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 137vs248...
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.
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.
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.
AAAAAAAAAB
AAAAAAAAAB
AAAABAAABB
AAAABAAABB
AAABBAABBB
AAABBAABBB
AABBBABBBB
AABBBABBBB
ABBBBBBBBB
ABBBBBBBBB
2. For portion A, move the whole...
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');
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...
Strictly speaking no, the R should be in series with L and C too. Then again, an R which is in parallel to the LC pair could be converted (with network transformation theorems) to being in series to it.
So I guess there is really no straight answer to this question.
I have copied-and-pasted the script on the Matlab (R11.1) command prompt and a figure appears. Nothing appears in the command line though.
How did you try to run the script?
3. Deactivate all independent sources. Insert a voltage source Vs between AB and measure the resultant current I; then RTh = Vs/I. Alternatively, insert a current source Is between AB and the resultant voltage V is measured; RTh = V/Is. You may choose any nonzero values for Vs and Is, e.g. Vs =...
Before you even start to use Matlab (or any other computing tools) to solve the problem, you should at least have a plan in mind about the approach/method to use.
You asked about how Matlab can be used to solve the problem; well, tell us something about your plan (a pseudocode or the sort)...
Since the resistors are connected in parallel to the voltage source, they do not affect the rest of the circuit voltage-wise (unless of course their resistances are set to 0).
You have missed out the (v3-v1)/1k current when you took the KCL on supernode 3-4. Apart from that, due the to 20V source, node 4 and the reference node will form a supernode, hence you'd have a (super-)supernode 3-4-ref.
Try again.
No, v4 is not 8V, but either one of (v4-v3) or (v3-v4) is and it is not difficult to figure out which one is.
The nodal equation for node 4 is incorrect. Please check that against your first post. You have above 3 equations with 3 unknowns (v2, v3, v4). It becomes thus a mathematical problem...