Particular equation and particular condition help?
914071
dy/dx +4y = 6\n\nSo dy/dx = 6 - 4y\n\nso dx/dy = 1/(6 - 4y)\n\nThus x = -¼ ln(6 - 4y) + c\nSo ln(6 - 4y) = -4x + k Where k = 4c\nThus 6 - 4y = e^(-4x + k) = Ae^-4x where A = e^k\nSo 4y = 6 - Ae^(-4x) \nie y = 1.5 - Be^(-4x) (where B = A/4)\n\nCheck dy/dx = 4Be^(-4x) = 4 (1.5 - y) = 6 - 4y\nNow y(0) = 0\nSo 0 = 1.5 - Be^0\nB = 1.5\n\nSo the solution being sought is:\n\ny = 1.5(1 - e^(-4x))
956102
Here is the equation\n\n(246,92)U ---> (4,2)He + (242,90)Th + 2Gamma
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
859744
(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
1112829
I hope I understood the statement of the problem correctly.\nYou are asking for a set of conditions when an area will be negative.\n\n1. Let f(x)<0 then the area must lie in the positive x domain (x>0).\n2. Let f(x)>0 the area must lie in the negative x domain (x<0).
332255
=IF(OR(AND(A1>=1,A1<=12),\n       AND(A1>=25,A1<=36)),1,\n   IF(OR(AND(A1>=13,A1<=24),\n         AND(A1>=37,A1<=48)),2,""))\n\n\n(make sure it's all on one line with no spaces)
928663
Are you asking for help or an answer?\n\nTake any 2 equations and multiply all of the terms or one or both by numbers that will make one of the terms in each equation have the same coefficient. Now subtract one from the other and the term will disappear. Now take the resulting equation and do the same with it against the third. Keep doing this until you are down to one term and solve for the variable. Now substitute the variable into the equation and solve for the other variables.
689918
use  power rule and product rule\nf'(x)=[1/5]*5*x^4 + 2x(x^2-x-9)+ \n(x^2+9)(2x-1) + 0=\n\nx^4 + 2x^3 - 2x^2 -18x + 2x^3 - x^2 +18x -9 =\n\nx^4 + 4x^3 - 3x^2 -9
243814
although the equation is not complete and therefore the full answer can not be given. i believe this equation will need to be differentiated using the product of a product rule.
156051
first let T =X^2\nthan we get a second degree equation\n2T^2-7T-4=0\nso you can solv it easyer
464039
Your equation would be:\n\n= ((1/p) - (1/q))/(1/pq)\n\n= ((q - p)/pq) * pq\n\n= q - p\n\nHope I am clear.
983225
remember the rule nx^n-1\n2x^2-1\n \nSo in this case it is 2x
443281
1.) A - 0.6666666\n2.) C - $12.26\n3.) B - 0.308\n4.) C - 19.87\n5.) A - $.65
794749
This shall work.\nhttp://pubs.logicalexpressions.com/pub0009/LPMArticle.asp?ID=305
1130821
so for the first part, f(x)=x \nthe derivative no matter what is 1 \n\nfor the second part f(x)=1\nthe derivative no matter what is 0 \n\nso you will have combinational solution
613986
=IF(E12=0,0,\nIF(AND(E12>0,E12<4),\n5.9,IF(E12>=4,12.9,\n"negative value")))\n\nAll on one line!\n\nEdit:  The >= (greater than or equal to) is required else the value of exactly 4 would return "negative value"
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.