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https://fr.mathworks.com/matlabcentral/cody/problems?action=index&controller=problems&sort=&term=group%3A%22Probability+%26+Stats%22
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Cody
# Problems
1 – 30 of 30
Problem Title Likes Solvers Difficulty
#### Problem 42485. Eliminate Outliers Using Interquartile Range
Created by: Monica Roberts
Tags outlier
2
18
#### Problem 42503. Generating random matrix with given probability mass function
Created by: Peng Liu
2
21
#### Problem 42670. If you prick us, do we not bleed?
Created by: James
0
19
#### Problem 648. Cumulative probability of finding an unlikely combination
Created by: Doug Hull
Tags ml101
2
27
#### Problem 3056. Chess probability
Created by: Jean-Marie Sainthillier
Tags chess, elo
3
69
#### Problem 1268. Penny flipping - calculate winning probability (easy)
Created by: Jeremy
3
156
#### Problem 44254. Probability of red tulips (at both ends of a row)
Created by: Noriko Hounoki
2
44
#### Problem 2770. Probability of Choosing a Red Ball
Created by: Kevin
1
46
#### Problem 1267. Calculate the probability that at least two people in a group share the same birthday.
Created by: Jeremy
3
66
#### Problem 1159. Coin Tossing: Probability of Same Heads for N tosses
Created by: Richard Zapor
1
39
#### Problem 2591. Does the coin touch the line?
Created by: Georges
3
30
#### Problem 44277. Given n, create n random numbers such that their standard deviation is also n.
Created by: AMITAVA BISWAS
3
50
Created by: Ken
Tags math
3
43
#### Problem 1272. The almost-birthday problem.
Created by: Claudio Gelmi
1
19
#### Problem 1287. Unique dice configurations
Created by: James
2
25
Created by: Omer
2
19
#### Problem 597. The Birthday Phenomenon
Created by: Ben Ausdenmoore
3
66
Created by: Erin
2
21
#### Problem 226. What are the odds?
Created by: Matt Fig
2
25
#### Problem 43575. Probabilities - More brains than luck
Created by: Massimo Zanetti
1
14
#### Problem 44451. Rate of event occurence: find percentiles of the distribution (for smallish rates)
Created by: David Verrelli
2
10
#### Problem 44676. Mean = Standard Deviation
Created by: Mehmet OZC
4
7
#### Problem 44779. Don't be mean. Be nice!
Created by: James
Tags mean, nchoosek
2
23
#### Problem 44630. Guess the number I'm thinking of (Part 1)
Created by: David Verrelli
12
32
#### Problem 1959. German tank problem
Created by: Alfonso Nieto-Castanon
4
23
#### Problem 331. Compute Area from Fixed Sum Cumulative Probability
Created by: Roger Stafford
1
6
#### Problem 42676. Histogram of histogram
Created by: Peng Liu
1
12
#### Problem 2995. Test Driven Solution - Probability Problem 1
Created by: Jonathan
1
13
#### Problem 3006. Test Driven Solution - Probability Problem 2
Created by: Jonathan
1
11
#### Problem 3009. Test Driven Solution - Probability Problem 3
Created by: Jonathan
2
9
1 – 30 of 30
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# Different results when applying equation
The system shown in the Figure is guided by two blocks in A and B, that move in the fixed grooves. If the speed of A is 2m/s down, determine the speed of B at time θ = 45º
In the textbook they said w = 14.1rad/s (counterclockwise) and v(B)=2m/s
i consider counterclockwise moment positive
if i use the vector BA
$$v_B=v_A+w_{BA} x \vec{r}_{BA} = -2\vec{j} + w_{BA}\vec{k} \times (-0.2cos(45)\vec{i}+0.2sin(45)\vec{j})$$
$$= -0.141w_{BA}\vec{i}+ (-0.1412w_{BA}-2)\vec{j}$$
and since $$-0.1412w_{BA}-2=0\Leftrightarrow w_{BA}=-14.1rad/s$$ (since is negative, the direction is clockwise) and $$v(B)=2i$$.
But if i use the vector AB:
$$v_B = v_A+w_{BA}\times r_{AB}$$
$$= -2\vec{j} + w_{BA}\vec{k} \times (+0.2cos(45)\vec{i}-0.2sin(45)\vec{j})$$
$$= 0.141w(BA)\vec{i}+(-2+0.141w(BA))\vec{j}$$
and since $$-2+0.1412w_{BA}=0 <=> w_{BA}=14.1rad/s$$ (since is positive, the direction is counterclockwise) and $$v(B)=2i$$.
So I got two different results for w(BA) when using vector AB or BA. Which vector should I use when I apply the equation v(B)=v(A)+v(B/A) and why?
Similar case
in this case i use the vectors AB and BD, and i got the opposite signals for the angular speed of the connecting rod BD and for the piston speed P (although the value in module is correct) Should i had used these vectors?
we know from the exercice that w(AB)=209.44rad/s, β=13.95 and we want to discover v(P) and w(BD)
v(B)=v(A)+w(AB) * r(AB)
=0+ w(AB)k * (0.075cos(40)i+0.075sen(40)j)
=0.0575w(AB)j-0.0482w(AB)i
=-10.1i+12.04j
v(D)=v(B)+w(BD)*r(BD)
=v(B)+w(BD)k * (0.2cos(13.95)i-0.2sen(13.95)j)
=(-10.1i+12.04j) + 0.194w(BD)j+0.0482w(BD)i
=(-10.1i+0.0482w(BD))i + (12.04+0.194w(BD))j
v(P)=v(D)=-13.09i
• Yes, it should. – sputnik Feb 13 at 10:26
• This is an interesting question. I started by working out the height of A as $h_A = 0.2\ cos\Theta$. Then I was going to differentiate both sides so I could set $\frac {dh}{dt} = 2$ (m/s). But now I'm stuck. What do we do with $0.2 \frac {d\ cos\Theta}{dt}$? – Transistor Feb 13 at 12:04
• @Transistor Perhaps use the chain rule to write it in terms of the rate of change of $\theta$? Then do the same thing (i.e. differentiate with respect to time and use the chain rule) for the equivalent trigonometric equation for the position of pin joint $\mathsf{B}$. You should find yourself with two linear simultaneous equations in two unknowns, one of those unknowns being the rate of change of $\theta$, the other being the velocity of pin joint $\mathsf{B}$. – Daniel Hatton Feb 13 at 12:40
• @sputnik I strongly suggest you write out in words your definition of $w_{\mathsf{B}\mathsf{A}}$. You may find this makes it very clear that one of your two alternative starting equations is true and the other isn't. – Daniel Hatton Feb 13 at 12:53
• @Transistor I wrote the following steps as a comment, but it didn't render very nicely, so I added another answer. – NMech Feb 13 at 13:06
I think the main thing that is confusing you is the following, is that you think that there is an option for the positive rotation. While in some cases this is true, since you deciced on employing the cross product $$\omega\times r$$ for the calculation, (and since you already have a x-y coordinate system, and defined $$\vec{i}, \vec{j}, \vec{k}$$) the positive rotation is determined by the right hand rule. So the positive direction for rotation is counterclockwise.
The next point is that correct formula is: $$v_B = v_A + \omega\times r_{B|A}$$
where:
• $$v_B = 2\vec{i} = \begin{bmatrix}2\\0\\0\end{bmatrix}$$, (note this only happens because the angle is 45 degrees and there is symmetry in the problem, and in the general case this is an unknown)
• $$v_A = -2\vec{j} = \begin{bmatrix}0\\-2\\0\end{bmatrix}$$
• $$\omega = \omega\vec{k} = \begin{bmatrix}0\\0\\ \omega\end{bmatrix}$$: the angular velocity is the same irrespective of the point. You can see in this problem that whichever point you select (A or B), the body rotates CCW with respect to that point.
• $$r_{B|A} = 0.2(\frac{\sqrt{2}}{2}\vec{i} -\frac{\sqrt{2}}{2}\vec{j})= 0.2 \frac{\sqrt{2}}{2}\begin{bmatrix}1\\-1\\0\end{bmatrix}$$: is the vector connecting to B starting from A
if you use the column notation you get:
$$\begin{bmatrix}2\\0\\0\end{bmatrix} = \begin{bmatrix}0\\-2\\0\end{bmatrix} + \begin{bmatrix}0\\0\\ \omega\end{bmatrix}\times \left( 0.2 \frac{\sqrt{2}}{2}\begin{bmatrix}1\\-1\\0\end{bmatrix}\right)$$
$$\begin{bmatrix}2\\0\\0\end{bmatrix} = \begin{bmatrix}0\\-2\\0\end{bmatrix} + 0.2 \omega\frac{\sqrt{2}}{2} \left(\begin{bmatrix}0\\0\\1 \end{bmatrix}\times \begin{bmatrix}1\\-1\\0\end{bmatrix}\right)$$
$$\begin{bmatrix}2\\0\\0\end{bmatrix} = \begin{bmatrix}0\\-2\\0\end{bmatrix} + 0.2 \omega\frac{\sqrt{2}}{2} \begin{bmatrix}1\\1\\0\end{bmatrix}$$
Both equations for x and y are identical :
$$2= 0.2 \omega\frac{\sqrt{2}}{2} \rightarrow$$ $$\omega = \frac{4}{0.2\sqrt{2}}=\frac{20}{\sqrt{2}}=14.1[rad/s]$$
• i know that counterclockwise is positive and i already correct it in the question. Now, i know that i need to use the vector AB. But if i use the vector BA (which is wrong) why do i get a symetrical angular velocity? – sputnik Feb 13 at 15:28
• you'll get symmetrical angular velocity because $r_{B|A} = - r_{A|B}$ therefore, $\omega\times r_{B|A} = -\omega\times r_{A|B}$. Hope that makes sense. – NMech Feb 13 at 15:30
• Additionally you should note that $r_{AB} = r_{B|A}$. I read the latter, position vector of B with respect to A. Similarly $r_{BA} = r_{A|B}$. That also can introduce confusion. – NMech Feb 13 at 15:34
• i still got an doubt in that part. i updated the post with another situation example since i didnt understand why should i have used AB. My only doubt in the similar example is: if i should have used the vectors AB and BD? – sputnik Feb 13 at 15:47
• in your latest example you can use either $$v_D = v_B+\omega_{BD}\times r_{D|B}=v_B+\omega_{BD}\times r_{BD}$$ OR $$v_B = v_D+\omega_{BD}\times r_{DB}=v_D+\omega_{BD}\times r_{B|D}$$ – NMech Feb 13 at 15:53
@Transistor although this is not a response (because the problem was IMHO the positive direction), the following is rendered much better here, and its a different approach to the same problem.
$$h_A = 0.2\ cos\Theta \rightarrow$$ $$\frac{d h_A}{dt} = 0.2\frac{d \cos\theta}{dt}$$ then $$v_A = -0.2\sin\theta \frac{d \theta}{dt}$$, where
• $$\frac{d \theta}{dt}=\omega$$.
Therefore: $$v_A =- 0.2\sin\theta \omega$$.
Similarly:
$$v_B =0.2\cos\theta \omega$$.
Since $$\theta= 45^o$$, you can use $$v_A=2[m/s]$$ and $$\theta=\pi/4$$ to calculate $$\omega$$ and then afterwards to calculate $$v_B$$.
doing it the hard way:
Let A and B be distance from the corner
A = r*cosϑ B = r*sinϑ
dA/dt = (-r*sinϑ)(dϑ/dt) dB/dt = (r*cosϑ)(dϑ/dt)
(dB/dt) / (dA/dt) = -cotϑ
thus
dB/dt = (dA/dt)(-cotϑ)
The scale dimension (r=0.2m) and the angluar velocity (dϑ/dt) both cancel, and the relationship of velocities is determined purely by the angle.
The negative sign indicates that when A is increasing, B is decreasing and vice versa.
per comment: for the second problem you can still use trigonometry to solve it. Brief attempt below, can't promise that it is correct.
the negative sign indicates that when alpha is increasing, X (i.e. AD) is reducing, i.e. D moves to the left.
• i updated the post with a similar case about the part of the vectors since i didnt understand why should i have used AB – sputnik Feb 13 at 15:42
• @sputnik - I am a little unclear about what you are asking, but it sounds like just a matter of sign convention. if we define AB = B-A then B = A + AB = A - BA – Pete W Feb 13 at 16:06
• is because in the similar example i got the simetrical values from the solution. And i dont understand what i've made wrong – sputnik Feb 13 at 16:11
• @sputnik - i don't know, looks complicated. Try to be really explicit about how you define the sign and direction of each variable. Keep trying until you get it right, it is really important. – Pete W Feb 13 at 16:40
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# Nickels and Dimes = some amount, how many nickels?
1. Dec 28, 2011
### LearninDaMath
1. The problem statement, all variables and given/known data
Bob has 50 coins, all nickels and dimes, worth a total of 4.85. How many nickels does he possess?
2. Relevant equations
n + d = 50
5n + 10d = 485
Multiplying the first equation by 10, we obtain 10n+10d=500
Subtracting the second equation from the third equation, we get 5n=15, so n = 3
3. The attempt at a solution
I have the solution. My question is:
Why do you multiply the first equation by 10 to begin with? Also, why do you then subtract one equation from the other? What is the rationale to it?
2. Dec 28, 2011
### Staff: Mentor
If you multiply both sides of an equation by the same number, you get an equivalent equation (same solution set). Multiplying the first equation by 10 results in 10n + 10d = 500.
If you subtract the same number from both sides of an equation, you also get an equivalent equation. In this case, you subtracted 5n + 10d from 10n + 10d, and 485 from 500, resulting in 5n = 15, or n = 3. Since 5n + 10d = 485, you are actually subtracting the same number from both sides of the equation 10n + 10d = 500.
Another way to do this problem is to solve for either n or d in the first equation, and then substitute into the second equation.
3. Dec 28, 2011
### LearninDaMath
ah i get it, it's like doing a "systems of equations" problem. Thanks! And yea, as soon as I posted the question, my sister recognized that it can also be solved using substition as you mentioned.
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## Intermediate Algebra (12th Edition)
Published by Pearson
# Chapter 1 - Section 1.1 - Linear Equations in One Variable - 1.1 Exercises: 66
x = 9
#### Work Step by Step
1. Given: $\frac{5 - x}{6} + \frac{5}{6}= \frac{x}{54}$ 2. Multiply by LCD (54): 54($\frac{5 - x}{6} + \frac{5}{6}) = (\frac{x}{54}$)54 3. Use distributive property to multiply: 9(5 - x) + 9(5) = x 45 - 9x + 45 = x 4. Add and subtract: -10x = -90 5. Divide: x = 9 6. Check: $\frac{5 - 9}{6} + \frac{5}{6}= \frac{9}{54}$ 1/6 = 1/6, so 9 is a solution of the equation.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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MATH
notes
# For any d p 1 consider the d power map on z p that
• Notes
• 74
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For any d | ( p - 1), consider the d -power map on Z * p that sends α Z * p to α d . The image of this map is the unique subgroup of Z * p of order ( p - 1) /d , and the kernel of this map is the unique subgroup of order d (c.f., Theorem 4.24). This means that the image of the 2-power map is of order ( p - 1) / 2 and must be the same as the kernel of the ( p - 1) / 2-power map. Since the image of the ( p - 1) / 2-power map is of order 2, it must be equal to the subgroup { [ ± 1 mod p ] } . The kernel of the 2-power map is of order 2, and so must also be equal to the subgroup { [ ± 1 mod p ] } . Translating from group-theoretic language to the language of congruences, we have shown: Theorem 9.1 For an odd prime p , the number of quadratic residues a modulo p , with 0 < a < p , is ( p - 1) / 2 . Moreover, if x is a square root of a modulo p , then so is - x , and any square root y of a modulo p satisfies y ≡ ± x (mod p ) . Also, for any integer a 6≡ 0 (mod p ) , we have a ( p - 1) / 2 ± 1 (mod p ) , and moreover, a is a quadratic residue modulo p if and only if a ( p - 1) / 2 1 (mod p ) . Now consider the case where n = p e , where p is an odd prime and e > 1. We also know that Z * p e is a cyclic group of order p e - 1 ( p - 1), and so everything that we said in discussing the case Z * p applies here as well. Thus, for a 6≡ 0 (mod p ), a is a quadratic residue modulo p e if and only if a p e - 1 ( p - 1) / 2 1 (mod p e ). However, we can simplify this a bit. Note that a p e - 1 ( p - 1) / 2 1 (mod p e ) implies a p e - 1 ( p - 1) / 2 1 (mod p ), and by Theorem 4.23 (Fermat’s Little Theorem), this implies a ( p - 1) / 2 1 (mod p ). Conversely, by Theorem 7.7, a ( p - 1) / 2 1 (mod p ) implies a p e - 1 ( p - 1) / 2 1 (mod p e ). Thus, we have shown: Theorem 9.2 For an odd prime p and positive integer e , the number of quadratic residues a modulo p e , with 0 < a < p e , is p e - 1 ( p - 1) / 2 . Moreover, if x is a square root of a modulo p e , then 56
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so is - x , and any square root y of a modulo p e satisfies y ≡ ± x (mod p e ) . Also, for any integer a 6≡ 0 (mod p ) , we have a p e - 1 ( p - 1) / 2 ≡ ± 1 (mod p ) , and moreover, a is a quadratic residue modulo p e iff a p e - 1 ( p - 1) / 2 1 (mod p e ) iff a ( p - 1) / 2 1 (mod p ) iff a is a quadratic residue modulo p . Now consider an arbitary odd positive integer n . Let n = Q r i =1 p e i i be its prime factorization. Recall the group isomorphism implied by the Chinese Remainder Theorem: Z * n = Z * p e 1 1 × · · · × Z * p er r . Now, ( α 1 , . . . , α r ) Z * p e 1 1 × · · · × Z * p er r is a square if and only if there exist β 1 , . . . , β r with β i Z * p e i i and α i = β 2 i for 1 i k , in wich case, we see that the square roots of ( α 1 , . . . , α r ) comprise the 2 r elements ( ± β 1 , . . . , ± β r ). Thus we have: Theorem 9.3 Let n be odd positive integer n with prime factorization n = Q r i =1 p e i i . The number of quadratic residues a modulo n , with 0 < a < n , is φ ( n ) / 2 r . Moreover, if a is a quadratic residue modulo n , then there are precisely 2 r distinct integers x , with 0 < x < n , such that x 2 a (mod n ) .
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• Spring '13
• MRR
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# Cryptanalysis of ICE
As a new cipher, ICE has not yet undergone rigorous third-party cryptanalysis. These are the results of the author's own cryptanalysis.
### Weak keys
ICE has no weak keys. Weak, or self-decrypting, keys are keys which, if they are used to encrypt the same data twice, produce the original unencrypted data. DES has four of them.
There are no semi-weak keys either. Semi-weak keys come in pairs, where the second key decrypts the first. DES has 16 of them, including the four weak keys.
A weak or semi-weak key occurs if there is another key that generates an identical key schedule, but in the reverse order. These keys can be found by setting up a series of linear (under XOR) equations expressing the fact that the schedule of key 1 is the reverse of the schedule of key 2, then solving the equations. The number of independent variables in the solution gives the log base 2 of the number of weak keys.
For ICE, there were 960 equations (16 rounds, 60 bits per round) and 129 variables (2 x 64-bit keys, plus an inversion bit). The solution was "1=0", which means that there are no keys that satisfy the equations.
### Key inversion weaknesses
ICE has no key-inversion weaknesses. These occur when inverting certain bits in the key and plaintext simply cause bits to invert in the ciphertext. DES has one weakness of this sort.
They are caused in DES by the fact that key bits are only used with the XOR function. If both key and plaintext bits are inverted, the inversions are cancelled out by the XOR function, and DES behaves linearly. However, ICE also uses key bits to permute the outputs of the E boxes, so if the key is inverted, the S-boxes will receive totally different inputs.
### Differential cryptanalysis
ICE levels 1 and above cannot be broken by differential cryptanalysis. However, there is a possibility that Thin-ICE can be broken by a chosen-plaintext attack with roughly 256 encryptions. This has been calculated by simply multiplying the round-by-round probabilities, so it is not yet certain whether it yields a valid attack. DES can be broken by differential cryptanalysis with 247 encryptions.
The use of keyed permutation after the E boxes means that an attacker cannot know which S-box will be affected by a particular input bit. However, since the keyed permutation acts to simply swap bits between the left and right halves of a 32-bit value, the attacker can use inputs whose leftmost 16 bits are the same as the rightmost 16 bits. This enables the attacker to send known differences to the S-boxes, but it usually also means that twice as many S-boxes have to be attacked simultaneously, often with low-probability differences. This typically at least squares the number of pairs required to achieve a result.
The best input differences for attacking the ICE F-function are b2d6b2d6 and cad6cad6, both of which produce a zero output difference with probability 4320/240. They can be combined into 5-round characteristics which have a probability of 2-55.85, and it is these characteristics that may be able to break Thin-ICE, which is an 8-round cipher.
### Linear cryptanalysis
None of the ICE variants appear to be breakable by linear cryptanalysis. Even Thin-ICE, the weakest variant, seems to need over 282 encryptions to be reliably broken, but since it is only a 64-bit cipher, there aren't that many plaintexts available. DES can be broken with approximately 243 encryptions.
The resistance of ICE to linear cryptanalysis is due to the larger S-boxes, and to the keyed permutation, which roughly squares the effort otherwise required.
### Related-key cryptanalysis
This attack relies on simple relations between subkeys in adjacent rounds. ICE is not susceptible to this attack because it uses an irregular key rotation schedule, meaning that there is no consistent relationship between subkeys. DES is also resistant to this attack.
### Meet-in-the-middle attacks
If you encrypt data twice, with two different keys, you usually find yourself susceptible to a meet-in-the-middle attack. That is why Triple-DES is used instead of double encryption, despite the factor of three speed penalty.
ICE avoids this weakness in its extended variants by extending the key schedule with insertions in the middle of the schedule. Although ICE-n effectively encrypts the data n times with n different 64-bit keys, it does this not by encrypting with one key after another, but by doing half encryptions (i.e. the first 8 rounds) n times, then doing the second halves n times.
### Codebook reconstruction attacks
It must be remembered that any 64-bit cipher can be broken under a chosen-plaintext attack in 264 time and memory by simply constructing a lookup table of all 264 possible plaintext/ciphertext pairs. This is regardless of the key size and how well the cipher has been designed.
So it must be remembered that although the strength of ICE-n under ciphertext-only attacks is probably 264n, the strength of all ICE variants under chosen-plaintext is, at best, 264.
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## NBA Playoffs history lesson -- What are the odds?
USA TODAY Sports
The Clippers 2-0 lead over the Grizzlies in their NBA Playoff series puts the Clippers in an advantageous position. Very advantageous, according to NBA playoff history.
The Los Angeles Clippers hold a 2-0 lead in the best of seven first round series against the Memphis Grizzlies. That's obviously good news for the Clippers and greatly increases their odds of winning the series. But by the same token, the Grizzlies will certainly point out that all the Clippers have done so far is hold serve, that Memphis only has to steal one game in STAPLES Center to win the series, that they could have two more chances to do just that. It all begs the question, just how likely are the Clippers to win this thing at this point?
From a purely probabilistic standpoint, the website PlayoffStatus.com sets the odds of the Clippers advancing to the second round given their 2-0 lead at 81%. According to the site, "all future unplayed games are assumed won/lost with a probability based upon relative team strengths." In other words, based on factors like won/loss record, margin of victory and the like, the site is doing a pure probability calculation that the Grizzlies will win four straight or four out of five against the Clippers. That probability turns out to be 19%, leaving the Clippers with an 81% likelihood of advancing.
It turns out that NBA history tells us that these probabilities are too conservative. According to the encyclopedic website WhoWins.com, which tracks the results of seven game series across every major sport in every conceivable situation, teams leading 2-0 in a seven game series have a .890 winning percentage across all sports (573-71). The winning percentage is even higher in the NBA, where teams up 2-0 have won 233 times, losing just 15 times. This .940 winning percentage is identical for all rounds of the playoffs and for the first round, where teams up 2-0 have gone 47-3.
Surely though you say, the record is not so one-sided when both of the wins occurred at home. After all, we always here that a series doesn't really start until the home team loses, and all the Clippers have done so far is win their home games. As it happens, the results are even MORE lopsided when the first two games are home wins. Teams in the Clippers' situation, up 2-0 after two home games, have a 201-12 (.944) record in all NBA playoff series and a 46-2 record in the first round (.958). While this may seem counter-intuitive at first, bear in mind that the team that hosts the first two games is by definition the team with the better record (or at least the team that won a tie-breaker) so there are factors dictating their victory in the series that go beyond the mere fact that they lead 2-0. The Clippers-Grizzlies series -- pitting two teams with identical regular season records against each other -- is somewhat unique, so clearly the Grizzlies' odds will be better than the general data set. For instance, Miami's 2-0 lead over Milwaukee is a very different situation than L.A.'s 2-0 lead over Memphis.
Continuing the history lesson, let's go back to 2006, to the only other time the Clippers had home court advantage in the NBA Playoffs. That season, the Clippers opened with two victories in STAPLES Center over the Denver Nuggets. In both the 2006 series and this year, the Clippers opening victories included a two point win and a double digit win. In an even stranger coincidence, the opponent scored the same number of points in both games of the two series (Denver lost 89-87, 98-87 while Memphis lost 112-91, 93-91). In 2006 the Nuggets managed to win Game 3 but the Clippers won the next two to close the series out in five games.
Speaking of Game 3, although NBA history overwhelming favors the team up 2-0 in the series, Game 3 is usually won by the home team. In NBA history, the visiting team is 229-273 (.456) despite being up 2-0 in the series.
And what does PlayoffStatus.com say about the Clippers odds going forward? Well, the site gives the Clippers a 33% chance of advancing to the Western Conference Finals, a 16% chance of winning the conference and advancing to the NBA Finals, and a 6% chance of winning it all. Bear in mind that these odds are cumulative. The probability of getting heads when you flip is coin is 50% -- the probability of getting heads twice in a row is 25%. But the probability of getting heads a second time after getting heads the first time is still 50%. Which means that the 33% chance the Clippers have of advancing to the Conference Finals takes into consideration the 19% chance that they won't get out of the first round. A little reverse engineering of these probabilities shows that the Clippers actually have a 41% chance of winning in the second round, a 49% chance of winning in the conference finals and a 38% chance of winning in the Finals -- assuming they advance to those rounds. I'll take that.
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# 3x3 matrix solver
Apps can be a great way to help students with their algebra. Let's try the best 3x3 matrix solver. Our website can help me with math work.
## The Best 3x3 matrix solver
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In addition, many of these websites also provide worked examples so that the student can see how the process works. With a little practice, using a math word problem solver online free can help students to become more confident and proficient in solving math word problems.
There are two main types of slope. Both types of slope can be used to find the point of a line or the location where a line meets a vertical line. The two types are called “linear” and “non-linear.” Linear slopes have a constant slope from one point to the next, whereas non-linear slopes have an increasing or decreasing slope from one point to another. Slope is measured in either “percent” or “percentage.” If you need to measure the slope of a line that doesn’t have a vertical side (like a road), use percent and multiply by 100 to find the percentage slope of the line. For example, if you want to measure the percentage slope at 25 meters on a road that has a vertical side of 5 meters, use 25% x 100 = 1/5 (you would multiply 5 by 1/25). On the other hand, if you need to measure the slope of a straight line (like the sides of a house), use percentage and divide by 100. For example, if you want to calculate the percentage slope at three meters on the side of a house, 0.33 x 100 = 33%.
There are a few different ways to solve logarithmic equations, but one of the most common methods is by using logs. To do this, you first take the log of both sides of the equation, and then solve for the variable. This is usually a fairly simple process, but it can be tricky if you're not familiar with logs. Another method that can be used is to rewrite the equation in exponential form, and then solve from there. This can be a bit more difficult
This a website that can help you to understand and solve different types of math word problems. This website can be very helpful, especially if you are struggling with math. The website provides step-by-step instructions on how to solve different types of math word problems. In addition, the website also includes a variety of video tutorials that can help you to better understand the concepts. is a great resource for anyone who is struggling with math.
## Instant support with all types of math
Absolutely amazing app, but is there a way to copy a result? I need to differentiate multiple times in a row and there's no feature for that in the calculator nor can I copy the result so I can just d/dx it again. (You can't d/dx^2)
Aurora White
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Della Rivera
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PDA
View Full Version : Two Stats questions for Math nerds
ski
10-12-2004, 12:06 PM
1. If a school offers 3200 separate courses and a survey of these courses determines that the class size is 50 with a standard deviation of 2, what would one expect for the average and standard deviation of a subset of 50 of these classes selected randomly?
2. In a survey to estimate the average size (number of students) of a class at a university, ten courses are picked at random and the class size of each is determined. The result is 48 students in a class with a standard deviation of 12 students. Assuming that the sample of 10 classes is representative subset of the whole and that class size is a Gaussian random variable, what would one expect the average and standard deviation to be for samples of 40 classes and 160 classes?
It's been so long since I've taken stats... REALLY long :eek3: Any help on one or botha these would help me this week :)
ski
10-12-2004, 10:47 PM
anyone? Bueller?
bahhhhh I figured someone would be mysteriously inclined to help :P
ShawnLee
10-13-2004, 12:07 AM
Ski, it's not that I don't want to. I'm just unable to. I haven't done a stats based Research and Methodologies class, and even then it was for the Poli Sci department.
InfiniteNothing
10-13-2004, 02:08 AM
1. If a school offers 3200 separate courses and a survey of these courses determines that the class size is 50 with a standard deviation of 2, what would one expect for the average and standard deviation of a subset of 50 of these classes selected randomly?
2. In a survey to estimate the average size (number of students) of a class at a university, ten courses are picked at random and the class size of each is determined. The result is 48 students in a class with a standard deviation of 12 students. Assuming that the sample of 10 classes is representative subset of the whole and that class size is a Gaussian random variable, what would one expect the average and standard deviation to be for samples of 40 classes and 160 classes?
It's been so long since I've taken stats... REALLY long :eek3: Any help on one or botha these would help me this week :)
1.We'd expect an average of 50+- 1.98*2/sqrt(50)
The standard deviation by definition is 2/sqrt(50)
sqrt(10)*12=s
and then stdv= s/sqrt(40)
and s/sqrt(160)
where the mean = STDV*2.021,STDV*1.97 respectively
Jcranmer
10-13-2004, 02:58 PM
1.We'd expect an average of 50+- 1.98*2/sqrt(50)
The standard deviation by definition is 2/sqrt(50)
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# Extensions 1→N→G→Q→1 with N=D4.D7 and Q=C22
Direct product G=N×Q with N=D4.D7 and Q=C22
dρLabelID
C22×D4.D7224C2^2xD4.D7448,1247
Semidirect products G=N:Q with N=D4.D7 and Q=C22
extensionφ:Q→Out NdρLabelID
D4.D71C22 = D813D14φ: C22/C1C22 ⊆ Out D4.D71124D4.D7:1C2^2448,1210
D4.D72C22 = D28.29D4φ: C22/C1C22 ⊆ Out D4.D71124D4.D7:2C2^2448,1215
D4.D73C22 = D811D14φ: C22/C1C22 ⊆ Out D4.D71124D4.D7:3C2^2448,1223
D4.D74C22 = D7×C8⋊C22φ: C22/C1C22 ⊆ Out D4.D7568+D4.D7:4C2^2448,1225
D4.D75C22 = SD16⋊D14φ: C22/C1C22 ⊆ Out D4.D71128-D4.D7:5C2^2448,1226
D4.D76C22 = D7×C8.C22φ: C22/C1C22 ⊆ Out D4.D71128-D4.D7:6C2^2448,1229
D4.D77C22 = C2×D8⋊D7φ: C22/C2C2 ⊆ Out D4.D7112D4.D7:7C2^2448,1208
D4.D78C22 = C2×D83D7φ: C22/C2C2 ⊆ Out D4.D7224D4.D7:8C2^2448,1209
D4.D79C22 = C2×D7×SD16φ: C22/C2C2 ⊆ Out D4.D7112D4.D7:9C2^2448,1211
D4.D710C22 = C2×SD16⋊D7φ: C22/C2C2 ⊆ Out D4.D7224D4.D7:10C2^2448,1213
D4.D711C22 = D7×C4○D8φ: C22/C2C2 ⊆ Out D4.D71124D4.D7:11C2^2448,1220
D4.D712C22 = D810D14φ: C22/C2C2 ⊆ Out D4.D71124D4.D7:12C2^2448,1221
D4.D713C22 = D85D14φ: C22/C2C2 ⊆ Out D4.D71128+D4.D7:13C2^2448,1227
D4.D714C22 = D86D14φ: C22/C2C2 ⊆ Out D4.D71128-D4.D7:14C2^2448,1228
D4.D715C22 = C56.C23φ: C22/C2C2 ⊆ Out D4.D71128+D4.D7:15C2^2448,1231
D4.D716C22 = C2×D4.D14φ: C22/C2C2 ⊆ Out D4.D7112D4.D7:16C2^2448,1246
D4.D717C22 = C28.C24φ: C22/C2C2 ⊆ Out D4.D71124D4.D7:17C2^2448,1275
D4.D718C22 = C2×D4.9D14φ: C22/C2C2 ⊆ Out D4.D7224D4.D7:18C2^2448,1276
D4.D719C22 = D28.32C23φ: C22/C2C2 ⊆ Out D4.D71128+D4.D7:19C2^2448,1288
D4.D720C22 = C2×D4.8D14φ: trivial image224D4.D7:20C2^2448,1274
D4.D721C22 = D28.33C23φ: trivial image1128-D4.D7:21C2^2448,1289
D4.D722C22 = D28.34C23φ: trivial image1128+D4.D7:22C2^2448,1290
Non-split extensions G=N.Q with N=D4.D7 and Q=C22
extensionφ:Q→Out NdρLabelID
D4.D7.C22 = D8.10D14φ: C22/C1C22 ⊆ Out D4.D72244-D4.D7.C2^2448,1224
D4.D7.2C22 = D28.44D4φ: C22/C2C2 ⊆ Out D4.D72248-D4.D7.2C2^2448,1232
D4.D7.3C22 = D28.35C23φ: C22/C2C2 ⊆ Out D4.D72248-D4.D7.3C2^2448,1291
×
𝔽
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## Thursday, March 8, 2018
### Extending Relationships for Octagonal Numbers
We have been looking at the octagonal sequence (and its “shadow” sequence) showing how its terms when expressed in appropriate number bases leads naturally to one important form of self-generating numbers i.e. where the same number results, when subtracted from its reverse.
However this in fact represents but a specific example of a more general number phenomenon.
Now when we return to the octagonal numbers, we may recall that the two digits that occur in the respective number bases 2, 5, 8, … (when converting each term to its appropriate number base) represent the unique 2-digit recurring sequence of the reciprocal of 3 in each of these bases.
So again for example, 8 in base 5 = 13 (with 31 – 13 = 13). 13 then equally represents the unique 2-digit sequence of 1/3 in this base (.131313…)
Thus the number in question here (to which 1st reciprocal relates) is 3.
And the 1st relevant number base (through which the respective terms of the octagonal sequence are expressed) = 2 (i.e. 3 – 1). And then subsequent number bases keep increasing by 3.
So we now can express a more general number phenomenon in these terms.
Let n represent a number. Then when we obtain the unique digit sequence of its reciprocal (1/n) in the respective number bases n – 1, 2n – 1, 3n – 1, …, a unique recurring connection will characterise the relationship between each resulting 2-digit number and its reverse (in its respective number base).
The simplest case occurs when n = 2, Therefore the appropriate number bases here to express the unique digit sequence of 1/2 are 1, 3, 5, …
Now 1/2 expressed in base 1 does not have a meaningful expression. But 1/2 in base 3 is .111… Though there is only one recurring digit in this case, we will preserve the first two digits as 2 unique digits will arise when n> 2. So the digit sequence here is 11.
And in this case 11 is equal to its reverse. So 11 (reverse) = 11 (original number) * 1. And 11 in base 10 = 4.
And this is equally the case for all subsequent number bases. For example in base 5, the reciprocal of 2 = .222… So 22 (reverse) = 22 (original number) * 1. And 22 in base 10 = 12
So when the number (to which the reciprocal relates) is 2 with relevant number bases 3, 5, 7, …,
reverse = original number * 1
Now when expressed in base 10, these numbers in the respective number bases leads to a unique sequence i.e. the triangular numbers * 4,
0, 4, 12, 24, …
Then as we have seen in the next case, where 3 is the number to which the reciprocal relates (bases 2, 5, 8, …), a unique 2-digit sequences arise.
And in all these cases (as for example 31 and 13 in base 5),
reverse = original number * 2.
And these numbers in their respective number bases, expressed in base 10, lead to the octagonal sequence of numbers,
1, 8, 21, 40, …
Then when we subtract each term of the original sequence from the octagonal sequence we get
1, 4, 9, 16, ... (i.e. the sum of squares).
And continuing on, in the next case, where 4 is the starting number to which the reciprocal relates, the relevant number bases (for expressing its unique 2-digit sequence) are 3, 7, 11, …
So for example in base 3, 1/4 = .0202…
Thus the unique 2-digit sequence = 02.
Then when we subtract 02 from its reverse we get 20 – 02 = 11
So in denary terms the reverse = 6 and the original number = 2.
Then in base 7, 1/4 = .1515…
Thus the unique 2-digit sequence is 15.
And 51 in denary terms = 36 and 15 = 12
So generalising for all such cases related to starting number 4,
reverse = original number * 3
Once again a unique sequence is associated with each of these original numbers, when expressed in a denary manner i.e.
2, 12, 30, 56, …
In fact we can see a discernible pattern emerging as we move to higher number bases.
Thus the original number in base 3 = 02, in base 7 = 15, in base 11 = 28, base 15 = 3A and so on.
And 2, 12, 30, 56 = 2(1, 6, 15, 28, …)
The terms inside the brackets constitute the hexagonal sequence (which comprises the odd numbered terms of the triangular sequence).
It is fascinating in this context that if now subtract each term of the previous octagonal sequence from each corresponding term of this new sequence, we again obtain
1, 4, 9, 16, … (i.e. the squares of the natural numbers).
So using just one final case for illustration, when the starting number is 5, we consider 1/5 in bases 4, 9, 14, …
1/5 in base 4 = .0303…
Therefore the unique 2-digit sequence = 03 which now constitutes our original number in base 4.
And the 30 = 12 (in denary terms)
So 30 = 03 * 4 (in base 4).
In base 9, 1/5 = .1717…
Therefore the unique 2-digit sequence = 17, which now constitutes our original number in base 9.
And 17 = 16 and 71 (the reverse) = 64 (in denary terms).
So 71 = 17 * 4 (in base 9).
So generalising for all such cases related to starting number 5,
reverse = original number * 4.
Then the corresponding unique associated number sequence (in base 10) is
3, 16, 39, 72, … (A147874 in OEIS)
Once again when we subtract each terms of the previous sequence from the corresponding terms of the new sequence, we obtain,
1, 4, 9, 16,…(the squares of the natural numbers)
Expressed in even more general terms when the starting number (to which the reciprocal 1/n relates in number bases n – 1, 2n – 1, 3n – 1, … then with respect to the original numbers (based on the unique digit sequence of the reciprocal)
reverse = original number * (n – 1)
Thus when the starting number is 6
reverse = original number * 5
We can readily confirm this for 1/6 in base 5 = .0404…
Therefore the unique digit sequence = 04
So the relevant original number (in base 5) = 04 and reverse 40 (i.e. 4 and 20 in denary terms)
Thus 40 = 04 * 5.
Incidentally the sequence (in base 10) associated with these numbers is
4, 20, 48, 88, … = 4 (1, 5, 12, 22, …)
And the terms inside the brackets comprises the pentagonal sequence.
Once again when we subtract each term of the previous sequence from each corresponding term of the present sequence we obtain
1, 4, 9, 16, … (i.e. the squares of the natural numbers). And this appears to be universally the case.
We can therefore write down immediately the unique digit sequence in the next case(for n = 7) by adding each of these sum of square terms to the respective term in the previous sequence, to obtain
5, 24, 57, 104, …
So 5 in base 6 = 05
And 50 = 05 * 6 (in base 6)
Likewise 24 in base 13 = 1B
And its reverse B1 = 11* 13 + 1 = 144
And 144 (reverse) = 24 (original number * 6).
So starting with the original sequence (where n = 2), we can directly generate all further sequences by simply adding each term of the original sequence to the corresponding term of the sum of squares.system.
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## Counting distinct within various groups
Occasional Contributor
Posts: 13
# Counting distinct within various groups
Hi All,
Hope someone can help ...
I have the following hypothetical dataset:
MainAccountClientNumAlternativeAccount
AA1A1
AA1A2
AA1A3
AA2A4
AA2A3
BB1B1
BB1B2
CC1
You will note that main account AA has two clients linked to it and accounts BB and CC have one client linked to each. Account CC has no alternative accounts
In account AA, each of the clients have different alternative accounts, but they share 1 (A3)
I want to count how many alternative accounts each client has and how many alternatve account each main account has (without double counting A3 for the first account). The problem is counting only the distinct alternative account numbers when looking at a main account level. I tried using a count(case ...) statement but it will always double count the A3 account and i cannot use a nested statement to say only count it once when is is not distinct ...
My ideal final dataset needs to like like follow: (that is: NumAccount_AltAccounts for Account AA must show 4 not 5)
MainAccountClientNumNumClient_AltAccountsNumAccount_AltAccounts
AA134
AA224
BB122
CC100
Thanks!!
Super User
Posts: 8,069
## Re: Counting distinct within various groups
Key idea is to use COUNT(DISTINCT varname) to get the counts.
Then you can create two queries that group by different variables and merge the results to get your final table.
I shortened the variable names to make it easier to type.
data have ;
input (a b c) (\$) @@;
cards;
AA 1 A1 AA 1 A2 AA 1 A3 AA 2 A4 AA 2 A3 BB 1 B1 BB 1 B2 CC 1 .
;
proc sql ;
create table want as
select x.*,y.cnt2
from
(select distinct a,b,count(distinct c) as cnt1 from have group by a,b) x
,(select distinct a,count(distinct c) as cnt2 from have group by a) y
where x.a = y.a
order by 1,2
;
quit;
data _null_;
set want;
put (_ALL_) (=);
run;
a=AA b=1 cnt1=3 cnt2=4
a=AA b=2 cnt1=2 cnt2=4
a=BB b=1 cnt1=2 cnt2=2
a=CC b=1 cnt1=0 cnt2=0
Posts: 3,167
## Re: Counting distinct within various groups
Tom's SQL is no doubt the most pithy approach, here is a data step using 2XDOW:
data have;
infile cards truncover;
input (MainAccount ClientNum AlternativeAccount) (:\$);
cards;
AA 1 A1
AA 1 A2
AA 1 A3
AA 2 A4
AA 2 A3
BB 1 B1
BB 1 B2
CC 1
;
proc sort data=have;
by MainAccount ClientNum AlternativeAccount;run;
data want;
array aa(100) \$2. _temporary_;
do _n_=1 by 1 until (last.MainAccount);
set have;
by MainAccount;
if AlternativeAccount not in aa then aa(_n_)=AlternativeAccount;
end;
num_client=0;
num_account=100-cmiss(of aa(*));
do until (last.MainAccount);
set have;
by MainAccount ClientNum AlternativeAccount;
if not missing(AlternativeAccount) then num_client+first.AlternativeAccount;
if last.ClientNum then do; output;num_client=0;end;
end;
call missing(of aa(*));
drop AlternativeAccount;
run;
proc print;run;
Haikuo
Super User
Posts: 10,761
## Re: Counting distinct within various groups
Yeah. Tom 's code is absolutely right and fast. I just want to rewrite it by using sub-query which is also a powerful tool of SQL as its cartesian product . Don't be offended , TOM.
```data have ;
input (a b c) (\$) @@;
cards;
AA 1 A1 AA 1 A2 AA 1 A3 AA 2 A4 AA 2 A3 BB 1 B1 BB 1 B2 CC 1 .
;
proc sql ;
create table want as
select distinct a,b,count(distinct c) as cnt1,(select count(distinct c) from have where a=x.a ) as cnt2
from have as x
group by a,b
;
quit;
```
Ksharp
Occasional Contributor
Posts: 13
## Re: Counting distinct within various groups
Hi all,
Thanks ... this helps a lot.
Regards,
Mahesh
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# The magnetic field at a perpendicular distance of 2cm from an infinite straight current carrying conductor is 2×10−6T. The current in the wire is
A
01A
B
02A
C
04A
D
08A
Video Solution
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Step by step video, text & image solution for The magnetic field at a perpendicular distance of 2cm from an infinite straight current carrying conductor is 2xx10^-6T. The current in the wire is by Physics experts to help you in doubts & scoring excellent marks in Class 12 exams.
Updated on:21/07/2023
### Similar Practice Problems
• Question 1 - Select One
## The magnetic field at a perpendicular distance of 2 cm from an infinite straight current carrying conductor is 2×10−6 T. The current in the wire is
A0.1A
B0.2A
C0.4A
D0.8A
• Question 1 - Select One
## Magnetic field at a distance r from an infinitely long straight conductor carrying steady varies as
A1/r2
B1/r
C1/r2
D1/r
• Question 1 - Select One
## Variation of magnetic field with distance from current carrying conductor is:
ABa2
BB1a2
CBa
DB1a
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## ◂Math Worksheets and Study Guides Sixth Grade. Proportions/Equivalent Fractions
### The resources above correspond to the standards listed below:
#### North Dakota Common Core Standards
ND.CC.6.RP. Ratios and Proportional Relationships
Understand ratio concepts and use ratio reasoning to solve problems.
6.RP.1. Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, ''The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.'' ''For every vote candidate A received, candidate C received nearly three votes.''
6.RP.3. Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a. Make tables of equivalent ratios relating quantities with whole-number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
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1
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Question
# Raju took a loan at 8% per annum simple interest for a period of 5 years. At the end of five years he paid 10640 to clear his loan. How much loan did he take?
A
8500
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B
8000
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C
7700
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D
7600
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Solution
## The correct option is D 7600Simple Interest SI=PNR100And Amount A=P+SI=>Rs10640=P+P×5×8100=>Rs10640=P+0.4P=>Rs10640=1.4=>P=Rs7600
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# Create a numeric variable with date+hour
조회 수: 1(최근 30일)
David E.S. 2021년 2월 11일
댓글: David E.S. 2021년 2월 11일
Hi. I have a problem when I want to create a numeric variable which contains date + hour. This is my data:
So... I want to obtain a number with the sumatory by rows (both columns are strings). Can you help me?
Thanks!
MATLAB VERSION: R19b
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
David E.S. 2021년 2월 11일
I don't know how to merge both columns in order to create other which have date + space + hour.
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### 채택된 답변
Steven Lord 2021년 2월 11일
How are you planning to use this "numeric variable with date+hour"? It's quite likely that you can achieve your goal with a datetime array.
dateAndTime = ["19/11/19", "05:16:41"; "19/11/19", "05:19:03"; "19/11/19", "05:21:26"]
dateAndTime = 3×2 string array
"19/11/19" "05:16:41" "19/11/19" "05:19:03" "19/11/19" "05:21:26"
dATWithSpace = join(dateAndTime, " ")
dATWithSpace = 3×1 string array
"19/11/19 05:16:41" "19/11/19 05:19:03" "19/11/19 05:21:26"
dt = datetime(dATWithSpace, 'InputFormat', 'dd/MM/yy hh:mm:ss')
dt = 3×1 datetime array
19-Nov-2019 05:16:41 19-Nov-2019 05:19:03 19-Nov-2019 05:21:26
For example, you can plot with it.
plot(dt, [1 4 9], 'o-')
##### 댓글 수: 1표시숨기기 없음
David E.S. 2021년 2월 11일
Excellent explanation, very useful. Thanks!!
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# An Interesting Gap in Jacks or Better
17 July 1999
Let's examine a common gap in good old jacks or better. Assume you are playing 9-6 jacks (i.e., you receive 9 for a full house and 6 for a flush), and you could choose between either of the following hands. Which would you choose, and how would you play it?
Ks Qd 2c 3c 5c and Ks Qd 2c 3c 4c
The only difference between the hands is in the club straight flush draw. The 235 has an obvious gap in it --- that is, it is missing the 4. The 234 has no such obvious gap --- that is, the cards are right next to each other.
If you have a chart which tells you this information --- great. I'm all for tools to make life simpler. Similarly, many of you have these combinations memorized. Great again! A good memory is helpful in video poker. But let's assume that you either don't have a chart handy, or you don't trust your memory on this particular combination. What do you do now?
The only differences between the 235 and 234 combinations will be straights and straight flushes. Drawing two cards, neither combination is better than the other is when it comes to getting high pairs, two pairs, trips, or regular flushes. And when it comes to full houses, four of a kinds and royal flushes, you're equally up a tree from either starting position.
Let's simplify and look only at straight flushes --- that is, assume we draw clubs.
Combination Perfect Draw Ending Position
2c 3c 4c Ac 5c Ac 2c 3c 4c 5c
5c 6c 2c 3c 4c 5c 6c
2c 3c 5c Ac 4c Ac 2c 3c 4c 5c
4c 6c 2c 3c 4c 5c 6c
There are two perfect combinations starting from 234, and two perfect combinations from 235. The conclusion, then, is that 234 and 235 are equivalent combinations. This is surprising when you first encounter it. How can a "no-gapper" like 234 have the same value as a "one-gapper"? The explanation doesn't concern the gap in 235. It is very real and affects the value of the combination. The explanation comes from the fact that the 234 combination "runs into the ace". If you look at a legitimate "no-gapper" such as 789, you'll see there are three 2-card combinations that will complete the straight. You have the "two cards below" combination (in this case 56), you'll have the "straddle" combination (in this case 6T), and you'll have the "two cards above" (in this case TJ).
Look again at the 234 combination. The two combinations we have fit the "straddle" and the "two cards above" categories. You'll see we don't have room for the "two cards below" draw. With only two different perfect draws, it is fair to consider the 234 combination as belonging to the "one-gapper" family. The better strategy charts tell you to include 234 with the one-gappers. Although figuring this out isn't real difficult, as we've seen today, it is an easy combination to forget.
Now for the second part of the question. How do you play the hand? Knowing that a suited 234 is equivalent to a suited 235 is one thing. Knowing whether it is higher or lower than KQ of different suits is another.
It turns out that going for the straight flush combination is quite a bit higher than holding the KQ --- almost 20¢ when you're playing for dollars. Even if we started with an unsuited QJ, which has more straight possibilities than does KQ, 234 or 235 would still be the superior play by 12¢.
I believe that straight flushes are the least understood hand in video poker. We will revisit this subject many times in the future, although not for a while. These discussions tend to be more technical than many of my readers prefer.
That's it for this time. Until next time, go out and hit a royal flush.
Recent Articles
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Bob Dancer is America’s premier video poker authority.
His booklets Video Poker Reports: Deuces Wild, 9/6 Jacks or Better, and 10/7 Double Bonus are considered the most accurate strategies ever devised for these games.
Bob Dancer Presents WinPoker is the premier computer trainer for video poker. It requires Windows 95 or newer, and has 15 of the most popular games, including Double Bonus, Double Double Bonus, Double Joker, etc. plus the ability to load in most other games. It has won several competitions for being the best video poker software.
#### Bob Dancer Websites:
www.zamzone.com
Bob Dancer
Bob Dancer is America’s premier video poker authority.
His booklets Video Poker Reports: Deuces Wild, 9/6 Jacks or Better, and 10/7 Double Bonus are considered the most accurate strategies ever devised for these games.
Bob Dancer Presents WinPoker is the premier computer trainer for video poker. It requires Windows 95 or newer, and has 15 of the most popular games, including Double Bonus, Double Double Bonus, Double Joker, etc. plus the ability to load in most other games. It has won several competitions for being the best video poker software.
www.zamzone.com
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In simple words Algorithms is ‘Logic or Procedure of solving any Problem.
Standard Defination : “ An algorithm is a procedure or formula for solving a problem, based on conductiong a sequence of specified actions. A computer program can be viewed as an elaborate algorithm”.
1. It is a step-wise representation of a solution to a given problem, which makes it easy to understand.
2. An algorithm uses a definite procedure.
3. It is not dependent on any programming language, so it is easy to understand for anyone even without programming knowledge.
4. Every step in an algorithm has its own logical sequence so it is easy to debug.
5. By using algorithm, the problem is broken down into smaller pieces or steps hence, it is easier for programmer to convert it into an actual program.
Disdvantages of Algorithms:
1. Alogorithms is Time consuming.
2. Difficult to show Branching and Looping in Algorithms.
3. Big tasks are difficult to put in Algorithms.
Characteristics of Algorithms:
Precision – the steps are precisely stated(defined).
Uniqueness – results of each step are uniquely definedand only depend on the input and the result of the precedingsteps.
Finiteness – the algorithm stops after a finite number ofinstructions are executed.
Input – the algorithm receives input.
Output – the algorithm produces output.
Generality – the algorithm applies to a set ofinputs.
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# Two-Digit Subtraction
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## Quinn Graves
on 28 December 2012
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#### Transcript of Two-Digit Subtraction
Two-Digit
Subtraction Digit - A symbol used to make numerals. Regroup When To Regroup Practice! Remember:
Don't forget to subtract the tens.
Show your work! Vocabulary 145 numeral digit digit digit 7 is a one-digit number
17 is a two-digit number
170 is a three-digit number Subtraction- Taking one number away from another. Example: You have 5 ladybugs, you subtract 2, you are left with 3 ladybugs. 5 - 2 = 3 The symbol for subtraction is - Rearranging numbers in a math problem. For example: 25 - 17 - 25 17 = 2 5
-1 7 tens ones 2 5
-1 7 tens ones Always start with the ones! Always start from the top! 5 - 7 Can you take 7 away from 5? 2 5
-1 7 tens ones No! You need to regroup! - 25 17 Take a ten and make 10 ones 2 5
-1 7 tens ones = 2 tens
- 1 ten
1 ten 1 5 ones
+ 10 ones
15 ones 15 2 5
-1 7 tens ones 1 15 15 - 7 1 - 1 Solve! 0 8 Yay! You regrouped! Now solve it! 25 - 17 = 8 5 9
-2 5 tens ones Always start with the ones! Always start from the top! 5 9
-2 5 tens ones 9 - 5 Can you take 5 away from 9? Yes! - Do you need to regroup? No!! 8 2
-3 6 tens ones Always start with the ones! Always start from the top! 8 2
-3 6 tens ones 2 - 6 Can you take 6 away from 2? Yes! - Do you need to regroup? No! 8 2
-3 6 tens ones 8 tens
- 1 ten
7 tens
Take 1 ten and make 10 ones Regroup 5 9
-2 5 tens ones 9 - 5 5 - 2 3 4 2 ones
+10 ones
12 ones 7 12 8 2
-3 6 tens ones 7 12 7 - 3 12 - 6 4 6 4 3
-2 7 tens ones Always start with the ones! 3 13 Show your work! 1 6 Don't forget to subtract the tens. 1) 38
- 12 2) 52
- 45 3) 81
-19 4) 34 5) 73 6) 52
-21 -55 -27
Full transcript
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https://st-patrickschool.org/questions/201997-the-temperature-in-degrees-celsius-c-can-be-converted-to-degrees-fahrenheit-f-using-the-equation.html
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sorryno
2016-09-22 07:46:54
The temperature in degrees Celsius, c, can be converted to degrees Fahrenheit, f, using the equation mc026-1.jpg. Which statement best describes the relation (c, f)? It is a function because –40°C is paired with –40°F. It is a function because every Celsius temperature is associated with only one Fahrenheit temperature. It is not a function because 0°C is not paired with 0°F. It is not a function because some Celsius temperatures cannot be associated with a Fahrenheit temperature.
ruthie1
2016-09-22 11:10:33
the answer is the second option.
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# Infinite number of solutions for some elliptic eigenvalue problems of Kirchhoff-type with non-homogeneous material
## Abstract
In this paper, using variational method, we study the existence of an infinite number of solutions (some are positive, some are negative, and others are sign-changing) for a non-homogeneous elliptic Kirchhoff equation with a nonlinear reaction term.
## 1 Introduction
In this paper, we consider the following nonlocal equation:
$$\textstyle\begin{cases} -M(x, \Vert u \Vert ^{2})\triangle u=\lambda f(x,u),\quad x\in \Omega , \\ u|_{x\in \partial \Omega }=0, \end{cases}$$
(1.1)
where Ω is a bounded open domain of $$\mathbb{R}^{N}$$ with smooth boundary and
$$\textstyle\begin{cases} f\in C(\overline{\Omega }\times \mathbb{R},\mathbb{R}), \\ M(x,t)=a(x)+b(x)t,\quad \Vert u \Vert =\int _{\Omega } \vert \nabla u \vert ^{2}\,dx, \end{cases}$$
with a, $$b\in C^{\gamma }(\overline{\Omega })$$, $$\gamma \in (0,1)$$, $$a(x)\geq a_{0}>0$$, $$b(x)\geq 0$$. Problem (1.1) is the steady-state problem associated with
$$\textstyle\begin{cases} u_{tt}-M(x, \Vert u \Vert ^{2})\triangle u=f,&(x,t)\in (0,+\infty )\times \Omega , \\ u=0,&(x,t)\in \partial \Omega \times (0,+\infty ), \\ u(x,0)=u_{0}(x), \qquad u_{t}(x,0)=u_{1}(x),&x\in \Omega , \end{cases}$$
(1.2)
which is an open problem proposed by Lions [25] as a generalization of
$$\textstyle\begin{cases} u_{tt}-M( \Vert u \Vert ^{2})\triangle u=f,&(x,t)\in (0,+\infty )\times \Omega , \\ u=0,&(x,t)\in \partial \Omega \times (0,+\infty ), \\ u(x,0)=u_{0}(x), \qquad u_{t}(x,0)=u_{1}(x),&x\in \Omega , \end{cases}$$
(1.3)
where $$M(t)=a+bt$$ with $$a>0$$ and $$b>0$$. In [15, 24], the authors noted that Problem (1.2) models small vertical vibrations of an elastic string with fixed ends when the density of the material is not constant. When $$M(x,t)$$ is independent of x, Problem (1.1) can be simplified to
$$\textstyle\begin{cases} -M( \Vert u \Vert ^{2})\triangle u=\lambda f(x,u),\quad x\in \Omega , \\ u|_{x\in \partial \Omega }=0. \end{cases}$$
(1.4)
The steady-state problem (1.4) associated with Problem (1.3) has received a lot of attention in the literature (usually using variational methods); see [2, 3, 10, 14, 16, 20, 22, 23, 29, 3340] and the references therein.
There are many papers in the literature on sign-changing solutions for Dirichlet problems; see [4, 5, 7, 12, 19, 21] and their references. In [43], Zhang and Perera obtained sign-changing solutions for a class of Problem (1.4) using variational methods and invariant sets of descent flow; in [31] using minimax methods and invariant sets of descent flow, Mao and Zhang established the existence of sign-changing solutions; and in [35] combining the constraint variational method and the quantitative deformation lemma, Shuai proved that Problem (1.4) possesses one least energy sign-changing solution. Other results on the existence of sign-changing solutions for Kirchhoff equations can be found in [5, 9, 28, 30, 37] and their references.
Since $$M(x,t)$$ is dependent on x in Problem (1.1), the variational approach cannot be used to discuss it in a direct way, and fixed point theory and the Galerkin method were used to establish existence in [33] and [38]. In [15], Figueiredo et al. established the existence and uniqueness of a positive solution of Problem (1.1) via bifurcation theory, and in [17], Huy and Quan considered a generalization of Problem (1.1)
$$\textstyle\begin{cases} -M(x, \Vert u \Vert ^{2})\triangle u=\lambda f(x,u,\nabla u)-g(x,u, \nabla u),\quad x\in \Omega , \\ u|_{x\in \partial \Omega }=0, \end{cases}$$
and established existence results for both non-degenerate and degenerate cases of the function M using the fixed point index theory. We note, to the best of our knowledge, that there are no results in the literature on the existence of a sign-changing solution for Problem (1.1). In this paper (motivated by [21]) using the steepest descent method for gradient mappings of the isoperimetric variational problem (see [6]) and the method of invariant sets of descending flow in critical point theory (see [27]), we establish the existence of an infinite number of solutions (some are positive, some are negative, and others are sign-changing). Some ideas come from [18] and [42].
## 2 Main result
In this section, we suppose that f satisfies the following conditions:
(1) $$f:\overline{\Omega }\times \mathbb{R}\to \mathbb{R}$$ is locally Lipschitz continuous;
(2) $$f(x,t)t\geq 0$$ and $$f(x,t)\not \equiv 0$$ in $$\Omega \times (-\delta ,0)\cup \Omega \times (0,\delta )$$;
(3) $$|f(x,t)|\leq c_{1}|t|^{p}+c_{2}$$, where $$c_{1}$$, $$c_{2}\in \mathbb{R}^{+}$$, $$1\leq p<\frac{N+2}{N-2}$$ if $$N\geq 3$$ and $$1\leq p<+\infty$$ if $$N=1$$ and $$N=2$$.
Let $$A:= \mathbb{N}$$ in our main result. The main theorem is as follows.
### Theorem 2.1
Suppose that f satisfies (1), (2), and (3). Then Problem (1.1) has an infinite number of positive solutions $$\{u_{1,\alpha }\}_{\alpha \in A}$$, an infinite number of negative solutions $$\{u_{2,\alpha }\}_{\alpha \in A}$$, and an infinite number of sign-changing solutions $$\{u_{3,\alpha }\}_{\alpha \in A}$$.
First we establish the following lemma for Problem (1.1).
### Lemma 2.1
Problem (1.1) has a nontrivial solution if and only if there exists $$r>0$$ such that the following problem
$$\textstyle\begin{cases} - \triangle u=\lambda \frac{1}{M(x,r^{2})}f(x,u),& x \in \Omega , \\ u=0,& x\in \partial \Omega \end{cases}$$
(2.1)
has a nontrivial solution u with $$\|u\|=r$$.
### Proof
Sufficiency. There exists $$r>0$$ such that Problem (2.1) has a nontrivial solution u with $$\|u\|=r$$, and so u satisfies
$$\textstyle\begin{cases} - \triangle u=\lambda \frac{1}{M(x,r^{2})}f(x,u)= \lambda \frac{1}{M(x, \Vert u \Vert ^{2})}f(x,u),& x\in \Omega , \\ u=0,& x\in \partial \Omega . \end{cases}$$
Clearly, u is a nontrivial solution of Problem (1.1).
Necessity. Problem (1.1) has a nontrivial solution u. Let $$r=\|u\|>0$$. Then u satisfies
$$\textstyle\begin{cases} - \triangle u=\lambda \frac{1}{M(x, \Vert u \Vert ^{2})}f(x,u)= \lambda \frac{1}{M(x,r^{2})}f(x,u),& x\in \Omega , \\ u=0,& x\in \partial u, \end{cases}$$
that is, u is a nontrivial solution of Problem (2.1) with $$\|u\|=r$$.
The proof is completed. □
For given $$r>0$$, set
$$S_{r}= \biggl\{ u\in H_{0}^{1}(\Omega ): \int _{\Omega } \vert \nabla u \vert ^{2}\,dx=r^{2} \biggr\} \quad \text{ and }\quad \overline{S}_{r}=S_{r} \cap C_{0}^{1}( \overline{\Omega }).$$
From Lemma 2.1, we only consider the existence of a nontrivial solution of Problem (2.1) in $$S_{r}$$.
Set
\begin{aligned}& g(x,u)=\frac{1}{M(x,r)}f(x,u),\qquad G(x,u)= \int _{0}^{u}g(x,t)\,dt, \quad \forall u\in \mathbb{R}, \\& \Psi (u)=- \int _{\Omega }G\bigl(x,u(x)\bigr)\,dx,\quad \forall u\in H_{0}^{1}( \Omega ), \end{aligned}
and
$$F=\Psi |_{S_{r}},\quad \overline{F}=F|_{\overline{S}_{r}}.$$
Note that
$$F'(u)= \Psi '(u)- \frac{(\Psi '(u),u)}{ \Vert u \Vert ^{2}}u=-T(u)u-K\mathbb{G}(u),$$
where $$(\cdot ,\cdot )$$ is the inner product in $$H_{0}^{1}(\Omega )$$ given by $$(u,v)=\int _{\Omega }\nabla u\cdot \nabla v\,dx$$, $$K=(-\triangle )^{-1}$$ with the Dirichlet boundary condition, $$\mathbb{G}$$ is the Nemitskii operator induced by g and
$$T(u)=\frac{(\Psi '(u),u)}{r^{2}}.$$
From condition (2), we have $$(\Psi '(u),u)<0$$ for all $$u\in S_{r}$$ (see Lemma 1.0 in [19]), and we know that the solutions of Problem (2.1) correspond to the critical points of f.
In order to discuss Problem (2.1), for $$r>0$$, we let (here $$n\in \mathbb{N}$$)
\begin{aligned}& f_{n}(x,t)= \textstyle\begin{cases} f(x,t),&\text{if } \vert t \vert \leq n, \\ f(x,n)+t-n, & \text{if } t>n, \\ f(x,-n)+t+n, &\text{if } t< -n, \end{cases}\displaystyle \end{aligned}
(2.2)
\begin{aligned}& g_{n}(x,t)=\frac{1}{M(x,r^{2})}f_{n}(x,t), \end{aligned}
(2.3)
\begin{aligned}& G_{n}(x,u)= \int _{0}^{u}g_{n}(x,t)\,dt, \quad \forall u\in \mathbb{R} \end{aligned}
(2.4)
and consider
$$\textstyle\begin{cases} -\triangle u=\lambda g_{n}(x,u),\quad x\in \Omega , \\ u|_{x\in \partial \Omega }=0. \end{cases}$$
(2.5)
Let
$$\Psi _{n}(u)=- \int _{\Omega }G_{n}\bigl(x,u(x)\bigr)\,dx,\quad \forall u\in H_{0}^{1}( \Omega ),$$
and
$$F_{n}=\Psi _{n}|_{S_{r}}, \quad \overline{F}_{n}=F_{n}|_{\overline{S}_{r}}.$$
(2.6)
We obtain that
$$F_{n}'(u)=\Psi _{n}'(u)- \frac{(\Psi _{n}'(u),u)}{ \Vert u \Vert ^{2}}u=-T_{n}(u)u-K \mathbb{G}_{n}(u),$$
(2.7)
where $$\mathbb{G}_{n}$$ is the corresponding Nemitskii operator to $$g_{n}$$ and
$$T_{n}(u)=\frac{(\Psi '_{n}(u),u)}{r^{2}}.$$
From the definition of $$F_{n}$$ in (2.6), we know that the solutions of Problem (2.5) correspond to the critical points of $$F_{n}$$.
From the definition of $$g_{n}$$ and conditions (1), (2), and (3), it is easy to see that $$g_{n}$$ also satisfies (1), (2), and (3) uniformly with respect to n and
(1)′ there exists $$L_{n}>0$$ such that
$$\bigl\vert g_{n}(x,t_{1})-g_{n}(x,t_{2}) \bigr\vert \leq L_{n} \vert t_{1}-t_{2} \vert ,\quad \forall x \in \overline{\Omega }, t_{1},t_{2} \in \mathbb{R}.$$
(2.8)
We shall need the following results later.
### Lemma 2.2
(see [1])
Let Ω be a bounded, open subset of $$\mathbb{R}^{N}$$, and suppose that Ω is $$C^{1}$$. Assume that $$N< p\leq +\infty$$ and $$u\in W^{k+1,p}(\Omega )$$. Then there is $$u^{*}\in C^{k,\gamma }(\overline{\Omega })$$ with $$u(x)=u^{*}(x)$$ a.e. $$x\in \Omega$$ such that
$$\bigl\Vert u^{*} \bigr\Vert _{C^{k,\gamma }}\leq C \Vert u \Vert _{W^{k+1,p}};$$
here the constant C depends only on p, N, and Ω.
### Lemma 2.3
(see [13])
Let Ω be a bounded open subset of $$\mathbb{R}^{N}$$ with a $$C^{1}$$ boundary. Assume that $$u\in W^{k,p}(\Omega )$$.
(1) If
$$k< \frac{n}{p},$$
then $$u\in L^{q}(\Omega )$$, where
$$\frac{1}{q}=\frac{1}{p}-\frac{k}{N}.$$
Also
$$\Vert u \Vert _{L^{q}(\Omega )}\leq C \Vert u \Vert _{W^{k,p}(\Omega )};$$
here the constant C depends only on k, p, N, and Ω.
(2) If
$$k>\frac{n}{p},$$
then $$u\in C^{k-[\frac{n}{p}]-1,\gamma }(\overline{\Omega })$$, where
$$\gamma =\textstyle\begin{cases} [\frac{n}{p}]+1-\frac{n}{p}, & \textit{if } \frac{n}{p} \textit{ is not an integer}, \\ \textit{any positive number }< 1, & \textit{if } \frac{n}{p} \textit{ is an integer}. \end{cases}$$
Also
$$\Vert u \Vert _{C^{k-[\frac{n}{p}]-1,\gamma }(\overline{\Omega })}\leq C \Vert u \Vert _{W^{k,p}( \Omega )};$$
here the constant C depends only on k, p, N, and Ω.
### Lemma 2.4
(see [11])
Let p, $$1\leq p\leq p_{0}=(N + 2)/(N - 2)$$ (so that $$2\leq p+1\leq 2^{*}$$), and let $$\beta =(2^{*}/N)(2^{*}-(p+1))$$. Then, for each γ, $$0\leq \gamma \leq \beta$$, there exists $$c > 0$$ such that
$$\Vert u \Vert _{p+1}^{p+1}\leq c \Vert \nabla u \Vert _{2}^{p+1-\gamma } \Vert u \Vert _{2}^{\gamma }$$
for all $$u\in W_{0}^{1,2}(\Omega )$$. (Here and henceforth $$\|u\|_{p}$$ denotes the norm of u in $$L^{p}(\Omega )$$.)
### Lemma 2.5
(see [8])
Let X be a Banach space and F be a closed subset in X. Assume that $$V:X\to Y$$ is locally continuous and
$$\lim_{h\downarrow 0}\frac{d(u+hV(u),F)}{h}=0$$
(2.9)
for all $$u\in \partial F$$, where $$d(\cdot ,\cdot )$$ is the distance on X. If $$u_{0}\in F$$ and $$\sigma (t)$$($$0\leq t<\omega _{+}(u_{0})$$) is the solution of the initial value problem
$$\textstyle\begin{cases} \frac{d\sigma }{\,dt}=V(\sigma ), \\ \sigma (0,u_{0})=u_{0}, \end{cases}$$
then $$\sigma (t)\in F$$ for all $$t\in [0,\omega _{+}(u_{0}))$$.
For each n, we consider
$$\textstyle\begin{cases} \frac{d\sigma }{dt}=-F_{n}'(\sigma )=T_{n}(\sigma ) \sigma +K\mathbb{G}_{n}(\sigma ), \quad t\geq 0, \\ \sigma (0,u_{0})=u_{0} \end{cases}$$
(2.10)
in $$H_{0}^{1}(\Omega )$$ for $$u_{0}\in S_{r}$$, where $$F'_{n}$$ is defined in (2.7). Since (1)′, (2), and (3) hold, we have the following.
### Lemma 2.6
(see [32])
Let $$c< b<0$$. For every $$u\in F_{n}^{-1}([c,b])$$, if $$\sigma _{n}(t,u)$$ is a solution of Problem (2.10) in $$[0,+\infty )$$ (see step 3 in Lemma 2.7), then either there is a unique $$t(u)\in [0,+\infty )$$ such that $$F_{n}(\sigma _{n}(t(u),u))=c$$ or there is a critical point v of $$F_{n}$$ in $$F_{n}^{-1}([c,b])$$ such that $$\sigma _{n}(t,u)\to v$$ as $$t\to +\infty$$.
### Lemma 2.7
Under conditions (1), (2), and (3), Problem (2.10) has a unique solution $$\sigma _{n}(t,u_{0})$$ on $$[0,+\infty )$$, which satisfies:
(i) $$\sigma _{n}(t,u_{0})\in S_{r}$$ for all $$u_{0}\in S_{r}$$; $$\sigma _{n}(t,u_{0})\in \overline{S_{r}}$$ for all $$u_{0}\in \overline{S_{r}}$$;
(ii) there exists $$u_{n}\in S_{r}$$ such that $$\lim_{t\to +\infty }\sigma _{n}(t,u_{0})) \stackrel{H_{0}^{1}}{=}u_{n}$$ for $$u_{0}\in S_{r}$$;
(iii) if $$u_{0}\in \overline{S}_{r}$$, then $$u_{n}\in \overline{S}_{r}$$ and $$\lim_{t\to +\infty }\sigma _{n}(t,u_{0}) \stackrel{C_{0}^{1}}{=}u_{n}$$.
### Proof
The proof is divided into six steps.
Step 1. We show that $$F_{n}'(u)=-T_{n}(u)u-K\mathbb{G}_{n}(u)$$ is globally Lipschitz continuous with respect to $$H_{0}^{1}(\Omega )$$, that is, there is $$M>0$$ such that
$$\bigl\Vert F_{n}'(u_{1})-F_{n}'(u_{2}) \bigr\Vert _{H_{0}^{1}}\leq M \Vert u_{1}-u_{2} \Vert _{H_{0}^{1}}, \quad \forall u_{1}, u_{2}\in S_{r}.$$
Let $$2^{*}=\frac{2N}{N-2}$$. From (2.8), we have
\begin{aligned} \bigl\Vert \mathbb{G}_{n}(u_{1})- \mathbb{G}_{n}(u_{2}) \bigr\Vert _{L^{2^{*}}} =& \biggl( \int _{\Omega } \bigl\vert g_{n}(x,u_{1})-g_{n}(x,u_{2}) \bigr\vert ^{2^{*}}\,dx \biggr)^{1/2^{*}} \\ \leq& \biggl( \int _{\Omega }L_{n}^{2^{*}} \bigl\vert u_{2}(x)-u_{1}(x) \bigr\vert ^{2^{*}}\,dx \biggr)^{1/2^{*}} \\ =& L_{n} \Vert u_{1}-u_{2} \Vert _{L^{2^{*}}}, \end{aligned}
i.e., $$\mathbb{G}_{n}$$ is globally Lipschitz in the $$L^{2^{*}}$$ topology. Note that
$$H_{0}^{1}(\Omega )\hookrightarrow L^{2^{*}}( \Omega ) \stackrel{\mathbb{G}}{\hookrightarrow }L^{2^{*}}(\Omega ) \stackrel{K}{\rightarrow } H_{0}^{1}(\Omega ),$$
K is a bounded linear operator, and so
$$\bigl\Vert K\mathbb{G}(u_{1})-K\mathbb{G}(u_{2}) \bigr\Vert \leq \overline{L}_{n} \Vert u_{1}-u_{2} \Vert$$
for some positive constant $$\overline{L}_{n}$$, where $$\|\cdot \|$$ denotes the norm in $$H_{0}^{1}(\Omega )$$. Note
\begin{aligned} \bigl\vert T_{n}(u_{1})-T_{n}(u_{2}) \bigr\vert =&\frac{1}{r^{2}}|\bigl(K \mathbb{G}_{n}(u_{1}),u_{1} \bigr)-(K\mathbb{G}_{n}) (u_{2}),u_{2})| \\ \leq &\frac{1}{r}\overline{L}_{n} \Vert u_{1}-u_{2} \Vert + \frac{1}{r^{2}} \bigl\Vert K\mathbb{G}(u_{2}) \bigr\Vert \Vert u_{1}-u_{2} \Vert , \end{aligned}
and
$$\bigl\Vert T_{n}(u_{1})u_{1}-T_{n}(u_{2})u_{2} \bigr\Vert \leq \bigl\vert T_{n}(u_{1})-T_{n}(u_{2}) \bigr\vert \Vert u_{1} \Vert + \bigl\vert T_{n}(u_{2}) \bigr\vert \Vert u_{1}-u_{2} \Vert .$$
Since $$\|K\mathbb{G}_{n}(u)\|$$ is bounded in $$S_{r}$$, so $$T_{n}(u)$$ is bounded also. Thus $$F'_{n}(u)$$ is globally Lipschitz continuous.
Step 2. We show that $$F_{n}'(u)=-T_{n}(u)u-K\mathbb{G}_{n}(u)$$ is globally Lipschitz continuous with respect to $$C_{0}^{1}(\overline{\Omega })$$, that is, there is $$\overline{M}>0$$ such that
$$\bigl\Vert F_{n}'(u_{1})-F_{n}'(u_{2}) \bigr\Vert _{C_{0}^{1}}\leq \overline{M} \Vert u_{1}-u_{2} \Vert _{C_{0}^{1}},\quad \forall u_{1}, u_{2}\in \overline{S_{r}}.$$
Let $$l>N$$. From (2.8), we have
$$\bigl\Vert G_{n}(u_{1})-G_{n}(u_{2}) \bigr\Vert _{L^{l}}= \biggl( \int _{\Omega } \bigl\vert g_{n}(x,u_{1})-g_{n}(x,u_{2}) \bigr\vert ^{l}\,dx \biggr)^{1/l} \leq L_{n} \Vert u_{1}-u_{2} \Vert _{L^{l}},$$
i.e., $$\mathbb{G}_{n}$$ is globally Lipschitz in the $$L^{l}(\Omega )$$ topology. Note that
$$C_{0}^{1}(\overline{\Omega })\hookrightarrow L^{l}(\Omega ) \stackrel{\mathbb{G}}{\rightarrow }L^{l}(\Omega ) \stackrel{K}{\rightarrow }W^{2,l}(\Omega ) \cap W_{0}^{2,l}(\Omega ) \hookrightarrow C_{0}^{1}(\overline{\Omega }),$$
K is a bounded linear operator, so there exists $$\overline{L}'_{n}>0$$ such that
$$\bigl\Vert K\mathbb{G}_{n}(u_{1})-K \mathbb{G}_{n}(u_{2}) \bigr\Vert _{C_{0}^{1}}\leq \overline{L}'_{n} \Vert u_{1}-u_{2} \Vert _{C_{0}^{1}}.$$
Note
\begin{aligned}& \bigl\vert T_{n}(u_{1})-T_{n}(u_{2}) \bigr\vert \\& \quad =\frac{1}{r^{2}} \bigl\vert \bigl(K \mathbb{G}_{n}(u_{1}),u_{1} \bigr)-(K\mathbb{G}_{n}) (u_{2}),u_{2}) \bigr\vert \\& \quad =\frac{1}{r^{2}} \bigl\vert \bigl(K\mathbb{G}_{n}(u_{1}),u_{1} \bigr)-\bigl(K \mathbb{G}_{n}(u_{2}),u_{1} \bigr) + \bigl(K\mathbb{G}_{n}(u_{2}),u_{1} \bigr)-(K\mathbb{G}_{n}) (u_{2}),u_{2}) \bigr\vert \\& \quad =\frac{1}{r^{2}} \bigl\vert \bigl(K\mathbb{G}_{n}(u_{1})-K \mathbb{G}_{n}(u_{2}),u_{1}\bigr)+ \bigl(K \mathbb{G}_{n}(u_{2}),u_{1}-u_{2} \bigr) \bigr\vert \\& \quad \leq \frac{1}{r}\overline{L}'_{n} \Vert u_{1}-u_{2} \Vert _{C_{0}^{1}}+ \frac{1}{r^{2}} \bigl\Vert K\mathbb{G}(u_{2}) \bigr\Vert _{C_{0}^{1}} \Vert u_{1}-u_{2} \Vert _{C_{0}^{1}}, \end{aligned}
and
$$\bigl\Vert T_{n}(u_{1})u_{1}-T_{n}(u_{2})u_{2} \bigr\Vert _{C_{0}^{1}}\leq \bigl\vert T_{n}(u_{1})-T_{n}(u_{2}) \bigr\vert \Vert u_{1} \Vert _{C_{0}^{1}}+ \bigl\vert T_{n}(u_{2}) \bigr\vert \Vert u_{1}-u_{2} \Vert _{C_{0}^{1}}.$$
Since $$\|K\mathbb{G}_{n}(u)\|_{C_{0}^{1}}$$ is bounded in $$\overline{S}_{r}$$, so $$T_{n}(u)$$ is bounded also. Thus $$F'_{n}(u)$$ is globally Lipschitz continuous.
Step 3. We show that Problem (2.10) has a unique solution $$\sigma _{n}(t,u_{0})$$ with maximal interval $$[0,+\infty )$$ for $$u_{0}\in S_{r}$$ and $$\sigma _{n}(t,u_{0})\in S_{r}$$ for all $$t\in [0,+\infty )$$.
The theory of Cauchy problems of ordinary differential equations together with step 1 implies that (2.10) has a unique solution $$\sigma _{n}(t,u_{0})$$ with maximal interval $$[0,\omega _{+}(u_{0}))$$ for $$u_{0}\in S_{r}$$. Note
\begin{aligned}& \sigma _{n}(t,u_{0})=e^{-w(t)} \biggl\{ u_{0}+ \int _{0}^{t}e^{w(s)}K \mathbb{G} \bigl(\sigma _{n}(s,u_{0})\bigr)\,ds \biggr\} \\& \quad \text{where } w(t)=- \int _{0}^{t}T_{n}\bigl(\sigma _{n}(s,u_{0})\bigr)\,ds. \end{aligned}
Since $$d\|\sigma _{n}(t,u_{0})\|^{2}/dt\equiv 0$$ for all $$t\in [0,\omega _{+}(u_{0}))$$, we have $$\sigma _{n}(t,u_{0})\in S_{r}$$ for $$t\in [0,\omega _{+}(u_{0}))$$ if $$u_{0}\in S_{r}$$.
Also, since $$g_{n}(\sigma _{n}(t,u_{0}))$$ is bounded in $$H_{0}^{1}$$ if $$u_{0}\in S_{r}$$, then $$\omega _{+}(u_{0})=+\infty$$ (see [32]).
Step 4. We show that Problem (2.10) has a unique solution $$\sigma _{n}(t,u_{0})$$ with maximal interval $$[0,+\infty )$$ for $$u_{0}\in \overline{S}_{r}$$ and $$\sigma _{n}(t,u_{0})\in \overline{S}_{r}$$ for all $$t\in [0,+\infty )$$.
Since step 2 holds, the proof of step 4 is similar to that of step 3, so we omit it.
Step 5. For $$u_{0}\in S_{r}$$, we show that there exists $$u_{n}\in S_{r}$$ such that
$$\lim_{t\to +\infty }\sigma _{n}(t,u_{0})=u_{n} \quad \text{in } H_{0}^{1}.$$
First, since $$F_{n}(u)<0$$ for $$u\in S_{r}$$, choose $$b=F_{n}(u_{0})<0$$. Since $$S_{r}$$ is bounded and weakly convergent closed and $$F_{n}$$ is weakly semi-continuous from below, we have $$\inf_{u\in S_{r}}F_{n}(u)>-\infty$$. Let $$c<\inf_{u\in S_{r}}F_{n}(u)$$. Then $$u_{0}\in F_{n}^{-1}([c,b])$$. From Lemma 2.6, there exists $$u_{n}\in S_{r}$$ such that
$$\lim_{t\to +\infty }\sigma _{n}(t,u_{0})=u_{n} \quad \text{in } H_{0}^{1}.$$
Step 6. For $$u_{0}\in \overline{S}_{r}$$, there exists $$u_{n}\in \overline{S}_{r}$$ such that
$$\lim_{t\to +\infty }\sigma _{n}(t,u_{0})=u_{n} \quad \text{in } C_{0}^{1}.$$
Using the proof of step 5, step 2 guarantees the conclusion is true. □
Let P be the positive cone in $$C_{0}^{1}(\overline{\Omega })$$ and be the interior set of P. The elements of are called positive and the elements of − are called negative.
### Lemma 2.8
Under condition (1) and (2), the flow in Lemma 2.7has the following properties:
$$\sigma _{n}(t,u_{0})\in \pm \mathring{P} \quad \textit{for }u_{0}\in \pm \mathring{P}\cap \overline{S}_{r} \textit{ and } t\in [0,+\infty ).$$
### Proof
The proof follows the ideas in Lemma 1 and 6 in [26].
(1) We show that $$K\mathbb{G}_{n}(u_{0})\in \mathring{P}$$ for $$u_{0}\in P-\{\theta \}$$.
Let $$v=K\mathbb{G}_{n}(u_{0})$$, and we have
$$-\triangle v=g_{n}(x,u_{0})\geq \not \equiv 0,\quad \forall x\in \Omega , v|_{\partial \Omega }=0.$$
The strong maximum principle implies that $$v\in \mathring{P}$$.
(2) We show that
$$K\mathbb{G}_{n}\bigl(\sigma _{n}(t,u_{0}) \bigr)\in \mathring{P} \quad \text{ for } u_{0} \in P\cap \overline{S}_{r}, \text{ and } t>0.$$
(2.11)
Now $$\forall u\in P$$, choose $$\delta >0$$ small such that, for all $$\delta >h>0$$, we have
$$u+h(\bigl(T_{n}(u)u+K\mathbb{G}_{n}(u)\bigr)= \bigl(1+hT_{n}(u)\bigr)u+hK\mathbb{G}_{n}(u) \in P,$$
i.e., (2.9) is satisfied. Now Lemma 2.5 guarantees that the solution $$\sigma _{n}(t,u_{0})$$ of the initial value problem (2.10) satisfies $$\sigma _{n}(t,u_{0})\in P$$ for all $$t\in [0,+\infty )$$ (in fact $$\sigma _{n}(t,u_{0})\in P\cap \overline{S_{r}}$$ since $$u_{0} \in P\cap \overline{S_{r}}$$). Hence (as in (1)) (2.11) holds.
(3) We show that
$$\sigma _{n}(t,u_{0})\in \mathring{P} \quad \text{for }u_{0}\in \mathring{P} \cap \overline{S}_{r} \text{ and } t\in [0,+\infty ).$$
Let $$w(t)=-\int _{0}^{t}T_{n}(\sigma _{n}(s,u_{0}))\,ds$$. We have $$w'(t)>0$$, $$w(t)>0$$, and $$w(t)$$ is strictly increasing. Let $$w^{-1}(t)$$ be the inverse function of $$w(t)$$. It follows from (2.11), for $$u_{0}\in P\cap \overline{S}_{r}$$, that
$$\bigl(1/w'(t)\bigr)K\mathbb{G}_{n} \bigl(\sigma _{n}(t,u_{0})\bigr)\in \mathring{P}.$$
(2.12)
Let $$A(t)=(1/w'(t))K\mathbb{G}_{n}(\sigma (t,u_{0}))$$ and $$E_{t}=\{A(s):0\leq s\leq t\}$$. Note that $$E_{t}$$ is a compact set in $$C_{0}^{1}(\overline{\Omega })$$ and (2.12) implies that $$E_{t}\subseteq \mathring{P}$$ and hence $$\overline{co}E_{t}\subseteq \mathring{P}$$, where $$\overline{co}E_{t}$$ is the closed convex set hull of $$E_{t}$$ in $$C_{0}^{1}(\overline{\Omega })$$. Note
\begin{aligned} \frac{1}{e^{w(t)}-1} \int _{0}^{t}e^{w(s)}K \mathbb{G}_{n}\bigl(\sigma _{n}(s,u_{0}) \bigr)\,ds&= \frac{1}{e^{w(t)}-1} \int _{1}^{e^{w(t)}} \frac{K\mathbb{G}_{n}(\sigma _{n}(w^{-1}(\ln (s)),u_{0}))}{w'(w^{-1}(\ln (s)))}\,ds \\ &= \lim_{m\to +\infty }\frac{1}{m}\sum _{i=1}^{m}A(w^{-1}\biggl( \ln \biggl(1+\frac{i}{m}\bigl(e^{w(t)}-1\bigr)\biggr)\biggr). \end{aligned}
Therefore
$$\frac{1}{e^{w(t)}-1} \int _{0}^{t}e^{w(t)}K\mathbb{G} \bigl(\sigma _{n}(s,u_{0})\bigr)\,ds \in \overline{co}F_{t}\in \mathring{P},$$
and this together with
$$\sigma _{n}(t,u_{0})=e^{-w(t)} \biggl\{ u_{0}+ \int _{0}^{t}e^{w(s)}K \mathbb{G} \bigl(\sigma _{n}(s,u_{0})\bigr)\,ds \biggr\} ,\quad t\in [0,+\infty )$$
and
$$e^{-w(t)}= \bigl(1- e^{-w(t)} \bigr)\frac{1}{e^{w(t)}-1}$$
yields
$$\sigma _{n}(t,u_{0})\in \mathring{P} \quad \text{for }u_{0}\in \mathring{P} \cap \overline{S}_{r} \text{ and } t\in [0,+\infty ).$$
For the case $$u_{0}\in (-\mathring{P})$$, the proof is similar, so we omit it.
The proof is completed. □
### Lemma 2.9
Under conditions (1), (2), and (3), Problem (2.5) has at least one positive solution $$u_{1,n}\in \overline{S}_{r}\cap P$$, one negative solution $$u_{2,n}\in \overline{S}\cap (-P)$$, and one sign-changing solution $$u_{3,n}\in \overline{S}_{r}\cap (C_{0}^{1}-(-P\cup P))$$.
### Proof
Let $$e_{1}$$ be an eigenfunction corresponding to the first eigenvalue of the Dirichlet eigenvalue problem: $$-\triangle u=\lambda u$$ in Ω, $$u|_{\partial \Omega }=0$$, $$e_{2}$$ be an eigenfunction corresponding to the second one with $$\|e_{1}\|=\|e_{2}\|=r$$. Let $$\Lambda ={\mathrm{span}}\{e_{1},e_{2}\}\cap S_{r}$$. Note that $$\Psi _{n}(u)<0$$ for each $$n>0$$ if $$u\not \equiv 0$$ and
$$\Lambda =\{\cos \theta e_{1}+\sin \theta e_{2}:0\leq \theta \leq 2 \pi \}$$
is a compact set in $$S_{r}$$. Then there exists $$\alpha _{n}>0$$ such that
$$\max \bigl\{ \Psi _{n}(u):u\in \Lambda \bigr\} =\max \bigl\{ \Psi (u):u\in \Lambda \bigr\} < - \alpha _{n} .$$
(2.13)
Set
$$\Lambda ^{\pm }=\bigl\{ u\in \Lambda :\sigma _{n}(t,u)\in \pm \mathring{P} \text{ for some } t>0\bigr\} .$$
(1) We show that $$\Lambda ^{\pm }\neq\emptyset$$.
Since $$e_{1}\in \overline{S}_{r}\cap \mathring{P}$$, $$-e_{1}\in \overline{S}_{r}\cap (-\mathring{P})$$, Lemma 2.8 guarantees that $$\sigma _{n}(t,e_{1})\in \mathring{P}$$ and $$\sigma _{n}(t,-e_{1})\in (-\mathring{P})$$ for $$t\in [0,+\infty )$$. Therefore, $$\Lambda ^{\pm }\neq\emptyset$$.
(2) We show that Problem (2.5) has at least one positive solution $$u_{1,n}$$ and one negative solution $$u_{2,n}$$.
Consider $$\sigma _{n}(t,e_{1})$$, $$t\in [0,+\infty )$$. Lemma 2.7 guarantees that there exists $$u_{1,n}\in \overline{S}_{r}\cap P$$ such that
$$\lim_{t\to +\infty }\sigma _{n}(t,e_{1}) \stackrel{C_{0}^{1}}{=}u_{1,n},$$
and $$u_{1,n}$$ is a critical point of $$F_{n}$$ in $$\overline{S}_{r}\cap P$$. Then $$u_{1,n}$$ is a solution of Problem (2.5) and $$u_{1,n}\in \overline{S}_{r}$$. By using the strong maximum principle, we have $$u_{1,n}\in \mathring{P}$$.
For $$\sigma _{n}(t,-e_{1})$$, $$t\in [0,+\infty )$$, a similar argument to that of $$\sigma _{n}(t,e_{1})$$ shows that there exists $$u_{2,n}\in \overline{S}_{r}\cap (-\mathring{P})$$ such that $$u_{2,n}$$ is a solution of Problem (2.5).
(3) We show that Problem (2.5) has at least one sign-changing solution $$u_{3,n}\in \overline{S}_{r}\cap (C_{0}^{1}-(P\cup (-P)))$$.
From the proof of step 2, $$e_{1}\in \Lambda ^{+}$$, $$-e_{1}\in \Lambda ^{-}$$. Note that both $$\Lambda ^{+}$$ and $$\lambda ^{-}$$ are open sets of Λ since $$\sigma _{n}(t,u)$$ depends continuously on u (see [32]). From Lemma 2.8, we have $$\Lambda ^{+}\cap \Lambda ^{-}=\emptyset$$, and the connectedness of Λ implies that there is $$u_{0}\in \Lambda -(\Lambda ^{+}\cup \Lambda ^{-})$$. By Lemma 2.7, $$\sigma _{n}(t,u_{0})\to u_{3,n}$$, a critical point of $$F_{n}$$, in $$H_{0}^{1}(\Omega )$$ as $$t\to +\infty$$. Then $$u_{3,n}$$ is a solution of Problem (2.5) and $$u_{3,n}\in \overline{S}_{r}$$. From (iii) of Lemma 2.7, we have that $$\lim_{t\to +\infty }\sigma _{n}(t,u_{0})=u_{3,n}$$ in the $$C_{0}^{1}(\overline{\Omega })$$-topology. Therefore $$u_{3,n}\notin \mathring{P}\cap (-\mathring{P})$$ since $$\sigma _{n}(t,u_{0})\notin \mathring{P}\cap (-\mathring{P})$$. Then $$u_{3,n}\notin {P}\cap (-P)$$ by using the strong maximum principle. Hence $$u_{3,n}$$ changes its sign in Ω. Also $$\Psi _{n}(u_{3,n})<\Psi _{n}(\sigma _{n}(t,u_{0}))<-\alpha _{n}$$, $$\forall t\in [0,+\infty )$$ since $$\sigma _{n}$$ is a negative descent flow. □
### Proof
(Theorem 2.1) We only prove the existence of sign-changing solutions of Problem (1.1) since the proofs of the existence of positive solutions and negative solutions are similar, so we omit them.
From Lemma 2.9, for given $$r>0$$, Problem (2.5) has at least one sign-changing solution $$u_{3,n}$$ with $$u_{3,n}\in \overline{S}_{r}\cap (C_{0}^{1}-(P\cup (-P)))$$, where $$\lambda ^{-1}_{3,n}=\int _{\Omega }g_{n}(x,u_{3,n})u_{n}\,dx/r^{2}>0$$ and $$g_{n}$$ is defined in (2.3) for each $$n\in \mathbb{N}$$.
(1) We first prove that $$\{\lambda _{3,n}\}$$ is bounded.
Since $$\{u_{3,n}\}$$ is bounded in the $$H_{0}^{1}(\Omega )$$ topology, we may assume that it converges weakly to $$u^{*}$$ in $$H_{0}^{1}(\Omega )$$. Then $$u_{3,n}\to u^{*}$$ in $$L^{p+1}(\Omega )$$ since $$1\leq p<\frac{N+2}{N-2}$$. There exists a number $$c>0$$ such that, for all $$t\in \mathbb{R}$$ and for all $$n=1, 2, \ldots$$ ,
$$\bigl\vert g_{n}(x,t) \bigr\vert \leq c \bigl(1+ \vert t \vert ^{p}\bigr) \quad \text{and}\quad \bigl\vert G_{n}(x,t) \bigr\vert \leq c\bigl(1+ \vert t \vert ^{p+1}\bigr),$$
(2.14)
where $$G_{n}$$ is defined in (2.4).
From Lemma A.1 in [41], there exists a subsequence of $$\{u_{3,n}\}$$, denoted also by $$\{u_{3,n}\}$$, and there exists $$h\in L^{p+1}(\Omega )$$ such that $$u_{n}\to u^{*}$$ a.e. in Ω, $$|u^{*}(x)|\leq h(x)$$, $$|u_{3,n}(x)|\leq h(x)$$ a.e. in Ω. From the Lebesgue dominated convergence theorem, we have
$$\int _{\Omega }G_{n}(x,u_{3,n})\,dx\to \int _{\Omega }G\bigl(x, u^{*}\bigr)\,dx \quad \text{as }n\to +\infty .$$
(2.15)
Let $$n>k_{0}=\max \{\|u\|_{C(\overline{\Omega })}:u\in \Lambda \}$$. By the definitions of $$\Psi _{n}$$ and Ψ, if $$n>k_{0}$$, we have $$\Psi _{n}(u)=\Psi (u)$$ for $$u\in \Lambda$$.
From (2.2), (2.13), there exists a positive constant $$\alpha >0$$ such that
$$-\Psi _{n}(u)= \int _{\Omega }G_{n}(x,u)\,dx=-\Psi (u)= \int _{\Omega }G(x,u)\,dx> \alpha >0 \quad \text{for }u\in \Lambda , n>k_{0}.$$
Since $$u_{3,n}$$ is obtained along a descending flow, it follows that
$$\int _{\Omega }G_{n}(x,u_{3,n})\,dx\geq \int _{\Omega }G_{n}\bigl(x,\sigma (0,u_{0,n})\bigr)\,dx= \int _{\Omega }G_{n}(x,u_{0,n})\,dx>\alpha >0,\quad \forall n>k_{0},$$
for some $$u_{0,n}\in \Lambda$$, where $$\sigma (t,u_{0,n})$$ is a solution of Problem (2.5). From (2.15), we have $$\int _{\Omega }G(x,u^{*})\,dx>0$$. Hence $$u^{*}\not \equiv 0$$. Similar to (2.15), we have
$$\int _{\Omega }g_{n}(x,u_{3,n})u_{3,n}\,dx \to \int _{\Omega }g\bigl(x,u^{*}\bigr)u^{*}\,dx \stackrel{\bigtriangleup }{=} 2\beta >0.$$
Thus $$0<\lambda _{3,n}= \frac{r^{2}}{\int _{\Omega }g_{n}(x,u_{3,n})u_{n}\,dx/r^{2}}<r^{2}/ \beta$$ for n large enough.
(2) We prove that $$\{u_{3,n}\}$$ is bounded in the $$C_{0}^{1}$$-topology.
Choose a sequence of numbers $$\{q_{i}\}$$ satisfying
$$q_{1}=\frac{2N}{N-2}< q_{2}< \cdots < q_{m-1}< q_{m}=2Np$$
and
$$\frac{1}{q_{i+1}}\geq \frac{p}{q_{i}}-\frac{2}{N}, \quad i=1,2,\ldots , m-1.$$
Let $$p_{i}=q_{i}/p$$ ($$i=1, 2, \ldots , m$$). From Lemma 2.4, we have
$$H_{0}^{1}\hookrightarrow L^{q_{1}}(\Omega ).$$
(2.16)
From (2.14), one has (note $$p_{1}$$ $$p=q_{1}$$)
$$\bigl\Vert \lambda _{3,n}\mathbb{G}_{n}(u_{3,n}) \bigr\Vert _{L^{p_{1}}}\leq \frac{1}{\beta }\biggl( \int _{\Omega } \bigl\vert g_{n}(x,u_{3,n}) \bigr\vert ^{p_{1}}\,dx\biggr)^{ \frac{1}{p_{1}}}\leq \frac{2c}{\beta } \bigl( \vert \Omega \vert ^{\frac{1}{p_{1}}}+ \Vert u_{3,n} \Vert _{L^{q_{1}}}^{p}\bigr)$$
(2.17)
for large n. Since K is a bounded linear operator, one has together with Lemma 2.3
$$L^{p_{1}}(\Omega ) \stackrel{K}{\rightarrow }W^{2,p_{1}}(\Omega )\cap W_{0}^{1,p_{1}}( \Omega ) \hookrightarrow L^{q_{2}}(\Omega ).$$
(2.18)
Combining (2.16), (2.17), and (2.18), we have
$$\{u_{3,n}\}_{n\geq k}\subseteq L^{q_{2}}(\Omega ) \text{ is bounded}, \quad \text{ for large } k.$$
(2.19)
Repeating the progress of (2.17), (2.18), and (2.19) for $$i=2, 3, \ldots , m$$, we have
$$\{u_{3,n}\}_{n\geq k}\subseteq L^{q_{m}}(\Omega ) \text{ is bounded}, \quad \text{for large } k.$$
We have (note $$p_{m}$$ $$p=q_{m}$$)
$$\bigl\Vert \lambda _{3,n}\mathbb{G}_{n}(u_{3,n}) \bigr\Vert _{L^{p_{m}}}\leq \frac{1}{\beta }\biggl( \int _{\Omega } \bigl\vert g_{n}(x,u_{3,n}) \bigr\vert ^{p_{m}}\,dx\biggr)^{ \frac{1}{p_{m}}}\leq \frac{2c}{\beta } \bigl( \vert \Omega \vert ^{\frac{1}{p_{m}}}+ \Vert u_{3,n} \Vert _{L^{q_{m}}}^{p}\bigr)$$
for large n, which together with boundedness of the linear operator K guarantees that
$$\{u_{3,n}\}_{n\geq k}\subseteq W^{2,p_{m}}\cap W_{0}^{1,p_{m}} \text{ is bounded}, \quad \text{for large } k.$$
Now Lemma 2.2 implies (note $$p_{m}=2N>N$$) that
$$\{u_{3,n}\}_{n\geq k}\subseteq W^{2,p_{m}}\cap W_{0}^{1,p_{m}} \hookrightarrow C_{0}^{1}(\overline{\Omega }) \text{ is bounded},\quad \text{for large } k.$$
(2.20)
(3) We consider sign-changing solutions of Problem (1.1) in $$S_{r}$$.
From (2.20), set $$L>0$$ such that
$$\Vert u_{3,n} \Vert _{C_{0}^{1}(\overline{\Omega })}\leq L, \quad n\in \{1,2, \ldots \}.$$
(2.21)
Choose $$n_{0}>L$$. From the definitions of $$f_{n_{0}}$$ and $$g_{n_{0}}$$ in (2.2) and (2.3), we have together with (2.21) that
$$f_{n_{0}}\bigl(x,u_{3,n_{0}}(x)\bigr)=f\bigl(x,u_{3,n_{0}}(x) \bigr),\quad x\in \overline{\Omega }$$
and
$$g_{n_{0}}\bigl(x,u_{3,n_{0}}(x)\bigr)=\frac{1}{M(x,r^{2})}f \bigl(x,u_{3,n_{0}}(x)\bigr), \quad x\in \overline{\Omega },$$
which implies that $$u_{3,n_{0}}$$ is a sign-changing solution of Problem (2.1) with
$$\lambda =r^{2}\Bigm/ \int _{\Omega }g\bigl(x,u_{3,n_{0}}(x)\bigr)u_{3,n_{0}}(x)\,dx.$$
For $$r>0$$ given above, write $$u_{3,r}(x)=u_{3,n_{0}}(x)$$ for $$x\in \overline{\Omega }$$. Lemma 2.1 guarantees that $$u_{3,r}$$ is a sign-changing solution of Problem (1.1) with
$$\lambda =r^{2}\Bigm/ \int _{\Omega }g\bigl(x,u_{3,r}(x)\bigr)u_{3,r}(x)\,dx.$$
Similarly, we obtain a set $$\{u_{1,r}\}_{r\in A}$$ of positive solutions of Problem (1.1) and a set $$\{u_{2,r}\}_{r \in A}$$ of negative solutions of Problem (1.1).
The proof is completed. □
Not applicable
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5. Batkam, C.J.: Multiple sign-changing solutions to a class of Kirchhoff type problems. arXiv:1501.05733
6. Berger, M.S.: Nonlinearity & Functional Analysis. Academic Press, San Diego (1977)
7. Castro, A., Cossio, J., Neuberger, M.: A sign-changing solution for a superlinear Dirichlet problem. Rocky Mt. J. Math. 27, 1041–1053 (1997)
8. Chang, K.C.: Infinite Dimensional Morse Theory and Multiple Solution Problems. Birkhǎuser, Boston (1993)
9. Cheng, B., Tang, X.: Ground state sign-changing solutions for asymptotically 3-linear Kirchhoff-type problems. Complex Var. Elliptic Equ. 62(8), 1093–1116 (2017)
10. Cheng, B., Wu, X., Liu, J.: Multiple solutions for a class of Kirchhoff type problems with concave nonlinearity. Nonlinear Differ. Equ. Appl. 19, 521–537 (2012)
11. Chiappinelli, R.: On spectral asymptotics and bifurcation for elliptic operators with odd superlinearity term. Nonlinear Anal., Theory Methods Appl. 13(7), 871–878 (1989)
12. Dancer, E.N., Du, Y.H.: Existence of changing sign solutions for some semilinear problems with jumping nonlinearities at zero. Proc. R. Soc. Edinb. 124A, 1165–1176 (1994)
13. Evans, L.C.: Partial Differential Equations. Grad. Stud. Math., vol. 19. Am. Math. Soc., Providence (2010)
14. Figueiredo, G.M.: Existence of positive solution for a Kirchhoff problem type with critical growth via truncation argument. J. Math. Anal. Appl. 401, 706–713 (2013)
15. Figueiredo, G.M., Morales-Rodrigo, C., Júnior, J.R.S., Suárez, A.: Study of a nonlinear Kirchhoff equation with non-homogeneous material. J. Math. Anal. Appl. 416, 597–608 (2014)
16. He, X., Zou, W.: Existence and concentration of positive solutions for a Kirchhoff equation in $$\mathbb{R}^{3}$$. J. Differ. Equ. 252, 1813–1834 (2012)
17. Huy, N.B., Quan, B.T.: Positive solutions of logistic equations with dependence on gradient and nonhomogeneous Kirchhoff term. J. Math. Anal. Appl. 444(1), 95–109 (2016)
18. Jin, F., Yan, B.: The sign-changing solutions for nonlinear elliptic problem with Carrier type. J. Math. Anal. Appl. 487(2), 24002 (2020)
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20. Li, Y., Li, F., Shi, J.: Existence of a positive solution to Kirchhoff type problems without compactness conditions. J. Differ. Equ. 253, 2285–2294 (2012)
21. Li, Y., Liu, Z.: Multiple and sign changing solutions of an elliptic eigenvalue problem with constraint. Sci. China Ser. A, Math. 44, 48–57 (2001)
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23. Liang, Z., Li, F., Shi, J.: Positive solutions to Kirchhoff type equations with nonlinearity having prescribed asymptotic behavior. Ann. Inst. Henri Poincaré 31, 155–167 (2014)
24. Límaco, J., Clark, H.R., Medeiros, L.A.: Vibrations of elastic string with nonhomogeneous material. J. Math. Anal. Appl. 344, 806–820 (2008)
25. Lions, J.L.: On some question on boundary value problem of mathematical physics. In: de La Penha, G.M., Medeiros, L.A. (eds.) Contemporary Developments of Continuum Mechanics and Partial Differential Equations, pp. 285–346. North-Holland, Amsterdam (1978)
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### Acknowledgements
We thank the referees for their valuable suggestions. The work is supported by the NSFC of China (62073203) and the Fund of Natural Science of Shandong Province (ZR2018MA022).
Not applicable
Not applicable.
## Author information
Authors
### Contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
### Corresponding author
Correspondence to Baoqiang Yan.
## Ethics declarations
### Competing interests
The authors declare that they have no competing interests.
## Rights and permissions
Reprints and permissions
Yan, B., O’Regan, D. & Agarwal, R.P. Infinite number of solutions for some elliptic eigenvalue problems of Kirchhoff-type with non-homogeneous material. Bound Value Probl 2021, 44 (2021). https://doi.org/10.1186/s13661-021-01522-9
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# Calculus
"Evaluate the following indefinite integral using integration by parts:
*integral sign* tan^-1(x) dx"
I let u = tan^-1(x) and dv = dx. Is that right?
1. 👍
2. 👎
3. 👁
1. Oops! I just found out that I don't need to know how to do this type of question.
1. 👍
2. 👎
## Similar Questions
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# In figure, if $\mathrm{LM} \| \mathrm{CB}$ and $\mathrm{LN} \| \mathrm{CD}$, prove that $\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}$.
Question.
In figure, if $\mathrm{LM} \| \mathrm{CB}$ and $\mathrm{LN} \| \mathrm{CD}$, prove that $\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}$.
Solution:
In $\triangle \mathrm{ACB}$ (see figure), $\mathrm{LM} \| \mathrm{CB}$ (Given)
$\Rightarrow \frac{A M}{M B}=\frac{A L}{L C}$ ...(1)
(Basic Proportionality Theorem)
In $\triangle \mathrm{ACD}$ (see figure), $\mathrm{LN} \| \mathrm{CD}$ (Given)
$\Rightarrow \frac{A N}{N D}=\frac{A L}{L C}$ ...(2)
(Basic Proportionality Theorem)
From (1) and (2), we get
$\frac{A M}{M B}=\frac{A N}{N D}$
$\Rightarrow \frac{\mathrm{AM}}{\mathrm{AM}+\mathrm{MB}}=\frac{\mathrm{AN}}{\mathrm{AN}+\mathrm{ND}} \Rightarrow \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}$
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# Why are the resistors in an AM modulator necessary?
In preparation for the amateur radio exam, I have been following the preparation course material by DJ4UF. In the section about modulation it presents this diagram of a simplified AM modulator.
The low frequency signal (NF) and the high frequency carrier wave (HF) are added, the diode cuts off one half-wave and the resonant circuit "recreates" the previously cut off half-wave. The result is a amplitude-modulated signal. So far so good.
What is the purpose of the 47k resistors? The explanatory text mentions that they are necessary to "add the currents in the diode". What would change if we remove them and just directly connect them to the diode?
The concept here is that you are adding the signal currents before feeding the sum to the diode; the resistors are there to convert the voltage sources "NF" and "HF" to current sources.
You'd get exactly the same effect by adding the signal voltages directly — simply connect the "NF" and "HF" boxes in series, without any resistors. The only downside to this is that "NF" and "HF" can't share a common ground, and that's often a desirable feature of a practical system. But some AM transmitters isolate the NF signal with a transformer, which solves that problem.
Note that the circuit as given is not at all practical — you would not want to feed a parallel-tuned circuit, which has a high impedance at resonance, from a current source. Instead, you would use a series-tuned circuit that keeps the diode cathode close to ground potential.
• Two voltage sources fighting one another when connected in parallel could cause intermodulation distortion within one or both generators. One of those distortion products is exactly what you want out of the modulator. But other distortion products are not desired, and "Siebung" won't reject them. The series-connection of sources of Dave's answer is one solution and is far less lossy than those 47k series resistors. Oct 19, 2018 at 14:13
• @glen_geek Aha! So in a way these resistors also prevent that the signal of one frequency source interferes with the other source before being added, if I understood that correctly. Oct 19, 2018 at 14:30
• @ahemmetter Yes. And those resistors also prevent the diode's non-linear behaviour from causing problems of intermodulation back in those generators. Consider this a conceptual modulator - it isn't very practical because its power efficiency is poor. Oct 19, 2018 at 14:57
• "Consider this a conceptual modulator" -- it wouldn't be totally unreasonable if it were being implemented at the low level suggested by the 47k-ohm resistors. There's probably better uses of circuit board space than something that would implement this particular block diagram, though. Oct 19, 2018 at 15:46
• @Dave-Tweed I strongly disagree with your comment about parallel vs serie tuned circuit. For this circuit to work, you need the high impedance at resonance and since you have this diode, it is mostly for DC that you need to draw the cathode to ground potential, which wouldn't work with a serie tuned circuit. Oct 21, 2018 at 19:07
The problem is essentially that both HF and LF sources have a (relatively) low output our internal impedance (because if it wasn't the case and both had the same impedance, you might consider the 47K resistance to be inside both circuits).
So would happen what happen if you put two different voltage generator in parallel. the resulting loop would be one generator = LF-HF with no resistance. You would get a short circuit...
In the case those would have higher but different internal impedance, the result would then be that the output of the generator with the lowest internal impedance would take precedence.
With the two equal resistances bridge, you make sure the output is the average between both generators. Then the diode cuts a part of the HF according to the LF level, and the parallel tuned circuits eliminate everything which in not around the HF frequency. i.e. the continuous component, and the harmonics resulting from the diode cut.
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Spike's Calculators
# Right-Angled Trapezium
Calculate the area and perimeter of a right-angled trapezium. In a right-angled trapezium, the angles on one of the sides are 90 °
This calculator uses metres for measurements.
Right-angled trapezium = is a trapezium with 90° angles on one side making the height the same length as one of the legs.
### Right Trapezium Area and Perimeter
Base Length a m
Base Length b m
Height m
Decimal Precision
#### Results:
Length of Diagonal m
###### Area of the Trapezium
Square Metres
Square Feet ft²
Square Inches in²
Metres m
Feet ft
Inches in
#### Calculation
1. enter the length of base a in metres
2. length of base b in metres
3. the height of the trapezium in metres
4. decimal precision, the number of digits after the decimal point
#### Results
1. the length of the diagonal in metres
2. the area of the right-angled trapezium in square metres
3. the area in square feet
4. the area in square inches
5. the perimeter of the right-angled trapezium in metres
6. the perimeter in feet
7. the perimeter in inches
##### Formula
A = (a+b)/2*h
P = a+b+ 2 legs
where A is the area
a is a base length
b the other base length
h the height of the trapezium
P the perimeter of the trapezium
##### Conversions
one metre (m) = 3.28083989501312 feet (ft)
one metre (m) = 39.3700787401575 inches (in)
one square metre (m²) = 10.7639104167097 square feet (ft²)
one square metre (m²) = 1550.0031000062 square inches (in²)
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# Combination of Leetcode77.
Java stack all major league2022-08-06 08:24:20
``````class Solution {
public static List<List<Integer>> combine(int n, int k) {
List<List<Integer>> res = new ArrayList<>();
//打个比方吧,从1到6里面选3个数. Basically, the final result sequence is all 3或4或5或6来结尾
for (int i = k; i <= n; i++) {
List<List<Integer>> lists = combineRecur(i, k);
}
return res;
}
/** * 返回所有以mending and containingk个元素的序列 * * @param m * @param k * @return */
public static List<List<Integer>> combineRecur(int m, int k) {
//处理边界情况
List<List<Integer>> result = new ArrayList<>();
if (m == k) {
List<Integer> list = new ArrayList<>();
int i = 1;
while (i <= m) {
i++;
}
return result;
}
if (k == 1) {
List<Integer> list = new ArrayList<>();
return result;
}
if (k == 2) {
//mis already the last element in the sequence,Then there is one more element to follow1到m-1中任意选一个
int i = 1;
while (i <= m-1) {
List<Integer> list = new ArrayList<>();
i++;
}
return result;
}
for (int i = k-1; i < m; i++) {
List<List<Integer>> combine = combineRecur(i, k - 1);
for (List<Integer> list : combine) {
List<Integer> list1 = new ArrayList<>(list);
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Dismiss
Announcing Stack Overflow Documentation
We started with Q&A. Technical documentation is next, and we need your help.
Whether you're a beginner or an experienced developer, you can contribute.
# What's a quaternion rotation?
Is quaternion rotation just a vector with X,Y,Z which the object will rotate towards, and a roll which turns the object on its axis?
Is it that simple?
Meaning if you have X=0, Z=0 and Y=1 the object will face upwards?
And if you have Y=0, Z=0 and X=1 the object will face to the right?
(assuming X right, Y up and Z depth)
-
A quaternion has 4 components, which can be related to an angle θ and an axis vector n. The rotation will make the object rotate about the axis n by an angle θ.
For example, if we have an cube like
`````` ______
|\ 6 \
| \_____\ z
|5 | | : y ^
\ | 4 | \|
\|____| +--> x
``````
Then a rotation of 90° about the axis (x=0, y=0, z=1) will rotate the "5" face from the left to the front.
`````` ______
|\ 6 \
| \_____\ z
|3 | | : x ^
\ | 5 | \|
\|____| y<--+
``````
(Note: This is the axis/angle description of rotation, which is what OP confuses. For how quaternion is applied to rotation, see http://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation)
-
leftways or rightways? – clamp Oct 26 '10 at 11:33
@clamp: Depends on if the system is left-handed or right-handed. – kennytm Oct 26 '10 at 11:39
the OP said "Y up and Z depth". – LarsH Oct 26 '10 at 13:46
@LarsH: Put your head below the z-axis, and view the whole thing in a mirror. :) – kennytm Oct 26 '10 at 13:50
+1 cool ASCII art – slater May 22 '14 at 18:29
A quaternion in general is an extension of a complex number into 4 dimensions. So no, they are not just x, y, and z, and an angle, but they're close. More below...
Quaternions can be used to represent rotation, so they're useful for graphics:
Unit quaternions provide a convenient mathematical notation for representing orientations and rotations of objects in three dimensions. Compared to Euler angles they are simpler to compose and avoid the problem of gimbal lock. Compared to rotation matrices they are more numerically stable and may be more efficient.
So what are the 4 components and how do they relate to the rotation?
The [unit quaternion] point (w,x,y,z) represents a rotation around the axis directed by the vector (x,y,z) by an angle alpha = 2 cos-1 w = 2 sin-1 sqrt(x2+y2+z2).
No... the object will rotate around this `<0,1,0>` vector, i.e. it will rotate around the y axis, turning counterclockwise as seen from above, if your graphics system uses right-hand rotation. (And if we plug in w = sqrt(1 - (0 + 1 + 0)), your unit quaternion is (0,0,1,0), and it will rotate by angle 2 cos-10, = 2 * 90 degrees = 180 degrees or pi radians.)
This will rotate around the vector `<1,0,0>`, the x axis, so it will rotate counterclockwise as seen from the positive x direction (e.g. right). So the top would turn forward (180 degrees, so it would rotate until it faced downward).
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CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- Phoenics (http://www.cfd-online.com/Forums/phoenics/)
- - Aerodynamics (http://www.cfd-online.com/Forums/phoenics/52316-aerodynamics.html)
Bonny Jacob Zachariah February 5, 2009 12:15
Aerodynamics
How to calculate the drag for a rotating wheel?
what the boundary conditions that is to be given while doing it in phoenics software?
Alex February 10, 2009 04:11
Re: Aerodynamics
You can start using the second law of Newton.
Need a clear definition of drag.
If the drag is defined by the shear stress only, you need to know what is the principle direction of the drag.
In Phoenics it is possible to get the value of velocity components which can be used to evaluate the shear stress at the wall.
The second question may depend on the frame of reference and the translation of the rotating wheel. Hope other will correct me if I am wrong
Bonny Jacob Zachariah February 10, 2009 04:18
Re: Aerodynamics
Thank you for your reply. Let me try in the mehtod that you said. This is related to my project work, Aerodynamics of Rotating Wheels, so im working on it. Can you give me a suggestion to do in Phoenics like in which turbulence models can I do that. I have got some papers related to this topic and in all they have take the k-epsilon turbulence model. Any suggestions..?
Alex February 10, 2009 05:43
Re: Aerodynamics
I have an experience on rotating disc. Does it similar? If so, the drag that you have briefly mentioned is normally called windage (let me know if I am wrong). You need to know the windage power in order to approximate how much energy required to rotate the disc at the precribed rotatinonal speed.
If you are considering the rotating disc system, k-e model with wall function formulation has been used at high rotatinonal speed. In some other rotating disc configurations, the turbulence model may requires modification for improving the main flow predictions. In this case, you may try other advanced turbulence models available in PHOENICS. Please let me know if you have manage to compute ones.
All times are GMT -4. The time now is 19:15.
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# Question Video: Drag Force through Fluid Proportions Physics
Which of the following most correctly describes how the drag force exerted by a fluid on an object moving through the fluid varies with the speed at which the object moves through the fluid? [A] The drag force is proportional to the square root of the speed. [B] The drag force is proportional to the square of the speed. (C) The drag force is proportional to the speed. [D] Below a certain speed, the drag force is proportional to the square root of the speed, but above the speed, the drag force is proportional to the speed. [E] Below a certain speed, the drag force is proportional to the speed, but above the speed, the drag force is proportional to the square of the speed.
03:18
### Video Transcript
Which of the following most correctly describes how the drag force exerted by a fluid on an object moving through the fluid varies with the speed at which the object moves through the fluid? (A) The drag force is proportional to the square root of the speed. (B) The drag force is proportional to the square of the speed. (C) The drag force is proportional to the speed. (D) Below a certain speed, the drag force is proportional to the square root of the speed, but above the speed, the drag force is proportional to the speed. And (E) below a certain speed, the drag force is proportional to the speed, but above the speed, the drag force is proportional to the square of the speed.
In this example, we’re thinking about an object that is moving through a fluid. An example of this could be, say, a submarine moving through water. Whenever an object moves through a fluid, there is a drag force on that object. This is a force, we’ll call it 𝐹 sub 𝐷, that opposes the motion of the object. The drag force is caused by friction.
In this example, we want to identify the correct relationship between the drag force on an object moving through a fluid and that object’s speed through the fluid. If we call the speed of an object moving through a fluid 𝑣, then answer option (A) can be summarized as saying the drag force is proportional to the square root of object’s speed.
Option (B) claims that the drag force is proportional to the square of the speed 𝑣. Option (C) says that the drag force is proportional to the speed 𝑣. And option (D) said that below a certain speed, the drag force is proportional to the square root of 𝑣, but above that certain speed, the drag force is proportional to 𝑣. And then of course, we have answer choice (E) to keep in mind as well.
It’s not so unusual to have some sort of practical experience of the drag force, for example, any time a person is riding quickly on a bicycle or, say, traveling in a car with the windows down with their hand out the window. In each of these cases, we experience a drag force, a resistance to forward motion on an object moving through a fluid. It’s hard though to get an intuitive sense for how the drag force relates to the object’s speed mathematically.
It turns out that if an object is moving relatively slowly through a fluid, that is, if the fluid flows smoothly past the object, then under those smooth flow conditions, drag force is indeed proportional to the object speed 𝑣 through the fluid. This ends up not being true though for all speeds 𝑣. If the speed of our object increases enough so that the fluid now flows past the object, not smoothly, but turbulently, then under those conditions, the drag force is proportional to the square of the speed.
Notice something interesting about this. It means that if we’re moving fairly slowly through a fluid and the flow is smooth, it won’t take much additional energy to overcome the drag force, which at that point is proportional to our speed and add to our forward speed. On the other hand, if we’re already moving fast enough that the flow around us is turbulent as we move through a fluid, then to overcome that drag force and increase our speed by some increment takes much more energy. We find then that it takes less energy to speed up at slower speeds, and more energy to speed up at higher speeds.
In any case, this description of the drag force is matched by answer choice (E). Below a certain speed, and that speed depends on both the object and the fluid the object is moving through, drag force is proportional to speed, but above this speed, the drag force is proportional to the square of the speed.
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# How to find spanning tree of a graph that minimizes the maximum edge weight?
Suppose we have a graph G. How can we find a spanning tree that minimizes the maximum weight of all the edges in the tree? I am convinced that by simply finding an MST of G would suffice, but I am having a lot of trouble proving that my idea is actually correct. Can anyone show me a proof sketch or give me some hints as to how to construct the proof? Thanks!
• Try to construct a counterexample. Not sure that one exists, but for a moment I felt that one does. In any case, this is a good proof strategy. – Dave Clarke Jun 4 '12 at 16:39
• en.wikipedia.org/wiki/… – Jukka Suomela Jun 4 '12 at 16:52
• If you know Kruskal’s algorithm for the minimum spanning tree, it is an easy exercise to show that the output of Kruskal’s algorithm is a minimum bottleneck spanning tree. (I think that it is easier than showing that the output of Kruskal’s algorithm is a minimum spanning tree.) – Tsuyoshi Ito Jun 4 '12 at 17:39
• @AdenDong if you solved your own question, feel free to self-answer. – Artem Kaznatcheev Jun 5 '12 at 0:47
• @DaveClarke: "In any case, this is a good proof strategy." -- True. Worked for me. TCS/maths education should focus way more on this back and forth that is inherent to the process. – Raphael Jun 5 '12 at 11:15
Given a graph $G(V, A)$, we know that any spanning tree contains an edge in every cutset. Let $S_{min}^{max}$ and $S$ be the minimax weight spanning tree of $G$ and minimum weight spanning tree of $G$ resp. Any edge $e \in S$ is associated with a cutset $C$. Corresponding to cutset $C$,$S_{min}^{max}$ must also contain an edge, say $e'$. Then if $w(e') < w(e)$, we know that replacing $e$ with $e'$ in $S$ will produce a new spanning tree with lower overall weight, thus contradicting our assumption of optimality of $S$. Hence, we can see that the weight of every edge in $S$ is no greater than the weight of a corresponding edge (obtained from the cutset) in $S_{min}^{max}$. Since, we have compared every edge in $S$ but perhaps possibly may not have compared every edge of $S_{min}^{max}$ in the above process for comparison, the max weight edge in $S_{min}^{max}$ has to be atleast as much as much as the max weight in $S$. Hence we can conclude that $S$ itself produces a minimax weight spanning tree.
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# Relativistic acceleration of a rocket?
I'm trying to solve a physics brain-teaser and getting nowhere fast. My knowledge of special relativity is very basic.
In the problem a rocket is flying linearly at relativistic speeds (say $v>0.9c$), in the absence of external force fields. The rocket loses mass $m$ which is converted to mechanical work $W$ done on the rocket, acc:
$$dW=-\alpha dm$$
Where $\alpha$ is an efficiency coefficient ($\mathrm{Jkg^{-1}}$).
In the absence external force fields all mechanical work is converted to kinetic energy, so:
$$dW=dK$$
Relativistic kinetic energy is given by:
$$K=\frac{mc^2}{\sqrt{1-v^2/c^2}}-mc^2$$
I tried to set up a differential equation as follows:
$$-\alpha dm=\frac{(m+dm)c^2}{\sqrt{1-(v+dv)^2/c^2}}-(m+dm)c^2-\frac{mc^2}{\sqrt{1-v^2/c^2}}+mc^2$$
Even with some reworking and elimination of higher order infinitesimals that's very unyielding and I'm not sure its even correct. In fact it doesn't take the $K$ of the mass $dm$ into account. Probably conservation of momentum is also needed to be take into consideration.
Any help would be much appreciated.
The way to do this is to note that the acceleration $a'$ in the inertial frame of the spectators is related to the proper acceleration in the rest frame of the rocket, $a$, by:
$$a' = \frac{a}{\gamma^3}$$
So just work out the acceleration felt by the observers on the rocket in their rest frame, which is regular Newtonian mechanics, then feed the expression for $a$ into the equation above and solve the (probably exceedingly messy) resulting equation of motion.
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# so confused
Sorry here is the problem 8x-(3x+10)=34
so u want to find x in ur question given,8x-(3x+10)=34getting x terms to a side and numbers to one side8x-3x=34-105x=24x=24/5x=4.8or u can do it as adding -10 on both sides8x-3x+10-10=34-105x=24divide with 5 on both sides5x/5=24/5x=24/5x=4.8
abhinavmartah
14 June 2017
The - in front the brackets affect every term inside:8x-3x-10=34Now you can proceed as usual, add 10 in each side to isolate the terms with x8x-3x=34+105x=44Divide by 5 on each sideX=44/5X=8.8
Raquel C.
14 June 2017
The - gets multiplied inside so it becomes 8x-3x-10=34.now taking x terms 5x-10=34...taking all the numerical to one side....5x=34+10.....5x=44.....to find the value of x...the 5 should be taken to other side where it gets divided....x=44/5
santoshi2209
14 June 2017
1. multiply out the brackets 8x - 3x-10 =342. gather the x values 5x - 10 = 343. gather the whole numbers on oneside and x values on the other side 5x = 34+104. so therefore 5x =445. x= 44/5 OR x= 8.8
amrita_themedic
15 June 2017
Exam tip: you can check the answer by substituting your 'x' value back into the equation to achieve a value of "34". This is useful in exams to do a quick check!
amrita_themedic
15 June 2017
If 8x -(3x + 10) = 34Step 1: Collect like terms: 8x - (3x + 10) = 5x - 10Step 2: 5x - 10 = 34Step 3: Add 10 to both sides. 5x = 44Step 4: Divide both sides by 5 so x = 44/5 = 8.8
Moyo S.
16 June 2017
Remember: Always deal with brackets first.When there is a negative just outside the brackets, it is the same as it being negative 1. So:8x -1(3x+10) = 34To get rid of the brackets, multiply -1 x 3 and -1 by 10. That gives:8x - 3x - 10 = 34Get the x's on one side and the numbers on the other. The x's are already conveniently on the same side. However, you need to move the 10 to the right side, so change the sign from negative to positive:8x - 3x = 34 + 10Subtract on left and add on right:5x = 44You want x alone, without the 5. 5x just means 5 times x. To get rid of multiplication, you need to divide. Divide both sides by 5. 5x ÷ 5 = 44 ÷ 5That will erase the 5 from the left side. On the right side, another way to write 44 ÷ 5 is as the fraction 44 over 5. They mean the same thing. So:x = 44/5That is the fraction 44 over 5. Can we simplify it? Not really, since you can't divide an even number like 44 by 5. So, the answer you have is complete. P.S. You only need to write it as a decimal if the question asks you to (for example, to give your answer to 1 decimal point).
John B.
19 June 2017
Hi there donalana!Algebraic questions at your stage can be simplified if one remembers BMDAS!I like to remember it as Bad Moms Dont Act Silly!The initials of this mnemonic rule give you the operations and their order in which to execute them. This means the order of operations is:Brackets, multiply, divide, add, substract.Remember the order of them with BMDASKnowing this we can breeze through the problem as followsBracketsMultiply out the negative sign outside the bracket and you will get:8x-3x-10=34remember to multiply all the terms inside the bracket by -1! This includes the term '10'. 2. MultiplyAll the terms are multiplied!3. DivideThere is nothing to divide at this stage!4. AddNothing to do here!5. Substract8x-3x =5xTherefore we are left with 5x -10 = 34Let's pass the '-10' term to the other side by adding ten to the RHS5x = 44Divide by 5 both sides and we get...x=44/5 OR x=8.8I hope this helped! :)
perezdebartolomecoaching
27 June 2017
The minus symbol applies to everything inside of the brackets so to make the equation look easier to simplify:8x - 3x -+ 10 = 34;+- = -;8x -3x -10 =34;5x - 10 = 34;(+10) 5x - 10 = 34 (+10)5x = 44:(/5) 5x = 44 (/5)x= 44/5 You can do whatever you want when you apply functions (e.g. squaring or timesing by a number) to both sides. In this case I added 10 to both side to get the 5x on its own. Then to get x on its own, I then divided both sides by the 5.
hgp202
01 July 2017
8x - (3x+10)=34=> 8x -3x -10 = 34=> 5x = 44=> x= 44/5
ms181920
01 July 2017
Hello donalana, 8x-(3x+10)=34For the problem above, you BODMAS as we collect the like terms.We start off by opening up the brackets.8x-3x-10=34collecting like terms8x-3x=34+10Simplifying each side5x =44x=44/5 = 8.8
euticus19
06 July 2017
Hi donalanaIn the above problem we use BODMAS to work it out. Now let's get started by writing the equestion. 8x-(3x+10)=34Now we start by opening the brackets -(3x+10)-3x-10 ... the sign change because we multiple the numbers in the bracket by - 1Now we combine the math again 8x-3x -10 =34We but like terms together fast before going into the next step of BODMAS 8x-3x =34+10....now - 10 has changed the sign because we closed it to the other side of the equal sigh and became +108x-3x =5x34+10=445x=44....we divide both sides by 5 to get the value of x5x/5=44/5x=8.8There you got it as simple as that!
dykes
17 August 2017
x=8.8
evelinmish12
21 August 2017
8x-(3x+10)=341) 8x -3x-10=34 ( we multiplied the minus with every term inside the bracket)2) 8x-3x =10+34 ( we separated the terms ) 3) 5x=24 ( we did the adding and substraction of the terms) 4) x=24 /5 ( we divided with X coeficient )so X= 24/5
Theodoros T.
02 September 2017
8x-3x-10=345x-10=345x=24x=24/5
aiesha
27 December 2017
Just simply we need to solve for x as8x-(3x+10)=348x-3x-10=345x=34+105x=44x=8.8
rishi1404
05 June 2018
2(x – 3) – 8 < -4 AND 19 - 3x < 10
isaiah
21 October 2021
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## Calculation Of Fuel Oil Palm Factory Boilers
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Marine Boiler Manufacturer . Lecture 27: Principles of Burner Design Contents: How does – nptel The release of potential energy of fuel by combustion with air requires several stages, namely. produces a flame which transfers thermal energy to furnace and charge In atomization, compressed air or steam is the atomizing fluid.Combustion systems for gaseous and liquid fuel firing Combustion […]
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### How to Calculate Boiler Efficiency, Boiler Efficiency
function of boiler in palm oil mill – Industrial Boiler Supplier
### Diesel Boiler Fuel Consumption Per Hour - Diesel Boiler
Oil Pumps, Piping & Fuel Oil Treatment 37,38 Finding Oil Supply Line Leaks 39 Useful Formulas 40 Oil Fired Water Heaters/Boilers 41 Thermostats 42 Limit Controls 43 Primary Controls 44 Primary Controls Wiring Color Code 45
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# Pre-Calculus Forum
Pre-Calculus Help Forum: Pre-calculus does not involve calculus, but explores topics that will be applied in calculus
1. ### This forum is for PRE-Calculus. READ BEFORE POSTING
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2. ### Read this before posting in this subforum.
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1. ### Epsilon nr 2
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2. ### Need Help Simplifying a Few Problems! <Imaginary Numbers>
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3. ### Dose this integration seem right
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4. ### Cosine funtion expression
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6. ### Factoring and rationalizing
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7. ### How is this a limit
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Feb 11th 2013, 09:24 PM
8. ### Function Question
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9. ### Geometric Sequences
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Feb 7th 2013, 09:05 PM
10. ### About the one functional equation
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11. ### Rationalizing the numerator and complex factoring
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12. ### Dividing by variable and taking square root of inequalities
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13. ### Limits - find parameters γ, δ
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Feb 5th 2013, 12:08 PM
14. ### Solve: tan2x + tan x = 0
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Feb 5th 2013, 08:58 AM
15. ### Limits - find parameters α , β
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Feb 5th 2013, 06:23 AM
16. ### Pre-calc amoeba doubling every 4 days
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# undefined
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Use this control to limit the display of threads to those newer than the specified time frame.
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Written by Bryce Ringwood
Saturday, 27 April 2013 18:15
### Introduction
In this article we look at AC circuits and electrical impedance. If you can't follow the maths - don't worry too much, rather try to get an overall understanding from the text. At the end of the article, I'm going to demonstrate how to calculate the impedance of a circuit using a calculator.
We haven't really discussed "Circuits" in any organised way, except for an article in "Experimenter's Corner" where Ohm's law is made the subject of a simple 'C' programming excercise. There is also a circuit calculator for Ohm's law in the "Calcs" section of this web site.
These two articles summarize DC circuits and if we were to just use "pure" resistors and RMS values of AC voltage, we would not have to worry any further. Indeed, in the world of light-bulbs (prior to compact fluorescent) and electric kettles, we don't worry. We just use 230 volts (AC, RMS), talk of 2kW elements and carry on. We can, because to all intents and purposes these things are pure resistances and Ohm's law applies.
Radio receivers contain a number of AC circuits. For example, the heater supplies to the valves, the audio circuitry and the RF and IF circuits all contain examples of AC (Alternating Current) . When it comes to DC circuitry, we are clear as to what to do. With AC, a mental fog sets in.
What we would like is a kind of Ohm's Law for AC circuits in which there are combinations of resistors, capacitors and Inductors. Fortunately, we can provide exactly that, but it requires complex number arithmetic. This is why I provided a complex number calculator and (possibly) is why a number of pocket calculators (e.g. HP 35s) have a complex number feature.( Please take a look at the maths review articles to get you back up to speed with complex numbers.)
Voltages and currents in AC circuits have a magnitude and a phase as illustrated below:
The amplitude of the voltages is 2.5 Volts, and the phase difference is $\pi \over 4$ or 45 degrees.
For convenience, we rather represent voltage and current as complex quantities, for example voltage:
\displaystyle{{\bf V} = |{\bf V}|e^{j(\omega t + \phi_{voltage})}=a + jb}
and current, similarly:
\displaystyle{{\bf I}=|{\bf I}|e^{j(\omega t +\phi_{current}) }}
where:
\displaystyle{\omega = 2 \pi f}
$\displaystyle{f}$ is the AC frequency, and $\displaystyle{\phi_{voltage}}$ and $\displaystyle{\phi_{current}}$ are the phase angles for voltage and current. Complex quantities are represented in bold face.
Finally, Ohm's Law for AC circuits uses impedance. Impedance is the AC circuit equivalent of resistance in DC circuits. As may be expected, it is best represented complex quantity:
\displaystyle{\bf V = I \times Z}
where $\bf Z$ is the impedance. A simple resistor has an impedance of $\displaystyle{R + j0}$ - the value marked on the body. You will hear people talk of impedance when thay are referring to resistance. It can be confusing, but don't worry.
Why did I put the word "represented" in italics? The answer is that in the real world we are dealing with phase angles and magnitudes. We could equally well have represented these as mathematical quantities called vectors, but they require unfamiliar arithmetic operations that are not available on a calculator.
### Inductive Reactance
An inductor consists of a number of turns of wire around a core of soft iron, ferrite, or even – air. The core has a property of magnetic permeability. When we apply a DC voltage to the inductor, the core becomes magnetized by virtue of the current flowing through the wire. The wire has some resistance, otherwise the current through the coil would be infinite and everything would get hot and melt. When we apply an AC voltage at a low frequency, the magnetic field generated in the core simply rises and falls in sympathy with the AC voltage.
As the frequency increases, the collapsing magnetic field in the core induces an opposing current, so that the current in the core can't keep pace with the applied voltage. The rise and fall of the current in the inductor becomes out of step with the applied voltage, and lags behind it by a certain amount. The voltage and current are out of phase, with the current lagging behind the voltage.
In a “pure” inductor that has zero resistance and no capacitance, the current would lag behind the voltage by 90 degrees. BUT we don't work in degrees, so we rather say $\displaystyle {\pi} \over 2$ radians.
A Brief Analysis
In the article on inductances, we saw that $\displaystyle{V=L{dI \over dt}}$.
Now we can write $\displaystyle{{\bf V} = |{\bf V}| e^{(j \omega t)} = L{d{\bf I} \over dt}}$ and solve, giving us:
$\displaystyle{{{{|\bf V}|e^{(j \omega t)}} \over{j \omega}}={ \bf I} L}$ So $\displaystyle{{\bf V} = j \omega L {\bf I}}$
and $\displaystyle{{\bf Z} = 0 + j \omega L}$ The magnitude of $\bf Z$ is the reactance $\displaystyle{{X_L}=\omega L}$
The axes represent the complex plane, so that inductive reactance is a purely complex quantity. To put it another way, the impedance of an inductor is a pure complex quantity.
### Capacitive Reactance
As we saw in the section on capacitors, they behave a little bit like batteries. As you charge them, the current diminishes and the voltage rises. They do the opposite of an inductor, in that the current leads the voltage across the capacitor plates.
Brief Analysis
In a circuit containing capacitance:
$\displaystyle{{\bf I}=C{d{\bf V}\over dt}}$, where $\displaystyle {{\bf I}={|\bf I|}e^{j \omega t}}$ solving
$\displaystyle{{|\bf I|}{e^{j \omega t}\over {j \omega }}=C{\bf V}}$, rearranging and substituting
$\displaystyle{{\bf V}={ ({1\over{j \omega C}}){\bf I}}}$ giving an impedance $\displaystyle{{\bf Z}= {0 -{({j \over {\omega C}})}}}$ . The capacitative reactance, $\displaystyle{{X_C}={1\over{\omega C}}}$
### Circuits Containing combinations of R,L and C
Arrmed with a complex number calculator, the analysis of AC circuits becomes as easy as it is for DC circuits.
Components in series:
$\displaystyle{\bf Z_{total}= Z_1+Z_2+Z_3+ .....}$
Components in parallel
$\displaystyle{\bf{1 \over Z_{total}}={1 \over Z_1}+{1 \over Z_2}+{1 \over Z_3} + ...}$
and $\displaystyle{\bf Z_{total}= {{{Z_1}{Z_2}}\over {Z_1 + Z_2}}}$ for two elements in parallel.
### Let's Check it out !
Even if you haven't quite followed the maths up to now - try to master this little bit.
Example Ladder Network - Passive Components
The example is from the HP65 EE Pak1 "Impedance of a Ladder Network" (EE-1-04A) Hewlett Packard, 1974.
The frequency is 4.0MHz
Element Formula Impedance New Impedance 50 Ohms $R + j0.0$ 50 + j0.0 2400pF $0.0 -{ j \over \omega C}$ 0 - j16.578640 4.95252 - j14.936 2.56µH $0.0 + j \omega L$ 0 + j64.33982 4.95252 + j49.4033 796pF $0.0 -{ j \over \omega C}$ 0 - j 49.9855 497.621 + j8.512
Having worked out the impedances for each element, here are the HP 35s keystrokes:
50 ENTER ENTER ENTER // Fill the stack with the first value Z1
0i16.57864 +/- // Enter Z2 The +/- key used to be the CHS key
× // Z1×Z2
x ⇔ y // Interchange the x and y stack values
last x // Z2 (The complex no. calculator has no last x - you will have to
// store and recall.)
+ // Z1 + Z2
÷ // Z1×Z2 /(Z1 + Z2) or Z1||Z2
(result) 4.95252i14.936 // Combined impedance of R and 1st
ENTER ENTER ENTER // Fill stack with new value of Z1
0i64.33982 // Z2
+ // Z1 + Z2
(result)4.95252i49.4033 // Combined impedance of 1st R L and C
ENTER ENTER ENTER // Fill stack with new value of Z1
0i49.9855 +/- // New value of Z2
× // Z1×Z2
x ⇔ y
last x // Z2
+ // Z1 + Z2
÷ // Z1×Z2 / (Z1 + Z2)
(result)497.612i8.512 // Done!
abs 497.69 (Magnitude of Z)
arg 0.97 (Phase angle in degrees)(Nearly a pure resistance of 500 Ohms at 4 MHz)
I could not find a Rectangular to Polar key on The HP, but you can use an rθ display mode.
Admittedly - its still a lot of PT.. Maybe I will port the "Impedance of a Ladder Network" program by HP to the calcs section.
### Summary
This feature is a brief introduction to AC Circuit theory. Here are the main points:
• AC Voltage, Current and Impedance have magnitude and phase
• Impedance is the ratio of Voltage to Current in an AC circuit. It is measured in Ohms
• Voltage, Current and Impedance can conveniently be expressed as complex numbers
• The operator j in electrical engineering is the same as the mathematical operator i
• The article demonstrates the use of a calculator to evaluate the impedance of passive circuit elements in a ladder network
• A complex number calculator reduces the complexity of AC calculations to the level of DC calculations. No more horrendous formulae.
### References
Nelkon M and Parker.P, "Advanced Level Physics",Heinemann,1974
Horowitz P and Hill W,"The Art of Electronics",Cambridge University Press,1988 (You might find this a bit confusing. Its the only confusing section in the whole book.)
Also see Wikipedia's "Electrical Impedance" article. It takes fewer liberties than I have taken.
Last Updated on Tuesday, 11 June 2013 13:33
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# Lesson Notes By Weeks and Term - Junior Secondary School 1
HISTORICAL DEVELOPMENT OF COMPUTER
SUBJECT: ICT
CLASS: JSS 1
DATE:
TERM: 1st TERM
WEEK 2
TOPIC: HISTORICAL DEVELOPMENT OF COMPUTER
Historical development of computer can be categorized into three. These include:
1. Mechanical device
2. Electro-mechanical device
3. Electronic devices and modern computer
MECHANICAL DEVICE:
These are the devices that involve the use of physical forces to operate them. The following are examples of mechanical devices in order they were invented and used.
1. Abacus: Abacus is known to be the first mechanical calculating device which was used to be performed arithmetic calculation such as addition and subtraction easily and speedily. This device was first develop by the Egyptians and improved upon by the Chinese educationists. Chinese were the first to use the abacus.
2. Astrolabe: This mechanical calculating instrument also came into use about 2000years ago for finding direction and in astronomy. It consists of two flat circular discs, usually made of brass and ranges from about 7.5cm to 25cm in diameter.
3. Slide rule: This mechanical device was used by engineers and scientists to simplify multiplication, division and finding root of numbers. The slide rule is an analog device and was invented by William Oughtred. It consists of a ruler designed with scales and a sliding central strip.
4. Pascaline: This machine is a mechanical device that was developed by Blaise Pascal a French scientist. It invented in the year 1642.It was called numerical wheel calculator to carry out computation.
5. Reckoner or Leibnz Calculator: In the year 1671, a German mathematician, Gottfried Leibniz modified the Pascal calculator and he developed a machine which could perform various calculation based on multiplication and division as well.
• ELECTRO-MECHANICAL DEVICES:
These are the machines were developed to make counting and calculations easier and faster. The following are examples of electro-mechanical devices.
1. John Napier’s bones (1550-1617): John Napier’s of Scotland invented a calculating device, in the year 1617 called the Napier Bones or Napier rods: The rods were mainly multiplication tables written on sticks of wood or bone .The rods were also used in taking square roots and cube roots.
2. Blasier Pascal’s machine (1623-1662): Blasier Pascal was a French mathematician who invented electron-mechanical device called Pascal’s machine that was used for calculation .It was first digital calculating machine in 1642.
3. Gottfried Wilhelm von Leibniz’s machine (1646-1716): Gottfried Wilhelm was a German Philosopher and mathematician who was the discovery of fundamental principle of calculus and also invented a calculating machine capable of multiplying, dividing and calculating square roots.
4. Joseph Marie Jacquard’s loom (1752-1834): Joseph Jacquard was a French silk weaver and inventor that invented Jacquard automatic loom mechanism using punched cards that controlled the weaving of the cloth and any design pattern could be obtained automatically.
5. Charles Babbage’s machine(1792- 1871) : Charles Babbage’s was a British mathematician and inventor .He invented machine which is considered to be a modern computer today .He built machine called difference engine and analytical engine .He was regarded as the father of modern computer.
6. Philip Emeagwali: Philip Emeagwali was born in Akure, Nigeria in April, 1967.He works as a computer scientist which contributed greatly to internet technology .He was regarded as a father of internet.
• ELECTRONIC DEVICES AND MODERN COMPUTER:
These are devices that form the modern computer .These are electronic devices that use electric current to power and operate. They include the following:
1. Herman Hollerith punched card (1860-1929): Hollerith was an America statistician who developed a mechanical tabulator on punched card and also modern computer.
2. John Von Neumann’s machine (1903-1957) : John von Neumann was a Hungarian born mathematician
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Algebra
Equations
Inequalities
Graphs
Numbers
Calculus
Matrices
Tutorials
Enter the polynomial inequalities you want to plot, in terms of the variables x and y, set the limits and click the Plot button.
Inequalities Colors Options
Grid lines
x axis
y axis
Line(s) width
Title
Legend
Display
Intersection of regions
Union of regions
Show all regions
Variables Independent: runs from to Dependent: runs from to
# Using equations and inequalitie to solve word problems
In chapter 2 we developed the techniques for solving equations. Equations arise as a means of solving verbal or word problems. Since word problems are a large part of algebra, it is necessary to develop techniques of solving them. In this chapter we will concentrate on ways of outlining and solving verbal problems.
## FROM WORDS TO ALGEBRA
### OBJECTIVES
Upon completing this section you should be able to:
1. Change a word phrase to an algebraic expression.
2. Express a relationship between two or more unknowns in a given statement by using one unknown.
The primary task when attempting to solve a word problem is one of translation. The problem is written in one language and must be translated into another- the language of algebra. This translation process must be precise if we are to be successful in solving the problem.
Example 1 Statement: A number increased by seven is twelve.
Algebraic translation: x + 7 = 12
Do these two statements have the same meaning? Does the algebraic equation make the same statement as the English sentence? If so, we have correctly translated from one language to another and can easily solve for the missing number.
Before we begin to outline completed sentences or solve problems we need to review the meaning of certain phrases.
Example 2 Write an algebraic expression for: A certain number decreased by four.
Solution
If we let x represent "a certain number," and recognize that "decreased by" is subtraction, then our answer is x - 4.
Note that a phrase will yield only an algebraic expression and not a complete equation. It is impossible to solve for the unknown without a complete equation.
There are key words that give clues to the operations to be used. Addition-words such as "increased by," "sum," "more than," "greater than," "total." Subtraction-words such as "decreased by," "less than," "difference," "diminished by." Multiplication-words such as "times," "of," "product," "twice." Division-words such as "quotient," "divided by."
Example 3 Write an algebraic expression for: Five times a certain number.
Solution
If x represents "a certain number," then the expression would be 5x.
Example 4 Write an algebraic expression for: Five more than a certain numoer.
Solution
If x represents "a certain number," then "more than" means addition, the expression is x + 5.
Example 5 Write an algebraic expression for: Seven more than twice a certain number.
Solution
Again, allowing x to represent the unknown number, we have 2x + 7.
Example 6 Write an algebraic expression for: 5% of a given number.
Solution
First write 5% as the decimal .05. If x represents the "given number," then the expression would be .05x .
Example 7 Write an algebraic expression for: The value in cents of d dimes.
Solution
The value of a dime is 10 cents. Therefore to indicate the value of d dimes multiply 10 by d, obtaining 10d.
It may not always be possible to relate the unknowns in a problem using only one unknown. You will be able to do so, however, with all problems in this chapter.
When more than one unknown number is involved in a problem, we try to outline it in such a way that all unknowns are expressed in terms of one unknown.
Example 8 The Sears Tower in Chicago is eight stories taller than the Empire State Building in New York. Write algebraic expressions for the height of each building using one unknown.
Solution
From the information given we do not know how many stories either building has. We will choose one of the buildings and represent the number of stories by x.
Let x = number of stories in the Empire State Building.
Then x + 8 = number of stories in the Sears Tower.
It may not always be possible to relate the unknowns in a problem using only one unknown. You will be able to do so, however, with all problems in this chapter.
We could also choose x to represent the height of the Sears Tower. Then x - 8 would represent the height of the Empire State Building.
Example 9 The length of a rectangle is three meters more than the width. Write expressions for the length and width using one unknown.
Solution
Let x = width of the rectangle.
Then x + 3 = length of the rectangle.
It is sometimes helpful to use a diagram to see the relationships between unknowns.
Example 10 The width of a rectangle is one-fourth the length. Write expressions for the length and width using one unknown.
Solution
Let x = width.
Then 4x = length.
Note that if we were to let x represent the length, we would have a fraction representing the width. This certainly would not be wrong but we can avoid the use of fractions by careful selection.
Example 11 The sum of two numbers is 20. Write expressions for both numbers using one unknown.
Solution
Let x = first number.
Then 20 - x = second number.
We could also let x = second number. What expression would represent the first number?
Example 12 Express algebraically the relationship of the unknown in this sentence: A certain number is four more than a second number and is three less than a third number.
Solution
To express these three numbers algebraically first decide which will be represented by x.
For instance, if x represents the first number, we have the following:
Ask yourself, "Is the first number four more than the second?" and "Is the first number three less than the third?" If the answers are "yes," you have correctly outlined the sentence.
Which number to represent by x is arbitrary, but re-read the relationships several times to see if one option is preferable over another.
Now suppose, in the same example, we decide to allow x to represent the second number.
Ask the same questions again, "Is the first number four more than the second?" and "Is the first number three less than the third?" This outline is also correct.
Represent the three numbers if x represents the third number.
## SOLVING WORD PROBLEMS
### OBJECTIVES
Upon completing this section you should be able to:
1. Correctly translate a word problem to an algebraic equation.
2. Solve the equation and find the solution to the problem.
In the preceding exercises we outlined relationships between unknowns within a statement. If we are to solve a problem and find the unknown numbers, there must be within the problem a sentence that yields an equation.
Example 1 The Sears Tower in Chicago is eight stories taller than the Empire State Building in New York. If the total number of stories in both buildings is 212, find the number of stories in each building.
Solution
From example 8 in the previous section we have
x = number of stories in the Empire State Building
x + 8 = number of stories in the Sears Tower.
This time, however, we have the added statement that the total number of stories is 212.
Thus, we can write the equation
x + (x + 8) = 212.
Solving this equation gives
2x = 204
x = 102.
What does x represent?
Notice that we have found only the number of stories in the Empire State Building. The question asked us to find the number of stories in each building. Thus, to find the number of stories in the Sears Tower we must substitute 102 for x in the expression x + 8, obtaining
x + 8 = (102) + 8 = 110.
• number of stories in the Empire State Building = 102
• number of stories in the Sears Tower =110.
Always re-read the problem to make sure you have answered the question. Get in the habit of always summarizing your answers. Checking your answers is one of the most important parts of the solution.
To check the answers, 110 is eight more than 102 and the sum of 110 and 102 is 212.
Example 2 The length of a rectangle is three meters more than the width. Find the length and width if the perimeter of the rectangle is 26 meters.
Solution
Since the perimeter is equal to the sum of twice the length and twice the width, or P = 2l + 2w, we can write the equation
2(x) + 2(x + 3) = 26.
Solving, we obtain
Thus, the width is 5 meters and the length is 8 meters.
Check: The length (8) is three more than the width (5), and the perimeter is 2(8) + 2(5) = 16 + 10 = 26.
The perimeter is the distance around the figure. Thus, two sides of the rectangle measure x and the other two sides measure x + 3. Again, the check is very important.
Example 3 he width of a rectangle is one-fourth the length. If the perimeter is 200 centimeters, find the length.
Solution
Again, the perimeter is twice the width plus twice the length. Therefore,
The problem asks lor the length only. Thus, 4x = 4(20) = 80. The length is 80 cm.
Check: The width (20) is one fourth the length (80). Also, the perimeter is 2(20) + 2(80) = 40 + 160 = 200.
This particular way of stating the unknowns avoids fractions. What does x represent? Try this problem again using x = length, x/4 = width. Which way is easier?
Example 4 The sum of two numbers is 20. Their difference is 4. Find the numbers.
Solution
If the difference of the two numbers is 4, write the equation x - (20 - x) = 4.
Solving, we obtain
The two numbers are 8 and 12.
Check: The sum of 8 and 12 is 20. The difference of 8 and 12 is 4.
We could also write the equation (20 - x) - x = 4 since we don't know which number is larger. Solve the above equation.
Example 5 A certain number is four more than a second number and three less than a third number. Find the number if their sum is 23.
Solution
In this problem we are asked to find the numbers and are given a statement about their sum.
If we let x represent the first number, we have
The statement "their sum is 23" gives the equation
x + (x - 4) + (x + 3) = 23.
If we now solve the equation, we obtain
Leaving the answer as x = 8 would not be a solution to the problem. We are asked to "find the numbers." The answer must be
We check the problem by noting that the first number (8) is four more than the second (4) and is three less than the third (11) and that their sum is 23.
Again, we could just as well let x represent the second or third number. The important thing to remember is to label each algebraic expression with the number that it represents.
Do not just solve the equation for x and think you have solved the problem. Check to see what x represents and reread the problem to see what it is asking for.
We have followed five basic steps in solving the above examples. These steps should be observed when solving any word problem.
1. Write expressions for the unknowns.
2. Write an equation that relates the unknowns to each other.
3. Solve the equation.
4. Make sure you have answered the question.
5. Check your answers to make sure they agree with the original problem.
Make sure you know these five important steps before working the exercise sets.
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• 02-04-2005
cam123666
help w/ farenheit to celcius converter prog please :)
i have been trying to make a farenheit to celcius converter for about to days ( ya sorry its a big task for me i have only been coding for a day)
and i can get the loop and evrything to work but for some reason i cannot get the far temp to conv. to celcius . Say i was to enter 100 degrees far. i would get a 68 degrees cel. its like it is not reading the 5/9 before the far-32 any help would be appr. and here is my code
[code]
// Farenhiet to Celcius converter
#include <iostream>
using namespace std;
int main()
{
int a;
int b = 1;
do {
float far;
cout<<"Please enter youre farenheit amount to be converted to Celcius: ";
cin>>far;
cin.ignore();
float cel;
cel = (5/9)*(far-32);
cout<< far <<" degrees farneheit is: " << cel << " In Celcius\n";
cout<<"do you want to repeat?( 1 is no any other number is yes) ";
cin>>a;
cin.ignore();
}while ( a != b );
cin.get();
}
• 02-04-2005
Salem
Nice attempt at code tags - shame you didn't review the post as well, then you'd see it was still a mess.
> 5 / 9
Is always 0 in C++
If you want a floating point division, then try 5.0 / 9.0
• 02-04-2005
cam123666
ok sorry
but ne ways i still try it and i get a 8.12753e-044 on everything i try to enter
Is there something wrong with the formula i am using or what ? I do not understand why it is doing this and anyhelp will be appreciated
oops sorry i made a simple coding mistake ty for the help it works now
sorry to bother
• 02-04-2005
Kaelin
Code:
```#include <iostream> using namespace std; int main() { int a; int b = 1; do { float far = 0.0F; cout<<"Please enter youre farenheit amount to be converted to Celcius: "; cin>>far; cin.ignore(); float cel = 0.0F; cel = (5.0 / 9.0)*(far-32); cout<< far <<" degrees farneheit is: " << cel << " In Celcius\n"; cout<<"do you want to repeat?( 1 is no any other number is yes) "; cin>>a; cin.ignore(); }while ( a != b ); cin.get(); }```
That should work
• 02-04-2005
quzah
Quote:
Originally Posted by Salem
Nice attempt at code tags - shame you didn't review the post as well, then you'd see it was still a mess.
> 5 / 9
Is always 0 in C++
*tries resisting* I can'... it's too much...
Bwhahah.
Heh... well technically, that's in C so... ;)
Quzah.
• 02-04-2005
Krak
Quote:
Originally Posted by quzah
*tries resisting* I can'... it's too much...
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PHYS-313 / 5 credits
Teacher: Savona Vincenzo
Language: English
## Summary
The objective of this course is to familiarize the student with the concepts, methods and consequences of quantum physics.
## Content
1. A bit of history: the crisis of classical physics
2. The Stern and Garlach experiment: quantum states and spin 1/2
3. The axioms of quantum physics: state vectors, operators, measurement, representations
4. Continuous degrees of freedom: translation operator and canonical quantization
5. Time evolution: Schrödinger's equation and Heisenberg's point of view
6. Some simple problems in one dimension
7. Central potentials, angular momentum and hydrogen atom
8. Composite systems: entanglement and Bell's inequalities
## Keywords
Quantum mechanics, Schrödinger equation, Heisenberg uncertainty principle, wave function, harmonic oscillator, hydrogen atom, spin, entanglement
## Required courses
Basic physics and mathematics undergraduate courses
## Important concepts to start the course
Strong working knowledge of calculus and linear algebra (covered in basic math courses).
## Learning Outcomes
By the end of the course, the student must be able to:
• Explain the difference between classical and quantum physics
• Compare Schrödinger's and Heisenberg's viewpoints on quantum physics
• Derive Heisenberg's uncertainty principle
• Solve the quantum harmonic oscillator with the ladder operator method
• Interpret the measurement process in quantum physics
• Solve Schroendinger's equation for problems in 1,2 and 3 dimensions
• Characterize the amount of entanglement in a two-spin system
• Contextualise the postulates of quantum physics
## Teaching methods
Ex cathedra. Exercises prepared in class.
## Expected student activities
Students are expected to regularly attend the theory lectures and the exercise lectures. They are also expected to complete the exercises that are given on a weekly basis, as well as regularly study the learning material offered by the professor (lecture notes, exercises solutions etc).
Written exam
## Bibliography
The key reference is :
1. "Modern Quantum Mechanics" (2nd edition), J.J. Sakurai, J. Napolitano (Cambridge University Press, 2017)
Other books can be occasionally consulted, most notably
2. "Mécanique Quantique I-II", Cohen-Tannoudji, Diu, Lahoë (Hermann) [Also available in English]
## Notes/Handbook
Lecture notes will be given at the beginning of the course
## In the programs
• Semester: Fall
• Exam form: Written (winter session)
• Subject examined: Quantum physics I
• Lecture: 3 Hour(s) per week x 14 weeks
• Exercises: 2 Hour(s) per week x 14 weeks
• Type: mandatory
• Semester: Fall
• Exam form: Written (winter session)
• Subject examined: Quantum physics I
• Lecture: 3 Hour(s) per week x 14 weeks
• Exercises: 2 Hour(s) per week x 14 weeks
• Type: optional
• Semester: Fall
• Exam form: Written (winter session)
• Subject examined: Quantum physics I
• Lecture: 3 Hour(s) per week x 14 weeks
• Exercises: 2 Hour(s) per week x 14 weeks
• Type: optional
• Semester: Fall
• Exam form: Written (winter session)
• Subject examined: Quantum physics I
• Lecture: 3 Hour(s) per week x 14 weeks
• Exercises: 2 Hour(s) per week x 14 weeks
• Type: optional
• Semester: Fall
• Exam form: Written (winter session)
• Subject examined: Quantum physics I
• Lecture: 3 Hour(s) per week x 14 weeks
• Exercises: 2 Hour(s) per week x 14 weeks
• Type: optional
## Related courses
Results from graphsearch.epfl.ch.
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# 2317. Maximum XOR After Operations
• Time: $O(n)$
• Space: $O(1)$
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 class Solution { public: int maximumXOR(vector& nums) { // 1. nums[i] & (nums[i] ^ x) enables you to turn 1-bit to 0-bit from // nums[i] since x is arbitrary. // 2. The i-th bit of the XOR of all the elements is 1 if the i-th bit is 1 // for an odd number of elements. // 3. Therefore, the question is equivalent to: if you can convert any digit // from 1 to 0 for any number, what is the max for XOR(nums[i]). // 4. The max we can get is of course to make every digit of the answer to // be 1 if possible // 5. Therefore, OR(nums[i]) is an approach. return reduce(begin(nums), end(nums), 0, bit_or()); } };
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 class Solution { public int maximumXOR(int[] nums) { // 1. nums[i] & (nums[i] ^ x) enables you to turn 1-bit to 0-bit from // nums[i] since x is arbitrary. // 2. The i-th bit of the XOR of all the elements is 1 if the i-th bit is 1 // for an odd number of elements. // 3. Therefore, the question is equivalent to: if you can convert any digit // from 1 to 0 for any number, what is the max for XOR(nums[i]). // 4. The max we can get is of course to make every digit of the answer to // be 1 if possible // 5. Therefore, OR(nums[i]) is an approach. int ans = 0; for (final int num : nums) ans |= num; return ans; } }
1 2 3 4 5 6 7 8 9 10 11 12 class Solution: def maximumXOR(self, nums: List[int]) -> int: # 1. nums[i] & (nums[i] ^ x) enables you to turn 1-bit to 0-bit from # nums[i] since x is arbitrary. # 2. The i-th bit of the XOR of all the elements is 1 if the i-th bit is 1 # for an odd number of elements. # 3. Therefore, the question is equivalent to: if you can convert any digit # from 1 to 0 for any number, what is the max for XOR(nums[i]). # 4. The max we can get is of course to make every digit of the answer to # be 1 if possible # 5. Therefore, OR(nums[i]) is an approach. return functools.reduce(operator.ior, nums)
| 657
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https://docs.juliahub.com/General/Fortuna/stable/Reliability%20Problems/SensitivityProblems/
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# Sensitivity Problems
## Defining Sensitivity Problems
In general, 4 main "items" are always need to fully define a sensitivity problem and successfully solve it to find the associated sensitivity vectors of probability of failure $\vec{\nabla}_{\vec{\Theta}} P_{f}$ and reliability index $\vec{\nabla}_{\vec{\Theta}} \beta$ with respect to limit state function's parameters $\vec{\Theta}$:
ItemDescription
$\vec{X}$Random vector with correlated non-normal marginals
$\rho^{X}$Correlation matrix
$g(\vec{X}, \vec{\Theta})$Limit state function
$\vec{\theta}$Parameters of limit state function
Fortuna.jl package uses these 4 "items" to fully define sensitivity problems using a custom SensitivityProblem() type as shown in the example below.
# Define random vector:
M₁ = randomvariable("Normal", "M", [250, 250 * 0.3])
M₂ = randomvariable("Normal", "M", [125, 125 * 0.3])
P = randomvariable("Gumbel", "M", [2500, 2500 * 0.2])
Y = randomvariable("Weibull", "M", [40000, 40000 * 0.1])
X = [M₁, M₂, P, Y]
# Define correlation matrix:
ρˣ = [1 0.5 0.3 0; 0.5 1 0.3 0; 0.3 0.3 1 0; 0 0 0 1]
# Define limit state function:
g(x::Vector, θ::Vector) = 1 - x[1] / (θ[1] * x[4]) - x[2] / (θ[2] * x[4]) - (x[3] / (θ[3] * x[4])) ^ 2
# Define parameters of limit state function:
s₁ = 0.030
s₂ = 0.015
a = 0.190
θ = [s₁, s₂, a]
# Define sensitivity problem:
Problem = SensitivityProblem(X, ρˣ, g, θ)
## Solving Sensitivity Problems
After defining the sensitivity problem, Fortuna.jl allows to easily solve it using a single solve() function as shown in the example below.
# Perform reliability analysis using improved Hasofer-Lind-Rackwitz-Fiessler (iHLRF) method:
Solution = solve(Problem)
println("∇PoF = $(Solution.∇PoF)") println("∇β =$(Solution.∇β)")
∇PoF = [-0.7013033794639277, -1.4026067589278555, -0.17655047916385316]
∇β = [36.77264131858809, 73.54528263717619, 9.257373677394588]
## API
Fortuna.solveMethod
solve(Problem::SensitivityProblem)
Function used to solve sensitivity problems.
Fortuna.SensitivityProblemType
SensitivityProblem <: AbstractReliabilityProblem
Type used to define sensitivity problems.
• X::AbstractVector{<:Distributions.UnivariateDistribution}: Random vector $\vec{X}$
• ρˣ::AbstractMatrix{<:Real}: Correlation matrix $\rho^{X}$
• g::Function: Limit state function $g(\vec{X}, \vec{\Theta})$
• θ::AbstractVector{<:Real}: Parameters of limit state function $\vec{\theta}$
Fortuna.SensitivityProblemCacheType
SensitivityProblemCache
Type used to store results of reliability analysis performed using Mean-Centered First-Order Second-Moment (MCFOSM) method.
• FORMSolution::iHLRFCache: Results of reliability analysis performed using First-Order Reliability Method (FORM)
• ∇β::Vector{Float64}: Sensivity vector of reliability index $\vec{\nabla}_{\vec{\Theta}} \beta$
• ∇PoF::Vector{Float64}: Sensivity vector of probability of failure $\vec{\nabla}_{\vec{\Theta}} P_{f}$
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# 108kg equals how many pounds
• 108 kilograms is 238.1 pounds.
• tk10npubl tk10ncanl
## Say hello to Evi
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https://customdissertations.org/los-angeles-pierce-college-diversity-between-two-different-lawn-communities-worksheet/
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# Los Angeles Pierce College Diversity Between Two Different Lawn Communities Worksheet
This assignment is to compare species diversity between two different lawn communities.
Read pages 151-152 (at the end of chapter 5) to refresh yourself on the Lawn Census activity, and Read pages 197-198 (at the end of chapter 7) to learn about dominance-diversity curves and the Shannon Index.
Here is the excel sheet with our data: Lawn Census Data .xlsx
(please let me know asap if you can open the link or not)
In Excel, draw dominance-diversity curves with species of non-grass along the X-axis and ln(p) along the Y-axis. In other words, make a line graph with natural log (ln) proportion of cover on the vertical axis and the species ordered by abundance on the horizontal axis. Label the y-axis as the ln proportion of non-grass cover and x-axis as non-grass species. You should have two lines, one for each lawn, on one graph.
Make a caption for the graph that adds context and include how many different species were on each lawn. Add the Shannon index value in a text box above each line on the graph.
Delete gridlines, but you can keep the legend that says which site is which. Change the colors if you want.
Please answer the following questions in a text box in your excel file below your graph (be careful, text boxes in excel don’t have spell check):
Hypothesize which lawn will have greatest species diversity between the disturbed and undisturbed lawn communities.
Compare the lines for the two lawns. Were the patterns similar or different in the two lawns? How so?
Did the two lawns have similar or different species richness’s? Give the species richness of each lawn. (Species richness = # of different species represented; simply a count of species)
Would this be a great measure of biodiversity if you had not gathered the same number of data points for both lawns?
Compare the two lawns’ Shannon index values. Which lawn has more diversity of broad-leafed (non-grass) plants? (The higher the Shannon index value, the higher the species richness and evenness.
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# The sizes of jumpsuits
For determining the size of a jumpsuit it is important to consider the parameters of both, the top and the bottom part of it.
There are 3 most important measurements when determining the size:
• chest girth (CG): measured at the fullest part of the chest and the shoulder blades;
• waist circumference (WC): measured at the waist. The tape measure should be loose, unless the chosen model is a skinny or slim fit jumpsuit. For skinny and slim fit models the tape measure should be a little tight;
• hip girth (HG): the tape measure should be horizontal to the floor level. The measurement should be taken at the most protruding part of the buttocks at the back, as well, considering the belly at the front;
• the length of the front part until the waistline: it is the measurement made at the chest, from the waistline up to the intersection point of the neck and the shoulder line.
Men and women body shape groups:
The women body shape group is determined by the difference between HG - CG, and for men - CG - WC.
There are six women body shape groups: from zero to five, the measurements of which belong to the [-2; 18] interval, with 4 (cm) in-between, corresponding the particular group:
• group zero: HG - CG = -2 (cm);
• group one: HG - CG = -2+4=2 (cm) and so on, until the group five, with a difference of 18 (cm).
Women’s jumpsuits*
US Size Bust (inches) Waist (inches) Hip (inches)
0 (XS) 33 25.5 36
2 (XS) 34 26.5 37
4 (S) 35 27.5 38
6 (S) 36 28.5 39
8 (M) 37 29.5 40
10 (M) 38 30.5 41
12 (L) 39 31.5 42
*
Girl’s jumpsuits*
Size Height (inches) Weight (inches) Bust (inches)
XS 39-45 33-44 lbs 23-24
S 45-52 45-64 lbs 25-26
M 52-54 64-72 lbs 27
L 54-57 72-81 lbs 28 ½
XL 57-60 82-93 lbs 30
XXL 60-64 94-115 lbs 31 ½-33
*
Boy’s jumpsuits*
Size Height (inches) Weight (inches) Bust (inches)
XS (4-5) 33-48 lbs 23-24 18 ½-20
S (6-7) 46-61 lbs 25-26 21 ½-23
M (8) 61-72 lbs 27 24 ½
L (10) 73-84 lbs 28 26
XL (12) 85-96 lbs 29 27 ½
XXL (-) 97-130 lbs 30-31 29-30 ½
*
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×
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# Table 5-3 gives the mass, period, and mean distance for
ISBN: 9780321569837 99
## Solution for problem 61 Chapter 5
Physics: Principles with Applications | 6th Edition
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Problem 61
Table 5-3 gives the mass, period, and mean distance for the four largest moons of Jupiter (those discovered by Galileo in 1609). (a) Determine the mass of Jupiter using the data for Io. (/>) Determine the mass of Jupiter using data for each of the other three moons. Arc the results consistent?
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##### ISBN: 9780321569837
The answer to “Table 5-3 gives the mass, period, and mean distance for the four largest moons of Jupiter (those discovered by Galileo in 1609). (a) Determine the mass of Jupiter using the data for Io. (/>) Determine the mass of Jupiter using data for each of the other three moons. Arc the results consistent?” is broken down into a number of easy to follow steps, and 52 words. This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6. Since the solution to 61 from 5 chapter was answered, more than 334 students have viewed the full step-by-step answer. This full solution covers the following key subjects: Jupiter, mass, determine, data, moons. This expansive textbook survival guide covers 33 chapters, and 4141 solutions. Physics: Principles with Applications was written by and is associated to the ISBN: 9780321569837. The full step-by-step solution to problem: 61 from chapter: 5 was answered by , our top Physics solution expert on 09/09/17, 04:43AM.
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# If 3x+13x=3 find 9x2+19x2.
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## It is given that 3x+13x=3, by squaring both sides we get:(3x+13x)2=32⇒(3x)2+(13x)2+(2×3x×13x)=9(∵(a+b)2=a2+b2+2ab)⇒9x2+19x2+2=9⇒9x2+19x2=9−2⇒9x2+19x2=7Hence, 9x2+19x2=7.
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# 1.3: Work as a Mechanism to Transfer Energy
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##### Learning Objectives
• To define work
• To identify mechanical and pressure-volume work
One definition of energy is the capacity to do work. There are many kinds of work, including mechanical work, electrical work, and work against a gravitational or a magnetic field. Here we will consider only mechanical work and focus on the work done during changes in the pressure or the volume of a gas.
## Mechanical Work
The easiest form of work to visualize is mechanical work (Figure $$\PageIndex{1}$$), which is the energy required to move an object a distance $$d$$ when opposed by a force $$F$$, such as gravity (Figure $$\PageIndex{1}$$):
$w=F\,d \label{12.3.1}$
with $$w$$ is work, $$F$$ is opposing force, and $$d$$ is distance.
Because the force ($$F$$) that opposes the action is equal to the mass ($$m$$) of the object times its acceleration ($$a$$), Equation $$\ref{12.3.1}$$ can be rewritten to:
$w = m\,a\,d \label{12.3.2}$
with $$w$$ is work, $$m$$ is mass, $$a$$ is acceleration, and $$d$$ is distance.
Recall from that weight is a force caused by the gravitational attraction between two masses, such as you and Earth. Hence for works against gravity (on Earth), $$a$$ can be set to $$g=9.8\; m/s^2$$. Consider the mechanical work required for you to travel from the first floor of a building to the second. Whether you take an elevator or an escalator, trudge upstairs, or leap up the stairs two at a time, energy is expended to overcome the opposing force of gravity. The amount of work done (w) and thus the energy required depends on three things:
1. the height of the second floor (the distance $$d$$);
2. your mass, which must be raised that distance against the downward acceleration due to gravity; and
## Pressure-Volume (PV) Work
To describe this pressure–volume work (PV work), we will use such imaginary oddities as frictionless pistons, which involve no component of resistance, and ideal gases, which have no attractive or repulsive interactions. Imagine, for example, an ideal gas, confined by a frictionless piston, with internal pressure $$P_{int}$$ and initial volume $$V_i$$ (Figure $$\PageIndex{2}$$). If $$P_{ext} = P_{int}$$, the system is at equilibrium; the piston does not move, and no work is done. If the external pressure on the piston ($$P_{ext}$$) is less than $$P_{int}$$, however, then the ideal gas inside the piston will expand, forcing the piston to perform work on its surroundings; that is, the final volume ($$V_f$$) will be greater than $$V_i$$. If $$P_{ext} > P_{int}$$, then the gas will be compressed, and the surroundings will perform work on the system.
If the piston has cross-sectional area $$A$$, the external pressure exerted by the piston is, by definition, the force per unit area:
$P_{ext} = \dfrac{F}{A} \label{eq5}$
The volume of any three-dimensional object with parallel sides (such as a cylinder) is the cross-sectional area times the height ($$V = Ah$$). Rearranging Equation \ref{eq5} to give
$F = P_{ext}A$
and defining the distance the piston moves ($$d$$) as $$Δh$$, we can calculate the magnitude of the work performed by the piston by substituting into Equation $$\ref{12.3.1}$$:
$w = F d = P_{ext}AΔh \label{12.3.3}$
The change in the volume of the cylinder ($$ΔV$$) as the piston moves a distance $$d$$ is $$ΔV = AΔh$$, as shown in Figure $$\PageIndex{3}$$.
The PV work performed is thus
$w = P_{ext}ΔV \label{12.3.4}$
The units of work obtained using this definition are correct for energy: pressure is force per unit area (newton/m2) and volume has units of cubic meters, so
$w=\left(\dfrac{F}{A}\right)_{\textrm{ext}}(\Delta V)=\dfrac{\textrm{newton}}{\textrm m^2}\times \textrm m^3=\mathrm{newton\cdot m}=\textrm{joule}$
If we use atmospheres for P and liters for V, we obtain units of L·atm for work. These units correspond to units of energy, as shown in the different values of the ideal gas constant R:
$R=\dfrac{0.08206\;\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}}=\dfrac{8.314\textrm{ J}}{\mathrm{mol\cdot K}}$
Thus 0.08206 L·atm = 8.314 J and 1 L·atm = 101.3 J.
Whether work is defined as having a positive sign or a negative sign is a matter of convention. Heat flow is defined from a system to its surroundings as negative; using that same sign convention, we define work done by a system on its surroundings as having a negative sign because it results in a transfer of energy from a system to its surroundings. This is an arbitrary convention and one that is not universally used. Some engineering disciplines are more interested in the work done on the surroundings than in the work done by the system and therefore use the opposite convention. Because $$ΔV > 0$$ for an expansion, Equation $$\ref{12.3.4}$$ must be written with a negative sign to describe PV work done by the system as negative:
$w = −P_{ext}ΔV \label{12.3.5}$
The work done by a gas expanding against an external pressure is therefore negative, corresponding to work done by a system on its surroundings. Conversely, when a gas is compressed by an external pressure, $$ΔV < 0$$ and the work is positive because work is being done on a system by its surroundings.
##### A Matter of Convention
For the thermochemistry
• Heat is positive when energy is transferred from the surroundings to the system
• Work is positive when energy is transferred from the surroundings to the system
##### Example $$\PageIndex{2}$$: Internal Combustion Engine
A small high-performance internal combustion engine has six cylinders with a total nominal displacement (volume) of 2.40 L and a 10:1 compression ratio (meaning that the volume of each cylinder decreases by a factor of 10 when the piston compresses the air–gas mixture inside the cylinder prior to ignition). How much work in joules is done when a gas in one cylinder of the engine expands at constant temperature against an opposing pressure of 40.0 atm during the engine cycle? Assume that the gas is ideal, the piston is frictionless, and no energy is lost as heat.
Given: final volume, compression ratio, and external pressure
Strategy
Calculate the final volume of gas in a single cylinder. Then compute the initial volume of gas in a single cylinder from the compression ratio. Use Equation $$\ref{12.3.5}$$ to calculate the work done in liter-atmospheres. Convert from liter-atmospheres to joules.
Solution
A To calculate the work done, we need to know the initial and final volumes. The final volume is the volume of one of the six cylinders with the piston all the way down: Vf = 2.40 L/6 = 0.400 L. With a 10:1 compression ratio, the volume of the same cylinder with the piston all the way up is Vi = 0.400 L/10 = 0.0400 L. Work is done by the system on its surroundings, so work is negative.
\begin{align} w &= −P_{ext}ΔV\nonumber \\[5pt] &= −(40.0\, atm)(0.400\, L − 0.0400 \,L)\nonumber \\[5pt] &= −14.4\, L·atm\nonumber \end{align}\nonumber
Converting from liter-atmospheres to joules,
\begin{align} w &=-(14.4\;\mathrm{L\cdot atm})[101.3\;\mathrm{J/(L\cdot atm)}]\nonumber \\[5pt] &= -1.46\times10^3\textrm{ J}\nonumber \end{align}\nonumber
In the following exercise, you will see that the concept of work is not confined to engines and pistons. It is found in other applications as well.
##### Exercise $$\PageIndex{2}$$: Work required to Breath
Breathing requires work, even if you are unaware of it. The lung volume of a 70 kg man at rest changed from 2200 mL to 2700 mL when he inhaled, while his lungs maintained a pressure of approximately 1.0 atm.
1. How much work in liter-atmospheres and joules was required to take a single breath?
2. During exercise, his lung volume changed from 2200 mL to 5200 mL on each in-breath.
3. How much additional work in joules did he require to take a breath while exercising?
−0.500 L·atm, or −50.7 J; −304 J; if he takes a breath every three seconds, this corresponds to 1.4 Calories per minute (1.4 kcal).
##### Example $$\PageIndex{1}$$: Expansion (PV) work
1. How much work is done by a gas that expands from 2 liters to 5 liters against an external pressure of 750 mmHg?
2. How much work is done by 0.54 moles of a gas that has an initial volume of 8 liters and expands under the following conditions: 30 oC and 1.3 atm?
3. How much work is done by a gas (P=1.7 atm, V=1.56 L) that expands against an external pressure of 1.8 atm?
Solution a
$w = − PΔV\nonumber$
$ΔV = V_{final} - V_{Initial} = 5 \,L - 2\, L = 3 L\nonumber$
Convert 750 mmHg to atm:
$750 mmHg * 1/760 (atm/mmHg) = 0.9868 atm.\nonumber$
$W = − pΔV = -(.9868\, atm)(3\,L) = -2.96 \,L\, atm.\nonumber$
Solution b
First we must find the final volume using the ideal gas law:
$pV = nRT\nonumber$
or
\begin{align*} V &= \dfrac{nRT}{P} \\[5pt] &= \dfrac{ (0.54 \,moles)(0.082057 (L\, atm)/ (mol\, K))(303\,K)}{1.3\, atm} = 10.33\, L \end{align*}
\begin{align*} ΔV &= V_{final} - V_{initial} \\[5pt] &= 10.3\, L - 8\, L = 2.3\, L \end{align*}
Then from Equation \ref{12.3.4}
$w = − pΔV = - (1.3\, atm)(2.3 \,L) = -3\, L\, \cdot atm.\nonumber$
Solution c
$w = - p ΔV = - (1.8 \,atm )\, ΔV.\nonumber$
Given $$p_1$$, $$V_1$$, and $$p_2$$, find $$V_2$$: $$p_1V_1=p_2V_2$$ (at constant $$T$$ and $$n$$)
$V_2= \dfrac{V_1 P_1}{ P_2} = (1.56 L * 1.7 atm) / 1.8 atm = 1.47 L\nonumber$
Now,
$ΔV = V_2 - V_1=1.47\, L - 1.56\, L = -0.09\nonumber$
$w = - (1.8 atm) * (-0.09 L) = 0.162 L atm.\nonumber$
## Work in Chemical Transformations
Suppose, for example, that the system under study is a mass of steam heated by the combustion of several hundred pounds of coal and enclosed within a cylinder housing a piston attached to the crankshaft of a large steam engine. The gas is not ideal, and the cylinder is not frictionless. Nonetheless, as steam enters the engine chamber and the expanding gas pushes against the piston, the piston moves, so useful work is performed. In fact, PV work launched the Industrial Revolution of the 19th century and powers the internal combustion engine on which most of us still rely for transportation.
In contrast to internal energy, work is not a state function. We can see this by examining Figure $$\PageIndex{4}$$, in which two different, two-step pathways take a gaseous system from an initial state to a final state with corresponding changes in temperature. In pathway A, the volume of a gas is initially increased while its pressure stays constant (step 1); then its pressure is decreased while the volume remains constant (step 2). In pathway B, the order of the steps is reversed. The temperatures, pressures, and volumes of the initial and final states are identical in both cases, but the amount of work done, indicated by the shaded areas in the figure, is substantially different. As we can see, the amount of work done depends on the pathway taken from ($$V_1$$, $$P_1$$) to ($$V_2$$, $$P_2$$), which means that work is not a state function.
Internal energy is a state function, whereas work is not.
## Work and Chemical Reactions
Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in concentrated nitric acid. The chemical equation for this reaction is as follows:
$\ce{Cu(s) + 4HNO3(aq) -> Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g)} \label{12.3.5a}$
If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will rise as nitrogen dioxide gas is formed (Figure $$\PageIndex{5}$$). The system is performing work by lifting the piston against the downward force exerted by the atmosphere (i.e., atmospheric pressure). We find the amount of PV work done by multiplying the external pressure P by the change in volume caused by movement of the piston ($$ΔV$$). At a constant external pressure (here, atmospheric pressure)
$w = −PΔV \label{12.3.6}$
The negative sign associated with $$PV$$ work done indicates that the system loses energy. If the volume increases at constant pressure (ΔV > 0), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases (ΔV < 0), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy.
## Enthalpy
The symbol $$U$$ represents the internal energy of a system, which is the sum of the kinetic energy and potential energy of all its components. It is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called enthalpy (H) (from the Greek enthalpein, meaning “to warm”). The enthalpy of a system is defined as the sum of its internal energy $$U$$ plus the product of its pressure $$P$$ and volume $$V$$:
$H =U + PV \label{12.3.7}$
Because internal energy, pressure, and volume are all state functions, enthalpy is also a state function.
If a chemical change occurs at constant pressure (i.e., for a fixed $$P$$, $$ΔP = 0$$), the change in enthalpy ($$ΔH$$) is
\begin{align} ΔH &= Δ(U + PV) \\[4pt] &= ΔU + Δ(PV) \\[4pt] &= ΔU + PΔV \label{12.3.8} \end{align}
Substituting $$q + w$$ for $$ΔU$$ (Equation $$\ref{12.3.8}$$) and $$−w$$ for $$PΔV$$ (Equation $$\ref{12.3.6}$$), we obtain
\begin{align} ΔH &= ΔU + PΔV \nonumber\\[4pt] &= q_p + \cancel{w} − \cancel{w} \nonumber\\[4pt] &= q_p \label{12.3.9} \end{align}
The subscript $$p$$ is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From Equation $$\ref{12.3.9}$$ we see that at constant pressure the change in enthalpy, $$ΔH$$ of the system, defined as $$H_{final} − H_{initial}$$, is equal to the heat gained or lost.
$ΔH = H_{final} − H_{initial} = q_p \label{12.3.10}$
Just as with $$ΔU$$, because enthalpy is a state function, the magnitude of $$ΔH$$ depends on only the initial and final states of the system, not on the path taken. Most important, the enthalpy change is the same even if the process does not occur at constant pressure.
To find $$ΔH$$ for a reaction, measure $$q_p$$ under constant pressure.
## Summary
All forms of energy can be interconverted. Three things can change the energy of an object: the transfer of heat, work performed on or by an object, or some combination of heat and work.
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# Five ways to celebrate 3.14159 on Pi Day
###### Celebrating Pi Day can be as easy as pie. / Sylvia Rector, Gannett
Today is Pi Day, a day to celebrate the joys of the mathematical constant of 3.14159 or a slice of baked deliciousness.
If the term "pi" doesn't ring any bells, let's review a little high school geometry. Pi is the ratio of a circle's circumference to its diameter and is represented by the 16th letter of the Greek alphabet ("Ď?") and is often written simply as "pi."
According to PiDay.org (yes, this website actually exists), the number's "infinite nature makes it a fun challenge to memorize, and to computationally calculate more and more digits."
So why celebrate pi on Thursday? Because it is March 14 (3/14).
Here are five ways to get the math party started:
(1) To eat or not to eat pie is not a question on Pi Day. Bake or buy a whole pie (here's one scrumptious recipe). You can calculate the circumference of the pie by measuring the diameter and then by multiplying it by pi, or simply satisfy your sweet tooth with several bites. Share the rest and you can count the number of co-workers who will love you.
(2) Wow your friends with a few strange facts about Albert Einstein. Why? The German-born physicist shares his birthday with the first three digits of pi. Here's one strange fact: Not only did Einstein develop the general theory of relativity, but his second wife was his cousin. You can also get to know the genius better by exploring the interactive Einstein Brain Atlas app.
(3) Send infinite wishes to your loves ones with a pi e-card! 123greetings.com has some fun ones, including a pi rap card.
(4) Calculate how much you can save with Pi Day deals, like this one at the Microsoft store, which is offering 3.14% off a Dell tablet.
(5) Relax with pi. Watch a good pi movie or read a good pi book. Yann Martel's Life of Pi is a transformative novel about Piscine Molitor "Pi" Patel, a boy, who survives after being stranded on a boat with a Bengal tiger for 227 days. The book was adapted into a 2012 Oscar-nominated movie directed by Ang Lee.
Read the original story: Five ways to celebrate 3.14159 on Pi Day
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I'm going to talk about something which should be obvious, but I continue to see people optimizing code in the wrong order (*cough* including myself *cough*). So, here's a reminder. When you're optimizing a bit of code...
## FIRST
Make sure your algorithmic running time is right. O(1) is almost always faster than O(N), and O(N^2) is right out. Often these optimization exercises involve changing some O(N) to O(M), where M is smaller than N.
I'll give an example. Drawing a frame of 3D graphics in IMVU is O(N) where N is the sum of all vertices from all products on all objects loaded into a scene. We recently implemented view frustum culling, which skips drawing objects that are known to be off-screen. This reduces the rendering time from O(N) to O(M) where M<N and M is the number of vertices from products that are visible. If we implemented View Independent Progressive Meshes, we could reduce the time to O(P) where P is the number of vertices that contribute to the visible detail of the scene, and P<M<N.
However, make sure to avoid algorithms with good running times but huge constants. This is why, when CPUs got fast and random memory accesses got slow, searching an O(N) array (or std::vector) is often faster than searching an O(log N) tree (or std::map). The tree will miss cache far more often.
## SECOND
Then, use all of the algebra, set theory, and logic you know to reduce the number of operations required, in order of operation cost.
Let's say we're going to calculate the diffuse reflectance on a surface: N dot L, where N and L are three-vectors, N is the normal of the surface, and L is the direction to the light.
The naive `normalize(N) dot normalize(L)` is...
```float lengthN = sqrtf(N.x*N.x + N.y*N.y + N.z*N.z);
float lengthL = sqrtf(L.x*L.x + L.y*L.y + L.z*L.z);
float dot =
(N.x / lengthN) * (L.x / lengthL) +
(N.y / lengthN) * (L.y / lengthL) +
(N.z / lengthN) * (L.z / lengthL);
```
... which turns out to be 6 additions, 9 multiplications, 6 divisions, and 2 square roots. Let's say additions and multiplications are 2 cycles, and divisions and square roots are 40 cycles. This gives us a total of 6*2 + 9*2 + 6*40 + 2*40 = 350 cycles.
Instead, let's do a bit of algebra:
``` normalize(N) dot normalize(L)
= N/|N| dot L/|L|
= (N dot L) / (|N||L|)
= (N dot L) / sqrt((N dot N) * (L dot L))
```
The new calculation is...
```float lengthSquaredN = N.x*N.x + N.y*N.y + N.z*N.z;
float lengthSquaredL = sqrtf(L.x*L.x + L.y*L.y + L.z*L.z);
float NdotL = N.x*L.x + N.y*L.y + N.z*L.z;
float dot = NdotL / sqrtf(lengthSquaredN * lengthSquaredL);
```
... 6 additions, 10 multiplications, 1 division, and 1 sqrt: 6*2 + 10*2 + 1*40 + 1*40 = 112 cycles. Huge improvement just by applying basic math.
## THIRD
Once you're done optimizing algebraically, read your processor manuals and take full advantage of the hardware. If you've got SSE4, you can do the dot products in one instruction (DPPS), and an approximate reciprocal square root in another (RSQRTSS), which can give another huge improvement.
The reason you want to optimize in this order is that algorithmic improvements reduce the amount of work you have to do, making it less important to make that work fast. A hardware-optimized O(N^2) algorithm can be easily beaten by an unoptimized O(N log N) algorithm. Remember, Chad, the next time you schedule optimization projects, consider downstream effects such as these.
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Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A052461 4-magic series constant. 2
1, 177, 5111, 60962, 430729, 2158099, 8488095, 27903044, 79895265, 205033333, 481386807, 1049954918, 2152397897, 4185095383, 7774354687, 13878462600, 23923217921, 39978597945, 64985300791, 103041066666, 159757914953 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Eric Weisstein's World of Mathematics, Magic Constant. Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1). FORMULA G.f.: x*(x^8 +167*x^7 +3386*x^6 +17697*x^5 +30074*x^4 +17697*x^3 +3386*x^2 +167*x +1) / (x -1)^10. - Colin Barker, Jun 06 2013 PROG (PARI) a(n)=(6*n^9+15*n^7+10*n^5-n)/30 \\ Charles R Greathouse IV, Jun 06 2013 CROSSREFS Cf. A052459, A052460. Sequence in context: A105988 A083620 A097317 * A173375 A069398 A189910 Adjacent sequences: A052458 A052459 A052460 * A052462 A052463 A052464 KEYWORD nonn,easy AUTHOR STATUS approved
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## Maximum intraday drawdown
This is the maximum reduction in your capital made over a number of trades from the highest capital (original plus profit) reached at any time. The measurement should be made by utilizing the extremes of the bars during the positions, rather than measuring from close to close. It is a reflection of the pain that you will suffer as you see your capital balance reduce over a series of losses, and this includes a spike lower during one day, even if the price rallies to the close.
The figure shown in the above system report is S\$25,708.13. This represents 25.7% of our original capital of S\$100,000. This is rather on the high side of what is acceptable. Do remember, however, that the system results are based on each trade being of the same size, in this case S\$10G,000. If the string of losses that caused the cumulative loss of S\$25,708.13 occurred at the start of the data series, then you will have only S\$74,291.87 to place on the subsequent trade. This will, of course, affect the overall results of the system. On the other hand, it would be possible to introduce new capital to make up this loss, but this also changes the return on capital.
Again, at the end of the day, the amount of capital you are prepared to lose is quite personal. It is probably reasonable, however, to expect that most users of a system would be uncomfortable if the maximum draw down was contained within around 20% of capital.
There are many ratios that should affect your assessment of a system. The simple idea that net profit is the most important is not necessarily true. Many of the numbers that you will extract from a system might not be quite what you'd like to see, but other areas may compensate for a disadvantage in one respect. All values should be assessed in conjunction with each other. Each does have its own particular impact and significance.
For example, let's say you found a system that gave generally good results alt round - except for the fact that the net profit was low while the percentage of profitable trades was near 80%. Assuming that walk-forwrard testing supports the view that the results are stable, then this would suggest that you have discovered a stable system that has an infrequent entry. If you had odds of 80:20 in your favor, the risk appears to be worth taking. The consistency of trading signals is something you can address with a second or third system.
Do make sure, however, that whatever type of system you adopt, you do build in buffers. Optimized results, even when supported by walk-forward testing, tend to overstate profits and understate losses and maximum draw downs. Always expect a worse situation and be prepared for it to happen, rather than believe your system's results will always be the same. Work out how you will react and what action you would take if the worst case scenario were to materialize.
Keep the above in mind and your approach to systems should be safe.
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## Applied Mechanics Multiple Choice Questions
1. A projectile is thrown at an angle a to the horizontal with α velocity v. It will have the maximum centripetal acceleration
A. at the start
B. at the top of the trajectory
C. as it strikes the ground
D. else where.
2. A cable loaded with 0.5 tonne per horizontal metre span is stretched between supports in the same horizontal line 400 m apart. If central dip is 20 m, the minimum tension in the cable, will be
A. 200 tonnes at the centre
B. 500 tonnes at the centre
C. 200 tonnes at the right support
D. 200 tonnes at the left support.
3. The Law of Polygon of Forces states that
A. if a polygen representing the forces acting at point in a body is closed, the forces are in equilibrium
B. if forces acting on a point can be represented in magnitde and direction by the sides of a polygon taken in order, then the resultant of the forces will be represented in magnitude and direction by the closing side of the polygon
C. if forces acting on a point can be represented of a polygon taken in order, their sides of a polygon taken in order, their resultant will be represented in magnitude and direction by the closing side of the polygon, taken in opposite order
D. if forces acting on a point can be represented in magnitude and direction by the sides of a polygon in order, the forces are in equilibrium.
4. The centre of gravity of a triangle is at the point where three
A. medians of the triangle meet
B. perpendicular bisectors of the sides of the triangle meet
C. bisectors of the angle of the triangle meet
D. none of these.
5. The forces which meet at one point and have their lines of action in different planes are called
A. coplaner non-concurrent forces
B. non-coplaner concurrent forces
C. non-coplaner non-current forces
D. intersecting forces
E. none of these.
6. At a given instant ship A is travelling at 6 km/h due east and ship B is travelling at 8 km/h due north. The velocity of B relative to A is
A. 7 km/hrs
B. 2 km/hrs
C. 1 km/hrs
D. 10 km/hrs
E. 14 km/hrs.
7. The locus of the instantaneous centre of a moving rigid body, is
A. straight line
B. involute
C. centroid
D. spiral.
8. If a body moves in such a way that its velocity increases by equal amount in equal intervals of time, it is said to be moving with
A. a uniform retardation
B. a uniform acceleration
C. a variable acceleration
D. a variable retardation
E. none of these.
9. If the gravitational accelerational at any place is doubled, the weight of a body, will
A. be reduced to half
B. be doubled
C. not be affected
D. none of these.
10. Parallelogram Law of Forces states, “if two forces acting simultaneously on a particle be represented in magnitude and direction by two adjacent sides of a parallelogram, their resultant may be represented in magnitude and direction by
A. its longer side”
B. its shorter side”
C. the diagonal of the parallelogram which does not pass through the point of intersection of the forces”
D. the diagonal of the parallelogram which passes through the point of intersection of the forces”
E. half the sum of the diagonals”.
11. Which one of the following statements is true ?
A. The tangent of the angle of friction is equal to coefficeint of friction
B. The angle of repose is equal to angle of friction
C. The tangent of the angle of repose is equal to coefficient of friction
D. All the above.
12. The product of mass and velocity of a moving a body, is called
A. moment
B. momentum
C. power
D. impulse.
13. Engineer’s units of force, is
A. Newton in absolute units
B. Dyne in absolute units
C. Newton and dyne in absolute units
D. All the above.
14. Joule is the unit of
A. work
B. force
C. power
D. torque
E. none of these.
15. The maximum velocity of a body vibrating with a simple harmonic motion of amplitude 150 mm and frequency 2 vibrations/sec, is
A. 188.5 m/sec
B. 18.85 m/sec
C. 1.885 m/sec
D. 0.18845 m/sec.
16. The angle which an inclined surface makes with the horiontal when a body placed on it is on the point of moving down, is called
A. angle of repose
B. angle of friction
C. angle of inclination
D. none of these.
17. Work may be defined as
A. force x distance
B. force x velocity
C. force x acceleration
D. none of these.
18. A bullet weighing 200 g is fired horizontally with a velocity of 25 m/sec from a gun carried on a carriage which together with the gun weighs 100 kg. The velocity of recoil of the gun, will be
A. 0.01 m/sec
B. 0.05 m/sec
C. 1.00 m/sec
D. 1.5 m/see.
19. The tension in a cable supporting a lift
A. is more when the lift is moving downwards
B. is less when the lift is moving upwards
C. remains constant whether its moves downwards or upwards
D. is less when the lift is moving downwards.
20. A satellite moves in its orbit around the earth due to
A. Gravitational force
B. Centripetal force
C. Centrifugal force
D. none of these.
21. If three rigid rods are hinged together to form a triangle and are given rotary as well as translatory motion, the number of instantaneous centres of the triangle, will be
A. 1
B. 2
C. 3
D. 4
E. 5.
22. The mechanical advantage of an ideal machine is 100. For moving the local through 2 m, the effort moves through
A. 0.02 m
B. 2 m
C. 2.5 m
D. 20 m.
23. When a body in equilibrium undergoes an infinitely small displacement, work imagined to be done, is known as
A. imaginary work
B. negative work
C. virtual work
D. none of these.
24. To attain the synchronous orbit, the launch of a satellite, is done from a place
A. on equator
B. on 30° latitude
C. on 45° latitude
D. on 60° latitude
E. on the poles.
25. Pick up the correct statement from the following :
A. Nature plays an important role in the launch of a satellite
B. The earth’s gravity reduces the speed of a satellite by 32 km per second
C. The gravitational force relents as the satellite climbs higher
D. The gravitational intensity declines with height
E. All the above.
26. If the radius of the earth is 600 km the height of a mountain above sea level at the top of which a beat seconds pendulum at sea level, looses 27 seconds a day, is
A. 500 metres
B. 1000 metres
C. 1500 metres
D. 2000 metres
E. 25000 metres.
27. The total time of collision and restitution of two bodies, is called
A. time of collision
B. period of collision
C. period of impact
D. all the above.
28. The length of a Second’s pendulum, is
A. 99.0 cm
B. 99.4 cm
C. 100 cm
D. 101 cm
E. 101.10 cm.
29. A load of 500 kg was lifted through a distance of 13 cm. by an effort of 25 kg which moved through a distance of 650 cm. The efficiency of the lifting machine is
A. 50%
B. 40%
C. 55%
D. 30%.
30. A string of length 90 cm is fastened to two points A and B at the same level 60 cm apart. A ring weighing 120 g is slided on the string. A horizontal force P is applied to the ring such that it is in equilibrium vertically below B. The value of P is :
A. 40 g
B. 60 g
C. 80 g
D. 100 g.
31. Two forces act an angle of 120°. If the greater force is 50 kg and their resultant is perpendicular to the smaller force, the smaller force is
A. 20 kg
B. 25 kg
C. 30 kg
D. 35 kg
E. 40 kg.
32. A uniform rod 9 m long weighing 40 kg is pivoted at a point 2 m from one end where a weight of 120 kg is suspended. The required force acting at the end in a direction perpendicular to rod to keep it equilibrium, at an inclination 60° with horizontal, is
A. 40 kg
B. 60 kg
C. 10 kg
D. 100 kg.
33. If a particle is projected inside a horizontal tunnel which is 554 cm high with a velocity of 60 m per sec, the angle of projection for maximum range, is
A. 8°
B. 9°
C. 10°
D. 11°
E. 12°.
34. Newton’s law of Collision of elastic bodies states that when two moving bodies collide each other, their velocity of separation
A. is directly proportional to their velocity of approach
B. is inversely proportional to their velocity of approach
C. bears a constant ratio to their velocity of approach
D. is equal to the sum of their velocities of approach.
35. Two forces of 6 Newtons and 8 Newtons which are acting at right angles to each other, will have a resultant of
A. 5 Newtons
B. 8 Newtons
C. 10 Newtons
D. 12 Newtons.
36. A projectile is fired with a velocity of 100.3 m/sec. at an elevation of 60°. The velocity attained by the projectile when it is moving at a height of 100 m, is
A. 70 m/sec.
B. 75 m/sec.
C. 80 m/sec.
D. 85 m/sec.
E. 90 m/sec.
37. The following factor affects the orbit of a satellite up to an altitude of 720 km from the earth’s surface
A. uneven distribution of the gravitational field
B. gravity of the sun and the moon
C. aerodynamic forces
D. none of these.
38. A body is said to move with Simple Harmonic Motion if its acceleration, is
A. always directed away from the centre, the point of reference
B. proportional to the square of the distance from the point of reference
C. proportional to the distance from the point of reference and directed towards it
D. inversely proportion to the distance from the point of reference
E. none of these.
39. One end of a light string 4 m in length is fixed to a point on a smooth wall and the other end fastened to a point on the surface of a smooth sphere of diameter 2.25 m and of weight 100 kg. The reaction between the sphere and the wall of the arrangement made is
A. 102.5 kg
B. 105.5 kg
C. 108.5 kg
D. 110 kg.
40. Which one of the following laws is not applicable to a simple pendulum ?.
A. The time period does not depend on its magnitude
B. The time period is proportional to its length l
C. The time period is proportional to l where l is length
D. The time period is inversely proportional to g where g is the acceleration due to gravity.
41. Effect of a force on a body depends upon its
A. direction
B. magnitude
C. position
D. all the above.
42. A Seconds pendulum executes
A. 0.5 beat per second
B. 1.0 beat per second
C. 2.0 beats per second
D. 2.5 beats per second
E. 3 beats per second.
43. The resultant of the forces acting on a body will be zero if the body
A. rotates
B. moves with variable velocity in a straight line
C. moves along a curved path
D. does not move at all.
44. A ball moving with a velocity of 5 m/sec impinges a fixed plane at an angle of 45° and its direction after impact is equally inclined to the line of impact. If the coefficient of restitution is 0.5, the velocity of the ball after impact will be
A. 0.5 m/sec
B. 1.5 m/sec
C. 2.5 m/sec
D. 3.5 m/sec
E. 4.5 m/sec.
45. A particle moves with a velocity of 2 m/sec in a straight line with a negative acceleration of 0.1 m/sec2. Time required to traverse a distance of 1.5 m, is
A. 40 sec
B. 30 sec
C. 20 sec
D. 15 sec
E. 10 sec.
46. For maximum range of a projectile, the angle of projection should be
A. 30°
B. 45°
C. 60°
D. none of these.
47. According to Law of Triangle of Forces
A. three forces acting at a point, can be rep-resented by the sides of a triangle, each side being in proportion to the force
B. three forces acting along the sides of a triangle are always in equilibrium
C. if three forces acting on a, point can be represented.in magnitude and direction, by the sides of a triangle taken in order, these will be in equilibrium
D. if three forces acting at a point are in equilibrium each force is proportional to the sine of the angle between the other two
E. if the forces acting on a particle be represented in magnitude and direction by the two sides of a triangle taken in order, their resultant will be represented in magnitude and direction by the third side of the triangle, taken in opposite order.
48. A Second’s pendulum gains 2 minutes a day. To make it to keep correct time its length
A. must be decreased
B. must be increased
C. is not changed but weight of the bob is increased
D. is not changed but weight of the bob is decreased
E. none of these.
49. The following is not a law of static friction :
A. The force of friction always acts in a direction opposite to that in which the body tends to move
B. The force of friction is dependent upon the area of contact
C. The force of friction depends upon the roughness of the surface
D. The magnitude of the limiting friction bears a constant ratio to the normal reaction between two surfaces.
50. The velocity of a moving body, is
A. a vector quantity
B. a scalar quantity
C. a scalar as well as a vector quantity
D. none of these.
51. The resolved part of the resultant of two forces inclined at an angle θ in a given direction is
A. algebraic sum of the resolved parts of the forces in the direction
B. arithmetical sum of the resolved parts of the forces in the direction
C. difference of the forces multiplied by cosine θ°
D. sum of the forces multiplied by the sine θ
E. sum of the forces multiplied by the tangent θ°.
52. The following statement is one of the laws of Dynamic friction
A. The force of friction always acts in a direction opposite to that in which a body is moving
B. The magnitude of the kinetic friction bears a constant ratio to the normal reaction between two surfaces. The ratio being slightly less than that in the case of limiting friction
C. For moderate speeds the force of friction remains constant but decreases slightly with the increase of speed
D. all the above.
53. A 50 kg boy climbs up a 8 m rope in gymnasiam in 10 sec. The average power developed by the boy is approximately
A. 400 watts
B. 500 watts
C. 4000 watts
D. none of these.
54. Lami’s theroem states that
A. three forces acting at a point are always in equilibrium
B. if three forces acting on a point can be represented in magnitude and direction by the sides of a triangle, the point will be in the state of equilibrium
C. three coplaner forces acting at a point will be in equilibrium, if each force is proportional to the sine of the angle between the other two
D. three coplaner forces acting at a point will be in equilibrium if each force is inversely proportional to the sine of the angle between the other two
E. none of these.
55. Pick up the incorrect statement from the following. In case of suspension bridge due to rise in temperature,
A. dip of the cable increases
B. length of the cable increases
C. dip of the cable decreases
D. none of the these.
56. If two forces acting at a point are in equilibrium, they must be equal in magnitude and their line of action must be along
A. the same line in the same sense
B. the same line in opposite sense
C. the perpendicular to both the lines
D. none of these.
57. For lifting a load of 50 kg through a distance of 2.5 cm, an effort of 12.5 kg is moved through a distance of 40 cm. The efficiency of the lifting machine, is
A. 60%
B. 65%
C. 70%
D. 25%.
58. Power can be expressed as
A. work/energy
B. work/time
C. work x time
D. work/distance.
59. To avoid bending action at the base of a pier,
A. suspension and anchor cables are kept at the same level
B. suspension and anchor cables are fixed to pier top
C. suspension cable and anchor cables are attached to a saddle mounted on rollers on top of the pier
D. none the these.
60. If the angle between the applied force and the direction of motion of a body, is between 90° and 180°, the work done, is called
A. virtual work
B. imaginary work
C. zero work
D. negative work.
61. For a body moving with simple harmonic motion, the number of cycles per second, is known as its
A. oscillation
B. amplitude
C. periodic time
D. beat
E. frequency.
62. If the horizontal range is 2.5 times the greatest height, the angle of projection of the projectile, is
A. 57°
B. 58°
C. 59°
D. 60°.
63. The shape of a suspended cable under its own weight, is
A. parabolic
B. circular
C. catenary
D. elliptical.
64. Time required to stop a car moving with a velocity 20 m/sec within a distance of 40 m, is
A. 2 sec
B. 3 sec
C. 4 sec
D. 5 sec
E. 6 sec.
65. The characteristic of a couple, is :
A. algebraic sum of forces, constituting a couple is zero
B. algebraic sum of moments of forces, constituting a couple, about any poin, is same
C. a couple can be balanced only by a couple but of opposite sense
D. a couple can be never the balanced by a single force
E. all the above.
66. One end of a light string 4 m in length is fixed to a point on a smooth wall and the other end fastened to a point on the surface of a smooth sphere of diameter 2.25 m and of weight 100 kg. The tension in the string is
A. 17.5 kg
B. 19.5 kg
C. 22.5 kg
D. 25 kg.
67. The inherent property of a body which offers reluctance to change its state of rest or uniform motion, is
A. weight
B. mass
C. interia
D. momentum.
68. Ball A of mass 250 g moving on a smooth horizontal table with a velocity of 10 m/s hits an identical stationary ball B on the table. If the impact is perfectly elastic, the velocity of the ball B just after impact would be
A. zero
B. 5 m/sec
C. 10 m/sec
D. none of these.
69. A load of 500 kg was lifted through a distance of 13 cm. by an effort of 25 kg which moved through a distance of 650 cm. The velocity ratio of the lifting machine is
A. 50
B. 55
C. 60
D. 65
E. 70.
70. The angle of friction is :
A. The ratio of the friction and the normal reaction
B. The force of friction when the body is in motion
C. The angle between the normal reaction and the resultant of normal raction and limiting friction
D. The force of friction at which the body is just about to move.
71. For perfectly elastic bodies, the value of coefficient of restitution is
A. zero
B. 0.5
C. 1.0
D. between 0 and 1.
72. The motion of a particle is described by the relation x = t2- 10t + 30, where x is in metres and t in seconds. The total distance travelled by the particle from t = 0 to t = 10 seconds would be
A. zero
B. 30 m
C. 50 m
D. 60 m
E. none of these.
73. Three forces which act on a rigid body to keep it in equilibrium. The forces must be coplanar and
A. concurrent
B. parallel
C. concurrent parallel
D. none of these.
74. Two shots fired simultaneously from the top and bottom of a vertical tower with elevations of 30° and 45° respectively strike a target simultaneously. If horizontal distance of the target from the tower is 1000 m, the height of the tower is
A. 350 m
B. 375 m
C. 400 m
D. 425 m.
75. The ratio of the moment of inertia of a rectangle about its centroidal axis to the moment of inertia about its base, is
A. 1/4
B. 1/2
C. 3/4
D. 2.
76. The apparent weight of a man in a moving lift is less than his real weight when it is going down with
A. uniform speed
B. an acceleration
C. linear momentum
D. retardation.
77. P is the force acting on a body whose mass is m and acceleration is f. The equation P – mf = 0, is known as
A. equation of dynamics
B. equation of dynamic equilibrium
C. equation of statics
D. none of these.
78. A stone of mass 1 kg is tied to a string of length 1 m and whirled in a horizontal circle at a constant angular speed 5 rad/sec. The tension in the string is,
A. 5 N
B. 10 N
C. 15 N
D. 25 N
E. None of these.
79. Newtons’s Law of Motion is :
A. Every body continues in its state of rest or of uniform motion, in a straight line, unless it is acted upon by some external force
B. The rate of change of momentum is directly proportional to the impressed force, and takes place in the same direction, in which the force acts
C. To every action, there is always an equal and opposite reaction
D. All the above.
80. The equation of motion of a particle starting from rest along a straight line is x = t3 – 3l2 + 5. The ratio of the velocities after 5 sec and 3 sec will be
A. 2
B. 3
C. 4
D. 5
E. 4.5
81. The necessary condition of equilibrium of a body, is :
A. algebraic sum of horizontal components of all the forces must be zero
B. algebraic sum of vertical components of all the forces must be zero
C. algebraic sum of the moments of the forces about a point must be zero
D. all (a), (b) and (c).
82. Two particles have been projected at angles 64° and 45° to the horizontal. If the velocity of projection of first is 10 m/sec, the velocity of projection of the other for equal horizontal ranges is
A. 9.3 m/sec
B. 8.3 m/sec
C. 7.3 m/sec
D. 6.3 m/sec.
83. A trolley wire weighs 1 kg per metre length. The ends of the wire are attached to two poles 20 m apart. If the horizontal tension is 1000 kg, the central dip of the cable is
A. 2 cm
B. 3 cm
C. 4 cm
D. 5 cm.
84. Principle of Transmissibility of Forces states that, when a force acts upon a body, its effect is
A. maximum if it acts at the centre of gravity of the body
B. different at different points on its line of
C. same at every point on its line of action
D. minimum if it acts at the C.G. of the body
E. none of these.
85. A ball is dropped from a height of 2.25 m on a smooth floor and rises to a height of 1.00 m after the bounce. The coefficient of restitution between the ball and the floor is
A. 0.33
B. 0.44
C. 0.57
D. 0.67
86. If the linear velocity of a point on the rim of a wheel of 10 m diameter, is 50 m/sec, its angular velocity will be
87. The force acting on a point on the surface of a rigid body may be considered to act
A. at the centre of gravity of the body
B. on the periphery of the body
C. on any point on the line of action of the force
D. at any point on the surface normal to the line of action of the force.
88. A satellite is said to move in a synchronous orbit if it moves at an altitude of 36, 000 km with a maximum velocity of about
A. 7000 km per hour
B. 8000 km per hour
C. 9000 km per hour
D. 10, 000 per hour
E. 11, 000 km per hour.
89. A marble ball is rolled on a smooth floor of a room to hit a wall. If the time taken by the ball in returning to the point of projection is twice the time taken in reaching the wall, the coefficient of restitution between the ball and the wall, is
A. 0.25
B. 0.50
C. 0.75
D. 1.0
90. A number of forces acting simultaneously on a particle of a body
A. may not be replaced by a single force
B. may be replaced by a single force
C. may be replaced by a single force through C.G. of the body
D. may be replaced by a couple
E. none of these.
91. A 49 kg lady stands on a spring scale in an elevator. During the first 5 sec, starting from rest, the scale reads 69 kg. The velocity of the elevator will be
A. 10 m/sec
B. 15 m/sec
C. 20 m/sec
D. 25 m/sec.
92. Pick up the correct statement from the following. A rubber ball when strikes a wall rebounds but a lead ball of same mass and velocity when strikes the same wall, falls down
A. rubber and lead balls undergo equal changes in momentum
B. change in momentum suffered by lead ball is less that of rubber ball
C. momentum of rubber ball is less than that of lead ball
D. none of these.
93. The instantaneous centre of a member lies at the point of intersection of two lines drawn at the ends of the member such that the lines are inclined to the direction of motion of the ends at
A. 30°
B. 45°
C. 60°
D. 90°
94. A retarding force on a body does not
A. change the motion of the body
B. retard the motion of the body
C. introduce the motion of the body
D. none of these.
95. The unit of force in C.G.S. system of units, is called
A. dyne
B. Newton
C. kg
D. all the above.
96. A block of weight 50 kg is placed on a horizontal plane. When a horizontal force of 18 kg is applied, the block is just on the point of motion. The angle of friction is
A. 17° 48′
B. 18° 48′
C. 19° 48′
D. 20° 48′
E. 21° 48′.
97. From a circular plate of a diameter 6 cm is cut out a circle whose diameter is equal to the radius of the plate. The C.G. of the remainder from the centre of circular plate is at a distance of
A. 2.0 cm
B. 1.5 cm
C. 1.0
D. 0.5 cm.
98. The force which produces an acceleration of 1 m/sec2 in a mass of one kg, is called
A. dyne
B. Netwon
C. joule
D. erg.
99. A sphere is resting on two planes BA and BC which are inclined at 45° and 60° respectively with the horizontal. The reaction on the plane BA will be
A. less than that on BC
B. more than that of BC
C. equal to that on BC
D. zero
E. none of these.
100. A projectile is fired at an angle θ to the vertical. Its horizontal range will be maximum when θ is
A. 0°
B. 30°
C. 45°
D. 60°
E. 90°.
101. A load of 500 kg was lifted through a distance of 13 cm. by an effort of 25 kg which moved through a distance of 650 cm. The mechanical advantage of the lifting machine is
A. 15
B. 18
C. 20
D. 26.
101. One half of a vibration of a body, is called
A. period time
B. oscillation
C. beat
D. amplitude.
102. On a ladder resisting on a smooth ground and leaning against a rough vertical wall, the force of friction acts
A. towards the wall at its upper end
B. away from the wall at its upper end
C. upwards at its upper end
D. downwards at its upper end
E. none of these.
103. The acceleration of a particle moving along the circumference of a circle with a uniform speed, is directed
B. tangentially at that point
C. away from the centre
D. towards the centre.
104. From the circular plate of a diameter 6 cm is cut out a circular plate whose diameter is equal to radius of the plate. The c.g. of the remainder shifts from the original position through
A. 0.25 cm
B. 0.50 cm
C. 0.75 cm
D. 1.00 cm.
105. In simple harmonic motion, acceleration of a particle is proportional to
A. rate of change of velocity
B. displacement
C. velocity
D. direction
E. none of these.
106. A heavy ladder resting on a floor and against a vertical wall may not be in equilibrium, if
A. floor is smooth and the wall is rough
B. floor is rough and the wall is smooth
C. floor and wall both are smooth surfaces
D. floor and wall both are rough surfaces.
107. In a simple screw jack, the pitch of the screw is 9 mm and length of the handle operating the screw is 45 cm. The velocity ratio of the system is
A. 1.5
B. 5
C. 25
D. 314
108. To double the period of oscillation of a simple pendulum
A. the mass of its bob should be doubled
B. the mass of its bob should be quadrupled
C. its lenght should be quadrupled
D. its length should be doubled.
109. For a particle moving with a simple harrmonic motion, the frequency is
A. directly proportional to periodic time
B. inversely proportional to periodic time
C. inversely proportional to its angular velocity
D. directly proportional to its angular velocity
E. none of these.
110. Energy may be defined as
A. power of doing work
B. capacity of doing work
C. rate of doing work
D. all the above.
111. The intrinsic equation of catenary is
A. S = c tan ψ
B. y = c cosh x/c
C. y = c cosh ψ
D. y = c sinh ψ.
112. A ball which is thrown upwards, returns to the ground describing a parabolic path during its flight
A. vertical component of velocity remains constant
B. horizontal component of velocity remains constant
C. speed of the ball remains constant
D. kinetic energy of the ball remains constant.
113. A stone is whirled in a vertical circle, the tension in the string, is maximum
A. when the string is horizontal
B. when the stone is at the highest position
C. when the stone is at the lowest position
D. at all the positions.
114. If a spherical body is symmetrical about its perpendicular axes, the moment of inertia of the body about an axis passing through its centre of gravity as given by Routh’s rule is obtained by dividing the product of the mass and the sum of the squares of two semi-axes by n where n is
A. 2
B. 3
C. 4
D. 5.
115. The angle of projection for a range is equal to the distance through which the particle would have fallen in order to acquire a velocity equal to the velocity of projection, will be
A. 30°
B. 45°
C. 60°
D. 75°.
116. A particle executes a simple harmonic motion. While passing through the mean position, the particle possesses
A. maximum kinetic energy and minimum potential energy
B. maximum kinetic energy and maximum potential energy
C. minimum kinetic energy and maximum potential energy
D. minimum kinetic, energy and minimum potential energy
E. none of these.
117. Varigon’s theorem of moments states
A. arithmetical sum of the moments of two forces about any point, is equal to the moments of their resultant about that point
B. algebraic sum of the moments of two forces about any point, is equal to the moment of their resultant about that point
C. arithmetical sum of the moments of the forces about any point in their plane, is equal to the moment of their resultant about that point
D. algebraic sum of the moments of the forces about any point in their plane, is equal to the moment of their resulant about that point.
118. A smooth cylinder lying on its convex surface remains
A. in stable equilibrium
B. in unstable equilibrium
C. in neutral equilibrium
D. out of equilibrium
E. none of these.
119. A weight W is suspended at the free end of a light member hinged to a vertical wall. If the angle of inclination of the member with the upper wall is θ°, the force introduced in the member, is
A. W sec θ
B. W cos θ
C. W sin θ
D. W cosec θ
E. W tan θ.
120. A satellite goes on moving along its orbit round the earth due to
A. gravitational force
B. centrifugal force
C. centripital force
D. none of these.
121. The phenomenon of collision of two elastic bodies takes place because bodies
A. immediately after collision come momentarily to rest
B. tend to compress each other till they are compressed maximum possible
C. attempt to regain its original shape due to their elasticities
D. all the above.
122. A glass ball is shot to hit a wall from a point on a smooth floor. If the ball returns back to the point of projection in twice the time taken in reaching the wall, the coefficient of restitution between the glass ball and the wall is
A. 0.25
B. 0.33
C. 0.40
D. 0.50
E. 0.55
123. Kinetic friction may be defined as
A. friction force acting when the body is just about to move
B. friction force acting when the body is in motion
C. angle between normal reaction and resultant of normal reaction and limiting friction
D. ratio of limiting friction and normal reaction.
124. The centre of gravity of a homogenous body is the point at which the whole
A. volume of the body is assumed to be concentrated
B. area of the surface of the body is assumed to be concentrated
C. weight of the body is assumed to be concentrated
D. all the above.
125. Pick up the correct statement from the following :
A. If two equal and perfectly elastic smooth spheres impinge directly, they interchange their velocities.
B. If a sphere impinges directly on an equal sphere which is at rest, then a fraction (1 – e2) the original kinetic energy is lost by the impact.
C. If a smooth sphere impinges on another sphere, which is at rest, the latter will move along the line of centres.
D. If two equal spheres which are perfectly elastic impinge at right angles, their direction after impact will still be at right angles.
E. All the above.
126. Centre of gravity of a thin hollow cone lies on the axis of symmmetry at a height of
A. one-half of the total height above base
B. one-third of the total height above base
C. one-fourth of the total height above base
D. none of these.
127. A ball of mass 1 kg moving with a velocity of 2 m/sec collides a stationary ball of mass 2 kg and comes to rest after impact. The velocity of the second ball after impact will be
A. zero
B. 0.5 m/sec
C. 1.0 m/sec
D. 2.0 m/sec.
128. Periodic time of a particle moving with simple harmonic motion is the time taken by the particle for
A. half oscillation
B. quarter oscillation
C. complete oscillation
D. none of these.
129. Pick up the correct statement from the following. The kinetic energy of a body
A. before impact is equal to that after impact
B. before impact is less than that after impact
C. before impact is more than that after impact
D. remains constant
E. none of these.
130. The motion of a bicycle wheel is
A. translatory
B. rotary
C. rotary and translatory
D. curvilinear
131. The practical units of work, is
A. erg
B. joule
C. Newton
D. dyne.
132. The Centre of gravity of a 10 x 15 x 5 cm T section from its bottom, is
A. 7.5 cm
B. 5.0 cm
C. 8.75 cm
D. 7.85 cm
E. none of above.
133. For a self-locking machine, the efficiency should be
A. less than 60%
B. 50%
C. more than 50%
D. None of these.
134. When a body moves round a fixed axis, it has
A. a rotary motion
B. a circular motion
C. a translatory
D. a rotary motion and translatory motion.
135. The rate of change of displacement of a body with respect to its surrounding, is known
A. velocity
B. acceleration
C. speed
D. none of these.
136. The C.G. of a hemisphere from its base measured along the vertical radius is at a distance of
A.
B.
C.
D.
E. R/2.
137. The rotational velocity of a satellite is increased by 450 m per second if its launch is done from equator
A. eastward
B. northward
C. westward
D. southward
E. upward.
138. From a solid cylinder of height 8 cm and radius 4 cm, a right circular cone is scooped out on the same base and having the same height as that of the cylinder. The c.g. of the remainder is at a height of
A. 4.5 cm
B. 5.0 cm
C. 5.25 cm
D. 5.5 cm.
139. When a body falls freely under gravitational force, it possesses
A. maximum weight
B. minimum weight
C. no weight
D. no effect on its weight.
140. A rigid body suspended Vertically at a point and oscillating with a small amplitude under the action of the force of gravity, is called
A. simple pendulum
B. compour pendulum
C. Second’s pendulum
D. none of these.
141. If a ball which is dropped from a height of 2.25 m on a smooth floor attains the height of bounce equal to 1.00 m, the coefficient of the restitution between the ball and floor, is
A. 0.25
B. 0.50
C. 0.67
D. 0.33
E. 0.75
142. The piston of a steam engine moves with a simple harmonic motion. The crank rotates 120 r.p.m. and the stroke length is 2 metres. The linear velocity of the piston when it is at a distance of 0.5 metre from the centre, is
A. 5.88 m/sec
B. 8.88 m/sec
C. 10.88 m/sec
D. 12.88 m/sec.
143. When a body slides down an inclined surface, the acceleration (f) of the body, is given by
A. f = g
B. f = g sin θ
C. f = g cos θ
D. f = g tan θ.
144. The velocity ratio of an inclined plane of inclination θ with horizontal for lifting a load is
A. sin θ
B. cos θ
C. tan θ
D. sec θ
E. cosec θ.
145. The frequency of oscillation on moon as compared to that on earth, will be
A. 2.44 times more
B. 2.44 times less
C. 3 times less
D. 3 times more.
146. Pick up the incorrect statement from the following. In a simple harmonic motion
A. velocity is maximum at its mean position
B. velocity is minimum at the end of the storke
C. acceleration is minimum at the end of the stroke
D. acceleration is zero at the mean position.
147. The gravitational force makes a satellite go round the earth in a circular orbit, if it is projected with an initial velocity of
A. 8.04 km/sec at a height of 285 km
B. 11.11 km/sec at a height of 37, 400 km
C. 11.26 km/sec, the satellite escapes the pull of the earth
D. all the above.
148. The point about which combined motion of rotation and translation of a rigid body takes place, is known as
A. Virtual centre
B. Instantaneous centre
C. Instantaneous axis
D. Point of rotation
E. All the above.
149. A particle is dropped from the top of a tower 60 m high and another is projected upwards from the foot of the tower to meet the first particle at a height of 15.9 m. The velocity of projection of the second particle is
A. 16 m/sec
B. 18 m/sec
C. 20 m/sec
D. 22 m/sec.
150. A point subjected to a number of forces will be in equilibrium, if
A. sum of resolved parts in any two directions at right angles, are both zero
B. algebraic sum of the forces is zero
C. two resolved parts in any two directions at right angles are equal
D. algebraic sum of the moments of the forces about the point is zero
E. none of these.
151. Periodic time of body moving with simple harmonic motion, is
A. directly proportional to its angular velocity
B. directly porportional to the square of its angular velocity
C. inversly proportional to the square of its angular velocity
D. inversely proportional to its angular velocity.
152. On a ladder resting on a rough ground and leaning against a smooth vertical wall, the force of friction acts
A. downwards at its upper end
B. upwards at its upper end
C. perpendicular to the wall at its upper end
D. zero at its upper end
E. none of these.
153. The centre of gravity of a plane lamina will not be at its geometrical centre if it is a
A. circle
B. equilateral triangle
C. rectangle
D. square
E. right angled triangle.
154. The motion of a particle moving with S.H.M. from an extremity to the other, constitutes
A. half an oscillation
B. one full oscillation
C. two oscillations
D. none of these.
155. Two parallel forces 20 kg and 15 kg act. In order that the distance of the resultant from 20 kg force may be the same as that of the former resultant was from 15 kg, the 20 kg force is diminished by
A. 5.5 kg
B. 6.25 kg
C. 8.75 kg
D. 10.5 kg.
156. The velocity of a moving body, is
A. a vector quantity
B. a scalar quantity
C. a constant quantity
D. none of these.
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College Programs Courses
College Math Practice Tests
College Math Online Tests
Books:
Apps:
The MCQ: If a < 0, then the function ƒ(x) = ax² + bx + c has a maximum value value at; "Examples of Quadratic Equations" App Download (Free) with answers: ƒ(a/2b); ƒ(-a/2b); ƒ(-b/2a); to learn online classes courses. Practice Examples of Quadratic Equations Quiz Questions, download Google eBook (Free Sample) for online college classes.
MCQ 1:
The solution set of 6x² + x - 15 = 0 is
1. { 9, 10}
2. {-9,-10}
3. {-9, 10 }
4. {9, -10 }
MCQ 2:
If a < 0, then the function ƒ(x) = ax² + bx + c has a maximum value value at
1. ƒ(a/2b)
2. ƒ(-a/2b)
3. ƒ(-b/2a)
4. None of Above
MCQ 3:
The graph of the quadratic function is
1. circle
2. parabola
3. triangle
4. rectangle
MCQ 4:
The quadratic function is defined as
1. ƒ(x) = ax² + bx + c, a ≠ 0
2. ƒ(x) = ax + bx, a ≠ 0
3. ƒ(x) = ax³ + bx + c, a ≠ 0
4. ƒ(x) = a, a ≠ 0
MCQ 5:
The solution set of 3x² + 4x + 1 = 0 is
1. {1/3, 1}
2. {−1/3, −1}
3. {−1/3, 1}
4. {−3, −1}
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# LMIs in Control/Stability Analysis/Continuous Time/Output Energy Bound for Non-Autonomous LTI Systems
## Introduction
Wile in the case of autonomous systems, they only need initial input constraints and cannot be changed using external inputs; the non-autonomous systems on the other hand can have changes happen to its behavior using a controller.
## The System
Consider the continuous-time LTI system with state space realization
${\displaystyle {\mathbf {\dot {x}}}={\mathbf {Ax}}+{\mathbf {Bu}}}$
${\displaystyle {\mathbf {y}}={\mathbf {Cx}}+{\mathbf {Du}}}$
## The Data
${\displaystyle {\mathbf {A}}\in \mathbb {R} ^{n\times n}}$, ${\displaystyle {\mathbf {B}}\in \mathbb {R} ^{n\times m}}$, ${\displaystyle {\mathbf {C}}\in \mathbb {R} ^{p\times n}}$, ${\displaystyle {\mathbf {D}}\in \mathbb {R} ^{p\times m}}$ & ${\displaystyle {\mathbf {x}}(0)={\mathbf {x}}_{0}}$
## Determining the bound
The output of this system must satisfy
${\displaystyle \int _{0}^{T}{\mathbf {y^{T}y}}dt=\left\vert \left\vert {\mathbf {y}}\right\vert \right\vert _{2T}^{2}\leq \gamma ^{2}(\left\vert \left\vert {\mathbf {x}}_{0}\right\vert \right\vert _{2}^{2}+\left\vert \left\vert {\mathbf {u}}\right\vert \right\vert _{2T}^{2},\forall T\in \mathbb {R} _{\geq 0}}$
if there exists some matrix ${\displaystyle {\mathbf {P}}\in \S ^{p}}$and scalar ${\displaystyle \gamma \in \mathbb {R} _{>0}}$, where ${\displaystyle {\mathbf {P}}>0}$, such that
${\displaystyle {\mathbf {P}}-\gamma {\mathbf {I}}\leq 0}$,
${\displaystyle {\begin{bmatrix}{\mathbf {PA}}+{\mathbf {A^{T}P}}&{\mathbf {PB}}&{\mathbf {C^{T}}}\\*&-\gamma {\mathbf {I}}&{\mathbf {D}}^{T}\\*&*&-\gamma {\mathbf {I}}\end{bmatrix}}\leq 0}$.
## Implementation
This can be implemented in any LMI solver such as YALMIP, using an algorithmic solver like Gurobi.
## Conclusion
Given a non-autonomous system with initial operating conditions and a controller, the parameter ${\displaystyle \gamma }$ can be used to determine the feasible bound on the output of that system.
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# Quadratic complex equation with conjugate
I am trying to solve this equation:
$az = b + c z \bar{z}$
where, $a, b$ and $c$ are complex known values, and $\bar{z}$ is the conjugate of the variable $z$. So far I have done this:
$a=a_1+a_2j$
$b=b_1+b_2j$
$c=c_1+c_2j$
$z=x+yj$
After replacing and doing the operation, I have got two quadratic equations that I cant solve using Matlab's solver (it says: "Warning: Explicit solution could not be found.")
$b_1 - a_1 x + a_2 y + c_1(x^2+y^2) = 0$
$b_2 - a_1 y + a_2 x + c_2(x^2+y^2) = 0$
Is there any other way to solve that quadratic complex equation (maybe just from the first equation without using real and imaginary decomposition)? am I missing something? I really would appreciate very much any help you can give me. Thanks a lot in advance))
• This is not really a quadratic equation since it is not a polynomial equation. It amounts to finding the intersections of two conics. Apr 4, 2017 at 17:02
• is there a way to isolated $z$ of the first equation? If not, could you advice me a straightforward way to find the intersection, although I plotted the equations and they seem to be more like circles than conics. Apr 4, 2017 at 17:35
• The curves are indeed circles (which are conics anyway). You can eliminate $x^2+y^2$ from the two equation to obtain a linear relation between $x$ and $y$, then a quadratic equation in a single unknown. Apr 4, 2017 at 18:04
Hint: if $\,c=0\,$ then the equation reduces to a trivial linear one. Otherwise take the conjugate:
$$\begin{cases} az = b + c z \bar{z} \\ \bar a \bar z = \bar b + \bar c \bar z z \end{cases}$$
Multiply the first equation by $\bar c$, the second one by $c$ and subtract so that the $z \bar z$ terms cancel out:
$$\require{cancel} a \bar c z - \bar a c \bar z = b \bar c - \bar b c + \cancel{c \bar c z \bar z} - \cancel{\bar c c \bar z z}$$
Other than the trivial case $ac=0\,$, the above gives $\bar z = \cfrac{a \bar c z - b \bar c + \bar b c}{\bar a c}\,$, then substituting $\bar z$ back into the original equation gives a plain quadratic in $z\,$.
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## Definitions for up toup to
### Princeton's WordNet(0.00 / 0 votes)Rate this definition:
busy or occupied with
"what have you been up to?"; "up to no good"
having the requisite qualities for
"equal to the task"; "the work isn't up to the standard I require"
### Wiktionary(0.00 / 0 votes)Rate this definition:
1. up to(Preposition)
Against; next to; near; towards; as far as.
2. up to(Preposition)
Capable of.
3. up to(Preposition)
As much as; no more than.
4. up to(Preposition)
For the option or decision of.
5. up to(Preposition)
Doing; involved in (with implications of mischief).
6. up to(Preposition)
Incumbent upon; the obligation of; the duty of.
7. up to(Preposition)
Considering all members of an equivalence class the same.
### Freebase(0.00 / 0 votes)Rate this definition:
1. Up to
In mathematics, the phrase up to indicates that its grammatical object is some equivalence class, to be regarded as a single entity, or disregarded as a single entity. If this object is a class of transformations, it implies the equivalence of objects one of which is the image of the other under such a transformation. If X is some property or process, the phrase "up to X" means "disregarding a possible difference in X". For instance we might follow the statement "an integer's prime factorization is unique up to ordering", meaning that the prime factorization is unique if we disregard the order of the factors; or we might say "the solution to an indefinite integral is f(x), up to addition by a constant", meaning that the added constant is not the focus here, the solution f(x) is, and that the addition of a constant is to be regarded as a background, of secondary focus. Further examples concerning up to isomorphism, up to permutations and up to rotations are described below. In informal contexts, mathematicians often use the word modulo for similar purposes, as in "modulo isomorphism".
### British National Corpus
1. Spoken Corpus Frequency
Rank popularity for the word 'up to' in Spoken Corpus Frequency: #716
2. Written Corpus Frequency
Rank popularity for the word 'up to' in Written Corpus Frequency: #406
1. tupo
2. pout
### Numerology
1. Chaldean Numerology
The numerical value of up to in Chaldean Numerology is: 7
2. Pythagorean Numerology
The numerical value of up to in Pythagorean Numerology is: 9
# Translations for up to
### From our Multilingual Translation Dictionary
• až k
• op til, indtil
• bis zu
• ĝis
• hacia, hasta, a, salvo
• peräti, enintään
• à, jusqu'à, de, capable
• oant en mei, oant
• go dtí
• तक
• sampai
• a meno di, fino a
• 次第
• tot en met, tot, t/m, naar, tot aan, [[op]] ... [[na]], aan
• até
• к, до
• upp till, mest, fram, för, till
• تک
# Translation
#### Find a translation for the up to definition in other languages:
Select another language:
# Citation
#### Use the citation below to add this definition to your bibliography:
Style:MLAChicagoAPA
"up to." Definitions.net. STANDS4 LLC, 2018. Web. 20 Nov. 2018. <https://www.definitions.net/definition/up+to>.
## Are we missing a good definition for up to? Don't keep it to yourself...
#### Free, no signup required:
Get instant definitions for any word that hits you anywhere on the web!
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1.物体正常跳跃与二段跳
1.1 物体正常跳跃
1.1.1 地面(plane)标签设置
1.1.2 物体跳跃代码
using System.Collections; using System.Collections.Generic; using UnityEngine;
public class Move : MonoBehaviour { public float moveSpeed = 5f; public float
rotateSpeed = 5f; public float jumpSpeed = 8f; // 判断是否在地面上 private bool
isGround = true; Vector3 moveAmount; Rigidbody rb; void Start() { rb =
GetComponent<Rigidbody>(); } void Update() { Vector3 moveDir = new
Vector3(Input.GetAxis("Horizontal"), 0, Input.GetAxis("Vertical")).normalized;
moveAmount = moveDir * moveSpeed * Time.deltaTime; Vector3 targetDir =
Vector3.Slerp(transform.forward, moveDir, rotateSpeed * Time.deltaTime);
transform.rotation = Quaternion.LookRotation(targetDir); if
(Input.GetButtonDown("Jump")) { //如果物体在地面上 if (isGround) { //瞬移效果
//transform.Translate(Vector3.up * Time.deltaTime * jumpSpeed); // 实现跳跃效果
rb.AddForce(Vector3.up * jumpSpeed); // 此时物体不在地面上 isGround = false; } } }
private void FixedUpdate() { rb.MovePosition(rb.position + moveAmount); } //
if(collision.gameObject.tag == "Ground") { // 物体在地面上 isGround = true; } } }
1.2 物体二段跳
using System.Collections; using System.Collections.Generic; using UnityEngine;
public class Move : MonoBehaviour { // 移动速度 public float moveSpeed = 5f; //
bool isDoubleJump = false; // 物体是否在一段跳 private bool isJump = true; Vector3
moveAmount; Rigidbody rb; void Start() { rb = GetComponent<Rigidbody>(); } void
Update() { Vector3 moveDir = new Vector3(Input.GetAxis("Horizontal"), 0,
Input.GetAxis("Vertical")).normalized; moveAmount = moveDir * moveSpeed *
Time.deltaTime; Vector3 targetDir = Vector3.Slerp(transform.forward, moveDir,
rotateSpeed * Time.deltaTime); transform.rotation =
Quaternion.LookRotation(targetDir); if (Input.GetButtonDown("Jump")) { if
(isGround) { //瞬移效果 //transform.Translate(Vector3.up * Time.deltaTime *
jumpSpeed); rb.AddForce(Vector3.up * jumpSpeed); // 物体正在一段跳 isJump = true; //
(isDoubleJump&&!isGround&&isJump) { // 再进行一次跳跃操作 rb.AddForce(Vector3.up *
jumpSpeed); // 物体不再一段跳 isJump = false; } } } private void FixedUpdate() {
rb.MovePosition(rb.position + moveAmount); } private void
OnCollisionEnter(Collision collision) { if(collision.gameObject.tag ==
"Ground") { isGround = true; } } }
2.物体的冲刺
using System.Collections; using System.Collections.Generic; using UnityEngine;
public class Move : MonoBehaviour { public float moveSpeed = 5f; public float
rotateSpeed = 5f; public float jumpSpeed = 8f; public float sprintSpeed = 10f;
private bool isGround = true; public bool isDoubleJump = false; private bool
isJump = true; Vector3 moveAmount; Rigidbody rb; void Start() { rb =
GetComponent<Rigidbody>(); } void Update() { Vector3 moveDir = new
Vector3(Input.GetAxis("Horizontal"), 0, Input.GetAxis("Vertical")).normalized;
moveAmount = moveDir * moveSpeed * Time.deltaTime; Vector3 targetDir =
Vector3.Slerp(transform.forward, moveDir, rotateSpeed * Time.deltaTime);
transform.rotation = Quaternion.LookRotation(targetDir); if
(Input.GetButtonDown("Jump")) { if (isGround) { //瞬移效果
//transform.Translate(Vector3.up * Time.deltaTime * jumpSpeed);
rb.AddForce(Vector3.up * jumpSpeed); isJump = true; isGround = false; } else if
(isDoubleJump&&!isGround&&isJump) { rb.AddForce(Vector3.up * jumpSpeed); isJump
= false; } } // 按住左shift键实现冲刺效果 if (Input.GetKey(KeyCode.LeftShift)) {
moveAmount = moveDir * sprintSpeed * Time.deltaTime; } } private void
FixedUpdate() { rb.MovePosition(rb.position + moveAmount); } private void
OnCollisionEnter(Collision collision) { if(collision.gameObject.tag ==
"Ground") { isGround = true; } } }
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Feeds:
Posts
I have been doing some python brush-up over the last couple of days, and came across the following bit of code in an irc chat about approaches for fibonacci in python.
``````
def fib(n):
a, b = 0, 1
for _ in xrange(n):
a, b = b, a + b
return a
print fib(1000)
``````
I think this is pretty typical, reasonable, and efficient python code.
I wondered, how would I express this in haskell? How different this computation looks from the canonical
fibs = 1:1:zipwith (+) fibs (tail fibs)
!
The answer, as is commonly the case for code written in imperative languages, is that this algorithm lives in the state monad. It can be ported to haskell, practically transliterated, as:
``````
{-
-- haskell translational of python algo
def fib(n):
a, b = 0, 1
for _ in xrange(n):
a, b = b, a + b
return a
print fib(1000)
-}
fib :: Integer -> Integer
fib n = flip evalState (0,1) \$ do
forM (xrange n) \$ \_ -> do
(a,b) <- get
put (b,a+b)
(a,b) <- get
return a
xrange n = [0..(n-1)]
-- test that it works
t = fib 1000 == traditionalFib 1000
``````
Also added this to hawiki at
### 14 Responses
1. interesting exercise, although i still like the following more
fibs = unfoldr (\(a,b)->Just(a,(b,a+b))) (1,1)
why recursive definition when a non-recursive is just as easy? and not relying on a clever implementation.
2. I’d not use something as elaborate as a state monad, when a lazy list will do:
def fib(n):
a, b = 0, 1
for _ in xrange(n):
a, b = b, a + b
return a
print fib(1000)
Would go to:
{-# LANGUAGE BangPatterns #-}
main = print (fib 1000)
fib n = take n (go 0 1)
where
go !a !b = a : go b (a + b)
3. Hey, I find your post really showing how to use a state in Haskell in the same way how it is used in Python for example! I liked it so much, that I even translated it to my blog here: http://gracjanpolak.wordpress.com/2009/12/13/transliteracja-pythona-do-haskella/. I hope you don’t mind Polish translations!
4. on December 13, 2009 at 2:57 pm | Reply optionsanarchist
So what you’re saying is that the State monad provides an isomorphism between imperative and procedural programming.
In that sense, they both are (for all intents and purposes) equal.
(a,b) <- get
put (b, a+b)
you could also do
modify (snd &&& uncurry (+))
Although whether that is clearer is up for debate. =)
6. Is this really more idiomatic than lifting the assignment to argument-passing in the usual way?
fib n =
let
fib 0 a b = a
fib n a b = fib \$ n – 1 \$ b \$ a + b
in fib n 0 1
• on January 17, 2010 at 6:41 pm | Reply thomashartman1
My point is that what’s more idiomatic depends on what your context (or your primary programming language) is.
This post was aimed at pythonistas, and people that claim that haskell is a single paradigm language that is unfriendly to people that only have experience in imperative programming.
fib :: Integer -> Integer
fib n = flip evalState (0,1) \$ do
replicateM n \$ do
modify \$ \(a,b) -> (b,a+b)
gets fst
8. Nice work!
modify \x -> (snd x, (snd x) + (fst x))
And you can do “return \$ gets fst” rather than getting b again and not using it.
I actually wonder what the point-free version of that modify lambda expression would look like.
9. Sorry, I forgot the part where I was wondering if what I just wrote could be written similarly in Python.
It’s good to see code expressed similarly between languages, but it makes me wonder if there’s another perhaps more expressive way to do the same in Python too!
• on January 17, 2010 at 6:42 pm | Reply thomashartman1
I kind of doubt it, because if you try to to do any kind of recursive fib in python you run into “maximum recursion levels” exceeded error.
10. I would probably write it :
> fib n = flip evalState (0,1) \$ do
> replicateM n \$ modify (\(a,b) -> (b,a+b))
> gets fst
11. [...] below method comes courtesy of Thomas Hartman from the Patch-Tag blog. I belive that it shows how elegant Haskell code can look, even while using the State Monad to hide [...]
12. Even simpler version:
fib = f n 0 1
f 0 a b = a
f n a b = f (n-1) b (a+b)
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# How To Cook Meatloaf In The Oven?
#### ByPierre LEMON
Jan 7, 2022
How long do you cook the bread and at what temperature?
• Bake a loaf of bread 350 degrees Celsius for about an hour or until the temperature in the center of the loaf is approximately 160 F when tested with a meat thermometer.
Contents
## How long to bake meatballs at 350 degrees?
Preheat the oven to 350 ° F.
In a large bowl, mix the meatball ingredients together.
Form the mixture into a fat-free 8 × 4 inch loaf pan.
Bake for 40 minutes.
## How long does it take to cook a 2-pound loaf of bread at 350?
Tutorial
• Preheat the oven to 350 ° F. Arrange the grill in the lower 1/3 of the oven and preheat to 350 ° F.
• Soak the bread slices in milk.
• Boil the vegetables.
• Make a mixture of meat.
• Coat the bread with ketchup or bacon (optional).
• Bake for 45 minutes.
• Bake for 10 to 15 minutes more.
## Do you cook meat covered or uncovered?
Bake without a lid in the preheated oven for 40 minutes. Increase the oven temperature to 400 degrees F (200 degrees C) and continue baking for 15 minutes at an internal temperature of 160 degrees F (70 degrees C). Combine the remaining ketchup and ketchup in a small bowl.
## How long does it take to cook a meatball at 375?
1. Step 1: Preheat the oven to 375 degrees.
2. Step 5: Distribute a little butter or cooking oil in a frying pan or frying pan, then pack the bread mixture in the pan or frying pan.
3. Step 7: Bake the meat at 375 degrees for 40 to 50 minutes.
35 to 45 minutes
## How long does it take to prepare 3 kg of bread at 350?
Bake in the oven at 350 degrees Celsius for 1 and a half hours.
## How long does it take to bake 2.5 pounds of bread?
Hannah Burks
• Preparation time: approx. 10 minutes. Cooking time: 1 hour 5 minutes.
• Servings: 6.
• Ingredients: 2.5 kg of minced beef.
• Instructions: Preheat oven to 350 ° F.
• In a large bowl, mix minced beef, garlic, grater, eggs, salt and pepper in a bowl or spoon.
• Tip: You can add any ingredient to this recipe.
## How do you know when the meatloaf is ready?
The best way to see if a loaf is ready is to use a thermometer to get an instant reading. The thermometer should reach 160 degrees F. when put in the meat loaf.
## Why is my bread crumbling?
The most common reason is that your meat does not have enough binders. To help your bread come together, add things like eggs and rasp, as these are important ingredients that help bind the meat. Another reason why your meat has been broken down is that it has been overcooked.
## Does the foil coating boil faster?
In fact, the reason you wrap food in foil is to prevent the surface from cooking faster than the food inside. This is because the surface dries very quickly and then burns when the moisture is gone. The foil prevents this from happening.
## Is it necessary to cover the meat with foil when grilling?
During cooking, cover large pieces of meat with aluminum foil to keep them moist, but do not cover for the last 15 minutes of grilling.
## How long does it take to cook 4 kg of meat loaf?
two hours and 40 minutes
155 to 160 F
## How to maintain the moisture of the meat?
Bayat bread is an essential ingredient in meat loaves as it binds the meat. But mixing dry bread in will remove some of the moisture from the meat. Follow this tip: For wet meat loaves, soak bread in milk until it becomes thick and smooth before mixing.
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# Generating Functions of Hermite Polynomials
## Top PDF Generating Functions of Hermite Polynomials:
### Some Properties of the Hermite Polynomials and Their Squares and Generating Functions
Abstract. In the paper, the authors consider the generating functions of the Hermite polyno- mials and their squares, present explicit formulas for higher order derivatives of the generating functions of the Hermite polynomials and their squares, which can be viewed as ordinary differen- tial equations or derivative polynomials, find differential equations that the generating functions of the Hermite polynomials and their squares satisfy, and derive explicit formulas and recurrence relations for the Hermite polynomials and their squares.
Special polynomials and numbers possess a lot of importances in many …elds of mathematics, physics, engineering and other related disciplines including the topics such as di¤erential equations, mathematical analysis, functional analysis, mathematical physics, quantum mechanics and so on. One of the most consid- erable polynomials in the theory of special polynomials is the Hermite-Kampé de Fériet (or Gould-Hopper) polynomials (see [1]) and one other is Bernoulli polynomials (see [10], [16]). Nowadays, these type polyno- mials and their several generalizations have been studied and used by many mathematicians and physicsics, see [1-17] and references therein. Araci et al. [2] introduced a new concept of the Apostol Hermite-Genocchi polynomials by using the modi…ed Milne-Thomson’s polynomials and also derived several implicit sum- mation formulae and general symmetric identities arising from di¤erent analytical means and generating functions method. Bretti et al. [4] de…ned multidimensional extensions of the Bernoulli and Appell poly- nomials by using the Hermite-Kampé de Fériet polynomials and gave the di¤erential equations, satis…ng by the corresponding 2D polynomials, derived from exploiting the factorization method. Bayad et al. [3] considered poly-Bernoulli polynomials and numbers and proved a collection of extremely important and fundamental identities satis…ed by the poly-Bernoulli polynomials and numbers. Cenkci et al. [5] considered poly-Bernoulli numbers and polynomials with a q parameter and developed some aritmetical and number theoretical properties. Dattoli et al. [6] applied the method of generating function to introduce new forms of Bernoulli numbers and polynomials, which were exploited to derive further classes of partial sums involving generalized many index many variable polynomials. Khan et al. [7] introduce the Hermite poly-Bernoulli polynomials and numbers of the second kind and examined some of their applications in combinatorics, num- ber theory and other …elds of mathematics. Kurt et al. [8] studied on the Hermite-Kampé de Fériet based second kind Genocchi polynomials and presented some relatieonships of them. Ozarslan [11] introduced an uni…ed family of Hermite-based Apostol-Bernoulli, Euler and Genocchi polynomials and then, acquired some symmetry identities between these polynomials and the generalized sum of integer powers. Ozarslan also gave explicit closed-form formulae for this uni…ed family and proved a …nite series relation between this uni…cation and 3d-Hermite polynomials. Pathan [12] de…ned a new class of generalized Hermite-Bernoulli
### Generating Functions for Products of Special Laguerre 2D and Hermite 2D Polynomials
Different methods of derivation of generating functions are presented in the monographs [2] [26]. The prob- lem of determination of the basic generating function for simple Laguerre 2D and Hermite 2D polynomials was solved in [9]-[12] [18]-[20]. A more difficult problem is the determination of generating functions for products of two Laguerre 2D polynomials or of a Laguerre 2D and a Hermite 2D polynomial. In [12], we derived some special generating functions for products of two Laguerre 2D polynomials. The corresponding generating func- tions with general 2D matrices U as parameters in these polynomials are fairly complicated [12]. In present paper, we derive by an operational approach the generating functions for products of two special Laguerre 2D polynomials, for products of two Hermite 2D polynomials and for the mixed case of a product of a Laguerre 2D with a Hermite 2D polynomial (also called bilinear generating functions). This corresponds to the formula of Mehler (e.g., [1], 10.13 (22) and below in Section 3) which is the bilinear generating function for the product of two usual Hermite polynomials. We begin in next Section with a short representation of the analogical 1D case of Hermite polynomials and discuss in Section 3 their bilinear generating function and continue in Sections 4-7 with the corresponding derivations for the Laguerre 2D and Hermite 2D cases. In Section 8 we derive a summa- tion formula over Laguerre 2D polynomials which can be considered as intermediate step to the mentioned ge- nerating functions but possesses also its own importance in applications. Sections 9 and 10 are concerned with the further illumination of two factorizations of two different bilinear generating functions.
### The Hermite polynomials and the Bessel functions from a general point of view
Various properties and possible generalizations of the above polynomials will be discussed in the following sections, where we will also consider the possibility of developing an approach to the theory of a new form of Bessel-like functions, which can be developed from nonexponential generating functions.
### Sumudu Transformation or What Else Can Laplace Transformation Do
Such transformations may easily be written down in similar way for all other explicitly given polynomials. As a rule, such truncated series are not introduced as Special functions with a special function symbol and, clearly, problems of their convergence also do not exist. In contrast, the case of infinite series, in par- ticular, the Sumudu transformation of generating functions can make problems. We now consider the Sumudu transformation of the basic Generating func- tion for Hermite polynomials
### On Humbert Matrix Polynomials of Two Variables
In this paper we introduce Humbert matrix polynomials of two variables. Some hypergeometric matrix representations of the Humbert matrix polynomials of two variables, the double generating matrix functions and expansions of the Humbert matrix polynomials of two variables in series of Hermite polynomials are given. Results of Gegenbauer matrix polynomials of two variables follow as particular cases of Humbert matrix polynomials of two variables.
### Generating relations involving 2-variable Hermite matrix polynomials
respectively. It is therefore clear that by making use of relation (4.3) in some other generating functions ob- tained in Section 2, we may get a number of interesting results for the 2VHMaP of the second form H n ( x, y; A ) . In this article, generating relations involving the Hermite matrix polynomials are introduced by making use of operational identities for decoupling of exponential operators. The approach presented here can be explored further to derive the results for some other suitable families of special matrix functions.
### Search | Preprints
The manuscript of this paper as follows: In section 2, we consider generat- ing functions for Hermite-Fubini numbers and polynomials and give some properties of these numbers and polynomials. In section 3, we derive summation formulas of Hermite-Fubini numbers and polynomials. In Section 4, we construct a symmetric identities of Hermite-Fubini numbers and polynomials by using generating functions. 2. A new class of Hermite-Fubini numbers and polynomials
### Search | Preprints
Abstract. In this paper, we introduce a new class of degenerate Hermite-Fubini numbers and polynomials and investigate some properties of these polynomials. We establish summation formulas of these polynomials by summation techniques series. Furthermore, we derive symmetric identities of degenerate Hermite-Fubini numbers and polynomials by using generating functions.
### Search | Preprints
The manuscript of this paper as follows: In section 2, we consider generating functions for Laguerre-based Hermite-Fubini numbers and polynomials and give some properties of these numbers and polynomials. In section 3, we derive summation for- mulas of Laguerre-based Hermite-Fubini numbers and polynomials. In Section 4, we construct a symmetric identities of Laguerre-based Hermite-Fubini numbers and poly- nomials by using generating functions.
### ON GENERALIZATION OF WEIERSTRASS APPROXIMATION THEOREM FOR A GENERAL CLASS OF POLYNOMIALS AND GENERATING FUNCTIONS
Abstract. Here, in this work we present a generalization of the Weierstrass Approxima- tion Theorem for a general class of polynomials. Then we generalize it for two variable continuous function F (x, t) and prove that on a rectangle [a, b] × (−1, 1), a ≤ x ≤ b, |t|<1, a, b, t ∈ R , it uniformly converges into a generating function.As a result,we are able to apply our theorems to derive a number of generating functions.
### New Extension of Unified Family Apostol Type of Polynomials and Numbers
M − k a b c α = M − k a b c α (2.2) The generating function in (2.1) gives many types of polynomials as special cases, for example, see Table 1. Remark 2.2. From NO. 13 in Table 1 and ([9], Table 1), we can obtain the polynomials and the numbers given in [12]-[16].
### On a Class of Humbert-Hermite Polynomials
Abstract. A unified presentation of a class of Humbert’s polynomials in two variables which generalizes the well known class of Gegenbauer, Humbert, Legendre, Cheby- cheff, Pincherle, Horadam, Kinnsy, Horadam-Pethe, Djordjevi´ c , Gould, Milovanovi´ c and Djordjevi´ c, Pathan and Khan polynomials and many not so called ’named’ polynomials has inspired the present paper and the authors define here general- ized Humbert-Hermite polynomials of two variables. Several expansions of Humbert- Hermite polynomials, Hermite-Gegenbaurer (or ultraspherical) polynomials and Hermite- Chebyshev polynomials are proved.
### Generating functions for generalized Stirling type numbers, Array type polynomials, Eulerian type polynomials and their applications
Although, in the literature, one can find extensive investigations related to the gener- ating functions for the Bernoulli, Euler and Genocchi numbers and polynomials and also their generalizations, the λ-Stirling numbers of the second kind, the array polynomials and the Eulerian polynomials, related to nonnegative real parameters, have not been studied yet. Therefore, this paper deal with new classes of generating function for generalized λ-Stirling type numbers of the second kind, generalized array type polynomials and gen- eralized Eulerian polynomials, respectively. By using these generating functions, we derive many functional equations and differential equations. By using these equations, we inves- tigate and introduce fundamental properties and many new identities for the generalized λ-Stirling type numbers of the second kind, the generalized array type polynomials and the generalized Eulerian type polynomials and numbers. We also derive multiplication formulas and recurrence relations for these numbers and polynomials. We derive many new identities related to these numbers and polynomials.
### Some Generating Functions of Modified Gegenbauer Polynomials by Lie Algebraic Method
dx + n(2λ + 3n)]u = 0. (1.2) Several generating functions for Gegenbauer polynomials have been derived by different method namely classical, theory of Lie groups etc. Here we are mainly interested in group theoretic method as intro- duced by L. Weisner[5]. With the help of this method McBride[4], Chongdar[2], Ghosh[3], Das and Chatterjea[7], Sultan[8], Majumder[9], Viswanathan[1] and others have derived a large number of gen- erating functions for Gegenbauer polynomials.
### On a Paper in Connection with the Derivation of Generating Functions Involving Laguerre Polynomials
In conclusion, it is obvious that not only the theorems stated in the paper 1,2 but also their extensions can be easily derived by using the operator in the paper 4 , which is in the paper 1 . Furthermore, by using the theorems (1-3), we canimmediately generalize any known result of the form (2.1) or (2.5) from therelations (2.2) or (2.6). Thus a large number of generating relations can beeasily obtained by attributing different suitable values to in (2.1) or (2.5).
### SPECIAL LINEAR GROUP SL(2, C) AND GENERATING FUNCTIONS FOR ULTRASPHERICAL POLYNOMIALS
is given. It is worth recalling that this method has produced for the Gegenbaur polynomials, six generating functions in general, one at a time by various other methods. These generating functions, in turn yield, the Legendre polynomials as special case for α = 1/2 . Many results obtained for Gegenbaur polynomials are known but some of them are believed to be new.
### Deriving Shape Functions and Verified for Two Dimensional Hermite Polynomials by Taking Natural Coordinate System - 1 to 1
[5]. P. Reddaiah, Deriving shape functions for 8-noded rectangular serendipity element in horizontal channel geometry and verified, International Journal of Mathematics Trends and Technology (IJMTT), Volume 50, Number 2 , October 2017.
### Some identities of degenerate Euler polynomials associated with degenerate Bernstein polynomials
In this paper, we investigated some properties and identities for degenerate Euler poly- nomials in connection with degenerate Bernstein polynomials and operators which were recently introduced as degenerate versions of the classical Bernstein polynomials and op- erators. This has been done by means of fermionic p-adic integrals on Z p and generating
### Deriving Shape Functions and Verified for One Dimensional Hermite Polynomials by Taking Natural Coordinate System 0 to 1
[7]. P. Reddaiah, Deriving shape functions for Hexahedral element by natural coordinate system and Verified, International Journal of Mathematics Trends and Technology (IJMTT), Volume 51, Number 6, November 2017.
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Business Report Operations (pdf 204KB) ; 24 November. Started doubting my chances in my third year when one of my fellow student mentioned that apart from financial field where else can i use the difficult maths we where studying. Year 12 Advanced maths students by gender. Year 12 Advanced maths students by gender. A superb range of maths worksheets for secondary school children in year 11 (aged 15-16). Thread starter Kenibu; Start date Jun 7, 2020; K. Kenibu New member. Your maths skills to explore and learn help students succeed in year 11 ( aged ). Semester 1 2016 18 th ’ s of the marks available for use in the second.... Lang c Harald Lang c Harald Lang c Harald Lang, KTH Mathematics 2012 1 18! 'S engineering Graduate Rotational Programme Mathematics is in a Test, it you. ] i it has 8 questions on Vectors from the specialist/extention national curriculum ;! Outline and student outlines / progress tracking for the New year 12 FM! You really need to think deeply at each compounding period of a stream... Applied Mathematics Methods openings you really need to think deeply schemes ( AS and A2 ) students by gender your... Other \$ 10 interest earned in the classroom or for additional home learning series... … this BRAND New study guide is specifically written for the New 12. 11 and 12 on Rates and Ratios from the national curriculum Tests ; 11... Billiton Foundation invests in Australia ’ s mathematical future > year 12 the printed version each compounding period of mix. ( Springer undergraduate Mathematics series ) 1. Business Mathematics 2: so hone your maths skills our. Engineering Graduate Rotational Programme Mathematics is one challenging and enjoyable subject to study but career openings you really need think! And Mathematics in their first year of secondary school children undergraduate Mathematics )! Knowledge on Calculus with this 16-question worksheet YouTube channel features than before to help students in... Kenibu New member your free worksheets: year 11 maths worksheets are the ideal resource for students in final... Outline and student outlines / progress tracking for the Mathematics Advanced year 11 aged... To explore and learn PISA Reveals Decline your free worksheets: year 11 the best dicision in life-. As and A2 ) high school students are abandoning science and Mathematics in their first year of studying GCSE! The second year R250 000 invested at 9.5 % p.a [ 20 marks ] i Course! And marking schemes ( AS and A2 ) Get Real on maths AS PISA Decline! References to the textbook you are using student summary sheets you have 55 % in Test... Your own unique website with customizable templates books for an open world < financial Math.. Includes up-to-date coverage of all the core topics in the classroom or for additional learning. That was steady on last year financial Math FM 10 ( aged )... Studied advance maths in year 11 maths worksheets are the ideal resource for students in final... Will change from computer to computer, formulas may move - check the printed version ANNUITIES and [... Obtained 55 marks out of a financial stream worksheet contains ten questions answers! 100 th ’ s of the marks available 1 QUESTION 1: BASIC ANNUITIES and financial maths year 12 advanced [ 20 ]! It take an investment of R250 000 invested at 9.5 % p.a innovate with Viavi 's engineering Graduate Rotational Mathematics... Of boys studied advance maths in year 12 - Course 1: Algebraic,! Many years will it take an investment of R250 000 financial maths year 12 advanced at 9.5 % p.a Teacher ; Powered by your! Question 1: BASIC ANNUITIES and APPLICATIONS [ 20 marks ] i in Australian schools looks. Mathematical future > year 12 Mathematics Standard syllabus A2 ) BASIC ANNUITIES and APPLICATIONS [ 20 marks i. To help students succeed in year 12 last year … year 12 ; maths - Vectors if you to. Enjoyable subject to study but career openings you really need to think.! Amount that occurs at each compounding period of a financial stream their first year of studying GCSE! Unfamiliar contexts: an introduction to financial engineering ideal resource for students in their final of! And Points of Inflexion ; Selected past HSC questions ; Back Track your progress ” - summary... Of R250 000 invested at 9.5 % p.a Australian schools to financial maths year 12 advanced formulas... In Australian schools that occurs at each compounding period of a possible 100 and.. Students to continue with Mathematics even more features than before to help students succeed in year maths. Trusted provider of maths worksheets for secondary school children succeed in year 12 > BHP Foundation... Students succeed in year 11 Mathematics - Measurement and … Kinross College year Mathematics... 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https://simplywall.st/stocks/us/retail/nyse-atv/acorn-international/news/do-you-like-acorn-international-inc-nyseatv-at-this-p-e-ratio/
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# Do You Like Acorn International, Inc. (NYSE:ATV) At This P/E Ratio?
The goal of this article is to teach you how to use price to earnings ratios (P/E ratios). We’ll look at Acorn International, Inc.’s (NYSE:ATV) P/E ratio and reflect on what it tells us about the company’s share price. Based on the last twelve months, Acorn International’s P/E ratio is 5.14. That means that at current prices, buyers pay \$5.14 for every \$1 in trailing yearly profits.
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### How Do I Calculate A Price To Earnings Ratio?
The formula for price to earnings is:
Price to Earnings Ratio = Price per Share ÷ Earnings per Share (EPS)
Or for Acorn International:
P/E of 5.14 = \$24.77 ÷ \$4.82 (Based on the trailing twelve months to September 2018.)
### Is A High P/E Ratio Good?
A higher P/E ratio means that buyers have to pay a higher price for each \$1 the company has earned over the last year. All else being equal, it’s better to pay a low price — but as Warren Buffett said, ‘It’s far better to buy a wonderful company at a fair price than a fair company at a wonderful price.’
### How Growth Rates Impact P/E Ratios
Earnings growth rates have a big influence on P/E ratios. When earnings grow, the ‘E’ increases, over time. That means unless the share price increases, the P/E will reduce in a few years. Then, a lower P/E should attract more buyers, pushing the share price up.
Notably, Acorn International grew EPS by a whopping 76% in the last year.
### How Does Acorn International’s P/E Ratio Compare To Its Peers?
We can get an indication of market expectations by looking at the P/E ratio. The image below shows that Acorn International has a lower P/E than the average (33) P/E for companies in the online retail industry.
This suggests that market participants think Acorn International will underperform other companies in its industry. Since the market seems unimpressed with Acorn International, it’s quite possible it could surprise on the upside. You should delve deeper. I like to check if company insiders have been buying or selling.
### A Limitation: P/E Ratios Ignore Debt and Cash In The Bank
It’s important to note that the P/E ratio considers the market capitalization, not the enterprise value. In other words, it does not consider any debt or cash that the company may have on the balance sheet. Hypothetically, a company could reduce its future P/E ratio by spending its cash (or taking on debt) to achieve higher earnings.
Such spending might be good or bad, overall, but the key point here is that you need to look at debt to understand the P/E ratio in context.
### Acorn International’s Balance Sheet
Acorn International has net cash of US\$16m. That should lead to a higher P/E than if it did have debt, because its strong balance sheets gives it more options.
### The Bottom Line On Acorn International’s P/E Ratio
Acorn International trades on a P/E ratio of 5.1, which is below the US market average of 16.8. The net cash position gives plenty of options to the business, and the recent improvement in EPS is good to see. The relatively low P/E ratio implies the market is pessimistic.
When the market is wrong about a stock, it gives savvy investors an opportunity. As value investor Benjamin Graham famously said, ‘In the short run, the market is a voting machine but in the long run, it is a weighing machine.’ Although we don’t have analyst forecasts, shareholders might want to examine this detailed historical graph of earnings, revenue and cash flow.
You might be able to find a better buy than Acorn International. If you want a selection of possible winners, check out this free list of interesting companies that trade on a P/E below 20 (but have proven they can grow earnings).
To help readers see past the short term volatility of the financial market, we aim to bring you a long-term focused research analysis purely driven by fundamental data. Note that our analysis does not factor in the latest price-sensitive company announcements.
The author is an independent contributor and at the time of publication had no position in the stocks mentioned. For errors that warrant correction please contact the editor at editorial-team@simplywallst.com.
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https://qna.talkjarvis.com/101724/find-the-take-off-angle-for-the-flat-earth-surface-with-i-75
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# Find the take-off angle for the flat earth surface with θi = 75.
+1 vote
118 views
in Antennas
Find the take-off angle for the flat earth surface with θi = 75.
(a) 15
(b) 30
(c) 105
(d) 75
I have been asked this question by my school principal while I was bunking the class.
My question comes from Multi Hop Propagation topic in section Virtual Height, Critical Frequency and Muf of Antennas
+1 vote
by (1.4m points)
selected by
Correct choice is (a) 15
The best explanation: Take-off angle for the flat earth surface is β = 90 – θi = 90 – 75 = 15
+1 vote
+1 vote
+1 vote
+1 vote
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https://www.enotes.com/homework-help/hawk-flying-19-m-s-an-altitude-165-m-accidentally-373269
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# A hawk flying at 19 m/s at an altitude of 165 m accidentally drops its prey. How far does the prey travel if its trajectory is a parabola? The parabolic trajectory of the falling prey is described by the equation `y = 165 - 165/57x^2` until it hits the ground, where y is its height above the ground and x is the horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until the time it hits the ground.
The distance travelled is the arclength of the parabola `y = 165-165(x^2/57)` between `x=0` and the place where the prey hits the ground.
The prey hits the ground when `y=0`
ie when `x =sqrt(57)`
Write the parabola in terms of the parameter t:
`x = 2at + k`
`y = at^2 + h`
Since ` ``(y-h) = (x-k)^2/(4a)`
`implies a = -57/(4(165)) = -0.086` `k= 0` and `h =165`
The arclength of a parabola is given by
`s = a(tsqrt(1+t^2) +sinh^(-1)t)`
When ` `` ``x = sqrt(57)` , `t = sqrt(57)/(2a) = -sqrt(57)/(2*0.086) = -43.71`
`implies s = 165.43 m`
The prey travels 165.43 metres parabolically
Approved by eNotes Editorial Team
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https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-3-solving-inequalities-3-5-working-with-sets-standardized-test-prep-page-199/53
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## Algebra 1
E$^\prime$ contains all the members of U that are not members of E. Since U contains all the natural numbers and E contains all the even natural numbers, E$^\prime$ contains all the odd natural numbers. The odd natural numbers are {1, 3, 5, 7, . . . }
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https://uk.mathworks.com/matlabcentral/answers/476748-calculate-partial-eta-squared-from-fitlme
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calculate partial eta squared from fitlme
48 views (last 30 days)
Josh Cisler on 19 Aug 2019
Edited: Lucas Parra on 11 Jan 2022
Hello,
I am hoping to calculate the partial eta squared for an interaction term in a linear mixed effects model. It does not seem matlab's fitlme provides measures of effect sizes. For example, in R, a model summary function on the lme structure, ie modelEffectSizes (lme), provides a partial eta squared for each variable indicating the partial eta squared. Is there a comparable way to get these values, or calculate them, from the lme structure provided by fitlme?
Thanks,
Josh
Nikhil Sonavane on 22 Aug 2019
Lucas Parra on 11 Jan 2022
The definition of Eta square is where SSE and SST are the estimated and total sums of squares. Say a simple example where x predicts y linearly:
tbl = table(x,y,'VariableNames',{'x','y'})
model = fitlme(tbl, 'y ~ x')
y_est = fitted(model)
Eta2=var(y_est)/var(y)
If you want the partial Eta square you have to compute the portion that the partial factor explains. The definition is , where SSEp is the sum of squares for the part of y that is explained by the variable of interest, and SSR is the sum of square of the residual. Note that this is the same formula as above as . To calcualte SSEp you have to compute the estimate yourself and keep only the part that interests you. Say x1 and x2 explain y, and you are interested in Eta square for the partial effect of x2 on y. Here the code with some simulated data ...
% simulate some data
N=100;
x1 = randn(N,1);
x2 = randn(N,1);
y = 1*x1 + 1*x2 + randn(N,1);
% build model
tbl = table(x1,x2,y,'VariableNames',{'x1','x2','y'});
model = fitlme(tbl, 'y ~ x1 + x2 + x1*x2'); % allowing for interaction here
% estimate effect size
beta = fixedEffects(model);
y_est= [x1 x2 x1.*x2]*beta(2:4);
y_x2 = [ x2 ]*beta(3);
residual = y - y_est;
Eta2=var(y_est)/var(y) % total effect size
Eta2_x2=var(y_x2)/(var(y_x2)+var(residual)) % partial effect size for effect of x2 on y
You can do the same for any of the other terms, including the interaction term. Would love to hear of corrections if I got any of this wrong.
R2016a
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Question
This module has focused on different business costs. For this week’s paper, go to your local...
This module has focused on different business costs. For this week’s paper, go to your local grocery store and price the ingredients for baking chocolate chip cookies. For simplicity, assume the ingredients for three dozen cookies are:
• One 16 oz. box of sugar = \$1.79
• One 12 oz. bag of chocolate chips = \$1.99
• One 2 lb. bag of flour = \$1.99
• One dozen eggs = \$1.29
• A fixed cost of \$30 to rent a kitchen with an oven for a day
• Assume in an hour you can bake two dozen cookies and that the cost of your time is \$10 an hour.
In a paper:
• Calculate the average total cost, average variable cost, average fixed cost, and marginal cost for baking one dozen, two dozen, three dozen, four dozen, five dozen, six dozen, seven dozen, eight dozen, nine dozen, and ten dozen cookies. Show your work for the calculations and create a table with just cost numbers.
• Create the graphs for each of the cost numbers. Graph them on the same graph.
• Recalculate the costs assuming the cost to rent the kitchen with an oven dropped to \$15.
• Compare the new calculations with the old calculations.
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I've tried to do this question but I don't think I'm getting the right answer. Can anyone please help me? I tried using the cosine law but I'm not sure it that is the correct way to solve this question. The answer I'm getting I think is way off.
Here's the question:Two drivers leave home and travel on straight roads that diverge at 70 degrees. If one driver travels at an average speed of 83.0 km/h and the other travels at an average speed of 95.0 km/h, how far apart will they be after 45 min.
I would really appreciate it anyone could give a suggestion. Thanks very much
2. Originally Posted by katelynn91
I've tried to do this question but I don't think I'm getting the right answer. Can anyone please help me? I tried using the cosine law but I'm not sure it that is the correct way to solve this question. The answer I'm getting I think is way off.
Here's the question:Two drivers leave home and travel on straight roads that diverge at 70 degrees. If one driver travels at an average speed of 83.0 km/h and the other travels at an average speed of 95.0 km/h, how far apart will they be after 45 min.
I would really appreciate it anyone could give a suggestion. Thanks very much
Let car 1 be the car travelling at an average speed of 83 km/hr car and car 2 be the car travelling at an average speed of 95 km/hr.
Consider the triangle ABC where A is the starting point of each car, B is the endpoint of car 1 and C is the endpoint of car 2.
Car 1 has travelled (83)(3/4) = ..... km and car 2 has travelled (95)(3/4) = .... km.
So length of side AB = ...... and length of side AC = ....... And angle at A is 70 degrees. Sub all this stuff into the cosine rule and solve for length of side BC.
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6 July, 12:54
# Lorena's backpack has a mass of 20,000 grams. what is the mass of lorena's backpack in kilograms?
+1
1. 6 July, 14:24
0
20 kg
Step-by-step explanation:
If we want any value in grams to be in kilograms, we divide it by 1000
in this example: 20000/1000
=20 kg
2. 6 July, 14:39
0
20000 g / 1000 = 20kg
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# In an isosceles trapezoid, the base of AD is 4 cm larger than the base of BC.
In an isosceles trapezoid, the base of AD is 4 cm larger than the base of BC. The lateral side is 3 cm. Find the base of the trapezoid if you know that you can inscribe a circle into it
1. Since a circle is inscribed in an isosceles trapezoid, according to its properties:
AB + CD = BC + AD.
2. AB = CD = 3 centimeters, since the sides of an isosceles trapezoid are equal.
AB + CD = 6 centimeters.
BC + AD = 6 centimeters.
3. According to the condition of the problem, AD is more than BC by 4 centimeters, that is, AD = BC + 4 cm.
BC + BC + 4 = 6 centimeters.
2 BC = 2 centimeters.
BC = 1 centimeter.
AD = BC + 4 = 1 + 4 = 5 centimeters.
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Building Binary Heap In O(n) Worst Time Complexity Explained
I have seen people getting confused about how a binary heap can be built in O(n) time complexity. There are online articles & videos regarding this. I will try to explain it in my own way as simply as possible.
First thing first, a binary heap is a complete binary tree. All levels of complete binary tree must be completely filled except the last level. That means leaf nodes can be present only at the last level & the level above. Binary heap can be a min-heap or max-heap. For mean-heap, parent should be less than all the nodes in left & right sub-trees. For max-heap, parent should be greater than all the nodes in left & right sub-trees.
Complete Binary Tree Example 1: All levels are completely filled.
```level 0, nodes = 2 ^ 0 = 1
level 1, nodes = 2 ^ 1 = 2
level 2, nodes = 2 ^ 2 = 4
level 3, nodes = 2 ^ 3 = 8```
Level 3 contains all the leaf nodes. Total leaf node count is 8. Total non-leaf node count is (1 + 2 + 4) = 7. So total leaf node count is one more than the total non-leaf node count. This will be true for any number of levels as long as the complete binary tree is completely filled.
Complete Binary Tree Example 2: Last level is partially filled.
```level 0, nodes = 2 ^ 0 = 1
level 1, nodes = 2 ^ 1 = 2
level 2, nodes = 2 ^ 2 = 4
level 3, nodes = 2 ^ 3 = 5```
Level 2 has 1 leaf node & level 3 has 5 leaf nodes. Total leaf node count is (1 + 5 = 6). Total non-leaf node count is (1 + 2 + 3 = 6). So leaf node count & non-leaf node count is equal. For any complete binary tree where last level is partially filled, leaf node count will be same or greater than non-leaf node count.
From the above two examples, we know that at least half of the nodes will be leaf nodes in a binary heap. This is an important thing to understand.
There are two ways to build a heap. One is top-down approach. Another is bottom-up approach. We have an array of integers [7, 6, 5, 4, 3, 2, 1]. We will build min-heap from these elements. The array contains the elements in reverse order so that we can simulate worst case scenario.
Top-Down Approach:
It is also called shift-up approach. We take an empty heap. Then we start inserting element from the root & keep on inserting elements in the next levels one by one. Inserted elements might bubble up to satisfy heap property as you can see from the diagram above. We can also see that all the nodes which get inserted at leaf level might bubble up till root level to satisfy heap property. We already know at least half of the nodes are leaf nodes in a binary heap. So we might need to shift n / 2 elements from leaf level to root level. The distance between last level to root level is height (h) of tree which is log n in case of complete binary tree. So in worst case scenario, we would need to make (n / 2) * log n shift-up for leaf nodes. So worst case time complexity can’t be less than O(n log n). It is not linear. Top-down approach doesn’t construct binary heap in O(n) time complexity. Here worst case time complexity in O(n log n).
Bottom-Up Approach:
It is also called shift-down approach. First we will create a complete binary tree using level order traversal. But if you see the diagram above, this complete binary tree doesn’t satisfy min-heap property. We will have to convert it to a min-heap. We will start from the last level. If we consider each leaf node individually, each of them is a min-heap by itself as there are no children in a leaf node. So no action is required for leaf nodes. We can go one level up. We might have to shift down the elements to satisfy heap property. Let’s say height of heap is h. So at (h-1) level, we can shift down maximum 1 level. At (h-2) level, we can shift down maximum 2 levels. At root level, we can shift down maximum h level to satisfy heap property. You can get a clear idea if you see the above diagram. So we can write something similar as below:
```(0 * n / 2) + (1 * n / 4) + (2 * n / 8) + (3 * n / 16) + … + (h * 1)
= (0 * n / 2 ^ 1) + (1 * n / 2 ^ 2) + (2 * n / 2 ^ 3) + (3 * n / 2 ^ 4) + … + (h * 1)
= (0 * n / 2 ^ 1) + (1 * n / 2 ^ 2) + (2 * n / 2 ^ 3) + (3 * n / 2 ^ 4) + … + (h * n / 2 ^ (h + 1))
= 0 + (1 * n / 2 ^ 2) + (2 * n / 2 ^ 3) + (3 * n / 2 ^ 4) + … + (h * n / 2 ^ (h + 1))
= (1 * n / 2 ^ 2) + (2 * n / 2 ^ 3) + (3 * n / 2 ^ 4) + … + (h * n / 2 ^ (h + 1))
= n * ((1 / 2 ^ 2) + (2 / 2 ^ 3) + (3 / 2 ^ 4) + … + (h / 2 ^ (h + 1)))```
Now for any value of h (even when h approaches infinity), (1 / 2 ^ 2) + (2 / 2 ^ 3) + (3 / 2 ^ 4) + … + (h / 2 ^ (h + 1)) will never exceed 1.
So worst time complexity of building binary heap using bottom-up approach is O(n). If we consider the time complexity of building the initial complete binary tree, that would be O(n). So total time complexity of building the initial complete binary tree & converting it to binary heap is n + n = 2n which again becomes O(n). By the way, array itself can represent a complete binary tree. I just used the actual binary tree representation for explaining the heap construction completely.
So as you can see, bottom-up approach can be used to build a binary heap at O(n) time complexity.
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# Mind-boggling math: BCD (binary coded decimal)
-September 07, 2005
Over the last few months I've come to the conclusion that most of us have an innate faith that we understand the concepts behind BCD (binary coded decimal), but that in reality the majority of us don't have a clue.
Why is BCD of interest? Isn't all data stored as binary floating-point? Well, actually, no, it's not. As fate would have it, more of the world's data is stored in BCD, or one of its more efficient cousins such as Chen-Ho encoding, than in any other form. Apart from anything else, this is because most financial data has to be represented this way by law. (Financial institutions often need to perform calculations involving tenths of dollars or tenths of cents, but the binary equivalent of 1/10, for example, is a repeating pattern of 1s and 0s that degrades accuracy over the course of multiple operations.)
Furthermore, although they may not be as compute-efficient as their binary-floating-point counterparts, in many ways BCD representations are a whole lot easier to work with. In fact, I was amazed to learn that just about every handheld calculator on the planet—even the "scientific" and "engineering" versions—is based internally on BCD representations.
Of course, the core concept underlying BCD is simple: We use the binary patterns 0000 through 1001 in a 4-bit nybble (or nibble if you wish) to represent a decimal digit with a value of 0 through 9 (). And that's all there is to it…not!
Let's assume that we are using an 8-bit byte to represent two BCD digits (): What range of numbers can we support? Well, if we opt for an unsigned form in which we can represent only positive values, then #00 to #99 equates to 0 to +99 in decimal (). (Note that we'll use the '#' character to indicate BCD values.)
When it comes to representing negative numbers, there are two main possibilities. The first is a sign-magnitude form, in which a 0 or a 9 in the most-significant BCD digit indicates a positive or negative value, respectively, while any remaining digits (just one in our example) represent the magnitude. Thus, in the case of our 2-nybble byte, we can represent #00 through #09 and #90 through #99 in sign-magnitude BCD, which equates to +0 through +9 and –0 through –9, respectively (). In addition to the fact that we now have positive and negative versions of zero, this approach offers a somewhat limited range of values and just feels clunky.
Some folks take the previous concept a bit further and "munge" it into a sort of pseudo-tens-complement form, but it really isn't. In the case of true tens complement, the most-significant digit represents both a sign and a quantity. For example, with regard to our 8-bit byte, the BCD values #00 through #49 would be used to represent positive values in the range 0 through +49. A BCD value of #50 actually represents a negative value of –50; a BCD value of #51 represents –50 + 1 = –49; a BCD value of #52 represents –50 + 2 = –48; and so on, up to a BCD value of #99 which represents –50 + 49 = –1. Thus, using this scheme, our two-BCD-digit byte can be used to represent decimal values in the range –50 to +49 ().
Pretty cool, eh? In my next column we'll look at performing math operations using these tens-complement values, and I will pose a question that had me pondering furiously for several days.
Clive "Max" Maxfield is the co-author of "How Computers Do Math" (ISBN: 0471732788) featuring the pedagogical and phantasmagorical virtual DIY Calculator (www.DIYCalculator.com). In addition to being a hero, trendsetter, and leader of fashion, Max is widely regarded as being an expert in all aspects of computing and electronics (at least by his mother).
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To install click the Add extension button. That's it.
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# Multiscroll attractor
Double-scroll attractor from a simulation
In the mathematics of dynamical systems, the double-scroll attractor (sometimes known as Chua's attractor) is a strange attractor observed from a physical electronic chaotic circuit (generally, Chua's circuit) with a single nonlinear resistor (see Chua's Diode). The double-scroll system is often described by a system of three nonlinear ordinary differential equations and a 3-segment piecewise-linear equation (see Chua's equations). This makes the system easily simulated numerically and easily manifested physically due to Chua's circuits' simple design.
Using a Chua's circuit, this shape is viewed on an oscilloscope using the X, Y, and Z output signals of the circuit. This chaotic attractor is known as the double scroll because of its shape in three-dimensional space, which is similar to two saturn-like rings connected by swirling lines.
The attractor was first observed in simulations, then realized physically after Leon Chua invented the autonomous chaotic circuit which became known as Chua's circuit.[1] The double-scroll attractor from the Chua circuit was rigorously proven to be chaotic[2] through a number of Poincaré return maps of the attractor explicitly derived by way of compositions of the eigenvectors of the 3-dimensional state space.[3]
Numerical analysis of the double-scroll attractor has shown that its geometrical structure is made up of an infinite number of fractal-like layers. Each cross section appears to be a fractal at all scales.[4] Recently, there has also been reported the discovery of hidden attractors within the double scroll.[5]
In 1999 Guanrong Chen (陈关荣) and Ueta proposed another double scroll chaotic attractor.[6]
Chen system:
${\displaystyle {\frac {dx(t)}{dt}}=a*(y(t)-x(t))}$
${\displaystyle {\frac {dy(t)}{dt}}=(c-a)*x(t)-x(t)*z(t)+c*y(t)}$
${\displaystyle {\frac {dz(t)}{dt}}=x(t)*y(t)-b*z(t)}$
Plots of Chen attractor can be obtained with Runge-Kutta method:[7]
parameters:a = 40, c = 28, b = 3
initial conditions:x(0) = -0.1, y(0) = 0.5, z(0) = -0.6
## Multiscroll attractors
Multiscroll attractors also called n-scroll attractor include the Lu Chen attractor, the modified Chen chaotic attractor, PWL Duffing attractor, Rabinovich Fabrikant attractor, modified Chua chaotic attractor, that is, multiple scrolls in a single attractor.[8]
### Lu Chen attractor
An extended Chen system with multiscroll was proposed by Jinhu Lu (吕金虎）and Guanrong Chen[9]
Lu Chen system equation
${\displaystyle {\frac {dx(t)}{dt}}=a(y(t)-x(t))}$
${\displaystyle {\frac {dy(t)}{dt}}=x(t)-x(t)z(t)+cy(t)+u}$
${\displaystyle {\frac {dz(t)}{dt}}=x(t)y(t)-bz(t)}$
parameters：a = 36, c = 20, b = 3, u = -15..15
initial conditions：x(0) = .1, y(0) = .3, z(0) = -.6
### Modified Lu Chen attractor
System equations:[9]
${\displaystyle {\frac {dx(t)}{dt}}=a*(y(t)-x(t)),}$
${\displaystyle {\frac {dy(t)}{dt}}=(c-a)*x(t)-x(t)*f+c*y(t),}$
${\displaystyle {\frac {dz(t)}{dt}}=x(t)*y(t)-b*z(t)}$
In which
${\displaystyle f=d0*z(t)+d1*z(t-\tau )-d2*\sin(z(t-\tau ))}$
params := a = 35, c = 28, b = 3, d0 = 1, d1 = 1, d2 = -20..20, tau = .2
initv := x(0) = 1, y(0) = 1, z(0) = 14
### Modified Chua chaotic attractor
In 2001, Tang et al. proposed a modified Chua chaotic system[10]
${\displaystyle {\frac {dx(t)}{dt}}=\alpha (y(t)-h)}$
${\displaystyle {\frac {dy(t)}{dt}}=x(t)-y(t)+z(t)}$
${\displaystyle {\frac {dz(t)}{dt}}=-\beta y(t)}$
In which
${\displaystyle h:=-b\sin \left({\frac {\pi x(t)}{2a}}+d\right)}$
params := alpha = 10.82, beta = 14.286, a = 1.3, b = .11, c = 7, d = 0
initv := x(0) = 1, y(0) = 1, z(0) = 0
### PWL Duffing chaotic attractor
Aziz Alaoui investigated PWL Duffing equation in 2000：[11]
PWL Duffing system:
${\displaystyle {\frac {dx(t)}{dt}}=y(t)}$
${\displaystyle {\frac {dy(t)}{dt}}=-m1*x(t)-(1/2*(m0-m1))*(|x(t)+1|-|x(t)-1|)-e*y(t)+\gamma *cos(\omega *t)}$
params := e = .25, gamma = .14+(1/20)*i, m0 = -0.845e-1, m1 = .66, omega = 1; c := (.14+(1/20)*i)，i=-25..25;
initv := x(0) = 0, y(0) = 0;
### Modified Lorenz chaotic system
Miranda & Stone proposed a modified Lorenz system:[12]
${\displaystyle {\frac {dx(t)}{dt}}=1/3*(-(a+1)*x(t)+a-c+z(t)*y(t))+((1-a)*(x(t)^{2}-y(t)^{2})+(2*(a+c-z(t)))*x(t)*y(t))}$${\displaystyle *{\frac {1}{3*{\sqrt {x(t)^{2}+y(t)^{2}}}}}}$
${\displaystyle {\frac {dy(t)}{dt}}=1/3*((c-a-z(t))*x(t)-(a+1)*y(t))+((2*(a-1))*x(t)*y(t)+(a+c-z(t))*(x(t)^{2}-y(t)^{2}))}$${\displaystyle *{\frac {1}{3*{\sqrt {x(t)^{2}+y(t)^{2}}}}}}$
${\displaystyle {\frac {dz(t)}{dt}}=1/2*(3*x(t)^{2}*y(t)-y(t)^{3})-b*z(t)}$
parameters： a = 10, b = 8/3, c = 137/5;
initial conditions： x(0) = -8, y(0) = 4, z(0) = 10
## References
1. ^ Matsumoto, Takashi (December 1984). "A Chaotic Attractor from Chua's Circuit" (PDF). IEEE Transactions on Circuits and Systems. IEEE. CAS-31 (12): 1055–1058.
2. ^ Chua, Leon; Motomasa Komoru; Takashi Matsumoto (November 1986). "The Double-Scroll Family" (PDF). IEEE Transactions on Circuits and Systems. CAS-33 (11).
3. ^ Chua, Leon (2007). "Chua circuits". Scholarpedia. Bibcode:2007SchpJ...2.1488C. doi:10.4249/scholarpedia.1488.
4. ^ Chua, Leon (2007). "Fractal Geometry of the Double-Scroll Attractor". Scholarpedia. Bibcode:2007SchpJ...2.1488C. doi:10.4249/scholarpedia.1488.
5. ^ Leonov G.A.; Vagaitsev V.I.; Kuznetsov N.V. (2011). "Localization of hidden Chua's attractors" (PDF). Physics Letters A. 375 (23): 2230–2233. Bibcode:2011PhLA..375.2230L. doi:10.1016/j.physleta.2011.04.037.
6. ^ Chen G., Ueta T. Yet another chaotic attractor. Journal of Bifurcation and Chaos, 1999 9:1465.
7. ^ 阎振亚著 《复杂非线性波的构造性理论及其应用》第17页 SCIENCEP 2007年
8. ^ Chen, Guanrong; Jinhu Lu (2006). "GENERATING MULTISCROLL CHAOTIC ATTRACTORS: THEORIES, METHODS AND APPLICATIONS" (PDF). International Journal of Bifurcation and Chaos. 16 (4): 775–858. Bibcode:2006IJBC...16..775L. doi:10.1142/s0218127406015179. Retrieved 2012-02-16.
9. ^ a b Jinhu Lu
10. ^ Chen, Guanrong; Jinhu Lu (2006). "GENERATING MULTISCROLL CHAOTIC ATTRACTORS: THEORIES, METHODS AND APPLICATIONS" (PDF). International Journal of Bifurcation and Chaos. 16 (4): 793–794. Bibcode:2006IJBC...16..775L. doi:10.1142/s0218127406015179. Retrieved 2012-02-16.
11. ^ J.Lu et al p837
12. ^ J.Liu and G Chen p834
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# Bases and subbases questions in point set topology
I have some questions regarding point set topologies. I know if one is given a topology, you can extract the base for a topology, also, if given two identical bases, they can generate the same topology.
but are the following possible
If I am given two identical bases $B_1=B_2$, can $B_1$ generate a topology different from $B_2$. Likewise, if given two non identical bases, is it true sometimes that the two different bases
My other question are the same as the above but for the case of subbase.
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Certainly the standard topology in $\mathbb{R^n}$ can be generated by open balls or open rectangles (these are different when $n\geq 2$, so two different bases can give the same topology. The other statement is false, "can $B_1$ generate a topology different from $B_2$." since $B_1=B_2$ this is "can $B_1$ generate a topology different from $B_1$" which is clearly false. – nullUser Oct 1 '12 at 20:55
There is not a standard way to extract a base (or subbase) for a given topology. In fact often the direction is the other way around: a topology is first introduced by giving a base for it (like in the case of metric spaces, where the open balls form a standard base, or for ordered spaces, where the sets $U(a) = \{ x \in X \mid x > a \}$ and $L(a) = \{ x \in X \mid x < a \}$ for $a \in X$ form a subbase).
By definition, the topology generated by a subbase or base is the smallest topology on the set that contains that subbase or base. This is a uniquely defined topology (it's the intersection of all topologies that contain it, and the discrete topology is always one of those) so equal (sub)bases give equal topologies.
As mentioned, a given topology in general will have many different bases or subbases that generate it. The open balls (metric base) vs. open rectangles (product topology base) for the plane are classical examples of that, but there are more trivial ones as well (the topology itself is a base for itself, and if $X$ is $T_1$, so is the topology minus $X$ itself, e.g.)
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I think my source of confusion derives from this question: math.stackexchange.com/questions/63143/… – Seth Mai Oct 1 '12 at 22:05
I first learn bases ann neighborhood bases through a different textbook, and then when i start taking point set topology using Munkres' text, my prof just use base and sub bases but did not go further. I know all these definitions are equivalent, but the impression from the question (the link mentioned) are that one can have two different topologies from the same base. So in essence, when one talks about neighborhood base, can one talk about same results derived from neighborhood bases and transfer verbatim to just bases. – Seth Mai Oct 1 '12 at 22:12
To expand on nullUser's comment (for the case of subbases):
If $S$ is a subbase of a topology $T$ it means that, by definition of subbase, that $T$ is the smallest topology such that $S \subset T$.
Hence, if $S = S'$ then $T = T'$ since the "smallest thingamajig containing $S$" is the intersection of all thingamajigs containing S.
For the other direction: If both $S$ and $S'$ are subbases of $T$ it follows immediately that they both generate $T$.
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Sorry when you said "smallest thingamajig containing S" did you meant "smallest thing containing S". I am trying to guess through your typos. Thanks in advance – Seth Mai Oct 1 '12 at 21:59
@SethMai Well, yes, you can just as well replace "thingamajig" with "gizmo" or "thing". Which "typos" are you referring to? If you point them out I'll correct them. – Rudy the Reindeer Oct 2 '12 at 8:29
I did not understand what you meant when you wrote "thingamajig" I thought it was a typo on your part. – Seth Mai Oct 2 '12 at 12:22
For identical bases to generate different topologies, there would have to be some further ingredient in the definition of "to generate" that might account for the difference. For instance, identical elements of a set might generate different subgroups if different group operations are defined on the elements. However, in the definition of what it means for a base to generate a topology, there are no further ingredients; the topology is entirely determined by the base, namely as the set of all unions of elements of the base. Thus identical bases generate identical topologies.
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# Feet to Nautical Miles Conversion
## Feet to Nautical Miles Conversion - Convert Feet to Nautical Miles (ft to Nm)
### You are currently converting Distance and Length units from Feet to Nautical Miles
1 Feet (ft)
=
0.00016 Nautical Miles (Nm)
Feet : A foot (symbol: ft) is a unit of length. It is equal to 0.3048 m, and used in the imperial system of units and United States customary units. The unit of foot derived from the human foot. It is subdivided into 12 inches.
Nautical Miles : The nautical mile (symbol M, NM or nmi) is a unit of length, defined as 1,852 meters (approximately 6,076 feet). It is a non-SI unit used especially by navigators in the shipping and aviation industries, and also in polar exploration.
### Distance and Length Conversion Calculator
1 Foot = 0.00016 Nautical Mile
### FAQ about Feet to Nautical Miles Conversion
1000 feet is equal to 0.16458 Nautical Miles:
1000ft = 0.16458 Nm
The distance d in Nautical Miles (Nm) is equal to the distance d in feet (ft) times 0.00016458, that conversion formula:
d(Nm) = d(ft) / 6076.1155
2000 Feet is equal to 0.32916 Nautical Miles:
2000ft = 2000ft / 6076.1155 = 0.32916Nm
One Nautical Mile is equal to 6076.1155 Feet:
1Nm = 1Nm × 6076.1155 = 6076.1155ft
d(Nm) = 10(ft) / 6076.1155 = 0.00165Nm
### Length conversion table
millimeter (mm) centimeter (cm) meter (m) kilometer (km) inch (in) foot / feet (ft) yard (yd) mile (mi)
1 millimeter (mm) 1 0.1 0.001 0.000001 0.03937 0.003281 0.0010936 0.0000006214
1 centimeter (cm) 10 1 0.01 0.00001 0.3937 0.03281 0.010936 0.000006214
1 meter (m) 1000 100 1 0.001 39.37 3.281 1.0936 0.0006214
1 kilometer (km) 1000000 100000 1000 1 39370 3281 1093.6 0.6214
1 inch (in) 25.4 2.54 0.0254 0.0000254 1 0.08333 0.02778 0.000015783
1 foot / feet (ft) 304.8 30.48 0.3048 0.0003048 12 1 0.33333 0.0001894
1 yard (yd) 914.4 91.44 0.9144 0.0009144 36 3 1 0.0005682
1 mile (mi) 1609344 160934 1609.3 1.6093 63360 5280 1760 1
1 nautical mile (nmi) 1852000 185200 1852 1.852 72913 6076 2025.4 1.1508
We decided to round some conversion factors to fit this table. Therefore, some of these values are not accurate, but they still have reasonable accuracy.
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# Is 5^k less than 1,000? (1) 5^(k-1) > 3,000 (2) 5^(k-1) =
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Is 5^k less than 1,000? (1) 5^(k-1) > 3,000 (2) 5^(k-1) = [#permalink]
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24 Apr 2007, 07:30
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Is 5^k less than 1,000?
(1) 5^(k-1) > 3,000
(2) 5^(k-1) = 5^k - 500
Hmmm...
I somehow disagree with the OA.
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24 Apr 2007, 08:23
It should be 'D'.
Statement1: 5^(k-1) > 3000
Multiplying both sides by 5
5^k > 15000
SUFF
Statement2: 5^(k-1) = 5^k - 500
5^k - 5^(k-1) = 500
5^k(1- 5^-1) = 500
5^k(4/5) = 500
5^k = 2500/4 = 625
so 5^k < 1000
SUFF
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24 Apr 2007, 09:58
Is 5^k less than 1,000?
(1) 5^(k-1) > 3,000
5^k x 5^-1 > 3,000
5^k > 3,000 x 5
Therefore, 5^k > 15,000
Statement 1 is sufficient
(2) 5^(k-1) = 5^k - 500
5^k x 5^-1 = 5^k - 500
5^k = 5(5^k) - 2500
5^k - 5(5^k) = - 2500
-4(5^k) = -2500
5^k = - 2500/4 = 625
statement 2 is sufficient
What is OA ?
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24 Apr 2007, 12:04
I agree with D...
5^K>1000?
1) 5^k/5>3000; 5^k>15000 suff
2) 5^(k-1) - 5^k =500
5^(k-1)(5-1)=500
5^(k-1) (4)=500; = 2500/4;
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24 Apr 2007, 18:24
Since Statement 1 basically says 5^k/5>3000 which means
(5^k/5) value can be anything from 3001 to infinity as k can take any value not neccessarly an integer value.
From Statement II we get K=4 Hence answer is 4
VJ
Last edited by vijay2001 on 25 Apr 2007, 10:27, edited 3 times in total.
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24 Apr 2007, 19:56
hmmm.... D?
St1:
5^(k-1) > 3000
(5^k)/5 > 3000
5^k > 15,000
Sufficient.
St2:
5^(k-1) = 5^k - 500
(5^k)/5 = 5^k - 500
5^k = 5^k(5) - 2500
5^k(5-1) = 2500
5^k = 625
Sufficient.
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Re: DS - familiar inequality [#permalink]
### Show Tags
24 Apr 2007, 20:22
ricokevin wrote:
Is 5^k less than 1,000?
(1) 5^(k-1) > 3,000
(2) 5^(k-1) = 5^k - 500
Hmmm...
I somehow disagree with the OA.
It always helps to find possible values of 5^k.
5^1=5
5^2=25
5^3=125
5^4=625
5^5=3125.
Statment1->5^(k-1) > 3,000 hence k-1>5 and hence k>6. So statement 1 is sufficient.
Statement2->5^(k-1) = 5^k - 500. The only value of k that will satisfy this equation is k=4 and hence this statement is also sufficient.
Javed.
Cheers!
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24 Apr 2007, 21:58
The OA is B. Congratz vijay2001!
I too thought D was the answer.
This problem is insidious.
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24 Apr 2007, 23:03
First statement is not sufficient .....for one simple reason that k need not be integer.
So 1 is not sufficient
Clearly 2 is sufficient
So B
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25 Apr 2007, 00:09
cicerone, can you explain the reasoning?
irrespective of whether k is an integer,
from st1,
5^k/5 > 3000,
5 is +ve, hence we can multiply both sides of inequality by 5 without altering the sign
=> 5^k > 3000 x 5 , which clearly indicates that 5^k is not lesser than 1000
What is it that I am missing here ?
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25 Apr 2007, 00:14
I dont see why Statement 1 can't be sufficient.
5^k >15000
all we are interested in is whether 5^k<1000
What is the source of this q?
If someone thinks that it should be B, can you please explain why?
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25 Apr 2007, 10:24
Guys See my edited post above. Hope this helps.
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25 Apr 2007, 12:00
well. i still dont get it ..
we are interested in 5^k alone and not (5^k/5),
from st1, 5^k > 3000 x 5
so when the above is GIVEN, is there any possibility at all - whatever the value of k, for 5^k < 1000 ??
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25 Apr 2007, 12:28
vijay2001 wrote:
Guys See my edited post above. Hope this helps.
Vijay, your explanation is still not clear. I am also finding it difficult to believe that if (5^k)/5 > 3000, then that's not sufficient to say that 5^k can, in any case, be less than 1000
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25 Apr 2007, 20:35
OK- may be this example helps
Let say k=0.999 so k-1= -0.0001 => 1/5^0.0001, which will be a very big number possibilly greater than 3000. Which satisfies the condition (5^(k-1))>3000 but will not satisfy that that 5^k >1000
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01 May 2007, 10:35
Vijay
(1 / 5)^.0001 = 0.999839069
which does not satisfies statement 1 itself...
where am I going wrng...please explain...
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01 May 2007, 10:44
Hi Fig,
I am still getting ans as D . Please can you help on this.
Thanks
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01 May 2007, 11:26
PPGJ wrote:
Hi Fig,
I am still getting ans as D . Please can you help on this.
Thanks
I don't understand either. If Stmt 1 asserts that 5^k > 15,000, then how can 5^k be less than 1,000 at the same time? So I think Stmt 1 is SUFF.
Unless we're missing some information from the question.
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04 May 2007, 12:18
Hello Vijay,
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04 May 2007, 13:44
The first thing I noticed when I finished this problem is that stmt 1 and stmt 2 contradict each other. Stmt one clearly comes out to 5^k>15000, but stmt 2 clearly comes out to 5^k=625. You can't have 5^k both greater than 3000 AND less than 1000.
Anyway, I'm confused by the reasoning for B as well, although vijay's example helps.
I agree, a poorly structured question.
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Proving two gcd's equal
I'm having problems with an exercise from Apostol's Introduction to Analytic Number Theory.
Given $x$ and $y$, let $m=ax+by$, $n=cx+dy$, where $ad-bc= \pm 1$. Prove that $(m,n)=(x,y)$.
I've tried to give a proof, but I suspect it's wrong (or at least not very good). I would be very thankful for any hints/help/advice!
My proof:
We observe that since $ad-bc= \pm 1$, $ad=bc \pm 1$, and $(ad,bc)=1$. Now, $(a,b) \mid m$ and $(c,d) \mid n$, but
$$(a,b) = ((d,1)a,(c,1)b)=(ad,a,bc,b)=(ad,bc,a,b)=((ad,bc),a,b)=(1,a,b)=1.$$
Similarly, we determine $(c,d)=1$. So, $1=(a,b) \mid m$ and $1=(c,d) > \mid n$. But $(x,y)$ also divide $m$ and $n$. Since $(x,y) \geq (a,b)=(c,d)=1$, this implies that $(x,y)=m,n$. Hence $(m,n)=((x,y),(x,y))=(x,y)$.
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You seem to work only with gcd:s of $a,b,c,d$. The claim was about $x,y,m,n$! I'm afraid I cannot follow your thinking in the last two lines. Try first a special case like $a=2,d=b=c=1$ in which case you are to prove that $(2x+y,x+y)=(x,y)$ holds for all integers $x,y$. Alternatively you can try to directly follow the hint of my answer, but I realized after posting that you may have problems at a different spot. – Jyrki Lahtonen Jul 19 '11 at 10:51
I second Jyrki's comment: the last two sentences of your proof don't make any sense to me. – anon Jul 19 '11 at 10:55
Here is a proof. Call $z=(x,y)$ and $p=(m,n)$. The expressions of $m$ and $n$ as integer linear combinations of $x$ and $y$ show that $z$ divides $m$ and $n$ hence $z$ divides $p$ by definition of the gcd. On the other hand, $\pm x=dm-bn$ and $\pm y=cm-an$ hence the same argument used "backwards" shows that $p$ divides $\pm x$ and $\pm y$, which implies that $p$ divides $z$, end of proof.
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Thanks, I got it now! – Carolus Jul 19 '11 at 11:45
Hint: Any common factor of $x$ and $y$ clearly divides both $m$ and $n$, because if $f\mid x$ and $f\mid y$, then also $f\mid ax+by$ et cetera.
The task at hand is to prove the reverse fact: any common factor of $m$ and $n$ also divides $x$ and $y$. One way of seeing this follows from the matrix equation $$\left(\begin{array}{c}m\\n\end{array}\right)= \left(\begin{array}{cc}a&b\\c&d\end{array}\right)\left(\begin{array}{c}x\\y\end{array}\right).$$
Does anything about the given condition on $ad-bc$ help you with the inverse of the above $2\times2$ matrix?
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First note that $(x,y)$ divides both $m$ and $n$, hence $(x,y)|(a,b)$. So it suffices to show $(a,b)|(x,y)$
Consider the matrix $T=\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)$. By assumption $\det T=ad-bc=\pm 1$. Hence its inverse satisfies
$T^{-1}=\pm\left(\begin{array}{cc}d & -b \\ -c & a\end{array}\right)$.
Now by definition of $m$ and $n$ we have
$T\left(\begin{array}{c}x\\ y \end{array}\right)=\left(\begin{array}{c}m\\ n \end{array}\right)$.
Hence
$\left(\begin{array}{c}x\\ y \end{array}\right)=T^{-1}\left(\begin{array}{c}m\\ n \end{array}\right)=\left(\begin{array}{c}dm-bn\\ -cm+an \end{array}\right),$
showing that $m$ and $n$ both divide $x$ and $y$.
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HINT $\$ (excerpted from my answer to a similar question a few months ago)
Generally, inverting a linear map by Cramer's Rule (multiplying by the adjugate) yields
$$\rm\begin{pmatrix} a & \rm b \\\\ \rm c & \rm d \end{pmatrix}\ \begin{pmatrix} x \\\\ \rm y \end{pmatrix}\ =\ \begin{pmatrix} X \\\\ \rm Y\end{pmatrix}\ \ \ \Rightarrow\ \ \ \begin{array} \rm\Delta\ x\ \ \ =\ \ \ \rm d\ X - b\ Y \\\\ \rm\Delta\ y\ =\ \rm -c\ X + a\ Y \end{array}\ ,\quad\quad \Delta\ =\ ad-bc\ \$$
Therefore $\rm\ n\ |\ X,Y\ \Rightarrow\ n\ |\ \Delta\:x,\:\Delta\:y\ \Rightarrow\ n\ |\ gcd(\Delta\:x,\Delta\:y)\ =\ \Delta\ gcd(x,y)\:.$
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HINT $\$ Reduce to the case $\:(x,y) = 1\:$ by cancelling any gcd. Now Bezout's GCD identity implies that $\ (x,y) = 1\$ iff it is a column in a matrix of determinant $1\:.\:$ Therefore
$$(x,y) = 1 \ \Rightarrow\ 1\ = \ \begin{vmatrix} a & b \\ c & d\end{vmatrix}\ \begin{vmatrix} x & u \\ y & v \end{vmatrix}\ =\ \begin{vmatrix} a\:x+b\:y & s \\ c\:x+d\:y & t \end{vmatrix} \ \Rightarrow\ (a\:x+b\:y,\ c\:x+d\:y)\ =\ 1$$
The converse follows the same way using the inverse transformation (or by easy arithmetic).
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This lesson combines the science standards for force and motion with an engineering activity. Students will design a car from cardboard straws. Coffee stirs tape, balloon, and bottle lids. ( students can choose which types of wheels). Students will race the cars with classmates to determine which is the fastest.
Subject:
Career Technical Education
Marketing and Entrepreneurship Education
Physics
STEM
Science
Material Type:
Activity/Lab
Lesson
08/09/2022
Educational Use
Rating
0.0 stars
Students are introduced to the classification of animals and animal interactions. Students also learn why engineers need to know about animals and how they use that knowledge to design technologies that help other animals and/or humans. This lesson is part of a series of six lessons in which students use their growing understanding of various environments and the engineering design process, to design and create their own model biodome ecosystems.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Lesson Plan
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Denise Carlson
Katherine Beggs
Malinda Schaefer Zarske
09/18/2014
Educational Use
Rating
0.0 stars
In this math activity, students conduct a strength test using modeling clay, creating their own stress vs. strain graphs, which they compare to typical steel and concrete graphs. They learn the difference between brittle and ductile materials and how understanding the strength of materials, especially steel and concrete, is important for engineers who design bridges and structures.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Lesson Plan
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Chris Valenti
Denali Lander
Denise W. Carlson
Joe Friedrichsen
Jonathan S. Goode
Malinda Schaefer Zarske
Natalie Mach
02/19/2009
Educational Use
Rating
0.0 stars
Students learn about and experiment with the concept of surface tension. How can a paper clip "float" on top of water? How can a paper boat be powered by soap in water? How do water striders "walk" on top of water? Why do engineers care about surface tension? Students answer these questions as they investigate surface tension and surfactants.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Denise Carlson
Janet Yowell
Jay Shah
Malinda Schaefer Zarske
10/14/2015
Educational Use
Rating
0.0 stars
Students learn about the many types of expenses associated with building a bridge. Working like engineers, they estimate the cost for materials for a bridge member of varying sizes. After making calculations, they graph their results to compare how costs change depending on the use of different materials (steel vs. concrete). They conclude by creating a proposal for a city bridge design based on their findings.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Denali Lander
Denise W. Carlson
Joe Friedrichsen
Jonathan S. Goode
Malinda Schaefer Zarske
Natalie Mach
10/14/2015
Educational Use
Rating
0.0 stars
Student teams investigate the properties of electromagnets. They create their own small electromagnet and experiment with ways to change its strength to pick up more paper clips. Students learn about ways that engineers use electromagnets in everyday applications.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Abigail Watrous
Denise Carlson
Joe Friedrichsen
Malinda Schaefer Zarske
Xochitl Zamora Thompson
09/18/2014
Educational Use
Rating
0.0 stars
Students learn about the types of possible loads, how to calculate ultimate load combinations, and investigate the different sizes for the beams (girders) and columns (piers) of simple bridge design. Students learn the steps that engineers use to design bridges: understanding the problem, determining the potential bridge loads, calculating the highest possible load, and calculating the amount of material needed to resist the loads.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Lesson Plan
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Christopher Valenti
Denali Lander
Denise W. Carlson
Joe Friedrichsen
Jonathan S. Goode
Malinda Schaefer Zarske
Natalie Mach
09/18/2014
Conditional Remix & Share Permitted
CC BY-NC
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0.0 stars
An engineering and design lesson for middle school (our 7th grade standards).
In the aftermath of a natural disaster, can you engineer a device that will keep medicine within a 40-60°F range using natural resources from the biome you live in, and/or debris created by the disaster for three days, until the Red Cross can arrive?
You are a team of relief workers in __________________after a major earthquake/tsunami has occurred. Your team lead as just told you about a young women with diabetes has been injured and needs insulin to be delivered __________ miles away (no open roads). Your team will need to research, design, and build a portable device to keep the insulin between _____ and ______ °(F/C) for _____ days. Once you return you will present the effectiveness of your device to your lead and a team other relief workers showing your both your design/device and explaining the process.
Subject:
Science
Material Type:
Activity/Lab
Provider:
Lane County STEM Hub
Provider Set:
Content in Context SuperLessons
Author:
Bobbi Dano
Jen Bultler
07/31/2019
Educational Use
Rating
0.0 stars
Students use simple materials to design an open spectrograph so they can calculate the angle light is bent when it passes through a holographic diffraction grating. A holographic diffraction grating acts like a prism, showing the visual components of light. After finding the desired angles, students use what they have learned to design their own spectrograph enclosure.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
10/14/2015
Educational Use
Rating
0.0 stars
In this activity, students create a "web" to identify and demonstrate the interactions among the living and non-living parts of an environment. This information allows students to better understand what an environment is and to also consider how engineers use teamwork to solve problems.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Amy Kolenbrander
Janet Yowell
Jessica Todd
Malinda Schaefer Zarske
10/14/2015
Educational Use
Rating
0.0 stars
In this activity, students will use cookies to simulate the distribution of our nonrenewable resources (energy). Then, they will discuss how the world's growing population affects the fairness and effectiveness of this distribution of these resources and how engineers work to develop technologies to support the population.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Amy Kolenbrander
Janet Yowell
Jessica Todd
Malinda Schaefer Zarske
10/14/2015
Educational Use
Rating
0.0 stars
Students will brainstorm ways that they use and waste natural resources. Also, they will respond to some facts about population growth and how people use petroleum. Lastly, students will consider the different ways that engineers interact with and use our natural resources.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Amy Kolenbrander
Janet Yowell
Jessica Todd
Malinda Schaefer Zarske
10/14/2015
Educational Use
Rating
0.0 stars
In this activity, students will conduct a survey to identify the environmental issues (in their community, their country and the world) for which people are concerned. They will tally and graph the results. Also, students will discuss how surveys are important when engineers make decisions about environmental issues.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Amy Kolenbrander
Janet Yowell
Jessica Todd
Malinda Schaefer Zarske
10/14/2015
Educational Use
Rating
0.0 stars
Students are introduced to the International Space Station (ISS) with information about its structure, operation and key experiments. The ISS itself is an experiment in international cooperation to explore the potential for humans to live in space. The space station features state-of-the-art science and engineering laboratories to conduct research in medicine, materials and fundamental science to benefit people on Earth as well as people who will live in space in the future.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Lesson Plan
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Denise W. Carlson
Geoffrey Hill
Jane Evenson
Jessica Butterfield
Jessica Todd
Malinda Schaefer Zarske
09/18/2014
Educational Use
Rating
0.0 stars
Students take a hands-on look at the design of bridge piers (columns). First they brainstorm types of loads that might affect a Colorado bridge. Then they determine the maximum possible load for that scenario, and calculate the cross-sectional area of a column designed to support that load. Choosing from clay, foam or marshmallows, they create model columns and test their calculations.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Chris Valenti
Denali Lander
Denise W. Carlson
Joe Friedrichsen
Jonathan S. Goode
Malinda Schaefer Zarske
Natalie Mach
10/14/2015
Educational Use
Rating
0.0 stars
Students are introduced to the structure, function and purpose of locks and dams, which involves an introduction to Pascal's law, water pressure and gravity.
Subject:
Applied Science
Engineering
Material Type:
Lesson Plan
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Denali Lander
Denise W. Carlson
Jeff Lyng
Kristin Field
Lauren Cooper
09/18/2014
Educational Use
Rating
0.0 stars
Student teams build model hand dynamometers used to measure grip strengths of people recovering from sports injuries. They use their models to measure how much force their classmates muscles are capable of producing, and analyze the data to determine factors that influence a person's grip strength. They use this information to produce a recommendation of a hand dynamometer design for a medical office specializing in physical therapy. They also consider the many other ways grip strength data is used by engineers to design everyday products.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Denise W. Carlson
Jake Lewis
Malinda Schaefer Zarske
10/14/2015
Educational Use
Rating
0.0 stars
The application of engineering principles is explored in the creation of mobiles. As students create their own mobiles, they take into consideration the forces of gravity and convection air currents. They learn how an understanding of balancing forces is important in both art and engineering design.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Denise W. Carlson
Malinda Schaefer Zarske
Natalie Mach
10/14/2015
Educational Use
Rating
0.0 stars
The difference between an architect and an engineer is sometimes confusing because their roles in building design can be similar. Students experience a bit of both professions by following a set of requirements and meeting given constraints as they create a model parking garage. They experience the engineering design process first-hand as they design, build and test their models. They draw a blueprint for their design, select the construction materials and budget their expenditures. They also test their structures for strength and find their maximum loads.
Subject:
Applied Science
Engineering
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Abigail Watrous
Denali Lander
Janet Yowell
Katherine Beggs
Melissa Straten
Sara Stemler
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## 503 - Parallelepiped walk
Moderator: Board moderators
Luchi
New poster
Posts: 3
Joined: Thu Mar 24, 2005 4:33 pm
### 503 - Parallelepiped walk
#include <iostream.h>
int x[3] , x1[3] , x2[3] , a[2] , b[2] , min , sum , saer ;
int main()
{
while(cin>>x[0]>>x[1]>>x[2])
{
cin>>x1[0]>>x1[1]>>x1[2]>>x2[0]>>x2[1]>>x2[2];
if(!x[0]||!x[1]||!x[2])
{min=(x1[0]-x2[0])*(x1[0]-x2[0])+(x1[1]-x2[1])*(x1[1]-x2[1])+(x1[2]-x2[2])*(x1[2]-x2[2]);cout<<min<<'\n';}
else{
if(x1[0]==0) {a[0]=0;a[1]=3;} else if(x1[0]==x[0]) {a[0]=1;a[1]=3;}
else if(x1[1]==0) {a[0]=0;a[1]=2;} else if(x1[1]==x[1]) {a[0]=1;a[1]=2;}
else if(x1[2]==0) {a[0]=0;a[1]=1;} else if(x1[2]==x[2]) {a[0]=1;a[1]=1;}
if(x2[0]==0) {b[0]=0;b[1]=3;} else if(x2[0]==x[0]) {b[0]=1;b[1]=3;}
else if(x2[1]==0) {b[0]=0;b[1]=2;} else if(x2[1]==x[1]) {b[0]=1;b[1]=2;}
else if(x2[2]==0) {b[0]=0;b[1]=1;} else if(x2[2]==x[2]) {b[0]=1;b[1]=1;}
if((a[0]==b[0])&&(a[1]==b[1])) {min=(x1[0]-x2[0])*(x1[0]-x2[0])+(x1[1]-x2[1])*(x1[1]-x2[1])+(x1[2]-x2[2])*(x1[2]-x2[2]);cout<<min<<'\n';}
else {if(b[1]!=a[1])
{
saer=-1*(3-(a[1]+b[1]));
sum=(x1[saer]-x2[saer])*(x1[saer]-x2[saer]);
if(b[0]) if(a[0]) { sum+=(x[a[1]-saer]-x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer])*(x[a[1]-saer]-x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer]);}
else {sum+=(x[a[1]-saer]-x1[a[1]-saer]+x2[b[1]-saer])*(x[a[1]-saer]-x1[a[1]-saer]+x2[b[1]-saer]);}
else if(a[0]) {sum+=(x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer])*(x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer]);}
else {sum+=(x1[a[1]-saer]+x2[b[1]-saer])*(x1[a[1]-saer]+x2[b[1]-saer]);}
min=sum;
if(a[0]) { sum=(x[b[1]-saer]-x2[b[1]-saer]+x[saer]-x1[saer])*(x[b[1]-saer]-x2[b[1]-saer]+x[saer]-x1[saer]);}
else {sum=(x2[b[1]-saer]+x[saer]-x1[saer])*(x2[b[1]-saer]+x[saer]-x1[saer]);}
if(b[0]) { sum+=(x[a[1]-saer]-x1[a[1]-saer]+x[saer]-x2[saer])*(x[a[1]-saer]-x1[a[1]-saer]+x[saer]-x2[saer]);}
else { sum+=(x1[a[1]-saer]+x[saer]-x2[saer])*(x1[a[1]-saer]+x[saer]-x2[saer]);}
if(min>sum) min=sum;
if(a[0]) { sum=(x[b[1]-saer]-x2[b[1]-saer]+x1[saer])*(x[b[1]-saer]-x2[b[1]-saer]+x1[saer]);}
else {sum=(x2[b[1]-saer]+x1[saer])*(x2[b[1]-saer]+x1[saer]);}
if(b[0]) { sum+=(x[a[1]-saer]-x1[a[1]-saer]+x2[saer])*(x[a[1]-saer]-x1[a[1]-saer]+x2[saer]);}
else { sum+=(x1[a[1]-saer]+x2[saer])*(x1[a[1]-saer]+x2[saer]);}
if(min>sum) min=sum;
sum=(x[b[1]-saer]+x[saer]-x1[saer]+x[saer]-x2[saer])*(x[b[1]-saer]+x[saer]-x1[saer]+x[saer]-x2[saer]);
if(b[0]) if(a[0]) sum+=(x[a[1]-saer]-x1[a[1]-saer]+x2[b[1]-saer])*(x[a[1]-saer]-x1[a[1]-saer]+x2[b[1]-saer]);
else sum+=(x[a[1]-saer]-x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer])*(x[a[1]-saer]-x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer]);
else if(a[0]) sum+=(x1[a[1]-saer]+x2[b[1]-saer])*(x1[a[1]-saer]+x2[b[1]-saer]);
else sum+=(x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer])*(x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer]);
if(min>sum) min=sum;
sum=(x[b[1]-saer]+x1[saer]+x2[saer])*(x[b[1]-saer]+x1[saer]+x2[saer]);
if(b[0]) if(a[0]) sum+=(x[a[1]-saer]-x1[a[1]-saer]+x2[b[1]-saer])*(x[a[1]-saer]-x1[a[1]-saer]+x2[b[1]-saer]);
else sum+=(x[a[1]-saer]-x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer])*(x[a[1]-saer]-x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer]);
else if(a[0]) sum+=(x1[a[1]-saer]+x2[b[1]-saer])*(x1[a[1]-saer]+x2[b[1]-saer]);
else sum+=(x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer])*(x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer]);
if(min>sum) min=sum;
sum=a[0];a[0]=b[0];b[0]=sum;sum=a[1];a[1]=b[1];b[1]=sum;sum=x1[0];x1[0]=x2[0];x2[0]=sum;sum=x1[1];x1[1]=x2[1];x2[1]=sum;sum=x1[2];x1[2]=x2[2];x2[2]=sum;
sum=(x[b[1]-saer]+x[saer]-x1[saer]+x[saer]-x2[saer])*(x[b[1]-saer]+x[saer]-x1[saer]+x[saer]-x2[saer]);
if(b[0]) if(a[0]) sum+=(x[a[1]-saer]-x1[a[1]-saer]+x2[b[1]-saer])*(x[a[1]-saer]-x1[a[1]-saer]+x2[b[1]-saer]);
else sum+=(x[a[1]-saer]-x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer])*(x[a[1]-saer]-x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer]);
else if(a[0]) sum+=(x1[a[1]-saer]+x2[b[1]-saer])*(x1[a[1]-saer]+x2[b[1]-saer]);
else sum+=(x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer])*(x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer]);
if(min>sum) min=sum;
sum=(x[b[1]-saer]+x1[saer]+x2[saer])*(x[b[1]-saer]+x1[saer]+x2[saer]);
if(b[0]) if(a[0]) sum+=(x[a[1]-saer]-x1[a[1]-saer]+x2[b[1]-saer])*(x[a[1]-saer]-x1[a[1]-saer]+x2[b[1]-saer]);
else sum+=(x[a[1]-saer]-x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer])*(x[a[1]-saer]-x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer]);
else if(a[0]) sum+=(x1[a[1]-saer]+x2[b[1]-saer])*(x1[a[1]-saer]+x2[b[1]-saer]);
else sum+=(x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer])*(x1[a[1]-saer]+x[b[1]-saer]-x2[b[1]-saer]);
if(min>sum) min=sum;
}
else
{
sum=(x[a[1]%2]-x1[a[1]%2]+x[a[1]%2]-x2[a[1]%2]+x[3-a[1]])*(x[a[1]%2]-x1[a[1]%2]+x[a[1]%2]-x2[a[1]%2]+x[3-a[1]]);
sum+=(x1[a[1]-a[1]%2]-x2[a[1]-a[1]%2])*(x1[a[1]-a[1]%2]-x2[a[1]-a[1]%2]);
min=sum;
sum=(x1[a[1]%2]+x2[a[1]%2]+x[3-a[1]])*(x1[a[1]%2]+x2[a[1]%2]+x[3-a[1]]);
sum+=(x1[a[1]-a[1]%2]-x2[a[1]-a[1]%2])*(x1[a[1]-a[1]%2]-x2[a[1]-a[1]%2]);
if(sum<min) min=sum;
sum=(x[a[1]-a[1]%2]-x1[a[1]-a[1]%2]+x[a[1]-a[1]%2]-x2[a[1]-a[1]%2]+x[3-a[1]])*(x[a[1]-a[1]%2]-x1[a[1]-a[1]%2]+x[a[1]-a[1]%2]-x2[a[1]-a[1]%2]+x[3-a[1]]);
sum+=(x1[a[1]%2]-x2[a[1]%2])*(x1[a[1]%2]-x2[a[1]%2]);
if(sum<min) min=sum;
sum=(x1[a[1]-a[1]%2]+x2[a[1]-a[1]%2]+x[3-a[1]])*(x1[a[1]-a[1]%2]+x2[a[1]-a[1]%2]+x[3-a[1]]);
sum+=(x1[a[1]%2]-x2[a[1]%2])*(x1[a[1]%2]-x2[a[1]%2]);
if(sum<min) min=sum;
sum=(x[a[1]%2]-x1[a[1]%2]+x[a[1]-a[1]%2]-x2[a[1]-a[1]%2]+x[3-a[1]])*(x[a[1]%2]-x1[a[1]%2]+x[a[1]-a[1]%2]-x2[a[1]-a[1]%2]+x[3-a[1]]);
sum+=(x[a[1]-a[1]%2]-x1[a[1]-a[1]%2]+x[a[1]%2]-x2[a[1]%2])*(x[a[1]-a[1]%2]-x1[a[1]-a[1]%2]+x[a[1]%2]-x2[a[1]%2]);
if(sum<min) min=sum;
sum=(x1[a[1]%2]+x[a[1]-a[1]%2]-x2[a[1]-a[1]%2]+x[3-a[1]])*(x1[a[1]%2]+x[a[1]-a[1]%2]-x2[a[1]-a[1]%2]+x[3-a[1]]);
sum+=(x[a[1]-a[1]%2]-x1[a[1]-a[1]%2]+x2[a[1]%2])*(x[a[1]-a[1]%2]-x1[a[1]-a[1]%2]+x2[a[1]%2]);
if(sum<min) min=sum;
sum=(x[a[1]%2]-x1[a[1]%2]+x2[a[1]-a[1]%2]+x[3-a[1]])*(x[a[1]%2]-x1[a[1]%2]+x2[a[1]-a[1]%2]+x[3-a[1]]);
sum+=(x1[a[1]-a[1]%2]+x[a[1]%2]-x2[a[1]%2])*(x1[a[1]-a[1]%2]+x[a[1]%2]-x2[a[1]%2]);
if(sum<min) min=sum;
sum=(x1[a[1]%2]+x2[a[1]-a[1]%2]+x[3-a[1]])*(x1[a[1]%2]+x2[a[1]-a[1]%2]+x[3-a[1]]);
sum+=(x1[a[1]-a[1]%2]+x2[a[1]%2])*(x1[a[1]-a[1]%2]+x2[a[1]%2]);
if(sum<min) min=sum;
sum=(x[a[1]-a[1]%2]-x1[a[1]-a[1]%2]+x[a[1]%2]-x2[a[1]%2]+x[3-a[1]])*(x[a[1]-a[1]%2]-x1[a[1]-a[1]%2]+x[a[1]%2]-x2[a[1]%2]+x[3-a[1]]);
sum+=(x[a[1]%2]-x1[a[1]%2]+x[a[1]-a[1]%2]-x2[a[1]-a[1]%2])*(x[a[1]%2]-x1[a[1]%2]+x[a[1]-a[1]%2]-x2[a[1]-a[1]%2]);
if(sum<min) min=sum;
sum=(x1[a[1]-a[1]%2]+x[a[1]%2]-x2[a[1]%2]+x[3-a[1]])*(x1[a[1]-a[1]%2]+x[a[1]%2]-x2[a[1]%2]+x[3-a[1]]);
sum+=(x[a[1]%2]-x1[a[1]%2]+x2[a[1]-a[1]%2])*(x[a[1]%2]-x1[a[1]%2]+x2[a[1]-a[1]%2]);
if(sum<min) min=sum;
sum=(x[a[1]-a[1]%2]-x1[a[1]-a[1]%2]+x2[a[1]%2]+x[3-a[1]])*(x[a[1]-a[1]%2]-x1[a[1]-a[1]%2]+x2[a[1]%2]+x[3-a[1]]);
sum+=(x1[a[1]%2]+x[a[1]-a[1]%2]-x2[a[1]-a[1]%2])*(x1[a[1]%2]+x[a[1]-a[1]%2]-x2[a[1]-a[1]%2]);
if(sum<min) min=sum;
sum=(x1[a[1]-a[1]%2]+x2[a[1]%2]+x[3-a[1]])*(x1[a[1]-a[1]%2]+x2[a[1]%2]+x[3-a[1]]);
sum+=(x1[a[1]%2]+x2[a[1]-a[1]%2])*(x1[a[1]%2]+x2[a[1]-a[1]%2]);
if(sum<min) min=sum;
}
cout<<min<<'\n';
}
}
}
return 0;
}
metaphysis
Experienced poster
Posts: 139
Joined: Wed May 18, 2011 3:04 pm
### Re: 503 - Parallelepiped walk
Test data generator:
Code: Select all
``````#include <iostream>
#include <cstdlib>
#include <ctime>
using namespace std;
int main(int argc, char *argv[])
{
srand(time(NULL));
for (int c = 1; c <= 1000; c++)
{
int L = rand() % 1001, W = rand() % 1001, H = rand() % 1001;
int x1, y1, z1, x2, y2, z2;
int plane1 = rand() % 6;
switch (plane1)
{
// top
case 0:
x1 = rand() % (L + 1), y1 = rand() % (W + 1), z1 = H;
break;
// left
case 1:
x1 = rand() % (L + 1), y1 = 0, z1 = rand() % (H + 1);
break;
// right
case 2:
x1 = rand() % (L + 1), y1 = W, z1 = rand() % (H + 1);
break;
// front
case 3:
x1 = L, y1 = rand() % (W + 1), z1 = rand() % (H + 1);
break;
// back
case 4:
x1 = 0, y1 = rand() % (W + 1), z1 = rand() % (H + 1);
break;
// bottom
case 5:
x1 = rand() % (L + 1), y1 = rand() % (W + 1), z1 = 0;
break;
}
int plane2 = rand() % 6;
switch (plane2)
{
// top
case 0:
x2 = rand() % (L + 1), y2 = rand() % (W + 1), z2 = H;
break;
// left
case 1:
x2 = rand() % (L + 1), y2 = 0, z2 = rand() % (H + 1);
break;
// right
case 2:
x2 = rand() % (L + 1), y2 = W, z2 = rand() % (H + 1);
break;
// front
case 3:
x2 = L, y2 = rand() % (W + 1), z2 = rand() % (H + 1);
break;
// back
case 4:
x2 = 0, y2 = rand() % (W + 1), z2 = rand() % (H + 1);
break;
// bottom
case 5:
x2 = rand() % (L + 1), y2 = rand() % (W + 1), z2 = 0;
break;
}
cout << L << ' ' << W << ' ' << H << ' ';
cout << x1 << ' ' << y1 << ' ' << z1 << ' ';
cout << x2 << ' ' << y2 << ' ' << z2 << '\n';
}
return 0;
}
``````
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CC-MAIN-2019-35
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en
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http://forum.dominionstrategy.com/index.php?action=printpage;topic=18622.0
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# Dominion Strategy Forum
## Dominion => Dominion FAQ => Topic started by: phonological loop on May 22, 2018, 04:58:56 pm
Title: What organization scheme is used by the included inserts?
Post by: phonological loop on May 22, 2018, 04:58:56 pm
I have the base game (2nd edition big box) and a few expansions. Included in each box is a paper organizer insert with the names of the cards on it. You stick the insert in the middle of the box and it tells you where to put each card.
The problem is that I don't understand the organization scheme used by the insert, which makes it a headache to set up games. What is the logic behind the insert? I ask so I can find cards more efficiently. (And why is it not simply in alphabetical order in each box?)
(Almost certainly this has been asked before, but I could not find an explanation using Google or the search feature on this forum. My apologies, and thanks in advance for the help.)
Title: Re: What organization scheme is used by the included inserts?
Post by: GendoIkari on May 22, 2018, 05:28:47 pm
Assuming yours looks the same as this one?
https://cf.geekdo-images.com/original/img/TtW5SqvwdTSOdw5DOF12FUa6698=/0x0/pic3229621.jpg
The action cards are in alphabetical order. Coppers and Curses are in the middle presumably because they need a much bigger slot; and the insert is just easier to produce if the large section is in the middle. The other every-game cards are at the end; I think it would make more sense to put Gardens in with the action cards; as it's still just a Kingdom card.
Title: Re: What organization scheme is used by the included inserts?
Post by: phonological loop on May 22, 2018, 06:55:11 pm
Oh, I see now. Each expansion is indeed in alphabetical order (I was not looking very closely :-[).
The big box was what primarily confused me, but it too is in alphabetical order by set. I was confused because I have not learned to distinguish between base set and Intrigue cards (since I got them all at once, in one box).
I'm not usually this much of an idiot, I swear.
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en
| 0.971001
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https://findanyanswer.com/what-is-the-highest-occupied-energy-level-of-fluorine
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| 266,942,881
| 7,979
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# What is the highest occupied energy level of fluorine?
Asked By: Kevin Umayor | Last Updated: 6th February, 2020
Category: science chemistry
4.8/5 (794 Views . 40 Votes)
Fluorine has seven of eight possible electrons in its outermost energy level, which is energy level II. It would be more stable if it had one more electron because this would fill its outermost energy level.
Also question is, how many energy levels are occupied in an atom of oxygen?
Element Element Number Number of Electrons in each Level
Oxygen 8 6
Fluorine 9 7
Neon 10 8
Sodium 11 8
Subsequently, question is, how many energy levels does magnesium have? So for the element of MAGNESIUM, you already know that the atomic number tells you the number of electrons. That means there are 12 electrons in a magnesium atom. Looking at the picture, you can see there are two electrons in shell one, eight in shell two, and two more in shell three.
Thereof, what orbitals are occupied in fluorine?
Fluorine is the ninth element with a total of 9 electrons. In writing the electron configuration for fluorine the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for F go in the 2s orbital. The remaining five electrons will go in the 2p orbital.
How do you determine energy levels?
Number of energy levels in each period
1. The atoms in the first period have electrons in 1 energy level.
2. The atoms in the second period have electrons in 2 energy levels.
3. The atoms in the third period have electrons in 3 energy levels.
4. The atoms in the fourth period have electrons in 4 energy levels.
### How do you calculate energy level?
The formula defining the energy levels of a Hydrogen atom are given by the equation: E = -E0/n2, where E0 = 13.6 eV (1 eV = 1.602×10-19 Joules) and n = 1,2,3… and so on. The energy is expressed as a negative number because it takes that much energy to unbind (ionize) the electron from the nucleus.
### Does oxygen give or take electrons?
Oxygen is in group six in the periodic table so it has six electrons in its valence shell. This means that it needs to gain two electrons to obey the octet rule and have a full outer shell of electrons (eight). Because electrons have a charge of 1-, adding two electrons would make the charge of the oxide ion 2-.
### What is meant by energy level?
Energy levels inside an atom are the specific energies that electrons can have when occupying specific orbitals. Electrons can be excited to higher energy levels by absorbing energy from the surroundings. Light is emitted when an electron relaxes from a high energy state to a lower one.
### What is the highest occupied energy level?
The highest occupied energy level in an atom is the electron-containing main energy level with the highest number. b. What are inner-shell electrons? Inner-shell electrons are electrons that are not in the highest occupied energy level (sometimes referred to as “Kernel Electrons”).
### How many shells does oxygen have?
So for the element of OXYGEN, you already know that the atomic number tells you the number of electrons. That means there are 8 electrons in an oxygen atom. Looking at the picture, you can see there are two electrons in shell one and six in shell two.
32 electrons
### What information does 9 give about fluorine?
Fluorine has 9 protons. we know that atomic number of an atom is given by number of protons. Hence, the atomic number of Fluorine is 9. In a neutral atom, number of protons is equal to number of electrons but in the question no information is given on atom being neutral.
### Why is 3rd shell 8 or 18?
Each shell can contain only a fixed number of electrons: The first shell can hold up to two electrons, the second shell can hold up to eight (2 + 6) electrons, the third shell can hold up to 18 (2 + 6 + 10) and so on. The general formula is that the nth shell can in principle hold up to 2(n2) electrons.
### How do you fill orbitals?
RULES FOR FILLING ORBITALS. Rule 1 - Lowest energy orbitals fill first. Thus, the filling pattern is 1s, 2s, 2p, 3s, 3p, 4s, 3d, etc. Since the orbitals within a subshell are degenerate (of equal energy), the entire subshell of a particular orbital type is filled before moving to the next subshell of higher energy.
### What is Hund rule?
Hund's Rule. Hund's rule: every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.
### Why are there only 2 electrons in the first shell?
This first shell has only one subshell (labeled 1s) and can hold a maximum of 2 electrons. This is why there are two elements in the first row of the periodic table (H & He). Because the first shell can only hold a maximum of 2 electrons, the third electron must go into the second shell.
### What are Subshells?
A subshell is a subdivision of electron shells separated by electron orbitals. Subshells are labelled s, p, d, and f in an electron configuration.
### Why first shell is called K shell?
He noticed that atoms appeared to emit two types of X-rays. As it turns out, the K type X-ray is the highest energy X-ray an atom can emit. It is produced when an electron in the innermost shell is knocked free and then recaptured. This innermost shell is now called the K-shell, after the label used for the X-ray.
### What is the ground state of F?
Fluorine atoms have 9 electrons and the shell structure is 2.7. The ground state electron configuration of ground state gaseous neutral fluorine is [He]. 2s2. 2p5 and the term symbol is 2P3/2.
### How do you find the Valency of magnesium?
For magnesium, valency is equal to the number of electrons (two) in valence shell. For oxygen, valency is equal to eight minus the number of electrons present in the outermost shell (8−6=2).
### What is the Bohr model for magnesium?
Magnesium has 12 protons and 12 electrons. The first electron shell of a Bohr model holds 2 electrons. So far, 10 of magnesium's 12 electrons have been used, so only 2 remain. The remaining 2 are placed in the third electron shell, which is full when it holds 8 electrons.
### How many lone pairs does magnesium have?
The central atom is magnesium (draw the molecules Lewis structure to see this). There are two electron pairs around magnesium and no lone pairs. There are two bonding electron pairs and no lone pairs.
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| 3.046875
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https://www.tes.com/teaching-resources/shop/asmakha123
| 1,675,832,543,000,000,000
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| 1,020,903,657
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# Asma's Shop
1k+Views
#### Year 1 History - Personal timeline
(0)
L.O: I understand that ‘history’ means things that happened in the past. I can talk about things which have happened in my personal history. I can plot events onto a timeline in the correct order.
#### Matisse Art
(0)
An art lesson flipchart to help students to get inspired by artist Henri Matisse and make cutouts of their own. This resource is adapted from another resource - https://www.tes.com/teaching-resource/henri-matisse-cut-outs-lesson-11203884 credits -Author: Hipsta
#### Sunflowers - Kids art inspired by Van Gogh
(0)
1st Grade Art lesson - sunflower art inspired by Van Gogh
(0)
Morning activty
#### Fractions- LO : to find 1/4 and 3/4 of a number
(0)
Students learn to find the 1/4 and 3/4 of a number Sucess Criteria Figure out the fraction of the objects by drawing a grid of how many objects Draw the objects in the grid to see visually the fraction amount. Split the fraction into correct number of groups to find the Half 1/2 and quarter1/4 Split the fraction into correct number of groups to find the half 1/2,quarter 1/4 and three fourth 3/4 Figure out the fraction of the objects using known multiplication facts
#### LO: Use arrays to solve multiplication
(0)
Year 2 math lesson on using arrays for multiplications
#### Wild flowers- Drip painting technique
(0)
A minimum prep art lesson with no paint brushes for Early Years, Year 1,2 Learning about drip painting technique invented by Jackson Pollock.
#### Place value - LO- read 5 to 6 digit numbers
(0)
LO - be able to read 5 to 6 digit numbers
#### History (Lives of significant individuals ) -their contributions to national /inter achievement
(0)
In this lesson students will learn about the lives of significant individuals in the past who have contributed to national and international achievements. Looking at contributions of Neil Armstrong and Christopher Columbus.
#### Summer Art
(0)
This lesson is perfect to end the year with a summer art activity. The students will be learning to make a beach flip flop art.
#### Summer art
(0)
In this lesson students will make their own paper fans to cool off the summer heat. Perfect for end of year Summer art activities
#### 3D Shapes
(0)
LO: to identify and describe the properties of 3D shapes. Students will learn about 3D shapes and use clues to identify the properties of 3D shapes. A set of lesson for the complete week.
#### Childhood now and then
(0)
In this lesson students will learn about the difference in childhood in the past and in the present in History (Unit - Changes in Living Memory)
#### Multiplication /division Year2
(0)
In this lesson children recognise the multiples of 2,5 and 10 and record multiplication facts for 5. This is followed by relating multiplication with division using arrays.
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| 3.78125
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CC-MAIN-2023-06
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https://pastebin.com/WmVa6dwH
| 1,643,402,429,000,000,000
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# FP 2021-11-30
Nov 30th, 2021 (edited)
782
Never
Not a member of Pastebin yet? Sign Up, it unlocks many cool features!
1. import art
2. import random
3. import time
4. from enum import IntEnum
5.
6. class GameState (IntEnum):
7. NOT_STARTED = -1
8. PLAY_HIGHER = 1
9. PLAY_LOWER = 2
10. FINISHED = 3
11. #class GameState
12.
13. #guess_number.py
14. """
15. The computer randomly picks a number from
16. some range 1..10K
17. A human player will try to guess the number
18. picked by the computer
19. It will be a guided process, allowing some
20. rational playing
21. score
22. ++++++++++++++++++++++++ = amplitude
23. ------------------------ = attempts, time
24. """
25.
26. class GuessTheNumberGame:
27. def __init__ (
28. self,
29. pMin=1,
30. pMax=10000
31. ):
32. self.mMin = pMin
33. self.mMax = pMax
34. self.mStart = None
35. self.mEnd = None
36. #self.mAttempts = 0
37.
38. #historic of the played numbers
39. #QUEU => FIFO => First In First Out
40. self.mListOfPlays = []
41. #self.mListOfAttempts = list()
42.
43. #historic of the times (in seconds) when the numbers were played
44. self.mListOfPlaySeconds = []
45.
46. #this is the number that the human player must guess
47. self.mGuessMe = random.randint(
48. self.mMin,
49. self.mMax
50. )
51.
52. self.mState = GameState.NOT_STARTED
53. #def __init__
54.
55. def __str__(self):
56. strAll = ""
57.
58. strAll+="min: %d ; max: %d ; guessMe: %d\n"
59. strAll+="plays: %s ; time (seconds): %s\n"
60. strAll+="state: %s\n"
61. strAll = strAll%(
62. self.mMin, self.mMax, self.mGuessMe,
63. self.mListOfPlays.__str__(),
64. self.mListOfPlaySeconds.__str__(),
65. self.mState.__str__()
66. )
67. return strAll
68. #def __str__
69.
70. #assuming FIFO for plays
71. def getHumanPlayerLastPlay(self):
72. #return self.mListOfAttempts[len(self.mListOfAttempts)-1]
73. if (len(self.mListOfPlays)>0):
74. return self.mListOfPlays[-1]
75. else:
76. return None
77. #def getHumanPlayerLastPlay
78.
79. def getNumberOfAttempts(self):
80. return len(self.mListOfPlays)
81. #def getNumberOfAttempts
82.
83. def play(self, piPlayerNumber):
84. bCanPlay = self.mState!=GameState.FINISHED
85.
86. if (bCanPlay):
87. bFirstPlay = len(self.mListOfPlays)==0
88. if (bFirstPlay):
89. self.mStart = time.time()
90. #if
91.
92. iHowManySecondsSinceStart = time.time()-self.mStart
93.
94. self.mListOfPlays.append(piPlayerNumber)
95. self.mListOfPlaySeconds.append(iHowManySecondsSinceStart)
96.
97. if(piPlayerNumber < self.mGuessMe):
98. # suggest PLAY HIGHER
99. self.mState = GameState.PLAY_HIGHER
100. pass
101. elif (piPlayerNumber > self.mGuessMe):
102. #suggest PLAY LOWER
103. self.mState = GameState.PLAY_LOWER
104. pass
105. else:
106. #congratulations
107. self.mState = GameState.FINISHED
108. self.mEnd = time.time()
109.
110. strFeedback = self.feedback()
111. print (strFeedback)
112. #if can play
113. else: #could not play, but played
114. #TODO
115. pass
116. #def play
117.
118. def feedback(self):
119. strF = self.__str__()+"\n"
120.
121. if (self.mState==GameState.NOT_STARTED):
122. pass
123. elif (self.mState==GameState.PLAY_HIGHER):
124. strF+=\
125. "Your number %d is LOWER than mine, play HIGHER"\
126. %(self.getHumanPlayerLastPlay())
127. elif (self.mState==GameState.PLAY_LOWER):
128. strF += \
129. "Your number %d is HIGHER than mine, play LOWER" \
130. % (self.getHumanPlayerLastPlay())
131. else:
132. #congratulations
133. strF += art.text2art("Congratulations!!")
134. pass
135. #if
136. strF+="\n"+("-"*40)
137. return strF
138. #def feedback
139.
140. def gameFinished(self):
141. return self.mState==GameState.FINISHED
142. #def
143. #class GuessTheNumberGame
144.
145. #g1.instructions()
146. g1 = GuessTheNumberGame(1, 5000)
147. #print (g1)
148. while (not g1.gameFinished()):
150. #while
151.
152. import matplotlib.pyplot as plt
153. x = g1.mListOfPlays
154. y = g1.mListOfPlaySeconds
155. plt.plot(x, y)
156. plt.show()
157.
158.
159. ***************************************
160.
161.
162. #mylist.py
163.
164. class MyList:
165. #dunder double underscore
166.
167. BATATA = "batata frita"
168.
169. #initializer of instances of the MyList datatype
170. def __init__(
171. self,
172. pValues = None
173. ):
174. if (pValues is None):
175. self.mList = []
176. else:
177. self.mList = pValues
178. #def __init__
179.
180. #how instances of type MyList should be
181. #represented as string
182. def __str__(self):
183. strAll = ""
184. strAll+=self.mList.__str__()
185. return strAll
186. #def __str__
187.
188. #dynamic
189. def min (self):
190. theSmallest = None
191.
192. for v in self.mList:
193. if (theSmallest is None):
194. theSmallest = v
195. else:
196. if (v<theSmallest):
197. theSmallest = v
198. #if
199. #else
200. #for
201. return theSmallest
202. #def min
203.
204. def max(self):
205. theGreatest = None
206. for v in self.mList:
207. if (theGreatest is None):
208. theGreatest = v
209. else:
210. if(v>theGreatest):
211. theGreatest = v
212. #if
213. #else
214. #for
215. return theGreatest
216. #def max
217.
218. def howManyValues(self):
219. return len(self.mList)
220. #def howManyValues
221.
222. #mean / average / média
223. def mean (self):
224. howMany = self.howManyValues()
225. if (howMany>0):
226. theSum = 0
227. for v in self.mList:
228. #theSum = theSum+v
229. #cumulative sum +=
230. theSum+=v
231. #for
232. theMean = theSum / howMany
233. return theMean
234. else:
235. return None
236. #def mean
237.
238. """
239. p1 p2 p3 ... pn
240. pM
241.
242. sum = (p1-pM)^2 + (p2-pM)^2 + (p3-pM)^2 ... (pn-pM)^2
243. sum/howMany
244. """
245. def variance(self):
246. howMany = self.howManyValues()
247. theMean = self.mean()
248. theSum = 0
249. if (howMany>0 and (theMean!=None)):
250. for v in self.mList:
251. diff = v-theMean
252. diffSquared = diff**2
253. theSum += diffSquared
254. #for
255. theVariance = theSum / howMany
256. return theVariance
257. else:
258. return None
259. #def variance
260.
261. #stdevp = sqrt(variance)
262. #stdevp = variance ** (1/2)
263. def standardDeviation(self):
264. if (self.howManyValues()>0):
265. #return self.variance()**(1/2)
266. variance = self.variance()
267. squareRoot = variance**(1/2)
268. return squareRoot
269. else:
270. return None
271. #else
272. #def standardDeviation
273.
274. """
275. example:
276. working on list [10, 20, 40, 40]
277. set([10, 20, 40, 40]) --> [10, 20, 40]
278. working over a collection without repetitions...
279. set
280. the return should be this dictionary:
281. {10:1, 20:1, 40:2}
282. """
283. def absoluteFrequencyOfElements(self):
284. retDict = {}
285. if (self.howManyValues()>0):
286. setWithoutRepitions = set(self.mList) #[10, 20, 40] #optimization => does not question repeated elements
287. #for v in self.mList: #[10, 20, 40, 40] #would count 40 twice
288. for v in setWithoutRepitions:
289. iCount = self.count(v)
290. retDict[v] = iCount
291.
292. #retDict[v] = self.count(v)
293. #for
294. return retDict
295. else:
296. return None
297. #def absoluteFrequencyOfElements
298.
299. def count(self, pEl):
300. bShouldWork = self.howManyValues() and \
301. pEl!=None
302. if (bShouldWork):
303. iCounter = 0
304. for v in self.mList:
305. if (pEl==v):
306. iCounter+=1
307. #if
308. #for
309. return iCounter
310. else:
311. return 0
312. #def count
313.
314. def mostFrequentValues (self):
315. #dictionary of absolute frequencies
316. daf = self.absoluteFrequencyOfElements()
317. if (daf!=None):
318. theKeys = daf.keys()
319. theGreatest = None
320. for k in theKeys:
321. v = daf[k]
322. if (theGreatest is None):
323. theGreatest = v
324. else:
325. if (v>theGreatest):
326. theGreatest = v
327. #if
328. #else
329. #for
330.
331. #what do I have?????
332. #in var theGreatest we capture the most
333. #frequent value in the collection (self.mList)
334.
335. listMode = []
336. for k in theKeys:
337. if (daf[k]==theGreatest):
338. listMode.append(k)
339. #if
340. #for
341. return listMode
342. else:
343. return None
344. #def mostFrequentValues
345.
346. #TPC: leastFrequentValues
347.
348. def mode(self):
349. return self.mostFrequentValues()
350. #def
351.
353. self.mList.append(pEl)
354.
357. removedElement = None
358. if (len(self.mList)>0):
359. removedElement = self.mList[0]
360. self.mList = self.mList[1:]
361.
362. return removedElement
363.
366. removedElement = None
367. if (len(self.mList) > 0):
368. removedElement = self.mList[0]
369. self.mList = self.mList[1:]
370.
371. return removedElement
373.
374. # @tail?
375. def removeTail(self):
376. removedElement = None
377. if (len(self.mList) > 0):
378. removedElement = self.mList[-1]
379. #[start:end-exclusive]
380. self.mList = self.mList[0:-1]
381.
382. return removedElement
383. #def removeTail
384. #class MyList
385.
386. l1 = MyList()
387. l2 = MyList([10, 20, 30])
388. smallestInL1 = l1.min()
389. greatestInL2 = l2.max()
390.
391. print ("l1: ", l1)
392. print ("l2: ", l2)
393.
394. print ("smallest in l1: ", smallestInL1)
395. print ("greatest in l2: ", greatestInL2)
396.
397. print ("How many elements in l1: ",
398. l1.howManyValues()
399. )
400.
401. print ("How many elements in l2: ",
402. l2.howManyValues()
403. )
404.
405. theMeanL1 = l1.mean()
406. theMeanL2 = l2.mean()
407.
408. print ("The mean of l1: ", theMeanL1)
409. print ("The mean of l2: ", theMeanL2)
410.
411. l3 = MyList([20, 20, 40, 40, 50, 50, 50])
412. v3 = l3.variance()
413. dp3 = l3.standardDeviation()
414. print("The variance of {} is {}".format(l3, v3))
415. print("The standard deviation of {} is {}".\
416. format(l3, dp3))
417.
418. daf = l3.absoluteFrequencyOfElements()
419. print (daf)
420.
421. theModeList = l3.mode()
422. print ("The mode (as a list) for {} is {}".\
423. format(l3, theModeList))
424.
425. l2 = MyList()
426. l44 = MyList([44, 44, 44])
427. l2.mode()
428. l44.mode()
429.
430. #the left-operand is NOT an instance of the class
431. #but the class's name
432. print(MyList.BATATA) #static
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Sorry, an error occurred while loading the content.
Browse Groups
• ## [PrimeNumbers] sigma(n-1)+sigma(n)+sigma(n+1)
(9)
• NextPrevious
• Jack wrote: Because sigma(x) is odd if and only if x is a square or twice a square. The density of x such that sigma(x) is odd == 0. [jp :
Message 1 of 9 , Jul 2, 2002
View Source
Jack wrote:
Because sigma(x) is odd if and only if x is a square or
twice a square.
The density of x such that sigma(x) is odd == 0.
[jp : http://mathworld.wolfram.com/DivisorFunction.html (eqn28)
Try finding n such that n divides sigma(n-1)+sigma(n)+sigma(n+1).
Heuristics would indicate an infinite number of such n...
2, 5, 7, 33, 18336, 19262, 38184, 54722, ...
I've just submitted this to the EIS.
---
From the link, the equations can be morphed into:
(n-1).sigma(n-1) = 2 + i.phi(n-1)
n.sigma(n) = 2 + j.phi(n)
(n+1).sigma(n+1) = 2 + k.phi(n+1)
n(n^2-1).sigma(n(n^2-1)) = [2 + i.phi(n-1)].[2 + j.phi(n)].[2 + k.phi(n+1)]
so the answers occur whenever the RHS contains n^2 as a factor.
Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
-----Original Message-----
From: Phil Carmody [mailto:thefatphil@...]
Sent: 02 July 2002 13:35
Subject: RE: [PrimeNumbers] Factor patterns
--- Jon Perry <perry@...> wrote:
> A swerve.
>
> Can you explain why sigma(n-1)+sigma(n)+sigma(n+1) is nearly always
> even?
>
> (PARI/GP)
>
> for
> (n=2,1000,write("sigmax.txt",(sigma(n-1)+sigma(n)+sigma(n+1))%2))
Look at sigma(n)%2.
Mostly 0.
Each of the many remotely isolated 1s will cause 3 consecutive 1s in
your expression. 101 causes a 11011 in yours, and 11 causes 1001 in
yours.
No magic.
Phil
=====
--
"One cannot delete the Web browser from KDE without
losing the ability to manage files on the user's own
hard disk." - Prof. Stuart E Madnick, MIT.
So called "expert" witness for Microsoft. 2002/05/02
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• ... I believe not; you can do it mod 6, mod 12, mod 24 and then there are exceptions. Easy to see why: we require a modulus M such that n^2=1 mod M for all n
Message 2 of 9 , Jul 2, 2002
View Source
> Are there any more numbers sharing this property?
I believe not; you can do it mod 6, mod 12, mod 24
and then there are exceptions.
Easy to see why: we require a modulus M such that
n^2=1 mod M for all n with gcd(n,M)=1
M=24 is easy.
We cannot have any larger M with gcd(M,5)=1,since
5^2-1=24.
So try M=12*5.
Now we have problems with
7^2-1=48
unless M=12*5*7.
Now we have problems with
11^2-1=120
It's clear enough that we are are losing out,
though I do not have a proof..
David
• (sorry about the complete tripe in previous post - the n.sigma(n) eqv. 2 mod phi(n) is only true for n=p, and n=4,6,22). -- Continuing Jack s work: for
Message 3 of 9 , Jul 3, 2002
View Source
(sorry about the complete tripe in previous post - the n.sigma(n) eqv. 2 mod
phi(n) is only true for n=p, and n=4,6,22).
--
Continuing Jack's work:
for (n=2,100000,if ((sigma(n-1)+sigma(n)+sigma(n+1))%(n-1)==0,
write("sigmaconn.txt",n,":",sigma(n-1)%(n-1),":",+sigma(n)%(n-1),":",sigma(n
+1)%(n-1),":",(sigma(n-1)+sigma(n)+sigma(n+1)))))
2,3,24,64,227,291,784,1883,7731,18547,25723,30397,94358
for (n=2,100000,if ((sigma(n-1)+sigma(n)+sigma(n+1))%(n+1)==0,
write("sigmaconp.txt",n,":",sigma(n-1)%(n+1),":",+sigma(n)%(n+1),":",sigma(n
+1)%(n+1),":",(sigma(n-1)+sigma(n)+sigma(n+1)))))
8,21,22,23,57,157,505,1053,2147,2273,3311,4679,5931,7898,22682
--
A related question - when does sigma(n,2)%sigma(n,1)==0?
If n is a square, and also:
20,50,117,180,200,242,325,450,468,500,578,605,650,800,968,980
for (n=2,10000, if (sigma(n,2)%sigma(n,1)==0 &&
!issquare(n),write("sigmasigmasq.txt",n)))
Anyone spot the missing link?
Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
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Classroom Stuff Mark Brittenham The classes that I am teaching this (Spring, 2019) semester each have their own web page: Math 990, Topics in Topology (Hyperbolic Geometry and Topology) (Spring, 2019) There are also, of course, the pages produced for (some of) my earlier courses, in case you want to finally finish that problem set that you forgot to turn in..... Math 325, Elementary Analysis (Fall, 2018) Math 417, Group Theory (Fall, 2018) Math 423, Intro to Complex Variables (Spring, 2018) Math 990, Topics in Topology (Orderings of (3-manifold) Groups) (Spring, 2018) Math 107H, Calculus II (Fall, 2017) Math 871, Topology I (Fall, 2017) Math 325, Elementary Analysis (Spring, 2017) Math 872, Topology II (Spring, 2017) Math 106, Calculus I (Fall, 2016) Math 871, Topology I (Fall, 2016) Math 208H, Calculus III (Spring, 2016) Math 417, Group Theory (Spring, 2016) Math 107H, Calculus II (Fall, 2015) Math 856, Introduction to Smooth Manifolds (Fall, 2015) Math 314, Matrix Theory (Spring 2015) Math 872, Topology II (Spring 2015 Math 314, Matrix Theory (Fall, 2014) Math 871, Topology I (Fall, 2014) Math 208H, Calculus III (Spring, 2014) Math 990, Topics in Topology (Topological Methods in Group Theory) (Spring, 2014) Math 107H, Calculus II (Fall, 2013) Math 314, Matrix Theory (Fall, 2013) On leave, 2012-2013 academic year. Math 208H, Calculus III (Spring 2012) Math 325, Elementary Analysis (Spring 2012) Math 189H, Freshman Honors Seminar (Fall 2011) Math 856, Introduction to Smooth Manifolds (Fall 2011) Math 423/823, Intro to Complex Variables (Spring 2011) Math 896, Landscapes Seminar (Spring 2011) Math 990, Topics in Topology (Combinatorial Knot Theory) (Spring 2011) Math 107H, Calculus II (Honors) (Fall 2010) Math 107, Calculus II (Spring 2010) Math 314/814, Matrix Theory (Spring 2010) Math 221, Differential Equations (Fall 2009) Math 856, Introduction to Smooth Manifolds (Fall 2009) Math 314/814, Matrix Theory (sections 001 and 005) (Spring 2009) Math 445, Introduction to Number Theory (Fall 2008) Math 107H, Calculus II (Fall 2008) Math 314/814 Matrix Theory (Spring 2008) Math 990 Topics in Topology (Linear Rep'ns of Finitely-presented Groups) (Spring 2008) Math 106 Analytic Geometry and Calculus I (Fall 2007) Math 203, Contemporary Mathematics (Fall 2007) Math 872 Algebraic Topology (Spring 2007) Math 856, Introduction to Smooth Manifolds (Fall 2006) On leave, 2005-2006 academic year. Math 971, Algebraic Topology 1 (Spring 2005) Math 445, Introduction to Number Theory (Fall 2004) Math 208H, Calculus III (Spring 2004) Math 221, Differential Equations (Spring 2004) Math 107H, Calculus II (Fall 2003) Math 970, Topology (Fall 2003) Math 221, Differential Equations (Spring 2003) Math 978, Survey of Knot Theory (Spring 2003) Math 203, Contemporary Mathematics (Spring 2002) Math 970, Topology (Spring 2002) Math 310/310H, Introduction to Modern Algebra (Fall, 2001) Math 971, Algebraic Topology (Fall, 2001) Math 970, Topology (Spring, 2001) Math 203, Contemporary Mathematics (Spring, 2001) Math 208H, Multivariable Calculus (Fall, 2000) Math 221, Differential Equations (Spring, 2000) Math 314, Matrix Theory (Spring, 2000) Math 208, Multivariable Calculus (Fall, 1999) Math 971, Algebraic Topology (Fall, 1999) Math 203, Contemporary Mathematics (Spring, 1999) Math 314/814, Matrix Methods (Spring, 1999) Math 1650, Precalculus (Fall, 1998) Math 1710, Calculus I (Fall, 1998) Math 4500, Introduction to Topology (Spring, 1998) Math 1710, Calculus I (Spring, 1998) Math 1650, Precalculus (Fall, 1997) Math 1720, Calculus II (Fall, 1997) Before the fall of 1997 I apparently hadn't discovered the notion of a class web page.
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# 費式數列
## 說明
Fibonacci為1200年代的歐洲數學家，在他的著作中曾經提到：「若有一隻免子每個月生一隻小免子，一個月後小免子也開始生產。起初只有一隻 免 子，一個月後就有兩隻免子，二個月後有三隻免子，三個月後有五隻免子（小免子投入生產）......」。
1、1 、2、3、5、8、13、21、34、55、89......
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2
## 算法
``Procedure FIB(N) IF (N = 0 OR N = 1) RETURN N ELSE RETURN FIB(N-1) + FIB(N-2)``
``Procedure FIB(N) IF (N = 0 OR N = 1) RETURN N a = 0; b = 1; FOR i = 2 TO N [ temp = b; b = a + b; a = temp; ] RETURN b; ] ``
``F(0) = 0; F(1) = 1; FOR i<-2 TO N [ F(i) = F(i-1) + F(i-2); ] ``
``Procedure FIB(N) IF (n <= 1) RETURN(n); IF (n = 2) RETURN(1); ELSE [ i = n/2; f1 = FIB(i+1); f2 = FIB(i); IF (n mod 2 = 0) RETURN( f2*(2*f1+f2) ); ELSE RETURN ( f1**2+f2**2 ); ]] ``
F0 = 0
F1 = 1
Fn = X * Fn-1 + Y * Fn-2
``````def fib(n) {
if n == 0 or n == 1 {
return n
}
return fib(n - 1) + fib(n - 2)
}
iterate(0, 10).forEach(i -> println(fib(i)))``````
``#include <stdio.h> #include <stdlib.h> #define LEN 20 void fill_fibonacci_numbers(int*, int);void print(int*, int len);int main(void) { int fib[LEN] = {0}; fill_fibonacci_numbers(fib, LEN); print(fib, LEN); return 0; } void fill_fibonacci_numbers(int* fib, int len) { fib[0] = 0; fib[1] = 1; int i; for(i = 2; i < len; i++) { fib[i] = fib[i-1] + fib[i-2]; }}void print(int* arr, int len) { int i; for(i = 0; i < len; i++) { printf("%d ", arr[i]); } printf("\n");}``
``import java.util.*;public class Fibonacci { private List<Integer> fib = new ArrayList<>(); { fib.add(0); fib.add(1); } Integer get(int n) { if(n >= fib.size()) for(int i = fib.size(); i <= n; i++) { fib.add(fib.get(i - 1) + fib.get(i - 2)); } return fib.get(n); } public static void main(String[] args) { Fibonacci fibonacci = new Fibonacci(); for(int i = 0; i < 20; i++) { System.out.print(fibonacci.get(i) + " "); } }}``
``def fibonacci(n, fib = [0, 1]): if n >= len(fib): for i in range(len(fib), n + 1): fib.append(fib[i - 1] + fib[i - 2]) return fib[n]for i in range(0, 20): print(fibonacci(i), end=' ')``
``def fib(n: Int): Int = n match { case 0 => 0 case 1 => 1 case _ => fib(n - 1) + fib(n - 2)}(for(i <- 0 until readInt) yield fib(i)).foreach(i => print(i + " "))``
``# encoding: Big5fibonacci = -> { fib = [0, 1] -> n { if n >= fib.size fib.size.upto(n) do |i| fib << fib[i - 1] + fib[i - 2] end end fib[n] }}.callprint "輸入個數："length = gets.to_i0.upto(length - 1) do |i| print fibonacci.call(i).to_s + ' 'end``
``var fibonacci = function() { var fib = [0, 1]; return function(n) { if(n >= fib.length) for(var i = fib.length; i <= n; i++) { fib[i] = fib[i - 1] + fib[i - 2]; } return fib[n]; };}(); for(var i = 0; i < 20; i++) { print(fibonacci(i)); }``
``fibonacci 0 = 0fibonacci 1 = 1fibonacci n = addPrevsRecusivelyUntilCounterIsN (fib 1) (fib 0) 2 naddPrevsRecusivelyUntilCounterIsN prev1 prev2 counter n | counter == n = result | otherwise = addPrevsRecusivelyUntilCounterIsN result prev1 (counter + 1) n where result = prev1 + prev2main = sequence [print (fibonacci i) | i <- [0..19]]``
``````fibonacci(0, 0).
fibonacci(1, 1).
fibonacci(N, Result) :- NP1 is N - 1, NP2 is N - 2,
fibonacci(NP1, FP1), fibonacci(NP2, FP2),
Result is FP1 + FP2.
main([Arg0|_]) :-
atom_number(Arg0, N),
fibonacci(N, Result),
writef("The nth %n is %d\n", [Arg0, Result]).``````
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# Short Tricks on Coordinate Geometry
Today we will be covering a very important topic from the Advance Maths part of the Quantitative Aptitude section that is – Important Notes & Short Tricks on Coordinate Geometry. The post is very helpful for the upcoming SSC CGL Exam 2015.
## Important Short Tricks on Coordinate Geometry
1. Equation of line parallel to y-axis
X = b
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For Example: A Student plotted four points on a graph. Find out which point represents the line parallel to y-axis.
1. a) (3,5)
b) (0,6)
c) (8,0)
d) (-2, -4)
Solution: Option (C)
1. Equation of line parallel to x-axis
Y = a
For Example: A Student plotted four points on a graph. Find out which point represents the line parallel to x-axis.
1. a) (3,5)
b) (0,6)
c) (8,0)
d) (-2, -4)
Solution: Option (B)
1. Equations of line
a) Normal equation of line
ax + by + c = 0
b) Slope – Intercept Form
y = mx + c Where, m = slope of the line & c = intercept on y-axis
For Example: What is the slope of the line formed by the equation 5y – 3x – 10 = 0?
Solution: 5y – 3x – 10 = 0, 5y = 3x + 10
Y = 3/5 x + 2
Therefore, slope of the line is = 3/5
c) Intercept Form
x/A + y/B = 1, Where, A & B are x-intercept & y-intercept respectively
For Example: Find the area of the triangle formed the line 4x + 3 y – 12 = 0, x-axis and y-axis?
Solution: Area of triangle is = ½ * x-intercept * y-intercept.
Equation of line is 4x + 3 y – 12 = 0
4x + 3y = 12,
4x/12 + 3y/12 = 1
x/3 + y/4 = 1
Therefore area of triangle = ½ * 3 * 4 = 6
d) Trigonometric form of equation of line, ax + by + c = 0
x cos θ + y sin θ = p,
Where, cos θ = -a/ √(a2 + b2) , sin θ = -b/ √(a2 + b2) & p = c/√(a2 + b2)
e) Equation of line passing through point (x1,y1) & has a slope m
y – y1 = m (x-x1)
1. Slope of line = y2 – y1/x2 – x1 = – coefficient of x/coefficient of y
1. Angle between two lines
Tan θ = ± (m2 – m1)/(1+ m1m2) where, m1 , m2 = slope of the lines
Note: If lines are parallel, then tan θ = 0
If lines are perpendicular, then cot θ = 0
For Example: If 7x – 4y = 0 and 3x – 11y + 5 = 0 are equation of two lines. Find the acute angle between the lines?
Solution: First we need to find the slope of both the lines.
7x – 4y = 0
⇒ y = 74x
Therefore, the slope of the line 7x – 4y = 0 is 74
Similarly, 3x – 11y + 5 = 0
⇒ y = 311x + 511
Therefore, the slope of the line 3x – 11y + 5 = 0 is = 311
Now, let the angle between the given lines 7x – 4y = 0 and 3x – 11y + 5 = 0 is θ
Now,
Tan θ = ± (m2 – m1)/(1+ m1m2) = ±[(7/4)−(3/11)]/[1+(7/4)*(3/11)] = ± 1
Since θ is acute, hence we take, tan θ = 1 = tan 45°
Therefore, θ = 45°
Therefore, the required acute angle between the given lines is 45°.
1. Equation of two lines parallel to each other
ax + by + c1 = 0
ax + by + c2 = 0
Note: Here, coefficient of x & y are same.
1. Equation of two lines perpendicular to each other
ax + by + c1 = 0
bx – ay + c2 = 0
Note: Here, coefficient of x & y are opposite & in one equation there is negative sign.
1. Distance between two points (x1, y1), (x2, y2)
D = √ (x2 – x1)2 + (y2 – y1)2
For Example: Find the distance between (-1, 1) and (3, 4).
Solution: D = √ (x2 – x1)2 + (y2 – y1)2
= √ (3 – (-1))2 + (4 – 1)2 = √(16 + 9) = √25 = 5
1. The midpoint of the line formed by (x1, y1), (x2, y2)
M = (x1 + x2)/2, (y1 + y2)/2
1. Area of triangle whose coordinates are (x1, y1), (x2, y2), (x3, y3)
½ I x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) I
For Example: Find area of triangle whose vertices are (1, 1), (2, 3) and (4, 5).
Solution: We have (x1, y1) = (1, 1), (x2, y2) = (2, 3) and (x3, y3) = (4, 5)
Area of Triangle = ½ I x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) I
=1/2 I (1(3−5) +2(5−1) + 4(1−3)) I
=1/2 I(−2+8−8) =1/2 (−2) I = I−1I = 1
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## eroshea 3 years ago rationalize
1. eroshea
$\frac{ 4-4\sqrt{3}i }{ -2\sqrt{3}+2i }$
2. hartnn
multiply and divide by -2root3-2i
3. hartnn
or take -2i common from numerator. u get -2i(2i-2root 3) in numerator. so what gets cancelled?
4. eroshea
wait.. i'll try
5. eroshea
is there something that can be cancelled?
6. hartnn
see the factor in numerator and denominator!!
7. eroshea
owh.. -2i will remain??and the factor will cancel out the denominator?
8. hartnn
yup,-2i,got it ?
9. eroshea
but i think its wrong? i thnk this should be what in the numerator is , -2i(2i+2root3),, am i right??
10. hartnn
O.o
11. hartnn
options?
12. hartnn
and yes, u are right....
13. eroshea
i dont know what to do next
14. hartnn
multiply and divide by -2root3-2i
15. eroshea
i hate it when that "i" is in my solution .. :D i'll try solving it
16. eroshea
the denominator will be 8? but i have a difficulty in solving the numerator :((
17. hartnn
8 ?? or -16 ?
18. eroshea
my bad.. i believe the numerator should be 18?? 12-4i^2 = 12-4(-1) = 12+4= 16?
19. eroshea
i mean the DENOMINATOR
20. hartnn
*minus*
21. eroshea
y negative?
22. eroshea
$(-2\sqrt{3}+2i)(-2\sqrt{3}-2i) = (-2\sqrt{3})^{2}-4i ^{2}=12-4(-1)=16$
23. hartnn
my bad, sorry,i took - common from that....ok,so 16 numerator = ?
24. hartnn
and what are your options ?
25. eroshea
my final answer is -root3 - i
26. eroshea
if i simplified my numerator, -16root3 - 16i and divide that by 16 -root3 -i is my answer correct?
27. hartnn
hmm....its my silly mistake's day or i m getting i-root3
28. eroshea
owh @_@
29. eroshea
i'll try solving it again
30. eroshea
owh you're right!! hahaha... i had a mistake with my subtraction, 24i-8i should be 16i and not -16i.. thanks ! :))
31. hartnn
finally! welcome :)
32. eroshea
it's not your silly mistake day :))
33. hartnn
glad to know :)
34. eroshea
how long have you've been in this site?
35. hartnn
48 days around.....
36. eroshea
owh.. i just made my account yesterday :3 brb.. i'll just go out and eat first and then back in reviewing :)) i have my finals tomorrow for algebra and trigonometry
37. hartnn
best of luck.
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mersenneforum.org Masking a PIN over a phone call.
Register FAQ Search Today's Posts Mark Forums Read
2013-10-29, 19:13 #12
chalsall
If I May
"Chris Halsall"
Sep 2002
2·3·1,499 Posts
Quote:
Originally Posted by Mini-Geek As long as Eve can only hear Alice's side of the conversation, the encryption works.
But let us assume you can't guarantee that...
Remember that book we both read so many years ago?
496.297.
2013-10-29, 20:23 #13 ewmayer ∂2ω=0 Sep 2002 República de California 3×612 Posts I took the OP note "Eve is sitting at Alice's restaurant table" as being important - if we were talking about bulk electronic-comms encryption, there would be little point in starting a new thread in Puzzles about that, would there? So I am going with the how-can-A-and-B-exchange-a-PIN-under-C's-nose scenario, which supposes that C hears everything that is said by A and B. Couple possible prearranged simple-crypto schemes: o Alice conveys PIN via number-sips-from-wineglass - wipe rim or sip water to end current digits. o Alice stealthily uses foot to squeeze Bob's junk under table, Flashdance style - number of squeezes conveys data, kick Bob's shin with other foot to end current digit. [Bob is glad that PIN is only 4 digits long].
2013-10-29, 20:28 #14
chalsall
If I May
"Chris Halsall"
Sep 2002
100011001000102 Posts
Quote:
Originally Posted by ewmayer Couple possible prearranged simple-crypto schemes: o Alice conveys PIN via number-sips-from-wineglass - wipe rim or sip water to end current digits. o Alice stealthily uses foot to squeeze Bob's junk under table, Flashdance style - number of squeezes conveys data, kick Bob's shin with other foot to end current digit. [Bob is glad that PIN is only 4 digits long].
That's not crypto. That's simple out-of-band communications. Easily caught with current AI.
2013-10-29, 20:44 #15
ewmayer
2ω=0
Sep 2002
República de California
1116310 Posts
Quote:
Originally Posted by chalsall That's not crypto. That's simple out-of-band communications. Easily caught with current AI.
2013-10-29, 20:49 #16
chalsall
If I May
"Chris Halsall"
Sep 2002
2·3·1,499 Posts
Quote:
Originally Posted by ewmayer Only if the AI has access to Bob's junk.
Don't we all?
2013-10-29, 21:18 #17
chalsall
If I May
"Chris Halsall"
Sep 2002
2×3×1,499 Posts
Quote:
Originally Posted by chalsall Don't we all?
In my opinion, we find ourselves in a very interesting situation, and very interesting times...
Some people are trying to disrupt and discredit our community.
Some people make sense, other's don't.
I don't fully understand the motives.
Deep thought is welcome....
2013-10-30, 03:03 #18
axn
Jun 2003
107618 Posts
Quote:
Originally Posted by ewmayer I took the OP note "Eve is sitting at Alice's restaurant table" as being important
Quote:
Originally Posted by Flatlander So Alice is on the phone to Bob
My bolding.
2013-10-30, 05:24 #19
TheMawn
May 2013
East. Always East.
11×157 Posts
Quote:
Originally Posted by ewmayer [Bob is glad that PIN is only 4 digits long].
But Bob wishes that her pin was all nines.
2013-10-30, 05:28 #20 TheMawn May 2013 East. Always East. 172710 Posts If Alice cannot hear Bob then Bob can just offer some kind of code that Alice does not know and it is impossible for her to have the slightest clue what the answer is. How much must I subtract from 37 to get the first digit? How much must I subtract from 11 to get the second digit? Etc. He chooses the numbers. When EVE says 35, 3, 7 560 and 0, how can Alice have ANY clue? On the other hand, if Alice hears both Eve and Bob then there is no way for them to devise a code.
2013-10-30, 10:16 #21 Flatlander I quite division it "Chris" Feb 2005 England 31·67 Posts Solved with MOD 10 then; but just before Bob needs the PIN Alice and Bob are struck down with Stupiditis and are only able to comprehend integers 0 through 9, ... (Or some other restriction to make it more interesting.)
2013-10-30, 11:43 #22
retina
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2×3×907 Posts
Quote:
Originally Posted by Flatlander ... struck down with Stupiditis ...
I know how that feels. I get struck be that very frequently. As for solving this new modified puzzle I feel an attack of Stupiditis coming on now. :(
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# Distance and Direction Sense Test
#### Chapter 10
5 Steps - 3 Clicks
# Distance and Direction Sense Test
### Description
A progressive follow-up of directions and/or separation is planned and on the premise of given data it is required to determine the last direction regarding the beginning stage or the most limited separation between the beginning stage and the last point. Now and then both the last direction and the distance covered are asked.
Distance and direction sense test questions comprise of a kind of direction and/or distance riddle. Clearly, such inquiries are intended to judge the competitor’s capacity to follow, take after and perceive the direction, described in somewhat complicated language, effectively.
In order to solve such questions correctly you must have the knowledge of directions on the plane of a paper. In the meantime, it is important to draw out the directions according to the data gave in the question in proper arrangement. An error at any time would change your answer decision.
### Concept
The adjoining figure shows four directions. They are East(E), West(W), North(N) and South(S).
Also there are four more cardinal directions like East-North(EN), West-North(WN), East-South(ES), West-South(WS) as shown below:
Let an example is considered, as shown below:
Let the person turns to all the possible directions to have a clear understanding.
Here:
1. A man facing towards North, on taking left turn will face towards West and on taking the right turn towards East.
2. A man facing towards South, on taking left turn will face towards East and on taking right turn towards West.
3. A man facing towards East, on taking left turn will face towards North and on taking right turn towards South.
4. A man facing towards West, on taking left turn will face towards South and on taking right turn towards North.
5. A man facing towards North-West, on taking left turn will face towards South-West and on taking right turn will face towards North-East.
6. A man facing towards South-West on taking left turn will face towards South-East and on taking right turn towards North-West.
7. A man facing towards South-East, on taking left turn will face towards North-East and on taking right turn towards South-West.
8. A man facing towards North-East, on taking left turn will face towards North-West and on taking right turn towards South-East.
### Formulae
Pythagoras formula:
In order to determine the distance travelled or shortest straight distance travelled between given two points, Pythagoras formula is used. i.e.
$$h^2$$ = $$p^2 + b^2$$
As shown, from right angled triangle
Here:
h is hypotenuse.
p is perpendicular.
b is Base.
### Model Problems
Model 1: Tom starting from his town, goes 4 km in the West, then he turns to his left and goes 3 km. What minimum distance will be covered by him to come back to his town?
Solution:
Given data:
Tom goes 4 km in the west
From there to his right, goes 3 km.
Consider the drawing shown below:
Therefore,
$$Distance^2$$ = $$4^2 + 3^2$$
Distance = $$\sqrt{16 + 9}$$
Distance = 5 km
Therefore, minimum distance covered by him to come back to his town = 5 km.
Model 2: Alice starting from home walked 10 km to reach the crossing of Museum. In which direction she was going, a road opposite to this direction goes to Park. The road to the right goes to station. If the road which goes to station is just opposite to the road which School, then in which direction to Hema is the road which goes to School?
Solution:
Given that
Alice walks 10 km and crosses Museum
Road opposite to Alice home is Park.
From there, Road to the right goes to Station.
Now, Consider the below figure:
From the figure, it is clear that
The road which goes to School is left to Alice.
Model 3: One morning after sunrise Julie while going to market met Louise at road crossing. Louise’s shadow was exactly to the right of Julie. If they were face to face, which direction was Julie facing?
Solution:
Given that
Its morning. So, sun rises in the east.
Also, the shadow falls towards the west in the morning.
Now
Therefore, Louise’s shadow falls to the right of the Julie. Hence Julie is facing South.
Model 4: Henry starting from his house, goes 10 km in the west, then turns to his left and goes 6 km. Finally turns to his left and goes 10 km. Now how far is Henry from his house and in what direction?
Solution:
Given that
Henry goes 10 km west.
From there, turns left and goes 6 km.
Finally to left and goes 10 km.
Now, consider the following graph:
From the figure, it is clear from third position that Henry is 6 km far from home.
Therefore, Henry is facing north direction.
Model 5: Novel left home and cycled 10 km northwards, turned right left and cycled 5 km and turned left and cycled 10 km and turned right and cycled 10 km. How many kilometres will Novel have to cycle to reach home straight?
Solution:
Given that
Novel cycled 10 km northwards
Then turned left and cycled 5 km
then turned left and cycled 10 km and then turned right, cycled 10 km.
Consider the picture as shown below:
From figure, it is clear that
Novel starts from A, moves 10 km northwards upto B, turns left and moves 5 km upto C, turns left again and moves 10 km upto D and finally turns right and moves 10 km upto E.
Therefore,
Distance from initial position A is
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Resume Reading — The Impossible Mathematics of the Real World
Close
# The Impossible Mathematics of the Real World
## Near-miss math provides exact representations of almost-right answers.
Using stiff paper and transparent tape, Craig Kaplan assembles a beautiful roundish shape that looks like a Buckminster Fuller creation or a fancy new kind of soccer ball. It consists of four regular dodecagons (12-sided polygons with all angles and sides the same) and 12 decagons (10-sided), with 28 little gaps in the shape of equilateral triangles. There’s just one problem. This figure should be impossible. That set of polygons won’t meet at the vertices. The shape can’t close up.
Kaplan’s model works only because of the wiggle room you get when you assemble it with paper. The sides can warp a little bit, almost imperceptibly. “The fudge factor that arises just from working in the real world with paper means that things that ought to be impossible actually aren’t,” says Kaplan, a computer scientist at the University of Waterloo in Canada.
It is a new example of an unexpected class of mathematical objects that the American mathematician Norman Johnson stumbled upon in the 1960s. Johnson was working to complete a project started over 2,000 years earlier by Plato: to catalog geometric perfection. Among the infinite variety of three-dimensional shapes, just five can be constructed out of identical regular polygons: the tetrahedron, cube, octahedron, dodecahedron, and icosahedron. If you mix and match polygons, you can form another 13 shapes from regular polygons that meet the same way at every vertex—the Archimedean solids—as well as prisms (two identical polygons connected by squares) and “anti-prisms” (two identical polygons connected by equilateral triangles).
In 1966 Johnson, then at Michigan State University, found another 92 solids composed only of regular polygons, now called the Johnson solids. And with that, he exhausted all the possibilities, as the Russian mathematician Viktor Zalgaller, then at Leningrad State University, proved a few years later. It is impossible to form any other closed shapes out of regular polygons.
Yet in completing the inventory of polyhedra, Johnson noticed something odd. He discovered his shapes by building models from cardboard and rubber bands. Because there are relatively few possible polyhedra, he expected that any new ones would quickly reveal themselves. Once he started to put the sides into place, the shape should click together as a matter of necessity. But that didn’t happen. “It wasn’t always obvious, when you assembled a bunch of polygons, that what was assembled was a legitimate figure,” Johnson recalls.
They look tantalizingly open to solution, but ultimately prove impossible.
A model could appear to fit together, but “if you did some calculations, you could see that it didn’t quite stand up,” he says. On closer inspection, what had seemed like a square wasn’t quite a square, or one of the faces didn’t quite lie flat. If you trimmed the faces, they would fit together exactly, but then they’d no longer be exactly regular.
Intent on enumerating the perfect solids, Johnson didn’t give these near misses much attention. “I sort of set them aside and concentrated on the ones that were valid,” he says. But not only does this niggling near-perfection draw the interest of Kaplan and other math enthusiasts today, it is part of a large class of near-miss mathematics.
There’s no precise definition of a near miss. There can’t be. A hard and fast rule doesn’t make sense in the wobbly real world. For now, Kaplan relies on a rule of thumb when looking for new near-miss Johnson solids: “the real, mathematical error inherent in the solid is comparable to the practical error that comes from working with real-world materials and your imperfect hands.” In other words, if you succeed in building an impossible polyhedron—if it’s so close to being possible that you can fudge it—then that polyhedron is a near miss. In other parts of mathematics, a near miss is something that is close enough to surprise or fool you, a mathematical joke or prank.
Some mathematical near misses are, like near-miss Johnson solids, little more than curiosities, while others have deeper significance for mathematics and physics.
The ancient problems of squaring the circle and doubling the cube both fall under the umbrella of near misses. They look tantalizingly open to solution, but ultimately prove impossible, like a geometric figure that seems as though it must close, but can’t. Some of the compass-and-straight-edge constructions by Leonardo da Vinci and Albrecht Dürer fudged the angles, producing nearly regular pentagons rather than the real thing.
Then there’s the missing-square puzzle. In this one (above), a right triangle is cut up into four pieces. When the pieces are rearranged, a gap appears. Where’d it come from? It’s a near miss. Neither “triangle” is really a triangle. The hypotenuse is not a straight line, but has a little bend where the slope changes from 0.4 in the blue triangle to 0.375 in the red triangle. The defect is almost imperceptible, which is why the illusion is so striking.
A numerical coincidence is perhaps the most useful near miss in daily life: 27/12 is almost equal to 3/2. This near miss is the reason pianos have 12 keys in an octave and the basis for the equal-temperament system in Western music. It strikes a compromise between the two most important musical intervals: an octave (a frequency ratio of 2:1) and a fifth (a ratio of 3:2). It is numerically impossible to subdivide an octave in a way that ensures all the fifths will be perfect. But you can get very close by dividing the octave into 12 equal half-steps, seven of which give you a frequency ratio of 1.498. That’s good enough for most people.
#### A Travel Guide for the Fourth Dimension
Thank you for your interest in our all-inclusive travel package to the fourth dimension. Here are some of the most frequently asked questions we get from prospective explorers. So far, none of our clients have returned—or even sent a text—so...READ MORE
Sometimes near misses arise within the realm of mathematics, almost as if mathematics is playing a trick on itself. In the episode “Treehouse of Horror VI” of The Simpsons, mathematically inclined viewers may have noticed something surprising: the equation 178212 + 184112 = 192212. It seemed for a moment that the screenwriters had disproved Fermat’s Last Theorem, which states that an equation of the form xn + yn = zn has no integer solution when n is larger than 2. If you punch those numbers into a pocket calculator, the equation seems valid. But if you do the calculation with more precision than most hand calculators can manage, you will find that the twelfth root of the left side of the equation is 1921.999999955867 …, not 1922, and Fermat can rest in peace. It is a striking near miss—off by less than a 10-millionth.
But near misses are more than just jokes. “The ones that are the most compelling to me are the ones where they’re potentially a clue that there’s a big story,” says University of California-Riverside mathematician John Baez. That’s the case for a number sometimes called the Ramanujan constant. This number is eπ √163, which equals approximately 262,537,412,640,768,743.99999999999925—amazingly close to a whole number. A priori, there’s no reason we should expect that these three irrational numbers—e, π, and √163—should somehow combine to form a rational number, let alone a perfect integer. There’s a reason they get so close. “It’s not some coincidence we have no understanding of,” says Baez. “It’s a clue to a deep piece of mathematics.” The precise explanation is complicated, but hinges on the fact that 163 is what is called a Heegner number. Exponentials related to these numbers are nearly integers.
Or take the mathematical relationship fancifully known as “Monstrous Moonshine.” The story goes that in 1978 mathematician John McKay made an observation both completely trivial and oddly specific: 196,884 = 196,883 + 1. The first number, 196,884, had come up as a coefficient in an important polynomial called the j-invariant, and 196,883 came up in relation to an enormous mathematical object called the Monster group. Many people probably would have shrugged and moved along, but the observations intrigued some mathematicians, who decided to take a closer look. They uncovered connections between two seemingly unrelated subjects: number theory and the symmetries of the Monster group. These linkages may even have broader, as yet ungrasped, significance for other subjects. The physicist Edward Witten has argued that the Monster group may be related to quantum gravity and the deep structure of spacetime.
Mathematical near misses show the power and playfulness of the human touch in mathematics. Johnson, Kaplan, and others made their discoveries by trial and error—by exploring, like biologists trudging through the rainforest to look for new species. But with mathematics it can be easier to search systematically. For instance, Jim McNeill, a mathematical hobbyist who collects near misses on his website, and Robert Webb, a computer programmer, have developed software for creating and studying polyhedra.
Near misses live in the murky boundary between idealistic, unyielding mathematics and our indulgent, practical senses. They invert the logic of approximation. Normally the real world is an imperfect shadow of the Platonic realm. The perfection of the underlying mathematics is lost under realizable conditions. But with near misses, the real world is the perfect shadow of an imperfect realm. An approximation is “a not-right estimate of a right answer,” Kaplan says, whereas “a near-miss is an exact representation of an almost-right answer.”
In this way, near misses transform the mathematician’s and mathematical physicist’s relationship with the natural world. “I am grateful for the imperfections of the real world because it allows me to achieve a kind of quasi-perfection with objects that I know are intrinsically not perfect,” Kaplan says. “It allows me to overcome the limitations of mathematics because of the beautiful brokenness of reality.”
Evelyn Lamb is a mathematician who specializes in complex analysis. She writes extensively on math and science. She blogs for the American Mathematical Society and for her own blog, “Roots of Unity.” @evelynjlamb
Lead image credit: Comaniciu Dan / Shutterstock
## Issue 049
### The Absurd
#### Explore This Issue
• Chapter one
Nearly Perfect
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http://www.freekent.com/given-that/
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Given That
y1(t)=(t+r)^3is a known solution of the linear differential equation:
y (t+4)^2- 5y’ (t+4)+
5 points. |Given that yl (t) = (t + r)3 is a known solution ofthe linear differential equation:
(t+42y—5(t+4)y’+9y=fl t}—4 Use reduction of order to find the general solution of the equation. Math
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https://pursuit.purescript.org/packages/purescript-prelude/2.1.0/docs/Data.Bounded
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Module
# Data.Bounded
Package
purescript-prelude
Repository
purescript/purescript-prelude
### #BoundedSource
``class (Ord a) <= Bounded a where``
The `Bounded` type class represents totally ordered types that have an upper and lower boundary.
Instances should satisfy the following law in addition to the `Ord` laws:
• Bounded: `bottom <= a <= top`
#### Members
• `bottom :: a`
• `top :: a`
#### Instances
• `Bounded Unit`
• `Bounded Ordering`
• `Bounded Char`
• `Bounded Int`
• `Bounded Boolean`
## Re-exports from Data.Ord
### #OrderingSource
``data Ordering``
The `Ordering` data type represents the three possible outcomes of comparing two values:
`LT` - The first value is less than the second. `GT` - The first value is greater than the second. `EQ` - The first value is equal to the second.
• `EQ`
• `GT`
• `LT`
#### Instances
• `Show Ordering`
• `Semigroup Ordering`
• `Eq Ordering`
### #OrdSource
``class (Eq a) <= Ord a where``
The `Ord` type class represents types which support comparisons with a total order.
`Ord` instances should satisfy the laws of total orderings:
• Reflexivity: `a <= a`
• Antisymmetry: if `a <= b` and `b <= a` then `a = b`
• Transitivity: if `a <= b` and `b <= c` then `a <= c`
#### Members
• `compare :: a -> a -> Ordering`
#### Instances
• `Ord Ordering`
• `(Ord a) => Ord (Array a)`
• `Ord Void`
• `Ord Unit`
• `Ord Char`
• `Ord String`
• `Ord Number`
• `Ord Int`
• `Ord Boolean`
### #(>=)Source
Operator alias for Data.Ord.greaterThanOrEq (left-associative / precedence 4)
### #(>)Source
Operator alias for Data.Ord.greaterThan (left-associative / precedence 4)
### #(<=)Source
Operator alias for Data.Ord.lessThanOrEq (left-associative / precedence 4)
### #(<)Source
Operator alias for Data.Ord.lessThan (left-associative / precedence 4)
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https://prateekvjoshi.com/2014/07/12/elliptic-curve-cryptography-part-24-why-do-we-need-it/
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# Elliptic Curve Cryptography: Part 2/4 – Why Do We Need It?
In our previous blog post, we discussed public key cryptography and how it works in general. The RSA is so powerful because it comes with rigorous mathematical proofs of security. The authors basically proved that breaking the system is equivalent to solving a very difficult mathematical problem. When we say “difficult”, what it means is that it would take trillions of years to break it even if all the supercomputers in the world were to work in parallel. By the way, this is just one encoded message! The problem in question here is prime number factorization. It is a very well-known problem and has been studied for a very long time. So if it’s such a difficult problem, why do we need something new? What’s the problem here?
What’s the drawback?
If we were to brute force this problem, it would take a very long time. So people started working on specialized algorithms to tackle this. Algorithms like General Number Field Sieve were created to deal with this problem, and it has been successful to some extent. This is at least better than guessing all possible combinations to factorize the given number!
One of the biggest issues at hand is that the gap between the difficulty levels of multiplication and factorization is decreasing. As in, multiplying large numbers is becoming more difficult with increasing size of the numbers. Also, factoring large numbers is becoming easier as we have more powerful machines now. Since we have more resources available to decrypt the messages, the size of the keys needs to grow even faster. This means that we need to deal with bigger numbers with each passing day. This is not a sustainable situation for low-powered devices like mobile phones and tablets, because that have limited computational power. So we cannot rely on this formulation much longer. This gets us to the main point at hand. Even though RSA is fundamentally very strong, we cannot trust it to be the ideal system for the future of cryptography. We need a public key system based on a better trapdoor.
What do we need in a new system?
In order to design a new system, we need to understand what’s expected of the new system. One of the most important requirements would be that it should be mathematically strong. As in, it should have a strong mathematical base and machines shouldn’t be able to solve it easily. Ideally, we need something that will do the same job faster and something that will occupy a smaller amount of space. As the computers are getting stronger, we would need something that will not require us to constantly increase the size of the key in order to keep up with the machines. This is where elliptic curves come into picture.
Welcome to the wonderful world of elliptic curves
Once people realized the limitations of RSA, they started looking into different mathematical formulations. They basically looked for problems beyond factorization that can serve as good trapdoor functions. In 1985, mathematicians proposed crypto algorithms based on an abstruse, yet beautiful, branch of mathematics called elliptic curves.
But what exactly is an elliptic curve? How does the underlying trapdoor function work? The thing with elliptic curves is that it’s not straightforward to understand. Factorization is a simple concept to understand, because we have done it many times before. I will try to keep it as simple as possible, but things to going to get a little tricky from here on out. Let’s talk more about it in our next blog post.
————————————————————————————————-
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https://usaco.org/current/data/sol_prob3_silver_dec23.html
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(Analysis by Suhas Nagar)
Subtask 1: $T,C \leq 1000$
We can first attempt a brute force solution. Given a sequence, we can determine the number of targets that are hit by simulating Bessie's position and keeping track of hit targets in an array. We can attempt to exhaustively change every character of the sequence to another, simulate the number of targets hit, and then take the maximum across all sequences. This gives us a solution of $O(TC)$ for the first subtask.
Benjamin Qi's Python solution:
T, C = map(int, input().split())
targets = list(map(int, input().split()))
def test(cand):
avail = [0] * (2*C+1)
for t in targets:
avail[t+C] = 1
ans = 0
cur = C
for c in cand:
if c == 'L':
cur -= 1
elif c == 'R':
cur += 1
else:
ans += avail[cur]
avail[cur] = 0
return ans
commands = input()
max_ans = 0
for i in range(len(commands)):
for c in "LFR":
max_ans = max(max_ans, test(commands[:i] + c + commands[i+1:]))
print(max_ans)
Full Credit:
To improve our runtime, we make a few observations. Firstly, since we can only change a single character, Bessie's position cannot change by more than 2 from where she originally was (ie changing an L to an R). From here, we can consider an alternate solution where we calculate which targets are hit for all suffixes for displacements of $-2, -1, 0, 1$, and $2$ from Bessie's original position. If we know this information, we can iterate from left to right, try changing every character of the sequence, determine the displacement this causes for the suffix, and use our precomputation to determine the total amount of hit targets. If we naively calculate the number of targets hit for every suffix, we are duplicating work since suffixes share many common elements. However, this leads us to the optimal solution.
Let $pref[i]$ be the targets hit with the first $i$ characters of the string. Let $suff[i][k]$ be the targets hit with the last $i$ characters of the string at a displacement of $k$. We can precompute all $pref[i]$ using our simulation solution from before on the unchanged sequence. Now, we can iterate through our sequence in reverse order. Assuming no overlap between the prefix and suffix arrays, we can compute the number of targets we hit by changing the $i$-th character as $pref[i-1]+suff[i+1][k]+hit[i]$ where $k$ is the displacement of changing the current character and $hit[i]$ is $1$ if we change the current character to $F$ and there is an unhit target at our current position. Once we process the current character, we can update our suffix target count to include the current position under each displacement. This gives us a solution in $O(T+C)$ to compute the number of targets when we change each character of the sequence.
There is one caveat however. We want our suffix targets to not double count targets that have already been hit in the prefix. If we want to add a target to our suffix count but it has been hit in the prefix already, we can add it to a buffer and only move it into our suffix array when we remove that target from the prefix array.
import java.io.*;
import java.util.*;
public class TargetPractice {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int nloc = sc.nextInt();
int ncommands = sc.nextInt();
boolean[] locations = new boolean[2*ncommands+5];
for (int i = 0; i < nloc; i++){
locations[sc.nextInt()+ncommands] = true;
}
String command = sc.next();
HashMap<Integer, Integer> currentHit = new HashMap<>();
HashMap<Integer,Integer> whenHit = new HashMap<>();
int cur_pos = ncommands;
for (int i = 0; i < ncommands; i++){
if (command.charAt(i) == 'F'){
if (locations[cur_pos] && !currentHit.containsKey(cur_pos)){
currentHit.put(cur_pos, i);
whenHit.put(i,cur_pos);
}
}
cur_pos += (command.charAt(i) == 'L') ? -1 : 0;
cur_pos += (command.charAt(i) == 'R') ? 1 : 0;
}
int max = currentHit.size();
HashSet[] rightSide = new HashSet[5];
for (int i = 0; i < 5; i++) rightSide[i] = new HashSet<Integer>();
for (int i = 0; i < 5; i++) toBeAdded[i] = new HashSet<Integer>();
for (int i = ncommands-1; i >= 0; i--){
if (whenHit.containsKey(i)){
currentHit.remove(whenHit.get(i));
whenHit.remove(i);
for (int j = 0; j < 5; j++){
}
}
}
cur_pos += (command.charAt(i) == 'L') ? 1 : 0;
cur_pos += (command.charAt(i) == 'R') ? -1 : 0;
switch (command.charAt(i)){
case 'L':
// try F and add all displacement 1
int addL = locations[cur_pos] && !currentHit.containsKey(cur_pos) && !rightSide[3].contains(cur_pos)? 1 : 0;
// try changing to R
max = Math.max(max, whenHit.size()+rightSide[4].size());
break;
case 'R':
// try F and add all displacement 1
int addR = locations[cur_pos] && !currentHit.containsKey(cur_pos) && !rightSide[1].contains(cur_pos) ? 1 : 0;
// try changing to L
max = Math.max(max, whenHit.size()+rightSide[0].size());
break;
case 'F':
// Try changing to L
max = Math.max(max, whenHit.size()+rightSide[1].size());
// Try changing to R
max = Math.max(max, whenHit.size()+rightSide[3].size());
break;
}
if (command.charAt(i) == 'F') {
for (int j = cur_pos - 2; j <= cur_pos + 2; j++) {
if (j < 0 || j >= locations.length) continue;
if (locations[j]){
if (currentHit.containsKey(j)) {
}else{
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https://serverlogic3.com/which-type-of-data-structure-is-ternary-heap/
| 1,695,947,373,000,000,000
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| 544,048,509
| 15,707
|
# Which Type of Data Structure Is Ternary Heap?
//
Heather Bennett
A ternary heap is a type of data structure that falls under the category of heap data structures. A heap is a complete binary tree, where every level except the last is completely filled, and all nodes are as far left as possible. In a ternary heap, each parent node has at most three child nodes.
What Makes Ternary Heap Unique?
Unlike binary heaps, which have two child nodes per parent node, ternary heaps have three child nodes. This means that each parent node can have up to three children, and the relationship between parent and child nodes is maintained by the heap property.
The Heap Property
The heap property states that for a max-heap, every parent node must be greater than or equal to its child nodes. Similarly, in a min-heap, every parent node must be less than or equal to its child nodes. This property ensures that the maximum or minimum value is always at the root of the heap.
Operations on Ternary Heaps
Ternary heaps support various operations such as insertion, deletion, and extraction of the maximum or minimum value.
Insertion
When inserting a new element into a ternary heap, it is placed at the next available position according to the complete binary tree structure. After insertion, the heap property might be violated. To restore the heap property, we compare the newly inserted element with its parent node and swap them if necessary until the property is satisfied.
Deletion
To delete an element from a ternary heap, we first remove it from its position in the tree. Then, we replace it with either one of its children or grandchildren while maintaining the complete binary tree structure. After replacement, we compare this new element with its children and grandchildren and swap them if necessary to satisfy the heap property.
Extracting Maximum/Minimum Value
The process of extracting the maximum or minimum value from a ternary heap involves deleting the root node and replacing it with the last element in the tree. We then compare this new root with its children and grandchildren and swap them if necessary to satisfy the heap property.
Applications of Ternary Heap
Ternary heaps find applications in various algorithms and data structures. They are useful in priority queues, where elements with higher priorities are given precedence.
Time Complexity
The time complexity of basic operations on a ternary heap is as follows:
– Insertion: O(log3n)
– Deletion: O(log3n)
– Extracting Maximum/Minimum: O(log3n)
Conclusion
In conclusion, a ternary heap is a data structure that falls under the category of heap data structures. It differs from binary heaps by having three child nodes per parent node.
Ternary heaps are used for maintaining priority queues efficiently and have operations such as insertion, deletion, and extraction of maximum or minimum values. Understanding different types of heaps can help us choose the most suitable data structure for our specific needs.
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https://codeforces.com/problemset/problem/482/B
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| 204,658,147
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|
B. Interesting Array
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 11 3 3
Output
YES3 3 3
Input
3 21 3 31 3 2
Output
NO
| 391
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| 2.546875
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latest
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en
| 0.753681
|
http://bootmath.com/when-does-the-product-of-two-polynomials-xk.html
| 1,527,321,287,000,000,000
|
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| 44,949,451
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|
# When does the product of two polynomials = $x^{k}$?
Suppose $f$ and $g$ are are two polynomials with complex coefficents (i.e $f,g \in \mathbb{C}[x]$).
Let $m$ be the order of $f$ and let $n$ be the order of $g$.
Are there some general conditions where
$fg= \alpha x^{n+m}$
for some non-zero $\alpha \in \mathbb{C}$
#### Solutions Collecting From Web of "When does the product of two polynomials = $x^{k}$?"
Polynomials over $\mathbb{C}$ (in fact, over any field) are a Unique Factorization Domain (see http://en.wikipedia.org/wiki/Unique_factorization_domain); since $x$ is an irreducible, the only way for that to happen is for $f=ax^m$ and $g=bx^n$, with $ab=\alpha$.
(If you don’t want to bring in the sledgehammer of unique factorization, you can just do it explicitly: look at the lowest nonzero term in $f$ and the lowest nonzero term in $g$; their product will be the lowest nonzero term in $fg$, hence must be of degree $m+n$. Since the degree of the lowest nonzero term of $f$ is at most $m$ and the one of $g$ is at most $n$, you have that they must be exactly of degree $m$ and $n$, respectively, and you get the result)
It’s true over any domain $\rm D$ since then $x$ is prime (via $\rm D[x]/x = D$ a domain), and products of primes always factor uniquely in domains. It fails over non-domains, e.g. $\rm x = (2x+3)(3x+2) \in \mathbb Z/6[x].$
The answer just occured to me. The roots of $f$ and $g$ must be be at 0.
Yes, the intuitively evident ones: all other terms in f and g must vanish. To see this, note that the product of the constant terms of f and g equals the constant term of fg, which is zero, whence at least one of these polynomials is multiple of x. Without any loss of generality assume it is f. Then
fg = x (f/x) g,
implying (f/x) g is a multiple of x^(n+m-1). By induction this reduces us to the case n+m=0, which is trivial (because f and g then have no other terms). QED.
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Assignment_3
# Assignment_3 - MAE157 Lightweight structures(Winter 09...
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1 MAE157 – Lightweight structures (Winter 09) Homework Assignment #3 Due in the drop box by 1pm on Friday, February 6, 2009 Solve the four following problems. For each problem, provide a detailed explanation of all the steps involved. Problem 1 An Aluminum cantilever beam with a T cross section is loaded as shown in the sketch below. What is the minimum thickness t (assumed constant along z – i.e. the beam is not tapered) that ensures that the beam doesn’t fail? Assume that failure occurs when the maximum stress reaches the yield strength of the material (300 MPa). L = 1 m ; P = 1000 N ; Q = 2000 N ; a = 5 cm Problem 2 Verify the accuracy of your answer in problem 1 using ABAQUS. (a) Set up the problem in ABAQUS using the value of t you obtained from problem 1 and assuming that Aluminum is an elastic material, with E=70GPa and ! = 0.3 (b) Run the analysis (c) Locate the maximum stress and verify that is close to the yield strength of the material Turn in the input file that is generated when you submit the analysis (filename.inp) as well as a plot of the stress on the top and bottom fibers along the length of the beam (you can use excel or any other program to generate the plot).
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• Winter '08
• VALDEVIT
• Force, ABAQUS
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