graph in the interval [0,5] indentify the end values and convert from parametric to rectangular form?
966138
Rectangular form first\n\nx=3t-2\nx+2=3t\n(x+2)/3=t\ny=t+2\ny=((x+2)/3) + 2\ny=x/3 +2/3 + 2\ny=x/3 + 2 2/3\n\nIn the interval x=0 to x=5, this should be a straight line with x-intercept at (0, 2 2/3) and slope 1/3.  at x=5, you get y=5/3 + 2 2/3 = 5/3 + 8/3 = 13/3 = 4 1/3.
158569
Here is a tutorial...you can change any of the parameters that you mentioned and look at the graph as it is made and changed...really cool if you ask me.\nhttp://www.analyzemath.com/logfunction/logfunction.html
608711
i think it's a parabola.. im not sure. u should try plotting it on a graph.. substitute  the x and y values and see what form comes out..
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.
6838
∫xSec(x²)²dx = .5Sin(x²)/Cos(x²)
535506
If on a TI-83:\n1) Put the equation into Y=\n2) Click 2nd+Window (Tblset) to select the start of the table and the intervals\n3) Click 2nd+Graph (Table)
900462
It really depends on how the equation of the path/curve is originally specified.\n\nIf it's specified in terms of x and y, an easy parameterization is:\n\nx = t\ny = f(x) = f(t)\n\nBut there are many other ways to do it.
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
372526
some calculators can do this\n\nmathematica software will do it
495873
Standard plot of what?  What is the analysis parameter you are interested in measuring?
971807
(0,0)\n(3,0)\ndraw the graphs and then find them out!\nif u can't, then mail me!
638682
Domain ~ x values\nRange ~ y values\n\nYou have to work the function and figure them out.
102977
symetric designs, ratios from diagram to real material
100190
plot these coordinates:\n(0,0),(22.5,-1),(45,0),(67.5,1),(360,0), etc., etc.\n\nconnect them together in a curve and you'll have it.\n\nsine graphs are a series of curved lines. they start at zero, go up to one, go down to zero, go down to -1, and go up to 0 again.\n\nthe negative in front of the sine means that the graph is reflected along the x-axis. so that the 0, 1, 0, -1, 0 sequence is switched to 0, -1, 0, 1, 0.\n\nthe 4 in front of the x means that the normal period for the graph to make a complete curve is divided by for. so instead of the graph following the sequence every 90 degrees, it is done every 22.5 degrees.
769460
I think that you want x element of [-6,4)\n\nYou wanted closed on the left and open on the right, is that true?\n\nJust remember that for intervals:\nopen   ( )\nclosed [ ]\n\nAnd you can match them up: [-6,4)
429405
go to your graphing calculator and hit the Y= button, then input your problem, it looks like an arch, that goes extremly high.. it starts on -3
402424
f+12<18  Subtract 12 from each side.\nf < 6\nThus, the solution is all numbers less than 6.\n(-infinity,6) is called interval notation.\nSet builder notation:\n{f|f<6}\nSet builder notation is always enclosed in braces {}, and always starts out with f|\nThe vertical bar is read "such that"\nThus, {f|f<6} is read "f such that f is less than 6."\nAlso, the graph is a number line with a parenthesis, ), at 6 and then shaded to the left of 6.