Ok, please help.. I'm stuck on this one. how do work it? What is the inteval for the solution?
961056
5z-(3+z)>=2(3z+1)\n5z-3-z>=6z+2\n4z-3>= 6z+2\n-3-2>=6z-4z\n-5>=2z\n-2.5>=z\nz<=-2.5
101802
I don't think there is a unique solution for it. You have more unknown variables than the equations.
778162
Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
796334
Consider the forces and torque in this problem. Setting up so that the x-axis is positioned parallel to the incline\n\nSFx = m*a = m*g*sin(q) - Fs\n\nwhere m is the mass of the cylinder, a is the acceleration, g is the acceleration due to gravity, q is the angle of the incline and Fs is the statis friction force. For the torque\n\nST = I*a/r = r*Fs\n\nso that we have for the statice friction force\n\nFs = I*a/r^2.\n\nPlugging this into the first equation\n\nm*a = m*g*sin(q) - I*a/r^2\n\nand solving for the acceleration\n\na = m*g*sin(q)/(m + I/r^2)\n\nNow, finding the time to travel a length L can be done in several ways. The first, and perhaps simplest is\n\nL = 0.5*a*t^2\n\nso that\n\nt = sqrt(2*L/a)\n\nAlso, you can consider how many revolutions of the cylinder is takes to achieve the same distance. This mean dividing the distance L by the circumference of the cylinder. Multiplying the result by 2*pi gives the angular travel so that the time can be determined by\n\n2*pi*L/(2*pi*r) = 0.5*alpha*t^2\n\nwhere alpha is the angular acceleration and is equal to a/r. \n\nConservation of energy can be employed as well\n\nm*g*h = 0.5*m*v^2 + I*w^2\n\nwhere h (= L*sin(q)) is the height, and w (= v/r) is the angular speed. That is to say that the potential energy at the start is equal to the kinetic energy after traveling through the distance L along the incline. Solving for v yields\n\nv = sqrt(2*m*g*L*sin(q)/(m + I/r^2))\n\nwhich is equal to the acceleration times the time traveled. From this \n\nt = sqrt(2*m*g*L*sin(q)/(m + I/r^2))/a
937830
T=m1*-9.8+m1*a\nT=m2*a--m2*-9.8\n\nset them equal and solve for a\n\ntrust me, i had a test on this today
421380
from first equation put the value of b in second and then form a second degree differential equation in "a"...good luck!
505936
What exactly are you supposed to do with this equation?  Are you suppose to graph it, find solutions, factor or what?  You need to be more specific.
614837
I already posted this solution (or how to get it).  Do a search of yahoo answers....
908083
Tried to do it algebraically, that was a booger.\n\nI plugged that into mathematica and it gave me an error normally generated when the inverse of a function is needed to represent a solution and the solution variable occurs in more than one argument of that function.\n\nThat one might not have a solution?
903948
find the slopes of both lines and if there answers are opposite recipricals, then they are parrallel\n\nheres adn ex:\nslope = 2 and -1/2\n\nin other workds, if you multiply the two slopes, u should get -1
774792
Let the length of side "b" be x ft.\nTherefore side a=2x and side c=2x+3\nAccording to the problem,\nx+2x+2x+3=93\n=>5x+3=93\n=>5x=93-3=90\n=>x=90/5=18\nTherefore.side a=2X18=36 and side c=36+3=39\nAnswer a=36ft. b=18 ft and c=39 ft
569320
v=vo + at\nv=0 + (9.8)(8)\nv=78.4\n\ny=1/2(vo + v)t\ny=1/2(78.4)(8)\ny=1/2(627.2)\ny=313.6m\n\nB
1082511
I think your question have a typo.  There are three variables and I am not sure which one are you trying to solve for.
233014
U R RIGHTT..HOW DID YA GET IT..PLZ CAN YA EXPLAIN..I'VE BEEN TRYING THIS PROBLEM FROM A LONGTIME.EVEN DE CALCULATOR SAYS ITZ "4"..I KNOW DE CALCULATORS WRONG TOO..PLZZ DO EXPLAIN IT TO ME ASAP
43253
Depending on what you need to solve, you need two angles and a side, or two sides and an angle. Hit me with an E-Mail if you are stuck with something.\n\nkeptcompany@yahoo.com
810099
Use distribution to simplify...\n\n6f + 27 = -7 + f\n\nNow isolate the variable and solve.
685042
A=2LW+2LH+2WH\n\nA-2LW=2LW-2LW+2LH+2WH\n\nA-2LW=2LH/2L+2WH\n\nA-2LW\n------------ = 2WH\n    2L\n\nA-2LW ~~~(1)  ~~~~~~(1)\n------------  -------- = 2WH   --------\n    2L~~~~~ (2W) ~~~~~(2W)\n\nA - 2LW\n------------- = H\n    2L(2W)