Guide :

finding length on a globe giving a parallel(with latitude and meridian)

latitude, meridians

Research, Knowledge and Information :


Finding your location throughout the world!


Finding your location throughout the world! ... • Latitude lines are parallel, ... meridian give locations west, ...
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Latitude - Wikipedia


The length of the meridian arc between two given latitudes is given by replacing the limits of the integral ... of a parallel of latitude ... (length) World Geodetic ...
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Latitude and Longitude (Meridians and Parallels)


Circles parallel to the Equator ... of latitude north or south of the Equator. ... and are at right angles to the Equator. The “Prime Meridian” which ...
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Latitude and Longitude - finding coordinates - Compass


Find latitude and longitude coordinates on ... Notice that each latitude is parallel to ... The longitude line that runs through it is called the Prime Meridian and ...
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latitude and longitude | geography | Britannica.com


latitude and longitude: ... Latitude is a measurement on a globe or map of location north or south ... The length of a degree of arc of latitude is approximately ...
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Earth, Parallels and Meridians, Latitude and Longitude [IGEO ...


Sep 24, 2013 · Earth, Parallels and Meridians, Latitude and ... on any plane parallel and meridian for the equator is ... (latitude 23 ° 27 ' S). Is the parallel further ...

Locating Points on a Globe | manoa.hawaii.edu ...


Locating Points on a Globe . NGSS ... Every point on a parallel of latitude is the same ... all lines of longitude are the same length. Since every meridian must ...
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Latitude and Longitude - My Illinois State


Lines of Latitude and Longitude. ... Two sets of lines encircle the globe: lines of latitude and lines of longitude. The parallel lines of latitude are horizontal, ...
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How To Use Your Globe - World Globes


because they are parallel to each other. Latitude ... a point on the exact opposite side of the globe. In giving a position, latitude ... How To Use Your Globe ...
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DIY: Measuring Latitude and Longitude - OpenLearn - Open ...


DIY: Measuring Latitude and Longitude ... and unlike latitude lines, they divide the globe into segments like those of an ... giving the time at the Prime Meridian.
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Suggested Questions And Answer :


finding length on a globe giving a parallel(with latitude and meridian)


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how do I calculate a perimeter

Two-sided object? A 4-sided object would be more sensible and you can find the area of such an object, but a 2-sided object would have an undefinable area. If you mean we have the length of two sides, then it does make sense and the answer would be that the object has two sides of the same length and another two sides of the same length, like a football field. We also have to assume that the object is a rectangle otherwise we can't define its area unless we know the height. If it's not a rectangle but a parallelogram (a skewed rectangle) you could find the area but not the perimeter. So it's a rectangle. Two sides of length 32 metres gives us 64m plus two of length 66m gives 64 + 132 metres which is 196 metres. The area is just the product of the two side lengths = 32*66 = 2,112 square metres.
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what are the properties of the quadrilateral whose vertices are the center of the squares?

The quadrilateral so formed is a rhombus (a parallelogram in which all the sides are of equal length). The interior angles of the rhombus are 90+x and 90-x where x is the angle of the parallelogram and where 90+x are the measures of opposite angles of the rhombus. The adjacent angles add up to 180, and 90+x+90-x=180. Call the parallelogram ABCD and let angle BAD be x. Because opposite side and opposite angles of a parallelogram are equal, we only need to consider two sides. Let AB=a and AD=b be two adjacent sides. Let the bisectors of these sides be points P and Q respectively. AP=PB=a/2 and AQ=QD=b/2. Outside ABCD, draw XP, length a/2, perpendicular to AB, and YQ, length b/2, perpendicular to AD. X and Y are the centres of the squares on AB and AD. One side of the new quadrilateral is defined by the line XY. So let's find its length. AX=a/sqrt(2) and AY=b/sqrt(2), because they are hypotenuses of triangles AXP and AYQ. XY=a^2/2+b^2/2-abcosXAY (cosine rule). Angle XAY=XAP+BAD+QAY=45+x+45=90+x. cosXAY=cos(90+x)=-sinx. So XY=a^2/2+b^2/2+absinx. Let's find the length of an adjacent side of the new quadrilateral. This time we have angle CDA=180-x. CD is bisected at R and WR is the perpendicular on to CD, and WR=RD=RC=a/2 because DC is parallel and equal in length to AB. Angle YDW=360-(RDW+180-x+QDY)=360-(45+180-x+45)=90+x, and cos(90+x)=-sinx. WY=DW^2+DY^2-2DWDYcosYDW=a^2/2+b^2/2+absinx. Therefore WY=XY. Since there is symmetry in the geometry the other two sides will be the same. And the opposite interior angles are equal. Therefore we have rhombus. When you draw the diagram you may find that XY is exterior to the parallelogram, and WY is interior. Nevertheless, since sin(180-x)=sinx, the lengths of the sides will still be equal. There are various other geometries that may apply, but you will find that you always get a combination of two angles of 45 degrees and the angle of the parallelogram (skew), but sinx will always be the result (for example, cos(90-x)=sinx, cos(45-x+45)=sin(x), cos(180-(90+x))=sinx, etc.).
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The perimeter of a rectangle swimming pool is 130 yards. Three times the length is equal to 10 times the width. Find the length and the width of the pool. Find the area of the pool.

The perimeter of a rectangle swimming pool is 130 yards. Three times the length is equal to 10 times the width. Find the length and the width of the pool. Find the area of the pool. Perimeter of a rectangle is 2*the length plus 2*the width P = 2L + 2W and P = 130, so L + W = 65 Three times the length is equal to 10 times the width, giving 3L = 10W Substituting into the above for L = 65 – W, 3(65 – W) = 10W 195 – 3W = 10W 195 = 13W W = 195/13 = 15 W = 15 yards, And L = (10/3)W = (10/3)*15 = 50 L = 50 yards
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how many 1/2" candies will fit in a 41/2" x 4" bowl?

Assuming that 4" is the height of the bowl, we can conclude that it is ellipsoidal, like a rugby football, with the shorter axis (x axis) being 4.5" in diameter and its longer axis (y axis) needing to be determined from the given figures. Two domes are cut off each end of the ellipsoid leaving a 3.5" diameter circular top and a 3" diameter circular base. The general equation of the ellipse that will be used to model the bowl is x^2/2.25^2+y^2/a^2 (2.25" being the "radius" of the x axis, and a the radius of the y axis). The ellipse has its centre at the x-y origin. To find a we work out in terms of a where the base of the top dome has a width of 3.5" and where the base of the bottom dome has a width of 3". Putting x=1.75 (half of 3.5) in the equation of the ellipse, we get y=sqrt(a^2(1-1.75^2/2.25^2))=4asqrt(2)/9. Similarly for the base dome: y=asqrt(1-1.5^2/2.25^2)=-asqrt(5)/3, negative because it's below the centre or origin. The difference between these two y values is the height of the bowl, 4", so 4asqrt(2)/9+asqrt(5)/3=4, and a=36/(4sqrt(2)+3sqrt(5))=2.9114 approx.  The volume of the bowl is found by considering thin discs of radius x and thickness dy so that we can use the integral of (pi)x^2dy (the volume of a disc) between the y limits imposed by the heights of the two domes. x^2=5.0625(1-y^2/a^2). Because a is a complicated expression, we'll just use the symbol for it. We can also write 5.0625 as 81/16. The integrand becomes (81(pi)/16)(1-y^2/a^2)dy, with limits y=-asqrt(5)/3 and 4asqrt(2)/9. The integrand is simply the sum of the volumes of the discs between the bases of the two domes. Integration gives us (81(pi)/16)(y-y^3/(3a^2)) between the limits, which I calculated to be 53.39 cu in approx. How many candies fit into this volume? We know the length of the candies is 1/2", but we don't know any other dimensions, so we don't know the volume of each candy. Also, the alignment of candies will improve the number of candies fitting into the bowl, but if this cannot be arranged, we have to assume they will be randomly orientated. We could make the assumption that they are cone-shaped with height 0.5" and base diameter 0.25" (radius 0.125" or 1/8") giving them a volume of (1/3)(pi)0.125^2*0.5=0.0082 cu in each approximately. So divide this into 53.39 and we get about 6,526, with very little space between the candies. A more realistic figure can be obtained by finding out how many candies fill a cubic inch when tightly packed. Two will fit lengthwise into an inch. Think of the candies as cuboids 1/4" square and 1/2" long (volume=1/4*1/4*1/2=1/32 cu in). That gives you 32 in a cubic inch, over 1,700 in a bowl. The cuboid is a sort of container that will hold just one candy and allow it some lengthwise freedom of movement. At the other extreme think of the candies as 1/2" cubes (volume=1/8 cu in) then only 8 will fit into a cubic inch and only about 427 will fit into the bowl. But the candies have greater freedom of movement in a cubic container. Let's say you can get 100 into one cubic inch packed tightly, then you would get about 5,340 in the bowl. Get a small box and pack in as many candies as you can. Measure the volume of the box (length*width*height). If N is the number of candies, then each has an average effective volume of (box volume)/N. Use this number to divide into the volume of the bowl.  
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How to find c and y so that each quadrilateral is a parallelogram

The opposite internal angles of a parallelogram are equal, and adjacent angles are supplementary, but which of the given angles are opposite and which are adjacent? We know that all angles must be positive, so 7x-11>0 and 5y-9>0 so x>11/7 or 1.57 and y>1.8. We also know that all the angles of a parallelogram can be determined if just one is known, because of the relationships. This means that if we take any pair of angles we know that they are (a) equal or (b) supplementary. Take the first pair: 5x+29 and 5y-9: if (a), 5x+29=5y-9, so 5(y-x)=38 and y=(38+5x)/5; or if (b), 5(y+x)=160, or y+x=32 and y=32-x. Also, the remaining pair 3y+15 and 7x-11: if (a), 7x-3y=26, y=(7x-26)/3; or if (b), 7x+3y=176 and y=(176-7x)/3. We can see that if (a) is applied to the first pair at least one of x or y will contain a fraction. If (a) is applied to the second pair, which can be written 2x-8+(x-2)/3, x must be x=5, 8, 11, ..., 3n+2 for y to be an integer (where n is a positive integer) and y is 3, 10, 17, ..., and if (b), which can be written 58-2x-(x-2)/3, x must be 3n+2 for y to be an integer (where n is an integer 0 Read More: ...

finding perimeter and area

Perimeter is straightforward: P = 18 + 6 + 11 + 9 = 44m To calculate the area of an irregular quadrilateral you need the side lengths and the diagonal lengths. Since the diagonal lengths are not provided I'll give the formula. For: a, b, c and d are lengths of sides s is the semiperimeter, s = (a + b + c + d)/2 p and q are the diagonals A = SQR[(s - a)(s - b)(s - c)(s - d) - 1/4(ac + bd + pq)(ac + bd - pq)]
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How do I find an inverse variation equation for a bridge-length experiment

The number of pennies, N, required to break the bridge length L is inversely proportional to the length. This is the basis for an equation. There may be a constant factor involved, so we call K this constant and we write N=p/L+K. Can we find a constant of proportionality to make this equation true for the values we have? Let's see. 24=p/4+K; 16=p/6+K; 13=p/8+K; 11=p/9+K; 9=p/10+K. We can eliminate K by subtracting one equation from another. We need to remember that the equation may only be approximate, a guide, if you like. Let's take the first two equations: 24-16=p(1/4-1/6), so 8=p/12, and p=96. Now take the last two: 11-9=p(1/9-1/10), so 2=p/90 making p=180. Let's take the first and last equations: 24-9=p(1/4-1/10), so 15=3p/20 and p=100. Take the second and third equations: 16-13=p(1/6-1/8), so 3=p/24 and p=72. So far p hasn't been particularly consistent having possible values between 72 and 180. Note also that if p=96, K=0; but if p=72, K=4. And K could be -1. Then there's something else: the breaking weight of pennies always involves a whole number of pennies, because they can't be split; so if N pennies is not quite enough to break the bridge, N+1 would certainly do it, so we can't expect N to be nicely rounded to a whole number. What we can say, though, is that if N is large it's more likely to be more accurate than if N is small. The first two equations show this. The third equation would be consistent with the first two if N=12 rather than 13. So we could go for N=96/L as the equation which fits the first two sets of figures and nearly fits the third. Let's use N=96/L and see what we get for (N,L). (24,4), (16,6), (12,8), (11,9), (10,10) and (24,4), (16,6), (13,7), (11,9), (9,11), depending on whether we start with N or L to find L or N. As a model, then, the formula gives reasonable results.
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how do you find height in trapeziod without knowing the area. legs are 29 and bases are 15 and 57

Question: how do you find height in trapeziod without knowing the area. legs are 29 and bases are 15 and 57 . Since the two legs are identical in length, then this trapezoid is symmetrical about a central axis. Let a be the length of the top base. Let b be the length of the bottom base. Let c be the length of each leg. Label the trapezoid A, B, C, D. Because of symmetrry, we can drop a perpindicular from each end of the top base, from A and B, down to the bottom base, intersecting it at E and F. This will give two congruent triangles at each side of the trapezoid, with the base of each triangle, DE and FC, having the same size. EF = AB = a DE = FC = (CD - EF)/2 = (b - a)/2 The height of the triangle is h = AE = sqrt(AD^2 - DE^2) h = sqrt(c^2 - (b - a)^2/4)
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Guyssss please help me "Create a problem situation in which only the positive solution is meaningful" what am I going to do? I can't understand it

First, you need to understand the question, right? You need to know what gives you more than one solution. The answer is an equation which contains a polynomial of at least degree 2. That means a quadratic equation or a cubic equation, or more, that has more than one solution. A typical example is a quadratic that has a positive root and a negative root: (x-a)(x+b)=0 or x^2+x(b-a)-ab=0, where a and -b are the roots. Next we need a situation which implies that we need a positive solution only. An example is a geometrical figure like a triangle or rectangle which must, of course, have sides of positive length. Now we have to think of a situation which introduces the square of a number. In the case of geometrical figures this is likely to be the area. Now we have to invent a problem. We start with a solution. Here's an example: A rectangle has dimensions 9 by 4 so its area is 36 (never mind the dimensions at the moment). Then we create an unknown, x. Now we use x to define the dimensions of the rectangle. If we say that the length is 5 more than the width and the area is 36, then the solution is: let x=width then length=x+5, then the area is x(x+5)=36. So x^2+5x=36, or the quadratic equation: x^2+5x-36=0=(x-4)(x+9). The roots are 4 and -9, but we can reject -9. Now we have the basics of the problem, let's make it more interesting. First add dimensions. Let's use feet. Replace the rectangle with a familiar rectangular object: a box, swimming pool, a garden, etc. OK here's the question. A swimming pool's length is 5 feet more than its width. What is its perimeter if the area of the base is 36 square feet? We know how to find the width (4 feet) from the solution above, so we find the length (4+5=9 feet). The perimeter adds an extra to the problem, because it uses the length and the width. Perimeter=2(length+width)=26 feet.
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