Equasion of a straight line?
600555
y = mx + b\n\n"m" is the slope, which is the rise over the run.  y2-y1 / x2/1 = (1-3)/(3-2) = -2/1\nSo, m = -2\n\nPlug in a set of coordinates to find b...\n3 = -2(2) + b\nb = 7\n\nSo, the equation is y = -2x + 7\n\nIn yout case, you want it as Ax+By+C = 0\nThis would be 2x + y - 7 = 0
412542
x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
911982
slope = f'(0),\nf'(x)= e^x 1/(x^2 +3) 2x + e^x ln( x^2 +e)\nf'(0)= e^0 1/3 (2(0)) + e^0 ln(e)\n= 1\n\nnow y=f(0)=e^0 ln( e) =1\nso the eq. of the tangent line is:\n\ny-1= x\ny=x+1\n\n\nfor question 2:\ne^{sin(0.001)} is very similar to e^0.001 which in turn is approx 1\ns
835719
~this is a horizontal line\nand the equation is x= -5
1146591
You can't really answer that question the way it was asked.  If you know a value of x or y, you can find the other value though.\n\nFor example, say you want to find the x-intercept (where y=0)\n\ny = 3x + 2\n0 = 3x + 2\n-2 = 3x\n-2/3 = x\n\nThere you go, you can do this for any point you know is on the line.
599765
When graphed, a linear equation will form a straight line while a non linear equation will not be a straight line
695930
<1=2(<2+<3)\n<1=2(30+<3)\n<1=180-(<2+<3)\nso equating\n60+2<3=180-30-<3\n3<3=90\n<3=30*\n<1=120*
97125
use the distance formula\nsqrt(3^2+(-sqr3)^2)=?
787261
Let the 2 points be (x1,y1)  and  (x2,y2)\n\neqn of line s given by\n\n(y-y1)/(y2-y1) = (x-x1)/(x2-x1)\n\nwhere  (y2-y1)/(x2-x1) is the slope of line
318213
eh? you using simultaneous equations with a straight line equation?\n\ny = mx + c is a straight line equation, where m is the gradient & c is where the line cuts the y axis.\n\nThere for lets say y = -2x + 8:\ny = mx + c\nCuts x-axis Put y = 0\n-2x + 8 = 0\n-2x = -8\nx = -8/-2\nx = 4\n\nHope that was correct :| lol
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
904469
Let u = e^x + 1.  Then du= ...\n\nHope that helps.
925629
polyeig(-5,-2,4,2) \n\nwill solve the equasion.  To learn about interpelations type:\n\nhelp interp1
610972
considering both y values are -2 for A and C....\n\nwe know that a straight line exists between these two points, which means that it HAS to be 1 of the 4 sides.\n\ngiven that, the coordinates or K and L HAVE to be (3, infinity (EXCEPT -2)) and (-4, infinity (EXCEPT -2))
725560
The  y coordinate is the same in both sets of coordinates the equation for this line is line y=-2. \n\nThis line is perfectly horizontal => the slope is 0.
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2