v = 4/3 pi r^3 dv = 4pi r^2 dr So, plug in your values: r=1.4 dr = 0.003
f'(x) = 5x^1/4 + 15x^1/2 + 9 now plug in 0 and 16 for x
That would be (7(a+h)^3 - 7a^3)/h Just expand the binomial, and the x^3 term will vanish, leaving evrything else with a power of h. Divide by h. Done.
1/x = 1/6 + 1/12 x = 4
what's the problem? Pick a value for x and use the relevant function. f(-1) = x^2+6x = -5 f(0) = x = 0
3500 <= n <= 4499 Let n = abcd If a=4, b+c+d=12 and h=2d Assuming c is not zero, c=4, since b≠c But 4848 is no good If a=3, c=3, so n = 3633 Note that if c=0, 4008 works as well
f(x) must be a monic cubic (why?), so f(x) = x^3+ax^2+bx+c f(x+1) = (x+1)^3 + a(x+1)^2 + b(x+1) + c = x^3 + (3+a)x^2 + (3+2a+b)x + (1+a+b+c) So, f(x)+(x+1)^3 = 2x^3 + (3+a)x^2 + (3+b)x + (1+c) 3+a = 2(3+a) 3+b = 2(3+2a+b) 1+c = 2(1+a+b+c) a = -3 b = 9 c = -13 f(x) = x^3 - 3x^2...
10 looks good to me. Of the 8 possible outcomes, 7 have at least one head 3 of those have exactly 2 heads.
I think a visit to http://davidmlane.com/hyperstat/z_table.html will help a lot here.
100(1+.08)^10 = 215.89
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