Number of results: 13
PLEASE HELP! Compute the difference quotient: [(f(x)- f(a))/(x-a)].. for f(x)= (-1/x^2)
Saturday, May 22, 2010 at 9:00pm by krista
y^2+4y-x+1=0 what is the vertex of this equation and the axis of symmetry for this parabola?
Saturday, March 5, 2011 at 11:12pm by anonymous
[-1/x^2 +1/a^2] / (x-a) [ x^2 - a^2 ]/ [x^2 a^2 (x-a)] [(x+a)(x-a)] / [x^2 a^2 (x-a)] (x+a)/x^2a^2 note for when you get to differentiation the limit of this as x-->a is 2x/x^4 = 2/x^3
Saturday, May 22, 2010 at 9:00pm by Damon
use the intermediate value theorem to show that f(x) has a zero in the given interval. f(x) = -x^5 -2x^4 + 5x^3 + 4; [-0.9, -0.8] Stuck!
Friday, December 9, 2011 at 4:38pm by james
Change the exponentional expression to an equivalent expression involving a logarithm. 7^c = 0.026
Saturday, December 17, 2011 at 8:39pm by anon
Y^2 + 4Y -X + 1 = 0. X = Y^2 + 4Y + 1 = 0, K = Yv = -b/2a = -4 / 2 = -2. h = Xv = (-2)^2 + 4*-2 + 1 = -3. V(-3,-2). Axis: Y = K = -2.
Saturday, March 5, 2011 at 11:12pm by Henry
Ok feeling kinda dumb because this isn't coming out right. I know it's simple. log6 (x -3) + log6(x -6) - log6(x + 1) = 2
Wednesday, December 21, 2011 at 2:28pm by Anon
f(-.9) = -.367 f(-.8) = +.948 since f is continuous, it must assume all values between -.367 and +.948 on the interval [-.9,-.8]. That includes f(x) = 0.
Friday, December 9, 2011 at 4:38pm by Steve
adding logs is multiplying arguments log 6 [(x-3)(x-6)(x+1)] = 2 6^log 6 [(x-3)(x-6)(x+1)] = 6^2 = 36 x^3 - 8 x^2 + 9 x + 18 = 36 x^3 - 8 x^2 + 9 x - 18 = 0 I do not see any easy factors. Have to solve by iteration.
Wednesday, December 21, 2011 at 2:28pm by Damon
Can someone help me with this prolem please? Use the information given about the angle Ɵ to find the exact value of: sin 2Ɵ. 1) sinƟ= 3/5, 0 < Ɵ < (π/2)
Wednesday, March 4, 2009 at 6:03pm by kayla
c log 7 = log 0.026 c = log 0.026 / log 7
Saturday, December 17, 2011 at 8:39pm by Damon
form a right-angled triangle in the first quadrant with the angle at the origin equal to Ɵ the opposite equal to 3 and the hypotenuse equal to 5 You should recognize the 3,4, 5 right-angled triangle so cosƟ = 4/5 sin2Ɵ = 2sinƟcosƟ = 2(3/5)(4/5) = 24/25...
Wednesday, March 4, 2009 at 6:03pm by Reiny
proof by mathematical induction
subject is PreCalulus. 2^(k+3) = and < (k+3)! i know how to do proving using mathematical induction when its just an equal sign, but I dont understand what to do when its an inequality. thank you!!! Here is what I would do Step 1: check if it is true for k=1 that is: let k=...
Friday, January 5, 2007 at 3:06pm by Cait