Guide :

# if 1 1/4 : 2 1/3 = p:q and q:r = 4 1/2 : 5 1/4 find p:r

continued proportion

## Research, Knowledge and Information :

### 1. If P=(2,?2) And Q=(?2,3), Find The Components O ... - Chegg

... and Q=(?2,3), find the components of PQ? 2. ... (3,?1),P=(?4,?4),Q=(1,?8) 3. Let R=(4,?1). Find the point P such that PR? has components <?1,2>. 4. ... Let a=<5,3 ...
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### If p:q=2:5, r:s=3:2 and p:r=1:3, how can you find out the ...

If p:q=2:5, r:s=3:2 and p:r=1:3, how can you find out the pqrs ratio? ... (2/1 * 3) 4(2/1 * 2) Thus,p:q:r:s = 2:5:6:4 Hope this helps :) 1.7k Views · 1 Upvote.
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### SOLUTIONS TO HOMEWORK ASSIGNMENT #2, Math 253

SOLUTIONS TO HOMEWORK ASSIGNMENT #2, ... 1;¡1), Q(1;¡1;2), and R(4;0;3). Set up two vectors: ... (1 6; 1 3; 1 2) . 5. Find the distance between the point (2;8;5) ...
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### Let P=(2,4,2), Q=(0,1,-1), R=(0,1,1) And S=(-1,3,0 ... - Chegg

Let P=(2,4,2), Q=(0,1,-1), R=(0,1,1) and S=(-1,3,0). ... Find your book. Need an extra hand? Browse hundreds of Algebra tutors. ABOUT CHEGG. Media Center; College ...
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### Name: Solutions to Practice Midterm 1. - Columbia University

Name: Solutions to Practice Midterm 1. (a) ... Are the points P= (1; 1;1), Q= (2;1; 1), R= (0;3; 1), S= ( 1;0;1) ... (1 3=5;1 1=5; 1=5) = (2=5;4=5; 1=5): 3.
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### Section 13.5 Equations of Lines and Planes - faculty.up.edu

Section 13.5 Equations of Lines and Planes ... Example 2.2. Find a vector equation for the plane which contains the points P(1,2,3), Q(2,4,1) and R ...
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### 1 Review of Probability - Columbia University

1 Review of Probability ... and Var(aX) = a2Var(X). (3) For any two r.v.s. X and Y E(X +Y) = E(X)+E(Y). (4) ... (1−p) p2 M(s) = pes 1−(1−p)es. 4
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### Truth table - Wikipedia

Xq 2 ¬p 3 ↛ 4 ¬q 5 ... The truth table for p OR q (also written as p ∨ q, Apq, p | ... A B C* | C R 0 0 0 | 0 0 0 1 0 | 0 1 1 0 0 | 0 1 1 1 0 ...
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## Suggested Questions And Answer :

### 55% of graduates find a job in field, probability that 1 of 7 random graduate find a job

wot yu need tu du is: chans 1 student will NOT find a job=45% chans 7 students will NOT find a job=(0.45)^7 =0.0037366945 or 0.0037 thus chans 1 av 7 students will find job=1-0.0037 =0.996263 or 99.6%
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### how do you get an equation of a tangent line to a parabola given the point of tangency

You need to find the derivative dy/dx of the curve and substitute x and y, if necessary, to find the slope at a particular given point. The slope is the tangent at the point. The standard linear equation y=ax+b can then be constructed, because the tangent will be a. To find b, the y intercept, substitute the coords of the given point into x and y and you'll be able to find b. That gives you the line. For example, let y=px^2+qx+r be the equation of a parabola, where p, q and r are constants, then dy/dx=2px+q. Let's say you're given the point on the curve where x=A, then dy/dx=2pA+q. This is a number because p, q and A are all known values (you'll be told what they are). This is your slope for the tangent, so a=2pA+q, and y=(2pA+q)x+b is the equation of the tangent line. To find b we first find out what y is on the curve when x=A. We know y=pA^2+qA+r. To make it easier to read, call this y coord B. So we substitute (A,B) in our linear equation to find b=B-A(2pA+q). It's much easier when you have actual numbers rather than symbols. In the same way you can find the equation of the perpendicular because its slope is related to the slope of the tangent. It's -1/(2pA+q), the negative reciprocal of the tangent. I hope this helps your understanding.
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### Twenty percent of a certain number is 16. find the number. And show your work.

Twenty percent of a certain number is 16. find the number. And show your work. 1. Twenty percent of a certain number is 16. Find the number. 2.The length of a rectangle is 5 meters more than twice the width. The perimeter of the rectangle is 46 meters. Find the length and width of the rectangle. 3. One angle of a triangle is 47 degrees of the other two angles, one of them is 3 degress less than three times the other angle. Find the measures of the two angles. Please show all work. 1. 0.2x = 16    5 * 0.2x = 16 * 5    x = 80 2. L = 2W + 5    P = 2L + 2W    46 = 2(2W + 5) + 2W    46 = 4W + 10 + 2W    46 = 6W + 10    36 = 6W    6 = W    L = 2W + 5    L = 2(6) + 5    L = 12 + 5    L = 17    P = 2L + 2W    P = 2(17) + 2(6)    P = 34 + 12    P = 46 3. a = 47    b + c = 180 - 47 = 133    b = 3c - 3    (3c - 3) + c = 133    4c - 3 = 133    4c = 136    c = 34    b = 3c - 3    b = 3(34) - 3    b = 102 - 3    b = 99    a + b + c = 180    47 + 99 + 34 = 180
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### Please help

You need to see through the problem and apply whatever is necessary to reduce the number of variables till eventually there's only one to find. Remember a simple fact: if you have two variables you always need two independent equations to find them; for three variables, three equations; four variables, four equations. You use the multiplication property if it helps you to eliminate a variable between two equations. Take some examples: x+y=10, x-y=3; simply adding these two equations will eliminate y and help you find x. 2x=13 so x=6.5 and y=3.5; 2x+y=10, x-2y=10; we could double one equation or the other so as to match the coefficients of one or other of the variables; but since it's easier to add two equations rather than to subtract them, where we have a minus in one equation and a plus in the other, we would prefer to use the multiplier for the relevant variable. So we double the first equation and add to the second: 4x+2y=20 PLUS x-2y=10: 5x=30, making x=6 and y=-2. The last pair of equations could have been written: 2x+y=x-2y=10, but it's still two equations. There is no one way to solve equations, and you can save yourself a lot of stress by not assuming you have to remember a rigid technique or formula as “The Way to do it”. You'll find mathematics is more fun when you intelligently try different methods and use your natural creativity to guide you. And here's another interesting thing. Those questions about finding a missing number in a series can be tackled in many cases as solving simultaneous equations. You only need n equations to find n variables, and a series can be seen as a set of terms generated by a function y=f(x) for different values of x (the position in the series), giving different values of y (the terms in the series). f(x) is a polynomial of the type ax^n+bx^(n-1)+cx^(n-2)+... If there are four given terms n=3 and the variables are a, b, c and d; if there are 3 given terms, n=2 and the variables are a, b and c. There is always a solution, we just have to work through and find it!
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### During Lunch, the cafeteria sells 12 sandwiches, 10 soups, and 9 salads

During Lunch, the cafeteria sells 12 sandwiches, 10 soups, and 9 salads Six students had sandwiches and soup, 4 students had sandwiches and salads, ... 5 students had soup and salad, and ... 2 students had all three. How many students ate lunch? We need a Venn diagram. This is the best I can do. Each rectangle holds the number of students who bought a particular item. The intersections of the rectangles (regions) will hold the number of students who bought the items covered by multiple rectangles.                                 Soup                     +----------------------+                     |               a             |                     |                              |                     |          +----------------------+                     |          |             f     |         |           +------|--------|-------+        |          |           |         |   d     |    g   |         |          |           |         |          |         |         |          | Sandwich           |        +--------|-------|-------+         |           |                    |         |                   | Salad |                    |   e    |         c         |           |       b           |         |                   |           |                   +-------|---------------+           |                             |          +-----------------------+ Let's define each of the regions. a is the number of students who bought only soup b is the number of students who bought only a salad c is the number of students who bought only a sandwich d is the number of students who bought only soup and a salad e is the number of students who bought only a salad and a sandwich f is the number of students who bought only soup and a sandwich g is the number of students who bought all three items Now, we show the aggregates of the regions representing the students who bought various combinations. cefg = 12 sandwiches (all four of those regions         are contained within the sandwich rectangle) adfg = 10 bought soup bdeg =  9 bought a salad fg      =  6 bought soup and a sandwich ge     =  4 bought a salad and a sandwich dg     =  5 bought soup and a salad g       =  2 bought all three items We were given the number of students in region g, those who bought all three items. We can insert the number 2 into that region.                                 Soup                     +----------------------+                     |               a             |                     |                              |                     |          +----------------------+                     |          |             f     |         |           +------|--------|-------+        |          |           |         |   d     |    g   |         |          |           |         |          |   2    |         |          | Sandwich           |        +--------|-------|-------+         |           |                    |         |                   | Salad |                    |   e    |         c         |           |       b           |         |                   |           |                   +-------|---------------+           |                             |          +-----------------------+ (I am restricted to 8000 characters, so I am forced to eliminate the remaining diagrams. I hope you can follow along without them.) By subtracting the smaller regions from larger combinations of regions, we can begin to determine the number of students in each region. Subracting g (2) from dg (5), we find that d represents 3 students who bought ONLY soup and a salad. Subracting g (2) from ge (4), we find that e represents 2 students who bought ONLY a salad and a sandwich. Subracting g (2) from fg (6), we find that f represents 4 students who bought ONLY soup and a sandwich. Subracting d (3), e (2) and g (2) from bdeg (9), we find that b represents 2 students who bought ONLY a salad. Subracting d (3), f (4) and g (2) from adfg (10), we find that a represents 1 student who bought ONLY soup. Finally, subracting e (2), f (4) and g (2) from cefg (12), we find that c represents 4 students who bought ONLY a sandwich. To find out how many students ate lunch (or at least bought lunch), we add the numbers from all of the regions in the Venn diagram. a + b + c + d + e + f + g = 1 + 2 + 4 + 3 + 2 + 4 + 2 a + b + c + d + e + f + g = 18 18 students ate lunch
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### help! Find all x-intercepts and y-intercepts, if no intercepts exist say so

Problem: help! Find all x-intercepts and y-intercepts, if no intercepts exist say so f(x)=5x-8     f(x)=3x ,  f(x)=5 i dont get how to solve these and find the intercepts.... In place of f(x), substitute y. To find the y-intercept, set x to zero. To find the x-intercept, set y to zero. f(x) = 5x - 8 y = 5x - 8 y = 5(0) - 8 y = -8, the y-intercept y = 5x - 8 0 = 5x - 8 5x = 8 x = 8/5, the x-intercept f(x) = 3x y = 3x y = 3(0) y = 0, the y-intercept y = 3x 0 = 3x 3x = 0 x = 0/3 = 0, the x-intercept f(x) = 5 y = 5, the y-intercept There is no x term, so there is no x-intercept This represents a horizontal line five units above the x-axis.
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### find the center of mass of the solid of uniform density biunded by the graphs of the equations

x^2+y^2=a^2 is a cylinder with its circular cross-section in the x-y plane; the cylinder is sliced at an angle by the plane z=cy, forming a wedge. y=0 is the x-z plane and z=0 is the x-y plane; these planes and the conditions that y and z are positive impose further constraints on the shape. The cylinder is sliced in half by the x-z plane so the cross-section will be a semicircle and only the upper half is required. The semicircular end rests on the z=0 plane, and doesn't extend beyond the vertex of the wedge. The length of the wedge is ac, because the slope of the angle of the wedge is tan^-1(c) to the y axis and the radius of the wedge is a. Viewed from the side, along the x axis, the wedge is a right-angled triangle with sides a (height), ac (length) and hypotenuse a√(1+c^2). In everyday terms, the wedge looks like the end of a lipstick that has been sliced across. Going back to the side view of the wedge, we can see that the flat end has a height a. As we move along the wedge towards the point the height changes and shrinks to zero when we reach the vertex. The length at a point P(y,z) is a-y. This height is the distance of a chord to the circumference, starting at height a. The first task is to work out the area of a segment. We do this by subtracting the area of a triangle from the area of a sector, given the distance of the chord from the circumference. This distance is a-y, so the distance from the centre of the circle is, conveniently, y. y/a=cosø where ø is half the angle subtended by the arc of the sector. So ø=cos^-1(y/a) and the area of the sector is given by 2ø/2π=ø/π=A/πa^2, or A=a^2ø (ø is measured in radians). The area of the isosceles triangle formed by joining the radial ends of the arc of the sector is B=aysinø=y√(a^2-y^2), and the area of the segment is A-B=a^2cos^-1(y/a)-y√(a^2-y^2). (Note that when y=0, this expression becomes πa^2/2, the area of the semicircle.) The volume of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz where infinitesimal dz is the thickness of the segment. The mass of the segment is directly proportional to its volume because the density is uniform, so the volume is sufficient to represent the mass. The centre of gravity (COG) of the wedge can be found by summing moments about a point G(x,y,z) which is the COG. This sum has to be zero. Since a semicircle is symmetrical in the x-y plane about the y axis, we know the x coord of G must be zero. We need G(0,g,h) where g is the height along the z axis of the COG and h the z position. We need to find the COG of a segment first. The length of the chord of a segment is 2√(a^2-y^2). Earlier we discovered this when we were finding out the areas of the sector and isosceles triangle. Consider just two dimensions. The length of the chord is one dimension and if we create a rectangle from this length and an infinitesimal width dy, the area will be 2√(a^2-y^2)dy. If we think of the rectangle as having mass, its mass is proportional to its area, so area is a sufficient measure for mass. Let's call the COG of the segment point C(x,y)=(0,q), then the moment of the rectangle about the COG=C(0,q) is mass times distance from C=2(q-y)√(a^2-y^2)dy. If we sum the rectangles over the interval y to a we have 2∫((q-y)√(a^2-y^2)dy)=0 for y≤y≤a. The interval looks strange, but it means that y is just a general value, which is determined when we return to the wedge. What we are trying to do is to find q relative to y. We need to evaluate the definite integral, so we split it: 2q∫(√(a^2-y^2)dy)-2∫(y√(a^2-y^2)dy. Call these (a) and (b). To evaluate (a), let y=asinu, then dy=acosudu; √(a^2-y^2)=acosu and the integral becomes 2a^2q∫(cos^2(u)du). Since cos(2u)=2cos^2(u)-1, cos^2(u)=½(cos(2u)+1), so we have a^2q∫((cos(2u)+1)du)=a^2q(sin(2u)/2+u). Since sin(2u)=2sinucosu, this becomes a^2q(sinucosu+u)=a^2q(y/a * √(1-(y/a)^2)+sin^-1(y/a))=qy√(a^2-y^2)+a^2qsin^-1(y/a). Now we apply the limits for y: πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a). To evaluate (b), let u=a^2-y^2, then du=-2ydy and the integral is +∫√udu = (2/3)u^(3/2)=(2/3)(a^2-y^2)^(3/2). Applying the limits we have: -(2/3)(a^2-y^2)^(3/2). (a)+(b)=πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a)-(2/3)(a^2-y^2)^(3/2)=0. Multiply through by 6: 3πa^2q-6qy√(a^2-y^2)+6a^2qsin^-1(y/a)-4(a^2-y^2)^(3/2)=0. From this q=4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a)). Fearsome though this looks, when y=0 and z=0, q=4a^3/(3πa^2)=4a/(3π), which is in fact the COG of a semicircle. When y=a and z=ac, q=0. Strictly speaking, we should write q(y) instead of just q, showing q to be a dependent variable rather than a constant. To find the COG of the wedge, we need to find the moments of the COGs of the segments. For a segment distance y above and cy along the z axis, we have the above expression for q. So q(y) is the y coord of the COG of the segment and cy is the z coord. The x coord is 0. The mass of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz, as we saw earlier, and this mass can be considered to be concentrated or focussed at the COG of the segment (0,q(y),cy). The COG of the wedge is at G(0,g,h) so the distance of the mass of the segment from G is g-q(y) and h-cy in the y and z directions. The y-moment is (g-4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a))(a^2cos^-1(y/a)-y√(a^2-y^2))dz and the z-moment is (h-cy)(a^2cos^-1(y/a)-y√(a^2-y^2))dz. The sums of these individual components are zero. These lead to rather complicated integrals, requiring considerably more space than is available to work out, even if I knew how to do so!
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### Find the equation? find the equation of a line through the point (-1,9) an perpendicular to 2x-4y =7

Find the equation? find the equation of a line through the point (-1,9) an perpendicular to 2x-4y =7 how do you find the equation of a line? We need the given equation in slope/intercept form. 2x - 4y = 7 -4y = -2x + 7 y = (-2/-4)x + 7/-2 y = 1/2 x - 3/5 A line perpendicular to this line has a slope that is the negative reciprocal of this line's slope. The negative reciprocal of 1/2 is -2/1, that is, -2. The new equation is y = -2x + b.  We need to calculate the y-intercept, b, for this new line. y = -2x + b b = y + 2x Plug in the x and y values of the given point. b = y + 2x b = 9 + 2(-1) b = 9 - 2 b = 7 Now, we can give the complete equation: y = -2x + 7
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### how do you find the equation of a line given a point and what it is perpendicular too?

how do you find the equation of a line given a point and what it is perpendicular too? i have a point and an equation in slope intercept form and i have to find the equation thats perpendicular to the eqution given . How do i do that? The equation of a line that is perpendicular to the given line has a slope that is the negative inverse of the given slope. E.G., if the given equation were y = 1/2 x + 7, the new equation would be y = -2x + b, understanding that b is the y-intercept of the new equation. Now, take the x and y values of the point you have been given and plug them into the new equation to find the value of b. If the point were (7, -10), you would have -10 = -2(7) + b. Solve that and you would find that b = 4. So, in this case, the equation would be y = -2x + 4. Follow the same steps, using the values from your equation and point. You will end up with the correct equation for your problem.
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