Write the equation using the function notation.?
818236
m= -2\ny=mx+b\n5= -2(2)+b\n5= -4+b\nb=9\ny= -2x+9
101134
A function always outputs a value. Suppose the function \nf(x) = 10x+20, what this means is that this function outputs different values for different input values of variable 'x'. For example:\nf(10) = 10X10+20 = 120\nf(20) = 10X20+20 = 220\ni.e. we substitute different values into x and get different outputs. \nThe important point is that every function outputs a value.\n\nComing to Equations. Equations define the relationship between one or more variables.\nSingle variable equation:\nx2 - 10x = 0\nMulti vairable equation:\n2x+3y = 10\nAn equation doesnt output a value. We can solve the equation and find out the values of the variables. The equation would be valid only for certain values of the variables.\n\nHope the difference is clear.
150092
t=to\n_____\n√1-v2/c2
889816
1/rt3 *1/rt2-1/rt3\n[1/rt3](1/rt2-1)\n[1/rt3](1-rt2/rt2)\nrationalising\n[(1-rt2)*rt6]/6
1044611
F = Gm1m2/ r^2\nF = ma  \na = v/t\nv = d/t\na =d/t^2 \nF = m d/t^2  = Gm1m2/ d^2\nm  = (m1 + m2)\nF = (m1 + m2) d/t^2 = Gm1m2 /d^2\nd/m2t^2 + d/m1t^2 = G/d^2\nd/m2 + d/m1 = Gt^2/d^2\nSQRT{ (1/G) d^2 (d/m2 +d/m1)} = t
695915
Break this down into three functions depending on the value of x:\n\nWhen x<-1 you have f(x) = (-x-1)(2-x) because (x+1) and (x-2) are negative.  This is the same as f(x) = (x+1)(x-2)\n\nWhen -1<=x<2 you have f(x) = (x+1)(2-x).\nWhen x>=2 then f(x) = (x+1)(x-2).\n\nSo f(x) = (x+1)(x-2) = x^2-x-2 outside -1<=x<2,\nand f(x) = -x^2+x+2 in -1<=x<2.\n\nf'(x) = 2x-1 outside -1<=x<2, and f'(x) = 1-2x in -1<=x<2.
246299
h= (C*I/A+T/k)*z*log(.7*y)
180190
let f(x) = (x^2) sin(x+2)\nthe derivitive of f(x) [ f'(x) = 2x sin(x+2) + (x^2) cos(x+2)]\n\nThis is done by taking the product rule {g(x)h(x)}' = g'(x)h(x) + g(x)h'(x)( the derivitive of the first times the second plus the derivitive of the second times the first) \nwhere g(x) = x^2 and h(x) = sin(x+2)
618071
simplify to f(x)=(sqrt(x-2)(x-3))/(x-4)(x+5)\nthen sub in 1/x\nto get (sqrt(1/x -2)(1/x -3))/(1/x -4)(1/x +5)
491386
Just substitute x+1 into f(x)\n\nf(x+1) = [ 6(x+1)^3 ] - [ 3(x+1)^2 ] + 4(x+1) - 9
741999
whats the equation or formula u need help with?
728450
G=(T+Mf)/M \ncross-multiply\nGM = T + Mf\nGM - Mf = T\nfactor out the M\nM(G - f) = T\nM = T(G-f)\n\nANSWER\nM = T(G-f)
870571
F(x)=2[sec(x)^2] * [tan(x)] / [xcos(x)+2sin(x)]\n\nremember:\nsec(x)=1/cos(x); tan(x)=[sin(x)/cos(x)]\n\nWe rewrite the function and get:\nF(x) = [2/cos(x)^2] * [sin(x)/cos(x)] / [xcos(x)+2sin(x)] <=>\n\nF(x) = [2 /cos(x)^2] * [sin(x) /{ cos(x) * [xcos(x)+2sin(x)]} <=>\n\nF(x)= 2sin(x) / [ x*cos(x)^4 + 2sin(x)cos(x) ] =>\n\nLimF(x)= 2sin(x)/[x+2sin(x)] \nx→0\nBecause all terms with cos(x)→1 IF x→0 =>\n\nLimF(x)=2sin(x)/[x+2sin(x)] = 2sin(x)/x \nx→0\nbecause 2sin(x)→0 IF x→0 =>\n\nLim F(x)=2 because sin(x)/x →1 IF x→0 \nx→0\n=============================\nRegard from Denmark
663777
SOrry, I can't help you. If I had the formulas I could do it, but there is no way I remember that off the top of my head! Good luck
840688
Not sure what you were doing. I would rewrite f(x) in this form:\n\nf(x) = (2x-1)^(-1/5)\n\nThen you're dealing with a u^n situation, so\n\nf'(x) = (-1/5)(2x-1)^(-6/5)*(2)\n\nf'(x) = (-2/5)(2x-1)^(-6/5) \n\nThat's equivalent to what you had, but a little simpler.\n\nLet the underboss know that this is a "site" and I did answer your question.
776321
the question is unclear.\nDo you want an explicit formula for Y[n or do you just want to write out the convolution]
844394
If you have Microsoft Office Discs, then load the Equation Editor. It will allow you to write all of these.