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**ap**^(bx + c) - d = 0**ap**^(bx + c) = d p^(bx + c) = d/a log (base p) of (d/a) = bx + c bx = (log (base p) of (d/a)) - c x = ((log (base p) of (d/a)) - c) / b2 Answers · Science & Mathematics · 27/10/2011

**Ap**calculus or**ap**statistics? Assuming that you have met all the prerequisites, both classes should be doable, but you might want to spend more time on math if you're going to take those classes.2 Answers · Science & Mathematics · 22/07/2012

lol I don't know how you are in any

**AP**courses if you can't even spell calculus or equivalent. But, anyway take calculus, it will help you a lot more for prerequisites.6 Answers · Science & Mathematics · 23/09/2011

...9,5,1] - [4,3,1] = [5,2,0] and .... B = [9,5,1] - 2[4,3,1] = [1,-1,-1]

**AP**= √(16 + 0 + 9) = √25 = 5 BP = √(0 + 9 + 16) = √25 = 52 Answers · Science & Mathematics · 24/04/2007

cotΘ=x/100 x=100cotΘ dx/dt=100(-csc^2Θ)dΘ/dt dx/dt=(-100/sin^2Θ)dΘ/dt dΘ/dt= (-sin^2Θ/100)dx/dt sinΘ=100/200=1/2 dΘ/dt=(-1/2)/100*8 dΘ/dt=-1/25 rad/sec Θ is decreasing at a rate of -1/25...

1 Answers · Science & Mathematics · 23/09/2012

If you are able to really understand mathematics well, then sure, give Calculus a shot. Calculus is A LOT different from your normal and easy math classes. For Calculus, you have use a lot of thinking and...

4 Answers · Science & Mathematics · 08/11/2014

y'(x) = 3x^2 + 2x - 1 = 0 at the tangents solve for the two values of x, (x = -1 and x = 1/3) plug them into y(x) and then subtract the differences

3 Answers · Science & Mathematics · 11/02/2013

(i)

**AP**+PB=20**AP**/PB=5/3 Therefore, 3AP=...throughout, 5PB+3PB=60 8PB=60 PB=60/8=15/2=7.5**AP**=5/3*PB=5/3*15/2=25/2=12.5. (ii) second question (2k-7a)/(3k...3 Answers · Science & Mathematics · 27/10/2011

Differentiate the function to get... x'(t) = 3/π * π/3 * sin(πt/3) = sin(πt/3) Gr

**ap**h that function, and we have.... →Decreasing at 3 < t < 4.5 →Increasing at 3 < t < 4.5 Good luck!2 Answers · Science & Mathematics · 13/11/2011

∫ 100 t² sin(√t) dt = pull the constant out: 100 ∫ t² sin(√t) dt = let (√t) = y t = y² dt = 2y dy substituting, you have: 100 ∫ t² sin(√t) dt = 100 ∫ (y²)² siny 2y dy = 100 (2) ∫ y^4 y siny dy = 200 ∫ y^5 siny dy = let: y^5 = u → 5y^4 dy...

1 Answers · Science & Mathematics · 19/05/2009

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