4 Hobbits in the Fellowship
4 fingers on the hand
The bonds should be listed in order from strongest to weakest: H-F, H-Cl, H-Br, H-I, H.
Do you mean y=3x or y=x3? I'll assume it's the latter. The first method to solving this is the easiest, the chain rule. Multiply the coefficient by the value of the exponent and reduce the exponent by 1. (f(x)=nAxn-1) You get y=3x2 Therefore, f(4)=3(42) f(4)=48 The longer method of solving this goes as follows: Your tangent formula is f'(x)=limh->0(f(x+h)-f(x))/h We know that f(x)=x3 so f(x+h)=(x+h)3 When we put this in the formula we have: f'(x)=limh->0((x+h)3-x3)/h f'(x)=limh->0((x+h)(x+h)(x+h)-x3)/h f'(x)=limh->0((x2+2hx+h2)(x+h)-x3)/h f'(x)=limh->0(x3+x2h+2x2h+2xh2+xh2+h3-x3)/h f'(x)=limh->0(3x2+3xh+h2) f'(x)=3x2+3x(0)+(02) f'(x)=3x2 And from there again we sub in 4 for x. f'(4)=3(42) f'(4)=48
4 he's a jolly good fellow
4 finger and a thumb on a hand
The two American doll codes are 1 4 7 j - 4 f a e - h 7 a g , the next code is r 3 9 4 - e f t h - 4 4 a g
Half & Huff.
+4,+1,+2,-4,-1,-2
The letter that have parallel line in it are H,I,F
f = 54
There are 16 possible outcomes, each of which is equally likely. So each has a probability of 1/16 HHHH: X = H*T = 4*0 = 0 HHHT: X = H*T = 3*1 = 3 HHTH: X = H*T = 3*1 = 3 HTHH: X = H*T = 3*1 = 3 THHH: X = H*T = 3*1 = 3 HHTT: X = H*T = 2*2 = 4 HTHT: X = H*T = 2*2 = 4 HTTH: X = H*T = 2*2 = 4 THHT: X = H*T = 2*2 = 4 THTH: X = H*T = 2*2 = 4 TTHH: X = H*T = 2*2 = 4 HTTT: X = H*T = 1*3 = 3 THTT: X = H*T = 1*3 = 3 TTHT: X = H*T = 1*3 = 3 TTTH: X = H*T = 1*3 = 3 TTTT: X = H*T = 0*4 = 0 So, the probability distribution function of X is f(X = 0) = 1/16 f(X = 1) = 4/16 = 1/4 f(X = 2) = 6/16 = 3/8 f(X = 3) = 4/16 = 1/4 f(X = 4 = 1/16 and f(X = x) = 0 for all other x