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how will you show that 9-23=1?

9-23=1? find the solution of 9-23=1?

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how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0

how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0 I think gaussian reduction is similar to back substitution, but with all the equation shaving different variables... just not sure how to do it. You create a matrix using the constants in the equations. Below, you will see approximately what the matrix would look like. Unfortunately, it is impossible to get it to display properly on this page. ┌                       ┐ │ 4   0   1  |   3  │ │ 2  -1  0   |  2   │ │ 0   3   2  |   0   │ └                        ┘ The idea is to perform the same math procedures on these matrix rows that you would perform on the full equations. We want the first row to be  1 0 0 | a, meaning that whatever value appears as the a entry is the value of x. The second row has to be  0 1 0 | b,  and the third row has to be  0 0 1 | c. Multiply row 2 by 2. ┌                      ┐ │ 4   0   1  |   3  │ │ 4  -2   0  |  4   │ │ 0   3   2  |   0   │ └                       ┘ Now subtract row 2 from row 1, and replace row 2 with the result. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  2   1   |  -1   │ │ 0   3   2  |   0   │ └                       ┘ Multiply row 2 by 2. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  4    2  |  -2   │ │ 0   3   2  |   0   │ └                       ┘ Subtract row 3 from row 2, replacing row 2. ┌                        ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   3   2  |   0   │ └                        ┘ Multiply row 2 by 3 and subtract row 3, replacing row 3. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   0  -2  |   -6  │ └                       ┘ Divide row 3 by -2. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   0   1  |   3   │ └                       ┘ Subtract row 3 from row 1, replacing row 1. ┌                      ┐ │ 4   0   0  |   0  │ │ 0  1    0  |  -2  │ │ 0   0  1  |   3   │ └                      ┘ Divide row 1 by 4. ┌                      ┐ │ 1   0   0  |   0  │ │ 0  1    0  |  -2  │ │ 0   0   1  |   3  │ └                      ┘ Row 1 shows that x = 0 Row 2 shows that y = -2 Row 3 shows that z = 3 Plug those values into the original equations to check the answer. 4x + z = 3 4(0) + 3 = 3 0 + 3 = 3 3 = 3 2x – y = 2 2(0) – (-2) = 2 0 + 2 = 2 2 = 2 3y + 2z = 0 3(-2) + 2(3) = 0 -6 + 6 = 0 0 = 0
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Find Intersections of Trig Functions with different periods

When the graphs intersect f(x)=g(x) so 600sin(2(pi)/3(x-0.25))+1000=600sin(2(pi)/7(x))+500. 600(sin(2(pi)/3(x-0.25))-sin(2(pi)/7(x)))+500=0. The 7 fundamental values of x satisfying this equation are to be found at the end of this answer. Trig identities: sin(p)=sin(p+2n(pi)) and cos(p)=cos(p+2m(pi)), where n and m are integers, so sin(2(pi)(x-0.25)/3)=sin(2(pi)(x-0.25)/3+2n(pi))=sin(2(pi)(x+3n-0.25)/3) and cos(2(pi)x/7)=cos(2(pi)x/7+2m(pi))=cos(2(pi)(x+7m)/7). For g(x) the value of the function repeats for x, x+7, x+14, etc.; and for f(x) it's x, x+3, x+6 etc. The repetition of intersections of f(x) and g(x) occur for x, x+21, x+42, etc., where x is a solution of the combined equation determining the intersection points. sin(A-B)=sinAcosB-cosAsinB and sin(A+B)=sinAcosB+cosAsinB. sin(A+B)-sin(A-B)=2cosAsinB. If X=A+B and Y=A-B, X+Y=2A so A=(1/2)(X+Y) and X-Y=2B so B=(1/2)(X-Y) sinX-sinY=2cos((X+Y)/2)sin((X-Y)/2). Using these identities X=2(pi)/3(x-0.25), Y=2(pi)x/7. (X+Y)/2=(pi)((x-0.25)/3+x/7)=(pi)(7x-1.75+3x)/21=(pi)(10x-1.75)/21 (X-Y)/2=(pi)((x-0.25)/3-x/7)=(pi)(7x-1.75-3x)/21=(pi)(4x-1.75)/21 600(sinX-sinY)+500=0 so sinX-sinY=-5/6=2cos((pi)(10x-1.75)/21)sin((pi)(4x-1.75)/21) cos((pi)(10x-1.75)/21)sin((pi)(4x-1.75)/21)=-5/12. Solutions for x: 1.67126, 3.03768, 7.82582, 9.27714, 14.13103, 15.28939, 16.90941, 22.67126,... There are 7 fundamental values for x and a series can be built on each one by adding (or subtracting) 21 (LCM of 3 and 7). The intersections of the sine waves repeat the earlier pattern indefinitely. Addendum The table below shows some points on the graphs of f(x) and g(x) for the purposes of comparison and to show roughly when  f(x) intersects g(x). The comparison column f(x)~g(x) shows whether f(x) is less than or greater than g(x). Where there is a change from < to >, or vice versa, an intersection point exists between the values of x listed. The table shows two whole cycles of f(x) (0 0.25 1000 633.5 > 0.5 1300 760.3 > 0.75 1519.6 874.1 > 1 1600 969.1 > 1.25 1519.6 1040.6 > 1.5 1300 1085 > 1.75 1000 1100 < 2 700 1085 < 2.25 480.6 1040.6 < 2.5 400 969.1 < 2.75 480.6 874.1 < 3 700 760.3 < 3.5 1300 500 > 4 1600 239.7 > 4.5 1300 30.9 > 5 700 -85 > 5.5 400 -85 > 6 700 30.9 > 6.5 1300 239.7 > 7 1600 500 >  
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How to solve a Logic problem?

Let q={ 0 1 4 9 16 25 36 }, r={ 1 3 5 7 9 11 }, p={ 0 2 4 6 8 10 } So q ^ r={ 1 9 }, p v q={ 0 1 2 4 6 8 9 10 16 25 36 }, p v r={ 0 1 2 3 4 5 6 7 8 9 10 11 } (p v q) ^ (p v r)={ 0 1 2 4 6 8 9 10 } p v (q ^ r)={ 0 1 2 4 6 8 9 10 }, so (p v q) ^ (p v r) = p v (q ^ r) (p v q) ^ r={ 1 9 } ≠ (p v q) ^ (p v r). Draw two intersecting circles representing sets q and r. Where they intersect is q ^ r (in the example the intersection would contain the numbers 1 and 9. Now consider augmenting the sets by the contents of p (this is the union of p with each of the two intersecting sets). This time the intersection would contain all the elements of p as well as 1 and 9. This demonstrates the first part of the question. But if the p elements are added only to q then the intersection only contains 1 9 because there are no more common elements in r than there were before. This demonstrates the second part of the question.
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Trace the conicoid –(x^2/4) +..... (Analytical Geometry)

The picture below shows an aerial view of the conicoid as seen looking down the y axis at the x-z plane. The outer ellipse is the way the conicoid appears at a distance of 3sqrt(5) from the origin, while the inner one is distance 3sqrt(2) from the origin. The origin itself is the point (0,3,0) as seen from above. The straight line is the line z=x, which is the edge view of the plane z=x. The origin is where the vertex of the conicoid is. The two vertical lines on the left are just markers showing the extent of the ellipse's radius in the x direction. The ellipse's radius in the z direction is 8 for the outer ellipse and 4 for the inner. The picture clearly shows that the plane z=x intersects the conicoid. This view is also the view looking towards the origin from the negative side of y, because the conicoid is in fact two shapes which are reflections of each other. The origin is the point (0,-3,0). Now we adjust our viewpoint and look along the other axes to get an idea of other aspects of the conicoid. Start with the x-y plane, viewed along the z axis. If we treat z in the equation as a constant we can see what the curve looks like for x and y. Initially put z=0: -x^2/4+y^2/9=1 is the equation of a hyperbola, with asymptotes given by x^2/4=y^2/9 or y=±3x/2. These are represented by two lines y=3x/2 and -3x/2, which lie on intersecting planes. When z^2/16+1=4, -x^2/4+y^2/9=4 and the asymptotes are x^2/16=y^2/36, y=3x/2 and -3x/2 as before, and z=4sqrt(3). No matter where we are along the z axis the asymptotes are the same, so the conicoid is contained within these planes. The picture below shows the view along the z axis. In this picture the line y=x (edge-on view of the plane) misses the conicoid completely, because it lies outside the asymptotes. You can also see the hyperbolas for increasing values of z. The inner hyperbolas are when z=0 and the outer ones are when z=4sqrt(3). By joining one side of each hyperbola to the other with a horizontal line you can visualise the edge-on appearance of the ellipses. These are x-diameters of each ellipse. Finally, the view from the x axis. Here the line y=z lies inside the asymptotes so the plane intersects the conicoid.
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Show that the conicoid x^2..... (Analytical Geometry)

Show that the conicoid S: x^2+2y^2+3z^2−2yz+4zx+6xy−2x−4y−6z+8=0 has a centre. Hence find the centre   If we have a conicoid, S’ ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0 then S' has a centre if the following set of equations has a solution ax + hy + gz + u = 0 hx + by + fz + v = 0 gx + fy + cz + w = 0 (see: http://www.scribd.com/doc/25330499/Unit-1, section 1.4, Reduction to Standard Form) In our conicoid S, a = 1, b = 2, c = 3 f = -1, g = 2, h = 3 u = -1, v = -2, w = -3 giving our system of simultaneous equations as, x + 3y + 2z - 1 = 0 3x + 2y – z – 2 = 0 2x – y + 3z – 3 = 0 The solution to this system of equations is x = 19/21, y = -4/21, z = 1/3 Since we have found a valid solution, then there is a centre and that centre is at (19/21, -4/21, 1/3)   Karan, I don't know if you want/need to show the above link in your submission. I have included it for your own interest.
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Find the equation of the cylinder having its base the circle

When z=1, x^2+y^2=4, which is a circle, centre at the origin and radius 2. x-2y=1 is a plane intercepting the x-y plane with y intercept=-1/2 and x intercept=1. The equation x^2+y^2+(z-1)^2=4 is a sphere, centre (0,0,1) and radius 2. The plane cuts through the sphere when (1+2y)^2+y^2=4; 1+4y+4y^2+y^2=4; 5y^2+4y-3=0, y=(-4±sqrt(16+60)/10=0.47178 ((-2+sqrt(19))/5) and -1.27178 ((-2-sqrt(19))/5). These values give x=1.94356 ((1+2sqrt(19))/5) and -1.54356 ((1-2sqrt(19))/5 respectively, by using the equation x=1+2y. So the points of intersection are (1.94356,0.47178) and (-1.54356,-1.27178). From these we can find the diameter of the cylinder: sqrt(76/25+304/25)=sqrt(380)/5=2sqrt(95)/5=3.9 approx. The radius is sqrt(95)/5. The sides of the cylinder are perpendicular to the diameter.  The slope of the diameter is the same as the slope of the plane x-2y=1, which is 1/2, so the slope of the perpendicular is -2. The equations of the sides of the cylinder where z=1 in the x-y plane are y=-2x+c where c is found by plugging in the intersection points: (-2+sqrt(19))/5=-2(1+2sqrt(19))/5+c; -2+sqrt(19)=-2-4sqrt(19)+5c. So c=sqrt(19) and y=sqrt(19)-2x as one side. Similarly, -2-sqrt(19)=-2+4sqrt(19)+5c, c=-sqrt(19) and y=-(sqrt(19)+2x) for the other side. Consider the view looking along the z axis. The circular cross-section of the cylinder will appear edge on, while the sides will have the slope -2 and will be spaced apart according to the value of z. When z=1±radius of cylinder=1±sqrt(95), the sides will appear to be as one line passing through the x-y plane's origin, the equation of the line being y=-2x when z=1±sqrt(95). The picture shows the view from z=1 looking at the x-y plane. The circle is the cross-section of the sphere and the line passing through (1,0) and (0,-1/2) is the edge of the plane x-2y=1. The diameter of the cylinder is constant and is shown by the two lines perpendicular to each end of the chord where the plane cuts the sphere. These lines represent the sides of the cylinder as they would appear at z=1. Parallel to them and passing through (0,0) is the single line that appears when z is at the extreme limits of the diameter of the cylinder. This line is also the central axis of the cylinder. (The vertical line at y=-2 is just a marker to show the the leftmost limit of the diameter of the sphere.) The general equation of a cylinder is the same as the 2-dimensional equation of a circle: x^2+y^2=a^2. This cylinder, radius a, has the z axis its central axis. x^2+z^2=a^2 is a cylinder with the y axis as its central axis. a=sqrt(95)/5 so a^2=95/25=19/5, making the equation of the cylinder 5x^2+5z^2=19. This is the same size as the cylinder in the problem, but with its central axis as the y axis. 5x^2+5(z-1)^2=19 is the equation of the cylinder with central axis passing through the centre of the sphere. If the cylinder is tilted so that its base coincides with the circle produced by the plane cutting through the original sphere, more calculations need to be made  to transform the coordinates. The picture shows the axial tilt of the cylinder. The central axis of the cylinder, y=-2x, bisects the chord on the line x-2y=1 when x-2(-2x)=1; 5x=1, x=1/5. So y=(x-1)/2=-2/5. If x-2y=1 represents the horizontal axis (we'll call X) and y=-2x represents the vertical axis (Y) we can see that, relative to these X-Y coordinates, the cylinder is upright. The  z value is unaffected by rotation. Take a point P(x,y) in the x-y plane. What are its coordinates in the X-Y plane? To find out we use geometry and trigonometry. The axial tilt of the X-Y axes is angle ø where tanø=1/2, so sinø=1/sqrt(5) and cosø=2/sqrt(5).  X=SP=QPcosø=2(x+y/2)/sqrt(5), Y=PRcosø=2(y-(x-1)/2)/sqrt(5). Z=z In the X-Z plane, 5X^2+5Z^2=19. This transforms to 4(x+y/2)^2+5z^2=19=(2x+y)^2+5z^2. In the X-Y plane the sides of the cylinder are the lines X=-a and X=a, where a^2=19/5. So X^2=19/5, which transforms to 4(x+y/2)^2 or (2x+y)^2=19/5; 2x+y=±sqrt(3.8), corresponding to the equation of two parallel, sloping line forming the sides of the cylinder.  
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show that Z15 is isomorphic to Z5 external Z3

Z15 can be represented as a pair of quantities a and b so that a has a range of 0 to 4 (Z5) and b has a range of 0 to 2 (Z3). So we have a scheme of representation: (0,0)=0 (1,1)=11 (2,2)=7 (0,3)=3 (1,4)=14 (2,0)=10 (0,1)=6 (1,2)=2 (2,3)=13 (0,4)=9 (1,0)=5 (2,1)=1 (0,2)=12 (1,3)=8 (2,4)=4. In general (X,Y) maps to (5X+6Y) modulo 15, where X is confined to modulo 3 and Y to modulo 5.  EXAMPLES:  ADDITION: (2,3)+(1,2)=(13+2) mod 15 = 15 mod 15 = 0 (2,3)+(1,2)=((2+1) mod 3, (3+2) mod 5)=(0,0)=0 (1,4)+(0,2)=(14+12) mod 15 = 26 mod 15 = 11 (1,4)+(0,2)=((1+0) mod 3, (4+2) mod 5)=(1,1)=11 SUBTRACTION: (2,3)-(1,2)=(13-2) mod 15 = 11 (2,3)-(1,2)=((2-1) mod 3, (3-2) mod 5)=(1,1)=11 (1,4)-(0,2)=(14-12) mod 15 = 2 (1,4)-(0,2)=((1-0) mod 3, (4-2) mod 5)=(1,2)=2 MULTIPLICATION (2,3)*(1,2)=(13*2) mod 15 = 26 mod 15 = 11 (2,3)*(1,2)=(2,3)+(2,3)=(1,1)=11 (1,4)*(0,2)=(14*12) mod 15 = 168 mod 15 = 3 (1,4)*(0,2)=((1,4)+(1,4)+(1,4))+((1,4)+(1,4)+(1,4))+((1,4)+(1,4)+(1,4))+((1,4)+(1,4)+(1,4)) =(0,2)+(0,2)+(0,2)+(0,2)=(0,3)=3 Or is this cheating?! The point is that the mapping shows for each of all the combinations of Z3 and Z5 there is one and only one Z15 element, and no elements of Z15 have been omitted, so demonstrating isomorphism.
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Algebra 2 homework, please help.

  First set of questions is all roots, or solutions, of the equations and second set is to find where the functions are zero, the rational roots only. The methods for roots and zeroes are basically the same. 1. 3x^2+10x+3=3x^2+9x+x+3=3x(x+3)+(x+3)=(3x+1)(x+3)=0; x=-1/3, -3 2. 5x^2+5x-x-1=5x(x+1)-(x+1)=(5x-1)(x+1)=0; x=1/5, -1 3. Cubic equation. Trial shows that x=-1 is a solution, so divide cubic by factor (x+1). Use synthetic division: -1 | 3 11  5 -3      | 3 -3 -8  3        3  8 -3 | 0 and  (x+1)(3x^2+8x-3)= (x+1)(3x^2+9x-x-3)=(x+1)(3x(x+3)-(x+3))=(x+1)(3x-1)(x+3)=0; x=-1, 1/3, -3 4. 3x^4-2x^2-5=(3x^2-5)(x^2+1)=0; 3x^2=5, so x=+/-sqrt(5/3) 5. Trial shows that 3x=-1 is a solution, i.e., x=-1/3. Use synthetic division: -1/3 | 3 10 30   9         | 3  -1 -3  -9           3   9 27 | 0      (3x+1)(x^2+3x+9)=0; x=-1/3 is the only real solution 6. (5x^2+7)(x^2-2)=0; x=+/-sqrt(2) 7. No real roots 8. (5x^2+2)(x^2-8)=0; x=+/-sqrt(8) or +/-2sqrt(2) 9. (5x^2+9)(x^2-4)=(5x^2+9)(x+2)(x-2)=0; x=+/-2 10. (5x^2+3)(x^2-8)=0; x=+/-sqrt(8) or +/-2sqrt(2) 11. f(x)=5x(x^2-1)-(x^2-1)=(5x-1)(x-1)(x+1); x=1/5, 1, -1 12. Try x=1/2, f(x)=0. So f(x)=(2x-1)(x^2-11x-1) has no more rational roots 13. f(x)=3x(x^2-1)-(x^2-1)=(3x-1)(x-1)(x+1); zeroes at x=1/3, 1, -1 14. f(x)=(x+2)(x^2+7x+3); only rational zero at x=-2 15. (x-3)(2x^2+7x-3); rational zero at x=3 16. (x-1)^2(4x-1); x=1, 1/4 17. (x+1)(2x-1)(2x+1); x=-1, +/-1/2 18. (x+1)^2(3x-1); x=-1, 1/3 19. (x-5)(3x^2-7x-75); x=5, no other rational roots 20. (x-3)(2x^2+11x-4); x=3, other roots irrational
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show that the marginal distribution of Y is a beta distribution

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(2x+1)(3x-1)=0

(2x+1)(3x-1)=0 can u please show me the steps to solve this problem thanks The first impulse would be to multiply those quantities, but that would be a waste of time, because you would then need to factor the result. You already have the factors staring you in the face. What the problem is telling you is that one, or both, of those factors has to be equal to zero. Remember, multiplying by zero gives a zero result. So, we set each quantity to zero and solve for x. 2x + 1 = 0 2x = -1 x = -1/2 3x - 1 = 0 3x = 1 x = 1/3 Now, if you want to, you can multiply the factors and use the resulting equation to check your answers. (2x + 1)(3x - 1) = 0 6x^2 + 3x - 2x - 1 = 0 6x^2 + x - 1 = 0 Plug in the values calculated for x. 6x^2 + x - 1 = 0 6(-1/2)^2 + (-1/2) - 1 = 0 6(1/4) - 1/2 - 1 = 0 1 1/2 - 1/2 - 1 = 0 0 = 0 That one checks. 6x^2 + x - 1 = 0 6(1/3)^2 + 1/3 - 1 = 0 6(1/9) + 1/3 - 1 = 0 2/3 + 1/3 - 1 = 0 0 = 0 That one checks, too. x = -1/2   and  x = 1/3
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