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at x=3, f(x)=5, with a slope of 1.1. So, f(x)=mx+b, 5=1.1(3)+b --> b = 5-3.3 = 1.7 f(x) = 1.1x+1.7 at x=3 g(x) = -4 with a slope of 0.7 g(x) = mx+b, -4 = 0.7(3)+b --> b=-4-2.1 = -6.1 g(x) = 0.7x - 6.1 f(x)*g(x) = (1.1x+1.7)(0.7x-6.1) = .77x^2-5.52x-10.37 [f(x)*g(x)]' = 1.54x-5.52 @x=3: 1.54(3)-5.52 = 4.62-5.52 = -0.9

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

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

Assume the lines intersect the parabola at the point (x, x^2 + x). It then follows that: dy/dx = 2x + 1 = (x^2 + x + 3)/(x - 2) 2x^2 - 3x - 2 = x^2 + x + 3 x^2 - 4x - 5 = 0 (x - 5)(x + 1) = 0 x = 5, -1

1 Answers · Science & Mathematics · 10/02/2013

4 + 0 + 4 + 8 + ........ + = 4 + an

**AP**with a = 0, d = 4, n = 199 but remember that you don't return after last post is fixed, so subtract 1/2 of the last term of 198*4 S = 4 + (199/2)(2*0 + 198*4) - 198*2 = 78412m = 78.412 km <-------1 Answers · Science & Mathematics · 10/02/2013

...the length of the corresponding side of BQP. Thus, we know

**AP**is 4 times as long as BP, which means AB...2 Answers · Science & Mathematics · 23/01/2013

... : y^4 + (4p+a) y^3 + (6p^2+3ap+b) y^2 + (4p^3+3ap^2+2bp+c) y + p^4+

**ap**^3+bp^2+cp+d if we take 4p+a=0 or p=-a/4, the coefficient of y^3 becomes...2 Answers · Science & Mathematics · 07/01/2013

a) is not an

**AP**since -6 - 8 ≠ 4 - (-6).3 Answers · Science & Mathematics · 03/01/2013

nth term of a

**AP**is Tn = a + (n -1)*d ; where a = first term, d is common...2 Answers · Science & Mathematics · 06/01/2013

...regardless of whether or no you're good at math. If you plan on taking

**AP**Chemistry, then you're going to be exceptionally good at math. ...3 Answers · Science & Mathematics · 16/12/2012

...3), the diagonal matrix of eigenvalues. Check that P^(-1)

**AP**= D. I hope this helps!1 Answers · Science & Mathematics · 05/12/2012

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