slope is 3. does that mean the equation in standard form is 3x+y=-8?? I also need help with.. if the slope = 3/4 and y-intercept= -2 what's the equation in standard form and how can you get 3/4 to not be a fraction or negative? same with slope= 3/5 and passes through 0, -6.. i think that means the y-int. is -6 but how do you graph slopes that are fractions and write them in an equation. sorry it's such a long question but please help me.. thanks!!
The slope intercept form of eqn.of a st.line is y=mx+c..where m is the slope and c is the y intercept.so your answer should be
..y=3x+8 and not y+3x=8
now slope=3/4.. we are given that y intercept is –2.. so the eqn.is..
..y=(3/4)x-2….to remove the fraction multiply both sides of the eqn.with 4 we get
4y=(3*4/4)x-2*4…or …4y=3x-8…if you so desire you put all of them on one side as
4y-3x+8=0
now slope is 3/5 and passes through 0, -6.. i think that means the y-int. is -6 YES CORRECT YOU ARE ..so proceed as above to get the eqn.and remove fraction in slope as is convenient.the resulting eqn.you should be able to graph easily by giving different values for x say 0,1,2,3..etc and finding corresponding values of y and plotting.
************
peter wanted to draw a line parallel to the line of a window with equation y=x+2.if the line is supposed to pass through point (2,3).what is the equation of the line desired by peter.
1 solutions
Answer 8914 by venugopalramana(345) About Me on 2005-11-06 00:23:15 (Show Source):
y=x+2....line parallel to this is given by y=x+k..it is passing through (2,3)
so substituting...3=2+k..or k=3-2=1
hence the eqn.of the parallel line is y=x+1
-------------------------------------------------------------------------
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!
SEE THE FOLLWING SOLVED EARLIER AND YOU CAN ANSWER YOUR QUESTION BY YOUR SELF!!!!
write an equation for the line, in point slope form, that passes through the points (-5,7) and (-4,-3)
1 solutions
Answer 9227 by venugopalramana(345) About Me on 2005-11-10 05:01:36 (Show Source):
write an equation for the line, in point slope form, that passes through the points (-5,7) and (-4,-3)
formula for eqn.of line connecting 2 points x1,y1 and x2,y2 is given by
y-y1=(x-x1)*{(y2-y1)/(x2-x1)}....and slope of the line =(y2-y1)/(x2-x1)
so eqn.of line is y-(-3)=(x-(-4))*{(-3-(7))/(-4-(-5))
y+3=(x+4)*(-10/1)=-10(x+4)=-10x-40
y=-10x-40-3
y=-10x-43
-------------------------------------------------------------------------------
i.. i need help with writing and solving an equation in standard form. for instance.. y- intercept=8 and the slope is 3. does that mean the equation in standard form is 3x+y=-8?? I also need help with.. if the slope = 3/4 and y-intercept= -2 what's the equation in standard form and how can you get 3/4 to not be a fraction or negative? same with slope= 3/5 and passes through 0, -6.. i think that means the y-int. is -6 but how do you graph slopes that are fractions and write them in an equation. sorry it's such a long question but please help me.. thanks!!
1 solutions
Answer 8968 by venugopalramana(345) About Me on 2005-11-07 01:15:23 (Show Source):
hi.. i need help with writing and solving an equation in standard form. for instance.. y- intercept=8 and the slope is 3. does that mean the equation in standard form is 3x+y=-8?? I also need help with.. if the slope = 3/4 and y-intercept= -2 what's the equation in standard form and how can you get 3/4 to not be a fraction or negative? same with slope= 3/5 and passes through 0, -6.. i think that means the y-int. is -6 but how do you graph slopes that are fractions and write them in an equation. sorry it's such a long question but please help me.. thanks!!
The slope intercept form of eqn.of a st.line is y=mx+c..where m is the slope and c is the y intercept.so your answer should be
..y=3x+8 and not y+3x=8
now slope=3/4.. we are given that y intercept is –2.. so the eqn.is..
..y=(3/4)x-2….to remove the fraction multiply both sides of the eqn.with 4 we get
4y=(3*4/4)x-2*4…or …4y=3x-8…if you so desire you put all of them on one side as
4y-3x+8=0
now slope is 3/5 and passes through 0, -6.. i think that means the y-int. is -6 YES CORRECT YOU ARE ..so proceed as above to get the eqn.and remove fraction in slope as is convenient.the resulting eqn.you should be able to graph easily by giving different values for x say 0,1,2,3..etc and finding corresponding values of y and plotting.
************
peter wanted to draw a line parallel to the line of a window with equation y=x+2.if the line is supposed to pass through point (2,3).what is the equation of the line desired by peter.
1 solutions
Answer 8914 by venugopalramana(345) About Me on 2005-11-06 00:23:15 (Show Source):
y=x+2....line parallel to this is given by y=x+k..it is passing through (2,3)
so substituting...3=2+k..or k=3-2=1
hence the eqn.of the parallel line is y=x+1
-------------------------------------------------------------------------
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!
SEE THE FOLLWING SOLVED EARLIER AND YOU CAN ANSWER YOUR QUESTION BY YOUR SELF!!!!
write an equation for the line, in point slope form, that passes through the points (-5,7) and (-4,-3)
1 solutions
Answer 9227 by venugopalramana(345) About Me on 2005-11-10 05:01:36 (Show Source):
write an equation for the line, in point slope form, that passes through the points (-5,7) and (-4,-3)
formula for eqn.of line connecting 2 points x1,y1 and x2,y2 is given by
y-y1=(x-x1)*{(y2-y1)/(x2-x1)}....and slope of the line =(y2-y1)/(x2-x1)
so eqn.of line is y-(-3)=(x-(-4))*{(-3-(7))/(-4-(-5))
y+3=(x+4)*(-10/1)=-10(x+4)=-10x-40
y=-10x-40-3
y=-10x-43
-------------------------------------------------------------------------------
i.. i need help with writing and solving an equation in standard form. for instance.. y- intercept=8 and the slope is 3. does that mean the equation in standard form is 3x+y=-8?? I also need help with.. if the slope = 3/4 and y-intercept= -2 what's the equation in standard form and how can you get 3/4 to not be a fraction or negative? same with slope= 3/5 and passes through 0, -6.. i think that means the y-int. is -6 but how do you graph slopes that are fractions and write them in an equation. sorry it's such a long question but please help me.. thanks!!
1 solutions
Answer 8968 by venugopalramana(345) About Me on 2005-11-07 01:15:23 (Show Source):
hi.. i need help with writing and solving an equation in standard form. for instance.. y- intercept=8 and the slope is 3. does that mean the equation in standard form is 3x+y=-8?? I also need help with.. if the slope = 3/4 and y-intercept= -2 what's the equation in standard form and how can you get 3/4 to not be a fraction or negative? same with slope= 3/5 and passes through 0, -6.. i think that means the y-int. is -6 but how do you graph slopes that are fractions and write them in an equation. sorry it's such a long question but please help me.. thanks!!
The slope intercept form of eqn.of a st.line is y=mx+c..where m is the slope and c is the y intercept.so your answer should be
..y=3x+8 and not y+3x=8
now slope=3/4.. we are given that y intercept is –2.. so the eqn.is..
..y=(3/4)x-2….to remove the fraction multiply both sides of the eqn.with 4 we get
4y=(3*4/4)x-2*4…or …4y=3x-8…if you so desire you put all of them on one side as
4y-3x+8=0
now slope is 3/5 and passes through 0, -6.. i think that means the y-int. is -6 YES CORRECT YOU ARE ..so proceed as above to get the eqn.and remove fraction in slope as is convenient.the resulting eqn.you should be able to graph easily by giving different values for x say 0,1,2,3..etc and finding corresponding values of y and plotting.
************
peter wanted to draw a line parallel to the line of a window with equation y=x+2.if the line is supposed to pass through point (2,3).what is the equation of the line desired by peter.
1 solutions
Answer 8914 by venugopalramana(345) About Me on 2005-11-06 00:23:15 (Show Source):
y=x+2....line parallel to this is given by y=x+k..it is passing through (2,3)
so substituting...3=2+k..or k=3-2=1
hence the eqn.of the parallel line is y=x+1
-------------------------------------------------------------------------
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!
SEE THE FOLLWING SOLVED EARLIER AND YOU CAN ANSWER YOUR QUESTION BY YOUR SELF!!!!
write an equation for the line, in point slope form, that passes through the points (-5,7) and (-4,-3)
1 solutions
Answer 9227 by venugopalramana(345) About Me on 2005-11-10 05:01:36 (Show Source):
write an equation for the line, in point slope form, that passes through the points (-5,7) and (-4,-3)
formula for eqn.of line connecting 2 points x1,y1 and x2,y2 is given by
y-y1=(x-x1)*{(y2-y1)/(x2-x1)}....and slope of the line =(y2-y1)/(x2-x1)
so eqn.of line is y-(-3)=(x-(-4))*{(-3-(7))/(-4-(-5))
y+3=(x+4)*(-10/1)=-10(x+4)=-10x-40
y=-10x-40-3
y=-10x-43
-------------------------------------------------------------------------------
i.. i need help with writing and solving an equation in standard form. for instance.. y- intercept=8 and the slope is 3. does that mean the equation in standard form is 3x+y=-8?? I also need help with.. if the slope = 3/4 and y-intercept= -2 what's the equation in standard form and how can you get 3/4 to not be a fraction or negative? same with slope= 3/5 and passes through 0, -6.. i think that means the y-int. is -6 but how do you graph slopes that are fractions and write them in an equation. sorry it's such a long question but please help me.. thanks!!
1 solutions
Answer 8968 by venugopalramana(345) About Me on 2005-11-07 01:15:23 (Show Source):
hi.. i need help with writing and solving an equation in standard form. for instance.. y- intercept=8 and the slope is 3. does that mean the equation in standard form is 3x+y=-8?? I also need help with.. if the slope = 3/4 and y-intercept= -2 what's the equation in standard form and how can you get 3/4 to not be a fraction or negative? same with slope= 3/5 and passes through 0, -6.. i think that means the y-int. is -6 but how do you graph slopes that are fractions and write them in an equation. sorry it's such a long question but please help me.. thanks!!
The slope intercept form of eqn.of a st.line is y=mx+c..where m is the slope and c is the y intercept.so your answer should be
..y=3x+8 and not y+3x=8
now slope=3/4.. we are given that y intercept is –2.. so the eqn.is..
..y=(3/4)x-2….to remove the fraction multiply both sides of the eqn.with 4 we get
4y=(3*4/4)x-2*4…or …4y=3x-8…if you so desire you put all of them on one side as
4y-3x+8=0
now slope is 3/5 and passes through 0, -6.. i think that means the y-int. is -6 YES CORRECT YOU ARE ..so proceed as above to get the eqn.and remove fraction in slope as is convenient.the resulting eqn.you should be able to graph easily by giving different values for x say 0,1,2,3..etc and finding corresponding values of y and plotting.
************
peter wanted to draw a line parallel to the line of a window with equation y=x+2.if the line is supposed to pass through point (2,3).what is the equation of the line desired by peter.
1 solutions
Answer 8914 by venugopalramana(345) About Me on 2005-11-06 00:23:15 (Show Source):
y=x+2....line parallel to this is given by y=x+k..it is passing through (2,3)
so substituting...3=2+k..or k=3-2=1
hence the eqn.of the parallel line is y=x+1
-------------------------------------------------------------------------
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