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lcm is 15. hence

Multiplying both sides with 15 , we get ....
-2x-10=1
-2x=1+10=11
x=11/2

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The equation of a circle with centre as (h,k) and radius r is given by the formula
.
So we have to write the given equation in this form to get the centre of the circle.
OR
...Note that we have added and subtracted 4 and 9 to make up the squares mentioned in brackets.
Hence
Hence (2,-3)is the centre of this circle .Similarly the centre of the second circle is obtained from
OR
...Note that we have added and subtracted 9 and 4 to make up the squares mentioned in brackets.
Hence
Hence the centre of the second circle is (-3,-2)
The equation of line joining 2 ponts (x1,y1) and (x2,y2)is given by the formula

Substituting (2,-3) and (-3,-2) in the above formula we get
Y-(-3)=(-2-(-3))*(X-2)/(-3-(-2))
Y+3=(X-2)/(-5)
-5Y-15=X-2
X+5Y+13=0

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X intercept of a line is the distance (to be measured along the positive direction of the X axis) from origin to the point where the line meets the X axis .As the intersecting point is on the X axis ,it is characterised by the fact that its Y coordinate is zero.From the table we find that when Y =0 , X = -2 . Hence the X intercept is -2.
Slope of a line is given by the formula
Difference in Y coordinates of any 2 points on the line / Difference in X coordinates of the same 2 points on the line .
If we take the 2 points as A (0,1) and B (-2,0) from the given ordered pairs...
slope = (0-1)/(-2-0)= -1/(-2)=1/2

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If the points are taken as P with coordinates (p,q) and A with coordinates (a,b)
then the slope of the line PA is given by the formula
Difference in Y coordinates / Difference in X coordinates
= (b-q)/(a-p)...It does not matter for this purpose whether we take the coordinates of A first or those of P first as long as we take the same order for calculating the difference of both Y and X cordinates of the 2 points.It is
easy to see that
Slope of PA = (b-q)/(a-p) = (-(q-b))/(-(p-a))= (q-b)/(a-p)= Slope of AP
Viewing it from the physical sence , slope of a line is the Tangent of the angle the line makes with X axis in the positive (anticlockwise)direction. The line afterall makes the same angle with X axis , whether we look at it as PA or AP and hence its tangent which is the slope of the line is also same whichever way we take the line segment.

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There seems to be an error in your problem .There are 2 basic (inter-related)rules governing the dimensions of the 3 sides of a triangle for it to exist,that is for you to physically draw a triangle with those dimensions.They are
1. Sum of any 2 sides shall be greater than the third side and
2. Difference of any 2 sides shall be less than the third side.
Otherwise, what happens is that when you attempt to draw such a triangle , you will not be able to get the third vertex after having started with any 2 sides and 2 vertices.
Hence ,in our present problem ,since 2 sides are 6 & 7 b,the third side x shall be greater than 13 or less than 1 (ofcourse it cannot be zero or negative).That is x should lie between 9 and 1 or should be greater than 13.
Hence it is not correct to say that x has a maximum value .We can however say it has a minimum of zero in the lower range of number line and in that vicinity a local maximum of one and then again a minimum of 13 in the upper range of number line with no limit on maximum higher value beyond that.

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SEE THE FOLLOWING SOLUTION GIVEN BY ME TO A SIMILAR PROBLEM AND WORK THE SOLUTION YOUR SELF.
Answer 7295 by venugopalramana(11) on 2005-10-10 05:14:13 (Show Source):
Hi Tutors:
I have working on this problem for some time but I am still having a hard time working this one:
I am suppose to solve this system using elimination method
10x+6y+z=7 (1)
5x-9y-2z=3 (2)
15x-12y+2z=-5 (3)
20x-21y=-2 (4)
Please show me how to work this problem from this point.
Thank you
Good you have proceeded correctly and infact on the way to solving the problem by your self..you only need a little guidance on the path you should follow to solve the problem..o.k. ..let us see you have added equations 2 and 3 to get equation 3 ,which has accomplished elimination of one unknown z . The basic procedure is , if we start with 3 equations in 3 unknowns ,we try to eliminate one unknown taking one pair of equations at a time to get 2 new equations in 2 unknowns only.Then we take those 2 new equations to eliminate one another unknown to get one more new equation , but this time with one unknown only.This we can easily solve to find the unknown.Now , we travel backwards along the same path as we travelled to find the 2 other unknowns one after another by substituting the known values every time.Let us illustrate the procedure now with this example.Now that you have already got one new equation 4 from 2 and 3 to eliminate z., let us take equations 1 and 2 to eliminate the same unknown z.For this we observe the coefficients of z in the two equations which are 1 and -2 respectively.So we multiply equation 1 with 2 and add it to equation 2.
Eqn.1 * 2 gives us ...20x+12y+2z=14 .....(5)
Eqn.2 is .............5x-9y-2z = 3........(6)
Eqn.5 + Eqn.6 gives us .....25x+3y = 17....(7)
but from Eqn.4 we have .....20x-21y=-2......(4)..proceeding on the same basis ,we eliminate y from these 2 equations.
Eqn.7 * 7 gives us .........175x+21y=119....(8)
Eqn.8 + Eqn.4 gives us .....195x=117 ..or x= 117/195 = 39/65 = 3/5.....now substitute this value of x in eqn.4 to get y
y=(20*(3/5)+2)/21=14/21=2/3…….now substitute these values of x and y in eqn.1 to get z.
z=(7-10*(3/5)-6*(2/3))=-3………….. As a check ,you can substitute these values of x,y,and z in the 3 given equations to
verify that your answer is correct.

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SEE THE FOLLOWING SOLUTION GIVEN BY ME TO A SIMILAR PROBLEM AND WORK THE SOLUTION YOUR SELF.
Answer 7295 by venugopalramana(11) on 2005-10-10 05:14:13 (Show Source):
Hi Tutors:
I have working on this problem for some time but I am still having a hard time working this one:
I am suppose to solve this system using elimination method
10x+6y+z=7 (1)
5x-9y-2z=3 (2)
15x-12y+2z=-5 (3)
20x-21y=-2 (4)
Please show me how to work this problem from this point.
Thank you
Good you have proceeded correctly and infact on the way to solving the problem by your self..you only need a little guidance on the path you should follow to solve the problem..o.k. ..let us see you have added equations 2 and 3 to get equation 3 ,which has accomplished elimination of one unknown z . The basic procedure is , if we start with 3 equations in 3 unknowns ,we try to eliminate one unknown taking one pair of equations at a time to get 2 new equations in 2 unknowns only.Then we take those 2 new equations to eliminate one another unknown to get one more new equation , but this time with one unknown only.This we can easily solve to find the unknown.Now , we travel backwards along the same path as we travelled to find the 2 other unknowns one after another by substituting the known values every time.Let us illustrate the procedure now with this example.Now that you have already got one new equation 4 from 2 and 3 to eliminate z., let us take equations 1 and 2 to eliminate the same unknown z.For this we observe the coefficients of z in the two equations which are 1 and -2 respectively.So we multiply equation 1 with 2 and add it to equation 2.
Eqn.1 * 2 gives us ...20x+12y+2z=14 .....(5)
Eqn.2 is .............5x-9y-2z = 3........(6)
Eqn.5 + Eqn.6 gives us .....25x+3y = 17....(7)
but from Eqn.4 we have .....20x-21y=-2......(4)..proceeding on the same basis ,we eliminate y from these 2 equations.
Eqn.7 * 7 gives us .........175x+21y=119....(8)
Eqn.8 + Eqn.4 gives us .....195x=117 ..or x= 117/195 = 39/65 = 3/5.....now substitute this value of x in eqn.4 to get y
y=(20*(3/5)+2)/21=14/21=2/3…….now substitute these values of x and y in eqn.1 to get z.
z=(7-10*(3/5)-6*(2/3))=-3………….. As a check ,you can substitute these values of x,y,and z in the 3 given equations to
verify that your answer is correct.

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Slope of a line is defined as the tangent of the angle the line makes with X axis in the positive (anticlockwise) direction.
if the line equation is given as ax + by + c = 0 , then the slope is given by
the formula m = slope = -a/b

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2x+3y=6 being a linear equation in x and y represents the graph of a straight line . we have to find its x and y intercepts ,that is where does the above graph cuts the X axis and Y axis.We can do it analytically by noting the fact that the graph intersects the X axis at a point where the point's Y coordinate is zero and ,similarly ,the graph intersects the Y axis at a point where the point's X coordinate is zero.Hence we get the X intercept by putting y=0 in the given equation and solving for x.Similarly we get the Y intercept by putting x=0 in the given equation and solving for y.
Substituting y=0 , we get
2x+3*0=6
2x=6
x=3 is the x intercept.
Substituting x=0 , we get
2*0+3y = 6
3y=6
y=2 is the y intercept.
.

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there is a formula for distance of a point (h,k) from a line given by the equation
a*x+b*y+c=0...the formula is
distance =
That is ,just substiute the given coordinates of the point (h,k)for x and y in the given equation of the line and devide it by
for the given problem we have line as 2x+y-2=0 and point as (-3,4).So a=2;b=1;c=-2;h=-3;k=4.
distance = =-4/

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the coordinates of point are x,y.since they are equal in absolute value let the absolute value of each be h.since the point is in second quadrant ,its x coordinate is negative and y coordinate is positive so x=-h and y=h.
now distance from origin is given by square root of (square of x coordinate + square of y coordinate)=square root of ((-h)^2+h^2)=square root of(h^2+h^2)=square root of(2*h^2)=2
hence h^2=2/2=1
hence h=1
hence the coordinates of the point are -1,1

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Help me please!
I don't understand what I am suppose to do.
Evaluate the determinant of the matrix.
items in a bracket. 3 -2 0
-1 5 3
2 7 1
The determinant of matrix

is evaluated in the usual manner as that of a determinant.IF IT IS NOT KNOWN TO YOU IT CAN BE EXPLAINED ON HEARING FROM YOU>
taking any one column or row .Expanding by using the first column of 3,-1,2 we get the value of the determinant as
=3*(5*1-3*7)-(-1)*((-2)*1-0*7)+2*((-2)*3-0*5)
=-48-2-12=-62

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Kramers rule is used to solve simultaneous equations.In the working given below let me use the notation to represent the determinant.However,in actual practice we should use |xxx|to indicate a determinant ,while (xxx)is used to represent a matrix.I shall check with the network authorities on how to represent determinants in their software.
In the present case there are 3 equations to solve for 3 unknowns.
The solution is given by Kramer's rule as follows.
x/=
y/=
z/=
1/
evaluating the determinants we get
x/(-3)=y/6=z/(-9)=1/(-3)
hence x=1
y=-2
z=3

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Hi Tutors:
I have working on this problem for some time but I am still having a hard time working this one:
I am suppose to solve this system using elimination method
10x+6y+z=7 (1)
5x-9y-2z=3 (2)
15x-12y+2z=-5 (3)
20x-21y=-2 (4)
Please show me how to work this problem from this point.
Thank you
Good you have proceeded correctly and infact on the way to solving the problem by your self..you only need a little guidance on the path you should follow to solve the problem..o.k. ..let us see you have added equations 2 and 3 to get equation 3 ,which has accomplished elimination of one unknown z . The basic procedure is , if we start with 3 equations in 3 unknowns ,we try to eliminate one unknown taking one pair of equations at a time to get 2 new equations in 2 unknowns only.Then we take those 2 new equations to eliminate one another unknown to get one more new equation , but this time with one unknown only.This we can easily solve to find the unknown.Now , we travel backwards along the same path as we travelled to find the 2 other unknowns one after another by substituting the known values every time.Let us illustrate the procedure now with this example.Now that you have already got one new equation 4 from 2 and 3 to eliminate z., let us take equations 1 and 2 to eliminate the same unknown z.For this we observe the coefficients of z in the two equations which are 1 and -2 respectively.So we multiply equation 1 with 2 and add it to equation 2.
Eqn.1 * 2 gives us ...20x+12y+2z=14 .....(5)
Eqn.2 is .............5x-9y-2z = 3........(6)
Eqn.5 + Eqn.6 gives us .....25x+3y = 17....(7)
but from Eqn.4 we have .....20x-21y=-2......(4)..proceeding on the same basis ,we eliminate y from these 2 equations.
Eqn.7 * 7 gives us .........175x+21y=119....(8)
Eqn.8 + Eqn.4 gives us .....195x=117 ..or x= 117/195 = 39/65 = 3/5.....now substitute this value of x in eqn.4 to get y
y=(20*(3/5)+2)/21=14/21=2/3…….now substitute these values of x and y in eqn.1 to get z.
z=(7-10*(3/5)-6*(2/3))=-3………….. As a check ,you can substitute these values of x,y,and z in the 3 given equations to
verify that your answer is correct.

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What would be an equation with the solutions 5+2i,5-2i?
There are 2 solutions given here.As per theory of equations ,if we have to have 2 solutions for an unknown variable called x ,then the equation has to be a polynomial, say f(x)=0 , of degree 2 in x.Further ,the fact that it has one or more solutions implies ,that on substitution of that value for x in f(x),it has to equal zero,that is if say x=a is a solution of f(x)=0 ,then f(a) should equal to zero.That is as per Remainder theorem (x-a)is a factor of f(x).Hence we have
f(x) has 2 factors namely x-(5+2i) and x-(5-2i) corresponding to the 2 solutions given.Hence
is the basic equation which gives us the given solutions

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This is an exponential notation where for the sake of convenience , we write the small numbers as in the above problem first as numbers starting with units place devided by the required power of 10 so as to equal the given number .Then we substitute the notation yy E xx where yy represents the number with units place mentioned first and xx represents the power of 10 obtained above.IT IS IMPORTANT TO NOTE THAT WE HAVE TO USE -ve OR +ve SIGN BEFORE XX NUMBER APPROPRIATELY DEPENDING ON WHETHER WE HAVE TO DEVIDE OR MULTIPLY THE POWER OF TEN TO GET THE GIVEN NUMBER. It ineffect means that the given number is equal to that given on the left of symbol E (yy) multiplied by 10 to the power of the number given to the right of E (xx).The following small example will illustrate the procedure.
0.32=3.2/10 =3.2 E -1 since 3.2 is devided by 10 to the power of one to get the given number.
Similarly 32 =3.2*10 =3.2 E 1 since 3.2 is multiplied by 10 to the power of one to get the given number.
Hence the solution to the given problems are
0.0009255==9.255 E -4
0.000000005317==5.317 E -9

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Given that
=,
we have to find the value of a when b=12
The formulae we use here are
1.
2. and
3.If , then
Using the first 2 formulae we simplify the left hand side of given equation as



=

==

Hence we have LHS==RHS
=
Now using the third formula , we have
=
Since it is given that b=12,substituting in the above ,we get
a=-5

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We are given the relation between x and y saying that .Then we are given a set of values for x and asked to calculate the corresponding values of y and tabulate them.
let us take the first value of x=2 as asked for to explain the calculation.We have to substitute this value of x in the given relation for y and x namely to get at the value of y corresponding to value of x=2 .
We get for x=2...... y=3*2*2-10=2. continuing in the same manner
x=1.......y=3*1*1-10=-7
x=0.......y=3*0*0-10=-10
x=10......y=3*10*10-10=290
x=5.......y=3*5*5-10=65