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Linear_Algebra/31747: Let P (n) represent the statment 2+6+10+...+(4n-2)=2n^2
In the proof that P (n) is true for all integers n, n>1, what term must be added to both sides of P (k) to show P (k+1) follows form p(k)?
(A) 4k+2
(B) P(k+1)
(C) 4k+6
(D) 4k-2
Thank you for your assistance!
1 solutions

Answer 20613 by venugopalramana(3286) on 2006-04-20 11:37:54 (Show Source):

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

Quadratic_Equations/34339: Can you show me how to simplify this problem?
3x+9/x^2-x-12 * x^2+5x+6/24x+48
1 solutions

Answer 20608 by venugopalramana(3286) on 2006-04-20 10:55:22 (Show Source):

You can put this solution on YOUR website!
{(3x+9)/(x^2-x-12)} * {(x^2+5x+6)/(24x+48)}
={3(X+3)/(X^2-4X+3X-12)}*{X^2+3X+2X+6)/24(X+2)}
={3(X+3)/(X(X-4)+3(X-4))}*{(X(X+3)+2(X+3))/24(X+2)}
={3(X+3)/(X+3)(X-4)}{(X+3)(X+2)/24(X+2)}
={3/(X-4)}{(X+3)/24}
=(X+3)/8(X-4)

Trigonometry-basics/34341: Find the projection on the x-axis if AB = 10 and theta = 30 degrees.
1 solutions

Answer 20607 by venugopalramana(3286) on 2006-04-20 10:46:02 (Show Source):

You can put this solution on YOUR website!
PROJECTION ON X AXIS OF AB = AB*COS(THETA)...WHERE THETA IS ANGLE BETWEEN X AXIA AND LINE AB.
SO PROJECTION =10*COS(30)=10/2=5

Quadratic_Equations/34352: (x^2+2x-15)/(6-2x) can you help simplify. I got -x-5/2
1 solutions

Answer 20606 by venugopalramana(3286) on 2006-04-20 10:43:30 (Show Source):

You can put this solution on YOUR website!
(x^2+2x-15)/(6-2x)
=(X^2+5X-3X-15)/2(3-X)={X(X+5)-3(X+5)}/2(3-X)=(X+5)(X-3)/{(-2)(X-3)
=-(X+5)/2
can you help simplify. I got -x-5/2...CORRECT..BUT PUT BRACKETS AS OTHERWISE IT MIGHT BE READ AS -X -(5/2)..WHERE AS IT IS ACTUALLY -(X+5)/2...OR...(-X-5)/2

Trigonometry-basics/34345: express in terms of theta: csc (90 degrees + theta) sin (180 degrees + theta) - tan (270 degrees - theta).
1 solutions

Answer 20604 by venugopalramana(3286) on 2006-04-20 10:37:49 (Show Source):

You can put this solution on YOUR website!
express in terms of theta: csc (90 degrees + theta) sin (180 degrees + theta) - tan (270 degrees - theta).
LET US USE A INSTEAD OF THETA.....USING ALL SILVER TEA CUPS FOR SIGN
=-SEC(A)*(-SIN(A))*(COT(A))=(1/COS(A))(SIN(A))(COS(A)/SIN(A))=1

Graphs/34244: Please help.
a)Find the equation of the line with a gradient of 3, passing through the points (4,-2).
b) Find the equation of the line containing the points (-2,4) and (5,3)
1 solutions

Answer 20586 by venugopalramana(3286) on 2006-04-20 03:30:09 (Show Source):

Graphs/34246: Determine whether the following pairs of lines are parallel, perpendicular or neither.
A) 2x + y = 1 and x-2y-1
b) 3x-y + 2= 0 and 3x + y = 0
C) y = 2x - 1 and 2x-y + 3=0
1 solutions

Answer 20585 by venugopalramana(3286) on 2006-04-20 03:29:21 (Show Source):

Trigonometry-basics/34256: find the eact value is 0 tan(x+y) if cscx=5/3 and cosy=5/13
I am aware that I have to use the sum idenity for the tangent function but I have no clue how to change csc and cos into a tangent idenity if that is even what I am supposed to do. I have tried a few times but I get different wrong answers each time.
1 solutions

Answer 20584 by venugopalramana(3286) on 2006-04-20 03:27:34 (Show Source):

You can put this solution on YOUR website!
tan(x+y)=(TAN(X)+TAN(Y))/(1-TAN(X)*TAN(Y))
if cscx=5/3=HYPOTENUSE/OPPOSITE SIDE....SO USING PYTHOGARUS
ADJ.SIDE^2+OPP.SIDE^2=HYP.^2.....ADJ.^2+3^2=5^2...ADJ.^2=25-9=16
ADJ.SIDE=4....TAN(X)=OPP.SIDE/ADJ.SIDE=3/4
and cosy=5/13 =ADJ.SIDE/HYP.
5^2+OPP.SIDE^2=13^2
OPP.SIDE^2=169-25=144
OPP.SIDE=12
TAN(Y)=OPP.SIDE/ADJ.SIDE=12/5
TAN(X+Y)=(3/4 + 12/5 )/(1 - (3/4)(12/5))
=(3*5+12*4)/(4*5-3*12)=(15+48)/(20-36)=-63/16

test/34327: when working alone, Gail can finish her paper route in 6hrs. When Mike substitutes for her, he can complete the job in 9 hrs. If they work together, how long should it take them ?
I dont quite get what does "paper route" mean? what is that?
1 solutions

Answer 20583 by venugopalramana(3286) on 2006-04-20 03:15:41 (Show Source):

You can put this solution on YOUR website!
SEE THE FOLLOWING AND TRY
Pump A can fill a tank in 6 hours. Pump A and B together can fill the same tank in 3.5 hours. How long will it take to pump B to fill the tank alone? Can you show me how to do this problem? Thank you.
1 solutions
Answer 20208 by venugopalramana(1495) About Me on 2006-04-17 01:14:18 (Show Source):
Pump A can fill a tank in 6 hours.
A FILLS IN 1 HOUR..............1/6 TANK
Pump A and B together can fill the same tank in 3.5 hours.
A&B FILL IN 1 HOUR..........1/3.5 TANK=10/35=2/7 TANK
SO IN 1 HOUR B FILLS 2/7 - 1/6 = (2*6-1*7)/(6*7)=5/42...TANK
SO PUMP B ALONE WILL TAKE 1/(5/42) HRS TO FILL THE TANK =42/5=8.4 HRS.
How long will it take to pump B to fill the tank alone?
Can you show me how to do this problem? Thank you.

Rational-functions/34335: rationalize the denominator, then simplify, if possible
sqrt (x)-2/sqrt (x)-5
1 solutions

Answer 20582 by venugopalramana(3286) on 2006-04-20 03:12:14 (Show Source):

You can put this solution on YOUR website!
rationalize the denominator, then simplify, if possible
(sqrt (x)-2)(SQRT(X)+5)/(SQRT(X)+5)(sqrt (x)-5)
=(X+5SQRT(X)-2SQRT(X)-10)/(5-5^2)
=-(X+3SQRT(X)-10)/20

logarithm/34336: This question is from textbook intermediate algebra
I am suppose to solve this with out a calcutlator:
log%2810%2C5sqrt%282%29%29

the sqrt kinda throws me off, i have a tutor who tried to explain it but they were also confused on what exactly to do. here is what we were trying:
log%2810%2C2%5E%281%2F5%29%29

log%2810%2C2%29%5E%281%2F5%29%2Blog%2810%2C1%29%5E%281%2F5%29

cancles out
answere:
3%5E%281%2F5%29

??????
we know we are going in the wrong direction with this but are very lost! please help! thanks!
1 solutions

Answer 20581 by venugopalramana(3286) on 2006-04-20 03:09:22 (Show Source):

You can put this solution on YOUR website!
I am suppose to solve this with out a calcutlator:
log%2810%2C5sqrt%282%29%29

FIRSTLY DO YOU MEAN 5*SQRT(2)....OR
FIFTH ROOT OF 2?
ASSUMING IT IS 5*SQRT(2)..DO AS FOLLOWS
1.FIND SQRT(2)=1.414..FROM CALCULATOR..STORE IT IN MEMORY
NOW FIND ITS LOG TO BASE 10 FROM CALCULATOR.RECALLING FROM MEMORY THE VALUE OF 1.414...YOU WILL GET THE ANSWER AS 0.1505
ON THE OTHER HAND IF IT IS FIFTH ROOT OF 2 ..
FIND 2^(1/5)=2^0.2.....USING X^Y FUNCTION IN THE CALCULATOR. YOU WILL GET
2^0.2=1.1487
AND LOG OF ABOVE NUMBER TO BASE 10 =0.0602
the sqrt kinda throws me off, i have a tutor who tried to explain it but they were also confused on what exactly to do. here is what we were trying:
log%2810%2C2%5E%281%2F5%29%29

cancles out
answere:
3%5E%281%2F5%29

??????
we know we are going in the wrong direction with this but are very lost! please help! thanks!

Linear_Algebra/34257: Problem: Construct an orthogonal basis for R^4 that has no zeros in it and demonstrate that your basis is orthogonal.
I don't understand what it is asking and I have no idea how to get the problem going and solved.
1 solutions

Answer 20580 by venugopalramana(3286) on 2006-04-20 02:05:05 (Show Source):

You can put this solution on YOUR website!
I AM GIVING BELOW ANSWER FOR R^4
X1 1 1 1 1
X2 1 -2 -1 1
X3 2 1 2 3
X4 1 0 0 0
Y1 1 1 1 1
SUM
X2.Y1 1 -2 -1 1 -1
Y1.Y1 1 1 1 1 4
Y2 1.25 -1.75 -0.75 1.25
X3.Y2 2.5 -1.75 -1.5 3.75 3
X3.Y1 2 1 2 3 8
Y2.Y2 1.5625 3.0625 0.5625 1.5625 6.75
Y3 -0.555555556 -0.222222222 0.333333333 0.444444444
X4.Y3 -0.555555556 0 0 0 -0.555555556
X4.Y2 1.25 0 0 0 1.25
X4.Y1 1 0 0 0 1
Y3.Y3 0.308641975 0.049382716 0.111111111 0.197530864 0.666666667
Y4 0.055555556 -0.111111111 0.166666667 -0.111111111
Y1.Y2 1.25 -1.75 -0.75 1.25 0
Y1.Y3 -0.555555556 -0.222222222 0.333333333 0.444444444 0
Y1.Y4 0.055555556 -0.111111111 0.166666667 -0.111111111 0
Y2.Y3 -0.694444444 0.388888889 -0.25 0.555555556 0
Y2.Y4 0.069444444 0.194444444 -0.125 -0.138888889 0
Y3.Y4 -0.030864198 0.024691358 0.055555556 -0.049382716 -9.02056E-17
-------------------------------------------------------------------------------
problem: Construct an orthogonal basis for R^4 that has no zeros in it and demonstrate that your basis is orthogonal.
I don't understand what it is asking and I have no idea how to get the problem going and solved.
OK..SHALL WE DO IT THIS WAY?I SHALL EXPLAIN THE MEANING AND METHODAND THE WORKING FOR R^3..YOU CONTINUE ON THAT FOR R^4 TO UNDERSTAND AND LEARN THE TEHNIQUE..OK!
-----------------------------------------------------------------------------
FIRSTLY WE ARE DEALING WITH VECTORS..YOU KNOW I THINK I,J,K AS UNIT VECTORS ALONG X,Y AND Z AXIS..
THEY CAN HAVE ANY DIMENSION ..THAT IS 1 OR 2 OR 3 OR 4 OR....N..IT MEANS FOR YOUR UNDERSTANDING
1 DIMENSION...ONLY ALONG A LINE SAY X AXIS..AS ALONG A CORNER LINE IN A ROOM...I IS THE EXAMPLE
2 DIMENSIONS...IN A PLANE SAY XY PLANE...AS IN A FLOOR OF A ROOM....I AND J ARE THE EXAMPLES
3 DIMENSIONS...IN A SPACE SAY SPANNED BY X,Y AND Z AXIS...AS IN A ROOM .....I,J AND K ARE THE EXAMPLES.
4 DIMENSIONS...IN A SPACE AND TIME SAY SPANNED BY X,Y,Z AND TIME AXIS...AS IN A ROOM AT DIFFERENT TIMES....I,J,K,T...COULD BE EXAMPLES
N DIMENSIONS....YOU HAVE TO ONLY IMAGINE THIS......
NOW NEXT IS BASIS...OR IN COMMON LANGUAGE SOME UNIT OF MEASUREMENT.AGAIN GOING BY DIMENSION...WE HAVE
1 DIMENSION..ALL POINTS OR SAY OUR HOUSES ARE ON ONE ROAD.WE ONLY NEED TO SAY
1 UNIT SAY 0.1 MILE FROM BEGINING OF THE ROAD,SEOND HOUSE IS 0.2 MILES FROM THE BEGINING ETC...WE DNOTE I AS UNIT VECTOR WHICH YOU MAY IMAGINE AS UNIT DISTANCE FROM START IN THAT DIRECTION.
2 DIMENSIONS...ALL HOUSES ARE NOW IN A RECTAGULAR ZONE..SO WE HAVE TO SPECIFY 2 DIRECTIONS ALONG MAIN ROAD AND CROSS ROAD.WE CAN SAY GO FROM START 0.1 MILE ALONG THE MAIN ROAD AND THEN GO 0.2 MILES ALONG A CROSS ROAD TO REACH A HOUSE...SO WE NEED 2 DIRECTIONS USUALLY CALLED X AND Y AXIS ..DENOTED BY I AND J VECTORS.
3 DIMENSIONS..HERE ALL HOUSES ARE IN APARTMENT BLOCKS SAY..SO WE HAVE TO REACH A BLOCK AS ABOVE AND THEN GO UP BY SOME DISTANCE TO REACH A PARTICULAR APARTMEN...SO WE NEED X,Y AND Z AXIS AND I,J,K AS 3 DIMENSIONED UNIT VECTORS
LIKE THIS WE CAN EXPAND THOUGH ,IT WOULD BE ONLY IMAGINATION AND IT IS DIFFICULT TO GIVE EXAMPLES AS ABOV.
SO THOSE MEASURING UNITS` IN DIFFERENT DIRECTIONS ARE CALLED BASIS...I,J,K..ETC.
OFCOURSE STRICTLY SPEAKING IN MATHS FOR THEM TO BE CALLED BASIS THEY SHOULD SATISFY 2 CRITERIA,...
1.THEY SHOULD BE INDEPENDENT....IT IS SO FOR I,J,K..SINCE A DISTANCE ALONG X AXIS CAN ONLY BE SHOWN BY I AND IT CANNOT BE REPRESENTED BY ANY COMBINATION OF J AND K..SIMILARLY FOR OTHER Y AND Z AXIS
2.TYHEY SHOULD BE SUFFIIENT TO SPAN ANY POINT / ECTOR IN THAT SPACE UNDER CONSIDERTATION....IT IS SO WITH I,J,K...FOR EXAMPLE TAKING THE EXAMPLE OF SPACE AS OUR ROOM..ANY POINT IN THAT CAN BE DESCRIBED AS GO FROM START SAY 2 UNITS ALONG X AXIS,THEN SAY 3 UNITS ALONG Y AXIS AND THEN SAY 5 UNITS ALONG Z AXIS ETC....
SO NOW WE HAVE DISCUSSED VECTORS,DIMENSIONS AD BASIS..NOW ON TO ORTHOGONAL BASIS..
IF BASIS VECTORS ARE ORTHOGONAL TO EACH OTHER WE SAY THEY FORM ORTHOGONAL BASIS..I,J,K ARE INDEED ORTHOGONAL SINCE THEY ARE ALOING X,Y,AND Z AXIS WHICH ARE MUTUALLY PERPENDICULAR TO EACH OTHER.AS YOU KNOW 2 VECTORS ARE ORTHOGONAL OR PE
RPENDICULAR IF THEIR DOT PRODUCT OR INNE OR SCALAR PRODUCT IS ZERO
WE FIND I.J=1*1*COS(90)=0..SO IS....J.K AND K.I ZERO
OK NOW WE GOT ALL THE DOUGH..SO LET US FIND HOW TO MAKE AN ORTHOGONAL BASIS IN 3 DIMENSIONS FROM ANY 3 GIVEN BASIS VECTORS IF THEY ARE NOT ORTHOGONAL..SUPPOSE I,J,K ARE GIVEN AS THE BASIS THEN THERE IS NOTHING ELSE TO BE DOBNE..THEY ARE ALREADY ORTHOGONAL AS ROVED ABOVE..SO LET US TAKE 3 VECTORS WHICH FORM A BASIS IN 3D (THAT IS THEY SATISFY THE 2 CONDITIONS MENTIONED ABOVE),BUT ARE NOT ORTHOGONAL AND HENCE WE HAVE TO MAKE A SET OF 3 ORTHONAL BAIS VECTORS FROM THEM..THAT IS AS A LINEAR COMBINATION OF THE 3 GIVEN VECTORS..OK..THE OBJECTIVE IS CLEAR I SUPPOSE!!
LET THE GIVEN BASIS BE...
X1=(1,1,1).............(THAT IS A VECTOR I+J+K..FOR YOUR UNDERSTANDING)
X2=(1,-2,1)
X3=(1,2,3)
WE TAKE IT THAT THEY ARE INDEPENDENT AND COVER ENTIRE SPAN OF R^3 AS THEY ARE GIVEN TO FORM BASIS..WE DONT PROVE IT..WE ONLY BUILD AN ORTHOGONAL BASIS WITH THESE 3 VECTORS
LET US SAY THE ORTHOGONAL BASIS BASED ON THESE IS Y1,Y2,Y3..WE HAVE TO FIND Y1,Y2,Y3 SO THAT Y1.Y2=Y2.Y3=Y3.Y1=0...AND THEY ARE ALL A LINEAR COMBINATION OF X1,X2 AND X3.
WE SAY FIRST ..Y1=X1...SO....Y1=*(1,1,1)
NOW FORMULA FOR Y2 IS
Y2=X2-Y1(Y1.X2)/(Y1.Y1)....
---------------------------------------------
{I SHALL SHOW THE PROOF HERE...THOUGH NOT ASKED FOR OR NEEDED...LET Y2=X2+AY1...SINCE Y1 AND Y2 ARE ORTHOGONAL...Y1.Y2=0...SO DOT BOTH SIDES WITH Y1...WE GET...
Y1.Y2=X2.Y1+AY1.Y1
0=X2.Y1+AY1.Y1...SO....A=-(X2.Y1)/(Y1.Y1)...OK...}
------------------------------------------------------------
Y2=(1,-2,1)-(1,1,1){(1,-2,1).(1,1,1)/(1,1,1).(1,1,1)}
=(1,-2,1)-(1,1,1){1*1+(-2*1)+1*1}/{1*1+1*1+1*1}
=(1,-2,1)-(12,1,1)*0/3=(1,-2,1)
NOW FORMULA FOR Y3 IS SIMILAR...
Y3=X3-Y2(Y2.X3)/(Y2.Y2) -Y1(Y1.X3)/(Y1.Y1).....YOU CAN WORK OUT AS ABOVE THE DOT PRODUCTS AND YOU WILL GET THE ANSWER AS
Y3=(1,2,3)-Y2(0/6)-(1,1,1)(6/3)
=(1,2,3)-(2,2,2)
=(-1,0,1)
SO THE ORTHOGONAL VECTORS CORRESPONDING TO THE GIVEN BASIS OF X1,X2 AND X3 ARE
Y1=(1,1,1)
Y2=(1,-2,1)
Y3=(-1,0,1)
YOU CAN CHECK THAT THEY ARE MUTUALLY ORTHOGONAL..THAT IS Y1.Y2=Y2.Y3=Y3.Y1=0
AND THEY ARE ALL OBTAINED AS LINEAR COMBINATIONS OF X1,X2 AND X3 AS THE FORMULAE FOR Y1 ,Y2 AND Y3 OBVIOUSLY PROVE...
NOW ON TO YOUR HOME WORK FOR R^4...HAPPY WORK..HOPE YOU WILL DO AND GET IT..IF STILL IN DOUBT OR DIFFICULTY PLEASE COME BACK....
venugopalramana

Linear_Algebra/34258: Problem: Is it possible for the columns of a 4x3 matrix to be linearly dependent? if so, give an example and demonstrate the dependence. If not, prove it.
I understand what linearly dependent means, but I don't know what kind of a matrix we should use to prove it without coming up with a clear example with numbers. Please help.
1 solutions

Answer 20579 by venugopalramana(3286) on 2006-04-20 00:54:43 (Show Source):

You can put this solution on YOUR website!
Problem: Is it possible for the columns of a 4x3 matrix to be linearly dependent?
YES ,WE CAN ALWAYS WRITE TO SUIT THE REQUIREMENT..FOR..EXAMPLE...LET THE MATRIX A BE EQUAL TO
1,2,3
2,4,6
3,6,9
4,8,12
THE COLUMNS ARE
C1...........1,2,3,4
C2...........2,4,6,8
C3...........3,6,9,12
WE FIND EACH ELEMENT OF C2=2*EACH CORRESPONDING ELEMENT OF C1
WE FIND EACH ELEMENT OF C3=3*EACH CORRESPONDING ELEMENT OF C1
SO THE COLUMNS ARE LINEARLY DEPENDENT....
ANOTHER WAY TO MAKE SUCH MATRIX IS SAY A IS.............
1,2,3
4,5,9
6,7,13
8,9,17
THE COLUMNS ARE
C1...1,4,6,8
C2...2,5,7,9
C3...3,9,13,17
WE FIND EACH ELEMENT OF C3=SUM OF EACH CORRESPONDING ELEMENT OF C1 AND C2
if so, give an example and demonstrate the dependence. If not, prove it.
I understand what linearly dependent means, but I don't know what kind of a matrix we should use to prove it without coming up with a clear example with numbers. Please help.
IN GENERAL JUST MAKE NUMBERS IN ONE COLUMN TO BE SUM OR DIFFERENCE OF ELEMENTS IN OTHER 2 COLUMNS...SAY..A..IS...
X,Y,X-Y
P,Q,P-Q
T,U,T-U
E,F,E-F...ETC...

Linear-equations/34292: This question is from textbook Algebra 1
I need to write an equation in slope-intercept form of the line that passes through the given point and is parallel to the graph of each equation. my problem is (2,3) y = x+5 Please help, I'm lost on these kinds of problems.
1 solutions

Answer 20578 by venugopalramana(3286) on 2006-04-20 00:36:19 (Show Source):

You can put this solution on YOUR website!
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.
Graphs/24461: find the slope,m,and the y intercept,b,of the line whose equation is 4x+2y+8=0
1 solutions
Answer 13032 by venugopalramana(1088) About Me on 2006-01-15 10:52:29 (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.

Linear-equations/34297: This question is from textbook Algebra 1
My problem is (-5, -4) 2x + 3y = -1 i have to write an equation in slope-intercept form of the line that passes through the given point and is paralle to ther graph of each equation. these make no sense, I've tried tofigure it out a few times. NO LUCK help please
1 solutions

Answer 20577 by venugopalramana(3286) on 2006-04-20 00:34:38 (Show Source):

Linear-equations/34299: This question is from textbook Algebra 1
Now this is a challenge for me. the question reads, write an equation in slope-intercept form of the line that passes through a given point and is perpendicular to the graph of each equation. problem #33 is (6, -13), 2x - 9y = 5
1 solutions

Answer 20576 by venugopalramana(3286) on 2006-04-20 00:33:47 (Show Source):

Geometry_Word_Problems/34303: Write and equation and solve the following problem.
The area or a rectangle is 130 square feet. Find the dimensions of the rectangle if the length is 3 feet greater than the width.
I know A=LW so: 130 = L+3 = w
Did I get the formula correct
1 solutions

Answer 20575 by venugopalramana(3286) on 2006-04-20 00:31:48 (Show Source):

You can put this solution on YOUR website!
SEE MY COMMENTS BELOW
The area or a rectangle is 130 square feet. Find the dimensions of the rectangle if the length is 3 feet greater than the width.
I know A=LW....OK
so: 130 = L+3 = w....NO....GO STEP BY STEP....LENGTH IS MORE THAN WIDTH...SO
L=W+3...OK...NOW AREA =LW=(W+3)W=130
SO WE GET
W^2+3W=130
W^3+3W-130=0
W^2+13W-10W-130=0
W(W+13)-10(W+13)=0
(W-10)(W+13)=0
W-10=0....OR...W=10
SO L=W+3=10+3=13
Did I get the formula correct

Graphs/34309: Geometry: Floor plans for a building have the four corners of a room located at the points (2,3), (11,6), (-3,18), and (8,21). Determine whether the side through the points (2,3) and (11,6) is perpendicular ot the side through the points (2,3) and (-3,18).
1 solutions

Answer 20574 by venugopalramana(3286) on 2006-04-20 00:02:17 (Show Source):

You can put this solution on YOUR website!
Geometry: Floor plans for a building have the four corners of a room located at the points (2,3)...A
, (11,6)....B
, (-3,18)....C
, and (8,21)...D
. Determine whether the side through the points (2,3)....A
and (11,6).....B
is perpendicular ot the side through the points (2,3).....A
and (-3,18)...........C.
SLOPE OF AB = (6-3)/(11-2)=3/9=1/3
SLOPE OF AC = (18-3)/(-3-2)=15/-5=-3
PRODUCT OF SLOPES OF AB AND AC =(1/3)(-3)=-1....
HENCE THE 2 LINES ARE PERPENDICULAR.

Graphs/34308: Find the slope of any line perpendicular to the line through points (0,5) and (-3,-4).
1 solutions

Answer 20573 by venugopalramana(3286) on 2006-04-19 23:56:41 (Show Source):

You can put this solution on YOUR website!
ok i am giving the answer below
Find the slope of any line perpendicular to the line through points A (0,5) and B(-3,-4).
SLOPE OF LINE AB=DIFFERNCE IN Y COORDINATES/DIFFERENCE IN X COORDINATES
=(5-(-4))/(0-(-3))=9/3=3
SLOPE OF PERPENCICULAR LINE = -1/SLOPE OF AB= -1/3

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.
Graphs/24461: find the slope,m,and the y intercept,b,of the line whose equation is 4x+2y+8=0
1 solutions
Answer 13032 by venugopalramana(1088) About Me on 2006-01-15 10:52:29 (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.

Graphs/34310: Write the equation of the line L satisfying the given geometric conditions:
L has y-intercept (0,-3) and is parallel to the line with equation y=2/3x + 1
1 solutions

Answer 20572 by venugopalramana(3286) on 2006-04-19 23:55:40 (Show Source):

You can put this solution on YOUR website!
OK I AM GIVING THE ANSWER BELOW
Write the equation of the line L satisfying the given geometric conditions:
L has y-intercept (0,-3) and is parallel to the line with equation y=2/3x + 1
EQN OF LINE PRALLEL TO GIVEN LINE IS
Y=2X/3 + K...IT HAS Y INTERCEPT (0,-3)...SO
-3=2*0/3+K
K=-3
SO EQN. OF LINE IS
Y=2X/3-3

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.
Graphs/24461: find the slope,m,and the y intercept,b,of the line whose equation is 4x+2y+8=0
1 solutions
Answer 13032 by venugopalramana(1088) About Me on 2006-01-15 10:52:29 (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.

Graphs/34311: Write the equatin of the line L satisfying the given geometric conditions:
L has y-intercept (0,2) and is perpendicular to the line with equation 2x-3y=6
1 solutions

Answer 20571 by venugopalramana(3286) on 2006-04-19 23:54:24 (Show Source):

You can put this solution on YOUR website!
OK I AM ANSWERING BELOW
Write the equatin of the line L satisfying the given geometric conditions:
L has y-intercept (0,2) and is perpendicular to the line with equation 2x-3y=6
EQN.OF LINE PERPENDICULAR TO ABOVE LINE IS
2Y+3X=K..(INTERCHANGE X AND Y COEFFICIENTS AND CHANGE SIGN AND EQUATE TO A CONSTANT TO BE FOUND)
IT PASSES THROUGH....(0,2)..SO
2*2+3*0=K
K=4
SO EQN. OF PERPENDICULAR LINE IS
2Y+3X=4
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.
Graphs/24461: find the slope,m,and the y intercept,b,of the line whose equation is 4x+2y+8=0
1 solutions
Answer 13032 by venugopalramana(1088) About Me on 2006-01-15 10:52:29 (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.

Linear-equations/34313: I need an equation that is parallel to the line y = -3x +2 and passes through the point (0,3)
1 solutions

Answer 20570 by venugopalramana(3286) on 2006-04-19 23:53:25 (Show Source):

Money_Word_Problems/34316: Give an example of using the distributive property for a negative monomial times a trinomial with different signs on the terms [for example: -3x(2xy + 3y – 2x)] and show each step of the distribution. Why do you think many students make sign errors on this type of problem? What would be your advice to a student who has trouble with the signs?
Can you help me with this please?
Thank you
1 solutions

Answer 20567 by venugopalramana(3286) on 2006-04-19 23:38:53 (Show Source):

You can put this solution on YOUR website!
Give an example of using the distributive property for a negative monomial times a trinomial with different signs on the terms [for example: -3x(2xy + 3y – 2x)] and
THE RULE IS SIMPLE..+ * + = +....- * - = + AND - * + = - = + * - ...THAT IS PRODUCT IS POSITIVE IF BOTH ARE OF SAME SIGN ...AND NEGATIVE IF BOTH ARE OF DIFFERENT SIGNS...APPLYING IN THE ABOVE EXAMPLE,WE GET
+(-3X)(2XY)+(-3X)(3Y)+(-3X)(-2X)=-6X^2*Y-6XY+6X^2.....SEE COMMENTS BELOW...
show each step of the distribution. Why do you think many students make sign errors on this type of problem?
1.NEGLIGENCE..
2.LACK OF AWARENESS OF THE ABOVE RULE OF PRODUCT OF SIGNS....
3.NOT USING APPROPRIATE BRACKETS OR USING WRONG BRACKETS...THIS IS THE MOST ..IMPORTANT ASPECT TO BE REMEDIED.
4.NOT FOLLOWING A SYSTEMATIC PROCEDURE..THAT IS OMITTING STEP BY STEP APPROACH WITHOUT HAVING THE EXPERIENCE.YOU CAN JUMP STEPS OR OMIT STEPS IF YOU ARE EXPERIENCED.IF NOT YOU ARE BOUND TO MAKE MISTAKES.
What would be your advice to a
student who has trouble with the signs?
Can you help me with this please?
REMEDY THE ABOVE DFECTS...
1...THIS IS IN YOUR OWN HANDS
2...NOW YOU KNOW THE RULE.
3...SEE THE USE OF BRACKETS IN THE ABOVE EXAMPLE BY ME.THAT CLEARLY SHOWS THE PRIORITY OR ORDER OF EXECUTION OF OPERATIONS.
4.ONE SYSTEMATIC PROCEDURE FOR BEGINERS IS TO PUT + SIGN BETWEEN EVERY 2 TERMS OF EXPANSION WHILE USING DISTRIBUTIVE PROPERTY OF MULTIPLICATION OVER ADDITION OR SUBTRACTION.
SEE IN THE ABOVE EXAMPLE.I PUT ALL + SIGNS ONLY ,THOUGH THERE IS A MINUS SIGN IN FRONT OF 3X BEFORE THE BRACKETS TO START WITH AND ALSO IN THE LAST TERM WITHIN THE BRACKETS NAMELY -2X.THIS - SIGN WHEREVER IT IS PRESENT, IS PUT WITHIN BRACKETS TO START WITH.THEN WHEN BRACKETS ARE OPENED BY MULTIPLICATION,WE FIRST USED THE ABOVE RULE OF SIGNS TO DETERMINE THE SIGN OF THE ANSWER AND THEN THE VALUE OF THE ANSWER IS CALCULATED AND PUT.
BY USING THIS SYSTEMATIC 2 STEP APPOACH,TILL YOU GAIN EXPERIENCE, THOUGH IT IS LITTLE TIME CONSUMING, WILL AVOID MISTAKES.ONCE YOU GAIN EXPERIENCE YOU CAN SKIP THE STEPS AND PUT THE FINAL ANSWER.
LAST WARNING...BE VERY CAREFULL IF THERE ARE MULTIPLE BRACKETS..SEE BELOW..
-3[5-2{4-3(2-3)}]...HERE OPEN BRACKETS ONE BY ONE STARTING FROM INNERMOST FIRST
=-3[5-2{4-3*(-1)}]
=-3[5-2{4+3}]
=-3[5-2*7]
-3[5-14]
=-3*(-9)
=+27

Square-cubic-other-roots/34217: the nth term of a different sequence is n-2/n2 write the first three terms of this sequence??
1 solutions

Answer 20520 by venugopalramana(3286) on 2006-04-19 12:37:18 (Show Source):

You can put this solution on YOUR website!
TN=(N-2)/N^2..IS WHAT YOU MEANT I SUPPOSE....YOU JUST PUT N=1,2,3...ETC...TO GET T1,T2,T3...ETC...
SO FOR
N=1.............T1=(1-2)/1^2=-1/1=-1
N=2.............T2=(2-2)/2^2=0/4=0
N=3.............T3==(3-2)/3^2=1/9=1/9....ETC...

Miscellaneous_Word_Problems/34218: What is the largest intger that n^3-n is divisable by k for all postive intgers n?
1 solutions

Answer 20519 by venugopalramana(3286) on 2006-04-19 12:34:18 (Show Source):

You can put this solution on YOUR website!
THE QUESTION IS NOT VERY CLEAR ..
IF IT REALLY MEANS WHAT IS WRITTEN THEN SINCE N IS AN INTEGER N^3-N IS AN INTEGER AND IT IS DIVISIBLE BY N^3-N WHICH IS THE LARGEST INTEGER.IF IT IS NEEDE TO FIND A NUMBER LESS THAN THIS THEN
N^3-N=N(N^2-1)=N(N+1)(N-1)...SO TAKING THE LARGER OF THE 3 FACTORS N(N+1)=N^2+N IS THE LARGEST INTEGER FOR ALL N.

logarithm/34221: This question is from textbook Intermediate Algebra
I am so lost with this, I don't even understand where to begin. We are supose to find each value of x. Here is the problem
log2 8=x The 2 is supose to be small and at the bottom of the g. Can someone please help me?
1 solutions

Answer 20518 by venugopalramana(3286) on 2006-04-19 12:28:35 (Show Source):

You can put this solution on YOUR website!
THIS MEANS LOG (8) TO BASE 2 =X...WHICH MEANS 2^X=8...WE HAVE TO FIND X ...WE SEE THAT 8=2*2*2=2^3
2^X=2^3...OR....X=3
IN GENERAL
IF ....B^P=N...WE SAY LOG OF N TO BASE B =P...WRITTEN AS LOG(N)TO BASE B=P

Radicals/34227: f(x)=thesquarerootof2x-7dividedbythesquareroot1-x?whatisthedomain
1 solutions

Answer 20517 by venugopalramana(3286) on 2006-04-19 12:25:15 (Show Source):

You can put this solution on YOUR website!
Y=SQRT(2X-7)/SQRT(1-X).....
DOMAIN IS THE VALUE OF X WHIH WILL MAKE THE FUNCTION MEANING FULL
HERE D.R.HAS SQRT(1-X)..WHICH IS ZERO AT X=1..........................I
SO THIS IS EXLUDED..
THEN SQRT(1-X)...WILL BE IMAGINARY IF 1-X IS NEGATIVE..SO 1-X>0..OR
1>X....OR.....X<1.....................................................II
NOW SQRT(2X-7)..ALSO NEEDS 2X-7>=0...OR....2X>=7...OR....X>=3.5...........III
NOW II AND III ARE CONTRADICTARY IF II IS TRUE THEN III IS FALSE AND VICEVERSA...HENCE DOMAIN OF X IS NULL SET..OR NO REAL VALUE

logarithm/34222: This question is from textbook
I feel stupid asking this question but, Is there a differnce between a logarithm of a product and the product of a logarithm? If there is a differnce can you explain it to me?
1 solutions

Answer 20515 by venugopalramana(3286) on 2006-04-19 12:17:21 (Show Source):

You can put this solution on YOUR website!
YA...LOG (10*10)=LOG(10)+LOG(10)..IS CORRECT SINCE LHS=LOG(100)=2...RHS=LOG(10)+LOG(10)=1+1=2
SO THE FORMULA IS LOG(A*B)=LOG(A)+LOG(B)

BUT LOG(10)*LOG(10)=1*1=1...I IS NOT LOG(10*10)..OR...ANY OTHER THING...THERE IS NO FORMULA FOR LOG(A)*LOG(B)

Mixture_Word_Problems/34225: How many oz. of pure chocolate must be added to 150 oz that is 60% chocolate to make a topping that is 75% chocolate?
1 solutions

Answer 20514 by venugopalramana(3286) on 2006-04-19 12:12:59 (Show Source):

You can put this solution on YOUR website!
PLEASE SEE THE FOLLOWING EXAMPLE AND TRY.IF STILL IN DIFFICULTY,PLEASE COME BACK
PLEASE HELP ASAP: Ziggy's famous yogurt blends regular yogurt that is 3% fat with its no fat yogurt to obtain low fat yogurt that is 1% fat. How many pounds of regular and how many pounds of non-fat yogurt should be mixed to obtain 60 pounds of lowfat yogurt.
PLEASE HELP ASAP. thank you
THESE ARE MATERIAL BALANCE PROBLEMS.THE PRINCIPLE IS TO APPLY
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS..
THIS PRINCIPLE CAN BE APPLIED TO TOTAL MIXTURE AS A WHOLE AS WELL AS INDIVIDUAL COMPONENTS OF THE MIXTURE.LET US SEE THE APPLICATION USING YOUR PROBLEM.
HERE THE MIXTURE COMPRISES 2 INPUTS-REGULAR YOGURT (RY) & NO FAT YOGURT (NFY)
AND ONE OUT PUT-LOW FAT YOGURT (LFY).THE COMPONENT OF IMPORTANCE IN THE MIXTURE IS FAT CONTENT.SO WE TAKE 2 BALANCES HERE ..ONE FOR THE TOTAL MIXTURE AND ANOTHER FOR COMPONENT OF FAT IN THE MIXTURE.
I..TOTAL BALANCE...
INPUTS
1.QTY.OF.RY=X POUNDS
2.QTY OF NFY=Y POUNDS
OUT PUT
1.QTY.OF LFY=60 POUNDS
SO APPLYING
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS.....WE GET
X+Y=60.............................I
II..COMPONENT BALANCE..HERE IT IS FAT .
INPUTS
1.QTY.OF FAT IN RY=X*3/100=3X/100 POUNDS
2.QTY OF FAT IN NFY=Y*0/100=0 POUNDS
OUT PUT
1.QTY.OF FAT IN LFY=60*1/100=60/100 POUNDS
SO APPLYING
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS.....WE GET
3X/100 + 0=60/100.............................II
3X=60
X=20 POUNDS. OF REGULAR YOGURT
Y=60-20=40 POUNDS OF NO FAT YOGURT.
Mixture_Word_Problems/30587: The amount (by weight) of gold, silver and lead in three alloys of these metals are in ratios:
4:3:2 - Alloy 1
3:5:1 - Alloy 2
2:2:5 - Alloy 3
It is desired to make a fourth alloy containing equal amounts of gold, silver and lead. How many grams each alloy should be used for every 10 grams of the new alloy?
1 solutions
Answer 17262 by venugopalramana(1167) About Me on 2006-03-18 04:34:35 (Show Source):
The amount (by weight) of gold, silver and lead in three alloys of these metals are in ratios:
4:3:2 - Alloy 1
3:5:1 - Alloy 2
2:2:5 - Alloy 3
It is desired to make a fourth alloy containing equal amounts of gold, silver and lead. How many grams each alloy should be used for every 10 grams of the new alloy?
LET X GMS OF ALLOY1 ,Y GMS OF ALLOY2 AND 10-X-Y GMS OF ALLOY3 BE USED TO GET
X+Y+10-X-Y=10 GMS OF ALLOY 4
SO.................GOLD..............SIVER............LEAD IN THE MIX IS GIVEN BY
X GMS A1...........4X/9..............3X/9..........2X/9
Y GMS A2...........3Y/9..............5Y/9...........Y/9....
10-X-Y GMS A3....(20-2X-2Y)/9 ......(20-2X-2Y)/9...(50-5X-5Y)/9
-------------------------------------------------------------------------------
10 GMS A4.......(20+2X+Y)/9........(20+X+3Y)/9.....(50-3X-4Y)/9
THESE ARE ALL EQUAL...HENCE
20+2X+Y = 20+X+3Y...OR......................X-2Y=0..............I
20+2X+Y = 50-3X-4Y...OR...5X+5Y=30....OR....X+Y=6......II
EQN.II - EQN I...GIVES
X+Y-X+2Y=6......OR 3Y=6.....Y=2
SO X=6-Y=6-2=4
Z=10-4-2=4...
HENCE 4 GMS OF A1,2 GMS OF A2 AND 4 GMS OF A3 ARE TO BE ADDED TO GET 10 GMS OF A4.

logarithm/34223: How would you go about solving these problems
2e^(x+1)-7=3

5log32^(x+1)=3 (note: 32 is the base)
1 solutions

Answer 20513 by venugopalramana(3286) on 2006-04-19 12:08:44 (Show Source):

You can put this solution on YOUR website!
How would you go about solving these problems
2e^(x+1)-7=3
E^(X+1)^2=3+7=10...TAKING LOGS
(X+1)^2*(LOG(E))=LOG(10)=1
(X+1)^2=10/LOG(E)
X+1=SQRT{10/LOG(E)}
X=SQRT{10/LOG(E)}-1=1.08

5log32^(x+1)=3 (note: 32 is the base)
LOG{(X+1)^5}TO BASE 32=3
(X+1)^5=32^3=(2^5)^3=(2^3)^5
X+1=2^3=8
X=8-1=7

Travel_Word_Problems/34210: I do not have a textbook because I am home schooling. If anyone can help I would be grateful>
Al goes 8mph; Ben goes 22mph in the opposite direction. How many hours has Ben traveled before they are 180 miles apart?
1 solutions

Answer 20500 by venugopalramana(3286) on 2006-04-19 08:43:54 (Show Source):

You can put this solution on YOUR website!
SEE THE FOLLOWING AND TRY
Travel_Word_Problems/29256: A freight train leaves Kansas and travels due west. A passenger train departs at exactly the same time on a parallel track and travels due east at a speed that is twice the speed of the freight train. After 7 hours the trains are 945 miles apart. How fast is each train traveling?
1 solutions
Answer 16084 by venugopalramana(1088) About Me on 2006-03-05 11:24:01 (Show Source):
A freight train leaves Kansas and travels due west. A passenger train departs at exactly the same time on a parallel track and travels due east at a speed that is twice the speed of the freight train. After 7 hours the trains are 945 miles apart. How fast is each train traveling?
TRAINS ARE TRAVELLING IN OPPOSITE DIRECTIONS..SO THEIR RELATVE SPEED =SUM OF THEIR INDIVIDUAL SPEEDS
SPEED OF ONE TRAIN IS X MPH....SAY..SECOND TRAIN SPEED =2X..MPH....SO
RELATIVE SPEED =X+2X=3X...MPH
TIME =7 HRS
INTIAL DISTANCE BETWEEN THEM =0
FINAL DISTANCE BETWEEN THEM =945 MILES
SO DISTANCE CHANGE =945-0=945 MILES..HENCE
945 =7*3X=21X
X=945/21=45 MPH
SO THE 2 TRAINS TRAVELLED AT 45 AND 90 MPH.

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