How do you find the coordinates of a figure?
82197
You must first of all fix a system of coordinates
986326
when x is 0, find y. plot the point (0, <<whatever you got>>).\nwhen y is 0, find x. plot the point (<<whatever you got>>, 0).\nConnect those two points, and extend the line.
426495
It depends on what shape you want to know the area.\n\nSimple shapes have formulas. (such as a circle's area is Pi*r*r)\n\nComplex shapes which have pattern.(you know the equation of lines that form the shape), calculus is used to find the area.\n\nIf a shape is not ordinary. You should draw lines to grid it and count the area square by square.\n\n-Hope this would help-
640114
Ellipse first:\nThe center is at the origin, because you have x^2 and y^2. If there is a translation, you would see something like (x+5)^2, that would be a translation of 5 units LEFT. The "a" value is always the bigger value. If it is under the x, then the ellipse is long the horizontal way, if it is under the y, then long the vertical way. From the center, count "a" units the long way, and "b" units the short way to find the shape of the ellipse. There is a formula for finding the focus points also. It is c^2=a^2-b^2, solve for c, and go that distance from the center along the long axis (or the major axis) of the ellipse.\n\nHyperbola:\nNote that the same translation holds for ellipse, but for the formula you put down, the center is at the origin. If the x^2 term comes first, the hyperbola opens left/right. If the y^2 term comes first, the hyperbola opens up/down. For a hyperbola, the "a" term is ALWAYS first, and "b" is next. Count "a" units left and right (if x^2 term is first) to find the vertices. Also, it is helpful to count "b" units up and down, and label those. Then, I have my students draw a rectangle between the a and the b markings, these are helper lines, and the diagonals of the rectangle are the asymptotes. Use basic slope-intercept to find the equations of the asymptotes. The focus has a formula, it is c^2=a^2+b^2, you can find "c" and label it in the bowl of the hyperbola.\n\nHope all that helped!
365741
I think all you have to do is plug in a few values for C to get a graph.  For example, try 0, 5, 10, 15, 20.  It's continuous so you can connect the dots.\n\n(0,0), (5,87.5), (10,350), (15,787.5), (20,1400)\n\nTotal labor for 20 is 1400.
475336
since you probably know how to get the X,Y coord relative to the form, all you need to do is add the X,Y coordinates of the form itself. that is, just add me.Location.X and me.Location.Y\n\nhere is something i wrote just now. when you click anywhere on the form, it changes the button text to be the x,y coordinates relative to the top left corner of the screen.\n\n    Private Sub Form1_MouseDown(ByVal sender As Object, ByVal e As System.Windows.Forms.MouseEventArgs) Handles MyBase.MouseDown\n        Button1.Text = (e.X + Me.Location.X) & " " & (e.Y + Me.Location.Y)\n    End Sub
131436
you can open your image in mspaint and click on SELECT TOOL now move your mouse where you want to start your map and read the X, Y coordinate reading in status bar.
1989
I assume you are talking about vector analysis.  The graphical method would be drawing the vectors head to tail and to scale. and then drawing in the resultant and using your ruler and protractor to measure the displacement and angle.
934709
Plot 3 points, connect them with lines. you get a triangle\n\nList most acute (sharp) angle 1st, most obtuse (blunt) angle last, and remaining one in the middle.\n\nIt looks like C will be the most obtuse
583383
if i understood correctly and b and -2 are the x and y coordinates i get x= 1/2
525749
The numbers you have are two points on a coordinate grid( the one with the x axis and the y axis. If you have some graph paper you could find the points. The first number in each set is the x-number and the second is the y number. You are trying to find the distance between those points. If you drew the dots on graph paper you could make a triangle out of them. One side is the x side and it is 3 units long(13 -10).  One side of the triangle is the y side and its 4 units long(20-16). Connecting those two points makes the third side or hypotenuse of the triangle. \nSo we use the Pythagorean theorem to find the distance.\nSo its 3squared + 4 squared =9 + 16 = 25.We must take the square root of 25 for the final answer, which is 5. Hope this helps.
271355
u could go to www.google.com or www.search.yahoo.com and enter figure scating and find out
373083
Well, I'm not going to answer the question for you, but I'll tell you how to do it.\n\nWhen you're graphing the function, you set up a table with f(x) and x as the two columns, then solve for x for any set of numbers, like -3, -2, -1, 0, 1, 2, 3, and that's probably more than enough. Graph those on a piece of graph paper (f(x)=y, and then x). \n\nNow, to find the INVERSE, you simple switch the X and the Y. Instead of solving for x, you solve for f(x). So, your problem would be written x= -2/3y + 5. An easier way to do it, take the table you made before, and switch the columns, so that the solutions for x are underneath f(x) and the numbers you picked for f(x) are underneath the x column. Graph those coordinates. When you look at the graph, the inverse should be a mirror reflection of the first one, with  the graph of y=x as the line of symmetry. \n\nThat's how you find the inverse of any problem. Sorry if anything's unclear, if you need a better visual, I'm sure you could google up this exact method.
979117
From a table -- see which y-value corresponds to x=0\nFrom a graph -- see where the graph crosses the y-axis\nFrom an equation -- get the equation in y=mx+b form; the y-interecept is b
778162
Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
1096150
To determine if it is a parallelogram:\nIf a quadrilateral has one pair of opposite sides which are parallel and equal in length, then the figure is a parallelogram.\n\nTo determine if it is a rectangle:\nIf a parallelogram has two pairs of opposite sides which are parallel and equal in length, then the figure is a rectangle.\n\nGraph the points and determine the distance between the points, or, in your case, two points have an X of 0 and two have X of 3\n\nBetween the (0, 2) and (0, 6), there is a distance of 4\nBetween the (3, -2) and (3,2) there is a distance of 4\n\nBetween the (0,6) and (3, 2) there is a distance of 3 horizontally.\nBetween the (0,2) and (3, -2) there is a distance of 3 horizontally.\n\nYour figure has two pairs of opposite sides that are equal in length and parallel.  You have a rectangle.
950608
get some graph paper. plug in numbers for x and calculate its equivalent y value. do that 5 times, and you should have your curved line. If not, use higher or lower numbers than you used the first time ( I like starting with 0) Then  draw a line perpendicular to the most bent part (aka axis of symmetry, aka cut the line in two so both sides of the new line look like mirror images), find 2 points on that cutting line, and there you go.