help me with these equations?
825795
OK,as the first one is already done,I am doing the rest for you\n4(2x-6)=x+2 =>8x-24=x+2  =>8x-x=2+24  =>7x=26  =>x=26/3=8 2/3  (3) 2(x+5)+1=3(x-1)+4   =>2x+10+1=3x-3+4 =>2x-3x=-3+4-10-1  => -x=-4  =>x=1  (5)x+6=4(4x-1)   =>x+6=16x-4  =>x-16x=-4-6  => -15x=-10  =>x=-10/-15=2/3 [after reducing]
589798
y' = 2x - 2\n2x - 2 = 0\n2x = 2\nx = 1
752959
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]
869887
If you have 2 equations, just solve for one variable for one of the equations. Take that equation and plug it into the same variable for the other equation. You will solve for that variable, and plug in that number to any of the 2 equations to solve for the other variable.
424681
substituiting your values you get (the values you gave are not very clear, can't make out whether positive or negative.\nAssuming all values are positive...)\ny-2 = 2(x-5)\ny-2 = 2x -10\n2x -y - 8 =0  is of the form ax + by + c =0
839940
1. 2C6H6 + 15O2 --> 12CO2 + 6HOH\n\n2. C12H22O11 + 12O2 --> 12CO2 + 11HOH\n\n3. CH3COCH3 + 4O2 --> 3CO2 + 3HOH\n\n4. C2H5OC2H5 + 6O2 --> 4CO2 + 5HOH\n\n5. C4H9OH + 6O2 --> 4CO2 + 5HOH
967876
1) 2a^3 - 13a^2b + 11ab^2 + 14b^3\n\n2) (2a-4b) (4a^2+8ab+16b^2)\n\n3)  (b-8)(b-1)=0\n      b=8   or    b=1\n\n4) 4m^3 - 26m^2 - 14m=0\n2m(2m^2 - 13m - 7)=0\n2m=0  or  2m^2 - 13m -7=0\nm=0  or    (2m+1)(m-7)=0\n                  m=-0.5   or  m=7\n\n so the final answer is m=0 and m=-0.5 and m=7
221530
Add them together and divide by 5.\n\nWow. I could seriously have you people do my homework for me. Every night. I'll type out the problems and see how you all do. :P JP
919864
1.  ax + xb\n2.  21x^3 + 24x*2y\n3.  24x^2 + 31x + 10\n4.  24x^2 - 8x - 6\n5.  40x^3 + 66x^2 + 38x + 6\n6.  x^2 - 1\n7.  x^3 - 8
289077
add all the equations, you get\n2(r+s+t) = 56\nso,\nr+s+t = 56/2 = 28\ndone!!
974863
1) .96 + .72(m - 1) <= 6\n\n2) 5f + 10t = 100, f + t = 12\n\n3) r = 2s - 4, r + s = 20\n\nI assume you do not need these equations to be solved for you.
789977
1.  15\n2.   a.  y=(2x-3)^2 + 4\n      b. (3/2;4)
502184
5H2O2(aq)+2KMnO4(aq)+\n3H2SO4(aq)-->\n2MnSO4(aq)+K2SO4(aq)+\n8H2O(l)+5O2(g)
388285
equate the two equations: \nsin 2x = pi/4 \n2x = inverse sine pi/4 + k2pi or (pi - inverse sine pi/4) + k2pi\n2x = 0.90 + k2pi or 2.24 + k2pi\nx = 0.45 + kpi or 1.12 + kpi\nx = 0.45, 1.12, 3.59, 4.26 ( k = 0,1)
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.
539827
(a) 2 NaClO3 -> 2 NaCl + 3 O2\n\n(b) 3 MnCl2 + 2 Al -> 3 Mn + 2 AlCl3\n\n(c) C4H10 + 13/2 O2 -> 5 H20 + 4 CO2 \n                  or\n(c) 2 C4H10 + 13 O2 -> 10 H20 + 8 CO2\n\nBalancing is quite simple for these simple equations. You just have to see the left hand side pick up an element, see how many atoms of that element are there in it and multiply the R.H.S compound containing that element with a number which will be equal to no. of atoms in the l.h.s. Then again repeat the procedure by picking another element.....see it's quite difficult to explain you with writing aneways i hope it helps u :)
279724
come on. all these questions are almost EXACTLY the same. \np1v1/t1 = p2v2/t2 = nr = constant.\n(0.998atm)(2.1L) / (36+273K) = (0.9atm)v2 / (28+273K)\nand then solve for v2. and u get\nv2 = 2.268L\n\n3. p1v1 = p2v2 = nr/t = constant\n(1.08atm)(14.57mL) = (1.43atm)v2\nv2 = 11.0039mL.