what is the range of the function c(x)=(500+3x)/x ??
469660
Find the inverse.\ny = (500+3x)/x \nSwitch x with y and solve for y\nx = (500+3y)/y  Multiply both sides by y\nxy = 500+3y  Subtract 3y from both sides\nxy - 3y = 500  Factor y from the left side\ny(x-3) = 500  Divide both sides by x-3\ny = 500/(x-3) \nThe domain of the inverse is the same as the range of the original function.\nThe domain of 500/(x-3) is found by setting the denominator equal to zero and solving for x.\nx-3 = 0 means x=3\nThus, the domain of the inverse is \nAll real numbers except 3.\nThus, the range of (500+3x)/x is\nAll real numbers except 3.
1029912
subtract 50 from both sides of the equation and you'll have \n3x=150\ndivide both sides by 3 and you'll have \nx=50
664975
Domain of the function is the value of x and the range is f(x)....\nfor f(-3) the domain is -3 and the range f(-3)=(-3)^2+2(-3)-48=9-6-48=-45  which is the range....\nsimilarly for f(a+1) the domain is a+1 and the range by substituting a+1 in the equation and factorising you get the range as (a+9)(a-5)...
124381
You can use the equation c = fλ.\nNow f = c/λ = 3 x 10*8/600 x 10-9 = 5 x 10*14 Hz
88674
F = 9/5 * C + 32\n\nF = C\n\nSolving, gives F = C = -40
239990
There are no values of x for which the function is less than 0.  The function value is between 0 and 1 when x<=0.\n\nPlot a few values to sketch the graph: x=-2,y=1/9; x=-1,y=1/3; x=0,y=1; x=1,y=3; x=2,y=9
22472
It's well known that lim x->0 sin(x)/x=1\nThen lim x->inf. sin(1/x)/(1/x)=1\nwhich is exactly what you need.
914486
just plug -1.5 in for x and see if it equals 1.5.       \n2 X -1.5 = -3                  -3 does not equal 1.5\nso the answer is false
99752
well, if you needed the "comnversion of C---> F\n\nthan its simply\n\nC = 5/9(f-32)
532934
INT[(1 - x)/x dx]\nINT[x^-1 (1-x) dx]\nINT[x^-1 - 1 dx]\nln|x| - x + C
836547
f(x)=rtx+C\nf(1)=1+C=2\nso C=1\nf(1/4)=rt(1/4)+1=3/2
311134
f(x) = 1/(x+5)(x+2)\n\nso -5 and -2 are restricted from the domain since then we would have the denominator = 0 and the function undefined.
506265
if f''(t) = -100 then \n\nf'(t) = -100t + C1  (we need to solve for C1)\n\nf(t) = -50t^2 + C1t + C2  (need to solve for C2)\n\nPlug in the values given and you'll get C1 and C2\n\n-20 = -100 * 0 + C1\nC1 = -20\n\n500 = -50 * 0^2 + -20 * 0 + C2\nC2 = 500\n\nthus f(t) = -50t^2 -20t + 500
1114168
Ok, here's how i think you do it.\n\nto get the horizontal asymptote, you have to get the limit of the functon when x tends to be infinite, so you would get something like this\n\nlim     (3x-x^2)/(4-x)\nx→∞\n\nthen you substitute ∞ on the x's and you get\n\n∞/∞, that's not defined, so what you use is l'hopital law where you derive both the nominator and the denominator of a fraction, so you will get\n\nlim    3-2x/-1\nx→∞\n\nnow you can substitute ∞ and you get\n\n(3-2*∞)/-1 =∞\n\nand you do the same with -∞ and you get as an answer -∞\n\nwhich means there is no horizontal asymptote.\n\nI hope this helps.
888701
ha! max is not a builtin function.\n\nbut Math.max is!\n\nalert( Math.max( 10,5,21,54,23) )\n\nworks!
1150392
No problem, i just finished a calc3 course in college.\n\nYour problem is the same as 125 * (6x^2 + 5x + 7)^(-1/3)\nYou have to use the chain rule so you get\n125 * (-1/3) * (12x + 5) * (6x^2 +5x + 7)^(-4/3)\n\nthen you can simplify how you like from there
605867
(x^2)/2 - 3(x^(3/2))/(3/2) + c = (x^2)/2 - 2x^(3/2) + c