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# given that 7/9 of a mass A is the same as 5/12 of a mass B, find the ration of mass A to mass B

given that 7/9 of a mass A is the same as 5/12 of a mass B, find the ration of mass A to mass B

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### B) 4.0 m/s C) 7.9 m/s D) 5.6 m/s - Iowa State University

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## Suggested Questions And Answer :

### given that 7/9 of a mass A is the same as 5/12 of a mass B, find the ration of mass A to mass B

start: (7/9)a=(5/12)b a=(9/7)*(5/12)b a=(45/84)b   =(15/28)b

### 3+i, 3 Lowest Degree

Question: 3+i, 3 Lowest Degree. Find a polynomial function of lowest degree with rational coefficients that has the given numbers as some of its zeros. When a polynomial has complex roots, they always come as a pair. So if x = 3+i is one root, then x = 3 - i is another root. (Two complex roots will always be of the form a + ib, a - ib. ) So we have at least three roots: x = 3, x = 3 + i, x = 3 - i. Creating the polynomial from these three roots. (x - 3)(x - (3 + i))(x - (3 - i)) = 0 (x - 3)(x^2 - 3x + 3i - 3x - 3i + 9 + 1) = 0 (x - 3)(x^2 - 6x + 10) = 0 x^3 - 6x^2 + 10x - 3x^2 + 18x - 30 = 0 x^3 - 9x^2 + 28x - 30 = 0

### problem solving in integral calculus about centoid

The centroid lies on the y-axis (x=0), because the semicircle is symmetrical about this axis. We need to find the y coord. The centroid is the "centre of mass" of the semicircle, and it can be considered to be made up of a number of very thin rectangles of width dx and length y or width dy and length x. The area of each rectangle is ydx (vertical) or xdy (horizontal) and the mass of the rectangle is proportional to this area. The area of the semicircle is 1/2 the area of its circle, which is (pi)r^2 where r is the radius, which is given by the equation of the circle x^2+y^2=4=r^2. The area of the semicircle is therefore 2(pi) and its mass is proportional to this area. The centroid is the average of the "moments" of the rectangles divided by the mass of the semicircle. The moment is mass*distance from the origin=xydx or yxdy for x and y cords respectively. The sum of these moments is the integral(xydx) (x coord) and integral(yxdy) (y coord) between the limits -2 and 2 for x, and 0 and 2 for y. These points on the axes are where the semicircle intersects. The centroid is the point (integral(xydx)/2(pi),integral(yxdy)/2(pi)). y=sqrt(4-x^2) or x=sqrt(4-y^2), and the integral(ysqrt(4-y^2)dy)/2(pi)=-(2/3)(4-y^2)^(3/2)/4(pi). This will give us the y coord of the centroid, and we already know the x coord is zero because of symmetry. The limits for y are 0 to 2: (2/3)4^(3/2)/4(pi)=(2/3)8/4(pi)=4/(3(pi))=0.4244. So the centroid is (0,0.4244). [integral(ysqrt(4-y^2)dy)/2(pi): let u=4-y^2, then du=-2ydy, so ydy=-du/2. The integral becomes integral(-u^(1/2)du)/4(pi)=-(2/3)u^(3/2)/4(pi)=-(2/3)(4-y^2)^(3/2)/4(pi).]

### is there a formula for finding a rational function of a given graph?

First you have to graph the lines.  Then you can see if there is any function to discribe them.     Since these lines do not have limits as the domain and range I would say that there is not a feunction to describe them.  Due to the fact that y has 2 values at all negative numbers and on the y-axis. f(x) = {y = -1/4x,   -3 <= x < 1}           {y = x - 5,     1 <= x < infinity.

### find a third degree polynomial equation with rational coefficients that has roots -4 and 6+i

A cubic equation always has three roots. These three roots are: 1) three real roots or, 2) 1 real root and two complex roots If one of the two complex roots is a + ib, then the other complex root is a - ib. We are given two of the roots as -4 and 6+i. Since one of the roots is complex and equals 6+i, the the other complex root is 6-i. Our three roots then are: -4, 6+i, 6-i. Our three solutions to the cubic equation are: x = -4, x = 6+i, x = 6-i. Which can be rewritten as: x + 4 = 0, x - (6+i) = 0, x - (6-i) = 0 Multiplying these together gives us the original cubic equation. (x + 4)(x - (6+i))(x - (6-i)) = 0 Multiplying this out, (x + 4)(x^2 - (6-i)x - (6+i)x + (6-i)(6+i)) = 0 (x + 4)(x^2 - 12x + 6^2 - i^2) = 0 (x + 4)(x^2 - 12x + 37) = 0 x^3 - 8x^2 - 11x +148 = 0

### A standard Hyperbola has its vertices ar 5,7 and 3,-1.Find its equation and its asymptotes

this equation expands toThe vertices share no common coordinate so the hyperbola cannot be orientated in the usual x-y form. The hyperbola must therefore be skew (tilted axis) and the line joining the vertices is sloping. This line is given by the equation y=((7+1)/(5-3))x+c where c is to be found. y=4x+c, and -1=12+c, plugging in (3,-1). c=-13. Therefore the axis of the hyperbola is y=4x-13. The centre or origin of the hyperbola lies midway between them at ((1/2)(5+3),(1/2)(7-1))=(4,3) and the equation of the conjugate axis is given by y=-x/4+k where k is found by plugging in the midpoint: 3=-4/4+k, so k=4 and y=4-x/4 is the equation of the conjugate axis. Let's define the axes of the hyperbola as X and Y instead of x and y. We can place the centre of the hyperbola at (0,0) in the X-Y plane so that in this plane the equation of the hyperbola is Y^2/A^2-X^2/B^2=1. The vertices are found by finding Y when X=0, so (0,A) and (0,-A) are the vertices. The asymptotes are given by (Y/A+X/B)(Y/A-X/B)=0. So Y=AX/B and Y=-AX/B are their equations. The distance between the vertices is 2A so we can work this out from the x-y coordinates of the vertices using Pythagoras' Theorem: 2A=sqrt((7+1)^2+(5-3)^2)=sqrt(68). A=sqrt(68)/2, A^2=68/4=17. Since the hyperbola is rectangular, B=A, and the equation of the hyperbola is Y^2-X^2=17. The asymptotes then become Y=X and Y=-X. Now we have to transform X-Y to x-y, so that the equation can be expressed in the x-y plane. We use geometry and trigonometry  for this. Let O'(4,3) be the location of the origin of the X-Y plane. Let P(x,y) be any point. In the X-Y plane this is P'(X,Y). Join O'P'=O'P. This line can be represented in both reference frames. O'P^2=X^2+Y^2=(x-4)^2+(y-3)^2 by Pythagoras. The gradient of the Y axis with respect to the x-y frame is 4, and we can represent this as tan(w) so w=tan^-1(4) or tan(w)=4; sin(w)=4/sqrt(17); cos(w)=1/sqrt(17). The angle z that O'P makes with the line y=3 is given by tan(z)=(y-3)/(x-4). sin(z)=(y-3)/O'P; cos(z)=(x-4)/O'P. Let v=90-w. sin(v+z)=Y/O'P; cos(v+z)=X/O'P. v+z=90-w+z=90-(w-z), so sin(v+z)=cos(w-z)=cos(w)cos(z)+sin(w)sin(z)=(x-4)/(O'Psqrt(17))+4(y-3)/(O'Psqrt(17))=(x+4y-16)/(O'Psqrt(17)). cos(v+z)=sin(w-z)=sin(w)cos(z)-cos(w)sin(z)=4(x-4)/(O'Psqrt(17))-(y-3)/(O'Psqrt(17))=(4x-y-13)/(O'Psqrt(17)). Y=O'P*sin(v+z)=(x+4y-16)/sqrt(17); X=O'P*cos(v+z)=(4x-y-13)/sqrt(17). So P maps to the x-y plane using these transformations. Y^2-X^2=17 becomes (x+4y-16)^2-(4x-y-13)^2=289, the equation of the hyperbola. This expands to x^2+8x(y-4)+16y^2-128y+256-(16x^2-8x(y+13)+y^2+26y+169)=289; -15x^2+16xy+15y^2+72x-154y-202=0. The asymptotes are x+4y-16+4x-y-13=0; 5x+3y-29=0 and x+4y-16-4x+y+13=0; 5y-3x-3=0.

### f(1)=x³+bx²-9x+c find the values of b and c

f(x)=x^3+bx^2-9x+c; f(1)=1+b-9+c. We are not given the value of f(1) so let's start by assuming f(1)=0 as a slight addition to the original question. This means that 1+b-9+c=0, b+c=8. f(1)=0 means that 1 is a zero of the expression and x^3+bx^2-9x+c=(x-1)(x^2+ax-c) where a needs to be found. This expands to: x^3+(a-1)x^2-(c+a)x+c so we can write b=a-1 and c+a=9. The quadratic x^2+ax-c can be written x^2+(9-c)x-c, so x=(c-9+sqrt(81-18c+c^2+4c))/2=(c-9+sqrt(c^2-14c+81))/2. The question offers no more information to determine unique values for b and c, but we can put in some constraints because b and c must be rational. c^2-14c+81=n^2, a perfect square. Completing the square for the left-hand side: (c^2-14c+49)-49+81=n^2; (c-7)^2=n^2-32; c-7=+sqrt(n^2-32). When n=6, c=7+2, so c=9 or 5 and b=-1 or 3, because b+c=8. Therefore f(x)=x^3-x^2-9x+9=(x-1)(x^2-9)=(x-1)(x-3)(x+3) or x^3+3x^2-9x+5=(x-1)(x^2+4x-5)=(x+5)(x-1)^2. Therefore the solutions are (b,c)=(-1,9) or (3,5).

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

### Find the center of the mass of a solid with constant density & bounded below by the disk R : x^2+y^2≤4 in the plane z=0 and above by the parabolaid z=4-x^2-y^2

The paraboloid equation can be written x^2+y^2=4-z. At z=0 this is the circle x^2+y^2=4 (centre at the origin and radius=2). As we move along the z axis the cross-section parallel to the x-y plane remains as a circle (x^2+y^2≤4) which reduces to a point at the vertex of the paraboloid at (0,0,4). The centre of gravity (COG) is therefore on the z axis at P(0,0,p) because the paraboloid is symmetrical, with the z axis acting like a spindle. We have to find p. We can consider the solid as a set of discs pierced through their centres by the z axis as a spindle. The volume of each disc is related to its mass by the constant density. The thickness of the disc is the infinitesimal dz. The radius of a disc is √(4-z), so its volume is π(4-z)dz. This expression can be taken to represent its mass. The mass of each disc is concentrated at its centre because of symmetry, so its COG is at the centre on the z axis. We need to consider the COG of all the discs together, so we need to define the moment of each disc about P. This is the product of mass and distance from P, which is p-z. The negative moments have to balance the positive moments. The sum total of the moments is zero. The expression for the moment of each disc is π(4-z)(p-z)dz and the sum of these must be zero, so π∫((4-z)(p-z)dz)=0 where the limits are 0≤z≤4. π∫((4p-z(p+4)+z^2)dz)=0; π[4pz-z^2(p+4)/2+z^3/3]=0, 0≤z≤4.  π(16p-8(p+4)+64/3)=0; 8p-32+64/3=0; 8p=32/3, p=4/3, so the COG is P(0,0,4/3).