You've given the answer yourself. d=6 and f=3.
If: df+10f = 3 Then: f(d+10) = 3 And: d = 3/f -10
Z if d is equally to f plus10
60 feet 6 inches equals the distance from the pitcher's mound to the home plate on a baseball diamond
There are 16C6 combinations = 16!/(6!(16-6)!) = 16!/(6!10!) = 8,008 possible combinations. Assuming the 16 numbers are the hexadecimal digits 0-F, they are: {0, 1, 2, 3, 4, 5}, {0, 1, 2, 3, 4, 6}, {0, 1, 2, 3, 4, 7}, ...., {0, 1, 2, 3, 4, E}, {0, 1, 2, 3, 4, F}, {0, 1, 2, 3, 5, 6}, ..., {0, 1, 2, 3, 5, F}, {0, 1, 2, 3, 6, 7}, ...., {0, 1, 2, 3, 6, F}, ...., {0, B, C, D, E, F}, {1, 2, 3, 4, 5}, ..., {9, A, B, C, D, E}, {9, A, B, C, D, F}, {9, A, B, C, E, F}, {9, A, B, D, E, F}, {9, A, C, D, E, F}, {9, B, C, D, E, F}, {A, B, C, D, E, F} I'll let you fill in the missing 7,990 possible combinations.
The original equation f(x) = 6/(x+3) can be rewritten as f(x) = 6(x+3)-1. Now derive the equation according the the power rule and the chain rule: y = 6 (x+3)-1 dy/dx = 6 (-1)(x+3)-2(1)* dy/dx = -6/(x+3)2 * by the chain rule, you must multiply by the derivative of which is simply one. Thus, the derivative of f(x) = 6/(x+3) equals -6/(x+3)2
If: df+10f = 3 Then: f(d+10) = 3 And: d = 3/f -10
Z if d is equally to f plus10
d/dx [f(x) + g(x)] = d/dx [f(x)] + d/dx [g(x)] or f'(x) + g'(x) when x = 3, d/dx [f(x) + g(x)] = f'(3) + g'(3) = 1.1 + 7 = 8.1 d/dx [f(x)*g(x)] = f(x)*d/dx[g(x)] + d/dx[f(x)]*g(x) when x = 3, d/dx [f(x)*g(x)] = f(3)*g'(3) + f'(3)*g(3) = 5*7 + 1.1*(-4) = 35 - 4.4 = 31.1
What does a F and a D- queal out as
I am assuming that k is the constant of proportionality so that f = k*a*t/b2 = 3*4*6/22 = 18
What does a F and a D- queal out as
60 feet 6 inches equals the distance from the pitcher's mound to the home plate on a baseball diamond
Mayberry R-F-D- - 1968 Hair 3-6 was released on: USA: 19 October 1970
Assuming its 100 % efficient, then f * d = f * d, then 2 * 6 = 4 * d so, (2 * 6) / 4 = d, then 3 = d so d = 3 metres
-2
24-8f = 6 -8f = 6-24 -8f = -18 f = 2.25
There are 16C6 combinations = 16!/(6!(16-6)!) = 16!/(6!10!) = 8,008 possible combinations. Assuming the 16 numbers are the hexadecimal digits 0-F, they are: {0, 1, 2, 3, 4, 5}, {0, 1, 2, 3, 4, 6}, {0, 1, 2, 3, 4, 7}, ...., {0, 1, 2, 3, 4, E}, {0, 1, 2, 3, 4, F}, {0, 1, 2, 3, 5, 6}, ..., {0, 1, 2, 3, 5, F}, {0, 1, 2, 3, 6, 7}, ...., {0, 1, 2, 3, 6, F}, ...., {0, B, C, D, E, F}, {1, 2, 3, 4, 5}, ..., {9, A, B, C, D, E}, {9, A, B, C, D, F}, {9, A, B, C, E, F}, {9, A, B, D, E, F}, {9, A, C, D, E, F}, {9, B, C, D, E, F}, {A, B, C, D, E, F} I'll let you fill in the missing 7,990 possible combinations.