how do you make an linear equasion?
540264
a linear equation is just x+y=a number, or y=x+a number.  It could also have a coefficient in front of x or y in either equation.  To graph it, either use the slope-intercept form, where you take y=x+a number, the coefficient in front of x is the slope, and the number after x is the y-intercept, so, you plot (0, the number) and use the slope to find another point, then draw your line.  The second is to find two or three points that fit the equation and plot them.  Just substitute a number in for x two or three times, find y each time, and plot those numbers, then draw a line through them.
1166133
Multiply both sides by R+r\nI(R+r) = E\nDistribute the I\nIR + Rr = E\nSubtract IR form each side\nRr = E - IR\nDivide both sides by R\nr = (E-IR)/R
778162
Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
859533
When the whole equation is greater than zero, graph it like normal. When it's less than zero, graph it's negative. For example:\n\ny = /x + 1/\n\nWithout the absolute value signs, this would be positive when x > -1. So, for x >= -1, graph it like normal. For x < -1, graph -x - 1. You should get a v-shape, with only positive values.\n\nIf you have values added on outside the absolute value, then you just raise and lower the graph accordingly.\n\nAnd if you're still confused... try setting the expression equal to zero, and find x. Then, since it's linear, just plug in a value both less than and greater than this x to get two more points. Graph the three points, and connect with lines to get a v-shape.\n\nHope this helps!
564627
change the equation to slope intercept form\nsolve the equations for y\nx+2y=3\n2y= 3- x\ny= -x/2 + 3/2\nDo the same thing for the other formula\nthe form is\ny = mx + b\nAfter that just plug in numbers for x to get y
366751
make x the width, and solve the two equations (the one given, and the one assumed about the rectangle where l*w=A)
1137447
2 linear equations\n\n X  + 3Y - 1 = 0\n4X + 7Y + 5 = 0\n\nFROM (1)   4X + 12Y - 4 = 0\nFROM (2)   4X +  7Y + 5 = 0\n\nSUBTRACT\n                          5Y - 9 = 0\n\n                          5Y = 9\n\n                            Y = 1.8\n \n  FROM (1)      X + 5.4 - 1 = 0\n\n                        X + 4.4 =   0\n\n                        X =  -4.4
398921
To make an augmented matrix, just take your coefficients and put them in a matrix, then draw a vertical line and place the constants on the right side, like this:\n\n| -2 0 9 | 2 |\n| 0 -2 9 | 22 |\n| 3 5 -2 | 10 |\n\nFill in 0's for any variables that aren't there (i.e. there's no y in the first equation, so there's a 0 in its place in the matrix).
621346
y=mx+b (this is known as the standard form of linear equations)\n\nm=slope\nx=x values\ny=y values\nb= y-intercept\n\nso the answer for you question is:\n\nslope = -1\ny-intercept = -3\n\nEXPLANATION:\n\ny = mx + b\ny = -x-3\n\nm= -1   and    b = -3\n \nslope = m    and   y-intercept = b\n\nthus, the slope is -1 and the y-intercept is -3\n\n\nNote: the slope is rise/run (rise over run) -- to make a graph use the y-intercept and the slope -- once you have the slope and the y-intercept, making this graph should be easy because it's a linear graph -- but if you still don't get it ask your teacher\n\n\nthus, the slope is not a-1 -----there is no "a"
554491
I LOVE U\n\nFirst line:\n= I [ (L+N)² - N(2L+N) ] O / L\n= I [ L² + 2LN + N² ) - ( 2LN + N² ) ] O / L\n= I [ L² ] O / L\n= I L O\n\nSecond line:\n= [(v+r)(v-r)][E/V(U³/U²)]  (Note: that is U^3/U^2)\n= [v² - r²] [E/V(U)]\n\nNow we know that we "r" (are) nothing without each other, so r = 0\n\n= [V²] [E/V(U)]\n= V E U\n\nAll together our equation is I LOVE U
1023953
If f is a linear function where f(x)=y, then it's of the form y = mx + b.\n\nYou know\n8 = 4m + b\n4 = 1m + b\nSubtract the second equation from the first.\n4 = 3m\n4/3 = m\nSubstitute this back into the second equation:\n4 = 4/3 + b\n8/3 = b\nSo y = 4/3x + 8/3.  Since for f(12) x=12,\ny = 4/3 (12) + 8/3\n   = 48/3 + 8/3\n   = 56/3, or 18 2/3
1140302
Here is an interactive page that discusses one way to do it:\n\nhttp://dilc.upd.edu.ph/loDeploy/math/math10a.html\n\nFor large numbers, bequalmings technique using logs is an easier way to go.
18487
A linear equation looks like a line when it is graphed.  The standard form of the equation is y=mx+b where m is the slope, and b is the value of y at the point where the line crosses the y-axis.\n\nA non-linear equation does not look like a line.  It may look like a curve.  A non-linear equation could possibly be a parabola, a circle, a wave, a hyperbola, etc... but not a line!
393240
Try here:\n\nhttp://www.extremepixels.com/forum/showthread.php?p=5144&mode=linear
364905
Use Charles' Law\n\nLet V initial = V\nLet T initial = T\n\nLet V final = V'\nLet T final = T'\n\n\nV/T = V'/T'\n\nSolve for V' you might have to convert from mL to L though.
978722
You can't, algebraically.  You have two equations and three variables.  Sorry.  You need one more equation, such as a = b.
268125
just investigate for example ODE y"=-y and then add to the left part increasingly small non-linearity +c*y^2 and find solution of the first order of c - you will see how different modes interact with each other.