Recent content by doodle

You are absolutely correct. :) Btw, here's my take on the christmas tree problem... Let the blue, red and white balls be labelled A, B, C, respectively. Weigh A1+B1 against A2+C1. If A1+B1 == A2+C1, weigh A1 versus A2. If A1 > A2, then the heavier balls are A1, B2 and C1; otherwise, it is...

I think I am able to get it in 4 weighings (but not less). The initial steps are as follows: Assuming that the coins are labelled 1 to 8, then the initial weighing is 12vs34. If 12=34, then the second weighing is 156vs278. Otherwise (whether 12>34 or 12<34), the second weighing is...

Oops, it should have been: Remove the path (say of length x) from the garden, we have the area equation (55)(40) - x = \sqrt{x^2 - 40^2} (40) and that gives x = 66.67 yards.[/color]

Answer is hidden. Remove the path (say of length x) from the garden, we have the area equation (55)(40) - x = 2 \sqrt{x^2 - 40^2} (40) and that gives x = 48.202 yards.[/color]

Hmm... how about paradox[/color]?

I am pretty positive the bolded part would not work.

Happy New Year to you too; yes, after much experimentation with 'staircase' cuts and with the initial assumption that the 8x1 is located at the center of the room.

Solution hidden below. As a hint, the 8x1 is placed exactly vertically at the center of the room. :) 1. Cut the 10x10 into 2 portions. [FONT="Courier New"]AAAAAAAAAB AAAAAAAAAB AAAABAAABB AAAABAAABB AAABBAABBB AAABBAABBB AABBBABBBB AABBBABBBB ABBBBBBBBB ABBBBBBBBB 2. For...

Answer is hidden. The probability of getting n heads and (N-n) tails in N toss is P(n) = \binom{N}{n} (0.5)^N and the corresponding multiplier of gamble is X(n) = (1.5)^n (0.6)^{N-n} = (0.6)^N (2.5)^n The mean of this measure is E{X} = \sum_{n=0}^N P(n) X(n) = (0.3)^N \sum_{n=0}^N...

You could do this... xbuff = []; for k = 1:(the number of tones) x = (insert your function here); xbuff = [xbuff; x(:)]; end sound(xbuff,fs); wavwrite(xbuff,fs,bits,'audio.wav');

No mistakes alright.

Express the convolution as an integral. Compare this integral with the problem statement and obtain a function of the form h(t-\lambda). It should not be too difficult then to determine h(t).

I think it is more of the latter, that the 2H inductor is not charged before t = 0. My reason for saying so is as follows: The voltage across the 2H is always 0V before the switch opens. Then Ldi/dt = 0, or in other words, there is no change in the current across this inductor, hence it remains...

Here's a screenshot...