url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://atcoder.jp/contests/abc217/tasks/abc217_c | 1,720,918,486,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00781.warc.gz | 96,002,066 | 6,220 | Contest Duration: - (local time) (100 minutes) Back to Home
C - Inverse of Permutation /
Time Limit: 2 sec / Memory Limit: 1024 MB
### 問題文
1,2,\dots,N1 回ずつ現れる長さ N の数列を「長さ N の順列」と呼びます。
• 全ての i (1 \leq i \leq N) に対して Qp_i 番目の要素が i である。
ただし、条件を満たす Q は必ずただ 1 つ存在することが証明できます。
### 制約
• 1 \leq N \leq 2 \times 10^5
• (p_1,p_2,\dots,p_N) は長さ N の順列である。
• 入力は全て整数である。
### 入力
N
p_1 p_2 \dots p_N
### 出力
q_1 q_2 \dots q_N
### 入力例 1
3
2 3 1
### 出力例 1
3 1 2
• i = 1 のとき p_i = 2, q_2 = 1
• i = 2 のとき p_i = 3, q_3 = 2
• i = 3 のとき p_i = 1, q_1 = 3
### 入力例 2
3
1 2 3
### 出力例 2
1 2 3
### 入力例 3
5
5 3 2 4 1
### 出力例 3
5 3 2 4 1
Score : 300 points
### Problem Statement
We will call a sequence of length N where each of 1,2,\dots,N occurs once as a permutation of length N.
Given a permutation of length N, P = (p_1, p_2,\dots,p_N), print a permutation of length N, Q = (q_1,\dots,q_N), that satisfies the following condition.
• For every i (1 \leq i \leq N), the p_i-th element of Q is i.
It can be proved that there exists a unique Q that satisfies the condition.
### Constraints
• 1 \leq N \leq 2 \times 10^5
• (p_1,p_2,\dots,p_N) is a permutation of length N (defined in Problem Statement).
• All values in input are integers.
### Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \dots p_N
### Output
Print the sequence Q in one line, with spaces in between.
q_1 q_2 \dots q_N
### Sample Input 1
3
2 3 1
### Sample Output 1
3 1 2
The permutation Q=(3,1,2) satisfies the condition, as follows.
• For i = 1, we have p_i = 2, q_2 = 1.
• For i = 2, we have p_i = 3, q_3 = 2.
• For i = 3, we have p_i = 1, q_1 = 3.
### Sample Input 2
3
1 2 3
### Sample Output 2
1 2 3
If p_i = i for every i (1 \leq i \leq N), we will have P = Q.
### Sample Input 3
5
5 3 2 4 1
### Sample Output 3
5 3 2 4 1 | 799 | 1,861 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-30 | latest | en | 0.358744 |
https://math.stackexchange.com/questions/574173/open-subsets-of-the-closure | 1,563,636,939,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526536.46/warc/CC-MAIN-20190720153215-20190720175215-00157.warc.gz | 468,317,057 | 36,560 | # Open subsets of the closure
I want to prove that every open subset of a topological subpace is an open subset of its closure.
Let $Y$ be a topological space and $X$ a subspace of $Y$. If $U$ is an open subset of $X$, we have $U=U\cap \overline X$, thus U is an open subset of $\overline X$ also.
Am I right?
• I don't understand : Consider $X=\Bbb Q\subset \Bbb R=Y$. Then $U=\Bbb Q_{+}^*$, the positive rational numbers, is open in $X$, and not in $Y$... – Olivier Bégassat Nov 20 '13 at 4:25
• Make sure you pay close attention to which set you're taking the closure of. I agree with Brian that the theorem is referring to the closure of $U$ in $X$. You would need to rewrite your proof to show that $U = U \cap \overline{U}$ (where $\overline{U}$ is the closure of $U$ in $X$) so $U$ is open in $\overline{U}$. – manthanomen Nov 20 '13 at 5:15
Yes, this is correct. More generally, if $U$ is an open subset of $X$, and $A$ is any subset of $X$ containing $U$, then $U\cap A=U$ is an open subset of $A$.
Added: By ‘this’ I meant the assertion that an open set is open in its closure; I was called away and didn’t look closely enough at the argument, which seems to go with the different (and false) proposition that if $U$ is an open subset of $X$, where $X\subseteq Y$, then $U$ is open in $\operatorname{cl}_YX$. A counterexample to this is to take $Y=\Bbb R$, $X=\Bbb Q$, and $U=\Bbb Q\cap(0,1)$. Then $U$ is relatively open in $\Bbb Q$, but $U$ is not an open set in $\operatorname{cl}_{\Bbb R}\Bbb Q=\Bbb R$.
• Does this apply to OP's question? Although $U$ is an open subset of $X$ and $U$ is contained in $\overline{X}$, $\overline{X}$ is only a subset of $X$ if $X$ is closed, so we can't conclude that $U$ is open in $\overline{X}$. – manthanomen Nov 20 '13 at 4:48
• @manthanomen: Yes, it applies. The theorem in the original question is that if $U$ is open in $X$, then $U$ is open in $\operatorname{cl}_XU$; I’ve simply replaced $\operatorname{cl}_XU$ by an arbitrary subset of $X$ containing $U$. – Brian M. Scott Nov 20 '13 at 4:52
• The statement's language is ambiguous, but from the proof attempt, I think the theorem refers to the closure of $X$ in $Y$, not the closure of $U$ in $X$. – manthanomen Nov 20 '13 at 5:01
• @manthanomen: I think it likelier that the proof attempt is based on a misunderstanding of the theorem. – Brian M. Scott Nov 20 '13 at 5:03
• Yes, that's likelier, I misunderstood and thought you were saying that OP's proof attempt was correct. – manthanomen Nov 20 '13 at 5:12
No, this is incorrect. An open subset $O$ of $X$ is an intersection $X\cap O'$ with $O'$ an open subset of $Y$. In general $O$ will not, in general, be open in $\overline{X}$, for there is no reason that there exists an open subset $O'\subset Y$ with both $$\begin{cases} O=X\cap O' \\ \qquad \text{and}\\ O=\overline{X}\cap O' \end{cases}$$
For instance, consider $X=\Bbb Q\subset \Bbb R=Y$. Then $O=\Bbb Q_{+}^*=(0,\infty)\cap\Bbb Q$, the positive rational numbers, is open in $X$, and not in $Y$. | 979 | 3,028 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-30 | latest | en | 0.899443 |
https://www.alignmentforum.org/posts/pHPmMGEMYefk9jLeh/llm-basics-embedding-spaces-transformer-token-vectors-are | 1,726,372,108,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651614.9/warc/CC-MAIN-20240915020916-20240915050916-00844.warc.gz | 604,068,694 | 124,218 | # 23
This post is written as an explanation of a misconception I had with transformer embedding when I was getting started. Thanks to Stephen Fowler for the discussion last August that made me realise the misconception, and others for helping me refine my explanation. Any mistakes are my own. Thanks to feedback by Stephen Fowler and JustisMills on this post.
TL;DR: While the token vectors are stored as n-dimensional vectors, thinking of them as points in vector space can be quite misleading. It is better to think of them as directions on a hypersphere, with a size component.
The I think of distance as the Euclidean distance, with formula:
Thus does not match up with the distance forumla used when calculating logits:
But it does match up with the cosine similarity forumula:
And so, we can see that the direction and size matter, but not the distance
# Introduction
In the study of transformers, it is often assumed that different tokens are embedded as points in a multi-dimensional space. While this concept is partially true, the space in which these tokens are embedded is not a traditional Euclidean space. This is because of the way probabilities of tokens are calculated, as well as how the behaviour of the softmax function affects how tokens are positioned in their space.
This post will have two parts. In the first part, I will briefly explain the relevant parts of the transformer, and in the second part, we will explore what is happening when a transformer moves from an input token to an output token explaining why tokens are better thought of as directions.
# Part 1: The Process of a Transformer
Here I will briefly describe how the relevant parts of the transformer work.
First, let's briefly explain the relevant parts at the start of the transformer. We will be studying the "causal" transformer model (ie: that given tokens, we want to predict the ()th token). The main "pieces" of a causal model are:
• The Tokeniser - turns "words" into "tokens" or "tokens" into "words"
• The Transformer - turn N tokens into a prediction for the (N+1)th token
• Input Embedding - turns "tokens" into "vectors"
• Many Decoder layers - turns input vectors into prediction vectors
• with a Self-Attention sub-layer - uses information from all states
• with a Feed Forward sub-layer - uses information from the current state
• Output Unembedding - turns prediction vectors into token probabilities
Note that I won't go into much depth about the positional and decoder layers in this post. If there is interest, I may write up another post explaining how they work if there is interest.
Also note, that for simplicity, I will initially assume that the input embedding and the output unembedding are the same. In this case, if you have a token, embed it, and then unembed it, you should get the same token out. The symmetry was true in the era of GPT-2, but nowadays, embedding and unembedding matrices are learned separately, so I will touch on some differences in the end.
Lastly, note that "unembedding" is actually a fake word, and usually it is just called the output embedding. I think unembedding makes more sense, so I will call it unembedding.
## 1. The Start of the Transformer Process
In the beginning of the transformer process, there is the Tokeniser (converting “words” into “tokens”) and the Token Embedding (converting “tokens” into “token vectors”/”hidden-state vectors”):
So to start:
2. The text is split into different “parts” called token ids
3. The token ids are converted into vectors using the embedding matrix
4. The vectors are passed on into the rest of the transformer
### 1.1 The Tokeniser
The tokeniser is a separate piece of language model training which is used to try to make language more machine readable. It is usually trained once, separately to training the model. I will explain some intuition here, but this hugging face article explains tokenisers in more detail, if you are interested.
When we read a word like “lighthouse”, we don’t think if it in terms of the individual letters “l” “i” “g” “h” “t” “h” “o” “u” “s” “e”, but think of it as a whole word or concept. A subword tokeniser tries to compress the letters into meaningful words (eg: “lighthouse”) or parts of words (eg: “light” + “house”) using magical statistics to try to compress it as much as possible given a fixed limit for number of tokens.
The tokeniser, as usually defined, needs to be able to be able to compress all of english language into a "token" vocabulary of approx 50000. (usually up to 50400 so that it can run on TPUs or something, but I will just write 50000 since the exact number varies). The tokeniser does this compression by constructing a one-to-one map of token strings (combinations of characters) to token IDs (a number).
One common restriction, is that token strings are case-sensitive, so if you want to encode a word with the first letter capitalised, or written fully capitalised, these need to be separate tokens.
Another restriction, is that we usually want to be able to return the tokens back into text, and so need to be able to handle spaces and punctuation. For example, while we might be able to converts strings like "the cat" into"the" + "cat", it is not certain how one should handle spaces and punctuation when converting it back from "the" + "cat" into "the cat".
We could solve the spaces problem by adding a "space" token between each word, but this would make the input approx 2x longer, so is quite inefficient. We could always add a space between tokens, but this only works if all words are made up of one token, and are always separated by spaced. The solution in GPT-2, is to instead make the space a part of most tokens, by having a space padded before the word. That is, instead of "cat" we would write " cat".
Here is some example of the variations that arise for a common word like "next" due to the restrictions with case-sensitivity, as well as restrictions from padding a space:
We could also have tokens for punctuation like comma (“,”) or full stop (“.”) or spaces (" ") for situations where they are needed, and in general, everything on the internet written in english should be able to be tokenised (though some things less efficiently than others).
Then, what the tokeniser does, is look at a string of text, split it into token strings, and turn that into a list of token IDs. This ie essentially a lookup table/dictionary: For example:
• Start with “ The cat went up the stairs.”
• Separate out [ “ The”, “ cat”, “ went”, “ up”, “ the”, “ stairs”, “.” ]
• Store as “Token IDs”: tensor([[ 20, 4758, 439, 62, 5, 16745, 4]]
For less common tokens, the tokeniser usually has to split up the word into multiple parts that might not make as much sense from just looking at it. A somewhat common thing to happen, is have a "space+first letter" token, then a "rest-of-the-word" token. For example. " Pickle" might become " P" + "ickle", but there are also other patterns.
While I gave some common restrictions above, you can have different restrictions, such as "make it so numbers are only tokenised by single digits". In addition, there are usually some "special tokens", such as "<|endoftext|>"x time. A useful way to get a better idea of tokenisation is the Redwood Research next-token prediction game.
### 1.2 The Input Embedding Matrix
The transformer, such as GPT-2 or Meta OPT, then uses an embedding matrix (WE) to convert its Tokens IDs into Token Vectors/Hidden State Vectors. For the smallest transformer models, the hidden state vector size for is usually 768, and for larger models goes up to 4096. I will pretend that we are using a nice small model (mostly OPT-125m).
The size of the embedding matrix (WE) is (50000 * 768) for small models, where there are 50000 tokens and 768 dimensions. The unembedding matrix, which in our case computes the left inverse of the embedding matrix , is (768 * 50000) in size.
One way to think of the embedding matrix, is as a list of 50000 token vectors, one corresponding to each token, and then we use the token ID to take the right token vector. Here is a diagram showing the embedding of token IDs:
The mathematical way we represent this conversion, is that we create a "token ID" vector that is the size of the token vocabulary, and make it 1.0 at the desired index, and 0 at all other indices. This makes it so that when we multiply the embedding matrix by this "token ID" vector, we get the vector for our token. This is just a different way choosing the vector from a list of vectors. The token ID is called a “hot-encoding” of the token ID vector, and corresponds to the index of the column in which the embedded token vector is stored. These extra steps are not important in the encoding, and are not physically done by machine (since it is essentially adding a lot of zeros), but are conceptually important for the decoding.
Here is an example for the token " the" (which in the case of OPT-125m has token ID 5), which gives the same result as before:
Here is an example of token embedding for the token " NEXT" (which has token ID 10000)
Now, with a symmetric unembedding, if we wanted to get back the original token, we could then look to see which of the rows in the unembedding the vector is most similar to. We can look at what happens when we turn the token ID into its vector of zeros and ones, then embed it into a hidden state vector, and unembed that hidden state vector again straight away.
The input token ID vector is a nice clean vector with all zeros, except for at the Token ID we are encoding (in this case, token 10000, which is “ NEXT”). The output from the left inverse gives us the same value 1.0 at the index 10000, but due to the fact that we crammed 50000 tokens into a vector space of 768, we see some noisiness when trying to restore the original token, and see non-zero values at all the other Token ID Indices. The graph below shows what the result of an inverse unembedding might look like:
Most of the other high peaks in the diagram correspond to variations on the word "next" in different formats. This makes sense, since we expect them to be in a similar direction in vector space:
Note that in reality, instead of the inverse, they usually just use the transpose, and the pattern looks similar but slightly different, and we don't get a nice peak of 1.0 at the value of the token we put in, but it is conceptually the same.
In some way, it is possible to think of space as having virtual points corresponding to each vector, and our real point vector starts off at one point corresponding to the vector that was input, and then after moving through the space, it finds a new point, which is the predicted output: the next token vector, decided by what the virtual tokens are. This is a useful picture, but lacks nuance, and it is better to think of the points more as “directions” in space instead of “points in space”, but I will describe the differnces in more detail after briefly describing the rest of the model.
### 1.3 Positional Encoding
In reality, I should also include positional encoding here, but it is not important for the discussion in this post, so I will skip it and bundle it into the "middle of the transformer process".
## 2. The Middle of the Transformer Process
We will not care too much about what happens in the middle of the transformer. We will just assume that it magically knows how to get to the output token based on the previous inputs. Here is a diagram of a transformer decoder for reference:
Some summary of steps that do happen in a typical next-token-prediction model:
• The Hidden State Vectors get a “positional encoding” to keep track of the positions of the tokens (since the order of words and symbols in a text matters in English)
• The “Transformer Model” had many decoder layers which gradually make changes to the Hidden State Vectors, which should slowly become more “close” to the next predicted token
• The Self-Attention layers look at hidden states for all tokens and make adjustments to the current token based on that information
• The Feed-Forward Multi-Layer-Perceptron (MLP) Layers only look at the current hidden state and make adjustments based on that information
One key bit of information, is that in a typical model (though not necessarily all models), the hidden state vectors are normalised before being input to the Self-Attention, and the MLP. The layer norm means that all the transformer sees is the direction of the vectors.
## 3. The End of the Transformer Process
With the description of the parts before, we can now look at what happens at the output of the transformer model.
Remember that we fed in a list of tokens into the transformer, and now we are trying to predict the th token.
The steps for next-token prediction in a transformer is then as follows:
1. In the output, we look at the final hidden state vector of the th token
2. The vector is multiplied by the unembedding matrix to get "logits," (unnormalised log probabilities), for each possible token
3. The logits are softmax-ed to get a normalised probability distribution over the possible tokens
4. A token is chosen based on the probabilities
5. The token is converted back into text using the tokeniser dictionary
If we are running in a practical setting like ChatGPT, then we can then add the new ()th token to the input and run the model again to predict the ()th token, and this way predict a lot of extra text.
Side Note:
Since the model only looks at previous tokens (ie: not future tokens) when calculating the hidden states, it is possible to save the outputs from before and use a lot less compute than if we were running the whole model.
In addition, in training, what actually happens is that each of the [1, 2, .., N] input tokens give N output hidden states, and should predict tokens [2, 3, ..., N+1] respectively. This means that in one run, you make N predictions instead of 1, which is why transformers are much easier to train than old recursive language models and scale much better.
### 3.1 Last hidden state vectors
As described before, for each token, we input a token hidden state vector, and get an output of a hidden state vector for that token. The optimiser is making it such that in the end, the model accurately predicts the next token.
### 3.2 The Output Unembed Matrix
We take the last hidden state vector, and multiply it by the unembed matrix. For symmetric models, it is simply given as the transpose:
If you wanted it to actually have the effect of being the inverse, and the tokens were not normalised, you would need to define it like below.
In practice, in symmetric models, they neither ensure normalisation, nor do they calculate the inverse like I have just shown. In more modern models, the unembed matrix is actually stored as a separate set of weights.
Then, the output is given by:
So what this formula means, is that the logit of each Token ID is just the dot product between the row of the unembed matrix () and the last hidden state vector (). So this formula for logits is how the "closeness to the output of a vector" is measured.
### 3.3 Softmax
Then, to get the actual probability of finding a token, we take the softmax, given by:
Where here is the "temperature". For temperature 0, this is usually defined to be argmax (that is, the largest value is 1, and all others are 0), and for , this becomes a uniform distribution.
The best token then has a probability that can range from to 1.0. Here is an example of the "best token" probability with a range of temperatures:
# Part 2: Misconception and Better Understanding
## "Points in Space" Explanation
Before thinking about it too much, I thought that the transformer:
• Encodes the initial vector that is the most recent word
• Pushes the hidden state vector towards the predicted word
• Chooses the token that is "closest" to that hidden state vector.
The process is true, but the word "closest" is misleading. In my mind, the word "closest" makes me think in terms of Euclidean distances. However, thinking in terms of Euclidean distances is is not an accurate way to think. In reality, the direction and size of the output vector matter more than its "position". When extracting the logits, there is no Euclidean distance used, but instead, just unembedding with a matrix, as described above.
I was implicitly thinking that "distance" meant Euclidean distance, even though I knew there was no use of the euclidean distance formula:
Here is a flawed diagram of how I was thinking:
## "Directions in Space" Explanation
While the tokens are in fact encoded as vectors in a matrix, and can be drawn as points in n-dimensional space, it is much better to think only of the directions of the tokens, and not the actual points. So it is much better to think of them as points on an (n-1)-dimensional hypersphere with a size component. The real distance function is actually:
Which is a lot more similar to cosine similarity, which cares only about direction:
Here is a slightly less flawed diagram:
In addition, most tokens are roughly the same size, so it can be easier to think of the vectors as all roughly having the same size. We see below a diagram for the distribution of token vector sizes in the matrix for Meta's OPT-125m model:
We do see that there is a tail of tokens for which the tokens are quite different in size compared to the normal ones. In the case of OPT-125m, the unused tokens are very large, in other cases, the tokens can be very small. For an exploration on these weird tokens, I recommend "SolidGoldMagikarp (plus, prompt generation)" and "SolidGoldMagikarp II: technical details and more recent findings"
## The Effect of Vector Length
While the direction explains what the "closest tokens" are, how does the size affect the outcomes? As we know, the softmax gives a normalised distribution of probabilities, adding to a total of 1. The effect that having a larger vector is essentially the same as multiplying the logits by a multiple. The size difference makes it so that the logits are more spread out, and so, the outcome is that the model is "more confident" in a particular outcome.
Here is a diagram of how size makes a difference:
In the limit of tokens being zero in size, the outcome is not pointing in any direction, so the probabilities of all tokens will be the same. With larger vectors, the model gets "more confident" in a particular outcome, as it is more able to distinguish the possible token directions. In essence, larger tokens => more "confidence".
In addition, when the tokens are not quite the same length, we can sort of account for the size discrepancy by thinking of the directions on the hypersphere being a larger target. In theory, it is possible for some of the target tokens directions to overshadow other possible token directions. This seems to be happening in the "SolidGoldMagikarp (plus, prompt generation)" post I mentioned before.
The size only showing the difference in certainty is why it seems like the size of token vectors seems to grow to large vector sizes in the models I have checked. In a euclidean way of thinking, the growing sizes wouldn't make sense, but with a direction-based thinking, it makes perfect sense. The plot below shows and example of vector norm size across layers for a few examples in OPT-125m
So our final explanation for how the tokens change through space might look like the diagram below (somewhat cluttered due to excessive details):
As a real example, we can look at a real model (OPT-125m) and look at the next-token prediction of:
"John and Mary went to the bar. John bought a drink for" -> " Mary"
(Example inspired by "Some Lessons Learned from Studying Indirect Object Identification in GPT-2 small). We see the same plot as before, but now in a real model.
The x and y coordinates are chosen to form a plane such that we can see the two tokens we are interested in. We see that there is some wandering in other dimensions for most of the start of the run initially (since I am only plotting 2 for the 768 dimensions), unit it finds out what it is supposed to be doing, and goes in the direction of the " Mary" token.
Do note though that it is difficult to draw a 768 dimensional space, so while the 2d image might give some insights, do not take it too seriously.
# Non-symmetric Embeddings
While before I had been talking only about symmetric embeddings, in most modern models, symmetric embeddings are no longer used. Instead, the input and output embeddings are trained separately. The reasoning initially might have been that "it would make sense for the tokens to have a consistent token space", but there are advantages to having a separate learned embedding.
One main thing, is that the requirement we have is not actually having a consistent token space, and being able to infer the input from the output. Instead, the task is next token prediction. Because of this, it makes sense for the output direction for likely next possible tokens to be in the same direction as the input direction. For example, in most contexts, a next word is not likely to be the previous word again, depending on the word, you can already do some bigram statistics on what the following word is.
For example: " what" -> " the" or " might" -> " be", or " How", " are".
This separation of embedding spaces might be particularly useful for when token are split up into multiple parts, for example, something like " Susp" -> "icious".
The key thing, is that the next predicted token is usually not the same as the input token, so it can be useful to have a separation.
The separation does mean, however, that when you are reasoning about what the vector directions are like in the middle of a transformer, that the hypersphere of directions may not correspond neatly to the input embedding space, or the output unembedding space, but might instead be something in between.
Previous research looking through the "logit lens" has mostly been looking at how the adjustments in the residual stream, and the activations in the residual stream, look from the point of view of the unembed matrix, so looking at the output logits seems more likely to be the more natural way of looking at things.
# Conclusion
While the tokens in transformers are stored as n-dimensional vectors, we should care not about the actual positions of the vectors in n-dimensional space, but instead, the directions of the vectors.
We can imagine the token vectors as directions on a (n-1)-hypersphere, with lengths showing how "large" the targets are on a hypersphere. The direction of the output decides what the next token should be, and the length decides the confidence.
1. ^
Feel free to look at how I generated some of the graphs in this python notebook.
New Comment | 5,052 | 22,887 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-38 | latest | en | 0.939637 |
https://philoid.com/question/34390-solve-each-of-the-following-quadratic-equations | 1,723,260,583,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00750.warc.gz | 340,971,664 | 8,326 | ##### Solve each of the following quadratic equations:
Given:
taking LCM
using (a - b)2 = a2 + b2 - 2ab
cross multiplying
18x2 - 48x + 130 = 105x - 140
18x2 - 153x + 270 = 0 taking 9 common
2x2 - 17x + 30 = 0
Using the splitting middle term - the middle term of the general equation is divided in two such values that:
Product = a.c
For the given equation a = 2 b = - 17 c = 30
= 2.30 = 60
And either of their sum or difference = b
= - 17
Thus the two terms are - 12 and - 5
Sum = - 12 - 5 = - 17
Product = - 12. - 5 = 60
2x2 - 17x + 30 = 0
2x2 - 12x - 5x + 30 = 0
2x(x - 6) - 5(x - 6) = 0
(x - 6) (2x - 5) = 0
(x - 6) = 0 or (2x - 5) = 0
Hence the roots of equation are
58 | 290 | 696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-33 | latest | en | 0.817792 |
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Topic: Random numbers
Replies: 64 Last Post: Dec 24, 2007 1:04 PM
Messages: [ Previous | Next ]
Marshall Posts: 1,936 Registered: 8/9/06
Re: Random numbers
Posted: Dec 21, 2007 2:53 PM
On Dec 21, 7:24 am, Pubkeybreaker <pubkeybrea...@aol.com> wrote:
> On Dec 21, 6:16 am, John <iamach...@gmail.com> wrote:
>
> > Given a function that returns a random number between 1-5, write one
> > that returns a random number between 1-7 for the case when it should
> > be integer and for the case it can be real.
>
> claims otherwise does not understand the question.
>
> Why is it impossible? Because you have failed to specify
> a *distribution* for either your original function or your
> new function. If you mean *uniformly* at random, you have failed
> to state it. What if "at random" meant "drawn at random from a
> truncated normal pdf"???
The OP also failed to state any constraints on running time
or memory consumption. Do such omissions make
a solution impossible as well?
That he didn't specify a distribution doesn't make a solution
impossible. It just means that, *if* the the OP has a specific
distribution in mind, then *since* he didn't state what it is, then
any possible solution *may not* have the distribution the OP
had in mind. By the same token, it very well might, also.
Of course, I suppose your real point was that it is extremely
unlikely that any real world application will be indifferent
to distribution, and so the question as asked is pretty crappy.
To which I would agree.
Marshall
Date Subject Author
12/21/07 Champ
12/21/07 quasi
12/21/07 quasi
12/21/07 Phil Carmody
12/21/07 quasi
12/21/07 quasi
12/21/07 Phil Carmody
12/21/07 quasi
12/21/07 Phil Carmody
12/21/07 Phil Carmody
12/21/07 quasi
12/21/07 quasi
12/21/07 Phil Carmody
12/21/07 quasi
12/21/07 Phil Carmody
12/21/07 Phil Carmody
12/21/07 quasi
12/21/07 Phil Carmody
12/21/07 quasi
12/21/07 Marshall
12/21/07 Phil Carmody
12/21/07 quasi
12/21/07 Phil Carmody
12/21/07 Marshall
12/21/07 briggs@encompasserve.org
12/21/07 William Elliot
12/21/07 quasi
12/22/07 William Elliot
12/21/07 Pubkeybreaker
12/21/07 b92057@yahoo.com
12/22/07 quasi
12/21/07 simple.popeye@gmail.com
12/21/07 simple.popeye@gmail.com
12/22/07 quasi
12/22/07 Gib Bogle
12/22/07 quasi
12/21/07 Marshall
12/22/07 simple.popeye@gmail.com
12/22/07 quasi
12/22/07 simple.popeye@gmail.com
12/22/07 quasi
12/22/07 quasi
12/22/07 quasi
12/22/07 simple.popeye@gmail.com
12/22/07 quasi
12/23/07 simple.popeye@gmail.com
12/23/07 simple.popeye@gmail.com
12/23/07 simple.popeye@gmail.com
12/23/07 simple.popeye@gmail.com
12/23/07 simple.popeye@gmail.com
12/22/07 simple.popeye@gmail.com
12/22/07 Herman Rubin
12/22/07 b92057@yahoo.com
12/22/07 quasi
12/23/07 b92057@yahoo.com
12/23/07 quasi
12/23/07 b92057@yahoo.com
12/24/07 quasi
12/24/07 quasi | 970 | 3,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-17 | longest | en | 0.89282 |
https://www.albert.io/learn/sat-chemistry-subject-test/question/calculate-grams-of-a-gas | 1,490,442,174,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188924.7/warc/CC-MAIN-20170322212948-00017-ip-10-233-31-227.ec2.internal.warc.gz | 880,948,613 | 98,883 | Limited access
How many grams of a gas are present in a 2.2 L sample of gas if the gas has a density of 1.8 g/L?
A
$1.22 \ g$
B
$4.00 \ g$
C
$3.96 \ g$
D
$0.02 \ g$
E
$0.82 \ g$
Select an assignment template | 85 | 219 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-13 | latest | en | 0.87791 |
https://www.physicsforums.com/threads/3-dimensional-center-of-mass-problem.200348/ | 1,480,927,871,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541556.70/warc/CC-MAIN-20161202170901-00198-ip-10-31-129-80.ec2.internal.warc.gz | 1,000,547,946 | 14,135 | # 3-dimensional center of mass problem
1. Nov 25, 2007
### brendan3eb
1. The problem statement, all variables and given/known data
In the ammonia (NH3) molecule, the three hydrogen (H) atoms form an equilateral triangle; the center of the triangle is 9.40x10^(-11) m from each hydrogen atom. The nitrogen (N) atom is at the apex of a pyramid, with the three hydrogen atoms forming the base. The nitrogen-to-hydrogen atomic mass ratio is 13.9, and the nitrogen-to-hydrogen distance is 10.4x10^(-11) m. Locate the center of the molecule relative to the nitrogen atom.
2. Relevant equations
Rcom=(1/M)(sum of MiRi from i=1 to n)
3. The attempt at a solution
I have read through the problem several times, and I started to try and solve the problem by just figuring out the three-dimensional coordinates, but it seems a bit odd for the book to give such a tedious problem. Is there some sort of symmetry or way of re-drawing the problem to eliminate the z-axis that I am not picking up on, or do I really need to use all three axis to figure out the answer?
2. Nov 25, 2007
### BlackWyvern
I think you're allowed to treat the 3 H atoms as one mass focused in the centre of the 3 H atoms. This would have a mass of triple one H atom, and it would be directly under the N atom. I think. | 338 | 1,288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2016-50 | longest | en | 0.931872 |
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# Crouching Kellogg, Hidden Wharton (article)
Author Message
TAGS:
### Hide Tags
SVP
Joined: 04 Dec 2007
Posts: 1689
Schools: Kellogg '11
Followers: 14
Kudos [?]: 198 [0], given: 31
Crouching Kellogg, Hidden Wharton (article) [#permalink]
### Show Tags
04 Apr 2009, 14:03
A few years old, but interesting read.
Quote:
Chris Schulze-Melander and Maya Lyttleton, who were married in December, are both in the first year of an MBA program-at different schools. Maya is at The Kellogg School; Chris is at Wharton. Both are from the U.K....The Kellogg Merger and the Wharton Journal spoke with Chris and Maya in order to get at what, exactly, are the differences between these business schools, seen by people who feel strongly about both schools - and each other.
Quote:
Current Student
Joined: 21 Aug 2008
Posts: 348
Schools: Fuqua '11
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Re: Crouching Kellogg, Hidden Wharton (article) [#permalink]
### Show Tags
06 Apr 2009, 08:37
Good find!
Manager
Joined: 09 Nov 2008
Posts: 148
Followers: 1
Kudos [?]: 11 [0], given: 3
Re: Crouching Kellogg, Hidden Wharton (article) [#permalink]
### Show Tags
06 Apr 2009, 20:05
Whartoners get Fridays off? I didn't know that. Wharton stock just rose a bit.
Re: Crouching Kellogg, Hidden Wharton (article) [#permalink] 06 Apr 2009, 20:05
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Display posts from previous: Sort by | 708 | 2,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-13 | longest | en | 0.882867 |
https://faculty.ksu.edu.sa/ar/shaaldossari/course/350415 | 1,695,575,753,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506658.2/warc/CC-MAIN-20230924155422-20230924185422-00713.warc.gz | 285,732,215 | 6,807 | # Math-106 Integral Calculus
Text:Calculus by Swokowski, Olinick, Pence (Sixth Edition).
Coures Contents: Sections,Topics:
4.1. Antiderivatives and indefinite integrals.
4.2. Change of variables in indefinite integrals.
4.3. Summation notation and area.
4.4. The definite integral.
4.5. Properties of definite integral.
4.6. The fundamental theorem of calculus.
4.7. Numerical integration.
6.2 The natural logarithm function.
6.3.The exponential function.
6.4. Integration using natural logarithm and exponential function.
6.5. General exponential function and logarithm function.
6.7.Inverse trigonometric functions.
6.8.Hyperbolic and inverse hyperbolic functions.
6.9. Indeterminate forms and L’Hopital’s rule
7.1. Integration by parts.
7.2.Trigonometric integrals.
7.3. Trigonometric substitutions.
7.4. Integrals of rational functions (Partial fractions).
7.5. Quadratic expressions and miscellaneous substitutions.
7.7. Improper integrals
5.1. Area between curves.
5.2. Volume (by disk or washer method).
5.3.Volume (by Cylindrical shells method).
5.5. Arc length and surface of revolution.
9.1. Parametric equations.
9.2 .Arc length and surface area.
9.3.Polar coordinates.
9.4.Integrals in polar coordinates.
المرفق الحجم
syllabus_math_106_.pdf 70.57 كيلوبايت
ملحقات المادة الدراسية | 353 | 1,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-40 | latest | en | 0.568043 |
https://www.australiancurriculum.edu.au/f-10-curriculum/mathematics/?year=11754&strand=Measurement+and+Geometry&capability=ignore&capability=Literacy&capability=Numeracy&capability=Information+and+Communication+Technology+%28ICT%29+Capability&capability=Critical+and+Creative+Thinking&capability=Personal+and+Social+Capability&capability=Ethical+Understanding&capability=Intercultural+Understanding&priority=ignore&priority=Aboriginal+and+Torres+Strait+Islander+Histories+and+Cultures&priority=Asia+and+Australia%E2%80%99s+Engagement+with+Asia&priority=Sustainability&elaborations=true&elaborations=false&scotterms=false&isFirstPageLoad=false | 1,619,089,285,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039603582.93/warc/CC-MAIN-20210422100106-20210422130106-00630.warc.gz | 731,456,698 | 14,907 | # Mathematics
Please select at least one year level to view the content
Please select at least one Strand to view the content
### Rationale
Learning mathematics creates opportunities for and enriches the lives of all Australians. The Australian Curriculum: Mathematics provides students with essential mathematical skills and knowledge in number and algebra, measurement and geometry, and statistics and probability.
### Aims
The Australian Curriculum: Mathematics aims to ensure that students:
are confident, creative users and communicators of mathematics, able to investigate, represent and interpret situations in their personal and work lives and as active citizens.
### Key ideas
In Mathematics, the key ideas are the proficiency strands of understanding, fluency, problem-solving and reasoning. The proficiency strands describe the actions in which students can engage when learning and using the content.
### Structure
The Australian Curriculum: Mathematics is organised around the interaction of three content strands and four proficiency strands.
The content strands are number and algebra, measurement and geometry, and statistics and probability. They describe what is to be taught and learnt.
### PDF documents
Resources and support materials for the Australian Curriculum: Mathematics are available as PDF documents.
Mathematics: Sequence of content
Mathematics: Sequence of achievement
## Year 3
### Year 3 Level Description
The proficiency strands understanding, fluency, problem-solving and reasoning are an integral part of mathematics content across the three content strands: number and algebra, measurement and geometry, and statistics and probability. The proficiencies reinforce the significance of working mathematically within the content and describe how the content is explored or developed. They provide the language to build in the developmental aspects of the learning of mathematics. The achievement standards reflect the content and encompass the proficiencies.
At this year level:
• understanding includes connecting number representations with number sequences, partitioning and combining numbers flexibly, representing unit fractions, using appropriate language to communicate times, and identifying environmental symmetry
• fluency includes recalling multiplication facts, using familiar metric units to order and compare objects, identifying and describing outcomes of chance experiments, interpreting maps and communicating positions
• problem-solving includes formulating and modelling authentic situations involving planning methods of data collection and representation, making models of three-dimensional objects and using number properties to continue number patterns
• reasoning includes using generalising from number properties and results of calculations, comparing angles and creating and interpreting variations in the results of data collections and data displays.
### Year 3 Content Descriptions
#### Using units of measurement
Measure, order and compare objects using familiar metric units of length, mass and capacity (ACMMG061 - Scootle )
Numeracy
Using measurement
• Estimate and measure with metric units
Literacy
Word Knowledge
• Understand learning area vocabulary
Comprehending texts through listening, reading and viewing
• Navigate, read and view learning area texts
• Comprehend texts
Composing texts through speaking, writing and creating
• Compose texts
• Compose spoken, written, visual and multimodal learning area texts
• recognising the importance of using common units of measurement
Critical and Creative Thinking
Inquiring – identifying, exploring and organising information and ideas
• Organise and process information
• Identify and clarify information and ideas
Numeracy
Using measurement
• Estimate and measure with metric units
• recognising and using centimetres and metres, grams and kilograms, and millilitres and litres
Literacy
Comprehending texts through listening, reading and viewing
• Interpret and analyse learning area texts
• Navigate, read and view learning area texts
Word Knowledge
• Understand learning area vocabulary
Critical and Creative Thinking
Inquiring – identifying, exploring and organising information and ideas
• Organise and process information
• Identify and clarify information and ideas
Numeracy
Using measurement
• Estimate and measure with metric units
Tell time to the minute and investigate the relationship between units of time (ACMMG062 - Scootle )
Literacy
Word Knowledge
• Understand learning area vocabulary
Composing texts through speaking, writing and creating
• Compose spoken, written, visual and multimodal learning area texts
• Compose texts
Comprehending texts through listening, reading and viewing
• Navigate, read and view learning area texts
• Comprehend texts
Numeracy
Using measurement
• Operate with clocks, calendars and timetables
• recognising there are 60 minutes in an hour and 60 seconds in a minute
Critical and Creative Thinking
Inquiring – identifying, exploring and organising information and ideas
• Identify and clarify information and ideas
• Organise and process information
Numeracy
Using measurement
• Operate with clocks, calendars and timetables
#### Shape
Make models of three-dimensional objects and describe key features (ACMMG063 - Scootle )
Numeracy
Using spatial reasoning
• Visualise 2D shapes and 3D objects
Literacy
Word Knowledge
• Understand learning area vocabulary
Composing texts through speaking, writing and creating
• Compose texts
• Compose spoken, written, visual and multimodal learning area texts
• exploring the creation of three-dimensional objects using origami, including prisms and pyramids
Critical and Creative Thinking
Inquiring – identifying, exploring and organising information and ideas
• Organise and process information
• Identify and clarify information and ideas
Intercultural Understanding
Recognising culture and developing respect
• Explore and compare cultural knowledge, beliefs and practices
Numeracy
Using spatial reasoning
• Visualise 2D shapes and 3D objects
#### Location and transformation
Create and interpret simple grid maps to show position and pathways (ACMMG065 - Scootle )
Critical and Creative Thinking
Generating ideas, possibilities and actions
• Consider alternatives
• Seek solutions and put ideas into action
• Imagine possibilities and connect ideas
Inquiring – identifying, exploring and organising information and ideas
• Identify and clarify information and ideas
• Organise and process information
Literacy
Composing texts through speaking, writing and creating
• Compose spoken, written, visual and multimodal learning area texts
• Compose texts
Comprehending texts through listening, reading and viewing
• Interpret and analyse learning area texts
• Navigate, read and view learning area texts
• Comprehend texts
Visual Knowledge
• Understand how visual elements create meaning
Numeracy
Using spatial reasoning
• Interpret maps and diagrams
• creating a map of the classroom or playground
Numeracy
Using spatial reasoning
• Interpret maps and diagrams
Critical and Creative Thinking
Generating ideas, possibilities and actions
• Consider alternatives
• Seek solutions and put ideas into action
• Imagine possibilities and connect ideas
Literacy
Visual Knowledge
• Understand how visual elements create meaning
Identify symmetry in the environment (ACMMG066 - Scootle )
Numeracy
Using spatial reasoning
• Visualise 2D shapes and 3D objects
• identifying symmetry in Aboriginal rock carvings or art
Critical and Creative Thinking
Inquiring – identifying, exploring and organising information and ideas
• Organise and process information
• Identify and clarify information and ideas
Intercultural Understanding
Recognising culture and developing respect
• Explore and compare cultural knowledge, beliefs and practices
Numeracy
Using spatial reasoning
• Visualise 2D shapes and 3D objects
• identifying symmetry in the natural and built environment
Critical and Creative Thinking
Inquiring – identifying, exploring and organising information and ideas
• Identify and clarify information and ideas
• Organise and process information
Numeracy
Using spatial reasoning
• Visualise 2D shapes and 3D objects
#### Geometric reasoning
Identify angles as measures of turn and compare angle sizes in everyday situations (ACMMG064 - Scootle )
Critical and Creative Thinking
Inquiring – identifying, exploring and organising information and ideas
• Organise and process information
• Identify and clarify information and ideas
Numeracy
Using spatial reasoning
• Visualise 2D shapes and 3D objects
• opening doors partially and fully and comparing the size of the angles created
Critical and Creative Thinking
Inquiring – identifying, exploring and organising information and ideas
• Identify and clarify information and ideas
• Organise and process information
Numeracy
Using spatial reasoning
• Visualise 2D shapes and 3D objects
• recognising that analogue clocks use the turning of arms to indicate time, and comparing the size of angles between the arms for familiar times
Critical and Creative Thinking
Inquiring – identifying, exploring and organising information and ideas
• Organise and process information
• Identify and clarify information and ideas
Numeracy
Using spatial reasoning
• Visualise 2D shapes and 3D objects
Using measurement
• Operate with clocks, calendars and timetables
### Year 3 Achievement Standards
By the end of Year 3, students recognise the connection between addition and subtraction and solve problems using efficient strategies for multiplication. They model and represent unit fractions. They represent money values in various ways. Students identify symmetry in the environment. They match positions on maps with given information. Students recognise angles in real situations. They interpret and compare data displays.
Students count to and from 10 000. They classify numbers as either odd or even. They recall addition and multiplication facts for single-digit numbers. Students correctly count out change from financial transactions. They continue number patterns involving addition and subtraction. Students use metric units for length, mass and capacity. They tell time to the nearest minute. Students make models of three-dimensional objects. Students conduct chance experiments and list possible outcomes. They conduct simple data investigations for categorical variables. | 1,967 | 10,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-17 | longest | en | 0.887383 |
https://methodsblog.com/2017/11/07/3d-fractal-geometry-irregular-organisms/ | 1,686,294,153,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655446.86/warc/CC-MAIN-20230609064417-20230609094417-00503.warc.gz | 440,978,331 | 33,046 | # The Problem with the Shape
More than anything else, the phenotype of an organism determines how it interacts with the environment. It’s subject to natural selection, and may help to unravel the underlying evolutionary processes. So shape traits are key elements in many ecological and biological studies.
Commonly, basic parameters like distances, areas, angles, or derived ratios are used to describe and compare the shapes of organisms. These parameters usually work well in organisms with a regular body plan. The shape of irregular organisms – such as many plants, fungi, sponges or corals – is mainly determined by environmental factors and often lacks the distinct landmarks needed for traditional morphometric methods. The application of these methods is problematic and shapes are more often categorised than actually measured.
As scientists though, we favour independent statistical analyses, and there’s an urgent need for reliable shape characterisation based on numerical approaches. So, scientists often determine complexity parameters such as surface/volume ratios, rugosity, or the level of branching. However, these parameters all share the same drawback: they are delineated to a univariate number, taking information from one or few spatial scales and because of this essential information is lost.
This problem is founded in the analytical tools we tend to use – the classical Euclidean geometry. Classical Euclidean geometry, as we know it from school, is based on a set of ideal shapes (e.g. rectangles, triangles, or circles), of which we can calculate distances, angles, or areas, based on simple mathematical formulas. Nature provides at best imperfect representations of these though, limiting the application of classical geometric approaches to natural structures and living organisms.
# A Potential Solution
One attempt to summarize the highly complex patterns found in nature comes from fractal geometry. If you take a closer look at a fern leaf, you’ll see that the little leaves have the same shape as the whole fern leaf. The same principle applies to fractals – mathematical sets that exhibit a repetitive pattern across all scales.
To understand the novelty of fractal geometry approaches we’ll go back to the fern leaf. Before we can start to measure the outline of that leaf, we have to define the scale we’re working at. If we’re interested in the shape of the entire leaf – for example, to address competition for space – we would probably use a cm scale and measure the outline of the big leaf. If we’re more interested in the smaller attributes – studying, for instance, habitat availability for small insects – we’d want to address the smaller leaves, and measure the outline on a mm scale.
These measurements are examples for classical Euclidian approaches. Fractal geometry, however, combines the information across spatial scales. The highly complex spatial patterns are summarized to a univariate or multivariate dataset. In our fern leaf example, this dataset now combines the information from all spatial scales that are within the preset frame of the analysis. This is a big advantage as we can then either analyse the whole fern at once or select the respective scales of interest in a second step.
The mathematical background behind this approach is derived from the hypothetical, self-similar shapes – the fractals. To determine their complexity, mathematicians have developed various methods to calculate the fractal dimension D. Mandelbrot’s famous synthesis of fractals (1983) introduced the measuring principle to almost all parts of science and has proven successful not only for true fractals but also for studying real-world phenomena. In biology for example, fractal dimension analyses are used to address ecological and evolutionary questions (e.g., as an objective, scale-independent descriptor of shape and measure of complexity).
# The Maths Behind the Measuring Principle of Fractal Dimensions
Different methods have been developed to determine the fractal dimension D (e.g., Box-counting, Minkowski-Bouligand dimension, or Hausdorff dimension; Schroeder 1991). Applying, for example, the Minkowski-Bouligand method at the 3D level, the object is dilated using spheres of every possible radius r. The influence volume V(r) of the spheres depends on the level of their interaction and produces a characteristic pattern for the object. These values are then either combined to a single value of fractal dimension: D; or given as a log-log curve of the influence volume V as a function of r: V(r). This means that, D includes information from various spatial scales (r) and synthesizes them to an absolute measure, with increasing D indicating a higher complexity.
# Applying 3D Fractal Dimension Analyses to Study the shape of Irregular Organisms
In ‘The power of 3D fractal dimensions for comparative shape and structural complexity analyses of irregularly shaped organisms’, we tested the potential of 3D fractal dimension analyses as a shape descriptor for irregular organisms. We characterised the shapes of a set of stony corals applying the Minkowski-Bouligand method. The models were generated using the handheld 3D scanner Artec Spider.
As model taxa we used six widespread species with varying complexity, belonging to the genera Acropora, Pocillopora, and Porites. For each species, we studied six to eight individuals, from three origin coral colonies, accounting for variation within species. To assess the performance of this novel approach, we addressed interspecific differences as well as changes in shape over a time of eleven weeks. We compared the performance of fractal dimensions to traditional measures such as surface-volume ratios and rugosity.
We were able to show that 3D fractal dimension analyses can be used to quantify both interspecific variations and changes over time in irregularly shaped organisms like stony corals. Compared to traditional methods, fractal dimensions performed at least as well at the interspecific level and considerably better at the intraspecific level over time.
# The Power of Infinity and its Application for Future Studies
In contrast to many other complexity and shape measures, fractal dimension analyses are easily computed and aren’t dependent on the orientation of the study object. This makes them less error-prone and more user-friendly. Based on our results, we can conclude that 3D fractal dimension analyses are an efficient and easily applicable method for shape quantification and complexity characterization of stony corals. It’s highly likely that this is will be the case for other irregularly shaped organisms, too. We see their strength in the application to detect small variations in shape, for example due to environmental change or competitive pressure. At the species level, fractal dimensions may provide new opportunities for semi-automatic or automatic species determination based on 3D morphological images.
The applications of 3D fractal dimension analyses are almost as infinite as a fractal itself. They can be used to study other taxa, including organisms with a more regular body plan such as insects or mollusks. Besides their application at the level of individual organisms, fractal analyses can be applied to the ecosystem scale for studying productivity, biodiversity, or resilience (Bradbury & Reichelt 1983; Martin-Garin et al. 2007; Thistle et al. 2010; Kamal et al. 2014).
We’d encourage everybody to explore the potential of fractal dimension analyses for answering ecological, taxonomic, and evolutionary questions in a wide range of organisms. Support for anxious anti-mathematicians could come from our fractal dimension toolbox. Ultimately, we see fractal dimension analyses as an essential compound to provide a more integrative understanding of morphological variation within ecological and evolutionary contexts.
To find out more about 3D fractal dimensions read our Methods in Ecology and Evolution article ‘The power of 3D fractal dimensions for comparative shape and structural complexity analyses of irregularly shaped organisms | 1,568 | 8,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-23 | latest | en | 0.935361 |
http://mathematicssolution.com/echelon-matrices-row-canonical-form-row-equivalence/ | 1,500,734,365,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424079.84/warc/CC-MAIN-20170722142728-20170722162728-00203.warc.gz | 196,850,090 | 11,289 | BIGtheme.net http://bigtheme.net/ecommerce/opencart OpenCart Templates
Saturday , July 22 2017
Home / Linear Algebra / Echelon matrices, Row canonical form, Row equivalence
# Echelon matrices, Row canonical form, Row equivalence
Echelon matrices: A matrix A is called an echelon matrix, or is said to be in echelon form, if the following two conditions:
(1) All zero rows, if any, are at the bottom of the matrix.
(2) Each leading non-zero entry in a row is to the right of leading non-zero entry in the preceding row.
That is, A=[aij] is an echelon matrix if there exist nonzero entries
a1j1, a2j2, ———–,arjr where j1>j2>—–>jr
with the property that
aij=0 for {i) i≤r, j<ji ii) i>r
The entries a1j1, a2j2, ———–,arjr which are the leading non-zero elements in their respective rows are called pivots of the echelon matrix.
## Row Canonical form:
A matrix A is said to be in row canonical form if it is an echelon matrix, that is, if it satisfies the above properties (1) and (2), and if it satisfies the following additional two properties:
(3) Each pivot is equal to 1.
(4) Each pivot is the only nonzero entry in its column.
Example:
## Row equivalence:
A matrix A is said to be row equivalent to a matrix B, written
A~B
if B can be obtained from A by a sequence of elementary row operations.
## Rank of a matrix:
The rank of a matrix A, written rank (A), is equal to the number of pivots in an echelon form of A.
Theorem: Let A be a square matrix. Then the following are equivalent
(i) A is invertible(nonsingular)
(ii) A is row equivalent to the identity matrix I.
(iii) A is a product of elementary matrices.
Proof: Suppose A is invertible and suppose A is row equivalent to matrix B in row canonical form. Then there exist elementary matrices E1, E2 ,……, Es such that Es, …..,E2, E1A=B. Since A is invertible and each elementary matrix is invertible, B is also invertible. But if B≠I, then B has a zero row; whence B is not invertible. Thus B=I, (i) implies (ii).
If (ii) holds, then there exist elementary matrices E1, E2 ,……, Es such that Es, …..,E2, E1A=I. Hence A=( E1, E2 ,……, Es)-1= E-11, E-2 2 ,……, E-1 s. But the
E-1i are also elementary matrices. Thus (ii) implies (iii) .
If (iii) holds, then A= E1, E2 ,……, Es. The Ei are invertible matrices; hence their product A is also invertible. Thus (iii) implies (i). (Proved )
## Prove that, every basis of a vector space V has the same number of elements.
Problem: Prove that, every basis of a vector space V has the same number of ... | 724 | 2,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-30 | latest | en | 0.854405 |
https://qna.carrieradda.com/32748/what-will-output-voltage-waveform-circuit-input-step-voltage-assume-that-initially-nulled | 1,720,876,699,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514494.35/warc/CC-MAIN-20240713114822-20240713144822-00508.warc.gz | 397,965,307 | 12,728 | # What will be the output voltage waveform for the circuit, R1×CF=1s and input is a step voltage. Assume that the op-amp is initially nulled.
+1 vote
What will be the output voltage waveform for the circuit, R1×CF=1s and input is a step voltage. Assume that the op-amp is initially nulled.
(a) Triangular function
(b) Unit step function
(c) Ramp function
(d) Square function
This question was addressed to me in my homework.
The doubt is from Integrator topic in division Operational Amplifier Applications of Linear Integrated Circuits
by (39.8k points)
The accurate answer is (c) Ramp function
Simply put: Input voltage Vin = 1.2v for 0≤t≤0.4ms. The output voltage at t=0.4ms is
Vo = (1/R×CF)×^^t∫0 Vindt+C =-(1/1) × ^^0.4∫01.2 dt
=> Vo =-[^^0.1∫01.2 dt + ^^0.2∫0.11.2 dt + ^^0.3∫0.21.2 dt + ^^0.4∫0.31.2 dt ] = -(1.2+1.2+1.2+1.2) = -4.8v
Therefore, the output voltage waveform is a ramp function.
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+1 vote | 355 | 1,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-30 | latest | en | 0.896235 |
https://www.astronomyclub.xyz/particle-acceleration/info-rgi.html | 1,585,704,748,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505359.23/warc/CC-MAIN-20200401003422-20200401033422-00323.warc.gz | 806,641,647 | 4,607 | ## Info
If the loss rate is p = -op (the generalization to different loss rates upstream and downstream is trivial) the basic equation becomes d
(4np2Lf)+ ^(o - 4np2/ (p)ol)= Q - ^np2/(p)- 4napAf (p)dL2 . (4.22.7) at dp dp
In the steady state and away from the source region this gives immediately the remarkably simple result for the logarithmic slope of the spectrum, d ln f = - 3(u1 - 4opL -op2 (dL1/dp)) (4 22 8)
The denominator goes to zero at the critical momentum pcr = (( - u2 )(3aL), (4.22.9)
where the losses exactly balance the acceleration. If the numerator at this point is negative, the slope goes to - ^ and there is no pile-up. However, the slope goes to + ^ and a pile-up occurs if u1 -4u2 + 3op2(dL1/dp)> 0 at p = pcr . (4.22.10)
In early analytic work (Webb et al., 1984; Bregman et al., 1981) the diffusion coefficient was taken to be constant, so dL1/ dp = 0 and this condition reduces to u > 4u2 . However, if, as in the work of Protheroe and Stanev (1998), the diffusion coefficient is an increasing function of energy or momentum the condition becomes less restrictive. For a power-law dependence of the form K ^ p0 the condition for a pile-up to occur reduces to u1 - 4u2 + 0(u1 - u2)āLā > 0. (4.22.11)
Drury et al. (1999) note that the equivalent criterion for the model used by Protheroe and Stanev (1998) is slightly different, namely, u1 - 4u2 + 0(u1 - u2 )> 0 (4.22.12)
because of their neglect of the additional loss process. For the case in which L\l L2 = u^ u1 and with 0 = 1 this condition predicts that shocks with compression ratios greater than about r = 3.45 will produce pile-ups whilst weaker shocks will not. | 510 | 1,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-16 | latest | en | 0.874237 |
https://guzintamath.com/appinfo/rattables_info.html | 1,506,294,597,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690228.57/warc/CC-MAIN-20170924224054-20170925004054-00048.warc.gz | 694,417,377 | 3,161 | ## Ratio Tables
Click on the image at right to get the lesson app with instructor notes.
In this lesson, students will learn about equivalent ratios, rates, and unit rates. They will use equivalent ratios to fill in missing numbers in tables and how to multiply and divide to determine equivalent ratios. The language "For every ___, there are ___" is used here to describe ratios. This language reinforces the idea that ratios are not necessarily descriptions of "fixed" amounts, but of amounts that can change together multiplicatively. Students will learn that rates are just ratios where the terms are given in different units. They will also briefly look at how a double number line can be used to model and find rates and unit rates. Finally, students will learn about kinetic energy, the energy of a body in motion, and the formula for calculating the kinetic energy of an object in motion, which relies on the velocity of the object.
## Module 1 Video
This video connects with the opening context in Ratio Names on identifying ratios and extends that lesson to include recording and reading ratios in tables and beginning to determine equivalent ratios.
The key concept for this lesson is that, if two values are in a ratio relationship, that means that if one value is multiplied or divided by a number, the other value is multiplied or divided by the same number.
The language "For every ___, there are ___" is important to convey this meaning of ratios.
## Module 2 Video (1 of 2)
This video introduces speed as a mathematical rate. To know the speed, one must know both the distance an object travels and the time it takes. The speed of the object is a ratio of distance to time, which is a rate.
ratio tables
Making and using tables of equivalent ratios.
6-RP.A.2, 3a | 366 | 1,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-39 | latest | en | 0.927621 |
https://www.nag.com/numeric/cl/nagdoc_cl24/html/G07/g07gac.html | 1,632,542,271,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00503.warc.gz | 928,973,336 | 5,530 | g07 Chapter Contents
g07 Chapter Introduction
NAG Library Manual
# NAG Library Function Documentnag_outlier_peirce (g07gac)
## 1 Purpose
nag_outlier_peirce (g07gac) identifies outlying values using Peirce's criterion.
## 2 Specification
#include #include
void nag_outlier_peirce (Integer n, Integer p, const double y[], double mean, double var, Integer iout[], Integer *niout, Integer ldiff, double diff[], double llamb[], NagError *fail)
## 3 Description
nag_outlier_peirce (g07gac) flags outlying values in data using Peirce's criterion. Let
• $y$ denote a vector of $n$ observations (for example the residuals) obtained from a model with $p$ parameters,
• $m$ denote the number of potential outlying values,
• $\mu$ and ${\sigma }^{2}$ denote the mean and variance of $y$ respectively,
• $\stackrel{~}{y}$ denote a vector of length $n-m$ constructed by dropping the $m$ values from $y$ with the largest value of $\left|{y}_{i}-\mu \right|$,
• ${\stackrel{~}{\sigma }}^{2}$ denote the (unknown) variance of $\stackrel{~}{y}$,
• $\lambda$ denote the ratio of $\stackrel{~}{\sigma }$ and $\sigma$ with $\lambda =\frac{\stackrel{~}{\sigma }}{\sigma }$.
Peirce's method flags ${y}_{i}$ as a potential outlier if $\left|{y}_{i}-\mu \right|\ge x$, where $x={\sigma }^{2}z$ and $z$ is obtained from the solution of
$Rm = λ m-n mm n-m n-m nn$ (1)
where
$R = 2 exp z2 - 1 2 1- Φz$ (2)
and $\Phi$ is the cumulative distribution function for the standard Normal distribution.
As ${\stackrel{~}{\sigma }}^{2}$ is unknown an assumption is made that the relationship between ${\stackrel{~}{\sigma }}^{2}$ and ${\sigma }^{2}$, hence $\lambda$, depends only on the sum of squares of the rejected observations and the ratio estimated as
$λ2 = n-p-m z2 n-p-m$
which gives
$z2 = 1+ n-p-m m 1-λ2$ (3)
A value for the cutoff $x$ is calculated iteratively. An initial value of $R=0.2$ is used and a value of $\lambda$ is estimated using equation (1). Equation (3) is then used to obtain an estimate of $z$ and then equation (2) is used to get a new estimate for $R$. This process is then repeated until the relative change in $z$ between consecutive iterations is $\text{}\le \sqrt{\epsilon }$, where $\epsilon$ is machine precision.
By construction, the cutoff for testing for $m+1$ potential outliers is less than the cutoff for testing for $m$ potential outliers. Therefore Peirce's criterion is used in sequence with the existence of a single potential outlier being investigated first. If one is found, the existence of two potential outliers is investigated etc.
If one of a duplicate series of observations is flagged as an outlier, then all of them are flagged as outliers.
## 4 References
Gould B A (1855) On Peirce's criterion for the rejection of doubtful observations, with tables for facilitating its application The Astronomical Journal 45
Peirce B (1852) Criterion for the rejection of doubtful observations The Astronomical Journal 45
## 5 Arguments
1: nIntegerInput
On entry: $n$, the number of observations.
Constraint: ${\mathbf{n}}\ge 3$.
2: pIntegerInput
On entry: $p$, the number of parameters in the model used in obtaining the $y$. If $y$ is an observed set of values, as opposed to the residuals from fitting a model with $p$ parameters, then $p$ should be set to $1$, i.e., as if a model just containing the mean had been used.
Constraint: $1\le {\mathbf{p}}\le {\mathbf{n}}-2$.
3: y[n]const doubleInput
On entry: $y$, the data being tested.
4: meandoubleInput
On entry: if ${\mathbf{var}}>0.0$, mean must contain $\mu$, the mean of $y$, otherwise mean is not referenced and the mean is calculated from the data supplied in y.
5: vardoubleInput
On entry: if ${\mathbf{var}}>0.0$, var must contain ${\sigma }^{2}$, the variance of $y$, otherwise the variance is calculated from the data supplied in y.
6: iout[n]IntegerOutput
On exit: the indices of the values in y sorted in descending order of the absolute difference from the mean, therefore $\left|{\mathbf{y}}\left[{\mathbf{iout}}\left[\mathit{i}-2\right]-1\right]-\mu \right|\ge \left|{\mathbf{y}}\left[{\mathbf{iout}}\left[\mathit{i}-1\right]-1\right]-\mu \right|$, for $\mathit{i}=2,3,\dots ,{\mathbf{n}}$.
7: nioutInteger *Output
On exit: the number of potential outliers. The indices for these potential outliers are held in the first niout elements of iout. By construction there can be at most ${\mathbf{n}}-{\mathbf{p}}-1$ values flagged as outliers.
8: ldiffIntegerInput
On entry: the maximum number of values to be returned in arrays diff and llamb.
If ${\mathbf{ldiff}}\le 0$, arrays diff and llamb are not referenced and both diff and llamb may be NULL.
9: diff[ldiff]doubleOutput
On exit: if diff is not NULL then ${\mathbf{diff}}\left[\mathit{i}-1\right]$ holds $\left|y-\mu \right|-{\sigma }^{2}z$ for observation ${\mathbf{y}}\left[{\mathbf{iout}}\left[\mathit{i}-1\right]-1\right]$, for $\mathit{i}=1,2,\dots ,\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left({\mathbf{ldiff}},{\mathbf{niout}}+1,{\mathbf{n}}-{\mathbf{p}}-1\right)$.
10: llamb[ldiff]doubleOutput
On exit: if llamb is not NULL then ${\mathbf{llamb}}\left[\mathit{i}-1\right]$ holds $\mathrm{log}\left({\lambda }^{2}\right)$ for observation ${\mathbf{y}}\left[{\mathbf{iout}}\left[\mathit{i}-1\right]-1\right]$, for $\mathit{i}=1,2,\dots ,\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left({\mathbf{ldiff}},{\mathbf{niout}}+1,{\mathbf{n}}-{\mathbf{p}}-1\right)$.
11: failNagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
On entry, argument $⟨\mathit{\text{value}}⟩$ had an illegal value.
NE_INT
On entry, ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{n}}\ge 3$.
NE_INT_2
On entry, ${\mathbf{p}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: $1\le {\mathbf{p}}\le {\mathbf{n}}-2$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
Not applicable.
## 8 Parallelism and Performance
Not applicable.
One problem with Peirce's algorithm as implemented in nag_outlier_peirce (g07gac) is the assumed relationship between ${\sigma }^{2}$, the variance using the full dataset, and ${\stackrel{~}{\sigma }}^{2}$, the variance with the potential outliers removed. In some cases, for example if the data $y$ were the residuals from a linear regression, this assumption may not hold as the regression line may change significantly when outlying values have been dropped resulting in a radically different set of residuals. In such cases nag_outlier_peirce_two_var (g07gbc) should be used instead.
## 10 Example
This example reads in a series of data and flags any potential outliers.
The dataset used is from Peirce's original paper and consists of fifteen observations on the vertical semidiameter of Venus.
### 10.1 Program Text
Program Text (g07gace.c)
### 10.2 Program Data
Program Data (g07gace.d)
### 10.3 Program Results
Program Results (g07gace.r) | 2,198 | 7,113 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 78, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-39 | latest | en | 0.547915 |
https://neveradulldayinpoland.com/qa/question-how-many-amps-does-the-average-computer-use.html | 1,621,384,060,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989874.84/warc/CC-MAIN-20210518222121-20210519012121-00064.warc.gz | 433,107,800 | 9,028 | Question: How Many Amps Does The Average Computer Use?
How many amps does a CPU draw?
100 AmpsA CPU can draw 100 Amps (just a random number, not necessarily related to reality), and it doesn’t go over the 62 Amps limit of your PSU.
Why.
Because 100 Amps at 1.2 Volts is MUCH LESS power drawn than your PSU can deliver: 62 Amps at 12 Volts..
How many amps does a hair dryer?
Usually, the old hair dryer consumes 15 – 20 amps.
How much voltage does a desktop need?
The typical PC power supply draws approximately 110 volts of alternating electrical current from the wall outlet, which converts to a much smaller unidirectional flow of electrical current.
How much power does a motherboard use?
Your motherboard probably uses 50 to 150 watts, and each stick of memory requires about 15 watts. The processor needs between 80 and 140 watts of power.
How many amps does a typical desktop computer use?
Ratings of commonly used household appliancesDomestic Portable ApplianceAmps UsedWatts UsedDesktop computer3.0700Television 42″ HD0.5120Games console0.86<200washing machine10220027 more rows
How many amps does it take to charge a computer?
Most likely, all you will need is 5 amps. Look at your laptop power adapter (the black brick). On the backside it should tell you what the power draw is in amps. Most likely its 1 to 2 amps.
How much current does a computer use?
A complete desktop uses an average of 200 Watt hours (Wh). This is the sum of the average consumption per hour of the computer itself (171 W), the internet modem (10 W), the printer (5 W) and the loudspeakers (20 W). Assuming that a computer is on for eight hours a day, the annual consumption comes to 600 kWh.
How many computers can you put on a 20 amp circuit?
5-6 computersYou can throw up to 5-6 computers onto a 20 Amp circuit without tripping the breaker.
What happens if amps are too high?
A higher voltage than a device is designed for will most likely blow up some components, but amperage is derived from how much current is actually drawn, based on the device’s resistance, so having “too much” is not a problem and will do no harm.
What will 10 amps run?
10 amps indicates how much you can use at once. 10 amps is a good supply and will run pretty much everything except a domestic microwave. It is not enough to run everything at once though. for example, if you use a domestic electric kettle and a heater together it may trip out.
How many watts is 15 amps?
1,800 watts15-amp 120-volt circuit: 15 amps x 120 volts = 1,800 watts.
Is it better to charge a battery at 2 amps or 10 amps?
It is best to slow charge the battery. Slow charging rates vary depending on the battery’s type and capacity. However, when charging an automotive battery, 10 amps or less is considered a slow charge, while 20 amps or above is generally considered a fast charge.
How many computers can run on a 15 amp circuit?
5 computersIf you are careful to avoid high power draining devices (heaters, motors, refrigerators, AC units, etc) and are sure that nothing else but your computers are on said 15 amp circuit (generally an entire room or sometimes even two rooms may be on the same circuit) then it is POSSIBLE to run 5 computers and 5 LCD monitors …
How many watts does a CPU use?
Standard CPUs use between 65 and 85 watts, while quad-core processors range from 95 to 140 watts.
How do I tell how much power my CPU is using?
To see which apps have used the most battery power on your PC, head to Settings > System > Battery.
Does a PC use a lot of electricity?
Most computers are built to use up to 400 kilowatts of electricity per hour, but they usually use less than that. … Any computer that’s running on a Pentium-type processor uses about 100 kWh. This is with the monitor off. Your computer’s monitor probably uses more electricity than the actual processor itself.
How much power does a Gaming PC draw?
Based on our actual measurements of gaming PCs with progressively more efficient component configurations, together with market data on typical patterns of use, we estimate that the typical gaming PC (including display) uses about 1400 kilowatt-hours of electricity per year.
Can I use a 19.5 V charger on a 19v laptop?
Voltage: 19.5V and 19V is a relatively small voltage difference. It is most probably safe to use as long as: The current of the charger is the same or higher than that of the laptop. … Laptops usually don’t draw the full power from the charger even when in use and being charged at the same time.
How many outlets can be on a 15 amp circuit?
8 outletsTechnically, you can have as many outlets on a 15 amp circuit breaker as you want. However, a good rule of thumb is 1 outlet per 1.5 amps, up to 80% of the capacity of the circuit breaker. Therefore, we would suggest a maximum of 8 outlets for a 15 amp circuit.
How do you tell if an outlet is 15 or 20 amp?
Look at your outlet. If the two slots are parallel, it is a 15 amp outlet. If one slot is ‘T’ shaped and the other is straight, it is a 20 amp outlet that will also accept 15 amp plugs. If the two slots are perpendicular the outlet is 20 amp only.
How many amps does a 1600 watt hair dryer pull?
Many hair dryers require about 1,875 watts, or 15 amps, of power. Because of this, it’s much easier to blow a circuit just by plugging your hair dryer in. Be aware of how many amps the circuits in your house can handle before you plug in too many devices to one outlet. | 1,289 | 5,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-21 | latest | en | 0.934381 |
https://origin.geeksforgeeks.org/sort-an-array-of-strings-according-to-the-number-of-vowels-in-them/ | 1,680,277,999,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949644.27/warc/CC-MAIN-20230331144941-20230331174941-00629.warc.gz | 496,495,344 | 43,342 | Open in App
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# Sort an Array of Strings according to the number of Vowels in them
• Difficulty Level : Hard
• Last Updated : 09 Mar, 2023
Given an array arr[] of N strings, the task is to sort these strings according to the numbers of vowels in them.
Examples:
Input: arr[] = { “geeks”, “for”, “coding” }
Output: for, coding, geeks
for -> o = 1 vowel
coding -> o, i = 2 vowels
geeks -> e, e = 2 vowels
Input: arr[] = { “lmno”, “pqrst”, “aeiou”, “xyz” }
Output: pqrst, xyz, lmno, aeiou
Approach: The idea is to store each element with its number of vowels in a vector pair and then sort all the elements of the vector according to the number of vowels stored. Finally, print the strings in order.
Below is the implementation of the above approach:
## C++
`// C++ implementation of the approach` `#include ` `using` `namespace` `std;` `// Function to check the Vowel` `bool` `isVowel(``char` `ch)` `{` ` ``ch = ``toupper``(ch);` ` ``return` `(ch == ``'A'` `|| ch == ``'E'` ` ``|| ch == ``'I'` `|| ch == ``'O'` ` ``|| ch == ``'U'``);` `}` `// Returns count of vowels in str` `int` `countVowels(string str)` `{` ` ``int` `count = 0;` ` ``for` `(``int` `i = 0; i < str.length(); i++)` ` ``if` `(isVowel(str[i])) ``// Check for vowel` ` ``++count;` ` ``return` `count;` `}` `// Function to sort the array according to` `// the number of the vowels` `void` `sortArr(string arr[], ``int` `n)` `{` ` ``// Vector to store the number of vowels` ` ``// with respective elements` ` ``vector > vp;` ` ``// Inserting number of vowels` ` ``// with respective strings` ` ``// in the vector pair` ` ``for` `(``int` `i = 0; i < n; i++) {` ` ``vp.push_back(` ` ``make_pair(` ` ``countVowels(` ` ``arr[i]),` ` ``arr[i]));` ` ``}` ` ``// Sort the vector, this will sort the pair` ` ``// according to the number of vowels` ` ``sort(vp.begin(), vp.end());` ` ``// Print the sorted vector content` ` ``for` `(``int` `i = 0; i < vp.size(); i++)` ` ``cout << vp[i].second << ``" "``;` `}` `// Driver code` `int` `main()` `{` ` ``string arr[] = { ``"lmno"``, ``"pqrst"``,` ` ``"aeiou"``, ``"xyz"` `};` ` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``sortArr(arr, n);` ` ``return` `0;` `}`
## Java
`// Java implementation of the approach ` `import` `java.util.*;` `import` `java.lang.*;` `import` `java.io.*;` `class` `GFG{` `static` `class` `pair` `{` ` ``int` `first;` ` ``String second;` ` ` ` ``pair(``int` `first,String second)` ` ``{` ` ``this``.first = first;` ` ``this``.second = second;` ` ``}` `}` `// Function to check the Vowel` `static` `boolean` `isVowel(``char` `ch)` `{` ` ``ch = Character.toUpperCase(ch);` ` ``return` `(ch == ``'A'` `|| ch == ``'E'` `|| ` ` ``ch == ``'I'` `|| ch == ``'O'` `||` ` ``ch == ``'U'``);` `}` `// Returns count of vowels in str` `static` `int` `countVowels(String str)` `{` ` ``int` `count = ``0``;` ` ``for``(``int` `i = ``0``; i < str.length(); i++)` ` ` ` ``// Check for vowel` ` ``if` `(isVowel(str.charAt(i))) ` ` ``++count;` ` ` ` ``return` `count;` `}` `// Function to sort the array according to` `// the number of the vowels` `static` `void` `sortArr(String arr[], ``int` `n)` `{` ` ` ` ``// Vector to store the number of vowels` ` ``// with respective elements` ` ``ArrayList vp = ``new` `ArrayList<>();` ` ``// Inserting number of vowels` ` ``// with respective strings` ` ``// in the vector pair` ` ``for``(``int` `i = ``0``; i < n; i++) ` ` ``{` ` ``vp.add(``new` `pair(countVowels(arr[i]),` ` ``arr[i]));` ` ``}` ` ``// Sort the vector, this will sort the pair` ` ``// according to the number of vowels` ` ``Collections.sort(vp, (a, b) -> a.first - b.first);` ` ` ` ``// Print the sorted vector content` ` ``for``(``int` `i = ``0``; i < vp.size(); i++)` ` ``System.out.print(vp.get(i).second + ``" "``);` `}` `// Driver code` `public` `static` `void` `main(String[] args) ` `{` ` ``String arr[] = { ``"lmno"``, ``"pqrst"``,` ` ``"aeiou"``, ``"xyz"` `};` ` ``int` `n = arr.length;` ` ` ` ``sortArr(arr, n);` `}` `}` `// This code is contributed by offbeat`
## Python3
`# Python3 implementation of the approach` `# Function to check the Vowel` `def` `isVowel(ch) :` ` ` ` ``ch ``=` `ch.upper();` ` ``return` `(ch ``=``=` `'A'` `or` `ch ``=``=` `'E'``or` `ch ``=``=` `'I'` `or` ` ``ch ``=``=` `'O'``or` `ch ``=``=` `'U'``);` `# Returns count of vowels in str` `def` `countVowels(string) :` ` ``count ``=` `0``;` ` ``for` `i ``in` `range``(``len``(string)) :` ` ` ` ``# Check for vowel` ` ``if` `(isVowel(string[i])) : ` ` ``count ``+``=` `1``;` ` ` ` ``return` `count;` `# Function to sort the array according to` `# the number of the vowels` `def` `sortArr(arr, n) :` ` ` ` ``# Vector to store the number of vowels` ` ``# with respective elements` ` ``vp ``=` `[];` ` ``# Inserting number of vowels` ` ``# with respective strings` ` ``# in the vector pair` ` ``for` `i ``in` `range``(n) :` ` ``vp.append((countVowels(arr[i]),arr[i]));` ` ``# Sort the vector, this will sort the pair` ` ``# according to the number of vowels` ` ``vp.sort()` ` ``# Print the sorted vector content` ` ``for` `i ``in` `range``(``len``(vp)) :` ` ``print``(vp[i][``1``], end``=` `" "``);` `# Driver code` `if` `__name__ ``=``=` `"__main__"` `:` ` ``arr ``=` `[ ``"lmno"``, ``"pqrst"``,``"aeiou"``, ``"xyz"` `];` ` ``n ``=` `len``(arr);` ` ``sortArr(arr, n);` `# This code is contributed by AnkitRai01`
## C#
`// C# implementation of the approach` `using` `System;` `using` `System.Collections.Generic; ` `class` `GFG ` `{` ` ` ` ``// Function to check the Vowel` ` ``static` `bool` `isVowel(``char` `ch)` ` ``{` ` ``ch = ``char``.ToUpper(ch);` ` ``return` `(ch == ``'A'` `|| ch == ``'E'` ` ``|| ch == ``'I'` `|| ch == ``'O'` ` ``|| ch == ``'U'``);` ` ``}` ` ` ` ``// Returns count of vowels in str` ` ``static` `int` `countVowels(``string` `str)` ` ``{` ` ``int` `count = 0;` ` ``for` `(``int` `i = 0; i < str.Length; i++)` ` ``if` `(isVowel(str[i])) ``// Check for vowel` ` ``++count;` ` ``return` `count;` ` ``}` ` ` ` ``// Function to sort the array according to` ` ``// the number of the vowels` ` ``static` `void` `sortArr(``string``[] arr, ``int` `n)` ` ``{` ` ` ` ``// Vector to store the number of vowels` ` ``// with respective elements` ` ``List> vp = ``new` `List>();` ` ` ` ``// Inserting number of vowels` ` ``// with respective strings` ` ``// in the vector pair` ` ``for` `(``int` `i = 0; i < n; i++)` ` ``{ ` ` ``vp.Add(``new` `Tuple<``int``, ``string``>(countVowels(arr[i]), arr[i]));` ` ``}` ` ` ` ``// Sort the vector, this will sort the pair` ` ``// according to the number of vowels` ` ``vp.Sort();` ` ` ` ``// Print the sorted vector content` ` ``for` `(``int` `i = 0; i < vp.Count; i++)` ` ``Console.Write(vp[i].Item2 + ``" "``);` ` ``}` ` ` ` ``// Driver code` ` ``static` `void` `Main()` ` ``{` ` ``string``[] arr = { ``"lmno"``, ``"pqrst"``,` ` ``"aeiou"``, ``"xyz"` `};` ` ``int` `n = arr.Length; ` ` ``sortArr(arr, n);` ` ``}` `}` `// This code is contributed by divyesh072019`
## Javascript
``
Output:
`pqrst xyz lmno aeiou`
Time Complexity: O(N*log N), for sorting the array.
Auxiliary Space: O(N), for storing the string in an extra array.
The approach sorts the input list of strings based on the number of vowels they contain using sorting function
## C++
`#include ` `#include ` `#include ` `using` `namespace` `std;` `// Function to count the number of vowels in a word` `int` `countVowels(string word)` `{` ` ``// A string of all vowels` ` ``string vowels = ``"aeiouAEIOU"``;` ` ``int` `count = 0;` ` ``for` `(``int` `i = 0; i < word.length(); i++) {` ` ``// If the character is a vowel, increment the` ` ``// count` ` ``if` `(vowels.find(word[i]) != string::npos) {` ` ``count++;` ` ``}` ` ``}` ` ``return` `count;` `}` `// Function to sort a vector of strings based on the` `// number of vowels in each string` `vector sortStrings(vector strings)` `{` ` ``// Sort the vector of strings based on the number of` ` ``// vowels in each string` ` ``sort(strings.begin(), strings.end(), [](``const` `string& s1, ``const` `string& s2) {` ` ``return` `countVowels(s1) < countVowels(s2);` ` ``});` ` ``return` `strings;` `}` `// Main function` `int` `main()` `{` ` ``// Input vector of strings` ` ``vector strings = {``"lmno"``, ``"pqrst"``, ``"aeiou"``, ``"xyz"``};` ` ``// Sort the vector of strings` ` ``vector sortedStrings = sortStrings(strings);` ` ``// Print the sorted vector of strings` ` ``for` `(``const` `auto``& s : sortedStrings) {` ` ``cout << s << ``" "``;` ` ``}` ` ``cout << endl;` ` ``return` `0;` `}`
## Java
`import` `java.util.*;` `public` `class` `GFG {` ` ``// Function to count the number of vowels in a word` ` ``public` `static` `int` `countVowels(String word)` ` ``{` ` ``// A string of all vowels` ` ``String vowels = ``"aeiouAEIOU"``;` ` ``int` `count = ``0``;` ` ``for` `(``int` `i = ``0``; i < word.length(); i++) {` ` ``// If the character is a vowel, increment the` ` ``// count` ` ``if` `(vowels.indexOf(word.charAt(i)) != -``1``) {` ` ``count++;` ` ``}` ` ``}` ` ``return` `count;` ` ``}` ` ``// Function to sort a list of strings based on the` ` ``// number of vowels in each string` ` ``public` `static` `List` ` ``sortStrings(List strings)` ` ``{` ` ``// Sort the list of strings based on the number of` ` ``// vowels in each string` ` ``Collections.sort(strings, ``new` `Comparator() {` ` ``@Override` ` ``public` `int` `compare(String s1, String s2)` ` ``{` ` ``return` `countVowels(s1) - countVowels(s2);` ` ``}` ` ``});` ` ``return` `strings;` ` ``}` ` ``// Main function` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``// Input list of strings` ` ``List strings = Arrays.asList(` ` ``"lmno"``, ``"pqrst"``, ``"aeiou"``, ``"xyz"``);` ` ``// Sort the list of strings` ` ``List sortedStrings = sortStrings(strings);` ` ``// Print the sorted list of strings` ` ``System.out.println(sortedStrings);` ` ``}` `}`
## Python3
`# Function to count the number of vowels in a word` `def` `count_vowels(word):` ` ``# A string of all vowels` ` ``vowels ``=` `"aeiouAEIOU"` ` ``# Return the count of vowels in the word` ` ``return` `sum``(c ``in` `vowels ``for` `c ``in` `word)` `# Function to sort a list of strings based on the number of vowels in each string` `def` `sort_strings(strings):` ` ``# Sort the list of strings based on the number of vowels in each string` ` ``return` `sorted``(strings, key``=``count_vowels)` `# Main function` `if` `__name__ ``=``=` `"__main__"``:` ` ``# Input list of strings` ` ``strings ``=` `[``"lmno"``, ``"pqrst"``, ``"aeiou"``, ``"xyz"``]` ` ``# Sort the list of strings` ` ``sorted_strings ``=` `sort_strings(strings)` ` ``# Print the sorted list of strings` ` ``print``(sorted_strings)`
## C#
`using` `System;` `using` `System.Collections.Generic;` `using` `System.Linq;` `public` `class` `GFG` `{` ` ``// Function to count the number of vowels in a word` ` ``public` `static` `int` `CountVowels(``string` `word)` ` ``{` ` ``// A string of all vowels` ` ``string` `vowels = ``"aeiouAEIOU"``;` ` ``int` `count = 0;` ` ``for` `(``int` `i = 0; i < word.Length; i++)` ` ``{` ` ``// If the character is a vowel, increment the` ` ``// count` ` ``if` `(vowels.IndexOf(word[i]) != -1)` ` ``{` ` ``count++;` ` ``}` ` ``}` ` ``return` `count;` ` ``}` ` ``// Function to sort a list of strings based on the` ` ``// number of vowels in each string` ` ``public` `static` `List<``string``>` ` ``SortStrings(List<``string``> strings)` ` ``{` ` ``// Sort the list of strings based on the number of` ` ``// vowels in each string` ` ``strings.Sort((s1, s2) => CountVowels(s1) - CountVowels(s2));` ` ``return` `strings;` ` ``}` ` ``// Main function` ` ``public` `static` `void` `Main(``string``[] args)` ` ``{` ` ``// Input list of strings` ` ``List<``string``> strings = ``new` `List<``string``>` ` ``{` ` ``"lmno"``, ``"pqrst"``, ``"aeiou"``, ``"xyz"` ` ``};` ` ``// Sort the list of strings` ` ``List<``string``> sortedStrings = SortStrings(strings);` ` ``// Print the sorted list of strings` ` ``Console.WriteLine(``string``.Join(``" "``, sortedStrings));` ` ``}` `}`
## Javascript
`// Javascript program for the above approach` `// Function to count the number of vowels in a word` `function` `count_vowels(word) {` ` ``// A string of all vowels` ` ``const vowels = ``"aeiouAEIOU"``;` ` ``// Return the count of vowels in the word` ` ``return` `Array.from(word).filter(c => vowels.includes(c)).length;` `}` `// Function to sort a list of strings based on the number of vowels in each string` `function` `sort_strings(strings) {` ` ``// Sort the list of strings based on the number of vowels in each string` ` ``return` `strings.sort((a, b) => count_vowels(a) - count_vowels(b));` `}` `// Main function` `const strings = [``"lmno"``, ``"pqrst"``, ``"aeiou"``, ``"xyz"``]; ``// Input list of strings` `const sorted_strings = sort_strings(strings); ``// Sort the list of strings` `console.log(sorted_strings); ``// Print the sorted list of strings` `// This code is contributed by princekumaras`
Output
`['pqrst', 'xyz', 'lmno', 'aeiou']`
Time Complexity: O(N log N)
Auxiliary Space: O(N)
My Personal Notes arrow_drop_up
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https://123deta.com/title/Lec%E6%9C%80%E8%BF%91%E3%81%AE%E6%9B%B4%E6%96%B0%E5%B1%A5%E6%AD%B4yyasudaswebsite-6 | 1,675,218,452,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499899.9/warc/CC-MAIN-20230201013650-20230201043650-00063.warc.gz | 93,644,943 | 22,376 | # Top PDF Lec6 最近の更新履歴 yyasuda's website
### Lec6 最近の更新履歴 yyasuda's website
A strategy in dynamic games is a complete action plan which prescribes how the player will act in each possible.. contingencies in future..[r]
### Lec6 最近の更新履歴 yyasuda's website
increases (resp. declines) as income increases, holding prices constant. ◮ A decrease in the own price of a normal good will cause quantity demanded to increase. If an own price decrease causes a decline in quantity demanded (known as Giffen’s paradox), the good must be inferior.
### Lec3 最近の更新履歴 yyasuda's website
payoff) while M gives 1 irrespective of player 1’s strategy. Therefore, M is eliminated by mixing L and R . After eliminating M , we can further eliminate D (step 2) and L (step 3), eventually picks up ( U , R ) as a unique outcome.
### PracticeF 最近の更新履歴 yyasuda's website
Q = K 1 =4 L 1 =8 Then, answer the following questions. (a) In the short run, the …rm is committed to hire a …xed amount of capital K(+1), and can vary its output Q only by employing an appropriate amount of labor L . Derive the …rm’s short-run total, average, and marginal cost functions. (b) In the long run, the …rm can vary both capital and labor. Derive the …rm’s
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(b) If consumer’s choice satis…es the weak axiom of revealed preferences, we can always construct a utility function which is consistent with such choice behav- iour. (c) If a consumer problem has a solution, then it must be unique whenever the consumer’s preference relation is convex.
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Exist exactly one for ANY exchange problem. Always Pareto efficient and individually rational[r]
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If the stage game has a unique NE, then for any T , the finitely repeated game has a unique SPNE: the NE of the stage game is played in every stage irrespective of the histor[r]
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3(a - e)/4, is greater than aggregate quantity in the Nash equilib- rium of the Cournot game, 2(a - e)/3, so the market-clearing price is lower in the Stackelberg game.. Thus, i[r]
### Lec8 最近の更新履歴 yyasuda's website
A tree starts with the initial node and ends at.. terminal nodes where payoffs are specified..[r]
### Lec7 最近の更新履歴 yyasuda's website
A strategy in dynamic games is a complete action plan which prescribes how the player will act in each possible.. contingencies in future..[r]
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elimination of strictly dominated strategies can never be selected (with positive probability) in a mixed-strategy Nash equilibrium.[r]
### Lec4 最近の更新履歴 yyasuda's website
Both the Bertrand and Cournot models are particular cases of a more general model of oligopoly competition where firms choose prices and quantities (or capacities.). Ber[r]
### Lec1 最近の更新履歴 yyasuda's website
Prisoners’ Dilemma: Analysis (3) (Silent, Silent) looks mutually beneficial outcomes, though Playing Confess is optimal regardless of other player’s choice! Acting optimally ( Confess , Confess ) rends up realizing!!
### EX3 最近の更新履歴 yyasuda's website
Solve the following problems in Snyder and Nicholson (11th):. 1.[r]
### Lec9 最近の更新履歴 yyasuda's website
3(a - e)/4, is greater than aggregate quantity in the Nash equilib- rium of the Cournot game, 2(a - e)/3, so the market-clearing price is lower in the Stackelberg game.. Thus, i[r]
### Lec10 最近の更新履歴 yyasuda's website
If the stage game has a unique NE, then for any T , the finitely repeated game has a unique SPNE: the NE of the stage game is played in every stage irrespective of the histor[r]
### Midterm2 最近の更新履歴 yyasuda's website
(a) If an agent is risk averse, her risk premium is ALWAYS positive. (b) When every player has a (strictly) dominant strategy, the strategy profile that consists of each player’s dominant strategy MUST be a Nash equilibrium. (c) If there are two Nash equilibria in pure-strategy, they can ALWAYS be Pareto
### PS3 最近の更新履歴 yyasuda's website
(c) Solve for the total saving S by all types who save and the total borrowing B.. by all types who borrow.[r]
### Final14 最近の更新履歴 yyasuda's website
5. Bayesian Nash Equilibrium (12 points) There are three different bills, \$5, \$10, and \$20. Two individuals randomly receive one bill each. The (ex ante) probability of an individual receiving each bill is therefore 1/3. Each individual knows only her own bill, and is simultaneously given the option of exchanging her bill for the other individual’s bill. The bills will be exchanged if and only if both individuals wish to do so; otherwise no exchange occurs. That is, each individuals can choose either exchange (E) or not (N), and exchange occurs only when both choose E. We assume that individuals’ objective is to maximize their expected monetary payoff (\$). | 1,430 | 4,914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-06 | longest | en | 0.860297 |
http://www.stat.wmich.edu/s216/book/node99.html | 1,544,790,476,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825728.30/warc/CC-MAIN-20181214114739-20181214140239-00443.warc.gz | 482,094,264 | 4,772 | Next: Type I and Type Up: Test of Significance involving Previous: Generalizing Tests of Significance
## Examples
Example 1. Suppose that an ad for Western Michigan University claims that a majority of WMU students come from outside the Southwest Michigan area'. A sample of 25 students revealed that only 9 out of 25, or 36%, come from outside Southwest Michigan. Can this just be chance error, or is this significant evidence that the advertisement is wrong?
Solution: The data consists of a sample percentage (or proportion); i.e. 36% of 25 students come from outside Southwest Michigan. Therefore, the ESTIMATE is a either a sample percentage or proportion (we will choose a proportion). At the end of the test, the conclusion will be one of two things: the data will provide evidence either for or against against the claim a majority of WMU students come from outside the Southwest Michigan area'. Therefore the competing hypotheses are versus p < .50. The hypothesis with the equal sign is null, so we have . The H0 Value is .50.
1.
Hypotheses: versus H1: p < .50
2.
Test Statistic:
3.
P-value: Presuming H0 is true, the likelihood of chance variation yielding a z-statistic as low as -1.40 is .08.
4.
Conclusion: Since P-value > .05, the observed sample value is not significantly different from .50. Hence, we do not reject . The sample does not provide enough evidence to prove the ad wrong.
Example 2 Is there "grade inflation" in WMU? A random sample of 100 student records from 10 years ago yields a sample average GPA of 2.90 with a standard deviation of .40. A random sample of 100 current students today yields a sample average of 2.98 with a standard deviation of .45. Is the difference between 2.90 and 2.98 just chance variation, or a statistically significant increase?
Solution: The evidence consists of two sample averages. The ESTIMATE of interest is the difference: 2.98-2.90=.08 (i.e average today minus average 10 years ago). At the end of the test, the conclusion will be one of two things: the data either support or provide evidence against the hypothesis of 'grade inflation'. Therefore the competing hypotheses are 'average grade today is higher than 10 years ago' or 'not'. In terms of the population averages, we write these as , or . The hypothesis with the equal sign is null, so we have . The H0 Value is 0.
1.
Hypotheses: versus
2.
Test Statistic:
3.
P-value: Presuming H0 is true, a z-statistic as large as 1.33 results from chance variation 9% of the time (P-value=.09).
4.
Conclusion: Since P-value > .05, the observed increase in average GPA is not statistically significant, and may be explained by chance variation. Do not reject H0.
If the sample sizes were small and the two variances were equal, a pooled-SD t-test would have been the more appropriate test. We illustrate the procedure here. The SD's for the two samples S1=.40 and S2=.45 are pooled into one SD:
The SE's for the two averages are and . Now we conduct the test.
1.
Hypotheses: versus
2.
Test Statistic:
3.
P-value: Presuming H0 is true, a t-statistic with 198 degrees of freedom will by chance be as large as 1.33 around 9% of the time (P-value=.09).
4.
Conclusion: Since P-value > .05, the observed increase in average GPA is not statistically significant, and may be explained by chance variation. Do not reject H0.
Example 3. The paired t-test Table 8.2 shows stock prices for six selected food service companies. Is there a significant difference between average Jan. 2002 and Jan. 2003 stock prices?
Solution:
At first glance, the evidence looks like it consists of two averages 31.12 and 28.07. However, remember that these two averages are not independent, because the data consists of two measurements on only ONE sample, rather than two separate samples. (Remember the Paired-t confidence interval?) The correct analysis here takes the differences between the two measurements, as shown on the table. If average 2002 and 2003 prices are significantly different, then the difference column will have an average that is significantly different from 0.
Ignoring the original two columns and focusing instead on the difference column, the evidence consists of one average. At the end of the test, the conclusion will be one of two things: the average difference is either significantly different from 0 or not ( versus ). Suppose we want a test that can detect either a positive or negative change in average stock prices, then we can choose to do a two-tailed test. (Of course, a one-tailed test that detects change in only one direction can also be done.)
1.
Hypotheses: versus
2.
Test Statistic:
3.
P-value: By chance alone, a t-statistic with 5 d.f. may get as large as 1.32 on either side of 0 around 24% of the time (P-value=.244).
4.
Conclusion: Since P-value > .05, the observed change in stock prices is not statistically significant, and may be explained by chance variation. Do not reject H0.
Example 4. Has retention rate at WMU been changing? Suppose that a random sample of 200 entering students in 1989 showed 74% were still enrolled 3 years later. Another random sample of 200 entering students in 1999 showed that 66% were still enrolled 3 years later. Is this drop of 8% statistical evidence of changing retention rate, or can it be just chance error (i.e luck of the draw in the students selected for the samples)?
Solution:
The evidence consists of two sample proportions , or more precisely, the difference between two sample proportions (.74-.66=.08). At the end of the test, the conclusion will be one of two things: the data either support or provide evidence against 'changing retention rates'. Since an increase or decrease will be interpreted as a change (this was discussed at the start of the study), we will conduct a two-tailed test. The competing hypotheses are 'change' versus 'no change'. In terms of the population, these may be written as versus p1-p2=0. The latter is assigned to H0 because it contains the equal sign. The H0 Value is 0.
1.
Hypotheses: H0: p1-p2=0 versus
2.
Test Statistic:
3.
P-value: Presuming H0 is true, a z-statistic as extreme as 1.78 on either side of 0 results from chance variation 7.5% of the time (P-value=.075).
4.
Conclusion: Since P-value > .05, the observed change in retention rate is not statistically significant, and may be explained by chance variation. Do not reject H0.
Next: Type I and Type Up: Test of Significance involving Previous: Generalizing Tests of Significance
2003-09-08 | 1,528 | 6,503 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2018-51 | latest | en | 0.923491 |
https://mathzsolution.com/lebesgue-measure-theory-vs-differential-forms/ | 1,674,831,372,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494986.94/warc/CC-MAIN-20230127132641-20230127162641-00591.warc.gz | 399,508,613 | 20,964 | # Lebesgue measure theory vs differential forms?
I am currently reading various differential geometry books. From what I understand differential forms allow us to generalize calculus to manifolds and thus perform integration on manifolds. I gather that it is, in general, completely distinct from Lebesgue measure theory and is more like a generalization of Riemann integration.
Ok so here’s the problem. I have always viewed Lebesgue measure theory as ‘solving the issues with Riemann integration’. For example, a big problem with Riemann integration is that the space of Riemann integral functions is not complete. The fact that $L^p$ spaces in the Lebesgue theory are complete seems like a huge improvement on the Riemann situation, and is vital for so many concepts in functional analysis, PDEs, operator theory, and numerical analysis.
So if we then consider differential geometry and integration via differential forms, unless I am misunderstanding something, we lose all the benefits of Lebesgue theory?
It seems like if do lose all those benefits we are in a very bad situation. For example, how are we supposed to rigorously define solution spaces for PDEs if we can’t use $L^p$ spaces and thus can’t use Sobolev spaces? How can we obtain acceptable convergence of some sequence that may arise during our work if we are operating in this generalized Riemann setting where we lack completeness?
In summary, if differential forms are a generalization of Riemann integration how are we supposed to perform analysis when we no longer have the power and utility of Lebesgue measure theory?
People use measure theory in tandem with differential forms all the time—there’s no contradiction whatsoever between the formalisms. Be aware, though, that the adjective “Riemannian” in the context of differential geometry refers to constructions depending on Riemannian metrics (which are “Riemannian” in the sense of originating in the work of Bernhard Riemann), not to Riemann integration.
Suppose that $$MM$$ a smooth $$nn$$-manifold. By definition, it’s locally diffeomorphic to $$\mathbb{R}^n\mathbb{R}^n$$, so that you can define a set $$S \subset MS \subset M$$ to be measurable if and only if $$x(S \cap U) \subset \mathbb{R}^nx(S \cap U) \subset \mathbb{R}^n$$ is Lebesgue measurable for every local coordinate chart $$x: U \to x(U) \subset \mathbb{R}^nx: U \to x(U) \subset \mathbb{R}^n$$. This gives you a $$\sigma\sigma$$-algebra of Lebesgue measurable sets on $$MM$$ that correctly completes the Borel $$\sigma\sigma$$-algebra generated by the open sets on $$MM$$ as a topological space. At this point, you have everything you need to define measurable functions, vector fields, differential forms, tensor fields, etc., in a manner compatible with calculations in local coordinates.
Now, suppose that $$MM$$ is a Riemannian manifold, so that it comes equipped with a Riemannian metric $$gg$$—again, the “Riemannian” here does not refer to Riemann integration, but to Riemann himself and his work on differential geometry. On any local coordinate chart $$x : U \to x(U) \subset \mathbb{R}^nx : U \to x(U) \subset \mathbb{R}^n$$, you can define a measure $$\lambda_{g,x}\lambda_{g,x}$$ on $$UU$$ by setting
$$\lambda_{g,x}(S \cap U) := \int_{x(S \cap U)} \sqrt{\det\left(g\left(\tfrac{\partial}{\partial x^i},\tfrac{\partial}{\partial x^j}\right)\right)} \,d\lambda \lambda_{g,x}(S \cap U) := \int_{x(S \cap U)} \sqrt{\det\left(g\left(\tfrac{\partial}{\partial x^i},\tfrac{\partial}{\partial x^j}\right)\right)} \,d\lambda$$
for any Lebesgue measurable $$S \subset MS \subset M$$, where $$\lambda\lambda$$ denotes Lebesgue measure on $$\mathbb{R}^n\mathbb{R}^n$$. By paracompactness of the manifold $$MM$$, one can cover $$MM$$ by a locally finite open cover of such local coordinate charts, and hence use a smooth partition of unity subordinate to this cover to patch these local scaled pullbacks of Lebesgue measure together into a single measure $$\lambda_g\lambda_g$$, the Riemannian measure [!] on $$MM$$ with respect to $$gg$$, which is a complete $$\sigma\sigma$$-finite measure on the $$\sigma\sigma$$-algebra of Lebesgue measurable sets in $$MM$$.
Let me now describe the basic properties of $$\lambda_g\lambda_g$$.
1. The measure $$\lambda_g\lambda_g$$ is compatible with calculations in local coordinates, in the precise sense that $$\lambda_g(S \cap U) = \lambda_{g,x}(S \cap U)\lambda_g(S \cap U) = \lambda_{g,x}(S \cap U)$$ for any Lebesgue measurable $$SS$$ and any local coordinate chart $$x : U \to x(U) \subset \mathbb{R}^nx : U \to x(U) \subset \mathbb{R}^n$$.
2. If $$g^\primeg^\prime$$ is any another Riemannian metric, then the Riemannian measures $$\lambda_g\lambda_g$$ and $$\lambda_{g^\prime}\lambda_{g^\prime}$$ will be mutually absolutely continuous $$\sigma\sigma$$-finite measures with smooth Radon–Nikodym derivative computable directly in terms of $$gg$$ and $$g^\primeg^\prime$$.
3. Suppose that $$MM$$ is orientable, and let $$\mathrm{vol}_g \in \Omega^n(M)\mathrm{vol}_g \in \Omega^n(M)$$ be the Riemannian volume form defined by $$gg$$. Then for any Riemann integrable $$ff$$ on $$MM$$,
$$\int_M f \, \mathrm{vol}_g = \int_M f \,d\lambda_g, \int_M f \, \mathrm{vol}_g = \int_M f \,d\lambda_g,$$
so that $$\lambda_g\lambda_g$$ really is the (completed) Radon measure on $$MM$$ corresponding to the positive functional $$C_c(M) \ni f \mapsto \int_M f \, \mathrm{vol}_gC_c(M) \ni f \mapsto \int_M f \, \mathrm{vol}_g$$ via the Riesz representation theorem. In other words, integration with respect to $$\lambda_g\lambda_g$$ really is the “Lebesgue-ification” of integration against the top-degree form $$\mathrm{vol}_g\mathrm{vol}_g$$.
Once you’ve constructed the Riemannian measure on your Riemannian manifold $$(M,g)(M,g)$$, the sky is now the limit—you can construct $$L^pL^p$$ and Sobolev spaces of functions, vector fields, differential forms, tensor fields, etc., and in particular, you can use them to study, for instance, the geometric partial differential operators (e.g., generalisations of the Laplacian and the Dirac operator) and their associated partial differential equations (e.g., heat equations) to great mathematical effect. As a mathematical researcher, I’m personally most familiar with the mathematical ecosystem centred around the Atiyah–Singer index theorem, which relates quantities from algebraic topology to functional-analytic computations on Riemannian manifolds, but you should be aware, for instance, that Perelman’s proof of the Poincaré conjecture involved the detailed analysis of a certain highly non-linear PDE for the Riemannian metric itself [!]. Perhaps the most accessible example of these methods in action is Hodge theory, which basically computes the cohomology of a compact manifold in terms of solutions of the Laplace equation (with respect to some Riemannian metric) on differential forms of various degrees.
P.S. People tend to take the extension of Lebesgue theory from $$\mathbb{R}^n\mathbb{R}^n$$ to manifolds more or less for granted, so precise accounts of this can be oddly hard to find. However, a precise if terse account of Lebesgue theory on manifolds can be found in Dieudonné’s Treatise of Analysis, Volume 3, Section 16.22 (especially Theorem 16.22.2 and the following discussion). Dieudonné doesn’t require a Riemannian metric, but the point is that Riemannian metric gives a canonical choice of Lebesgue measure in the sense of Dieudonné, in exactly the same way that it gives a canonical volume form in the orientable case. In fact, Lebesgue measures in the sense of Dieudonné can be identified with nowhere vanishing $$11$$-densities, and the construction of the Riemannian measure $$\lambda_g\lambda_g$$ is really the construction of the canonical $$11$$-density $$\lvert \mathrm{vol}_g \rvert\lvert \mathrm{vol}_g \rvert$$ associated to $$gg$$.
One can define a measurable $$kk$$-form on $$MM$$ to be a map $$\omega : M \to \wedge^k T^\ast M\omega : M \to \wedge^k T^\ast M$$, such that the following hold.
1. For every $$x \in Xx \in X$$, $$\omega(m) \in \wedge^k T^\ast M_m\omega(m) \in \wedge^k T^\ast M_m$$ (i.e., $$\omega\omega$$ is a set-theoretic section of $$\wedge^k T^\ast M\wedge^k T^\ast M$$).
2. For every local coordinate chart $$x : U \to x(U) \subset \mathbb{R}^nx : U \to x(U) \subset \mathbb{R}^n$$, the pullback $$(x^{-1})^\ast \omega : x(U) \to \wedge^k \mathbb{R}^n(x^{-1})^\ast \omega : x(U) \to \wedge^k \mathbb{R}^n$$ defined by
$$(x^{-1})^\ast\omega := \sum_{i_1 < \cdots < i_k} \omega\left(\tfrac{\partial}{\partial x^{i_1}},\dotsc,\tfrac{\partial}{\partial x^{i_k}}\right) dx^{i_1} \wedge \cdots \wedge dx^{i_k} (x^{-1})^\ast\omega := \sum_{i_1 < \cdots < i_k} \omega\left(\tfrac{\partial}{\partial x^{i_1}},\dotsc,\tfrac{\partial}{\partial x^{i_k}}\right) dx^{i_1} \wedge \cdots \wedge dx^{i_k}$$
(with the usual abuses of notation) is measurable; this turns out to be equivalent to requiring that $$\omega(X_1,\dotsc,X_k) : M \to \mathbb{R}\omega(X_1,\dotsc,X_k) : M \to \mathbb{R}$$ be measurable (in the above sense) for any smooth vector fields $$X_1,\dots,X_k \in \mathfrak{X}(M)X_1,\dots,X_k \in \mathfrak{X}(M)$$.
Now, suppose that $$NN$$ is an oriented $$kk$$-dimensional submanifold of $$MM$$ (compact and without boundary, for simplicity), and let $$x : U \to x(U) \subset \mathbb{R}^nx : U \to x(U) \subset \mathbb{R}^n$$ be a local coordinate chart of $$MM$$, such that $$x(N \cap U) = V_{x,N} \times \{0\}x(N \cap U) = V_{x,N} \times \{0\}$$ for some open $$V_{x,N} \subset \mathbb{R}^kV_{x,N} \subset \mathbb{R}^k$$, and such that restriction of $$xx$$ to a diffeomorphism $$N \cap U \to V_{x,N}N \cap U \to V_{x,N}$$ is orientation-preserving. Then we can define
$$\int_{N \cap U} \omega := \int_{V_{x,N}} \omega\left(\tfrac{\partial}{\partial x^{1}},\dotsc,\tfrac{\partial}{\partial x^{k}}\right) \lambda_{\mathbb{R}^k} \int_{N \cap U} \omega := \int_{V_{x,N}} \omega\left(\tfrac{\partial}{\partial x^{1}},\dotsc,\tfrac{\partial}{\partial x^{k}}\right) \lambda_{\mathbb{R}^k}$$
whenever the Lebesgue integral on the right-hand side exists (with $$\lambda_{\mathbb{R}^k}\lambda_{\mathbb{R}^k}$$ the Lebesgue measure on $$\mathbb{R}^k\mathbb{R}^k$$). We can then define $$\omega\omega$$ to be integrable on $$NN$$ whenever it's integrable in this way on $$N \cap UN \cap U$$ for any suitable local coordinate chart $$x : U \to \mathbb{R}^nx : U \to \mathbb{R}^n$$, and then, by exactly the same arguments as in the Riemann integral case, patch these local integrals into a global Lebesgue integral $$\int_N \omega\int_N \omega$$, which turns out to be independent of all the choices of local coordinate chart and partition of unity made along the way. | 3,157 | 10,715 | {"found_math": true, "script_math_tex": 2, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 86, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-06 | latest | en | 0.931016 |
https://www.teachstarter.com/us/teaching-resource/time-conversion-math-maze-worksheets/ | 1,713,016,294,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816734.69/warc/CC-MAIN-20240413114018-20240413144018-00039.warc.gz | 952,444,004 | 74,010 | teaching resource
# Time Conversion Math Maze Worksheets
• Updated: 14 Mar 2024
Convert units of time while finding your way through this set of 3 Time Conversion math mazes.
• Non-Editable: PDF
• Pages: 1 Page
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teaching resource
# Time Conversion Math Maze Worksheets
• Updated: 14 Mar 2024
Convert units of time while finding your way through this set of 3 Time Conversion math mazes.
• Non-Editable: PDF
• Pages: 1 Page
Convert units of time while finding your way through this set of 3 Time Conversion math mazes.
## Practice Converting Days to Hours, Months to Years, and More!
With this math worksheet bundle, students will work their way through 3 different mazes. To find their way through each maze, students will convert between hours, minutes, days, and more! Students will practice the following skills:
• Converting Weeks to Days
• Converting Hours to Minutes
• Converting Days to Hours
• Converting Months to Years
• Converting Years to Days
• Converting Months to Days
• And more!
Students will convert the unit within a rectangle and follow the path with the correct answer. Students can shade the path with colored pencils, crayons, etc. But watch out! Dead ends are waiting for your students in some of the mazes.
If you are looking to incorporate a little competitive fun into your classroom, consider assigning this resource to a pair of students. When someone says, “Go!” students can race each other through the maze and check their answers at the end with the provided answer key.
Use the dropdown icon on the Download button to choose between the PDF or editable Google Slides version of this resource.
Because this resource includes an answer sheet, we recommend you print one copy of the entire file. Then, make photocopies of the blank worksheet for students to complete.
This resource was created by Brittany Collins, a teacher in Indiana and Teach Starter collaborator.
Don’t stop there! We’ve got more activities to shorten your lesson planning time:
### teaching resource
#### Converting Units of Length – Error Analysis Worksheet
Analyze length conversions to identify and correct mistakes with this error analysis worksheet.
### teaching resource
#### Converting Units of Length - Math Mazes
Strengthen measurement conversion skills while converting units of length with this set of math mazes.
### teaching resource
#### Riddle Me! Converting Units of Length – Worksheet
Use ratio reasoning and measurement conversion skills to complete a set of 3 riddle worksheets. | 550 | 2,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-18 | latest | en | 0.863202 |
https://www.scribd.com/document/7900368/Digital-Electronics | 1,516,539,501,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890582.77/warc/CC-MAIN-20180121120038-20180121140038-00190.warc.gz | 984,569,787 | 33,609 | # INTERFACING.
PPT(10/01/2008)
2.1
INTERFACING.PPT(10/01/2008)
2.2
Lecture 2
Synchronous Bit-Serial Transmission
FRAME
Synchronous Bit-Serial Interfacing
A
Objectives
CLOCK
DATA
B
Explain how data is sent between two digital systems using a synchronous bit-serial protocol – Synchronous: same clock at transmitter & receiver – Bit-serial: Only one bit sent at a time – Protocol: The procedure for exchanging information Explain the meaning of setup and hold times Investigate the timing constraints in a transmission system
CLOCK FRAME DATA B senses
1 0 0 0 0 1 1 0 0 0 0 1 0 0 1 0 134 18 0 1 0
• •
Transmitting 8 bit values from A to B: – FRAME indicates the first bit of each value; the other 7 bits follow on consecutive clock cycles. The FRAME signal is often called a frame sync pulse. – DATA changes on the falling CLOCK edge – Propagation delays are often omitted from diagram. – DATA is sensed by system B on the rising CLOCK edge to maximise tolerance to timing errors. We must always clock a flipflop at a time when its DATA input is not changing.
Page 2.1
Interfacing
E.2.12/ISE 2.2 – Digital Electonics II
INTERFACING.PPT(10/01/2008)
2.3
INTERFACING.PPT(10/01/2008)
2.4
Transmission Delays
Timing Specifications
A DA tD DB CB B
Propagation speed = (L0C0)–½ where L0 and C0 are inductance and capacitance per unit length. For a uniform line this gives a total delay of (LC)½ where L and C are the total inductance and capacitance. Any additional load capacitance will increase delay. Signal speed can be expressed in terms of: – the speed of light (c = 30 cm/ns) – the geometry of the wiring – the relative permittivity of the insulator:
CLOCK
Time: CA CB DA DB Time:
CA
1D
C1
C1
tC
0
tP
½T
0
tP+tD
½T+tC
Examples: – Coax cable: c × εr–½ ⇒ 20 cm/ns for εr =2.3 (teflon) – PCB with ground plane: 1.4c × (1.4+εr )– ½ cm/ns ⇒ 17 cm/ns for εr =5 (fibreglass)
tP T t C, t D
Propagation delay for device A. Clock Period. Transmission line delays for CLOCK and DATA
For Device B: • • Data input changes at time tP+tD Clock input changes ↑ at time ½T+tC
Rule-of-thumb: Data travels along typical wires and circuit board tracks at about 15 cm/ns: half the speed of light.
Page 2.2
Interfacing
E.2.12/ISE 2.2 – Digital Electonics II
INTERFACING.PPT(10/01/2008)
2.5
INTERFACING.PPT(10/01/2008)
2.6
Setup and Hold Times
A
Timing Constraints
DA tD DB CB B
The DATA input to a flipflop or register must not change at the same time as the CLOCK.
CLOCK
tH tS
CA
1D
C1
C1
tC
0 CA DA ½T T
DATA CLOCK
1D C1
Q
CLOCK DATA Q tP
DB CB
Setup Time: DATA must reach its new value at least tS before the CLOCK↑ edge. Hold Time: DATA must be held constant for at least tH after the CLOCK↑ edge.
For Device B: • •
tP+tD
½T+tC
T+tP+tD
Data input (DB) changes at tP+tD (and T+ tP+tD ) Clock↑ (CB) at time ½T+tC
Typical values for a register: tS = 5 ns, tH =3 ns The setup and hold time define a window around each CLOCK ↑ edge within which the DATA must not change. If these requirements are not met, the Q output may oscillate for many nanoseconds before settling to a stable value.
For reliable operation:
• • Setup Requirement: tP + tD + tS < ½T + tC Hold Requirement: ½T+tC + tH < T + tP + tD
Get a pair of inequalities for each flipflop/register in a circuit. You never get both tS and tH in the same inequality.
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Example Values
A CA CLOCK
C1
Propagation Delay Constraint Inequalities
DB CB B CA CLOCK A
C1
DA
tD
1D
DA
tD
C1
DB CB
1D
B
C1
tC
tC
For Motorola 56001 27MHz DSP processor:
0 < tP < 50 ns, tS = 12 ns, tH = 27 ns Suppose differential delay: –10 < (tD – tC ) < +10
Find maximum CLOCK frequency (min CLOCK period):
When do they arise
• Whenever a flipflop’s clock and data input signals originate from the same ultimate source. Here CB and DB both originate from CLOCK. You normally get two inequalities for each flipflop in a circuit.
max (tP + tD ) + tS < min ( ½T + tC ) 50 + 10 + 12 < ½T + 0 (tD =10, tC =0) ½T > 12 + 50 + (+10) = 72 ⇒ T > 144 ns max ( ½T+tC) + tH < min( T + tP + tD ) ½T + 10 + 27 < T + 0 + 0 (tD =0, tC =10) ½T > 27 + 10 = 37 ⇒ T > 74 ns
Relationship beween setup and hold inequalities:
• • • Setup Requirement: Hold Requirement:
tP + tD + tS < ½T + tC ½T+tC + tH < tP + tD + T
To get the Hold inequality you change tS to tH , swap the sides of the other terms and add T onto the right side. No. For a <, take max of everything on the left and min of everything on the right. max = most positive: for example, max(–7,–2) = –2
Are both tS and tH ever in the same inequality?
Hence fCLOCK < 1/144 = 7 MHz To test for worst case: make the left side of the inequality as big as possible and the right side as small as possible.
How do you decide to take the max or the min?
• •
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Quiz Questions
1. What is a bit-serial transmission system? 2. What is a synchronous transmission system? 3. In a synchronous transmission system in which the transmitted data changes on the rising edge of the CLOCK, why is it normal for the receiver to sense the data on the falling edge of the CLOCK ? 4. What is the purpose of the frame sync signal In a synchronous bit-serial transmission system? 5. How far does a signal travel along a typical wire in one nanosecond? 6. What do the terms setup time and hold time mean? 7. Why do you get a pair of timing inequalities for each flipflop or register in a circuit? 8. In formulating the timing inequalities, how do you choose what to use for a quantity whose value may lie anywhere within a particular range?
Answers are all in the notes.
Lecture 3
Asynchronous Bit-Serial Interfacing
Objectives
• • • •
Explain how data is sent between two digital systems using an asynchronous bit-serial protocol Explain why it is necessary to include START and STOP bits in an asynchronous protocol. Explain the circuitry needed for an asynchronous bitserial receiver Derive the tolerances for the transmitter and receiver clocks in an asynchronous bit-serial system
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Asynchronous Bit-Serial Transmission
Combine timing and data into a single signal to circumvent differential delays over long distances (saves wires too). RS232: Serial Port on a PC
10T
Good time resolution ⇒ use a master clock period of T/16:
DATA CLOCK cycle: 0 24 40 56 72 88 104 120 136 (152→0)
• Middle of STOP bit is after 9½ bitcells ⇒ 9½ × 16 = 152 master clock cycles.
0 0 1 Stop Bit
1 Start Bit
0
1
1
0
• Use ÷152 counter but hold it at 0 until the START bit arrives ⇒ inhibit counting whenever CT=0 and DATA=1. • Counter will increment either if DATA=0 or if CT is already non-zero. • We use a clock enable input, G1, to control whether or not the counter increments; much better design than using gates to modify the clock signal.
CTR DIV 152 CLOCK (16 × baud rate) DATA 1+ CT=0 G1 CT0:7
8 Data Bits (LSB first)
• Idle state has signal = 1. START bit indicates new byte. • Data transmitted LSB first (above example equals 14110) • Bit cell duration is T: e.g. T=52 µs, 1/T = 19200 baud • STOP bit needed to ensure signal goes to 1 before the next START bit which might follow immediately. • Signal is decoded by sampling each bit in the centre of its cell an appropriate time after the start bit:
&
ZERO
Time/T: 0
1½ 2½ 3½ 4½ 5½ 6½ 7½ 8½
The CT=0 output from counter goes high when the contents of the counter, CT, are zero. Generate this signal using a NOR gate connected to all 8 counter outputs.
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DATA CLOCK cycle: 0 24 40 56 72 88 104 120 136 (152)
DATA
Double Buffering
151
} 0
CLOCK cycle:
0
24 40 56 72 88 104 120 136
24 40
• Use an 8-bit shift register to store the data value. Only allow it to shift when CT = 24, 40, … , 136. • The decode logic output, MID, goes high when the counter has one of these values (all odd multiples of 8 ⇒ four LSBs = 10002). • Notation: – The shift register clock has two functions separated by a /: 2C1 clocks first bit, 2→ shifts the rest. – Both these functions are controlled by the clock enable input, G2.
SRG
CLOCK (16 × baud rate) MID 2C1/2→ CT=24,...,136 CT0:7 see previous slide for details Decode Logic
• The 8 data bits only stay in the shift register for 3T before they get shifted out again by the next data byte. • Host microprocessor must respond to an interrupt within this time and retrieve the data. • Use a second register to grab the data at T=151 and keep it for a whole 10T. This gives the µP more time.
TRANSFER
SRG
CLOCK (16 × baud rate) 2C1/2→ 2C1
Timing Circuit
G2 1D
Q7 Q6 Q5 Q4 Q3 Q2 Q1 Q0
CT=151 CT=24,...,136 Timing Circuit & Decode Logic MID
G2 1D
Q7 Q6 Q5 Q4 Q3 Q2 Q1 Q0
G2 1D
Z7 Z6 Z5 Z4 Z3 Z2 Z1 Z0
DATA
DATA
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Timing Errors
• Ideal situation: – Receiver clock period P = T/16 – Counter starts counting exactly on DATA falling edge
CT will change to 1 on the first CLOCK ↑ edge after DATA goes to 0: P≈T/16 time = 0 Δ
DATA CLOCK CT 0 1 2 3
Neglecting logic propagation delays, 0 < Δ < P where P is receiver clock period. – Count 0→1 a time Δ after the START bit – Count n → n+1 a time nP+ Δ after the START bit • Real situation: – Receiver clock period not exactly T/16 – Counter starts with some delay
• On first rising edge of P after DATA goes low
Timing in the last (MSB) bit cell:
8T DATA Counter: CT MID 135 136 137 Sample Instant 9T
P slightly too small
136P+Δ
We will sample the correct bit cell if: 8T < 136P+ Δ < 9T
8T < 136P+0 136P+P < 9T
P much too small
⇒ T/P < 17 ⇒ T/P > 15.2
Hence T/P = 16 +6.3% –5.0% which implies a clock accuracy of around ± 2.5% at transmitter and receiver.
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Quiz Questions
1. How can you be sure that in the RS232 protocol there will always be a high-to-low transition at the beginning of each transmitted byte ? 2. What is the function of the clock enable input on a counter or register? 3. What logic gate is needed to detect when the contents of a counter is equal to zero? 4. If an asynchronous protocol has one START bit, eight data bits and one STOP bit, how may bitcell periods is it from the beginning of the START bit until the centre of the STOP bit? 5. What is the function of an input pin that is labelled 2C1/2→? 6. If the CLOCK input of a counter has period P, what is the range of possible delays between the counter’s enable pin going high and the counter incrementing? 7. What is the purpose of double buffering the data in an asynchronous bit-serial receiver? 8. How can you tell if a binary number is an odd multiple of 16?
Answers are all in the notes.
Lecture 4
Microprocessor to Memory Interface
Objectives
• • •
Explain how memory is connected to a microprocessor Describe the sequence of events in reading from and writing to a static RAM Describe the structure and input/output signals of a static RAM
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Microprocessor Memory Map
A typical 8-bit microprocessor has • A 16-bit address bus, A15:0 – Can have up to 216=65536 memory locations – Value is usually written in hexadecimal often with \$ prefix:
• e.g. \$1000 = 212 = 4k = 4096
Microprocessor ↔ Memory Interface
µP A A15:0 16 16 A Memory
An 8-bit data bus, D7:0 – Each data word in memory has 28 = 256 possible values
D7:0 D∇
\$FFFF \$F000 \$EFFF ROM 16k words Input/Output
8
D∇ 8
We can tell which region of memory an address is in by inspecting the top few bits: A15:12
CLOCK
Control Signals
Control Signals
Unused
\$7FFF
RAM 32k words
\$0000
F: E: D: C: B: A: 9: 8: 7: 6: 5: 4: 3: 2: 1: 0:
1111 1110 1101 1100 1011 1010 1001 1000 0111 0110 0101 0100 0011 0010 0001 0000
Input/Output ROM ROM ROM ROM
During each memory cycle: • A15:0 selects one of 216 possible memory locations
RAM RAM RAM RAM RAM RAM RAM RAM
• D7:0 transfer one word (8 bits) of information either to the memory (write) or to the microprocessor (read). • D7:0 connections to the microprocessor are tri-state (∇): they can be: – “logic 0”, “logic 1” or “high impedance” (inputs) • The control signals tell the memory what to do and when to do it.
INOUT = ROM = RAM =
A15·A14 ·A13 ·A12 A15·A14 ·!(A13 ·A12) + A15·!A14·A13·A12 !A15
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Memory Interface Control Signals
Memory Chip Selection
µP A A15:0 16 16 A M em ory
D 7:0 D∇ 8 M CLO CK 8
D∇
W RITE C LO C K
Control signals vary between microprocessors but all have: • Each memory circuit has a “chip enable” input (CE) • The “Decoder” uses the top few address bits to decide which memory circuit should be enabled. Each one is enabled only for the correct address range: RAM = !A15 ROM = A15·(A14 ·!(A13 ·A12) +!A14·A13·A12) INOUTx = A15·A14·A13·A12·!A11·A10·!A9·A8· !A7·A6·A5·A4·!A3·A2 • INOUTx responds to addresses: \$F574 to \$F577 other I/O circuits will have different addresses
D7:0
• A clock signal to control the timing (can be the same as the system CLOCK) • A signal to say whether the microprocessor wants to read from memory or to write to memory – Must make sure that D7:0 is only driven at one end
Read Cycle MCLOCK A15:0 WRITE from µP from mem Write Cycle
• Low n address bits select one of locations within each memory circuit (value of n depends on memory size)
2n
D7:0 from memory only allowed when MCLOCK·!WRITE true
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Memory Circuit Control Signals
Static RAM: Dynamic RAM: Data stored in bistable latches Data stored in charged capacitors: retained for only 2ms. Less circuitry ⇒ denser ⇒ cheaper.
8k × 8 Static RAM
RAM 8192 × 8
Read Cycle MCLOCK A15:0 WRITE from µP D7:0 from mem Write Cycle
CE 0 1 1 1
OE ? 0 1 ?
WR ? 0 0 1
D0:7 Hi Z Hi Z Out In
A12:0 WR OE CE
A WR OE CE ∇ D7:0
• Output enable: OE = MCLOCK·!WRITE turns on the D7:0 output from the memory • Write enable: WE = MCLOCK·WRITE writes new information into the selected memory location with data coming frommicroprocessor • Chip enable: comes from the decoder and makes sure the memory only responds to the correct addresses
Tri-state output: Low, High or Off (High Impedance). Allows outputs from several chips to be connected; Designer must ensure only one is enabled at a time. Chip Enable: disabling chip cuts power by 80%. Output Enable: Turns the tri-state outputs on/off. Address: selects one of the 213 8-bit locations. Write: stores new data in selected location Data in for write cycles or out for read cycles.
CE OE A12:0 WR D7:0
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8k × 8 Static RAM
The 64k memory cells are arranged in a square array:
256 Cells D7 D6 D5 D4 D3 D2 D1 D0
Quiz Questions
1. What is the memory map of a microprocessor system 2. Why do all microprocessor systems include some readonly memory (ROM) 3. What does it mean if a digital device has a tri-state output? When are such outputs necessary ? 4. What is the difference between the chip enable and the output enable inputs of a static RAM? 5. If a static RAM has n address inputs and m data outputs, how many bits of information does it store? 6. What is the binary value of the three most significant address bits for the hexadecimal address \$BC37 ?
8 × 32 = 256 cells
32 cells
Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Bit 0
For each output bit, an 8192-way multiplexer selects one of the cells. The control signals, OE, CE and WR determine how it connects to the output pin via buffers:
256 cells
32 cells
Answers are all in the notes.
(8192-way multiplexer)
A12:0 CE•WR
1
1
Dn
CE•OE•!WR
Occasionally DIN and DOUT are separate but ⇒ more pins
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Lecture 5
8k × 8 Static RAM
Microprocessor to Memory Interface
Objectives
A12:0 A
RAM 8192 × 8
CE 0 1 1 1
OE ? 0 1 ?
WR ? 0 0 1
D0:7 Hi Z Hi Z Out In
Investigate the timing constraints for a microprocessor when reading from or writing to memory.
WR OE CE
WR OE CE
D7:0
Tri-state output: Low, High or Off (High Impedance). Allows outputs from several chips to be connected; Designer must ensure only one is enabled at a time. Address: selects one of the 213 8-bit locations. Data in for write cycles or out for read cycles. Chip Enable: disabling chip cuts power by 80%. Output Enable: Turns the tri-state outputs on/off. Write: stores new data in selected location
A12:0 D7:0 CE OE WR
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Memory Write Cycle
>5 >30 >5
CE•OE•!WR A12:0 D7:0 >5 <35
High, Low Note: Time axis not to scale
>5 <10 (20) <20 (35)
Input { Don't CareOutput Unknown
A12:0 CE•WR CE•OE•!WR D7:0 <10 >15 >2 >5
Constant, Hi Z
A write cycle happens whenever CE•WR is true. CE•WR must go high for at least 30 ns. To avoid writing to unwanted locations, the address, A12:0 must remain constant for at least 5 ns at both ends of the write pulse. Input data D0:7 only matters at the end of the write pulse. Setup & hold times of 15 ns & 2 ns define a window within which it must not change. Input When CE•OE•!WR goes high, the memory reverts to read mode. The input data must be removed from D7:0 before this happens. • Timing specifications that end on an output are guarantees from the chip manufacturer (shown in black). • Timing specifications that end on an input are requirements that the designer must meet (shown blue).
A read cycle happens when CE•OE•!WR is true. If A12:0 changes, D7:0 remains for at least 5 ns and goes to new value within 35 ns. Rubbish in between even if new and old locations contain the same value. If a read cycle ends due to OE going low, the outputs go Hi-Z within 10 ns If a read cycle starts due to OE going high, D7:0 stays Hi-Z for at least another 5 ns and the selected word appears within 20ns You can use CE instead of OE but it is slower: 20 ns to turn off and 35 ns to turn on (in parentheses on timing diagram). When reading data, the propagation delay to the D7:0 outputs is called the RAM’s access time: 35 ns from A12:0 and 20 ns from OE.
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Microprocessor ↔ Memory Interface
µP A A15:0 16 A15:13 A12:0 13 A RAM 8k × 8
Microprocessor Write to Memory
0 MCLOCK A12:0 WRITE D0:7 OE WR >15 >2 >5
RAM 8k × 8
<181
250
255 <378
505 500 >533
D∇
8
D7:0
8
D∇
WRITE
1
&
1
OE CE WR
RAM Requirements:
µP A D∇
>5
>30
A12:0 D7:0
MCLOCK
&
A D∇
OE = MCLOCK • !WRITE
WR = MCLOCK • WRITE
WRITE
1
& &
OE
• Reading or writing takes place during the second half of the clock cycle when MCLOCK is high.
MCLOCK
WR
• WRITE output from µP determines whether WR or OE goes high. Assume NAND gate delay = 5 ns.
0 MCLOCK >33 A15:0 WRITE WR or OE 0 5 255 505 <181 250 500
• µP emits data within 128ns of MCLOCK↑ • Requirements: – Addr→WR setup: 181+5 < 255 – Data →!WR setup: 378+15 < 505 – WR pulse: – Addr hold: – Data hold:
255+30 < 505 505+5 < 533 505+2 < 533
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0 MCLOCK A12:0 WRITE WR OE D0:7 µP Requirements: >30 >10,<83 <181 250 255 <275 >505 500 505 <515
0 MCLOCK A12:0 WRITE WR OE D0:7 µP Requirements: >30 >10,<83 <181 250 255 <275 >505 500 505 <515
• Data Setup time: 30 ns before MCLOCK↓. – Three paths must be satisfied – Check each one individually
µP A D∇ WRITE A12:0 D7:0 A D∇ RAM 8k × 8
• D7:0 must go tristate 10 to 83 ns after MCLOCK↓. – MCLOCK path is the only relevant one – OE↓ to tristate delay varies between 0 and 10 ns
µP A D∇ WRITE A12:0 D7:0 A D∇ RAM 8k × 8
1
& &
OE
1
& &
OE
WR MCLOCK
MCLOCK
WR
• Requirements: – Addr to Data setup:
• Requirements: – Min hold: – Max hold:
181+35+30 < 500 – WRITE to Data setup: 181+5+5+20+30 < 500 – MCLOCK to Data setup: 250+5+20+30 < 500
505 > 500+10 515 < 500+83
May need to add some delay to !OE signal to meet min hold
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Quiz Questions
1. What is the access time of a static RAM? 2. When writing to a static RAM, why is does the state of the data inputs matter only at the end of the write pulse? 3. How do you check timing constraints if the manufacturer specifies a maximum propagation delay but no minimum ? 4. How do you check timing constraints if the validity of an output depends on several of the input signals ?
Answers are all in the notes.
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How Many Steps Is 25 Meters? Update
# How Many Steps Is 25 Meters? Update
Let’s discuss the question: how many steps is 25 meters. We summarize all relevant answers in section Q&A of website 1st-in-babies.com in category: Blog MMO. See more related questions in the comments below.
Table of Contents
## How many steps is a meter?
›› More information from the unit converter
See also How To Install Apk Files On Iphone Without Jailbreak? Update
How many step in 1 meters? The answer is 1.3123359580052. We assume you are converting between step and metre.
## How many steps is 15 meters walking?
The simplest mental act to do, is to divide by 4 (divide by two and again divide by two) and then multiply by three. If you walk 8 steps you walk 8/4*3=6 meters; 12 steps, you get 9 meters; 20 steps shall be 15 meters and so on.
### How far IS 10 metres?
How far IS 10 metres?
How far IS 10 metres?
## How many Metres is 1 foot steps?
1 metre is equal to 1 m, or 1.3123359580052 steps.
## How long is a step?
An average person has a stride length of approximately 2.1 to 2.5 feet. That means that it takes over 2,000 steps to walk one mile and 10,000 steps would be almost 5 miles. A sedentary person may only average 1,000 to 3,000 steps a day.
## How do you measure 20 meters in feet?
You can easily convert meters into feet by using a formula and conversion factor. 20 meters x 3.28 = 65.6 feet.
See also How Long Does It Take For Paper Towels To Dissolve? New
## How long does 200 Metres take to walk?
Although a little slower than normal, this is a reasonable time period for someone to walk 200 metres and therefore Mr X can complete the activity in a reasonable time period’. This again does not make sense, as at 30 minutes per minute it would take Mr X at least 6.66 minutes to walk 200 meters.
## How many steps am I supposed to take a day?
How many steps should you take a day? A 2011 study found that healthy adults can take anywhere between approximately 4,000 and 18,000 steps/day, and that 10,000 steps/day is a reasonable target for healthy adults.
### Physics 8 Work, Energy, and Power (20 of 37) Climbing Stairs
Physics 8 Work, Energy, and Power (20 of 37) Climbing Stairs
Physics 8 Work, Energy, and Power (20 of 37) Climbing Stairs
## How many steps is 1km?
1 km =1312.33595801 steps.
## How many feet is a step?
According to the University of Iowa, the average person’s walking step length is 2.5 feet (30 inches), so the average stride length would be approximately 5 feet (60 inches).
## How long is the average step in meters?
How many Meters are in a Step? The answer is one Step is equal to 0.762 Meters.
See also How To Apply Foot Powder? Update New
## How do I calculate my walking steps?
Count your steps for that length and then multiple by ten, if counting for a tenth of a mile, and you will have your distance for a mile. If counting for a quarter of a mile then multiply that number by four and you will have your steps for a full mile.
## How long is a stair step?
Most stair treads measure between 8 and 10 inches long, with longer treads being used on larger staircases. Nine-inch stair treads are the most common lengths.
## How long does it take to walk 10m?
For example: if you did a 10-meter gait speed test and it took you 7 seconds, the equation would like: 10 meters / 7 seconds = 1.4 meters per second.
Normative Values.
Age (Men and Women) Average Walking Speed (m/s)
40-49 1.43-1.39 m/s
50-59 1.43-1.31 m/s
60-69 1.34-1.24 m/s
70-79 1.26-1.13 m/s
### 11.8 Seconds – 25 Meter Freestyle Sprint
11.8 Seconds – 25 Meter Freestyle Sprint
11.8 Seconds – 25 Meter Freestyle Sprint
## What is considered walking distance?
Walking distance is an important concept in the fields of transportation and public health. A distance of 0.25 miles is often used as an acceptable walking distance in U.S. research studies.
## How many meters can you walk in 10 minutes?
It is the standard used in real estate listings. For example, if a building is a 10-minute walk from a particular park or train station, it is 800 meters away.
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## Information related to the topic how many steps is 25 meters
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You have just come across an article on the topic how many steps is 25 meters. If you found this article useful, please share it. Thank you very much. | 1,242 | 4,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-40 | latest | en | 0.941048 |
https://www.jobilize.com/course/section/circle-the-right-word-toys-by-openstax?qcr=www.quizover.com | 1,579,270,928,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250589560.16/warc/CC-MAIN-20200117123339-20200117151339-00537.warc.gz | 950,947,872 | 19,023 | # 3.1 Toys
Page 1 / 1
## Circle the right word.
The bear costs a little / much money
The doll costs a little / much money
The train costs a little / much money
## Which toys do you like best?
Draw a picture of any of the old toys.
## My puppet:
• Draw a face for your puppet.
• Use different shapes, like a triangle, circle and square.
## Assessment
Learning Outcomes(LO’s) EMSLO 1 THE ECONOMIC CYCLE The learner will be able to demonstrate knowledge and understanding of the economic cycle within the context of ‘the economic problem’. We know this when the learner: 1.4 starts understanding that goods (toys, food, etc. ) have a price. ARTS AND CULTURELO 1 CREATING, INTERPRETING AND PRESENTING The learner will be able to create, interpret and present work in each of the art forms. We know this when the learner: 1.10 ( visual arts ) - discovers simple geometric shapes such as circles, lines, triangles and squares and combines and arranges them in patterns.
## Memorandum:
Toys may be exhibited in class.
Make sure that you have enough magazines available in class.
ENRICHMENT:
• Show the learners the video of Toy Story. Let them draw their favourite character.
• Discuss the characteristics of the character (friendly, inconsiderate, caring, etc.)
• Visit a toy shop.
• Compare the prices of the different toys.
• Arrange a fashion show with Barbie or other toys.
• Allow the learners to dress in the clothes of different characters, e.g. Superman, Cowboy, etc.
• Arrange a toy car race – work in groups.
• Organise a series of game stations, e.g. top throwing, rope skipping, hop scotch, etc. Learners move on in a circuit, getting the opportunity to play each of the games for a specific time.
• Group games, e.g. drop-the-handkerchief (Afr. vroteier), blind-man’s buff, cat and mouse, etc.
• Make a puppet and have your own puppet theatre.
what is the stm
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
Got questions? Join the online conversation and get instant answers! | 1,081 | 4,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-05 | longest | en | 0.896477 |
https://forums.theregister.co.uk/post/reply/3948934 | 1,582,773,004,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146643.49/warc/CC-MAIN-20200227002351-20200227032351-00137.warc.gz | 372,781,964 | 14,906 | # Reply to post: Re: 2nd question
### H0LiCOW: Cosmoboffins still have no idea why universe seems to be expanding more rapidly than expected
#### Re: 2nd question
If my crap maths is anything to go by, the answer is approximately 11.25km...
73km/sec/megaparsec is expansion rate
1 megaparsec is 3.08*10^19 km
1000 years = 3.16*10^10 seconds (with some rounding to 2dp)
Expansion = 73.3 km * 3.16E+10 seconds * (150,000,000 / 3.08E+19) megaparsecs
So, ignoring the fact that the earth and the sun are linked via gravity cos they're relatively close, and so this probably doesn't apply, this is saying that the distance between the earth and the sun (~147 million km) will increase by 11.25km over the period of 1000 years due to the universe expanding
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Biting the hand that feeds IT © 1998–2020 | 275 | 978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-10 | latest | en | 0.933592 |
https://www.blogbyben.com/2015/01/the-most-powerful-computer-youll-never.html | 1,713,492,255,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817253.5/warc/CC-MAIN-20240419013002-20240419043002-00116.warc.gz | 600,813,372 | 23,154 | ## Monday, January 12, 2015
### The Most Powerful Computer You'll Never Own
Inspired by the movie the Imitation Game, I just had build my own Turing Machine (I know, I should have built one of these! Some day). While folks have built actual physical machines, I took the programmer's way out and made mine strictly virtual.
A Turing Machine is the theortical device Alan Turing devised in his 1936 paper on computability. To this day, Computer Science students are made to do battle with this machine and the theories that stem from it.
While it's been years since I've done any CS homework that involved Turing Machines, I can't recall actually ever implementing one. What a shame.
The web is filled with such simulators including: here, here and here. These were invaluable for refreshing my memory as to how a Turing Machine operates.
Implementing a Turing Machine was oddly satisfying. One has to contend with the fact that a program handed to a Turing Machine may never complete. I dealt with this by allowing the user to execute the machine in terms of blocks of cycles. So you create the machine, run it for 100 cycles, and if you want, run it 100 more.
Here's the code to create a machine. The program is actually Alan Turing's first example:
```(define r1 '(((b _) (c 0 >))
((c _) (e _ >))
((e _) (f 1 >))
((f _) (b _ >))))
(define tm1 (make-tm 'b '() '() r1))
```
The rules specified by r1 are in the format:
``` (((current-state current-symbol) (new-state new-symbol new-direction))
...)
```
Invoking make-tm creates the new Turing Machine. You provide as arguments the initial state, initial tape-contents, list of halting states and the transition rules.
Once you've got the machine (tm1, above), you can invoke it with an integer to say how many cycles of the machine should execute (e.g., (tm1 10)).
Scheme truly shined as a host language: I was trivially able to make the machine operate on arbitrary states and tape contents. Also sexpr's were ideal for defining the state transition rules.
All of this may seem fairly unimpressive until you ponder this thought:
However odd it may sound for so simple a machine, the Turing Machine is the most powerful computing model known to computer scientists. In this context, "powerful" refers only to what it is capable of doing, not to how fast or efficiently it does it. It has been proven that a TM is capable of performing any computation that a modern computer can, given enough time. Infact, it is technically MORE powerful than modern computers, since it has no storage limitations.
Here's the complete source code for my simulator.
Update: Programming Praxis tackled the Turing Machine almost 5 years ago. I've updated my code below with their example.
```; A Turing Machine Impl
(define (range lower upper)
(if (>= lower upper) '()
(cons lower (range (+ 1 lower) upper))))
(define (show . words)
(for-each display words))
(define tape-cell '#(_))
(define tape-padding '#(_ _ _ _ _))
(define (make-tape contents)
(cons 0 (vector-append (list->vector contents) tape-padding)))
(car t))
(define (tape-items t)
(cdr t))
(define (tape-write t x)
(vector-set! (tape-items t) (tape-head t) x)
t)
(define (tape-move t direction)
(case direction
((< L) -1)
((> R) 1)
((- N) 0)
(else (error "Unknown tape movement: " direction))))))
(cond ((= index -1)
(set-cdr! t (vector-append tape-cell (tape-items t)))
(set! index 0))
((= index (vector-length (tape-items t)))
(set-cdr! t (vector-append (tape-items t) tape-padding))))
(set-car! t index)
t))
(define (display-tape t)
(for-each (lambda (i)
(let ((v (vector-ref (tape-items t) i)))
(show "[" v "]"))
(else (show v)))
(show " ")))
(range 0 (vector-length (tape-items t)))))
(define (tm-find state tape rules)
(let ((needle (list state (tape-read tape))))
(let loop ((rules rules))
(cond ((null? rules) #f)
((equal? needle (caar rules)) (cadar rules))
(else
(loop (cdr rules)))))))
(define (make-tm initial-state initial-tape halting-states rules)
(let ((state initial-state)
(tape (make-tape initial-tape)))
(lambda (cycles)
(let loop ((cycle 0))
(show state ": ")
(display-tape tape)
(newline)
(cond ((= cycle cycles) #t)
((member state halting-states) state)
((tm-find state tape rules) =>
(lambda (dest)
(let* ((new-state (car dest))
(set! state new-state)
(tape-write tape new-symbol)
(tape-move tape new-direction)
(loop (+ 1 cycle)))))
(else 'no-matching-rules))))))
; Binary Counting - http://aturingmachine.com/examples.php
(define r0 '(((walk 0) (walk 0 >))
((walk 1) (walk 1 >))
((walk _) (count _ <))
((count 0) (walk 1 >))
((count 1) (count 0 <))
((count _) (walk 1 >))))
(define tm0 (make-tm 'walk '(0) '() r0))
; Turing's First Example
; http://en.m.wikipedia.org/wiki/Turing_machine_examples
(define r1 '(((b _) (c 0 >))
((c _) (e _ >))
((e _) (f 1 >))
((f _) (b _ >))))
(define tm1 (make-tm 'b '() '() r1))
; http://programmingpraxis.com/2009/3/27/a-turing-machine-simulator/
(define r2 '(((0 1) (0 1 R))
((0 +) (1 1 R))
((1 1) (1 1 R))
((1 _) (2 _ L))
((2 1) (H 1 N))))
(define tm2
(make-tm 0
'(1 1 1 + 1 1 1 1)
'(H)
r2))
``` | 1,365 | 5,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.947071 |
https://www.physicsforums.com/threads/help-with-this-notation-some-sort-of-norm.906435/ | 1,508,683,058,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825264.94/warc/CC-MAIN-20171022132026-20171022152026-00259.warc.gz | 1,006,099,435 | 17,465 | # Help with this notation -- some sort of norm?
1. Mar 4, 2017
I need help understanding this notation, what does this mean?
Squared of 2-norm?
1. The problem statement, all variables and given/known data
Thanks
2. Mar 4, 2017
### StoneTemplePython
yes. Say $\mathbf b = \mathbf{y - Xw}$. I assume that b is real valued for this example. Say you want to minimize the euclidean length of $\mathbf b$. You write that as $min \big(\mathbf b^T \mathbf b\big)^\frac{1}{2}$. But square roots are unpleasant to work with, so you then recognize that if you minimize the squared euclidean length of $\mathbf b$ then that also must minimize the euclidean length of $\mathbf b$. (Why must this be the case?). Hence you recover the problem above that reads as: $min \big(\mathbf b^T \mathbf b\big)$.
What you have there is the setup for the Normal Equations, and doing ordinary least squares estimations. There are two approaches to deriving the solution for an over-determined system of equations -- one involves calculus and the other involves wielding orthogonality. Both approaches are worth understanding and thinking on.
3. Mar 5, 2017
I am learning Machine Learning by myself. I have BSEE but I am encountering many symbols/notations that I don't understand.
For example, what does the "1 with a vertical line through its back" mean?
I know as far as E stands for expected value.
Thanks
4. Mar 5, 2017
### StoneTemplePython
You'
My recommendation is to find a good text like "Learning From Data" and learn from that text (plus its echapters). (The book is quite cheap at $30 in the US, though due to peculiarities with licensing,$100 in Canada?) There is an associated free course with the same title at work.caltech.edu, and also on itunes store.
More to the point: a good text will have an appendix that lists and defines all the notation that it uses. Unfortunately notation is not standardized or uniform between authors.
A $\mathbf 1$ will tend to mean a ones vector or an indicator function or sometimes even the identity matrix. Here it is an indicator function. I personally prefer a $\mathbf 1$ to mean ones vector, $\mathbf I$ to mean identity matrix, and $\mathbb I(Y = 1)$ to denote an indicator function, but the fundamental issues is non-homogeneity of notation in the space --- again my solution is that you can homogenize things when starting off by picking one good source to learn from that has its own consistent notation. (Then once you've mastered that one source, you can much more easily infer / guess other people's notation as your branch out.)
Good luck. | 623 | 2,595 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-43 | longest | en | 0.916688 |
https://www.physicsforums.com/threads/getting-in-a-muddle.159658/ | 1,477,040,825,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718034.77/warc/CC-MAIN-20161020183838-00514-ip-10-171-6-4.ec2.internal.warc.gz | 976,735,878 | 14,880 | # Getting in a muddle
1. Mar 7, 2007
### Trail_Builder
hi, soz, this is way below my level but i havnt done these in age and are tired and getting in a muddle. thnx for clarifying things...
1. The problem statement, all variables and given/known data
Two mathematically similar frustums have heights of 20cm and 30cm
The surface area of the smaller frustum is 450cm^2
qu) calculate the surface area of the larger frustum
2. Relevant equations
surface area of a frustum = 2pi x dia. x (square root)[h^2 + d^2]
3. The attempt at a solution
right, i am being a moron i know, but i forget where the dimensional adjustment to
20/450 = 30/ans
is amplied. I was thinking of something like:
20/450^2 = 30/ans^2
but then i tried using my own example of stuff like cubes and swaures and stuff and it didnt work.
then i tried
20^2/450 = 30^2/ans
but then using my own dimensional examples it didnt work.
i also tried using the surface area formula for frustums, but that didnt work either.
i know i am being a dumbhead, but hey, i guess its got i havn't answered these qus in years lol. and it called a frustum for a reason ;)
thnx
2. Mar 7, 2007
### HallsofIvy
Staff Emeritus
Irrelevant! You are told that these figures are "mathematically similar" which means that they have the same "shap": all lengths in the large one are 30/20= 3/2 the corresponding small one. Since the "area" of anything depends upon multiplying two lengths together, area always is proportion to the square of a length. If in going from the smaller to the larger, all lengths are multiplied by 3/2, what is the area multiplied by?
3. Mar 7, 2007
### Trail_Builder
would it be 9/4? | 466 | 1,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2016-44 | longest | en | 0.936951 |
https://www.stefanobattarola.com/uncategorized/live-nfl-betting-odds-lines-point-spreads-totals-3/ | 1,721,372,893,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00499.warc.gz | 859,192,784 | 110,156 | When it comes to this odds format, 2.00 will represent the even-money informative post equivalent. Therefore, as long as the value of the odds is higher than this, the price will be odds against. If the value is smaller than this, the price will be odds on. The approximate percent of hitting a set on the flop of a hand is 12% which isn’t too bad, but certainly nothing that you want to rely on while playing the game. If you’re planning on winning with a flush and you have four spades, then there will be nine spades left in the deck. If you have two suited cards, you will flop a flush roughly one time in 118.
What Does +400 Mean In A Bet?
As it suggests, every seven units you stake will get you four units. If the first number in this odds format is greater than the second, this is said to be “odds against”. What it means is that the potential winnings will be much bigger than the amount of money you need to stake.
How Sportsbooks Set The Cricket Betting Odds
This format is considered to be the most popular of all the previously mentioned ones. It is very easy to operate with decimal odds, because they How to make money Rest Gaming Losers For your Perk On Betfair indicate how much money you’ll get from a one unit-stake. So in order to find out how much you’ve won, you simply have to multiply the odds by the amount of your stake.
If you think Miami Dolphins will win, the +130 means that for every £100 you stake on them, you will win £230 (£130 plus your £100 stake). A £10 bet would mean you will win £23 – your £10 stake and £13 profit (calculated as £10/£100 x 130). Now you know odds are set out with two numbers separated by a forward slash, you can use them to work out the probability of an event happening.
The first is that, when possible, you want to find outcomes that are correlated such that if one thing happens it is likely that the other result will happen. If you like to live dangerously in pursuit of big payouts, betting parlays in soccer might be perfect for you. Most lines will have a half goal added rather than a whole number. Win amounts are equal to the difference between the total number of goals and the purchased amount, multiplied by the stake. Bets like these are long odds bets that you should go after when you are trying to get lucky and win big. Strategically, you might find that more goals are scored in the latter portion of games where players are more physically spent and therefore more prone to error.
They usually don’t have to do with who wins the game in the end. This is a simple wager on which team will win a particular game. Odds compilers will weigh up the relative strengths and weaknesses of each team, factoring in head-to-head records, injuries, fatigue, motivation levels, home advantage and so on. They will then assign a set of sports betting odds to each team. Also known as UK odds, fractional odds are perhaps the oldest type of betting odds in sports betting history.
You only need to consider the possibility of a push when betting against the spread or placing a totals bet. Your payout would be \$23.75 (\$10 initial stake + your winnings of \$13.75). Implied probability is useful because if your estimate of the probability of an event occurring is different than a sportsbook’s you can and should adjust your bet accordingly. Therefore, if you wager \$40 on Pittsburgh, you will win \$30.77, and your payout would be \$70.77 . | 752 | 3,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.959671 |
https://www.coursemerit.com/solution-details/2297/Statistics-1024b-Section-570-Homework-assignment-4--Complete-Solution | 1,521,518,196,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647280.40/warc/CC-MAIN-20180320033158-20180320053158-00432.warc.gz | 761,490,513 | 10,324 | Question DetailsNormal
\$ 18.00
Statistics 1024b Section 570 Homework assignment 4 | Complete Solution
Question posted by
Statistics 1024b
Section 570 Homework assignment 4
The responses to these question must be handed in at the beginning of the class on Thursday, March 26. At most, you will be expected to hand in your work on ONE SIDE OF A SINGLE SHEET of paper. The work will be graded out of 8 marks, with a minimum of 2 marks for handing in something.
1. (a) A physiology researcher took a random sample of 100 students and found that their average blood pressure was 122 mmHg. Assume that the population standard deviation is 10 mmHg, and nd a 95% con dence interval for , the population mean. (2 marks) (b) Why is 95% a reasonable number? (1 mark)
2. Suppose that previous records show that the average height of King's students is 172 cm. You suspect that the current students are taller than that average. You take a random sample of 25 students, and get a sample mean of 176 cm. You know that the population standard de- viation is 10 cm. Using an level of 0.05, conduct a test of hypothesis, and come to a conclusion about the average height of current students.
(3 marks)
3. Use the con dence interval in 1)(a) to test the hypothesis that the average blood pressure is not 120 mmHg. (2 marks)
Available Solution
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Statistics 1024b Section 570 Homework assignment 4 | Complete Solution
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https://zivgitlab.uni-muenster.de/pymor/pymor/-/commit/a040d4b106ef134306e67cda3e175604ad4aac5e | 1,618,815,227,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038878326.67/warc/CC-MAIN-20210419045820-20210419075820-00383.warc.gz | 1,150,179,026 | 27,452 | Commit a040d4b1 by Tim Keil
### [docs/tutorials] last comments from stephan
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... ... @@ -183,7 +183,7 @@ to resolve the data structure in the diffusion. This suggests to use an even finer mesh. However, for enabling a faster runtime for this tutorial, we stick with this mesh and remark that refining the mesh does not change the interpretation of the methods that are discussed below. It rather furhter improves the speedups achieved by model reduction. It rather further improves the speedups achieved by model reduction. Before we discuss the first optimization method, we define helpful functions for visualizations. ... ... @@ -360,7 +360,7 @@ a perfect surrogate model in the sense that a low error tolerance for the :func:~pymor.algorithms.greedy.rb_greedy already suffices to converge to the same minimum. In our case we choose atol=1e-2 and yield a very low dimensional space. In general, however, it is not a priorily clear how to choose atol In general, however, it is not a priori clear how to choose atol in order to arrive at a minimum which is close enough to the true optimum. ... ... @@ -484,7 +484,7 @@ equation, i.e. .. math:: r_\mu^{\text{pr}(u)[v] := l_\mu(v) - a_\mu(u, v) &&\text{for all }v \in V r_\mu^{\text{pr}}(u)[v] := l_\mu(v) - a_\mu(u, v) &&\text{for all }v \in V A major issue of this approach is that the computation of the full gradient requires :math:P solutions of :math:\eqref{sens}. ... ... @@ -603,7 +603,7 @@ output functional. report(opt_rom_result, opt_rom_minimization_data, reference_mu) The online phase is even slightly faster than before but the offline phase is obviously still the same as before. We also conclude that the phase is still the same as before. We also conclude that the ROM model eventually gives less speedup by using a better optimization method for the FOM and ROM. ... ... @@ -697,8 +697,9 @@ for the gradients since we compute the dual solutions with the ROM. Adaptively enriching along the path ----------------------------------- This gives rise to another idea where we only enrich if it is necessary. For example it could be the case that the model is already good at In order to further speedup the above algorithm, we enhance it by only adaptive enrichments of the model. For instance it may happen that the model is already good at the next iteration, which we can easily check by evaluating the standard error estimator which is also used in the greedy algorithm. In the next example we will implement this adaptive way of enriching and set a ... ... @@ -821,21 +822,17 @@ the traditional offline/online splitting by only enriching the model along the path of optimization or (even better) only enrich the model if the standard error estimator goes above a certain tolerance. A main drawback of the content in this tutorial was that the choice of the tolerance atol that has been used to build the RB spaces cannot be known a priorily. This shows the need for certified and robust reduced methods. For some standard literature for faster and robust optimization tools we refer to CGT00 __ and In this tutorial we have only covered a few basic approaches to combine model reduction with optimization. For faster and more robust optimization algorithms we refer to the textbooks CGT00 __ and NW06 __. For recent research on using trust-region methods for MOR of PDE-constrained optimization problems, we refer to For recent research on combining trust-region methods with model reduction for PDE-constrained optimization problems we refer to YM13 __, QGVW17 __ and KMSOV __ where for the latter, pyMOR has been used for the numerical experiments. KMSOV20 __ where for the latter a pyMOR implementation is available as supplementary material. Download the code: :jupyter-download:script:tutorial_optimization ... ...
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Finish editing this message first! | 914 | 4,007 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-17 | latest | en | 0.848551 |
https://math.answers.com/other-math/What_two_numbers_equal_150 | 1,708,917,397,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474650.85/warc/CC-MAIN-20240226030734-20240226060734-00287.warc.gz | 385,748,612 | 45,231 | 0
What two numbers equal 150?
Updated: 4/28/2022
Wiki User
6y ago
15 and 10 if you multiply them.
140 and 10 if you add them.
160 and 10 if you subtract them.
1500 and 10 if you divide them.
Wiki User
6y ago | 70 | 217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-10 | latest | en | 0.817894 |
http://hades.mech.northwestern.edu/index.php?title=A*_Graph_Search_Project&direction=next&oldid=24629 | 1,596,940,351,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738380.22/warc/CC-MAIN-20200809013812-20200809043812-00135.warc.gz | 48,613,260 | 7,766 | # A* Graph Search Project
This page describes the A* Graph Search Project from the Coursera course "Modern Robotics, Course 4: Robot Motion Planning and Control."
First you should read about the V-REP simulator scene "CSV Motion Planning Kilobot" on the Modern Robotics wiki. You should then download the associated .ttt V-REP scene file and the .zip file with sample input to the scene. When you extract the .zip file, you will have a folder with four files inside: nodes.csv (describing the nodes of the graph), edges.csv (describing the edges of the graph), path.csv (describing a solution path), and obstacles.csv (describing cylindrical obstacles, just for visualization purposes). Read the wiki (and the files themselves, which have comments) for more information on the formats of these files.
Play the scene in V-REP and choose the folder with the four files as the input. When you press the "play" button on the interface, you will see a display of the graph, the obstacles, and a path through the graph. You will also see a small mobile robot, a swarm robot called a kilobot, travel along the path. Make sure you can do this before moving on with the rest of the assignment.
In this assignment, you will create a program that implements A* search to find a minimum-cost path through an undirected graph. You will use the V-REP scene to visualize the graph and the solution path found by your program. Although the nodes of your graph are assigned (x,y) coordinates in a plane for visualization purposes (and the solution is animated by a planar mobile robot), the A* graph search you will implement is for fully general undirected weighted graphs. As such, the nodes could represent configurations in arbitrary C-spaces, such as the six-dimensional C-space of the rigid chassis of a spacecraft flying among asteroids. We visualize the nodes of the graph as points in a plane, for simplicity.
You will write a program that takes two files as input, nodes.csv and edges.csv, and produces a single file as output, path.csv. You can test your program on the nodes.csv and edges.csv file that you downloaded from the Modern Robotics wiki, to make sure that your code is working properly.
The sample nodes.csv and edges.csv files on the wiki have comments in them. All lines beginning with the hashtag character # are ignored. You can write your software to handle comments such as these, or you can assume that the input files have no comments. In the latter case, you will need to delete the comments from the nodes.csv and edges.csv file before using them to test your code.
You can write your program in any language, provided your program is clearly structured, reasonably modular, easy to understand for a reader with no experience in your programming language, and with sufficient comments. Your program can span multiple files, or it can be a single file, if appropriate.
If there are N nodes, numbered 1...N in nodes.csv, you can assume that node 1 is the start node and node N is the goal node. Your output file path.csv should be a single line of the form
``` 1, node2, node3, ..., N
```
indicating that the solution path goes from nodes numbered 1, to node2, to node3, etc., finally to N. Your program must find the proper node numbers for node2, node3, etc. The solution path should minimize the sum of the edge costs among all possible paths from node 1 to node N, where the edge costs are given in the edges.csv file. If there is no solution, your output file can be simply
``` 1, 1
```
indicating that the robot does not go anywhere.
To see the results of your planner, call the V-REP scene with the nodes.csv and edges.csv file you used as input, the path.csv file that your program wrote as output, and an obstacles.csv file (which is not needed by your program). The obstacles .csv file can be empty if you don't want to visualize obstacles.
Data structures, algorithms, and computational efficiency
This project is a more extensive programming project than you have done before in the Coursera Modern Robotics specialization. Even if you have not had a lot of programming experience, you should still be able to complete the project based on the A* description in the book.
You will have to make decisions about how to implement data structures such as OPEN, CLOSED, the graph edges, the past costs, and the search trees. You will also have to decide how to implement certain algorithms on these data structures, such as inserting nodes into the sorted list OPEN. Decisions you make about this will affect the efficiency of your implementation (e.g., the amount of memory needed to store your data, or the amount of time it takes to perform certain operations). However, your assignment will not be graded on efficiency. You are free to simply use two-dimensional matrices and one-dimensional arrays to represent your data and to use inefficient algorithms on them (e.g., for inserting nodes in the sorted list OPEN).
For large search problems, efficiency matters, but this is not a computer science course, and there are no prerequisites in data structures or algorithms, so our focus is on understanding the ideas of A* graph search and correctness of the implementation, not efficiency.
You may also have to decide the largest problem your search can handle (e.g., in terms of the number of nodes in the graph). You can make any reasonable assumption (e.g., at least 100 nodes).
What to submit
Once your program is producing correct output, you are ready to assemble your documents for submission. You will submit a single .zip file of a directory with the following contents:
1. Your commented code in a directory called "code." Your code should be lightly commented, so it is clear to the reader what the code is doing. No need to go overboard, but keep in mind your reviewer may not be fluent in your programming language. Your code comments must include an example of how to use the code. You can write your program in any language, provided it is clearly structured, reasonably modular, easy to understand for a reader with no experience in your programming language, and with sufficient comments. Your program can span multiple files, or it can be a single file, if appropriate. If your code is in Mathematica, turn in (a) your .nb notebook file and (b) a .pdf printout of your code, so a reviewer can read your code without having to have the Mathematica software.
2. A directory called "results" with four files in it: nodes.csv, edges.csv, obstacles.csv, and path.csv. The nodes.csv file and edges.csv file should be readable by your program. The nodes.csv file should be identical to the sample nodes.csv file you downloaded, except you can delete the comments if your code does not handle files with comments. The edges.csv file should also be identical to the sample edges.csv file (perhaps without the comments) except the edge from node 4 to node 7 should be deleted, to change the graph from the original graph. The obstacles.csv file should be identical to the sample file. The path.csv file should be the unmodified output of your program when taking nodes.csv and edges.csv as input.
3. A V-REP video animating your solution path on the graph from the bullet above. (See the V-REP description in Week 1 of Course 1, or go directly to V-REP Introduction, to learn about making videos with V-REP.) Your video should be a "reasonable" size (e.g., a few MB, less than 10 MB) and use a standard codec (e.g., some variant of .mp4) that others can view. Your video should be taken from a virtual camera angle that makes it easy to see the graph and kilobot motion.
4. (OPTIONAL) A plain text file called "README.txt." This has any other information that may help the reviewer understand your submission. | 1,656 | 7,762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-34 | latest | en | 0.89656 |
https://au.mathworks.com/help/gads/how-the-genetic-algorithm-works.html | 1,726,824,448,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00858.warc.gz | 93,727,374 | 19,899 | ## How the Genetic Algorithm Works
### Outline of the Algorithm
The following outline summarizes how the genetic algorithm works:
1. The algorithm begins by creating a random initial population.
2. The algorithm then creates a sequence of new populations. At each step, the algorithm uses the individuals in the current generation to create the next population. To create the new population, the algorithm performs the following steps:
1. Scores each member of the current population by computing its fitness value. These values are called the raw fitness scores.
2. Scales the raw fitness scores to convert them into a more usable range of values. These scaled values are called expectation values.
3. Selects members, called parents, based on their expectation.
4. Some of the individuals in the current population that have lower fitness are chosen as elite. These elite individuals are passed to the next population.
5. Produces children from the parents. Children are produced either by making random changes to a single parent—mutation—or by combining the vector entries of a pair of parents—crossover.
6. Replaces the current population with the children to form the next generation.
3. The algorithm stops when one of the stopping criteria is met. See Stopping Conditions for the Algorithm.
4. The algorithm takes modified steps for linear and integer constraints. See Integer and Linear Constraints.
5. The algorithm is further modified for nonlinear constraints. See Nonlinear Constraint Solver Algorithms for Genetic Algorithm.
### Initial Population
The algorithm begins by creating a random initial population, as shown in the following figure.
In this example, the initial population contains `20` individuals. Note that all the individuals in the initial population lie in the upper-right quadrant of the picture, that is, their coordinates lie between 0 and 1. For this example, the `InitialPopulationRange` option is `[0;1]`.
If you know approximately where the minimal point for a function lies, you should set `InitialPopulationRange` so that the point lies near the middle of that range. For example, if you believe that the minimal point for Rastrigin's function is near the point `[0 0]`, you could set `InitialPopulationRange` to be `[-1;1]`. However, as this example shows, the genetic algorithm can find the minimum even with a less than optimal choice for `InitialPopulationRange`.
### Creating the Next Generation
At each step, the genetic algorithm uses the current population to create the children that make up the next generation. The algorithm selects a group of individuals in the current population, called parents, who contribute their genes—the entries of their vectors—to their children. The algorithm usually selects individuals that have better fitness values as parents. You can specify the function that the algorithm uses to select the parents in the `SelectionFcn` option. See Selection Options.
The genetic algorithm creates three types of children for the next generation:
• Elite children are the individuals in the current generation with the best fitness values. These individuals automatically survive to the next generation.
• Crossover children are created by combining the vectors of a pair of parents.
• Mutation children are created by introducing random changes, or mutations, to a single parent.
The following schematic diagram illustrates the three types of children.
Mutation and Crossover explains how to specify the number of children of each type that the algorithm generates and the functions it uses to perform crossover and mutation.
The following sections explain how the algorithm creates crossover and mutation children.
#### Crossover Children
The algorithm creates crossover children by combining pairs of parents in the current population. At each coordinate of the child vector, the default crossover function randomly selects an entry, or gene, at the same coordinate from one of the two parents and assigns it to the child. For problems with linear constraints, the default crossover function creates the child as a random weighted average of the parents.
#### Mutation Children
The algorithm creates mutation children by randomly changing the genes of individual parents. By default, for unconstrained problems the algorithm adds a random vector from a Gaussian distribution to the parent. For bounded or linearly constrained problems, the child remains feasible.
The following figure shows the children of the initial population, that is, the population at the second generation, and indicates whether they are elite, crossover, or mutation children.
### Plots of Later Generations
The following figure shows the populations at iterations 60, 80, 95, and 100.
As the number of generations increases, the individuals in the population get closer together and approach the minimum point ```[0 0]```.
### Stopping Conditions for the Algorithm
The genetic algorithm uses the following options to determine when to stop. See the default values for each option by running ```opts = optimoptions('ga')```.
• `MaxGenerations` — The algorithm stops when the number of generations reaches `MaxGenerations`.
• `MaxTime` — The algorithm stops after running for an amount of time in seconds equal to `MaxTime`.
• `FitnessLimit` — The algorithm stops when the value of the fitness function for the best point in the current population is less than or equal to `FitnessLimit`.
• `MaxStallGenerations` — The algorithm stops when the average relative change in the fitness function value over `MaxStallGenerations` is less than ```Function tolerance```.
• `MaxStallTime` — The algorithm stops if there is no improvement in the objective function during an interval of time in seconds equal to `MaxStallTime`.
• `FunctionTolerance` — The algorithm runs until the average relative change in the fitness function value over `MaxStallGenerations` is less than ```Function tolerance```.
• `ConstraintTolerance` — The `ConstraintTolerance` is not used as stopping criterion. It is used to determine the feasibility with respect to nonlinear constraints. Also, `max(sqrt(eps),ConstraintTolerance)` determines feasibility with respect to linear constraints.
The algorithm stops as soon as any one of these conditions is met.
### Selection
The selection function chooses parents for the next generation based on their scaled values from the fitness scaling function. The scaled fitness values are called the expectation values. An individual can be selected more than once as a parent, in which case it contributes its genes to more than one child. The default selection option, `@selectionstochunif`, lays out a line in which each parent corresponds to a section of the line of length proportional to its scaled value. The algorithm moves along the line in steps of equal size. At each step, the algorithm allocates a parent from the section it lands on.
A more deterministic selection option is `@selectionremainder`, which performs two steps:
• In the first step, the function selects parents deterministically according to the integer part of the scaled value for each individual. For example, if an individual's scaled value is 2.3, the function selects that individual twice as a parent.
• In the second step, the selection function selects additional parents using the fractional parts of the scaled values, as in stochastic uniform selection. The function lays out a line in sections, whose lengths are proportional to the fractional part of the scaled value of the individuals, and moves along the line in equal steps to select the parents.
Note that if the fractional parts of the scaled values all equal 0, as can occur using `Top` scaling, the selection is entirely deterministic.
For details and more selection options, see Selection Options.
### Reproduction Options
Reproduction options control how the genetic algorithm creates the next generation. The options are
• `EliteCount` — The number of individuals with the best fitness values in the current generation that are guaranteed to survive to the next generation. These individuals are called elite children.
When `EliteCount` is at least 1, the best fitness value can only decrease from one generation to the next. This is what you want to happen, since the genetic algorithm minimizes the fitness function. Setting `EliteCount` to a high value causes the fittest individuals to dominate the population, which can make the search less effective.
• `CrossoverFraction` — The fraction of individuals in the next generation, other than elite children, that are created by crossover. The topic "Setting the Crossover Fraction" in Vary Mutation and Crossover describes how the value of `CrossoverFraction` affects the performance of the genetic algorithm.
Because elite individuals have already been evaluated, `ga` does not reevaluate the fitness function of elite individuals during reproduction. This behavior assumes that the fitness function of an individual is not random, but is a deterministic function. To change this behavior, use an output function. See `EvalElites` in The State Structure.
### Mutation and Crossover
The genetic algorithm uses the individuals in the current generation to create the children that make up the next generation. Besides elite children, which correspond to the individuals in the current generation with the best fitness values, the algorithm creates
• Crossover children by selecting vector entries, or genes, from a pair of individuals in the current generation and combines them to form a child
• Mutation children by applying random changes to a single individual in the current generation to create a child
Both processes are essential to the genetic algorithm. Crossover enables the algorithm to extract the best genes from different individuals and recombine them into potentially superior children. Mutation adds to the diversity of a population and thereby increases the likelihood that the algorithm will generate individuals with better fitness values.
See Creating the Next Generation for an example of how the genetic algorithm applies mutation and crossover.
You can specify how many of each type of children the algorithm creates as follows:
• `EliteCount` specifies the number of elite children.
• `CrossoverFraction` specifies the fraction of the population, other than elite children, that are crossover children.
For example, if the `PopulationSize` is `20`, the `EliteCount` is `2`, and the `CrossoverFraction` is `0.8`, the numbers of each type of children in the next generation are as follows:
• There are two elite children.
• There are 18 individuals other than elite children, so the algorithm rounds 0.8*18 = 14.4 to 14 to get the number of crossover children.
• The remaining four individuals, other than elite children, are mutation children.
### Integer and Linear Constraints
When a problem has integer or linear constraints (including bounds), the algorithm modifies the evolution of the population.
• When the problem has both integer and linear constraints, the software modifies all generated individuals to be feasible with respect to those constraints. You can use any creation, mutation, or crossover function, and the entire population remains feasible with respect to integer and linear constraints.
• When the problem has only linear constraints, the software does not modify the individuals to be feasible with respect to those constraints. You must use creation, mutation, and crossover functions that maintain feasibility with respect to linear constraints. Otherwise, the population can become infeasible, and the result can be infeasible. The default operators maintain linear feasibility: `gacreationlinearfeasible` or `gacreationnonlinearfeasible` for creation, `mutationadaptfeasible` for mutation, and `crossoverintermediate` for crossover.
The internal algorithms for integer and linear feasibility are similar to those for `surrogateopt`. When a problem has integer and linear constraints, the algorithm first creates linearly feasible points. Then the algorithm tries to satisfy integer constraints by rounding linearly feasible points to integers using a heuristic that attempts to keep the points linearly feasible. When this process is unsuccessful in obtaining enough feasible points for constructing a population, the algorithm calls `intlinprog` to try to find more points that are feasible with respect to bounds, linear constraints, and integer constraints.
Later, when mutation or crossover creates new population members, the algorithms ensure that the new members are integer and linear feasible by taking similar steps. Each new member is modified, if necessary, to be as close as possible to its original value, while also satisfying the integer and linear constraints and bounds. | 2,452 | 12,879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-38 | latest | en | 0.885056 |
https://www.airmilescalculator.com/distance/ulk-to-tbs/ | 1,606,813,200,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141672314.55/warc/CC-MAIN-20201201074047-20201201104047-00392.warc.gz | 573,552,726 | 115,286 | # Distance between Lensk (ULK) and Tbilisi (TBS)
Flight distance from Lensk to Tbilisi (Lensk Airport – Tbilisi International Airport) is 3126 miles / 5030 kilometers / 2716 nautical miles. Estimated flight time is 6 hours 25 minutes.
Driving distance from Lensk (ULK) to Tbilisi (TBS) is 4547 miles / 7317 kilometers and travel time by car is about 101 hours 10 minutes.
## Map of flight path and driving directions from Lensk to Tbilisi.
Shortest flight path between Lensk Airport (ULK) and Tbilisi International Airport (TBS).
## How far is Tbilisi from Lensk?
There are several ways to calculate distances between Lensk and Tbilisi. Here are two common methods:
Vincenty's formula (applied above)
• 3125.779 miles
• 5030.454 kilometers
• 2716.228 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 3117.543 miles
• 5017.199 kilometers
• 2709.071 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Lensk Airport
City: Lensk
Country: Russia
IATA Code: ULK
ICAO Code: UERL
Coordinates: 60°43′14″N, 114°49′33″E
B Tbilisi International Airport
City: Tbilisi
Country: Georgia
IATA Code: TBS
ICAO Code: UGTB
Coordinates: 41°40′9″N, 44°57′16″E
## Time difference and current local times
The time difference between Lensk and Tbilisi is 5 hours. Tbilisi is 5 hours behind Lensk.
+09
+04
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 349 kg (770 pounds).
## Frequent Flyer Miles Calculator
Lensk (ULK) → Tbilisi (TBS).
Distance:
3126
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
3126
Round trip? | 513 | 1,867 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-50 | latest | en | 0.788428 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/5225/2/a/l/1/1/ | 1,726,855,762,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00540.warc.gz | 795,565,220 | 83,291 | # Properties
Label 5225.2.a.l.1.1 Level $5225$ Weight $2$ Character 5225.1 Self dual yes Analytic conductor $41.722$ Analytic rank $1$ Dimension $6$ CM no Inner twists $1$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [5225,2,Mod(1,5225)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(5225, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([0, 0, 0]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("5225.1");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$5225 = 5^{2} \cdot 11 \cdot 19$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 5225.a (trivial)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: yes Analytic conductor: $$41.7218350561$$ Analytic rank: $$1$$ Dimension: $$6$$ Coefficient field: 6.6.7281497.1 comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{6} - 2x^{5} - 5x^{4} + 7x^{3} + 6x^{2} - 2x - 1$$ x^6 - 2*x^5 - 5*x^4 + 7*x^3 + 6*x^2 - 2*x - 1 Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 1045) Fricke sign: $$+1$$ Sato-Tate group: $\mathrm{SU}(2)$
## Embedding invariants
Embedding label 1.1 Root $$-0.326248$$ of defining polynomial Character $$\chi$$ $$=$$ 5225.1
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
$$f(q)$$ $$=$$ $$q-1.73890 q^{2} +0.901323 q^{3} +1.02379 q^{4} -1.56731 q^{6} -2.22757 q^{7} +1.69754 q^{8} -2.18762 q^{9} +O(q^{10})$$ $$q-1.73890 q^{2} +0.901323 q^{3} +1.02379 q^{4} -1.56731 q^{6} -2.22757 q^{7} +1.69754 q^{8} -2.18762 q^{9} -1.00000 q^{11} +0.922763 q^{12} +3.64023 q^{13} +3.87353 q^{14} -4.99943 q^{16} -0.644551 q^{17} +3.80406 q^{18} -1.00000 q^{19} -2.00776 q^{21} +1.73890 q^{22} +6.15983 q^{23} +1.53003 q^{24} -6.33001 q^{26} -4.67572 q^{27} -2.28056 q^{28} -1.88204 q^{29} -0.183675 q^{31} +5.29846 q^{32} -0.901323 q^{33} +1.12081 q^{34} -2.23966 q^{36} +4.11428 q^{37} +1.73890 q^{38} +3.28102 q^{39} -1.53371 q^{41} +3.49130 q^{42} +1.53577 q^{43} -1.02379 q^{44} -10.7113 q^{46} -1.75837 q^{47} -4.50610 q^{48} -2.03793 q^{49} -0.580948 q^{51} +3.72682 q^{52} +2.81491 q^{53} +8.13063 q^{54} -3.78139 q^{56} -0.901323 q^{57} +3.27269 q^{58} -3.90068 q^{59} +0.734057 q^{61} +0.319392 q^{62} +4.87307 q^{63} +0.785358 q^{64} +1.56731 q^{66} +1.30264 q^{67} -0.659883 q^{68} +5.55199 q^{69} +10.5493 q^{71} -3.71357 q^{72} +5.17599 q^{73} -7.15433 q^{74} -1.02379 q^{76} +2.22757 q^{77} -5.70538 q^{78} +2.89974 q^{79} +2.34852 q^{81} +2.66698 q^{82} -13.8209 q^{83} -2.05552 q^{84} -2.67055 q^{86} -1.69633 q^{87} -1.69754 q^{88} -6.39884 q^{89} -8.10886 q^{91} +6.30635 q^{92} -0.165550 q^{93} +3.05763 q^{94} +4.77562 q^{96} -6.23352 q^{97} +3.54376 q^{98} +2.18762 q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$6 q + 2 q^{2} + q^{3} + 4 q^{4} - 5 q^{7} + 12 q^{8} + q^{9}+O(q^{10})$$ 6 * q + 2 * q^2 + q^3 + 4 * q^4 - 5 * q^7 + 12 * q^8 + q^9 $$6 q + 2 q^{2} + q^{3} + 4 q^{4} - 5 q^{7} + 12 q^{8} + q^{9} - 6 q^{11} - q^{12} + 5 q^{13} - 8 q^{14} + 4 q^{16} - q^{17} - 6 q^{18} - 6 q^{19} - 21 q^{21} - 2 q^{22} - 4 q^{23} - q^{24} - 14 q^{26} + 16 q^{27} - 10 q^{28} - 9 q^{29} - 21 q^{31} + q^{32} - q^{33} - 28 q^{36} + 3 q^{37} - 2 q^{38} + 20 q^{39} - 23 q^{41} - q^{42} - 7 q^{43} - 4 q^{44} - 12 q^{46} + 18 q^{47} - 3 q^{49} - 16 q^{51} - 13 q^{52} + 17 q^{53} + q^{54} - 2 q^{56} - q^{57} - 23 q^{58} - 29 q^{59} + 17 q^{61} - 2 q^{62} - 6 q^{63} - 18 q^{64} - 8 q^{67} + q^{68} - 38 q^{69} - 12 q^{71} - 13 q^{72} - 2 q^{73} - 37 q^{74} - 4 q^{76} + 5 q^{77} - q^{78} + 3 q^{79} - 2 q^{81} - 24 q^{82} + 11 q^{83} - 3 q^{84} - 12 q^{86} + 12 q^{87} - 12 q^{88} - 22 q^{89} - 18 q^{91} + 15 q^{92} - 18 q^{93} + 22 q^{94} - 17 q^{96} + 2 q^{97} + q^{98} - q^{99}+O(q^{100})$$ 6 * q + 2 * q^2 + q^3 + 4 * q^4 - 5 * q^7 + 12 * q^8 + q^9 - 6 * q^11 - q^12 + 5 * q^13 - 8 * q^14 + 4 * q^16 - q^17 - 6 * q^18 - 6 * q^19 - 21 * q^21 - 2 * q^22 - 4 * q^23 - q^24 - 14 * q^26 + 16 * q^27 - 10 * q^28 - 9 * q^29 - 21 * q^31 + q^32 - q^33 - 28 * q^36 + 3 * q^37 - 2 * q^38 + 20 * q^39 - 23 * q^41 - q^42 - 7 * q^43 - 4 * q^44 - 12 * q^46 + 18 * q^47 - 3 * q^49 - 16 * q^51 - 13 * q^52 + 17 * q^53 + q^54 - 2 * q^56 - q^57 - 23 * q^58 - 29 * q^59 + 17 * q^61 - 2 * q^62 - 6 * q^63 - 18 * q^64 - 8 * q^67 + q^68 - 38 * q^69 - 12 * q^71 - 13 * q^72 - 2 * q^73 - 37 * q^74 - 4 * q^76 + 5 * q^77 - q^78 + 3 * q^79 - 2 * q^81 - 24 * q^82 + 11 * q^83 - 3 * q^84 - 12 * q^86 + 12 * q^87 - 12 * q^88 - 22 * q^89 - 18 * q^91 + 15 * q^92 - 18 * q^93 + 22 * q^94 - 17 * q^96 + 2 * q^97 + q^98 - q^99
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ −1.73890 −1.22959 −0.614795 0.788687i $$-0.710761\pi$$
−0.614795 + 0.788687i $$0.710761\pi$$
$$3$$ 0.901323 0.520379 0.260190 0.965558i $$-0.416215\pi$$
0.260190 + 0.965558i $$0.416215\pi$$
$$4$$ 1.02379 0.511894
$$5$$ 0 0
$$6$$ −1.56731 −0.639853
$$7$$ −2.22757 −0.841943 −0.420971 0.907074i $$-0.638311\pi$$
−0.420971 + 0.907074i $$0.638311\pi$$
$$8$$ 1.69754 0.600171
$$9$$ −2.18762 −0.729206
$$10$$ 0 0
$$11$$ −1.00000 −0.301511
$$12$$ 0.922763 0.266379
$$13$$ 3.64023 1.00962 0.504809 0.863231i $$-0.331563\pi$$
0.504809 + 0.863231i $$0.331563\pi$$
$$14$$ 3.87353 1.03525
$$15$$ 0 0
$$16$$ −4.99943 −1.24986
$$17$$ −0.644551 −0.156327 −0.0781633 0.996941i $$-0.524906\pi$$
−0.0781633 + 0.996941i $$0.524906\pi$$
$$18$$ 3.80406 0.896625
$$19$$ −1.00000 −0.229416
$$20$$ 0 0
$$21$$ −2.00776 −0.438129
$$22$$ 1.73890 0.370736
$$23$$ 6.15983 1.28441 0.642206 0.766532i $$-0.278019\pi$$
0.642206 + 0.766532i $$0.278019\pi$$
$$24$$ 1.53003 0.312316
$$25$$ 0 0
$$26$$ −6.33001 −1.24142
$$27$$ −4.67572 −0.899842
$$28$$ −2.28056 −0.430985
$$29$$ −1.88204 −0.349487 −0.174743 0.984614i $$-0.555910\pi$$
−0.174743 + 0.984614i $$0.555910\pi$$
$$30$$ 0 0
$$31$$ −0.183675 −0.0329889 −0.0164945 0.999864i $$-0.505251\pi$$
−0.0164945 + 0.999864i $$0.505251\pi$$
$$32$$ 5.29846 0.936644
$$33$$ −0.901323 −0.156900
$$34$$ 1.12081 0.192218
$$35$$ 0 0
$$36$$ −2.23966 −0.373276
$$37$$ 4.11428 0.676383 0.338192 0.941077i $$-0.390185\pi$$
0.338192 + 0.941077i $$0.390185\pi$$
$$38$$ 1.73890 0.282088
$$39$$ 3.28102 0.525384
$$40$$ 0 0
$$41$$ −1.53371 −0.239525 −0.119763 0.992803i $$-0.538213\pi$$
−0.119763 + 0.992803i $$0.538213\pi$$
$$42$$ 3.49130 0.538720
$$43$$ 1.53577 0.234202 0.117101 0.993120i $$-0.462640\pi$$
0.117101 + 0.993120i $$0.462640\pi$$
$$44$$ −1.02379 −0.154342
$$45$$ 0 0
$$46$$ −10.7113 −1.57930
$$47$$ −1.75837 −0.256484 −0.128242 0.991743i $$-0.540933\pi$$
−0.128242 + 0.991743i $$0.540933\pi$$
$$48$$ −4.50610 −0.650400
$$49$$ −2.03793 −0.291133
$$50$$ 0 0
$$51$$ −0.580948 −0.0813491
$$52$$ 3.72682 0.516817
$$53$$ 2.81491 0.386658 0.193329 0.981134i $$-0.438071\pi$$
0.193329 + 0.981134i $$0.438071\pi$$
$$54$$ 8.13063 1.10644
$$55$$ 0 0
$$56$$ −3.78139 −0.505309
$$57$$ −0.901323 −0.119383
$$58$$ 3.27269 0.429726
$$59$$ −3.90068 −0.507825 −0.253913 0.967227i $$-0.581718\pi$$
−0.253913 + 0.967227i $$0.581718\pi$$
$$60$$ 0 0
$$61$$ 0.734057 0.0939863 0.0469932 0.998895i $$-0.485036\pi$$
0.0469932 + 0.998895i $$0.485036\pi$$
$$62$$ 0.319392 0.0405629
$$63$$ 4.87307 0.613949
$$64$$ 0.785358 0.0981698
$$65$$ 0 0
$$66$$ 1.56731 0.192923
$$67$$ 1.30264 0.159143 0.0795717 0.996829i $$-0.474645\pi$$
0.0795717 + 0.996829i $$0.474645\pi$$
$$68$$ −0.659883 −0.0800226
$$69$$ 5.55199 0.668381
$$70$$ 0 0
$$71$$ 10.5493 1.25197 0.625983 0.779837i $$-0.284698\pi$$
0.625983 + 0.779837i $$0.284698\pi$$
$$72$$ −3.71357 −0.437648
$$73$$ 5.17599 0.605804 0.302902 0.953022i $$-0.402044\pi$$
0.302902 + 0.953022i $$0.402044\pi$$
$$74$$ −7.15433 −0.831674
$$75$$ 0 0
$$76$$ −1.02379 −0.117437
$$77$$ 2.22757 0.253855
$$78$$ −5.70538 −0.646007
$$79$$ 2.89974 0.326246 0.163123 0.986606i $$-0.447843\pi$$
0.163123 + 0.986606i $$0.447843\pi$$
$$80$$ 0 0
$$81$$ 2.34852 0.260947
$$82$$ 2.66698 0.294518
$$83$$ −13.8209 −1.51704 −0.758522 0.651647i $$-0.774078\pi$$
−0.758522 + 0.651647i $$0.774078\pi$$
$$84$$ −2.05552 −0.224276
$$85$$ 0 0
$$86$$ −2.67055 −0.287973
$$87$$ −1.69633 −0.181866
$$88$$ −1.69754 −0.180958
$$89$$ −6.39884 −0.678275 −0.339138 0.940737i $$-0.610135\pi$$
−0.339138 + 0.940737i $$0.610135\pi$$
$$90$$ 0 0
$$91$$ −8.10886 −0.850040
$$92$$ 6.30635 0.657483
$$93$$ −0.165550 −0.0171667
$$94$$ 3.05763 0.315371
$$95$$ 0 0
$$96$$ 4.77562 0.487410
$$97$$ −6.23352 −0.632918 −0.316459 0.948606i $$-0.602494\pi$$
−0.316459 + 0.948606i $$0.602494\pi$$
$$98$$ 3.54376 0.357974
$$99$$ 2.18762 0.219864
$$100$$ 0 0
$$101$$ −16.3563 −1.62751 −0.813756 0.581206i $$-0.802581\pi$$
−0.813756 + 0.581206i $$0.802581\pi$$
$$102$$ 1.01021 0.100026
$$103$$ −0.645519 −0.0636049 −0.0318025 0.999494i $$-0.510125\pi$$
−0.0318025 + 0.999494i $$0.510125\pi$$
$$104$$ 6.17943 0.605943
$$105$$ 0 0
$$106$$ −4.89487 −0.475432
$$107$$ 17.5900 1.70049 0.850243 0.526390i $$-0.176455\pi$$
0.850243 + 0.526390i $$0.176455\pi$$
$$108$$ −4.78694 −0.460624
$$109$$ 19.4359 1.86162 0.930808 0.365507i $$-0.119104\pi$$
0.930808 + 0.365507i $$0.119104\pi$$
$$110$$ 0 0
$$111$$ 3.70829 0.351976
$$112$$ 11.1366 1.05231
$$113$$ 5.88383 0.553504 0.276752 0.960941i $$-0.410742\pi$$
0.276752 + 0.960941i $$0.410742\pi$$
$$114$$ 1.56731 0.146792
$$115$$ 0 0
$$116$$ −1.92681 −0.178900
$$117$$ −7.96342 −0.736219
$$118$$ 6.78291 0.624417
$$119$$ 1.43578 0.131618
$$120$$ 0 0
$$121$$ 1.00000 0.0909091
$$122$$ −1.27645 −0.115565
$$123$$ −1.38237 −0.124644
$$124$$ −0.188044 −0.0168868
$$125$$ 0 0
$$126$$ −8.47381 −0.754907
$$127$$ −15.9676 −1.41690 −0.708450 0.705762i $$-0.750605\pi$$
−0.708450 + 0.705762i $$0.750605\pi$$
$$128$$ −11.9626 −1.05735
$$129$$ 1.38422 0.121874
$$130$$ 0 0
$$131$$ −3.93173 −0.343517 −0.171759 0.985139i $$-0.554945\pi$$
−0.171759 + 0.985139i $$0.554945\pi$$
$$132$$ −0.922763 −0.0803162
$$133$$ 2.22757 0.193155
$$134$$ −2.26517 −0.195681
$$135$$ 0 0
$$136$$ −1.09415 −0.0938226
$$137$$ −15.5422 −1.32786 −0.663929 0.747796i $$-0.731112\pi$$
−0.663929 + 0.747796i $$0.731112\pi$$
$$138$$ −9.65438 −0.821836
$$139$$ −6.69900 −0.568202 −0.284101 0.958794i $$-0.591695\pi$$
−0.284101 + 0.958794i $$0.591695\pi$$
$$140$$ 0 0
$$141$$ −1.58486 −0.133469
$$142$$ −18.3441 −1.53941
$$143$$ −3.64023 −0.304411
$$144$$ 10.9368 0.911404
$$145$$ 0 0
$$146$$ −9.00056 −0.744891
$$147$$ −1.83683 −0.151499
$$148$$ 4.21215 0.346236
$$149$$ 0.783045 0.0641496 0.0320748 0.999485i $$-0.489789\pi$$
0.0320748 + 0.999485i $$0.489789\pi$$
$$150$$ 0 0
$$151$$ 8.46749 0.689075 0.344537 0.938773i $$-0.388036\pi$$
0.344537 + 0.938773i $$0.388036\pi$$
$$152$$ −1.69754 −0.137689
$$153$$ 1.41003 0.113994
$$154$$ −3.87353 −0.312138
$$155$$ 0 0
$$156$$ 3.35907 0.268941
$$157$$ −8.27024 −0.660037 −0.330019 0.943974i $$-0.607055\pi$$
−0.330019 + 0.943974i $$0.607055\pi$$
$$158$$ −5.04237 −0.401150
$$159$$ 2.53715 0.201209
$$160$$ 0 0
$$161$$ −13.7215 −1.08140
$$162$$ −4.08385 −0.320858
$$163$$ −23.9030 −1.87223 −0.936113 0.351699i $$-0.885604\pi$$
−0.936113 + 0.351699i $$0.885604\pi$$
$$164$$ −1.57019 −0.122612
$$165$$ 0 0
$$166$$ 24.0333 1.86534
$$167$$ 19.4405 1.50435 0.752176 0.658962i $$-0.229004\pi$$
0.752176 + 0.658962i $$0.229004\pi$$
$$168$$ −3.40825 −0.262952
$$169$$ 0.251253 0.0193272
$$170$$ 0 0
$$171$$ 2.18762 0.167291
$$172$$ 1.57230 0.119887
$$173$$ 11.6888 0.888682 0.444341 0.895858i $$-0.353438\pi$$
0.444341 + 0.895858i $$0.353438\pi$$
$$174$$ 2.94975 0.223620
$$175$$ 0 0
$$176$$ 4.99943 0.376847
$$177$$ −3.51577 −0.264261
$$178$$ 11.1270 0.834001
$$179$$ −6.77452 −0.506351 −0.253176 0.967420i $$-0.581475\pi$$
−0.253176 + 0.967420i $$0.581475\pi$$
$$180$$ 0 0
$$181$$ −4.34551 −0.322999 −0.161500 0.986873i $$-0.551633\pi$$
−0.161500 + 0.986873i $$0.551633\pi$$
$$182$$ 14.1005 1.04520
$$183$$ 0.661622 0.0489085
$$184$$ 10.4565 0.770867
$$185$$ 0 0
$$186$$ 0.287876 0.0211081
$$187$$ 0.644551 0.0471342
$$188$$ −1.80020 −0.131293
$$189$$ 10.4155 0.757616
$$190$$ 0 0
$$191$$ −1.37612 −0.0995728 −0.0497864 0.998760i $$-0.515854\pi$$
−0.0497864 + 0.998760i $$0.515854\pi$$
$$192$$ 0.707861 0.0510855
$$193$$ −5.15711 −0.371217 −0.185609 0.982624i $$-0.559426\pi$$
−0.185609 + 0.982624i $$0.559426\pi$$
$$194$$ 10.8395 0.778231
$$195$$ 0 0
$$196$$ −2.08641 −0.149029
$$197$$ −12.0998 −0.862072 −0.431036 0.902335i $$-0.641852\pi$$
−0.431036 + 0.902335i $$0.641852\pi$$
$$198$$ −3.80406 −0.270343
$$199$$ −13.2948 −0.942445 −0.471223 0.882014i $$-0.656187\pi$$
−0.471223 + 0.882014i $$0.656187\pi$$
$$200$$ 0 0
$$201$$ 1.17410 0.0828149
$$202$$ 28.4420 2.00117
$$203$$ 4.19239 0.294248
$$204$$ −0.594768 −0.0416421
$$205$$ 0 0
$$206$$ 1.12250 0.0782080
$$207$$ −13.4753 −0.936601
$$208$$ −18.1991 −1.26188
$$209$$ 1.00000 0.0691714
$$210$$ 0 0
$$211$$ −26.0866 −1.79588 −0.897939 0.440120i $$-0.854936\pi$$
−0.897939 + 0.440120i $$0.854936\pi$$
$$212$$ 2.88188 0.197928
$$213$$ 9.50828 0.651497
$$214$$ −30.5873 −2.09090
$$215$$ 0 0
$$216$$ −7.93722 −0.540059
$$217$$ 0.409148 0.0277748
$$218$$ −33.7971 −2.28903
$$219$$ 4.66524 0.315248
$$220$$ 0 0
$$221$$ −2.34631 −0.157830
$$222$$ −6.44837 −0.432786
$$223$$ −13.8606 −0.928175 −0.464087 0.885789i $$-0.653618\pi$$
−0.464087 + 0.885789i $$0.653618\pi$$
$$224$$ −11.8027 −0.788600
$$225$$ 0 0
$$226$$ −10.2314 −0.680583
$$227$$ 12.9104 0.856893 0.428446 0.903567i $$-0.359061\pi$$
0.428446 + 0.903567i $$0.359061\pi$$
$$228$$ −0.922763 −0.0611115
$$229$$ 7.93710 0.524499 0.262249 0.965000i $$-0.415536\pi$$
0.262249 + 0.965000i $$0.415536\pi$$
$$230$$ 0 0
$$231$$ 2.00776 0.132101
$$232$$ −3.19484 −0.209752
$$233$$ −17.6265 −1.15475 −0.577374 0.816480i $$-0.695922\pi$$
−0.577374 + 0.816480i $$0.695922\pi$$
$$234$$ 13.8476 0.905248
$$235$$ 0 0
$$236$$ −3.99347 −0.259953
$$237$$ 2.61360 0.169772
$$238$$ −2.49669 −0.161836
$$239$$ −18.6909 −1.20901 −0.604507 0.796600i $$-0.706630\pi$$
−0.604507 + 0.796600i $$0.706630\pi$$
$$240$$ 0 0
$$241$$ −0.877285 −0.0565109 −0.0282555 0.999601i $$-0.508995\pi$$
−0.0282555 + 0.999601i $$0.508995\pi$$
$$242$$ −1.73890 −0.111781
$$243$$ 16.1439 1.03563
$$244$$ 0.751518 0.0481110
$$245$$ 0 0
$$246$$ 2.40381 0.153261
$$247$$ −3.64023 −0.231622
$$248$$ −0.311795 −0.0197990
$$249$$ −12.4571 −0.789438
$$250$$ 0 0
$$251$$ −0.448818 −0.0283291 −0.0141646 0.999900i $$-0.504509\pi$$
−0.0141646 + 0.999900i $$0.504509\pi$$
$$252$$ 4.98899 0.314277
$$253$$ −6.15983 −0.387265
$$254$$ 27.7662 1.74221
$$255$$ 0 0
$$256$$ 19.2311 1.20194
$$257$$ 7.85809 0.490174 0.245087 0.969501i $$-0.421183\pi$$
0.245087 + 0.969501i $$0.421183\pi$$
$$258$$ −2.40703 −0.149855
$$259$$ −9.16484 −0.569476
$$260$$ 0 0
$$261$$ 4.11719 0.254848
$$262$$ 6.83691 0.422386
$$263$$ −0.585215 −0.0360859 −0.0180429 0.999837i $$-0.505744\pi$$
−0.0180429 + 0.999837i $$0.505744\pi$$
$$264$$ −1.53003 −0.0941669
$$265$$ 0 0
$$266$$ −3.87353 −0.237502
$$267$$ −5.76742 −0.352960
$$268$$ 1.33363 0.0814645
$$269$$ −3.59078 −0.218934 −0.109467 0.993990i $$-0.534914\pi$$
−0.109467 + 0.993990i $$0.534914\pi$$
$$270$$ 0 0
$$271$$ −23.4466 −1.42428 −0.712139 0.702038i $$-0.752274\pi$$
−0.712139 + 0.702038i $$0.752274\pi$$
$$272$$ 3.22239 0.195386
$$273$$ −7.30870 −0.442343
$$274$$ 27.0264 1.63272
$$275$$ 0 0
$$276$$ 5.68406 0.342140
$$277$$ 19.2912 1.15909 0.579547 0.814939i $$-0.303229\pi$$
0.579547 + 0.814939i $$0.303229\pi$$
$$278$$ 11.6489 0.698656
$$279$$ 0.401809 0.0240557
$$280$$ 0 0
$$281$$ −5.82550 −0.347520 −0.173760 0.984788i $$-0.555592\pi$$
−0.173760 + 0.984788i $$0.555592\pi$$
$$282$$ 2.75592 0.164112
$$283$$ 15.5166 0.922368 0.461184 0.887305i $$-0.347425\pi$$
0.461184 + 0.887305i $$0.347425\pi$$
$$284$$ 10.8002 0.640874
$$285$$ 0 0
$$286$$ 6.33001 0.374301
$$287$$ 3.41645 0.201667
$$288$$ −11.5910 −0.683006
$$289$$ −16.5846 −0.975562
$$290$$ 0 0
$$291$$ −5.61842 −0.329357
$$292$$ 5.29912 0.310107
$$293$$ 12.8089 0.748305 0.374152 0.927367i $$-0.377934\pi$$
0.374152 + 0.927367i $$0.377934\pi$$
$$294$$ 3.19407 0.186282
$$295$$ 0 0
$$296$$ 6.98415 0.405945
$$297$$ 4.67572 0.271313
$$298$$ −1.36164 −0.0788777
$$299$$ 22.4232 1.29677
$$300$$ 0 0
$$301$$ −3.42103 −0.197185
$$302$$ −14.7242 −0.847280
$$303$$ −14.7423 −0.846923
$$304$$ 4.99943 0.286737
$$305$$ 0 0
$$306$$ −2.45191 −0.140166
$$307$$ 14.0052 0.799321 0.399661 0.916663i $$-0.369128\pi$$
0.399661 + 0.916663i $$0.369128\pi$$
$$308$$ 2.28056 0.129947
$$309$$ −0.581821 −0.0330987
$$310$$ 0 0
$$311$$ 4.10325 0.232674 0.116337 0.993210i $$-0.462885\pi$$
0.116337 + 0.993210i $$0.462885\pi$$
$$312$$ 5.56966 0.315320
$$313$$ −21.5394 −1.21748 −0.608740 0.793370i $$-0.708325\pi$$
−0.608740 + 0.793370i $$0.708325\pi$$
$$314$$ 14.3812 0.811576
$$315$$ 0 0
$$316$$ 2.96872 0.167004
$$317$$ 8.36917 0.470059 0.235030 0.971988i $$-0.424481\pi$$
0.235030 + 0.971988i $$0.424481\pi$$
$$318$$ −4.41186 −0.247405
$$319$$ 1.88204 0.105374
$$320$$ 0 0
$$321$$ 15.8542 0.884898
$$322$$ 23.8603 1.32968
$$323$$ 0.644551 0.0358638
$$324$$ 2.40439 0.133577
$$325$$ 0 0
$$326$$ 41.5650 2.30207
$$327$$ 17.5180 0.968746
$$328$$ −2.60353 −0.143756
$$329$$ 3.91689 0.215945
$$330$$ 0 0
$$331$$ −29.1194 −1.60055 −0.800274 0.599635i $$-0.795312\pi$$
−0.800274 + 0.599635i $$0.795312\pi$$
$$332$$ −14.1497 −0.776565
$$333$$ −9.00046 −0.493222
$$334$$ −33.8052 −1.84974
$$335$$ 0 0
$$336$$ 10.0377 0.547600
$$337$$ −3.36844 −0.183491 −0.0917453 0.995783i $$-0.529245\pi$$
−0.0917453 + 0.995783i $$0.529245\pi$$
$$338$$ −0.436905 −0.0237645
$$339$$ 5.30323 0.288032
$$340$$ 0 0
$$341$$ 0.183675 0.00994653
$$342$$ −3.80406 −0.205700
$$343$$ 20.1326 1.08706
$$344$$ 2.60702 0.140561
$$345$$ 0 0
$$346$$ −20.3257 −1.09272
$$347$$ −3.97272 −0.213267 −0.106633 0.994298i $$-0.534007\pi$$
−0.106633 + 0.994298i $$0.534007\pi$$
$$348$$ −1.73668 −0.0930959
$$349$$ 1.51616 0.0811581 0.0405790 0.999176i $$-0.487080\pi$$
0.0405790 + 0.999176i $$0.487080\pi$$
$$350$$ 0 0
$$351$$ −17.0207 −0.908496
$$352$$ −5.29846 −0.282409
$$353$$ −4.76876 −0.253816 −0.126908 0.991915i $$-0.540505\pi$$
−0.126908 + 0.991915i $$0.540505\pi$$
$$354$$ 6.11359 0.324934
$$355$$ 0 0
$$356$$ −6.55105 −0.347205
$$357$$ 1.29410 0.0684912
$$358$$ 11.7802 0.622605
$$359$$ 0.542750 0.0286453 0.0143226 0.999897i $$-0.495441\pi$$
0.0143226 + 0.999897i $$0.495441\pi$$
$$360$$ 0 0
$$361$$ 1.00000 0.0526316
$$362$$ 7.55643 0.397157
$$363$$ 0.901323 0.0473072
$$364$$ −8.30176 −0.435130
$$365$$ 0 0
$$366$$ −1.15050 −0.0601375
$$367$$ 26.1792 1.36654 0.683271 0.730165i $$-0.260557\pi$$
0.683271 + 0.730165i $$0.260557\pi$$
$$368$$ −30.7956 −1.60533
$$369$$ 3.35517 0.174663
$$370$$ 0 0
$$371$$ −6.27042 −0.325544
$$372$$ −0.169488 −0.00878755
$$373$$ 12.7807 0.661760 0.330880 0.943673i $$-0.392655\pi$$
0.330880 + 0.943673i $$0.392655\pi$$
$$374$$ −1.12081 −0.0579558
$$375$$ 0 0
$$376$$ −2.98490 −0.153934
$$377$$ −6.85107 −0.352848
$$378$$ −18.1115 −0.931557
$$379$$ −26.0931 −1.34031 −0.670157 0.742219i $$-0.733773\pi$$
−0.670157 + 0.742219i $$0.733773\pi$$
$$380$$ 0 0
$$381$$ −14.3920 −0.737325
$$382$$ 2.39295 0.122434
$$383$$ 12.7812 0.653091 0.326545 0.945182i $$-0.394115\pi$$
0.326545 + 0.945182i $$0.394115\pi$$
$$384$$ −10.7821 −0.550224
$$385$$ 0 0
$$386$$ 8.96773 0.456445
$$387$$ −3.35967 −0.170781
$$388$$ −6.38180 −0.323987
$$389$$ −8.97979 −0.455293 −0.227647 0.973744i $$-0.573103\pi$$
−0.227647 + 0.973744i $$0.573103\pi$$
$$390$$ 0 0
$$391$$ −3.97032 −0.200788
$$392$$ −3.45946 −0.174729
$$393$$ −3.54376 −0.178759
$$394$$ 21.0403 1.06000
$$395$$ 0 0
$$396$$ 2.23966 0.112547
$$397$$ −26.8417 −1.34715 −0.673573 0.739120i $$-0.735242\pi$$
−0.673573 + 0.739120i $$0.735242\pi$$
$$398$$ 23.1184 1.15882
$$399$$ 2.00776 0.100514
$$400$$ 0 0
$$401$$ −9.25518 −0.462182 −0.231091 0.972932i $$-0.574229\pi$$
−0.231091 + 0.972932i $$0.574229\pi$$
$$402$$ −2.04165 −0.101828
$$403$$ −0.668617 −0.0333062
$$404$$ −16.7454 −0.833114
$$405$$ 0 0
$$406$$ −7.29016 −0.361804
$$407$$ −4.11428 −0.203937
$$408$$ −0.986183 −0.0488233
$$409$$ 26.0835 1.28975 0.644874 0.764289i $$-0.276910\pi$$
0.644874 + 0.764289i $$0.276910\pi$$
$$410$$ 0 0
$$411$$ −14.0085 −0.690990
$$412$$ −0.660875 −0.0325590
$$413$$ 8.68904 0.427560
$$414$$ 23.4323 1.15164
$$415$$ 0 0
$$416$$ 19.2876 0.945652
$$417$$ −6.03796 −0.295680
$$418$$ −1.73890 −0.0850526
$$419$$ −32.5181 −1.58861 −0.794307 0.607517i $$-0.792166\pi$$
−0.794307 + 0.607517i $$0.792166\pi$$
$$420$$ 0 0
$$421$$ 3.16934 0.154464 0.0772320 0.997013i $$-0.475392\pi$$
0.0772320 + 0.997013i $$0.475392\pi$$
$$422$$ 45.3621 2.20819
$$423$$ 3.84664 0.187030
$$424$$ 4.77843 0.232061
$$425$$ 0 0
$$426$$ −16.5340 −0.801074
$$427$$ −1.63516 −0.0791311
$$428$$ 18.0084 0.870469
$$429$$ −3.28102 −0.158409
$$430$$ 0 0
$$431$$ −1.19700 −0.0576575 −0.0288288 0.999584i $$-0.509178\pi$$
−0.0288288 + 0.999584i $$0.509178\pi$$
$$432$$ 23.3759 1.12468
$$433$$ −23.4218 −1.12558 −0.562789 0.826601i $$-0.690272\pi$$
−0.562789 + 0.826601i $$0.690272\pi$$
$$434$$ −0.711469 −0.0341516
$$435$$ 0 0
$$436$$ 19.8982 0.952950
$$437$$ −6.15983 −0.294664
$$438$$ −8.11241 −0.387626
$$439$$ −21.4961 −1.02595 −0.512975 0.858403i $$-0.671457\pi$$
−0.512975 + 0.858403i $$0.671457\pi$$
$$440$$ 0 0
$$441$$ 4.45821 0.212295
$$442$$ 4.08001 0.194066
$$443$$ −23.6133 −1.12190 −0.560950 0.827850i $$-0.689564\pi$$
−0.560950 + 0.827850i $$0.689564\pi$$
$$444$$ 3.79650 0.180174
$$445$$ 0 0
$$446$$ 24.1023 1.14128
$$447$$ 0.705777 0.0333821
$$448$$ −1.74944 −0.0826533
$$449$$ −5.35004 −0.252484 −0.126242 0.991999i $$-0.540292\pi$$
−0.126242 + 0.991999i $$0.540292\pi$$
$$450$$ 0 0
$$451$$ 1.53371 0.0722196
$$452$$ 6.02379 0.283335
$$453$$ 7.63194 0.358580
$$454$$ −22.4499 −1.05363
$$455$$ 0 0
$$456$$ −1.53003 −0.0716503
$$457$$ −39.7929 −1.86143 −0.930716 0.365742i $$-0.880815\pi$$
−0.930716 + 0.365742i $$0.880815\pi$$
$$458$$ −13.8019 −0.644919
$$459$$ 3.01374 0.140669
$$460$$ 0 0
$$461$$ −3.21566 −0.149768 −0.0748840 0.997192i $$-0.523859\pi$$
−0.0748840 + 0.997192i $$0.523859\pi$$
$$462$$ −3.49130 −0.162430
$$463$$ −16.7065 −0.776416 −0.388208 0.921572i $$-0.626906\pi$$
−0.388208 + 0.921572i $$0.626906\pi$$
$$464$$ 9.40915 0.436809
$$465$$ 0 0
$$466$$ 30.6507 1.41987
$$467$$ 37.8565 1.75179 0.875894 0.482503i $$-0.160272\pi$$
0.875894 + 0.482503i $$0.160272\pi$$
$$468$$ −8.15285 −0.376866
$$469$$ −2.90173 −0.133990
$$470$$ 0 0
$$471$$ −7.45416 −0.343469
$$472$$ −6.62156 −0.304782
$$473$$ −1.53577 −0.0706146
$$474$$ −4.54481 −0.208750
$$475$$ 0 0
$$476$$ 1.46994 0.0673744
$$477$$ −6.15796 −0.281953
$$478$$ 32.5017 1.48659
$$479$$ −16.6101 −0.758936 −0.379468 0.925205i $$-0.623893\pi$$
−0.379468 + 0.925205i $$0.623893\pi$$
$$480$$ 0 0
$$481$$ 14.9769 0.682888
$$482$$ 1.52552 0.0694853
$$483$$ −12.3675 −0.562739
$$484$$ 1.02379 0.0465358
$$485$$ 0 0
$$486$$ −28.0727 −1.27341
$$487$$ −29.2975 −1.32760 −0.663798 0.747912i $$-0.731056\pi$$
−0.663798 + 0.747912i $$0.731056\pi$$
$$488$$ 1.24609 0.0564079
$$489$$ −21.5443 −0.974267
$$490$$ 0 0
$$491$$ −0.806717 −0.0364066 −0.0182033 0.999834i $$-0.505795\pi$$
−0.0182033 + 0.999834i $$0.505795\pi$$
$$492$$ −1.41525 −0.0638045
$$493$$ 1.21307 0.0546340
$$494$$ 6.33001 0.284800
$$495$$ 0 0
$$496$$ 0.918269 0.0412315
$$497$$ −23.4992 −1.05408
$$498$$ 21.6617 0.970686
$$499$$ −30.7212 −1.37527 −0.687634 0.726058i $$-0.741351\pi$$
−0.687634 + 0.726058i $$0.741351\pi$$
$$500$$ 0 0
$$501$$ 17.5222 0.782833
$$502$$ 0.780451 0.0348332
$$503$$ 13.0527 0.581991 0.290995 0.956724i $$-0.406014\pi$$
0.290995 + 0.956724i $$0.406014\pi$$
$$504$$ 8.27223 0.368475
$$505$$ 0 0
$$506$$ 10.7113 0.476178
$$507$$ 0.226460 0.0100574
$$508$$ −16.3475 −0.725302
$$509$$ −22.6615 −1.00445 −0.502226 0.864736i $$-0.667485\pi$$
−0.502226 + 0.864736i $$0.667485\pi$$
$$510$$ 0 0
$$511$$ −11.5299 −0.510052
$$512$$ −9.51581 −0.420544
$$513$$ 4.67572 0.206438
$$514$$ −13.6645 −0.602713
$$515$$ 0 0
$$516$$ 1.41715 0.0623865
$$517$$ 1.75837 0.0773329
$$518$$ 15.9368 0.700222
$$519$$ 10.5354 0.462451
$$520$$ 0 0
$$521$$ 19.0608 0.835070 0.417535 0.908661i $$-0.362894\pi$$
0.417535 + 0.908661i $$0.362894\pi$$
$$522$$ −7.15940 −0.313358
$$523$$ −15.1864 −0.664054 −0.332027 0.943270i $$-0.607732\pi$$
−0.332027 + 0.943270i $$0.607732\pi$$
$$524$$ −4.02526 −0.175844
$$525$$ 0 0
$$526$$ 1.01763 0.0443709
$$527$$ 0.118388 0.00515704
$$528$$ 4.50610 0.196103
$$529$$ 14.9435 0.649716
$$530$$ 0 0
$$531$$ 8.53319 0.370309
$$532$$ 2.28056 0.0988748
$$533$$ −5.58305 −0.241829
$$534$$ 10.0290 0.433997
$$535$$ 0 0
$$536$$ 2.21129 0.0955132
$$537$$ −6.10603 −0.263495
$$538$$ 6.24402 0.269199
$$539$$ 2.03793 0.0877798
$$540$$ 0 0
$$541$$ 41.1432 1.76888 0.884442 0.466650i $$-0.154539\pi$$
0.884442 + 0.466650i $$0.154539\pi$$
$$542$$ 40.7714 1.75128
$$543$$ −3.91671 −0.168082
$$544$$ −3.41513 −0.146422
$$545$$ 0 0
$$546$$ 12.7091 0.543901
$$547$$ −40.7714 −1.74326 −0.871630 0.490164i $$-0.836937\pi$$
−0.871630 + 0.490164i $$0.836937\pi$$
$$548$$ −15.9119 −0.679722
$$549$$ −1.60583 −0.0685354
$$550$$ 0 0
$$551$$ 1.88204 0.0801778
$$552$$ 9.42473 0.401143
$$553$$ −6.45938 −0.274681
$$554$$ −33.5455 −1.42521
$$555$$ 0 0
$$556$$ −6.85835 −0.290859
$$557$$ 14.2315 0.603009 0.301504 0.953465i $$-0.402511\pi$$
0.301504 + 0.953465i $$0.402511\pi$$
$$558$$ −0.698708 −0.0295787
$$559$$ 5.59054 0.236454
$$560$$ 0 0
$$561$$ 0.580948 0.0245277
$$562$$ 10.1300 0.427308
$$563$$ 11.1511 0.469963 0.234982 0.972000i $$-0.424497\pi$$
0.234982 + 0.972000i $$0.424497\pi$$
$$564$$ −1.62256 −0.0683220
$$565$$ 0 0
$$566$$ −26.9819 −1.13414
$$567$$ −5.23149 −0.219702
$$568$$ 17.9078 0.751393
$$569$$ 11.4022 0.478005 0.239003 0.971019i $$-0.423180\pi$$
0.239003 + 0.971019i $$0.423180\pi$$
$$570$$ 0 0
$$571$$ 12.5111 0.523573 0.261787 0.965126i $$-0.415688\pi$$
0.261787 + 0.965126i $$0.415688\pi$$
$$572$$ −3.72682 −0.155826
$$573$$ −1.24033 −0.0518156
$$574$$ −5.94088 −0.247967
$$575$$ 0 0
$$576$$ −1.71806 −0.0715860
$$577$$ −7.95520 −0.331179 −0.165590 0.986195i $$-0.552953\pi$$
−0.165590 + 0.986195i $$0.552953\pi$$
$$578$$ 28.8390 1.19954
$$579$$ −4.64822 −0.193174
$$580$$ 0 0
$$581$$ 30.7871 1.27726
$$582$$ 9.76989 0.404975
$$583$$ −2.81491 −0.116582
$$584$$ 8.78645 0.363586
$$585$$ 0 0
$$586$$ −22.2735 −0.920109
$$587$$ −45.7680 −1.88905 −0.944524 0.328443i $$-0.893476\pi$$
−0.944524 + 0.328443i $$0.893476\pi$$
$$588$$ −1.88052 −0.0775515
$$589$$ 0.183675 0.00756818
$$590$$ 0 0
$$591$$ −10.9058 −0.448604
$$592$$ −20.5691 −0.845383
$$593$$ 26.3613 1.08253 0.541264 0.840853i $$-0.317946\pi$$
0.541264 + 0.840853i $$0.317946\pi$$
$$594$$ −8.13063 −0.333604
$$595$$ 0 0
$$596$$ 0.801672 0.0328378
$$597$$ −11.9829 −0.490429
$$598$$ −38.9917 −1.59449
$$599$$ 17.3368 0.708361 0.354180 0.935177i $$-0.384760\pi$$
0.354180 + 0.935177i $$0.384760\pi$$
$$600$$ 0 0
$$601$$ −15.7946 −0.644274 −0.322137 0.946693i $$-0.604401\pi$$
−0.322137 + 0.946693i $$0.604401\pi$$
$$602$$ 5.94884 0.242457
$$603$$ −2.84969 −0.116048
$$604$$ 8.66891 0.352733
$$605$$ 0 0
$$606$$ 25.6355 1.04137
$$607$$ 1.80754 0.0733659 0.0366829 0.999327i $$-0.488321\pi$$
0.0366829 + 0.999327i $$0.488321\pi$$
$$608$$ −5.29846 −0.214881
$$609$$ 3.77869 0.153120
$$610$$ 0 0
$$611$$ −6.40086 −0.258951
$$612$$ 1.44357 0.0583529
$$613$$ −10.9246 −0.441239 −0.220619 0.975360i $$-0.570808\pi$$
−0.220619 + 0.975360i $$0.570808\pi$$
$$614$$ −24.3538 −0.982838
$$615$$ 0 0
$$616$$ 3.78139 0.152357
$$617$$ 2.68742 0.108191 0.0540957 0.998536i $$-0.482772\pi$$
0.0540957 + 0.998536i $$0.482772\pi$$
$$618$$ 1.01173 0.0406978
$$619$$ 18.2439 0.733284 0.366642 0.930362i $$-0.380507\pi$$
0.366642 + 0.930362i $$0.380507\pi$$
$$620$$ 0 0
$$621$$ −28.8016 −1.15577
$$622$$ −7.13515 −0.286094
$$623$$ 14.2539 0.571069
$$624$$ −16.4032 −0.656655
$$625$$ 0 0
$$626$$ 37.4550 1.49700
$$627$$ 0.901323 0.0359954
$$628$$ −8.46697 −0.337869
$$629$$ −2.65186 −0.105737
$$630$$ 0 0
$$631$$ 9.12012 0.363066 0.181533 0.983385i $$-0.441894\pi$$
0.181533 + 0.983385i $$0.441894\pi$$
$$632$$ 4.92243 0.195804
$$633$$ −23.5125 −0.934537
$$634$$ −14.5532 −0.577981
$$635$$ 0 0
$$636$$ 2.59750 0.102998
$$637$$ −7.41852 −0.293932
$$638$$ −3.27269 −0.129567
$$639$$ −23.0777 −0.912940
$$640$$ 0 0
$$641$$ −2.90773 −0.114848 −0.0574241 0.998350i $$-0.518289\pi$$
−0.0574241 + 0.998350i $$0.518289\pi$$
$$642$$ −27.5690 −1.08806
$$643$$ 3.35603 0.132349 0.0661745 0.997808i $$-0.478921\pi$$
0.0661745 + 0.997808i $$0.478921\pi$$
$$644$$ −14.0479 −0.553563
$$645$$ 0 0
$$646$$ −1.12081 −0.0440978
$$647$$ −26.5428 −1.04351 −0.521753 0.853096i $$-0.674722\pi$$
−0.521753 + 0.853096i $$0.674722\pi$$
$$648$$ 3.98670 0.156613
$$649$$ 3.90068 0.153115
$$650$$ 0 0
$$651$$ 0.368774 0.0144534
$$652$$ −24.4716 −0.958381
$$653$$ −28.6844 −1.12251 −0.561253 0.827644i $$-0.689681\pi$$
−0.561253 + 0.827644i $$0.689681\pi$$
$$654$$ −30.4621 −1.19116
$$655$$ 0 0
$$656$$ 7.66768 0.299373
$$657$$ −11.3231 −0.441756
$$658$$ −6.81110 −0.265524
$$659$$ −39.0719 −1.52203 −0.761013 0.648737i $$-0.775297\pi$$
−0.761013 + 0.648737i $$0.775297\pi$$
$$660$$ 0 0
$$661$$ −21.1901 −0.824199 −0.412099 0.911139i $$-0.635204\pi$$
−0.412099 + 0.911139i $$0.635204\pi$$
$$662$$ 50.6358 1.96802
$$663$$ −2.11478 −0.0821314
$$664$$ −23.4616 −0.910486
$$665$$ 0 0
$$666$$ 15.6509 0.606462
$$667$$ −11.5931 −0.448885
$$668$$ 19.9030 0.770069
$$669$$ −12.4929 −0.483003
$$670$$ 0 0
$$671$$ −0.734057 −0.0283379
$$672$$ −10.6380 −0.410371
$$673$$ 24.6931 0.951849 0.475924 0.879486i $$-0.342114\pi$$
0.475924 + 0.879486i $$0.342114\pi$$
$$674$$ 5.85739 0.225618
$$675$$ 0 0
$$676$$ 0.257230 0.00989345
$$677$$ −5.41516 −0.208122 −0.104061 0.994571i $$-0.533184\pi$$
−0.104061 + 0.994571i $$0.533184\pi$$
$$678$$ −9.22180 −0.354161
$$679$$ 13.8856 0.532881
$$680$$ 0 0
$$681$$ 11.6364 0.445909
$$682$$ −0.319392 −0.0122302
$$683$$ 13.7885 0.527604 0.263802 0.964577i $$-0.415023\pi$$
0.263802 + 0.964577i $$0.415023\pi$$
$$684$$ 2.23966 0.0856354
$$685$$ 0 0
$$686$$ −35.0087 −1.33664
$$687$$ 7.15389 0.272938
$$688$$ −7.67796 −0.292719
$$689$$ 10.2469 0.390377
$$690$$ 0 0
$$691$$ −31.5750 −1.20117 −0.600586 0.799560i $$-0.705066\pi$$
−0.600586 + 0.799560i $$0.705066\pi$$
$$692$$ 11.9668 0.454911
$$693$$ −4.87307 −0.185113
$$694$$ 6.90818 0.262231
$$695$$ 0 0
$$696$$ −2.87959 −0.109150
$$697$$ 0.988554 0.0374442
$$698$$ −2.63645 −0.0997912
$$699$$ −15.8871 −0.600906
$$700$$ 0 0
$$701$$ −17.4459 −0.658924 −0.329462 0.944169i $$-0.606867\pi$$
−0.329462 + 0.944169i $$0.606867\pi$$
$$702$$ 29.5973 1.11708
$$703$$ −4.11428 −0.155173
$$704$$ −0.785358 −0.0295993
$$705$$ 0 0
$$706$$ 8.29242 0.312089
$$707$$ 36.4348 1.37027
$$708$$ −3.59940 −0.135274
$$709$$ −38.7266 −1.45441 −0.727205 0.686421i $$-0.759181\pi$$
−0.727205 + 0.686421i $$0.759181\pi$$
$$710$$ 0 0
$$711$$ −6.34352 −0.237901
$$712$$ −10.8623 −0.407081
$$713$$ −1.13140 −0.0423714
$$714$$ −2.25032 −0.0842162
$$715$$ 0 0
$$716$$ −6.93567 −0.259198
$$717$$ −16.8465 −0.629145
$$718$$ −0.943791 −0.0352220
$$719$$ −28.9402 −1.07929 −0.539644 0.841894i $$-0.681441\pi$$
−0.539644 + 0.841894i $$0.681441\pi$$
$$720$$ 0 0
$$721$$ 1.43794 0.0535517
$$722$$ −1.73890 −0.0647153
$$723$$ −0.790717 −0.0294071
$$724$$ −4.44888 −0.165341
$$725$$ 0 0
$$726$$ −1.56731 −0.0581685
$$727$$ 0.331738 0.0123035 0.00615174 0.999981i $$-0.498042\pi$$
0.00615174 + 0.999981i $$0.498042\pi$$
$$728$$ −13.7651 −0.510169
$$729$$ 7.50533 0.277975
$$730$$ 0 0
$$731$$ −0.989879 −0.0366120
$$732$$ 0.677361 0.0250360
$$733$$ 30.5241 1.12743 0.563716 0.825969i $$-0.309371\pi$$
0.563716 + 0.825969i $$0.309371\pi$$
$$734$$ −45.5231 −1.68029
$$735$$ 0 0
$$736$$ 32.6376 1.20304
$$737$$ −1.30264 −0.0479835
$$738$$ −5.83432 −0.214764
$$739$$ −8.65314 −0.318311 −0.159155 0.987254i $$-0.550877\pi$$
−0.159155 + 0.987254i $$0.550877\pi$$
$$740$$ 0 0
$$741$$ −3.28102 −0.120531
$$742$$ 10.9037 0.400286
$$743$$ −33.9919 −1.24704 −0.623521 0.781807i $$-0.714298\pi$$
−0.623521 + 0.781807i $$0.714298\pi$$
$$744$$ −0.281028 −0.0103030
$$745$$ 0 0
$$746$$ −22.2244 −0.813694
$$747$$ 30.2349 1.10624
$$748$$ 0.659883 0.0241277
$$749$$ −39.1829 −1.43171
$$750$$ 0 0
$$751$$ −50.2581 −1.83394 −0.916972 0.398951i $$-0.869374\pi$$
−0.916972 + 0.398951i $$0.869374\pi$$
$$752$$ 8.79085 0.320569
$$753$$ −0.404530 −0.0147419
$$754$$ 11.9133 0.433859
$$755$$ 0 0
$$756$$ 10.6633 0.387819
$$757$$ −10.7948 −0.392344 −0.196172 0.980570i $$-0.562851\pi$$
−0.196172 + 0.980570i $$0.562851\pi$$
$$758$$ 45.3734 1.64804
$$759$$ −5.55199 −0.201525
$$760$$ 0 0
$$761$$ 15.5661 0.564272 0.282136 0.959374i $$-0.408957\pi$$
0.282136 + 0.959374i $$0.408957\pi$$
$$762$$ 25.0263 0.906608
$$763$$ −43.2947 −1.56737
$$764$$ −1.40886 −0.0509707
$$765$$ 0 0
$$766$$ −22.2254 −0.803035
$$767$$ −14.1994 −0.512709
$$768$$ 17.3334 0.625465
$$769$$ 19.6520 0.708671 0.354335 0.935118i $$-0.384707\pi$$
0.354335 + 0.935118i $$0.384707\pi$$
$$770$$ 0 0
$$771$$ 7.08267 0.255076
$$772$$ −5.27979 −0.190024
$$773$$ 41.0208 1.47541 0.737707 0.675121i $$-0.235908\pi$$
0.737707 + 0.675121i $$0.235908\pi$$
$$774$$ 5.84214 0.209991
$$775$$ 0 0
$$776$$ −10.5817 −0.379859
$$777$$ −8.26048 −0.296343
$$778$$ 15.6150 0.559825
$$779$$ 1.53371 0.0549509
$$780$$ 0 0
$$781$$ −10.5493 −0.377482
$$782$$ 6.90401 0.246887
$$783$$ 8.79991 0.314483
$$784$$ 10.1885 0.363874
$$785$$ 0 0
$$786$$ 6.16226 0.219801
$$787$$ −25.2114 −0.898688 −0.449344 0.893359i $$-0.648342\pi$$
−0.449344 + 0.893359i $$0.648342\pi$$
$$788$$ −12.3876 −0.441289
$$789$$ −0.527467 −0.0187783
$$790$$ 0 0
$$791$$ −13.1066 −0.466018
$$792$$ 3.71357 0.131956
$$793$$ 2.67213 0.0948902
$$794$$ 46.6752 1.65644
$$795$$ 0 0
$$796$$ −13.6111 −0.482432
$$797$$ 23.4053 0.829057 0.414528 0.910036i $$-0.363947\pi$$
0.414528 + 0.910036i $$0.363947\pi$$
$$798$$ −3.49130 −0.123591
$$799$$ 1.13336 0.0400953
$$800$$ 0 0
$$801$$ 13.9982 0.494602
$$802$$ 16.0939 0.568294
$$803$$ −5.17599 −0.182657
$$804$$ 1.20203 0.0423924
$$805$$ 0 0
$$806$$ 1.16266 0.0409530
$$807$$ −3.23645 −0.113929
$$808$$ −27.7655 −0.976786
$$809$$ −25.3235 −0.890325 −0.445163 0.895450i $$-0.646854\pi$$
−0.445163 + 0.895450i $$0.646854\pi$$
$$810$$ 0 0
$$811$$ −1.24610 −0.0437566 −0.0218783 0.999761i $$-0.506965\pi$$
−0.0218783 + 0.999761i $$0.506965\pi$$
$$812$$ 4.29211 0.150624
$$813$$ −21.1329 −0.741165
$$814$$ 7.15433 0.250759
$$815$$ 0 0
$$816$$ 2.90441 0.101675
$$817$$ −1.53577 −0.0537296
$$818$$ −45.3567 −1.58586
$$819$$ 17.7391 0.619854
$$820$$ 0 0
$$821$$ 20.4113 0.712359 0.356180 0.934418i $$-0.384079\pi$$
0.356180 + 0.934418i $$0.384079\pi$$
$$822$$ 24.3595 0.849634
$$823$$ 15.4655 0.539093 0.269546 0.962987i $$-0.413126\pi$$
0.269546 + 0.962987i $$0.413126\pi$$
$$824$$ −1.09579 −0.0381738
$$825$$ 0 0
$$826$$ −15.1094 −0.525723
$$827$$ 46.2008 1.60656 0.803280 0.595602i $$-0.203087\pi$$
0.803280 + 0.595602i $$0.203087\pi$$
$$828$$ −13.7959 −0.479440
$$829$$ 2.91356 0.101192 0.0505961 0.998719i $$-0.483888\pi$$
0.0505961 + 0.998719i $$0.483888\pi$$
$$830$$ 0 0
$$831$$ 17.3876 0.603168
$$832$$ 2.85888 0.0991139
$$833$$ 1.31355 0.0455117
$$834$$ 10.4994 0.363566
$$835$$ 0 0
$$836$$ 1.02379 0.0354084
$$837$$ 0.858810 0.0296848
$$838$$ 56.5459 1.95334
$$839$$ 28.2887 0.976633 0.488317 0.872667i $$-0.337611\pi$$
0.488317 + 0.872667i $$0.337611\pi$$
$$840$$ 0 0
$$841$$ −25.4579 −0.877859
$$842$$ −5.51117 −0.189928
$$843$$ −5.25066 −0.180842
$$844$$ −26.7072 −0.919299
$$845$$ 0 0
$$846$$ −6.68893 −0.229970
$$847$$ −2.22757 −0.0765402
$$848$$ −14.0730 −0.483268
$$849$$ 13.9855 0.479981
$$850$$ 0 0
$$851$$ 25.3432 0.868755
$$852$$ 9.73446 0.333497
$$853$$ −41.9234 −1.43543 −0.717715 0.696337i $$-0.754812\pi$$
−0.717715 + 0.696337i $$0.754812\pi$$
$$854$$ 2.84339 0.0972989
$$855$$ 0 0
$$856$$ 29.8597 1.02058
$$857$$ −52.4352 −1.79115 −0.895576 0.444909i $$-0.853236\pi$$
−0.895576 + 0.444909i $$0.853236\pi$$
$$858$$ 5.70538 0.194778
$$859$$ −21.2305 −0.724374 −0.362187 0.932105i $$-0.617970\pi$$
−0.362187 + 0.932105i $$0.617970\pi$$
$$860$$ 0 0
$$861$$ 3.07932 0.104943
$$862$$ 2.08147 0.0708952
$$863$$ −1.75582 −0.0597690 −0.0298845 0.999553i $$-0.509514\pi$$
−0.0298845 + 0.999553i $$0.509514\pi$$
$$864$$ −24.7741 −0.842832
$$865$$ 0 0
$$866$$ 40.7282 1.38400
$$867$$ −14.9480 −0.507662
$$868$$ 0.418881 0.0142177
$$869$$ −2.89974 −0.0983670
$$870$$ 0 0
$$871$$ 4.74192 0.160674
$$872$$ 32.9931 1.11729
$$873$$ 13.6366 0.461528
$$874$$ 10.7113 0.362317
$$875$$ 0 0
$$876$$ 4.77622 0.161373
$$877$$ −27.7873 −0.938309 −0.469155 0.883116i $$-0.655441\pi$$
−0.469155 + 0.883116i $$0.655441\pi$$
$$878$$ 37.3796 1.26150
$$879$$ 11.5450 0.389402
$$880$$ 0 0
$$881$$ 38.8712 1.30960 0.654802 0.755800i $$-0.272752\pi$$
0.654802 + 0.755800i $$0.272752\pi$$
$$882$$ −7.75239 −0.261037
$$883$$ 1.20467 0.0405404 0.0202702 0.999795i $$-0.493547\pi$$
0.0202702 + 0.999795i $$0.493547\pi$$
$$884$$ −2.40212 −0.0807922
$$885$$ 0 0
$$886$$ 41.0612 1.37948
$$887$$ 44.0947 1.48056 0.740278 0.672301i $$-0.234694\pi$$
0.740278 + 0.672301i $$0.234694\pi$$
$$888$$ 6.29497 0.211245
$$889$$ 35.5691 1.19295
$$890$$ 0 0
$$891$$ −2.34852 −0.0786784
$$892$$ −14.1903 −0.475127
$$893$$ 1.75837 0.0588415
$$894$$ −1.22728 −0.0410463
$$895$$ 0 0
$$896$$ 26.6475 0.890230
$$897$$ 20.2105 0.674809
$$898$$ 9.30320 0.310452
$$899$$ 0.345683 0.0115292
$$900$$ 0 0
$$901$$ −1.81436 −0.0604450
$$902$$ −2.66698 −0.0888006
$$903$$ −3.08345 −0.102611
$$904$$ 9.98803 0.332197
$$905$$ 0 0
$$906$$ −13.2712 −0.440907
$$907$$ −36.2297 −1.20299 −0.601493 0.798878i $$-0.705427\pi$$
−0.601493 + 0.798878i $$0.705427\pi$$
$$908$$ 13.2175 0.438638
$$909$$ 35.7813 1.18679
$$910$$ 0 0
$$911$$ −5.41417 −0.179380 −0.0896898 0.995970i $$-0.528588\pi$$
−0.0896898 + 0.995970i $$0.528588\pi$$
$$912$$ 4.50610 0.149212
$$913$$ 13.8209 0.457406
$$914$$ 69.1960 2.28880
$$915$$ 0 0
$$916$$ 8.12591 0.268488
$$917$$ 8.75822 0.289222
$$918$$ −5.24060 −0.172966
$$919$$ 54.0931 1.78437 0.892183 0.451675i $$-0.149173\pi$$
0.892183 + 0.451675i $$0.149173\pi$$
$$920$$ 0 0
$$921$$ 12.6232 0.415950
$$922$$ 5.59172 0.184153
$$923$$ 38.4017 1.26401
$$924$$ 2.05552 0.0676217
$$925$$ 0 0
$$926$$ 29.0510 0.954674
$$927$$ 1.41215 0.0463811
$$928$$ −9.97193 −0.327345
$$929$$ 52.0785 1.70864 0.854320 0.519747i $$-0.173974\pi$$
0.854320 + 0.519747i $$0.173974\pi$$
$$930$$ 0 0
$$931$$ 2.03793 0.0667904
$$932$$ −18.0457 −0.591108
$$933$$ 3.69835 0.121079
$$934$$ −65.8288 −2.15398
$$935$$ 0 0
$$936$$ −13.5182 −0.441857
$$937$$ −35.9861 −1.17561 −0.587807 0.809002i $$-0.700008\pi$$
−0.587807 + 0.809002i $$0.700008\pi$$
$$938$$ 5.04584 0.164752
$$939$$ −19.4140 −0.633551
$$940$$ 0 0
$$941$$ 11.9972 0.391098 0.195549 0.980694i $$-0.437351\pi$$
0.195549 + 0.980694i $$0.437351\pi$$
$$942$$ 12.9621 0.422327
$$943$$ −9.44739 −0.307649
$$944$$ 19.5012 0.634709
$$945$$ 0 0
$$946$$ 2.67055 0.0868271
$$947$$ −37.1663 −1.20774 −0.603871 0.797082i $$-0.706376\pi$$
−0.603871 + 0.797082i $$0.706376\pi$$
$$948$$ 2.67577 0.0869051
$$949$$ 18.8418 0.611630
$$950$$ 0 0
$$951$$ 7.54332 0.244609
$$952$$ 2.43730 0.0789933
$$953$$ −16.5500 −0.536106 −0.268053 0.963404i $$-0.586380\pi$$
−0.268053 + 0.963404i $$0.586380\pi$$
$$954$$ 10.7081 0.346687
$$955$$ 0 0
$$956$$ −19.1355 −0.618886
$$957$$ 1.69633 0.0548345
$$958$$ 28.8834 0.933180
$$959$$ 34.6213 1.11798
$$960$$ 0 0
$$961$$ −30.9663 −0.998912
$$962$$ −26.0434 −0.839673
$$963$$ −38.4801 −1.24000
$$964$$ −0.898154 −0.0289276
$$965$$ 0 0
$$966$$ 21.5058 0.691939
$$967$$ −37.7443 −1.21377 −0.606887 0.794788i $$-0.707582\pi$$
−0.606887 + 0.794788i $$0.707582\pi$$
$$968$$ 1.69754 0.0545610
$$969$$ 0.580948 0.0186628
$$970$$ 0 0
$$971$$ −40.2912 −1.29301 −0.646504 0.762911i $$-0.723769\pi$$
−0.646504 + 0.762911i $$0.723769\pi$$
$$972$$ 16.5280 0.530134
$$973$$ 14.9225 0.478393
$$974$$ 50.9455 1.63240
$$975$$ 0 0
$$976$$ −3.66987 −0.117470
$$977$$ −1.24031 −0.0396812 −0.0198406 0.999803i $$-0.506316\pi$$
−0.0198406 + 0.999803i $$0.506316\pi$$
$$978$$ 37.4635 1.19795
$$979$$ 6.39884 0.204508
$$980$$ 0 0
$$981$$ −42.5182 −1.35750
$$982$$ 1.40280 0.0447653
$$983$$ −8.75471 −0.279232 −0.139616 0.990206i $$-0.544587\pi$$
−0.139616 + 0.990206i $$0.544587\pi$$
$$984$$ −2.34662 −0.0748077
$$985$$ 0 0
$$986$$ −2.10942 −0.0671775
$$987$$ 3.53038 0.112373
$$988$$ −3.72682 −0.118566
$$989$$ 9.46005 0.300812
$$990$$ 0 0
$$991$$ 49.7066 1.57898 0.789491 0.613763i $$-0.210345\pi$$
0.789491 + 0.613763i $$0.210345\pi$$
$$992$$ −0.973192 −0.0308989
$$993$$ −26.2460 −0.832891
$$994$$ 40.8629 1.29609
$$995$$ 0 0
$$996$$ −12.7534 −0.404108
$$997$$ −46.3005 −1.46635 −0.733176 0.680039i $$-0.761963\pi$$
−0.733176 + 0.680039i $$0.761963\pi$$
$$998$$ 53.4211 1.69102
$$999$$ −19.2372 −0.608638
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
## Twists
By twisting character
Char Parity Ord Type Twist Min Dim
1.1 even 1 trivial 5225.2.a.l.1.1 6
5.4 even 2 1045.2.a.f.1.6 6
15.14 odd 2 9405.2.a.z.1.1 6
By twisted newform
Twist Min Dim Char Parity Ord Type
1045.2.a.f.1.6 6 5.4 even 2
5225.2.a.l.1.1 6 1.1 even 1 trivial
9405.2.a.z.1.1 6 15.14 odd 2 | 22,586 | 41,302 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.486753 |
https://mapleprimes.com/questions/102148-Finite-Difference-Implicit-Option-Pricing-Mehtod | 1,726,705,599,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651944.55/warc/CC-MAIN-20240918233405-20240919023405-00144.warc.gz | 343,436,501 | 25,593 | # Question:Finite Difference Implicit Option Pricing Mehtod
## Question:Finite Difference Implicit Option Pricing Mehtod
Can somebody show me how to input this into maple please
f_i+1,j := (a_j)(f_i,j-1) + (b_j)(f_i,j) + (c_j)(f_i,j+1)
where
a_j := [1/2 (r-q) j Delta t - 1/2 sigma^2 j^2 Delta t]
b_j := [1 + sigma^2 j^2 Delta t + r Delta t]
c_j := [-1/2 (r-q) j Delta t - 1/2 sigma^2 j^2 Delta t]
with r:= 0.05, Delta t:= T/N , Delta S:=S/M, T:=0.4, q:=0, K:=1.1, S:=2, M=N:=10
I then want to define a tridiagonal matrix |i-j|>1, a_i,j =0
[1 0 0 0 0 0 0 0 0 0 0 0 0 0 ]
[a_1 b_1 c_1 0 0 0 0 0 0 0 ]
[0 a_2 b_2 c_2 0 0 0 0 0 0 0]
[. . . . . . . . . . . . . . . . . . ]
[. . . . . . . . . . . . . . . . . . ]
[ 0 0 0 0 0 0 0 0 0 0 0 0 a_m-1 b_m-1 c_m-1 ]
[ 0 0 0 0 0 0 0 . . . . . . . . . . . . . . . . . . . . 1]
With this matrix multiplied by the vector
[f_i,0]
[f_i,1]
[. . .]
[. . . ]
[. . . ]
[f_i,m]
which then equals the vector
[f_i+1,0]
[f_i+1,1]
[. . .]
[. . . ]
[. . . ]
[f_i+1,m]
I was told to start with i:=N-1 and then to define this vector in maple
[f_N,0]
[f_N,1]
[. . .]
[. . . ]
[. . . ]
[f_N,m]
which would equal a vector containing the max(K-j Delta S, 0) for 1<= j <= M. I was told to note that this matrix is not dependent on i. I need to find the matrix inverse and premultiply the vector by the inverse N times.
I am able to do everything up until the Matrix part which is where I am stuck.
Can anbody help me with this?
| 627 | 1,489 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-38 | latest | en | 0.811283 |
iching.egoplex.com | 1,621,065,253,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991378.48/warc/CC-MAIN-20210515070344-20210515100344-00514.warc.gz | 28,153,827 | 11,087 | ## Do the Eight Palaces Hold A Solution To King Wen's Sequence?
Previously, I analyzed the King Wen pairs as they played across the Jing Fang arrangement of the i-Ching, also known as the Eight Palaces arrangement. I charted the King Wen pairs as they are arranged in Jing Fang, and the result was striking.
The Eight Palaces is based on eight rows, each "ruled" by one of the eight trigrams. Each row starts with doubled trigram (1, Heaven over Heaven, 2, Earth over Earth, etc.). One line changes in each hexagram as you progress from left to right through the seventh column. From the seventh to the eight, the bottom trigram changes to its complement (yin-yang opposite). For the entire eight-by-eight square, the top half is the complement of the bottom half.
So the Jing Fang sequence is a structured sequence with logical rules, and the King Wen pairs overlaid on the diagram also follow rules, and rather interesting ones at that. The pairs are evenly distributed in two patterns -- half are vertical lines and half are diagonal lines, with the diagonals all passing through a single vertical line.
The vertical lines include the following hexagram pairs: 1-2, 5-6, 7-8, 11-12, 13-14, 17-18, 27-28, 29-30, 31-32, 35-36, 41-42, 51-52, 53-54, 57-58, 61-62, 63-64. The astute observer will note that this includes all of the doubled-trigrams (1-2, 29-30, 51-52, 57-58), but it also includes an assortment of other hexagrams whose pattern was not immediately clear to me. Eventually I figured it out.
Here's the trick. Imagine the Fu Xi circular diagram not as an eight-point compass, but as two four-point compasses. For the sake of coherence, I am just going to call them "red" and "blue" for now, but we'll work on that later. As seen in the compass at right, the Red Compass, or axis, consists of the trigrams Heaven-Water-Earth-Fire, and the Blue Compass consists of Lake-Wind-Mountain-Thunder.
In the vertical line pairs on the Jing Fang diagram above, each hexagram is drawn exclusively from one compass or the other, i.e., the hexagram is either all red or all blue, and never a mix of the two. So Heaven over Earth, for instance, is Red, while Lake over Mountain is Blue. In other words, there is a distinct structural difference between the diagonals and the verticals. The diagonals are all "mixed" hexagrams -- drawing one trigram from the red compass and one from the blue compass. So, for instance, let's look at hexagrams 9 through 12. Nine and 10 are diagonals, 11 and 12 are verticals.
Because we'll be switching diagrams shortly, let's set some terminology. The vertical pairs -- which do not mix red and blue trigrams -- we will refer to as "polar," because they conform to a single compass. The diagonal lines mix between the red and blue compasses, so let's call them bipolar.
All this might say a lot about Jing Fang's sequence, but the question this article seeks to answer is: Does it say about King Wen's sequence? I believe it does. Below is the King Wen sequence laid out as an 8x8 grid. I've highlighted the polar pairs.
The significance of this pattern may not be evident unless we rotate the diagram 90 degrees.
What we have here is a broken symmetry -- the polar pairs form a symmetrical grid across King Wen's sequence, or rather they would, if not for two pairs that are "misplaced" -- 29-30 and 39-40.
In the rotated version of the diagram above, the symmetry runs down the vertical center of the diagram -- mirror image left and right. If you exchange the two cited pairs, the symmetry becomes perfect and the weight of each line becomes consistent -- with four polar pairs and four bipolar pairs each. The diagram is show below.
And here's the same diagram rotated 90 degrees to the right to make the symmetry more visible.
Although I am using the word "misplaced" here, it is not my intention to say the King Wen ordering is wrong. Rather, the opposite -- I believe the symmetry break is deliberate. I think the key to understanding the logic of the King Wen sequence lies in understanding the effect of the symmmetry break on the ordering of the sequence as a whole.
The easiest way to proceed is to examine the "corrected" symmetrical version. The human mind can more easily identify patterns that are symmetrical and possess some degree of regularity. Once we've picked apart the symmetrical version, we might be able to identify how the pattern is modified in the assymetrical original sequence.
Let's look at the polar hexagram pairs alone. I've arranged the polars below according to quadrant -- i.e., the arrangement below is what you would get if you pulled out all of the bipolar hexagrams from the "corrected" King Wen, then slid the tiles together.
The first and fourth quadrants are made up entirely of complementary pairs -- pairs in which the first hexagram is the yin-yang opposite of the second hexagram. This suggests we should be thinking about Wen's structure in terms of quadrants rather than a linear sequence of 1, 2, 3, 4, etc. So let's go back to the "corrected" Wen grid. We can now identify a structural symmetry across the center line, in which the distribution of polar pair sets on the top and bottom reflect each other along the horizontal axis. We have another, different, structural symmetry in which the complementary pairs appear only in the top left and bottom right quadrants. The two corners anchoring the complement symmetry are, at top left are "Initiating" and "Responding," and bottom right, "Already Completed" and "Not Yet Completed." These two pair sets, the first and last, are one of the few traditional elements that even hint at any sort of cohesive structural order in the King Wen sequence.
To my mind, this echoes the directional numbering found in the traditional Fu Xi circular arrangement, as shown in the illustration below from Alfred Huang's The Numerology of the i Ching.
A few words on the "correction" of pairs 29-30 and 39-40. As I said before, I don't mean to indicate that I think the King Wen sequence is wrong. Rather, I want to examine the pristine symmetry and compare it to the broken symmetry to discover why it has been broken. Hopefully this will eventually lead to an actual solution for the whole set -- a goal of which this effort still falls short. I think the pattern here is strong evidence for a structural design in King Wen, but it's not yet an explanation of that design.
The key, I believe, is that the i-Ching is a dynamic system of processes. In previous analyses, I've compared the Earlier Heaven and Later Heaven compasses by analyzing how the King Wen pairs chart using each. What I found was that the Earlier Heaven, a carefully balanced primal ordering of the trigrams, yields a crystalline and largely stable mandala. The Later Heaven, which is unbalanced and generally taken to represent temporal processes, yields an unbalanced but much more dynamic mandala.
In my opinion, the balanced structure is primal and largely abstract, representing universal rules of organization. The unbalanced structure represents temporal processes -- processes as they unfold in spacetime, or rather, in the real world.
In science, balanced equations are very good at describing the action of principles -- such as Newton's famous axiom, "For every action, there is an equal and opposite reaction." But these balanced and linear equations don't hold up well in describing organic processes, such as the development of life. For this purpose, you need imbalanced equations -- fractals.
I see the balanced grid produced by the Earlier Heaven arrangement as the equivalent of Newton's third law. I see the imbalanced grid produced by the Later Heaven arrangement as more akin to chaos theory.
Seen in this light, the break in King Wen's symmetry becomes of paramount importance. A balanced arrangement, in the end, will play out the same exact cycle over and over again, endless repetition. An imbalanced arrangement, however, will produced different results from cycle to cycle. The imbalance is what allows complexity to develop, and the integration of that imbalance into King Wen's sequence -- in a very careful and precise manner -- allows the sequence to produce complexity.
I will have more to say about this, obviously, if and when I crack the actual rules that dictate the sequence. For now, I will close this installment by pointing out a possible textual clue within the "misplaced" hexagrams themselves. As previously noted, we're talking about four hexagrams here -- 29, 30, 39 and 40. The following text is from The Original I-Ching Oracle by Ritsema and Sabbadini:
29: The Gorge
Repeated gorge, possessing conformity. Holding-fast the heart. Growing movement possesses honor.
Streams reiterated culminate. Repeated gorge. A jun-zi uses constancy in actualizing dao to move. And repetition to teach affairs.
The radiance. Harvesting. Trial. Growing. Accumulating fertile cattle. Significant.
Brightness doubled arouses the radiance. The great in the person uses consecutive brightening to illuminate tending-towards the four sides.
39: Limping
Limping. Harvesting: Western South. Not Harvesting: Eastern North. Harvesting, viewing the great in the person. Trial. Significant.
The mountain possesses stream above. Limping. A jun zi uses reversing individuality to renovate actualizing-dao.
40: Unraveling
Unraveling. Harvesting: Western South. Without a place to go: One's coming return, significant. Possessing directed going: daybreak, significant.
Thunder and rain arousing. Unraveling. A jun zi uses forgiving excess to pardon offenses.
I'm not going to attempt to parse every word of these four passages at this time, but I did want to note a couple of interesting turns of phrase. "Renovation" in Hex 39 is the only occurrence of this word in the entire i-Ching, according to concordance in the The Original I-Ching Oracle. The original Chinese word is "XIU" and it can be read as "repair, mend, improve, adjust or regulate." The concept of "mending" the broken symmetry is very appealing here, and as noted, this is a unique word within the text.
The idea of "regulating actualizing-dao" fits in with my comments above regarding primal vs. temporal organization. The swapped hexagrams might represent a sort of timing mechanism, a beat counter, or the iterations of a fractal. In that case, we should look for the structure spreading out from the center of the 8x8 grid and not sequentially from Number One.
Without trying to explain it right now, I will say I was also struck by the phrase "The great in the person uses consecutive brightening to illuminate tending-towards the four sides." It strikes me, on a mostly intuitive level, as being a clue to the structure of what's going on here.
One final note: When correcting" the grid I simply exchanged the two pairs (29-30 and 39-40) without changing their internal order. A "correct correction" might place 29-30 in its new location as 30-29, for instance. So if you're looking for the pattern, keep that in mind.
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How many weeks in a year
How many weeks in a year
Everyone knows that there are 365 days in a non-leap year, but some people do not the answer the question – how many weeks are there in a year? At the same time, many peoples are interested in leap year information for various reasons, for example, for accurate planning of cases, working weeks, and so on.
How many weeks in a year? In fact, this calculation is easy enough. As we know that there are 365 days in a non leap year and the number of days in a week is seven, it is necessary to divide 365 by 7.
52 weeks One year has 52 weeks and 1 day – if the year is a normal year and 52 weeks and 2 days – if the year is a leap year. To calculate the number of weeks in a year, you need to divide the number of days in a year by the number of days in a week. For a leap year (2012, 2016, 2020 – every fourth year) has 366 days, the number of weeks equals 52 weeks and two days. Thus, in a year there are 53 weeks – 52 complete weeks and 1 incomplete week comprises of 1 or 2 days. Maybe it’s time to think about how to change something in yourself and your life, plan interesting events that will make life more interesting and colourful, and start rejoice not only every new week, but every day ? A week is a equals to seven days. There are 24 hours a day. To always be aware of the current time. A year equal to 365 or 366 days and associated with the total period of the Earth’s revolution around the Sun.
How many weeks in a year? Calculating the number of weeks in a year Typical year: 365/7 = 52.14 – 52 weeks and 1 day (52 * 7 = 364, 365-364 = 1). Leap year: 366/7 = 52.29, – 52 weeks and 2 days (52 * 7 = 364, 366-364 = 2).
Calculating the number of full weeks in a year 2020 or How many weeks in a year 2020?current year 2020 starts on Wednesday (minus 5 days) and ends on Thursday (minus 4 days). The current year is a leap year with 366 days. (366-5-4) / 7 = 51. This means there are 51 full weeks in 2020.
UPSC, IAS, IPS, IFS, PCS, UPPSC, SSC CGL, BANK, RAILWAYS, RRB NTPC, LIC AAO, Patwari, TET, BED, etc. Examinations are going to start after a few months. In these examinations, many questions are coming from Art and Culture. So, therefore, How many weeks in a year is a important subject in all these examinations. We will provide you the Indian History PDF books and notes at our Website which include How many weeks in a year. Which is helpful for all those aspirants who preparing for all these examinations at various level. Previous years questions are placed in How many weeks in a year, which has been put together in most examinations; you can download these PDF Notes simply by clicking on the Download Button at the bottom.
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## Finance
Basic Finance Corporate Finance Financial Management Financial Econometrics Portfolio Management Risk Management Public Finance Business Law & Ethics
problem 1:
Mr Smart desires to buy a Blackberry costing Rs8000. He is willing to wait a year to get it. The bank pays 10% annual interest compounded monthly. How much money should Mr Smart put aside each month starting at the start of this one, to buy the Blackberry at the end of year.
problem 2:
You’ve just celebrated your twentieth birthday and you think you will spend Rs120, 000 on your 60th birthday and the same amount on each of upcoming 9 birthdays. If the interest rate is 10% per annum, how much must you save each year, supposing that annual savings start on the eve of your 21st birthday, till retirement to support your consumption pattern after retirement?
problem 3:
A Mauritian couple living in England plans to pay their child’s tuition fees for 4 years starting 18 years from now. The recent annual cost of university is £7,000, and they anticipate this cost to rise at an annual rate of 5%. In their planning, they suppose which they can earn 6% yearly. How much must they put aside each year, starting upcoming year, if they plan to make 17 equal payments.
problem 4:
You are encountered with a choice between two investments, A and B. Investment A is expected to generate Rs1, 000 every year till perpetuity. Investment B is also perpetuity and will generate a cash flow of Rs800 in year 1 which will grow at a constant rate each year forever. If your required rate of return is 10% per annum, what rate of raise in Investment B’s cash flows is essential to make it as desirable as Investment A.?
problem 5:
On 1 January 2011, Rs500, 000 was borrowed by the Mr Needy from bank for a period of 10 years at a fixed annual rate of interest of 12%. Mr Needy will reimburse interest and principal by equivalent annual payments. Compute the annual payment the first annual payment is effected on 31 December 2011.
Basic Finance, Finance
• Category:- Basic Finance
• Reference No.:- M98905
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Describe what you learned about the impact of economic, social, and demographic trends affecting the US labor environmen | 1,292 | 6,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-13 | longest | en | 0.956284 |
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## Why Mechanical Engineering Compressors, Gas Dynamics and Gas Turbines?
In this section you can learn and practice Mechanical Engineering Questions based on "Compressors, Gas Dynamics and Gas Turbines" and improve your skills in order to face the interview, competitive examination and various entrance test (CAT, GATE, GRE, MAT, Bank Exam, Railway Exam etc.) with full confidence.
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## How to solve Mechanical Engineering Compressors, Gas Dynamics and Gas Turbines problems?
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### Exercise :: Compressors, Gas Dynamics and Gas Turbines - Section 1
1.
In a four stage compressor, if the pressure at the first and third stage are 1 bar and 16 bar, then the delivery pressure at the fourth stage will be
A. 1 bar B. 16 bar C. 64 bar D. 256 bar
Explanation:
No answer description available for this question. Let us discuss.
2.
The compressed air may be used
A. in gas turbine plants B. for operating pneumatic drills C. in starting and supercharging of I.C. engines D. all of the above
Explanation:
No answer description available for this question. Let us discuss.
3.
A rocket works with maximum overall efficiency when air-craft velocity is __________ the jet velocity.
A. equal to B. one-half C. double
Explanation:
No answer description available for this question. Let us discuss.
4.
The volumetric efficiency for reciprocating air compressors is about
A. 10 to 40% B. 40 to 60% C. 60 to 70% D. 70 to 90%
Explanation:
No answer description available for this question. Let us discuss.
5.
The maximum temperature in a gas turbine is
A. 200°C B. 500°C C. 700°C D. 1000°C | 628 | 2,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-39 | latest | en | 0.868864 |
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# Quantitative Aptitude Quiz For FCI Phase I 2022- 09th November
Directions (1-5): – Table given below gives information about Cost price, Selling price, profit percent and discount percentage of four article A, B, C and D. Some data is missing, calculate the data and answer the following questions.
Q1. If cost price of article D is 10% more than cost price of article A, find Market price of article D?
(a) Rs. 1000
(b) Rs. 1100
(c) Rs. 900
(d) Rs. 950
(e) Rs. 1050
Q2. Average of market price of article A and cost price of article B is how much percent more than cost price of article C?
(a) 75%
(b) 80%
(c) 60%
(d) 90%
(e) 95%
Q3. If discount percentage is twice of profit percentage for article C, find ratio of cost price of article A to market price of article C?
(a) 2:3
(b) 1:2
(c) 2:5
(d) 5:6
(e) None of these.
Q4. Selling price of article C is what percent of market price of article B?
(a) 100%
(b) 75%
(c) 50%
(d) 25%
(e) 60%
Q5. If for article D difference between profit earned and discount given is Rs. 440, find selling price of the D?
(a) 550
(b) 650
(c) 308
(d) 450
(e) 605
Directions (6-10):- The bar graph given below shows number of students (in ’00) of five different school (A, B, C, D and E) and no. of passed student (in %) in each school.
Study the given graph carefully and answer the following questions.
Q6. Find the ratio between the number of students of school B and no. of passed students from school D.
(a) 2∶1
(b) 1∶3
(c) 3∶1
(d) 4∶3
(e) 2∶3
Q7. Find the difference between the average no. of passed students from school A and C to that of total number of students of school B.
(a) 1500
(b) 2000
(c) 2500
(d) 3000
(e) 1800
Q8. Students failed from school C is what percent of passed students from school E?
(a) 150%
(b) 66⅔%
(c) 56%
(d) 125%
(e) 69%
Q9. In which school number of passed students is maximum?
(a) B
(b) D
(c) A
(d) E
(e) C
Q10. What is average no. of students passed from school A, C and E?
(a) 5200
(b) 4800
(c) 5000
(d) 5600
(e) 4600
Directions (11-15) :- The line graph shows the no. of passengers who travels from Metro in 7 days of a given week. Study the graph carefully and answer the following question.
Q11. What is the average no. of passenger travels from Tuesday to Saturday?
(a) 64000
(b) 67000
(c) 72000
(d) 70000
(e) 75000
Q12. Passenger who travel on Thursday are what percentage of passengers who travels on Saturday?
(a) 120%
(b) 71.4%
(c) 140%
(d) 87.5%
(e) 175%
Q13. Find the ratio of no. of passenger travels on Monday and Thursday together to Tuesday and Sunday together.
(a) 33 : 32
(b) 33 : 34
(c) 17 : 16
(d) 1 : 1
(e) 3 : 2
Q15. Passenger travels on Wednesday and Friday together are how much more or less than that on Sunday and Monday together?
(a) 30000
(b) 50000
(c) 35000
(d) 40000
(e) 45000
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Lesson 1: Lines, Rays, and Angles
Lesson 2: Classify Triangles by Angles
Lesson 3: Parallel Lines and Perpendicular Lines
Mid-Chapter Checkpoint
Lesson 5: Line Symmetry
Lesson 6: Find and Draw Lines of Symmetry
Lesson 7: Problem Solving • Shape Patterns
Review/Test
### Common Core – New – Page No. 553
Lines, Rays, and Angles
Draw and label an example of the figure.
Question 1.
obtuse ∠ABC
Think: An obtuse angle is greater than a right angle. The middle letter, B, names the vertex of the angle.
_________
Explanation:
An obtuse angle is greater than a right angle. The middle letter, B, names the vertex of the angle.
Question 2.
$$\overrightarrow{G H}$$
_________
Explanation:
GH is a ray that has one endpoint and continues without an end in one direction.
Question 3.
acute ∠JKL
_________
Explanation:
Angle JKL is an acute angle that is less than a right angle.
Question 4.
$$\overline{B C}$$
Explanation:
BC is a line that continues without an end in both directions.
Use the figure for 5–8.
Question 5.
Name a line segment.
line segment EF
Explanation:
EF line is a straight path of points that continues without an end in both directions.
Question 6.
Name a right angle.
∠ _____
∠EJF
Explanation:
EJF is a right angle that forms a square corner.
Question 7.
Name an obtuse angle.
obtuse ∠ _____
∠CEJ
Explanation:
CEJ is an obtuse angle that is greater than a right angle.
Question 8.
Name a ray.
Ray JD
Explanation:
JD is a ray that has one endpoint and continues without an end in one direction.
Problem Solving
Use the figure at the right for 9–11.
Question 9.
Classify ∠AFD
_________
Obtuse Angle
Explanation:
AFD is an obtuse angle that is greater than a right angle.
Question 10.
Classify ∠CFE.
_________
Right Angle
Explanation:
∠CFE is a right angle that forms a square corner.
Question 11.
Name two acute angles.
acute ∠ _____ acute ∠ _____
∠AFB and ∠DFE
Explanation:
∠AFB and ∠DFE are two acute angles with less than a right angle.
### Common Core – New – Page No. 554
Lesson Check
Question 1.
The hands of a clock show the time 12:25.
Which best describes the angle between the hands of the clock?
Options:
a. acute
b. right
c. obtuse
d. straight
c. obtuse
Explanation:
The hands of the time 12:25 are forming at greater than a right angle. So, the answer is the Obtuse angle.
Spiral Review
Question 3.
Jan’s pencil is 8.5 cm long. Ted’s pencil is longer. Which could be the length of Ted’s pencil?
Options:
a. 0.09 cm
b. 0.8 cm
c. 8.4 cm
d. 9.0 cm
d. 9.0 cm
Explanation:
9 ones is greater than 8 ones. So, 9.0 cm > 8.5 cm
Question 4.
Kayla buys a shirt for $8.19. She pays with a$10 bill. How much change should she receive?
Options:
a. $1.81 b.$1.89
c. $2.19 d.$2.81
a. $1.81 Explanation: Kayla buys a shirt for$8.19. She pays with a $10 bill. To find the change she received,$10 – \$8.19 = 1.81
Question 6.
Jose jumped 8 $$\frac{1}{3}$$ feet. This was 2 $$\frac{2}{3}$$ feet farther than Lila jumped. How far did Lila jump?
Options:
a. 5 $$\frac{1}{3}$$
b. 5 $$\frac{2}{3}$$
c. 6 $$\frac{1}{3}$$
d. 11
b. 5 $$\frac{2}{3}$$
Explanation:
Jose jumped 8 $$\frac{1}{3}$$ feet. This was 2 $$\frac{2}{3}$$ feet farther than Lila jumped.
8 $$\frac{1}{3}$$ – 2 $$\frac{2}{3}$$ = $$\frac{25}{3}$$ – $$\frac{8}{3}$$ = $$\frac{7}{3}$$ = 5 $$\frac{2}{3}$$
### Page No. 557
Classify each triangle. Write acute, right, or obtuse.
Question 2.
_____
Obtuse triangle;
Angle B and Angle C are both acute.
Angle A is obtuse.
Explanation:
From triangle ABC, Angle B, and Angle C are both acute with less than a right angle. Angle A is obtuse angle that is greater than a right angle.
Question 3.
_____
Obtuse triangle;
Angle A and Angle C are both acute.
Angle B is obtuse.
Explanation:
From triangle ABC, Angle A, and Angle C are both acute with less than a right angle. Angle B is an obtuse angle that is greater than a right angle. A triangle with an obtuse angle is called an obtuse triangle.
Question 4.
_____
Acute triangle;
Angle A, Angle B, and Angle C are acute angles.
Explanation:
From triangle ABC, Angle A, Angle B, and Angle C are acute angles with less than a right angle. A triangle with three acute angles called an acute triangle. So, the given triangle is an acute triangle.
Question 6.
_____
Acute triangle;
Angle A, Angle B, and Angle C are acute angles.
Explanation:
From triangle ABC, Angle A, Angle B, and Angle C are acute angles with less than a right angle. A triangle with three acute angles called an acute triangle. So, the given triangle is an acute triangle.
Question 7.
_____
Right Triangle;
∠A and ∠C are acute angles.
∠B is Right angle
Explanation:
∠A and ∠C are acute angles with less than a right angle. ∠B is the Right angle that forms a square corner. A triangle that has one right angle is called a right triangle.
Question 8.
Cross out the figure that does not belong. Explain.
Type below:
_________
Explanation:
From the given image, 1, 3, and 4 have two acute angles, and one obtuse angle. 2 have three acute angles.
### Page No. 558
Use the Venn diagram for 9–10.
Question 9.
Which triangles do NOT have an obtuse angle? Explain.
_______ triangles
4 triangles;
Triangle DEF, Triangle SPN, Triangle ABC, and Triangle GHP are don’t have an obtuse angle. Triangle DEF, Triangle SPN are acute angles. An acute triangle is a triangle with three acute angles. Triangle ABC, and Triangle GHP are right angles. A right triangle is a triangle with one right angle. The sum of the triangle is 180 degrees. A right triangle has 90 degrees. So, the remaining angles must be acute angles.
Question 10.
How many triangles have at least two acute angles? Explain.
_______ triangles
4 triangles;
Triangle DEF, Triangle SPN, Triangle ABC, and Triangle GHP at least two acute angles. Triangle DEF, Triangle SPN are acute angles. An acute triangle is a triangle with three acute angles. Triangle ABC, and Triangle GHP are right angles. A right triangle is a triangle with one right angle and two acute angles.
Question 11.
Use the square shown at the right. Draw a line segment from point M to point P. Name and classify the triangles formed by the line segment.
Type below:
_________
Angle MNP and Angle MQP
Explanation:
The line segment from M to P forms Angle MNP and Angle MQP.
Question 12.
Write the letter of the triangle under its correct classification.
Type below:
_________
Explanation:
Triangle A and triangle B have three acute angles. So, they are acute triangles. Triangle D and triangle F have one obtuse angle. So, they are obtuse triangles.
Triangle C and triangle E have one right angle. So, they are right triangles.
### Common Core – New – Page No. 559
Classify Triangles
Classify each triangle. Write acute, right, or obtuse.
Question 1.
Think: Angles A and C are both acute.
Angle B is obtuse.
Obtuse triangle;
Angle A and Angle C are both acute.
Angle B is obtuse.
Explanation:
From triangle ABC, Angle A, and Angle C are both acute with less than a right angle. Angle B is an obtuse angle that is greater than a right angle.
Question 4.
_________
Obtuse triangle;
Angle L and Angle N are both acute.
Angle M is obtuse.
Explanation:
From triangle LMN, Angle L and Angle N are both acute with less than a right angle. Angle M is an obtuse angle that is greater than a right angle. A triangle with an obtuse angle is called an obtuse triangle.
Problem Solving
Question 5.
Use figure ABCD below. Draw a line segment from point B to point D. Name and classify the triangles formed.
Two _________ triangles
△ _________
△ _________
Two Acute triangles.
△ ABD
△ BCD
Explanation:
If we draw a line segment from point B to point D, then there are two traingles formed with less than right angles. They are △ ABD and △ BCD.
Question 6.
Use figure ABCD below. Draw a line segment from point A to point C. Name and classify the triangles formed.
Two _________ triangles
△ _________
△ _________
Two Acute triangles.
△ ABC
Explanation:
If we draw a line segment from point A to point C, then there are two traingles formed with less than right angles. They are △ ABC and △ ADC.
### Common Core – New – Page No. 560
Lesson Check
Question 1.
Stephen drew this triangle. How many obtuse angles does the triangle have?
Options:
a. 0
b. 1
c. 2
d. 3
a. 0
Explanation:
The given image has three acute angles. So, there are 0 obtuse angles.
Spiral Review
Question 3.
Oliver drew the figure below to show light traveling from the sun to Earth. Name the figure he drew.
Options:
a. segment SE
b. ray SE
c. line SE
d. ray ES
b. ray SE
Explanation:
SE is a ray that has one endpoint and continues without an end in one direction.
Question 6.
A basketball team averaged 105 points per game. How many points did the team score in 6 games?
Options:
a. 605 points
b. 630 points
c. 900 points
d. 6,030 points
b. 630 points
Explanation:
A basketball team averaged 105 points per game.
They score in 6 games = 6 x 105 = 630 points.
### Page No. 563
Question 1.
Draw and label $$\overline{Q R} \| \overline{S T}$$.
Think: Parallel lines never intersect. Parallel line segments are parts of parallel lines.
Type below:
_________
$$\overline{Q R} \| \overline{S T}$$
Explanation:
Parallel lines never intersect. Parallel line segments are parts of parallel lines.
Use the figure for 2 and 3.
Question 2.
Name two line segments that appear to be parallel.
Type below:
_________
Line Segment CB and Line Segment DF.
Explanation:
Line Segment CB and Line Segment DF are parallel lines. The both lines never intersect and are always the same distance apart.
Question 3.
Name two line segments that appear to be perpendicular.
Type below:
_________
Line Segment CB and Line Segment BF are perpendicular lines.
Explanation:
Line Segment CB and Line Segment DF are perpendicular lines. Both lines intersect to form four right angles.
Use the figure for 4–5.
Question 4.
Name a pair of lines that are perpendicular.
Type below:
_________
FJ and HG are perpendicular lines.
Explanation:
FJ and HG lines intersect each other and form four right angles.
Question 5.
Name a pair of lines that appear to be parallel.
Type below:
_________
DC and FJ are parallel lines.
Explanation:
DC and FJ are never intersected and are always the same distance apart.
Question 6.
$$\overline{R S} \| \overline{T U}$$
Type below:
_________
$$\overline{R S} \| \overline{T U}$$
Explanation:
Parallel lines never intersect. Parallel line segments are parts of parallel lines.
Question 9.
$$\overset { \longleftrightarrow }{ JK }$$ ⊥ $$\overset { \longleftrightarrow }{ LM }$$
Type below:
_________
$$\overset { \longleftrightarrow }{ JK }$$ ⊥ $$\overset { \longleftrightarrow }{ LM }$$
Explanation:
JK and LM are two lines that intersected each other to form right angles.
Question 10.
$$\overset { \longleftrightarrow }{ ST }$$ intersecting $$\overset { \longleftrightarrow }{ UV }$$ at point X
Type below:
_________
Explanation:
ST and UV are two lines intersecting at point X.
Question 11.
$$\overset { \longleftrightarrow }{ AB }$$ || $$\overset { \longleftrightarrow }{ FG }$$
Type below:
_________
$$\overset { \longleftrightarrow }{ AB }$$ || $$\overset { \longleftrightarrow }{ FG }$$
Explanation:
Parallel lines never intersect. Parallel line segments are parts of parallel lines.
Use the figure for 12–13.
Question 12.
Dan says that $$\overset { \longleftrightarrow }{ HL }$$ is parallel to $$\overset { \longleftrightarrow }{ IM }$$. Is Dan correct? Explain.
_____
No; HL and IM are not parallel lines. Parallel lines are always the same distance apart. But from the given image, the H and I may intersect if the line is extended.
Question 13.
Name two intersecting line segments that are not perpendicular.
Type below:
_________
JM and KG are two intersecting line segments and are also not perpendicular.
### Page No. 564
Use the house plan at the right for 14–16.
Question 14.
What geometric term describes a corner of the living room?
_________
The corner of the living room is a vertex. Corners of any shape are vertexes.
Question 15.
Name three parts of the plan that show line segments.
_________
Kitchen, Living Room, and Master Bedroom.
Question 16.
Name a pair of line segments that appear to be parallel
_________
Two sidelines of Living Room are parallel.
Two sidelines of Master Bedroom are parallel.
Use the map at the right for 17–19.
Question 17.
Name a street that appears to be parallel to S 17th Street.
_________
S 18th Street
Explanation:
S 18th Street is parallel to S 17th Street. They never meet each other and are always the same distance apart.
Question 18.
Use Diagrams Name a street that appears to be parallel to Vernon Street.
_________
Perry Street
Explanation:
Perry Street is parallel to Vernon Street. They never meet each other and are always the same distance apart.
Question 19.
Name a street that appears to be perpendicular to S 19th Street.
_________
Austin Street
Explanation:
Austin Street is perpendicular to S 19th Street. They are intersect with each other and form four right angles.
Question 20.
Choose the labels to make a true statement.
Type below:
_________
Line AB is perpendicular to Line EF.
### Common Core – New – Page No. 565
Parallel Lines and Perpendicular Lines
Use the figure for 1–3.
Question 1.
Name a pair of lines that appear to be perpendicular.
Think: Perpendicular lines form right angles.
$$\overset { \longleftrightarrow }{ AB }$$ and $$\overset { \longleftrightarrow }{ EF }$$ appear to form right angles.
$$\overset { \longleftrightarrow }{ AB }$$ and $$\overset { \longleftrightarrow }{ EF }$$
$$\overset { \longleftrightarrow }{ AB }$$ and $$\overset { \longleftrightarrow }{ EF }$$
Explanation:
Perpendicular lines form right angles.
$$\overset { \longleftrightarrow }{ AB }$$ and $$\overset { \longleftrightarrow }{ EF }$$ appear to form right angles.
$$\overset { \longleftrightarrow }{ AB }$$ and $$\overset { \longleftrightarrow }{ EF }$$
Question 2.
Name a pair of lines that appear to be parallel.
_____ and _____
$$\overset { \longleftrightarrow }{ AB }$$ and $$\overset { \longleftrightarrow }{ CD}$$
Explanation:
Parallel lines never interest each other. $$\overset { \longleftrightarrow }{ AB }$$ and $$\overset { \longleftrightarrow }{ CD}$$ are parallel lines.
Question 3.
Name another pair of lines that appear to be perpendicular.
_____ and _____
$$\overset { \longleftrightarrow }{ CD }$$ and $$\overset { \longleftrightarrow }{ EF }$$
Explanation:
Perpendicular lines form right angles.
$$\overset { \longleftrightarrow }{ CD }$$ and $$\overset { \longleftrightarrow }{ EF }$$ appear to form right angles.
$$\overset { \longleftrightarrow }{ CD }$$ and $$\overset { \longleftrightarrow }{ EF }$$
Draw and label the figure described.
Question 6.
$$\overset { \longleftrightarrow }{ FH }$$ ⊥ $$\overset { \longleftrightarrow }{ JK }$$
Explanation:
FH and JK are two lines that intersect each other to form four right angles.
Problem Solving
Use the street map for 7–8.
Question 7.
Name two streets that intersect but do not appear to be perpendicular.
Type below:
_________
Maple and Oak or Oak and Birch
Explanation:
Maple and Oak or Oak and Birch; They are intersecting with each other and not perpendicular.
Question 8.
Name two streets that appear to be parallel to each other.
Type below:
_________
Maple and Birch
Explanation:
Maple and Birch are streets and do not intersect with each other. They appear to be parallel to each other.
### Common Core – New – Page No. 566
Lesson Check
Question 1.
Which capital letter appears to have perpendicular line segments?
Options:
a. N
b. O
c. T
d. V
c. T
Explanation:
T has two lines and is interesting to form four right angles.
Question 2.
In the figure, which pair of line segments appear to be parallel?
Options:
a. $$\overline{F G} \text { and } \overline{G H}$$
b. $$\overline{F J} \text { and } \overline{G H}$$
c. $$\overline{F G} \text { and } \overline{J H}$$
d. $$\overline{J H} \text { and } \overline{F J}$$
c. $$\overline{F G} \text { and } \overline{J H}$$
Explanation:
$$\overline{F G} \text { and } \overline{J H}$$ are parallel lines that never intersect
Spiral Review
Question 3.
Nolan drew a right triangle. How many acute angles did he draw?
Options:
a. 0
b. 1
c. 2
d. 3
c. 2
Explanation:
A triangle with one right angle will have two acute angles.
Question 4.
Mike drank more than half the juice in his glass. What fraction of the juice could Mike have drunk?
Options:
a. $$\frac{1}{3}$$
b. $$\frac{2}{5}$$
c. $$\frac{3}{6}$$
d. $$\frac{5}{8}$$
d. $$\frac{5}{8}$$
Explanation:
Mike drank more than half the juice in his glass. He drunk $$\frac{5}{8}$$ of the juice.
Question 5.
A school principal ordered 1,000 pencils. He gave an equal number to each of 7 teachers until he had given out as many as possible. How many pencils were left?
Options:
a. 2
b. 4
c. 6
d. 142
c. 6
Explanation:
A school principal ordered 1,000 pencils. He gave an equal number to each of 7 teachers until he had given out as many as possible. He shared 142 pencils for each of 7 teachers. So, 142 X 7 = 994. The remaining pencils are 6.
Question 6.
A carton of juice contains 64 ounces. Ms. Wilson bought 6 cartons of juice. How many ounces of juice did she buy?
Options:
a. 364 ounces
b. 370 ounces
c. 384 ounces
d. 402 ounces
c. 384 ounces
Explanation:
A carton of juice contains 64 ounces. Ms. Wilson bought 6 cartons of juice. 64 X 6 = 384 ounces of juice she can buy.
### Page No. 569
Question 1.
Tell whether the quadrilateral is also a trapezoid, parallelogram, rhombus, rectangle, or square.
Think:
____ pairs of parallel sides
____ sides of equal length
____ right angles
Quadrilateral ABCD is also a __________
__________
2 pairs of parallel sides
4 sides of equal length
0 right angles.
Quadrilateral ABCD is also a Rhombus
Explanation:
A Rhombus is a quadrilateral that has 2 pairs of parallel sides and 4 sides of equal lengths.
Classify each figure as many ways as possible. Write quadrilateral, trapezoid, parallelogram, rhombus, rectangle, or square.
Question 4.
_________
_________
_________
Explanation:
2 pairs of parallel sides
4 sides of equal length
0 right angles.
Given quadrilateral is Rhombus and Parallelogram.
A Rhombus is a quadrilateral that has 2 pairs of parallel sides and 4 sides of equal lengths, and 0 right angles.
Question 5.
_________
_________
Explanation:
2 pairs of parallel sides
2 pairs of sides of equal length
0 right angles.
A Parallelogram is a quadrilateral that has 2 pairs of parallel sides and 2 pairs of sides of equal lengths, and 0 right angles.
Question 6.
_________
_________
_________
Explanation:
2 pairs of parallel sides
4 sides of equal length
4 right angles.
A Square is a quadrilateral that has 2 pairs of parallel sides and 4 sides of equal lengths, and 4 right angles.
Question 7.
_________
_________
Explanation:
1 pair of parallel sides
0 sides of equal length
0 right angles.
A Square is a quadrilateral that has 1 pair of parallel sides and 0 sides of equal lengths, and 0 right angles.
### Page No. 570
Question 9.
Classify the figure. Select all that apply.
Options:
b. trapezoid
c. parallelogram
d. rectangle
e. rhombus
f. square
b. trapezoid
c. parallelogram
Explanation:
A Parallelogram is a quadrilateral that has 2 pairs of parallel sides and 2 pairs of sides of equal lengths, and 0 right angles.
The Louvre Museum is located in Paris, France. Architect I. M. Pei designed the glass and metal structure at the main entrance of the museum. This structure is called the Louvre Pyramid. Below is a diagram of part of the entrance to the Louvre Pyramid.
Question 10.
Describe the quadrilaterals you see in the diagram.
_________
_________
Trapezoid and Rhombus
Explanation:
There are 2 quadrilaterals available in the given image. One is a Trapezoid with 1 pair of parallel sides. Another one is Rhombus is with 2 pairs of parallel sides and 4 sides of equal lengths, and 0 right angles.
Question 11.
How many triangles do you see in the diagram? Explain.
______ triangles
11 triangles
Explanation:
The given image has 11 triangles
### Common Core – New – Page No. 571
Classify each figure as many ways as possible. Write quadrilateral, trapezoid, parallelogram, rhombus, rectangle, or square.
Question 1.
Think: 2 pairs of parallel sides
4 sides of equal length
0 right angles
Explanation:
2 pairs of parallel sides
4 sides of equal length
0 right angles
Question 2.
Type below:
_________
Explanation:
2 pairs of parallel sides
2 pairs of sides of equal length
4 right angles
Question 3.
Type below:
_________
Explanation:
1 pair of parallel sides
2 sides of equal length
0 right angles
Question 4.
Type below:
_________
Explanation:
0 pair of parallel sides
0 sides of equal length
0 right angles
Question 5.
Type below:
_________
Explanation:
2 pairs of parallel sides
4 sides of equal length
0 right angles
Question 6.
Type below:
_________
Explanation:
1 pair of parallel sides
0 sides of equal length
2 right angles
Question 7.
Type below:
_________
Explanation:
2 pairs of parallel sides
2 pairs of sides of equal length
0 right angles
Problem Solving
### Common Core – New – Page No. 572
Lesson Check
Question 1.
Joey is asked to name a quadrilateral that is also a rhombus. What should be his answer?
Options:
a. square
b. rectangle
c. parallelogram
d. trapezoid
a. square
Explanation:
The quadrilateral square is also called a rhombus. Both square and rhombus have 2 pairs of parallel sides and 4 sides of equal length.
Question 2.
Which quadrilateral has exactly one pair of parallel sides?
Options:
a. square
b. rhombus
c. parallelogram
d. trapezoid
d. trapezoid
Explanation:
A trapezoid has exactly one pair of parallel sides.
Spiral Review
Question 3.
Terrence has 24 eggs to divide into equal groups. What are all the possible numbers of eggs that Terence could put in each group?
Options:
a. 1, 2, 3, 4
b. 2, 4, 6, 8, 12
c. 1, 2, 3, 4, 6, 8, 12, 24
d. 24, 48, 72, 96
c. 1, 2, 3, 4, 6, 8, 12, 24
Explanation:
Terrence has 24 eggs to divide into equal groups. Terence could put in each group in 1, 2, 3, 4, 6, 8, 12, 24 ways.
Question 4.
In a line of students, Jenna is number 8. The teacher says that a rule for a number pattern is add 4. The first student in line says the first term, 7. What number
should Jenna say?
Options:
a. 31
b. 35
c. 39
d. 43
b. 35
Explanation:
In a line of students, Jenna is number 8. The teacher says that a rule for a number pattern is add 4. The first student in line says the first term, 7.
7 + 4 = 11
11 + 4 = 15
15 + 4 = 19
19 + 4 = 23
23 + 4 = 27
27 + 4 = 31
31 + 4 = 35.
Jenna says 35.
Question 5.
Lou eats $$\frac{6}{8}$$ of a pizza. What fraction of the pizza is left over?
Options:
a. $$\frac{1}{8}$$
b. $$\frac{1}{4}$$
c. $$\frac{1}{2}$$
d. $$\frac{3}{4}$$
b. $$\frac{1}{4}$$
Explanation:
Lou eats $$\frac{6}{8}$$ of a pizza. So, 6 parts of pizza is finished and remaining 2 parts of pizza is remained. So, the left over pizza is $$\frac{2}{8}$$ = $$\frac{1}{4}$$.
Question 6.
Which capital letter appears to have parallel lines?
Options:
a. D
b. L
c. N
d. T
c. N
Explanation:
N has two parallel lines that never intersect each other.
### Page No. 573
Choose the best term from the box to complete the sentence.
Question 1.
A _______ is part of a line between two endpoints.
_________
line segment
Question 2.
A _______ forms a square corner.
_________
Right angle
Question 3.
An _______ is greater than a right angle and less than a straight angle.
_________
Obtuse angle
Question 4.
The two-dimensional figure that has one endpoint is a ________.
_________
ray
Question 5.
An angle that forms a line is called a _______.
_________
straight line
Question 6.
On the grid below, draw a polygon that has 2 pairs of parallel sides, 2 pairs of sides equal in length, and 2 acute and 2 obtuse angles. Tell all the possible names for the figure.
Type below:
_________
Parallelogram
Explanation:
The possible polygon that has 2 pairs of parallel sides, 2 pairs of sides equal in length, and 2 acute and 2 obtuse angles is Parallelogram.
Draw the figure.
Question 7.
parallel lines
Type below:
_________
Explanation:
QR and ST are two parallel lines. they never intersect each other.
Question 8.
obtuse ∠ABC
Type below:
_________
Explanation:
From triangle, ABC, Angle A, and Angle C are both acute with less than a right angle. Angle B is an obtuse angle that is greater than a right angle.
Question 9.
intersecting lines that are not perpendicular
Type below:
_________
Explanation:
ST and UV are two lines intersecting at point X.
Question 10.
acute ∠RST
Type below:
_________
### Page No. 574
Question 14.
What is the correct name of the figure shown?
_________
Ray
Explanation:
EF is a ray that has one endpoint and continues without an end in one direction.
Question 15.
Describe the angles of an obtuse triangle.
Type below:
_________
An obtuse triangle (or obtuse-angled triangle) is a triangle with one obtuse angle (greater than 90°) and two acute angles.
### Page No. 577
Tell whether the parts on each side of the line match. Is the line a line of symmetry? Write yes or no.
Question 1.
____
Yes;
Explanation:
The line of symmetry divides a shape into two parts that are the same size and shape.
Question 2.
____
Yes;
Explanation:
The line of symmetry divides a shape into two parts that are the same size and shape.
Question 3.
____
Yes;
Explanation:
The line of symmetry divides a shape into two parts that are the same size and shape.
Question 4.
____
No;
Explanation:
The line of symmetry divides a shape into two parts that are not with the same size and shape.
Tell if the blue line appears to be a line of symmetry. Write yes or no.
Question 5.
____
Yes;
Explanation:
The line of symmetry divides a shape into two parts that are the same size and shape.
Question 6.
____
No;
Explanation:
The line of symmetry divides a shape into two parts that are not the same size and shape.
Question 7.
____
No;
Explanation:
The line of symmetry divides a shape into two parts that are not the same size and shape.
Question 8.
____
Yes;
Explanation:
The line of symmetry divides a shape into two parts that are the same size and shape.
Tell if the blue line appears to be a line of symmetry. Write yes or no.
Question 9.
____
No;
Explanation:
The line of symmetry divides a shape into two parts that are not the same size and shape.
Question 10.
____
Yes;
Explanation:
The line of symmetry divides a shape into two parts that are the same size and shape.
Question 11.
____
No;
Explanation:
The line of symmetry divides a shape into two parts that are not with the same size and shape.
Question 12.
____
Yes;
Explanation:
The line of symmetry divides a shape into two parts that are the same size and shape.
Question 13.
Which best describes the symmetry in the letter I?
Type below:
________
The two parts of the folded I match exactly. The fold line is a line of symmetry.
Explanation:
Take the Horizontal line in the middle of the Letter I. Cut out the tracing. Fold the tracing over a horizontal line. The two parts of the folded I match exactly. The fold line is a line of symmetry.
### Page No. 578
Question 14.
Which shape has a correctly drawn line of symmetry?
a. What do you need to find?
Type below:
________
Find the shape that has an exact line of symmetry.
Question 14.
b. How can you tell if the line of symmetry is correct?
Type below:
________
If the two parts of the folded match exactly, then the line is a line of symmetry.
Question 14.
c. Tell how you solved the problem.
Type below:
________
From fig 1 to 4, the fig 2 is has a line of symmetry that can exactly separate the two parts equally.
Question 14.
d. Circle the correct shape above.
Type below:
________
Question 15.
Reason Abstractly Draw a line of symmetry in the figure shown.
Question 16.
Evie’s birthday is on the 18th of May. Since May is the 5th month, Evie wrote the date as shown.
Evie says all the numbers she wrote have line symmetry. Is she correct? Explain.
No; The number 5 doesn’t have a line of symmetry. So, Evie explanation is wrong.
### Common Core – New – Page No. 579
Line Symmetry
Tell if the dashed line appears to be a line of symmetry. Write yes or no.
Question 1.
yes
Yes;
Explanation:
The line of symmetry divides a shape into two parts that are the same size and shape.
Question 2.
____
No;
Explanation:
The line of symmetry divides a shape into two parts that are not with the same size and shape.
Question 3.
____
Yes;
Explanation:
The line of symmetry divides a shape into two parts that are the same size and shape.
Question 4.
____
No;
Explanation:
The line of symmetry divides a shape into two parts that are not with the same size and shape.
Question 5.
____
No;
Explanation:
The line of symmetry divides a shape into two parts that are not with the same size and shape.
Question 6.
____
Yes;
Explanation:
The line of symmetry divides a shape into two parts that are the same size and shape.
Question 7.
____
No;
Explanation:
The line of symmetry divides a shape into two parts that are not with the same size and shape.
Question 8.
____
Yes;
Explanation:
The line of symmetry divides a shape into two parts that are the same size and shape.
Complete the design by reflecting over the line of symmetry.
Question 9.
Question 10.
Problem Solving
Question 11.
Kara uses the pattern below to make paper dolls. The dashed line represents a line of symmetry. A complete doll includes the reflection of the pattern over the line of symmetry. Complete the design to show what one of Kara’s paper dolls looks like.
### Common Core – New – Page No. 580
Lesson Check
Question 1.
Which best describes the line of symmetry in the letter D?
Options:
a. horizontal
b. vertical
c. diagonal
d. half turn
a. horizontal
Explanation:
The horizontal line of symmetry in the letter D can exactly separate two parts equally.
Question 2.
Which shape has a correctly drawn line of symmetry?
Options:
a.
b.
c.
d.
b.
Explanation:
Image b has a line of symmetry that separates two parts equally.
Spiral Review
Question 3.
The class has 360 unit cubes in a bag. Johnnie divides the unit cubes equally among 8 groups. How many unit cubes will each group get?
Options:
a. 40
b. 44
c. 45
d. 48
c. 45
Explanation:
The class has 360 unit cubes in a bag. Johnnie divides the unit cubes equally among 8 groups. 360/8= 45.
Question 4.
There are 5,280 feet in one mile. How many feet are there in 6 miles?
Options:
a. 30,680
b. 31,260
c. 31,608
d. 31,680
d. 31,680
Explanation:
There are 5,280 feet in one mile. So, for 6 miles = 6 x 5, 280 = 31,680.
### Page No. 583
Question 1.
The shape at the right has line symmetry. Draw the 2 lines of symmetry.
Type below:
_________
Tell whether the shape appears to have zero lines, 1 line, or more than 1 line of symmetry. Write zero, 1, or more than 1.
Question 2.
_________
more than 1
Explanation:
There is more than 1 line of symmetries that separates two parts equally.
Question 3.
_________
more than 1
Explanation:
There is more than 1 lines of symmetries that separates two parts equally.
Question 4.
_________
1 line
Explanation:
There is 1 line of symmetry that separates two parts equally.
Question 5.
_________
zero lines
Explanation:
There is no line of symmetries that separates two parts equally.
Tell whether the shape appears to have zero lines, 1 line, or more than 1 line of symmetry. Write zero, 1, or more than 1.
Question 6.
_________
more than 1
Explanation:
There is more than 1 lines of symmetries that separate two parts equally.
Question 7.
_________
zero lines
Explanation:
There is no line of symmetries that separates two parts equally.
Question 8.
_________
zero lines
Explanation:
There is no line of symmetries that separates two parts equally.
Question 9.
_________
1 line
Explanation:
There is 1 line of symmetry that separates two parts equally.
Practice: Copy and Solve Does the design have line symmetry?
Write yes or no. If your answer is yes, draw all lines of symmetry.
Question 10.
____
Yes;
Question 11.
_____
No;
Question 12.
_____
Yes;
Question 13.
_____
No;
Question 14.
Draw a figure that has 5 sides and exactly 1 line of symmetry.
Type below:
_________
Explanation:
the above 5 sides shape has only 1 line symmetry
### Page No. 584
Use the chart for 15–17.
Question 15.
Which letters appear to have only 1 line of symmetry?
Type below:
_________
A, B, C, D, E, T, U, V, W
Explanation:
The letters A, B, C, D, E, T, U, V, W have only 1 line of symmetry.
Question 16.
Which letters appear to have zero lines of symmetry?
Type below:
_________
J, N, S
Explanation:
The letters J, N, S have only zero lines of symmetry.
Question 17.
The letter C has horizontal symmetry. The letter A has vertical symmetry. Which letters appear to have both horizontal and vertical symmetry?
Type below:
_________
H and I
Explanation:
The letters H and I have both horizontal and vertical symmetry.
Question 18.
Verify the Reasoning of Others Jeff says that the shape has only 2 lines of symmetry.
Does his statement make sense? Explain.
Type below:
_________
No; Jeff’s explanation is wrong. Because the given shape has only 2 lines of symmetry.
Question 19.
Match each figure with the correct number of lines of symmetry it has.
Type below:
_________
### Common Core – New – Page No. 585
Find and Draw Lines of Symmetry
Tell whether the shape appears to have zero lines, 1 line, or more than 1 line of symmetry. Write zero, 1, or more than 1.
Question 1.
1
more than 1
Explanation:
There is more than 1 line of symmetry that separates two parts equally.
Question 2.
________
more than 1
Explanation:
There is more than 1 line of symmetry that separates two parts equally.
Question 3.
________
Zero
Explanation:
There are 0 lines of symmetries.
Question 4.
________
more than 1
Explanation:
There is more than 1 line of symmetry that separates two parts equally.
Does the design have line symmetry? Write yes or no.
Question 5.
_____
Yes;
Question 6.
_____
Yes;
_____
No;
Question 8.
______
Yes;
Draw a shape for the statement. Draw the line or lines of symmetry.
Question 9.
zero lines of symmetry
Question 10.
1 line of symmetry
Question 11.
2 lines of symmetry
Problem Solving
Use the chart for 12–13.
Question 12.
Which number or numbers appear to have only 1 line of symmetry?
_____
3
Explanation:
The number 3 has only 1 line of symmetry.
Question 13.
Which number or numbers appear to have 2 lines of symmetry?
_____
0 and 8
Explanation:
The numbers 0 and 8 appear to have 2 lines of symmetry.
### Common Core – New – Page No. 586
Lesson Check
Question 1.
How many lines of symmetry does this shape appear to have?
Options:
a. 0
b. 2
c. 6
d. 12
c. 6
Explanation:
The given shape has 6 lines of symmetries.
Question 2.
Which of the following shapes appears to have exactly 1 line of symmetry?
Options:
a.
b.
c.
d.
d.
Explanation:
The trapezoid has exactly 1 line of symmetry.
Spiral Review
Question 3.
Richard practiced each of 3 piano solos for $$\frac{5}{12}$$ hour. How long did he practice in all?
Options:
a. $$\frac{2}{3}$$ hours
b. 1 $$\frac{1}{4}$$ hours
c. 1 $$\frac{1}{3}$$ hours
d. 1 $$\frac{5}{12}$$ hours
b. 1 $$\frac{1}{4}$$ hours
Explanation:
Richard practiced each of 3 piano solos for $$\frac{5}{12}$$ hour. $$\frac{5}{12}$$ hour = 1 $$\frac{1}{4}$$ hours hours.
Question 6.
Kevin draws a figure that has four sides. All sides have the same length. His figure has no right angles. What figure does Kevin draw?
Options:
a. square
b. trapezoid
c. rhombus
d. rectangle
c. rhombus
Explanation:
### Page No. 589
Question 1.
Marisol is making a pattern with blocks. What might the missing shape be?
First, look at the blocks.
Next, describe the pattern.
Type below:
_________
The first image 1 has three line segments. The next shape will have four line segments.
Question 1.
Finally, draw the missing shape.
Type below:
_________
Question 2.
Use the shapes to write a number pattern. Then describe the pattern in the numbers.
Shape 1: Triangle
Shape 2: Square
Shape 3: Pentagon
Shape 4: Hexagon
Shape 5: Heptagon
Question 3.
What if the pattern continued? Write an expression to describe the number of sides the sixth shape has in Marisol’s pattern.
Type below:
_________
If the pattern continued, then the next shape will have one more extra line segment to it. The sixth shape will become the octagon.
Shape 6: Octagon
Question 4.
Sahil made a pattern using circles. The first nine circles are shown. Describe the pattern. If Sahil continues the pattern, what might the next three circles be?
Type below:
_________
The pattern is repeated for every three circles. One big circle followed by two small circles.
### Page No. 590
Use the toy quilt designs for 5–6.
Question 5.
Lu is making a quilt that is 20 squares wide and has 24 rows. The border of the quilt is made by using each toy design equally as often. Each square can hold one design. How many of each design does she use for the border?
______ times
The border will have 20 squares two times, and 24 squares two times as well, that is the perimeter or the border, because a quilt has 4 sides:
20 X 2 + 24 X 2 = 40 + 48 = 88
So, the border will have 88 squares in total. So if the border can have only one design, Lu can use any toy design 88 times if she wants the border to have the same toy design in it.
Question 6.
Communicate Starting in the first square of her quilt, Lu lined up her toy designs in this order: plane, car, fire truck, helicopter, crane, and wagon. Using this pattern unit, which design will Lu place in the fifteenth square? Explain how you found your answer.
_________
The answer is fire truck. As the pattern repeats, the fifteenth square will fire the truck.
Question 8.
Norris drew the pattern shown.
Label the circles to show the colors in the fourth figure of the pattern.
Type below:
_________
### Common Core – New – Page No. 591
Problem Solving Shape Patterns
Solve each Problem.
Question 1.
Marta is using this pattern to decorate a picture frame. Describe the pattern. Draw what might be the next three figures in the pattern.
Possible answer: the pattern repeats: one trangle followed by two squares.
The pattern repeats one triangle followed by two squares.
Question 2.
Describe the pattern. Draw what might be the next three figures in the pattern. How many circles are in the sixth figure in the pattern?
_____ circles
Add one more column with 1 more circle than in the previous column; 21.
Question 3.
Larry stencils this pattern to make a border at the top of his bedroom walls. Describe the pattern. Draw what might be the missing figure in the pattern.
2 triangles placed side to side followed by 2 sets of 2 triangles placed vertex to vertex
### Common Core – New – Page No. 592
Lesson Check
Question 1.
What might be the next three figures in this pattern?
Options:
a.
b.
c.
d.
a.
Explanation:
the pattern has odd numbers of up arrows then even number of down arrows. So, the next three figures are .
Question 2.
Which might be the missing figure in the following pattern?
Options:
a.
b.
c.
d.
a.
Explanation:
From the pattern, the missing image will have a vertical rectangle with the circle and X mark in it.
Spiral Review
Question 4.
Olivia finished a race in 40.64 seconds. Patty finished the race in 40.39 seconds. Miguel finished the race in 41.44 seconds. Chad finished the race in 40.46 seconds. Who finished the race in the least time?
Options:
a. Olivia
b. Patty
c. Miguel
b. Patty
Explanation:
Patty finished the race in 40.39 seconds which is the least time compared to others.
Question 5.
Justin bought 6 ribbons for an art project. Each ribbon is $$\frac{1}{4}$$ yard long. How many yards of ribbon did Justin buy?
Options:
a. $$\frac{2}{3}$$ yard
b. 1 $$\frac{1}{4}$$ yards
c. 1 $$\frac{1}{2}$$ yards
d. 1 $$\frac{3}{4}$$ yards
c. 1 $$\frac{1}{2}$$ yards
Explanation:
Justin bought 6 ribbons for an art project. Each ribbon is $$\frac{1}{4}$$ yard long. So, 6 X $$\frac{1}{4}$$ = $$\frac{3}{2}$$ = 1 $$\frac{1}{2}$$ yards.
Question 6.
Kyle and Andrea were asked to make a list of prime numbers.
Kyle: 1, 3, 7, 19, 23
Andrea: 2, 3, 5, 7, 11
Whose list is correct?
Options:
a. Only Kyle’s list
b. Only Andrea’s list
c. Both lists are correct.
d. Neither list is correct.
b. Only Andrea’s list
Explanation:
1 is not a prime number. So, the answer is Only Andrea’s list is correct.
### Page No. 593
Question 1.
Gavin is designing a kite. He sketched a picture of the kite.
How many right angles does the kite appear to have?
_____ right angles
0 right angles
Explanation:
There is no right angles in the given shape.
Question 2.
Write the letter of the triangle under its correct classification.
Explanation:
C and F are Acute angles with less than right angles.
B and D are Obtuse Angles with more than right angles.
A and E are Right Angles.
Question 3.
Select the angles that identify an obtuse triangle. Mark all that apply.
Options:
a. acute, acute, acute
b. acute, acute, obtuse
c. right, acute, acute
d. obtuse, right, acute
b. acute, acute, obtuse
Explanation:
An obtuse triangle will have one obtuse angle and two acute angles.
### Page No. 594
Question 4.
Write the word that describes the part of Figure A written below.
$$\overline{E B}$$ _________
$$\overset { \longleftrightarrow }{ AB }$$ _________
$$\overrightarrow{G A}$$ _________
∠EBG _________
∠CGB _________
$$\overline{E B}$$ line segment.
$$\overset { \longleftrightarrow }{ AB }$$ Line.
$$\overrightarrow{G A}$$ Ray.
∠EBG right angle.
∠CGB acute angle.
Explanation:
$$\overline{E B}$$ is a line segment that has two endpoints connected to form a line.
$$\overset { \longleftrightarrow }{ AB }$$ is a Line that continues without an end in both directions.
$$\overrightarrow{G A}$$ is a Ray that has one endpoint and continues without an end in one direction.
∠EBG right angle.
∠CGB is an acute angle with less than the right angle.
Question 5.
What term best describes the figure shown below?
perpendicular lines
Explanation:
The lines are forming four right angles they form squares. So, both lines are perpendicular lines.
Question 6.
Naomi leaves for her trip to Los Angeles on the 12th day of August. Since August is the 8th month, Naomi wrote the date as shown.
Naomi says all the numbers she wrote have line symmetry. Is she correct? Explain your thinking.
_______
Naomi is incorrect. The number 2 does not have a line of symmetry because if it were cut out, there would be no way to fold it in half so that the two parts matched exactly.
### Page No. 595
Question 7.
Max made a pennant that looks like a triangle. How can you classify the triangle based on its angles?
The triangle is a(n) ____________ triangle.
The triangle is an acute triangle.
Explanation:
The triangle is an acute triangle. because it has angles with less than right angles.
Question 8.
Choose the labels to make a true statement.
_____ is parallel to ______
Line AB is parallel to line CD.
Explanation:
From the given image, Line AB is parallel to line CD.
Question 9.
Classify the figure. Select all that apply.
Options:
b. trapezoid
c. parallelogram
d. rectangle
e. rhombus
f. square
b. trapezoid
c. parallelogram
d. rectangle
Explanation:
The given image has 2 parallel sides, 2 pairs of sides of length, and four right angles. So, the possible answers are quadrilateral, trapezoid, parallelogram, and rectangle.
### Page No. 596
Question 11.
Match each figure with the correct number of lines of symmetry it has.
Explanation:
Image 1: 1 line of symmetry
Image 2: 2 lines of symmetry
Image 3: 0 lines of symmetry
Image 4: More than 2 lines of symmetry.
Question 12.
Barb drew the pattern shown.
Use the square shown to draw the missing pattern. □
Explanation:
The fourth shape must consist of one extra square box in the top line and bottom line.
Question 13.
Claudia drew the figure below. Draw a line of symmetry on Claudia’s figure.
Explanation:
The image can have one line symmetry.
Question 14.
Write the word or words that best describe this figure.
_________
Ray
Explanation:
The ray has one endpoint and continues without an end in one direction.
### Page No. 597
Question 16.
Mike drew a figure with opposite sides parallel. Write the pairs of parallel sides. What figure is it?
Line DG is parallel to Line FE and Line DE is parallel to Line GF; the figure is a parallelogram.
Question 17.
Circle the letter that does not have line symmetry.
Explanation:
The S does not have line symmetry.
Question 18.
Joseph made a pattern using ovals and rectangles. The first four figures of his pattern are shown. Draw the next figure in the pattern.
Question 19.
Jeremy drew Figure 1 and Louisa drew Figure 2.
Part A
Jeremy says both figures are rectangles. Do you agree with Jeremy?
_____
Yes; both figures have 2 pairs of parallel sides, opposite sides that are equal in length, and 4 right angles.
Question 19.
Part B
Louisa says both figures are rhombuses. Do you agree with Louisa?
_____
No; figure 2 is a rhombus since it has 2 pairs of parallel sides and 4 sides of equal length. Figure 1 does not have 4 sides of equal length so it cannot be a rhombus.
### Page No. 598
Question 20.
Veronica found the number of lines of symmetry for the figure below. How many lines of symmetry does it have?
______ lines of symmetry
2 lines of symmetry
Explanation:
the given shape can have 2 lines of symmetry.
Question 21.
Jordan drew the pattern below.
Part A
Describe the pattern.
Each figure has 2 more squares than the preceding figure.
Question 21.
Part B
Write a rule using numbers to find the number of squares in any figure in the pattern.
multiply the figure number by 2.
Question 21.
Part C
Draw Figure 5.
### Page No. 603
Tell what fraction of the circle the shaded angle represents.
Question 1.
$$\frac{□}{□}$$
$$\frac{1}{2}$$
Explanation:
Half of the part is shaded out of the circle. So, the answer is 1/2.
Question 2.
$$\frac{□}{□}$$
$$\frac{1}{4}$$
Explanation:
$$\frac{1}{4}$$ part of the circle is shaded out of the circle.
Question 3.
$$\frac{□}{□}$$
$$\frac{3}{4}$$
Explanation:
3 parts of the circle is shaded out of the circle
Question 4.
$$\frac{□}{□}$$
$$\frac{1}{1}$$ = 1
Explanation:
Question 5.
$$\frac{□}{□}$$
$$\frac{1}{2}$$
Explanation:
Half of the part is shaded out of the circle. So, the answer is 1/2.
Question 6.
$$\frac{□}{□}$$
$$\frac{1}{12}$$
Tell whether the angle on the circle shows a $$\frac{1}{4}, \frac{1}{2}, \frac{3}{4}$$, or 1 full turn clockwise or counterclockwise.
Question 7.
Type below:
________
$$\frac{3}{4}$$; counterclockwise
Explanation:
The image shows the counterclockwise and formed fraction of $$\frac{3}{4}$$.
Question 8.
Type below:
________
$$\frac{1}{2}$$; clockwise
Explanation:
The image shows the clockwise and formed fraction of $$\frac{1}{2}$$.
Question 9.
Type below:
________
$$\frac{1}{4}$$; clockwise
Explanation:
The image shows the clockwise and formed fraction of $$\frac{1}{4}$$.
Question 10.
Susan watched the game from 1 p.m. to 1:30 p.m. Describe the turn the minute hand made.
Type below:
________
The minute hand made a $$\frac{1}{2}$$ turn clockwise.
### Page No. 604
Question 12.
Malcolm drew this angle on the circle. Which of the following describes the angle? Mark all that apply.
Options:
a. $$\frac{3}{4}$$ turn
b. $$\frac{1}{4}$$ turn
c. clockwise
d. counterclockwise
a. $$\frac{3}{4}$$ turn
d. counterclockwise
Explanation:
The image show the $$\frac{3}{4}$$ turn and also the counterclockwise.
Sense or Nonsense?
Question 13.
Whose statement makes sense? Whose statement is nonsense? Explain your reasoning.
Type below:
__________
The girl’s statement makes sense. The boy’s statement makes non-sense. Because from the figure it is clearly shown that the shaded part is $$\frac{1}{4}$$ of the circle. There is no particular direction given to measure the shaded part. | 12,983 | 49,221 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-38 | latest | en | 0.857472 |
https://www.thenational.academy/pupils/lessons/fractions-divide-a-proper-fraction-by-an-integer-6tjkgd/video | 1,716,657,922,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058830.51/warc/CC-MAIN-20240525161720-20240525191720-00230.warc.gz | 907,328,526 | 27,423 | # Lesson video
In progress...
This is our third lesson in our fractions topic.
Today, we'll be dividing a proper fraction by an integer.
It's just a pencil and piece of paper lesson.
So pause the video now and get your things if you haven't done so already.
Then we'll be looking at the connection between fractions and division before deepening our understanding of dividing a fraction by an integer then you'll complete some independent work and a final quiz.
Pause the video now and complete the quiz and click restart once you're finished.
Good job.
Now we're starting by looking today at the connection between fractions and division.
So we've got a word problem here, 12 bars of chocolate are shared equally between four people, how much chocolate does each person get? So we're going to look at how to represent this pictorially.
So we've got 12 bars of chocolate, shared between four people.
So here are our four groups.
And we can see from this representation that each of those four people will receive three bars of chocolate.
Or we know that sharing 12 wholes into four equal groups gives us three, so four equal groups of three, and that's the same as 12 divided by four.
But we also can remember from our previous lesson that if we're dividing by four, we're finding a quarter of something.
And if we're finding a quarter of something, that is the same as multiplying by one quarter.
So finding a quarter of 12 is the same as doing one quarter multiplied by 12.
So we're linking here division and multiplication of fractions.
Let's have a look at another one.
So here, the question is we have six bars of chocolate and they're shared equally between two people, how much chocolate does each person get? So I wouldn't need to pause the video and create your own pictorial representation of this problem.
And you can always go back to our previous slide to see how that might look, pause the video now.
So you had your six bars of chocolate, so you may have drawn a bar model divided into six equal parts, and you're sharing them between two people so you're sharing them into two groups and you can see that each of two groups will get three bars of chocolate.
So you're sharing six wholes into two equal groups of three, which is the same as six divided by two.
And then if we link it back to our multiplication from yesterday, if we're dividing something by two, we're finding a half and finding a half of six is the same as doing a half multiplied by six.
And if you remember back to yesterday, we could turn this integer into a fraction over one, one multiplied by six is six, two multiplied by one is two and if I convert that back to in fact, it turns into an integer because there are three groups of two in six.
How're we going to deepen our understanding of this by applying this two division questions.
So these are the sorts of questions that you may have seen before.
So we've got three quarters divided by five.
What we're doing here is we're dividing three quarters, which is the whole here into five equal groups.
So we're dividing three quarters into fifths.
If we divide something by five, we are finding fifths.
What we're actually doing here is we're finding a fifth of three quarters.
And that is the same as three quarters times a fifth or a fifth times three quarters.
So remember that we multiply numerators, which is three and denominators, which is 20.
So three quarters divided by five is equal to 3/20, is the same if we're dividing by five, it's the same as finding one fifth.
Let's look at another one.
Omar has one third of a bottle of juice left.
So the whole is a third.
He shares the juice equally between five people.
And we want to know what fraction of the juice does each person get.
So what we're actually being asked here to do is take our whole here which is a third and divide it by five.
So let's look at it pictorially.
We've got a third and we've divided it into five equal parts.
So we're dividing it into fifths, okay? We're trying to find a fifth of a third.
So this is how we can represent it, we can represent it how we did up here, a third divided by five, or we can think about it as a fifth of a group of one third, which is the same as a third times a fifth, one times one is one and three times five is 1/15, So a third divided by five is equal to 1/15.
Now you're going to look at one independently I'll read it out to you first.
So Novie has three fifths of a pack of stickers left, So that's the whole, she wants to share the remaining stickers equally with her friend Elizabeth.
Now, what we're saying here is that she wants to share them between herself and Elizabeth, so think about what you're dividing your fraction by.
and I would like you to represent this pictorially and then if you think back to our previous one, I'll just pop back.
If you think about representing it in these three different ways, pause the video now and represent the problem.
So you knew that the whole was three fifths and it was being divided into two equal parts between Novie herself and Elizabeth.
So if you're dividing something by two, you're finding a half, and the ways that you could have represented that was by thinking about finding three fifths divided by two, which is the same as finding a half of three fifths, which is the same as a half multiplied by three fifths.
One times three is equal to three and two times five is equal to 10.
So you're going to complete some independent work.
Now, the main thing that you need to think of doing is converting your division question into a multiplication question by thinking about what am I actually being asked to do here.
And as always, it's very helpful to draw yourself a picture to make that connection.
So pause the video now and complete the task, then click restart once you're finished and we'll work through those solutions together.
So for the first question, you were asked to find 9/10 divided by three.
Remember that if we're dividing something by three, we're finding a third.
So you were doing a third of 9/10, which is the same as 9 /10 times one third, which is equal to 9/30, which simplifies to 3/10.
And we're looking at the same procedure here.
So we're looking at dividing 10/12, sorry by five, which is the same as finding a fifth of 10/12, which is the same as 10/12 multiplied by one fifth, 10 times one is 10, 12 times five is 60 and that simplifies to one sixth.
If we're dividing by four, we're finding a quarter.
So if finding a quarter of 8/11, which is the same as 8/11 times one quarter, which is equal to 8/44, which simplifies to 2/11 by dividing by four, then you'll find a one.
If we're dividing by two, we're finding a half.
So we're doing seven eighths multiplied by one half, which is equal to 7/16.
In question two, we had a word problem.
Sally has forgotten to take her phone charger away for the weekend.
And she has four fifths of her battery life remaining.
She has to make it last for three days.
So how much of the battery can she use each day, if she wants to use an equal amount each day? So you are being asked to divide four-fifths by three, which is the same as four fifths multiplied by one third, because you're finding four fifths or third or four fifths.
And I've also got a representation here, pictorially to think of it in a different way.
So we know that Sally has four fifths of her battery left and over three days that can be divided into 15 parts of battery life.
So she has 12 parts left because she's already used 3/15.
So she's got 12/15, which is coloured in and split equally over those three equal days means that she has 4/15 to use each day for 15 days.
So for question three, dividing by two is the same as finding a half of something.
So I'm thinking what multiply by one half is equal to 3/8, I know the three times on is three and four times two is eight.
So three quarters times a half is three eighth so a half of three quarters is three eighths.
For the next one, we're dividing something by three to find one ninth.
I know that dividing by three, it's the same as finding one third of something.
So I know that one times one is one and three times three is nine.
So a third of a third is a ninth.
Question four, this is a multi step problem.
So the field is two thirds of a kilometres long by one ninth of a kilometre wide.
Hopefully you drew yourself an image there so the area of the field is split into three equal parts.
And you have to find the area as a fraction of each section.
So I gave you a hint to find the area first We know the area is length times width, so the area was 2/27 of a kilometre squared and then I'll just put my units in there.
And then the field was divided into three equal sections.
So we know that the whole area is 2/27 and we needed to divide that by three.
So 2/27 divided by three is the same as finding a third of 2/27, which is the same as 2/27 times one third, which is equal to 2/81 kilometres squared So each section had an area of 2/81 of a kilometre squared.
And question five, you were asked to find the value of the calculation when A equals three and B equals nine So you just needed to substitute those numbers in.
So A is three, then it's three fifths divided by B is nine so divided by nine so you're dividing by nine.
You're finding a ninth of three fifths, three fifths times one ninth is equal to 3/45, which can be divided by three and simplified to 1/15.
Now it's time for your final quiz, so pause the video and complete the quiz and click restart once you're finished.
Great work today in our next lesson, we'll be looking at both multiplying and dividing proper fractions, I'll see you then. | 2,231 | 9,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2024-22 | latest | en | 0.971739 |
https://givemeasine.blogspot.com/2018/02/ | 1,642,955,019,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304287.0/warc/CC-MAIN-20220123141754-20220123171754-00210.warc.gz | 333,164,358 | 16,416 | ## Tuesday, February 27, 2018
### Which Ratio? Activity for Right Triangle Trig
One of the things my geometry kids consistently struggle with each year is deciding just which ratio to use for each problem. They can reliably go from there, but when you get stuck on the set up, it can be a SINE that you're lost!
I tried something new this week to combat this issue and so far, it's been really effective. My lab students are especially benefiting as I see them consistently being able to identify the ratio and therefore gain an entry point into the problem.
Pam Wilson wrote a great blog post about an activity she did with her students- read it here- and I stole the document she created to adapt to my kiddos. I put each one of these on a slide and students are asked to analyze (think-pair-share) style which ratio is represented in each picture.
My favorite part is I gave each student a set of these paddles to participate. They were quick to make....run an exacto knife through some cardboard and hot glue popsicle sticks to the back of each (how most people spend their Sunday nights, right?!). Instant feedback for me, fun for them....an all around win. I've been just leaving these on the tables as we work and I'll ask the kids to periodically show me which ratio they think we should use before we attempt a problem. This has been especially helpful with word problems because I do a quick analysis before moving on to make sure we've "got it" and know how to approach it.
Also, I know this could totally be done in a Kahoot or another way, but something about the hands on "auction" style get the kids super engaged and I love that I can continue to use them throughout the unit. A nice tool to keep around whenever necessary, not just on "Kahoot" day.
A small thing but so far, especially for me struggling learners, it's made a big difference!
## Thursday, February 8, 2018
### Differential Equations CSI
Differential Equations are a topic with so many applications in the real world. Knowing how rough the Potato Question was for AP Calculus AB students last year (49% of students earned 0/9 points), I wanted to bring some more of these applications to my kiddos within our unit. Luckily, I'm married to a scientist and he's overflowing with ideas to help me out!
First, we looked at rate laws in Chemistry. First order rate laws are a direct application of the law of exponential change. Honestly, these have more to explore when we get into definite integrals, but I wanted students seeing them for the first time now so we can continue to work with them when we start definite integrals. Since about half of my class is enrolled in AP Chem, this was an easy connection for them to make. We also got to discuss why they're called Integrated Rate Laws (oh, THAT'S what the chem teacher was talking about!) and most were able to easily tie this to the derivation we always do for this.
After this and some more examples, I handed out the lab for the day-
Determining Time of Death with Newton's Law of Cooling.
I modified it from an old Houghton Mifflin activity, adding in the information about forensic science. I used this article on using ODE's to determine time of death as a resource for what I added, if you're interested. I felt like I'd scaffolded it enough and was excited to get my kids working.
When they walked in, I had the room set up like a crime scene. My goal was that they'd solve the murder before they left. I had a "breaking news" picture of the murderer poised and ready....it was obviously the Night King from Game of Thrones. But I'll be honest....I should have left more time for this. It felt rushed, so we didn't finish it. My students this year struggle a lot more with fundamentals than in years past, so properties of logarithms that should be 2nd nature were taking way more explanation than the time I'd allotted. We got through the first page- the separate and solving for C. We analyzed what C represented. The rest we will save for next class.
I still love the idea and the intrigue it created for the kids- walking into a "crime scene" and being responsible for solving it. Next year, hopefully with less snow days, we will leave ourselves more time to make sure our algebra skills are on point and we have time to really solve the mystery! Next year I also want to tie in temperature probes and have students actually gather their own data, instead of using the data provided in the problem.
## Sunday, February 4, 2018
### Hands On Slope Fields
Slope fields have always been a weird topic for me...I never loved my approach to them. They weren't hard for my kids, but they weren't particularly engaging and I always saw my student getting frustrated by the "stakes" of doing them on paper and feeling like they were doing it wrong before they got a real feel for it. I wanted to have my kids be able to experiment with it in a hands on way before having them commit anything to paper, so I've been tweaking this approach for the past few years and felt like this week it finally went the way I'd like it to go!
I started the class by using Rebecka Peteron's Slope Field Activity, where each student is given a coordinate point and then asked to come up to the board and draw a small segment with that slope. We used these discussion questions pictured to talk about isoclines and beginning to see patterns beyond just "plug and chug" methods of determining information from the field itself.
From there, we did some low-stakes practice with the interactive slope fields I built. To make them, I went to Staples and bought washi tape, brass paper fasteners, and a roll of packing cardboard. I cut out 1 ft x 1 ft squares of cardboard, used washi tape to make the axis, and put the fasteners through the back. In class, I showed the students a differential equation and they modelled the slope field for me. We were able to work out kinks and make sure we all understand what they should look like before going any further. Then, I would ask the students to trace a particular solution using their Twizzlers through a point. We were able to discuss the general vs particular solution, the patterns we saw, and more. All of this was followed by a small group exploration where students played with dependence on x vs y, determining particular solutions, and relating differential equations and slope fields in different representations.
.
I still want to work on the type of practice we do after the exploration and relate it to some more applications, but this approach was way effective for my kids than normal. I have a group this year who really benefit from more visual and hands on approaches anyway.
Do you have any other approaches to slope fields that you love? Share in the comments! | 1,459 | 6,769 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-05 | longest | en | 0.973855 |
https://advantekservice.com/scratches/how-many-volts-does-a-refrigerator-use-solution-found.html | 1,660,924,403,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573699.52/warc/CC-MAIN-20220819131019-20220819161019-00356.warc.gz | 102,130,236 | 12,744 | # How Many Volts Does A Refrigerator Use? (Solution found)
Assuming that most refrigerators in the United States and Canada operate on 120 volts, dividing the average wattage of 25.91 by 120 volts will get the refrigerator’s typical power consumption.
In order for a refrigerator to operate, what voltage is required?
• Typically, batteries with voltages ranging from 12v through 24v and 48v are available for a variety of purposes. Then we’ll need to know how many amp-hours the battery has. The battery should be capable of providing enough energy to ensure that your refrigerator runs smoothly on solar power without interruption.
## Are refrigerators 120v or 240?
Refrigerator — The majority of refrigerators are powered by a conventional 100-120 volt electrical socket and outlet combination. There are three prongs on the electrical plug, and this indicates that you will require an electrical outlet with three inserts. The 240 volt circuits are distributed across two wires, one of which is neutral and the other of which is ground.
## Can you plug a fridge into a regular outlet?
Refrigerators do not require specific electrical outlets. So long as the socket is three-pronged, they may be plugged into a conventional 110-120 volt outlet without issue. But having your refrigerator on its own dedicated circuit is the best option in most cases. In this way, you can guarantee that your refrigerator receives the electricity it requires to function correctly.
## Is a refrigerator 120v?
Almost any refrigerator that plugs into a regular wall outlet operates at a voltage of 120 volts, which is essentially the same voltage as 125 volts or even 115 or 110 volts, depending on the appliance you are looking at. Furthermore, each particular refrigerator consumes more current and electricity when it is turned on for the first time than when it is running.
## Do fridges use 110v?
A conventional domestic refrigerator is connected to a regular household socket through an electrical cord. The device is powered by 110 volts of energy provided by the wall socket. The refrigerator must be connected to a dedicated circuit according to most building requirements. This means that a circuit breaker in the breaker box is only capable of supplying electricity to that particular item and nothing else.
## How do you tell if you have 240 or 120?
Find the breaker in your electrical panel that is connected to your thermostat using Method One. It is likely that you have 120V if you have a single breaker switch, such as the one seen in the figure below. If you notice a double breaker, such as the ones shown in the images below, you most likely have 240V.
## How do I know if I have a 240 volt outlet?
What is the best way to tell if an outlet is 240-volts? In comparison to 120-volt outlets, 240-volt outlets have rounder tops with three or four holes, and they are bigger in size. When looking at an earlier three-prong 240-volt plug, the top hole appears like a backward ‘L,’ while the other two holes are diagonally located on either side of the plug.
## Should a fridge be on its own circuit?
The recommended best practice for homeowners is to install the refrigerator on a separate dedicated circuit from the rest of the house. Refrigerators and freezers should be put on a dedicated 120 volt circuit with a power rating of 15-20 amps. This will prevent an electrical overload from occurring as a result of your present wiring being unable to handle the increased power.
## Can a refrigerator share an outlet with a TV?
In such case, unless the outlet is connected to an uninterruptible power supply or a line voltage stabilizer that is too tiny, there is no overload. Check the labels on the refrigerator and television to see how many AMPS or WATTS they consume; they should be fewer than 8 amps combined, or less than 1000 watts combined, if possible.
## Can I plug a fridge into an Extension Cord?
There are several reasons why you should not plug a refrigerator into an extension cable. Extension cables contain shorter wire than conventional home sockets, which makes them more energy efficient. These characteristics make extension cords particularly vulnerable to failure and fire, particularly if the equipment hooked into them consume more electricity than the extension cable is designed to handle.
## What is a 110 outlet?
A 110-volt outlet is the industry standard for electrical house wiring in the United States. It looks somewhat like this: When it comes to electricity, the voltage “110” is just as prevalent in speech as the voltages 115, 117, and 120. See It has been my experience that the terms “110 volt outlet” and “120 volt outlet” are used interchangeably when talking about voltage in a US household.
## Can you plug a freezer into a regular outlet?
Because it can be plugged into a conventional wall electrical socket, a freezer is one of the most straightforward household equipment to set up. Because a freezer has one of the longest predicted life cycles of any kitchen equipment, it is a long-term investment in terms of installation. | 1,053 | 5,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2022-33 | latest | en | 0.921204 |
https://www.unix.com/man-page/centos/3/cgehd2.f/ | 1,524,497,462,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946077.4/warc/CC-MAIN-20180423144933-20180423164933-00515.warc.gz | 901,761,094 | 6,526 | Home Search Forums Register Forum Rules Man Pages FAQ Members Search Today's Posts Mark Forums Read
# CentOS 7.0 - man page for cgehd2.f (centos section 3)
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```cgehd2.f(3) LAPACK cgehd2.f(3) NAME cgehd2.f - SYNOPSIS Functions/Subroutines subroutine cgehd2 (N, ILO, IHI, A, LDA, TAU, WORK, INFO) CGEHD2 reduces a general square matrix to upper Hessenberg form using an unblocked algorithm. Function/Subroutine Documentation subroutine cgehd2 (integerN, integerILO, integerIHI, complex, dimension( lda, * )A, integerLDA, complex, dimension( * )TAU, complex, dimension( * )WORK, integerINFO) CGEHD2 reduces a general square matrix to upper Hessenberg form using an unblocked algorithm. Purpose: CGEHD2 reduces a complex general matrix A to upper Hessenberg form H by a unitary similarity transformation: Q**H * A * Q = H . Parameters: N N is INTEGER The order of the matrix A. N >= 0. ILO ILO is INTEGER IHI IHI is INTEGER It is assumed that A is already upper triangular in rows and columns 1:ILO-1 and IHI+1:N. ILO and IHI are normally set by a previous call to CGEBAL; otherwise they should be set to 1 and N respectively. See Further Details. 1 <= ILO <= IHI <= max(1,N). A A is COMPLEX array, dimension (LDA,N) On entry, the n by n general matrix to be reduced. On exit, the upper triangle and the first subdiagonal of A are overwritten with the upper Hessenberg matrix H, and the elements below the first subdiagonal, with the array TAU, represent the unitary matrix Q as a product of elementary reflectors. See Further Details. LDA LDA is INTEGER The leading dimension of the array A. LDA >= max(1,N). TAU TAU is COMPLEX array, dimension (N-1) The scalar factors of the elementary reflectors (see Further Details). WORK WORK is COMPLEX array, dimension (N) INFO INFO is INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value. Author: Univ. of Tennessee Univ. of California Berkeley Univ. of Colorado Denver NAG Ltd. Date: September 2012 Further Details: The matrix Q is represented as a product of (ihi-ilo) elementary reflectors Q = H(ilo) H(ilo+1) . . . H(ihi-1). Each H(i) has the form H(i) = I - tau * v * v**H where tau is a complex scalar, and v is a complex vector with v(1:i) = 0, v(i+1) = 1 and v(ihi+1:n) = 0; v(i+2:ihi) is stored on exit in A(i+2:ihi,i), and tau in TAU(i). The contents of A are illustrated by the following example, with n = 7, ilo = 2 and ihi = 6: on entry, on exit, ( a a a a a a a ) ( a a h h h h a ) ( a a a a a a ) ( a h h h h a ) ( a a a a a a ) ( h h h h h h ) ( a a a a a a ) ( v2 h h h h h ) ( a a a a a a ) ( v2 v3 h h h h ) ( a a a a a a ) ( v2 v3 v4 h h h ) ( a ) ( a ) where a denotes an element of the original matrix A, h denotes a modified element of the upper Hessenberg matrix H, and vi denotes an element of the vector defining H(i). Definition at line 150 of file cgehd2.f. Author Generated automatically by Doxygen for LAPACK from the source code. Version 3.4.2 Tue Sep 25 2012 cgehd2.f(3)```
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All times are GMT -4. The time now is 11:31 AM. | 1,168 | 3,750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-17 | latest | en | 0.635219 |
https://elteoremadecuales.com/ores-theorem/?lang=fr | 1,679,921,047,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00425.warc.gz | 282,938,243 | 13,150 | # Ore's theorem
Ore's theorem For Ore's theorem in ring theory, see Ore condition. A graph meeting the conditions of Ore's theorem, and a Hamiltonian cycle in it. There are two vertices with degree less than n/2 in the center of the drawing, so the conditions for Dirac's theorem are not met. Cependant, these two vertices are adjacent, and all other pairs of vertices have total degree at least seven, the number of vertices.
Ore's theorem is a result in graph theory proved in 1960 by Norwegian mathematician Øystein Ore. It gives a sufficient condition for a graph to be Hamiltonian, essentially stating that a graph with sufficiently many edges must contain a Hamilton cycle. Spécifiquement, the theorem considers the sum of the degrees of pairs of non-adjacent vertices: if every such pair has a sum that at least equals the total number of vertices in the graph, then the graph is Hamiltonian.
Contenu 1 Déclaration formelle 2 Preuve 3 Algorithm 4 Résultats associés 5 References Formal statement Let G be a (finite and simple) graph with n ≥ 3 sommets. We denote by deg v the degree of a vertex v in G, c'est à dire. the number of incident edges in G to v. Alors, Ore's theorem states that if deg v + deg w ≥ n for every pair of distinct non-adjacent vertices v and w of G () then G is Hamiltonian.
Proof Illustration for the proof of Ore's theorem. In a graph with the Hamiltonian path v1...vn but no Hamiltonian cycle, at most one of the two edges v1vi and vi − 1vn (shown as blue dashed curves) can exist. Pour, if they both exist, then adding them to the path and removing the (rouge) edge vi − 1vi would produce a Hamiltonian cycle.
It is equivalent to show that every non-Hamiltonian graph G does not obey condition (). Accordingly, let G be a graph on n ≥ 3 vertices that is not Hamiltonian, and let H be formed from G by adding edges one at a time that do not create a Hamiltonian cycle, until no more edges can be added. Let x and y be any two non-adjacent vertices in H. Then adding edge xy to H would create at least one new Hamiltonian cycle, and the edges other than xy in such a cycle must form a Hamiltonian path v1v2...vn in H with x = v1 and y = vn. For each index i in the range 2 ≤ i ≤ n, consider the two possible edges in H from v1 to vi and from vi − 1 to vn. At most one of these two edges can be present in H, for otherwise the cycle v1v2...vi − 1vnvn − 1...vi would be a Hamiltonian cycle. Ainsi, the total number of edges incident to either v1 or vn is at most equal to the number of choices of i, which is n − 1. Par conséquent, H does not obey property (), which requires that this total number of edges (deg v1 + deg vn) be greater than or equal to n. Since the vertex degrees in G are at most equal to the degrees in H, it follows that G also does not obey property ().
Algorithm Palmer (1997) describes the following simple algorithm for constructing a Hamiltonian cycle in a graph meeting Ore's condition.
Arrange the vertices arbitrarily into a cycle, ignoring adjacencies in the graph. While the cycle contains two consecutive vertices vi and vi + 1 that are not adjacent in the graph, perform the following two steps: Search for an index j such that the four vertices vi, vi + 1, vj, and vj + 1 are all distinct and such that the graph contains edges from vi to vj and from vj + 1 to vi + 1 Reverse the part of the cycle between vi + 1 and vj (inclusive).
Each step increases the number of consecutive pairs in the cycle that are adjacent in the graph, by one or two pairs (depending on whether vj and vj + 1 are already adjacent), so the outer loop can only happen at most n times before the algorithm terminates, where n is the number of vertices in the given graph. By an argument similar to the one in the proof of the theorem, the desired index j must exist, or else the nonadjacent vertices vi and vi + 1 would have too small a total degree. Finding i and j, and reversing part of the cycle, can all be accomplished in time O(n). Par conséquent, the total time for the algorithm is O(n2), matching the number of edges in the input graph.
Related results Ore's theorem is a generalization of Dirac's theorem that, when each vertex has degree at least n/2, the graph is Hamiltonian. Pour, if a graph meets Dirac's condition, then clearly each pair of vertices has degrees adding to at least n.
In turn Ore's theorem is generalized by the Bondy–Chvátal theorem. One may define a closure operation on a graph in which, whenever two nonadjacent vertices have degrees adding to at least n, one adds an edge connecting them; if a graph meets the conditions of Ore's theorem, its closure is a complete graph. The Bondy–Chvátal theorem states that a graph is Hamiltonian if and only if its closure is Hamiltonian; since the complete graph is Hamiltonian, Ore's theorem is an immediate consequence.
Woodall (1972) found a version of Ore's theorem that applies to directed graphs. Suppose a digraph G has the property that, for every two vertices u and v, either there is an edge from u to v or the outdegree of u plus the indegree of v equals or exceeds the number of vertices in G. Alors, according to Woodall's theorem, G contains a directed Hamiltonian cycle. Ore's theorem may be obtained from Woodall by replacing every edge in a given undirected graph by a pair of directed edges. A closely related theorem by Meyniel (1973) states that an n-vertex strongly connected digraph with the property that, for every two nonadjacent vertices u and v, the total number of edges incident to u or v is at least 2n − 1 must be Hamiltonian.
Ore's theorem may also be strengthened to give a stronger conclusion than Hamiltonicity as a consequence of the degree condition in the theorem. Spécifiquement, every graph satisfying the conditions of Ore's theorem is either a regular complete bipartite graph or is pancyclic (Bondy 1971).
References Bondy, J. UN. (1971), "Pancyclic graphs I", Journal de théorie combinatoire, Série B, 11 (1): 80–84, est ce que je:10.1016/0095-8956(71)90016-5. Meiniel, M. (1973), "Une condition suffisante d'existence d'un circuit hamiltonien dans un graphe orienté", Journal de théorie combinatoire, Série B (en français), 14 (2): 137–147, est ce que je:10.1016/0095-8956(73)90057-9. Minerai, Ø. (1960), "Remarque sur les circuits de Hamilton", Mensuel mathématique américain, 67 (1): 55, est ce que je:10.2307/2308928, JSTOR 2308928. Palmer, E. M. (1997), "The hidden algorithm of Ore's theorem on Hamiltonian cycles", Computers & Mathematics with Applications, 34 (11): 113–119, est ce que je:10.1016/S0898-1221(97)00225-3, M 1486890. Woodall, ré. R. (1972), "Sufficient conditions for circuits in graphs", Actes de la London Mathematical Society, Third Series, 24: 739–755, est ce que je:10.1112/plms/s3-24.4.739, M 0318000. Catégories: Extremal graph theoryTheorems in graph theoryHamiltonian paths and cycles
Si vous voulez connaître d'autres articles similaires à Ore's theorem vous pouvez visiter la catégorie Extremal graph theory.
Monter
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https://download.cnet.com/Surveying/3000-20412_4-78600486.html | 1,656,476,826,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00638.warc.gz | 267,362,152 | 96,927 | X
# Surveying for iOS
Free
## Developer's Description
This free Surveying App covers most important topics in simple English and diagrams for a quick study and revisions at the time of Exams, Viva, Assignments and Job interviews. It is the most useful App for last minute preparations.
Important Topics on Surveying
1. surveying in civil engineering
2. land surveying
3. surveying equipment
4. Levelling
5. Contouring
6.Simple Curve
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1. INTRODUCTION
2. BASIC MEASUREMENTS
3. CONTROL NETWORKS
4. LOCATING POSITION
5. PLOTTING DETAIL
7. Error and uncertainty
8. SIGNIFICANT FIGURES
9. ERRORS IN MEASUREMENT
10. Probability
11. INDICES OF PRECISION
12. COMBINATION OF ERRORS
13. Uncertainty of addition or subtraction
14. LEVELLING
15. CURVATURE AND REFRACTION
16. EQUIPMENT
18. Automatic level
19. PRINCIPLE OF LEVELLING
20. SOURCES OF ERROR
21. LEVELLING APPLICATIONS
22. Direct and Indirect contouring
23. RECIPROCAL LEVELLING
24. PRECISE LEVELLING
25. Parallel plate micrometer
26. Field procedure
27. Booking and computing
28. DIGITAL LEVELLING
29. Factors affecting the measuring procedure
30. TRIGONOMETRICAL LEVELLING
31. Contouring
32. HEIGHTING WITH GPS
33. TAPES
34. FIELD WORK
35. Measuring in catenary
37. Sag
38. ERRORS IN TAPING
39. Tension, Sag and Slope
40. ELECTROMAGNETIC DISTANCE MEASUREMENT (EDM)
42. FIELD PROCEDURE
43. Setting up using the optical plumb-bob
44. MEASURING ANGLES
45. Measurement by directions
46. SOURCES OF ERROR
47. PLANE RECTANGULAR COORDINATES
48. TRAVERSING
49. Reconnaissance
50. Sources of error
51. Traverse computation
52. TRIANGULATION
53. Resection and intersection
54. Resection
55. NETWORKS
56. PRINCIPLE OF LEAST SQUARES
57. LINEARIZATION
58. NETWORKS
59. PRINCIPLE OF LEAST SQUARES
60. LEAST SQUARES APPLIED TO SURVEYING
61. LINEARIZATION
62. PARAMETER VECTOR
63. DESIGN MATRIX AND OBSERVATIONS VECTOR
64. Plan network
65. Distance equation
66. Direction & Angle equation
67. WEIGHT MATRIX
68. ERROR ANALYSIS
69. VARIANCE-COVARIANCE MATRIX OF THE PARAMETERS
70. ERROR ELLIPSES
71. Eigenvalues, eigenvectors and error ellipses
72. BLUNDER DETECTION
73. RELIABILITY OF THE OBSERVATIONS
74. PRACTICAL CONSIDERATIONS
75. ESTIMATION IN THREE DIMENSIONS
76. REFERENCE ELLIPSOID
77. The geoid
78. The ellipsoid
79. COORDINATE SYSTEMS
80. Geodetic coordinates
81. Cartesian coordinates
82. Plane rectangular coordinates
83. Height
84. LOCAL SYSTEMS
85. Deviation of the vertical
86. COMPUTATION ON THE ELLIPSOID
87. DATUM TRANSFORMATIONS
88. ORTHOMORPHIC PROJECTION
89. ORDNANCE SURVEY NATIONAL GRID
90. (t - T) correction
91. PRACTICAL APPLICATIONS
92. Introduction of Satellite positioning
93. GPS SEGMENTS
94. GPS
95. SATELLITE ORBITS
96. BASIC PRINCIPLE OF POSITION FIXING
97. DIFFERENCING DATA
98. GPS OBSERVING METHODS
99. Kinematic positioning
Surveying is part of Civil engineering education in various colleges and univeristies globally.
## Full Specifications
### General
Release July 15, 2020
Version 2
### Operating Systems
Operating Systems iOS
Additional Requirements Requires iOS 8.2 or later. Compatible with iPhone, iPad, and iPod touch.
Report Software
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### BookBuddy - Book Library Manager
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Index Card Magic | 1,077 | 4,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-27 | latest | en | 0.736372 |
http://encyclopedia2.thefreedictionary.com/Exawatt | 1,545,025,870,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828318.79/warc/CC-MAIN-20181217042727-20181217064727-00194.warc.gz | 84,762,480 | 11,105 | # watt
(redirected from Exawatt)
Also found in: Dictionary, Thesaurus, Medical, Acronyms.
## watt
[for James WattWatt, James,
1736–1819, Scottish inventor. While working at the Univ. of Glasgow as an instrument maker, Watt was asked to repair a model of Thomas Newcomen's steam engine.
], abbr. W, unit of power, or work done per unit time, equal to 1 joule per second. It is used as a measure of electrical and mechanical power. One watt is the amount of power that is delivered to a component of an electric circuit when a current of 1 ampere flows through the component and a voltage of 1 volt exists across it. The derivative units are kilowatt (1,000 W; kW) and megawatt (1,000,000 W; MW), used in electric power systems, and milliwatt (0.001 W; mW) and microwatt (0.000001 W; μW), used in electronics.
## watt
(wot) Symbol: W. The SI unit of power, defined as the power resulting from the dissipation of one joule of energy in one second.
## watt
[wät]
(physics)
The unit of power in the meter-kilogram-second system of units, equal to 1 joule per second. Symbolized W.
## watt
A unit of power; the power required to do work at the rate of 1 joule per second, which is equal to the power dissipated in an electric circuit in which a potential difference of 1 volt causes a current of 1 ampere to flow.
## watt
the derived SI unit of power, equal to 1 joule per second; the power dissipated by a current of 1 ampere flowing across a potential difference of 1 volt. 1 watt is equivalent to 1.341 × 10--3 horsepower.
## Watt
James. 1736--1819, Scottish engineer and inventor. His fundamental improvements to the steam engine led to the widespread use of steam power in industry
## watt
The standard unit of measurement of electrical power. One watt is one ampere of current flowing at one volt. Watts are typically rated as AMPS x VOLTS or VOLT-AMP (V-A). However, this rating is only equivalent to watts when it applies to devices that absorb all the energy, such as electric heating coils or incandescent light bulbs. With computer power supplies, the actual watt rating is only 60% to 70% of the VOLT-AMP rating.
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Open / Close | 555 | 2,167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2018-51 | latest | en | 0.920298 |
https://dsp.stackexchange.com/questions/27077/creating-odd-harmonics-without-the-fundamental?noredirect=1 | 1,618,211,285,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066613.21/warc/CC-MAIN-20210412053559-20210412083559-00145.warc.gz | 335,712,332 | 41,859 | # Creating odd harmonics without the fundamental
I'm interested in creating a series of odd only harmonics, without the fundamental, for audio use.
As an example the input could be a sine wave at specific frequency $f$ and output is $3f$, $5f$, $7f$, etc. - but without the original $1f$.
Since the $1f$ (in a real audio system) is unknown and the input is multi-tone, just filtering the $1f$ out is not an option.
Any idea for replacing an input of $1f$ with an output of $3f$, $5f$, $7f$, etc. ?
Edit: Additional information from the deleted post
Example: Let's assume the fundamental to be anything possible between 100 and 4000 Hz. Whats more, what if the input is multitone (as most music is...). 50 and 425 Hz input simulatenously ??
Im looking for a mathematical function which can recreate any fundamentals odd harmonics, in a multitone input...
• is your question about how to make a device where a single sinusoid goes in and what comes out is an (even or odd) harmonic only? the answer is "Tchebyshev polynomials". that gets you 1 harmonic. then you have to add up the outputs of several of these to get you a collection of harmonics. – robert bristow-johnson Nov 16 '15 at 6:24
• dsp.stackexchange.com/q/5959/29 – endolith Nov 16 '15 at 16:46
• actually my question is how to produce only harmonics 3,5,7 and cancel out the original F, without knowing what it might be (so I can;t just add a high pass filter). – Malcolm Rest Nov 16 '15 at 18:03
• you need to be much more clear about technically what you have for an input and what you are looking for in an output. if you want to "create" harmonics from where there were none, it's gonna be a non-linear function. if you want to remove the fundamental and, presumably, all even-order harmonics, that would be a special kind of tuned filter (which would need a pitch detector to tune it). so which it is? – robert bristow-johnson Nov 16 '15 at 22:32
Im looking for a mathematical function which can recreate any fundamentals odd harmonics, in a multitone input...
(emphasis by me)
What you ask for is not possible. If you demand multiple tones and complex tones as they appear in music, then your problem leads to a conflict. A complex tone comes with overtones. Even if you have a process that adds odd harmonics to each sinusoidal component, then the even overtones of the fundamental frequency will contribute even harmonics. That is because an odd number multiplied with an even number is again even.
So circumvent that, you would have to identify the fundamental frequency of complex tones first, leading you directly to the problem of polyphonic pitch detection which remains mostly unsolved.
Maybe you should explain what you're really trying to accomplish, as in what the application is that you have in mind. With that information it's quite likely that we can come up with a relaxed set of requirements that can actually be implemented.
• $y(t) = x(t)^3 - \frac 3 4 x(t)$ (3rd harmonic)
• $y(t) = x(t)^5 - \frac {10} {16} x(t)$ (harmonics 3, 5)
• $y(t) = x(t)^7 - \frac {35} {64} x(t)$ (harmonics 3, 5, 7)
• etc.
Presumably this is a sampled system, so you could choose N based on your Nyquist frequency?
There's a pattern to the coefficients that cancel out the fundamental but I don't know what it is. :)
• Note that with these formulas the different harmonics have different scale factors. If you want the same scaling for all harmonics you just have to sum the appropriate Chebyshev polynomials. E.g., for the 3rd and 5th harmonic you'd get $y(t)=T_3(x(t))+T_5(x(t))$. Also note that this will only work for a sinusoidal input. – Matt L. Nov 16 '15 at 15:30
• In addition to only working for sinusoidal input, the amplitude of the input sinusoid must also be constant and equal to 1. – Jazzmaniac Nov 16 '15 at 16:34
• Yes. I pretty went off using Chebyshev for this problem, as the ouptut scaling (e.g. harmonics) are not linearly scaled to the input, The result si really bad sound. – Malcolm Rest Nov 16 '15 at 18:05
(Note that this answer was written before the question was edited in a way which made clear that the input is not sinusoidal, and that filtering out the fundamental frequency is also no option. Anyway, for the time being I'll leave it as is.)
One easy way to achieve this is to symmetrically clip the sinusoid, which will create all odd multiples of the fundamental frequency, and then use a high-pass filter to filter out the fundamental frequency.
• actually my question is how to produce only harmonics 3,5,7 and cancel out the original F, without knowing what it might be (so I can;t just add a high pass filter). – Malcolm Rest Nov 16 '15 at 18:03
• @Downvoter: If you check the edit history of the question, you'll see that with the initial version of the question (which I tried to answer) it was not clear that the input is multi-tone, and that filtering is not an option. – Matt L. Nov 16 '15 at 20:27
The function should be something like this:
SR=sampling rate
T=duration in second of the output
t=0:(1/SR):T
X=fft(x)
freq=index of max(mod(X)) from 0 to SR/2 (in [Hz])
ai=[1 1 0.5 ... ] %harmonics amplitude
y=zeros(length(t))
for i=1:n %n harmonics, n=floor(SR/2/(freq*2))-1 if you want all the harmonics
y=y+ai(i)*sin(2*pi*(2*i+1)*freq*t)
end
EDIT:
If x is a sin and you don't want to use FFT you can find the freq quite well in this way $O(n)$:
sgn=sign(x);
i=1;
periods=0;
Psmp=0;
while i<length(t)
if (sgn(i)~=sgn(i+1) && sgn(i+1)~=0) || sgn(i)==0
for j=(i+1):length(t)-1
if (sgn(j)~=sgn(j+1) && sgn(j+1)~=0) || sgn(j)==0
Psmp=Psmp+(j+j+(sgn(j)~=0))/2-(i+i+(sgn(i)~=0))/2;
periods=periods+1;
break;
end
end
i=j;
end
i=i+1;
end
if periods>0
Psmp=Psmp/periods/2;
P=Psmp/SR;
freq=1/P;
end
It makes a linear interpolation of each couples of samples about 0 and finds the average of the periods.
• Hi,thanks for the answer, but I'm looking for a plain math/trigonometric function for the harmonics re-creation, not FFT, for obvious reasons. – Malcolm Rest Nov 16 '15 at 13:07
• Have you tried your "algorithm" for finding the period of a sinusoid? The result might be surprising ... – Matt L. Nov 16 '15 at 13:40
• I've wrote a new solution to find period that works quite well – Andrea Nov 16 '15 at 20:44 | 1,745 | 6,245 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-17 | latest | en | 0.895681 |
https://mw-live.lojban.org/index.php?title=Lojban_Wave_Lessons/14&oldid=124294 | 1,696,220,347,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510967.73/warc/CC-MAIN-20231002033129-20231002063129-00255.warc.gz | 450,438,544 | 8,992 | # Lojban Wave Lessons/14
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Lojban Wave Lessons: Foreword | ← Lesson 13 | Lesson 14 | Lesson 15 →
## Lesson 14: Lojban sumti 1ː LE and LA
If you have read and understood the content of all the lessons until now, you have amassed a large enough knowledge of Lojban so that it doesn't matter in which order you learn the rest. As a result, the order of the next lessons will be a mixture of sorted by increasing difficulty and sorted by importance in ordinary Lojban conversation.
One of the biggest constrains on your speak now is your limited knowledge on how to make sumti. So far, you only know ti and lo SELBRI, which doesn't take you far considering how important sumti are in Lojban. This lesson as well as the following two will be about the Lojban sumti. For now, we focus on the descriptive-like sumti, the ones made with articles like lo
Articles are in lojban called gadri, and all the ones discussed in this lesson are terminated by ku, except the combination LA CMEVLA. We will begin by describing three simple kinds of descriptive sumti, and then immediately find that they are not so simple after all:
lo = gadri: generic, converts selbri to sumti.
le = gadri: specific and descriptive, converts selbri to sumti.
la = gadri: naming article, convert selbri or cmevla to sumti.
You are already familiar with lo and what it does. lo creates a sumti referring to what fits in the first place of the following selbri.
This may be contrasted with le, which is specific and descriptive. Saying le gerku says that you have one or more specific objects in mind, and you use the selbri gerku to describe it, so that the listener may identify those specific objects. This means that le has two important differences from lo: Firstly, it cannot refer to objects in general, but always refers to specific objects. Secondly, le gerku explicitly hints that the selbri is only meant to help the speaker identify what the description refers to, whether or not it actually satisfies the selbri. Perhaps the speaker is referring to a hyena, but is not familiar with them and thinks dog is a good enough approximation to be understood; however, this non-veridicality is perhaps over-emphasized in many texts. The best way to describe a dog is usually to describe it as being a dog, and unless there is a good reason not to, le gerku is usually presumed to refer to something which is also lo gerku.
In translation, lo gerku can be any of the dog, the dogs, a dog, some dogs, or dogs in general, while le gerku is the dog or the dogs. Even better for le gerku would be the dog(s)
Last of the three basic gadri, there is la, the naming gadri, which strips the following selbri of its usual meaning and refers to someone or something whose name is that selbri. If I in English refer to a person called Innocent by her name, I neither describe her as being innocent, nor do I state that she is. I only state that by convention, that object is referred to by that selbri or cmevla. Note that la and the gadri derived from it can convert cmevla to sumti unlike any other gadri. Also: Be cautious: Other texts do not mention that names can be formed from ordinary selbri using the gadri la. But those heretics must burn; selbri names are as good as they get, and many a proud Lojbanist have them.
la is a bit eccentric, since it always marks the beginning of a name. Unlike the other gadri, anything that can be grammatically placed after la and its sisters are considered part of the name. For example, le mi gerku is "my dog", but la mi gerku is someone or something called "My Dog".
These three basic gadri can be expanded with three more, which correspond to the previous:
loi = gadri: generic, mass of individuals
lei = gadri: specific and descriptive, mass of individuals
lai = gadri: mass of named individuals
These are the same as the first three articles in all aspects except for one: they wrap the sumti into masses. Masses are used to abstract multiple individuals into single entities for ease of expression. For example, a football team can be described as a mass of its members, or an animal as a mass of cells.
mivysle = x1 is a biological cell of organism x2
remna = x1 is a human
loi mivysle cu remna - "Masses of cells are humans". Again, none of the cells are humans. Indeed, the cells have very few human traits, but the cells considered as a whole makes up a human. This example also shows that the selbri following loi must be satisfied by the members of the mass, but not by the mass itself: a human isn't a cell either.
A mass made with lei, such as lei gerku, refers to a mass formed by a group of specific individuals, which the speaker refers to as le gerku.
It needs to be noted that regular sumti built using lo are usually sufficient even when a selbri is meant to apply collectively. Constructing masses is mainly necessary when talking about multiple groups of individuals.
sruri = x1 flanks/encircles/encloses x2 in line/plane/directions x3
lo prenu cu sruri lo zdaniPeople surround the house. The Lojban sentence shares the obvious meaning of the English one, namely that people are arranged around the house, not necessarily that individual people are stretched around it.
But what if we want to explicitly say that people surround the house from head to toe? In order to be explicit about a selbri distributing across individuals, one needs lo, le or la with an outer quantifier. The subject of quantifiers will be considered later, in lesson twenty-two.
Lastly, there are the (only two) generalizing gadri:
lo'e = gadri: veridical convert selbri to sumti. Sumti refers to the archetype of lo {selbri}.
le'e = gadri: Descriptive convert selbri to sumti. Sumti refers to the described/perceived archetype of le {selbri}.
Of which there is no la-equivalent.
So, what is actually meant by the archetype? For lo'e tirxu, it is an ideal, imagined big cat, which has all the properties which best exemplifies big cats. It would be wrong to say that this includes having a striped fur, since a big systematic subgroup of the members of the set of big cats do not have striped fur, such as the leopards and the jaguars. Likewise, the typical human does not live in Asia even though a lot of humans do. However, if sufficiently many humans have a trait, for instance being able to speak, we can say:
kakne = x1 is capable of doing/being x2 under circumstance x3
lo'e remna cu kakne lo nu tavla - The typical human being can speak.
le'e then, is the ideal object as perceived or described by the speaker. This need not be factually correct, and is often translated as the stereotype, even though the English phrase have some unpleasant negative connotations, which the Lojban does not. In fact, everyone has a stereotypical archetypichal image of any category. In other words, lo'e remna is the archetype which best exemplifies all lo remna, while the archetype le'e remna best exemplifies all le remna.
The eleven gadri can be seen in the diagram below.
Generic Masses Generalizing lo loi lo'e le lei le'e la lai does not exist
Note: Earlier, there was a word xo'e for the generic gadri. However, the rules and definitions for gadri were changed in late 2004, and the current set of rules on this topic, nicked xorlo has replaced the old way. Now, lo is generic, and xo'e is used as an elliptical digit (lesson nineteen). There are also gadri for building sets (lo'i, le'i, la'i), but they are exceedingly rare in post-xorlo usage.
Lojban Wave Lessons: Foreword | ← Lesson 13 | Lesson 14 | Lesson 15 → | 1,809 | 7,622 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-40 | latest | en | 0.919343 |
https://ptcouncil.net/51-acres-to-sq-ft/ | 1,653,701,567,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663011588.83/warc/CC-MAIN-20220528000300-20220528030300-00087.warc.gz | 546,746,066 | 4,536 | To calculation a acre worth to the matching value in square feet, simply multiply the amount in acre through 43560 (the conversion factor). Below is the formula:
You are watching: .51 acres to sq ft
Suppose you want to convert 0.51 acre right into square feet. Making use of the conversion formula above, you will get:
Value in square feet = 0.51 × 43560 = 22215.6 square feet
How countless acres room in 0.51 square feet?0.51 acre is equal to how numerous square feet?How lot is 0.51 acre in square feet?How to convert acres to square feet?What is the conversion aspect to convert from acres to square feet?How to transform acre in square feet?What is the formula to convert from acre to square feet? amongst others.
Acres to square feet counter chart
0.42 acre = 18300 square feet
0.43 acre = 18700 square feet
0.44 acre = 19200 square feet
0.45 acre = 19600 square feet
0.46 acre = 20000 square feet
0.47 acre = 20500 square feet
0.48 acre = 20900 square feet
0.49 acre = 21300 square feet
1/2 acre = 21800 square feet
0.51 acre = 22200 square feet
Acres come square feet conversion chart
0.51 acre = 22200 square feet
0.52 acre = 22700 square feet
0.53 acre = 23100 square feet
0.54 acre = 23500 square feet
0.55 acre = 24000 square feet
0.56 acre = 24400 square feet
0.57 acre = 24800 square feet
0.58 acre = 25300 square feet
0.59 acre = 25700 square feet
0.6 acre = 26100 square feet
See more: Which Of The Answers Best Describes Chronology ? Which Of The Answers Best Describes Chronology | 433 | 1,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2022-21 | latest | en | 0.748391 |
https://achtungbeauty.com/qa/question-what-day-is-it-going-to-be-in-13-days.html | 1,603,662,484,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890028.58/warc/CC-MAIN-20201025212948-20201026002948-00403.warc.gz | 204,123,759 | 7,366 | Question: What Day Is It Going To Be In 13 Days?
What day is it going to be in 1000 days?
1000 Days From Today.
– Today is : Saturday, May 30, 2020.
– The date after 1000 days is : Friday, February 24, 2023..
What date is 3 weeks from now?
3 Weeks Today. In 3 weeks time the date will be 31st July 2020 , that’s also 504 hours away.
Whats 14 days from now?
To get exactly fourteen weekdays from now, you actually need to count 20 total days (including weekend days). That means that 14 weekdays from today would be July 30, 2020.
How long was 1000 days ago?
If you multiply 1000 by 24, then you will get how many hours since 1000 days ago: 1000 days ago is 24000 hours ago.
What day is 51 days away?
To get exactly fifty-one weekdays from now, you actually need to count 71 total days (including weekend days). That means that 51 weekdays from today would be October 2, 2020.
What date will it be 2 weeks today?
Two Weeks Today In 2 weeks time the date will be 18th June 2020 , that’s also 336 hours away.
What day is it going to be in 50 days?
That means that 50 weekdays from today would be August 16, 2020.
How much is a 1000 days?
1,000 days to years. This conversion of 1,000 days to years has been calculated by multiplying 1,000 days by 0.0027 and the result is 2.7397 years.
How many days is 100 days today?
100 Days From Today. – Today is : Saturday, July 4, 2020. – The date after 100 days is : Monday, October 12, 2020.
What day is 85 days away?
To get exactly eighty-five weekdays from now, you actually need to count 119 total days (including weekend days). That means that 85 weekdays from today would be November 5, 2020.
How long is 150 days today?
That means that 150 weekdays from today would be January 21, 2021. | 490 | 1,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-45 | latest | en | 0.979559 |
https://solvedlib.com/howell-corporation-produces-an-executive-jet-for,18704 | 1,723,122,224,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640728444.43/warc/CC-MAIN-20240808124926-20240808154926-00225.warc.gz | 424,585,853 | 17,409 | # Howell Corporation produces an executive jet for which it currently manufactures a fuel valve; the cost...
Howell Corporation produces an executive jet for which it currently manufactures a fuel valve; the cost of the valve is indicated below: Variable costs Cost per Unit Direct material $920 Direct labor 600 Variable overhead 300 Fixed costs Depreciation of equipment 500 Depreciation of building 250 Supervisory salaries 300 The company has an offer from Duvall Valves to produce the part for$2,000 per unit and supply 1,000 valves (the number needed in the coming year). If the company accepts this offer and shuts down production of valves, production workers and supervisors will be reassigned to other areas. The equipment cannot be used elsewhere in the company, and it has no market value. However, the space occupied by the production of the valve can be used by another production group that is currently leasing space for $55,000 per year. What is the incremental savings of buying the valves? (The answer should be stated in a per-unit format and is a positive number) Round to two decimal places. Your Answer: ## Answers #### Similar Solved Questions 5 answers ##### Question 252 ptsThis question requires _ numerical answer: Type only the numerical value: Do NOT include units and do NOT use scientific notation; Give vour answer to two decimal places. If Canvas rounds to fewer that is OK but you should not round to fewer on your own; A gas mixture in a rigid container has total pressure Of 2.5 atm; One gas let$ call it X has 0.85 moles and accounts for 1.2 atm of pressure: What Is the mole fraction of the other Bases (all gases that are not X) in the contai
Question 25 2 pts This question requires _ numerical answer: Type only the numerical value: Do NOT include units and do NOT use scientific notation; Give vour answer to two decimal places. If Canvas rounds to fewer that is OK but you should not round to fewer on your own; A gas mixture in a rigid co...
##### Problem 4Consider a world with two states of nature State 1 and State 2 at each date t_ This means that at each point in time the eCOnomy Can either be in State 1 O1" in State 2. The following matrix gives the one-period Arrow-Debreu prices (at each date t)Time t +1 State State 2 Time t State 1 0.45 0.51 State 2 0.43 0.49For example; assume that at time t we are in State 1. Then, the price at time t of an Arrow Debreu security paying unit at time t + 1 if State OCCurs is 0.51. Similarly; as
Problem 4 Consider a world with two states of nature State 1 and State 2 at each date t_ This means that at each point in time the eCOnomy Can either be in State 1 O1" in State 2. The following matrix gives the one-period Arrow-Debreu prices (at each date t) Time t +1 State State 2 Time t State...
##### Directions: Questons and 3 are long; free-response questions that should require about 15 minutes each t0 answer: Questions 2 and are short free-response questions that should require about 5 minutes each to answer: Read each question carefully and write your response_ Answers must be written out Oudline fomm is not acceptable: It is important thatyou read each question completely before you begin to wite_Question 1: Conservation of Matter and the Law of Definite Proportions (20 points}a) Tetrap
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##### Two long thin parallel wires 13.0 cm apart carry 24-Acurrents in the same direction. Part A:...
Two long thin parallel wires 13.0 cm apart carry 24-Acurrents in the same direction. Part A: Determine the magnitude of the magnetic field vector at a point 10.0 cm from one wire and 6.0 cm from the other (Figure 1) . Part B: Determine the direction of the magnetic field vector at that point. degree...
##### 5_ Consider the differential equation y" = 9y_ (a) Calculate the Wronskian of the following sets of functions, and hence determine which are fundamental sets of solutions to the above differential equation:S1 = {cosh 31,sinh 3x}. (ii) Sz = {1,e3r , e 31 X(iii) S3 = {e35 , cosh 32 , sinh 3x}(b) Use the sets which were fundamental sets of solutions to write down the general solution to this differential equation, and hence solve the DE when y(0) = 2 and y (0) = 3. (c) You are given that S4 =
5_ Consider the differential equation y" = 9y_ (a) Calculate the Wronskian of the following sets of functions, and hence determine which are fundamental sets of solutions to the above differential equation: S1 = {cosh 31,sinh 3x}. (ii) Sz = {1,e3r , e 31 X (iii) S3 = {e35 , cosh 32 , sinh 3x} ...
##### Question 3: [25 pts] Answer the following questions. a) What is the first derivative of y...
Question 3: [25 pts] Answer the following questions. a) What is the first derivative of y = 2n=o 2"x" ? b) What is the second derivative of y = 2n=0 c) Write the first four non-zero terms of the derivatives y' and y". n! 2x n!...
##### Question 12) Compute the joint density function for S= 4X +9Y and T= 3X - 7Y. Assume that both X and Y areindependent normal random variables. Please explain clearly!
Question 12) Compute the joint density function for S= 4X + 9Y and T= 3X - 7Y. Assume that both X and Y are independent normal random variables. Please explain clearly!...
##### Why is it important for caregivers who work with alzheimer patients to spot opportunitys and risk...
why is it important for caregivers who work with alzheimer patients to spot opportunitys and risk in the workplace? give examples what are the opportunitys what are the risks? also, is there a cure for alzheimers?...
##### A boat is held stationary by a cable attached to a pier. A firehose directs a...
A boat is held stationary by a cable attached to a pier. A firehose directs a spray of 5°C water at a speed of V=60 m/s. If the allowable load on the cable is 10 kN, calculate the mass flow rate of the water jet. What is the corresponding diameter of the water jet?...
##### Shown below are two carts connected by cord that passes Over small frictionless pulley- Each cart rolls freely with negligible friction. Calculate the magnitude of the acceleration (in m/s?) of the carts and the tension (in N) in the cord. (Due the nature of this problem, do not use rounded intermediate values your calculations_ including answers submitted in WebAssign)14 kg
Shown below are two carts connected by cord that passes Over small frictionless pulley- Each cart rolls freely with negligible friction. Calculate the magnitude of the acceleration (in m/s?) of the carts and the tension (in N) in the cord. (Due the nature of this problem, do not use rounded intermed...
##### Question 5Mass m 0.1 kg moves to the right with speed 0.58 ns and collides with an equal mass initially atrest After this inelastic collision the system retains fraction 0.61 afits original kinetic encrgy: How much impulse (in units of sec) does the mass originally at rest receive during the collision?Hints: All motion Is in 10_ Ignare friction between the masses and tlie horizontal surface You will probably need to use the quadratic formula to solve the resulting equations. Va must be greater t
Question 5 Mass m 0.1 kg moves to the right with speed 0.58 ns and collides with an equal mass initially atrest After this inelastic collision the system retains fraction 0.61 afits original kinetic encrgy: How much impulse (in units of sec) does the mass originally at rest receive during the collis...
##### Let f(x, y) = x + y 2 . Compute SS D f dA, whereD is the plane region in the first quadrant bounded by the curvesof equation y = x and y = x 3
Let f(x, y) = x + y 2 . Compute SS D f dA, where D is the plane region in the first quadrant bounded by the curves of equation y = x and y = x 3...
##### For the matrix9a) Find the rank and b) Find basis nullity of A vectors for the range space of A and also for the null space of _ Consider the linear equation, Y=Ax Find the general solution ofthis equation if d) Determine whether this equation x = [1 0 4]" has a solution if y = [I 4 0j" ?
For the matrix 9 a) Find the rank and b) Find basis nullity of A vectors for the range space of A and also for the null space of _ Consider the linear equation, Y=Ax Find the general solution ofthis equation if d) Determine whether this equation x = [1 0 4]" has a solution if y = [I 4 0j"...
##### Evaluate tne sunace integral:x2z2 dSS is the part of the cone 22 = x2that lies between the planesand 2
Evaluate tne sunace integral: x2z2 dS S is the part of the cone 22 = x2 that lies between the planes and 2...
##### 2. Name three structures that you see in the model that are involved in regulating temperature....
2. Name three structures that you see in the model that are involved in regulating temperature. Explain how these structures facilitate thermoregulation. UUUUUUUUU 3. What is the effect of contraction of the arrector pili muscles on the hair?...
##### 1) Provide the products produced from each of the following reactions. a) H2 Lindlar's Catalyst b) CH3CH2ONa c) 1) LiAIH4 2) H20 но d) 8.25% NaClO NaHSO4 но e) 1 NaBHA 2) H20 но 1) Provide t...
1) Provide the products produced from each of the following reactions. a) H2 Lindlar's Catalyst b) CH3CH2ONa c) 1) LiAIH4 2) H20 но d) 8.25% NaClO NaHSO4 но e) 1 NaBHA 2) H20 но 1) Provide the products produced from each of the following reactions. a) H2 Lind...
##### FODUVah 102 Grades avall2017- Vnlc ~haihjoo-LocFCrondmirQ1 (10 points)Let f (x) be given byZar? 'sin(,} ) - barctande-V), ifx < 0; JG) 24+1, ifx =o; sin(axNt, ifx > 0 What are the valea of a and Kich ensure that f () continuous on R Be sure exdlain your answorDrag and drop your images click t0 Lro30points)Find the vertica and hontonte asymplotos ol the graph Inte function f(x) arctan ( ; 1 -Dragtop your ImwnosCackIC Doteupoints)sror Ulat tartInaato vahm:quanvalRisucn Iat 4 =sin(4)_008
FODU Vah 102 Grades avall 2017- Vnlc ~haihjoo-LocF Crondmir Q1 (10 points) Let f (x) be given by Zar? 'sin(,} ) - barctande-V), ifx < 0; JG) 24+1, ifx =o; sin(axNt, ifx > 0 What are the valea of a and Kich ensure that f () continuous on R Be sure exdlain your answor Drag and drop your ima...
##### A 15g bullet is fired at an unknow speed into a 6.25 kg blockthat sits on the edge of a 75cm tall table. The bullet get stuck inthe block, and they go off the table together. If the block plusthe bullet land 1.25m from the edge of the table, what was themuzzle velocity (speed of the bullet) before It collided with theblock?
a 15g bullet is fired at an unknow speed into a 6.25 kg block that sits on the edge of a 75cm tall table. The bullet get stuck in the block, and they go off the table together. If the block plus the bullet land 1.25m from the edge of the table, what was the muzzle velocity (speed of the bullet) befo...
##### Edit View History Bookmarks People Window HelpDo Homework Elsa Koloa Secure https: |lwwwmathxlcom/Student PlayerHomeworkaspr?homeworkld-4688Spring'18-Online-College Algebra-3CO76Homework: Section 13.3 Homework Score: 0 of 1 pt 4 of 14 (2 complete)13.3.19Solve the following system by the substitution method2y = -34How many solulions are there?Enter your answer the answer box and then click Check AnsworClearpan remaining
Edit View History Bookmarks People Window Help Do Homework Elsa Koloa Secure https: |lwwwmathxlcom/Student PlayerHomeworkaspr?homeworkld-4688 Spring'18-Online-College Algebra-3CO76 Homework: Section 13.3 Homework Score: 0 of 1 pt 4 of 14 (2 complete) 13.3.19 Solve the following system by the su...
##### 28. Stocks are bought and sold i markets: a. Equity b. Debt c. Derivatives d. Foreign...
28. Stocks are bought and sold i markets: a. Equity b. Debt c. Derivatives d. Foreign exchange...
##### I need help determining the present value of the costs for machine A. Esquire Company needs...
I need help determining the present value of the costs for machine A. Esquire Company needs to acquire a molding machine to be used in its manufacturing process. Two types of machines that would be appropriate are presently on the market. The company has determined the following: Machine A could be ...
##### JO-2014_content_id=1450872_1&step=null laps News OC) A carbon radical reacts with Cl2. OD) A carbon radical reacts with...
JO-2014_content_id=1450872_1&step=null laps News OC) A carbon radical reacts with Cl2. OD) A carbon radical reacts with a chlorine radical. E) Two chlorine radicals combine. QUESTION 18 2 points Save Answer The stereochemical configuration of D-Threonine is (2R, 35). Which of the following struc...
##### What causes Acidification of Oceans?Answer:List 2 places carbon is found in nature? (1Opts) List 1 way humans are increasing the amount of Carbon (1 Opts)
What causes Acidification of Oceans? Answer: List 2 places carbon is found in nature? (1Opts) List 1 way humans are increasing the amount of Carbon (1 Opts)...
##### E The Information predict the genome NOIls3no M newly peiscorvergd ! 8 cpeoihes H and - in the genome: tound to have 1696 Adenine nucleotides: Given that
E The Information predict the genome NOIls3no M newly peiscorvergd ! 8 cpeoihes H and - in the genome: tound to have 1696 Adenine nucleotides: Given that...
##### Review Problem 5.042 A pollutant in smog is nitrogen dioxide, NO2. The gas has a reddish brown color and is responsible...
Review Problem 5.042 A pollutant in smog is nitrogen dioxide, NO2. The gas has a reddish brown color and is responsible for the red-brown color associated with this type of air pollution. Nitrogen dioxide is also a contributor to acid rain because when rain passes through air contaminated with NO2, ...
##### 6. Which pair of functions below cannot be a fundamental set of solutions? * (8 Puan)...
6. Which pair of functions below cannot be a fundamental set of solutions? * (8 Puan) O 1, x, x? Inx O 5e + 5, 2eX + 1 0 2x, cos 3x + sin 3x, 1 3 cost + 6 sint , 5 cost + 10 sint O e-31, te-31...
##### Find the surface area and volume of the right cylindersurface areaJmvolumeh=Find tc height ind surface Theh] of the right circular Gylinder with he height equal 10 tnc dlmcidibeight3unss?r4V-1281cm !
Find the surface area and volume of the right cylinder surface area Jm volume h= Find tc height ind surface Theh] of the right circular Gylinder with he height equal 10 tnc dlmcidi beight 3unss?r4 V-1281cm !...
##### [0/3 Points]DETAILSPREVIOUS ANSWERSSCALC8 13.3.04EFind the vectors T, N, and B at the given point, r(t) (3 cos(t) , 3 sin(t) , 3 In(cos(t))) , (3, 0, 0)Need Help?Read It{Talk to TutorSubmit Answer
[0/3 Points] DETAILS PREVIOUS ANSWERS SCALC8 13.3.04E Find the vectors T, N, and B at the given point, r(t) (3 cos(t) , 3 sin(t) , 3 In(cos(t))) , (3, 0, 0) Need Help? Read It {Talk to Tutor Submit Answer...
##### Saved Help Save & E ises (1 p Check Jurvin Enterprises is a manufacturing company that...
Saved Help Save & E ises (1 p Check Jurvin Enterprises is a manufacturing company that had no beginning inventories. A subset of the transactions that it recorded during a recent month is shown below. a. $75,100 in raw materials were purchased for cash. 6.$72100 in raw materials were used in pr...
##### Please complete part C only! Thank you in advance! s Question Help Credit terms Purchases made...
please complete part C only! Thank you in advance! s Question Help Credit terms Purchases made on credited in by the end of the big period. Many forms extend a discount for payment made in the first part of the ring period. The original invoice contains type of shorthand nation that the creditor... | 4,188 | 15,745 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.920135 |
https://katalistiwa.id/en/3669/volume-bangun-diatas-adalah.html | 1,679,308,283,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943471.24/warc/CC-MAIN-20230320083513-20230320113513-00085.warc.gz | 387,010,219 | 15,957 | # The build volume above is
Welcome to catalyst, blog to discuss about the discussion of subject matter from Higher Education, High School, Middle School and Elementary Schools. This time Katalistiwa will discuss a question that is often asked in school exams. The questions are: The build volume above is
## The build volume above is
The formula
v = pxlx t
The solution
v = (8 x 8 x 8 ) + ( 17 x 8 x 12 )
= 512 + 1632
= 2.144 cm
### Detailed discussion of wake volume
The volume of a cuboid is the area of the base times the height of the cuboid.
The formula is:
Volume = Base Area x Height
Example:
If the area of the base of a cuboid is 100 cm rectangle and the height of the block is 50 cm, then the volume of the block is 100 x 50 = 5000 cubic cm.
This is the discussion that we have summarized by the Katalistiwa.id Team from various learning sources. Hopefully this discussion is useful, don't forget if you have other answers, you can contact the admin. Thank you | 247 | 986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2023-14 | latest | en | 0.918017 |
https://pe.mathgames.com/skill/5.51-increasing-growth-patterns | 1,713,724,666,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817790.98/warc/CC-MAIN-20240421163736-20240421193736-00689.warc.gz | 396,123,091 | 152,345 | More Ways To Use Math Games
Game Quickplay
Math Games for Teachers
Grade 5 - Patterns and Relations
Standard 5.PR.1 - Analyze the increasing growth pattern to find the next number in the sequence.
Included Skills:
Determine the pattern rule to make predictions about subsequent elements.
Achievement Indicators
Students who have achieved this outcome(s) should be able to:
- Extend a given pattern with and without concrete materials, and explain how each element differs from the preceding one.
- Describe, orally or in writing, a given pattern using mathematical language, such as one more, one less, five more.
- Write a mathematical expression to represent a given pattern, such as r + 1, r – 1, r + 5.
- Describe the relationship in a given table or chart using a mathematical expression.
- Determine and explain why a given number is or is not the next element in a pattern.
- Predict subsequent elements in a given pattern.
- Solve a given problem by using a pattern rule to determine subsequent elements.
- Represent a given pattern visually to verify predictions.
If you notice any problems, please let us know. | 235 | 1,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-18 | latest | en | 0.86122 |
http://online.wsj.com/news/articles/SB10001424052970204349404578100942150867894?mod=_newsreel_2&mg=reno64-wsj&url=http%3A%2F%2Fonline.wsj.com%2Farticle%2FSB10001424052970204349404578100942150867894.html%3Fmod%3D_newsreel_2 | 1,394,828,093,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678694628/warc/CC-MAIN-20140313024454-00080-ip-10-183-142-35.ec2.internal.warc.gz | 107,698,767 | 27,132 | Roughly 1 in 12 men suffer from some degree of color blindness, and 1 in 4 women are carriers of the color blindness gene. New tests are being developed to discern degree of the condition. Vision scientist Dr. Jay Neitz and WSJ's Melinda Beck discuss on Lunch Break. Photo: Testing Color Vision.
For people who are colorblind, life involves little workarounds and big compromises alike. Daily challenges range from not knowing whether meat is fully cooked to not being able to read whether a horizontal traffic light is showing green or red. More serious repercussions include being shut out of a dream job, like piloting planes, because misreading landing-strip lights can have life-or-death consequences.
#### What Color-Blind People See
Compare what the same image looks like to people with red, green and blue colorblindness.
#### Are You Color Blind?
Look for the numerals in each circle of dots and see the answers below. Note: Colors may vary slightly on individual screens.
These are samples from a test with up to 104 questions. The full test, with more detailed analysis, is available at www.testingcolorvision.com for \$29.50 -- or \$5 if you use the code "WSJ."
Even people who are color blind should be able to see a 16 in this demonstration plate.
If you can't see the numeral 2, you may have red-green color blindness.
If you can't see the numeral 6, you may have red-green color blindness.
If you can't see the numeral 7, you may have blue-yellow color blindness.
If you can't see the numeral 5, you may have red-green color blindness.
If you can't see the numeral 42, you may have red-green color blindness
If you don't see the numeral 9, you may have red-green color blindness.
If you can't see the numeral 74 -- or you see a 71 or 21 -- you may have red-green color blindness.
Source: TestingColorVision.com
#### More About Vision Problems
Apps for Eyes: Colorblind Vision, lets users choose their specific deficiency and compensates colors accordingly. Colorblind Vision
#### Color Challenges
Occupations that are usually closed to colorblind people for safety reasons:
• Pilots: They must recognize colorcoded instruments, maps, landing-strip approach lights
• Air traffic controllers: Colors represent different altitudes on radar screens.
• TSA officers: Colors are crucial to explosive wires and X-ray screens.
• Law enforcement officers: Identifications often rely on color of suspects' clothes, cars and hair.
• Railroad engineers: Signal lights are color-coded.
• Lab technicians: Colors can be critical on slides, samples, litmus tests.
Now, a host of new research and tools promise to improve life for the estimated 32 million Americans—8% of men and 0.5% of women—who have some degree of colorblindness. For many, getting through the day—avoiding wardrobe perils and worse—has often involved bringing in a second pair of eyes. But new websites and smartphone apps offer to help identify or enhance hard-to-see colors. Videogame manufacturers are increasingly including "colorblind" modes in their games. And researchers are homing in on more specific vision tests that may allow mildly colorblind people to qualify for jobs that, until now, have been closed to them.
A genetic test, made by Genevolve Vision Diagnostics, will soon be available that can identify the exact type of colorblindness someone has, which the company hopes could pave the way for customized tools.
A cure for colorblindness might even be in the offing. Vision scientist Jay Neitz and his colleagues at the University of Washington are building on their 2009 breakthrough in which they restored red-green vision in two colorblind squirrel monkeys by inserting the missing gene into a virus and injecting it into their retinas. Four years later, the monkeys, Sam and Dalton, still pass daily vision tests, identifying colors on a computer screen correctly. They also have a newfound liking for green M&M's, Dr. Neitz says.
#### "So What's It Like Being Colorblind?"
He and his colleagues are working on a similar therapy for humans, but many hurdles remain. "We know it's effective. The issue is whether it's safe," says Dr. Neitz.
Many colorblind people aren't even aware they have a "color-vision deficiency," as it's officially known, unless they apply for a job that requires precise color recognition. Even people with mild colorblindness can be barred from being pilots, air-traffic controllers, police officers, lab technicians and electricians—usually for safety reasons.
The term "colorblindness" is actually a misnomer. "People think you're living in a black-and-white TV show and that's not true. There are all different degrees, from mild to severe. And you can see colors—they're just different," says Terrance Waggoner, an ophthalmologist consulting on color vision for the U.S. Navy.
But the impact does go beyond missing just one color. "A colorblind person who can't see red can't see the red in purple—he just sees blue," says Dr. Neitz. Since red and green make brown, people with red-green blindness often have trouble telling the three colors apart.
The vast majority of colorblind people have trouble seeing red or green, due to a genetic defect in the color-sensing cells, called cones, at the back of the eye. About 75% of them are specifically green-deficient; another 20% are red-deficient. Either way, the impact on their vision is so similar that it's considered one disorder, red-green colorblindness, which is the most common single-gene disorder in humans, affecting 1 in 8 men and 1 in 230 women of Northern European descent world-wide (and slightly fewer in other racial groups).
Blue-yellow colorblindness is rarer and develops later in life, often brought on by aging, illness, medication or head injuries. Rarer still is achromatopsia, the inability to see any color.
Red-green colorblindness is far more common in men than women because it's a recessive gene carried on the X chromosome. Men have one X and one Y, so a defective gene on the X shows itself readily. Women have two X chromosomes, and a normal copy of the gene will override the defective one. But women who have one defective X will be carriers of colorblindness. Each of their children has a 50% chance of inheriting the defective gene. Within families, red-green colorblindness typically travels from the maternal grandfather to grandson. A woman can only exhibit colorblindness if she receives a defective X chromosome from a colorblind father and another from a carrier mother.
But it does happen—and it can come as a surprise if previous generations didn't notice or didn't discuss their colorblindness. Ingrid Perri, a life coach in Melbourne, Australia, discovered she was mildly colorblind at age 47. "My family howled, 'That explains so much!'" says Ms. Perri.
Experts recommend that children have regular eye exams, including color-vision tests, starting between ages 3 and 5. Some children with color vision problems are labeled "learning-disabled" if they can't follow instructions. That happened with Dr. Waggoner's son at age 6, prompting the doctor to develop pediatric color-vision tests using shapes instead of numbers, now used by about 15% of school nurses.
Many colorblind people say they have no problem recognizing pure, strong colors. It is the blended and muted colors in between that are often difficult to tell apart.
The DanKam, an augmented-reality app for iPhone and Android, works on the same principle. Users look through the phone's camera and the program converts all the reds and greens in view into pure, basic versions that are easier for colorblind people to see. "It is like having magic eyes," says Andy Baio, a writer and programmer in Portland, Ore., who is red-green colorblind. "It doesn't make me see red or green the way you see them, but it makes it blazingly obvious the difference between them."
Contact lenses and glasses for colorblind people, which can cost up to \$700, use tinted lenses to alter light coming into the eye. Manufacturers say they make hard-to-see colors brighter. Many employers don't let applicants use them during vision tests. Other online tools, such as Color Name & Hue, can identify colors by name or numerical codes, so that graphic artists, say, can "see" what colors they are using. Apps such as Colorblind Avenger let people do the same thing through smartphone cameras. Other apps, such as HueVue and Colorblind Helper, assist with matching or harmonizing colors.
Some tools, like Colorblind Vision, help businesses and marketers see what their materials look like to colorblind people by simulating various color-vision deficiencies.
To help colorblind users in the trading world, Bloomberg financial-data terminals include plus and minus symbols or up and down arrows to indicate the direction of the market or stock price, says a spokeswoman for Bloomberg LP. After speaking with many traders with color-vision deficiencies, designers of DJ FX Trader, a foreign exchange tool offered by Dow Jones, the publisher of The Wall Street Journal, say they made a point of using icons to supplement color. FactSet Research Systems Inc., FactSet Research Systems Inc. U.S.: NYSE \$103.67 +0.04 +0.04% March 14, 2014 4:00 pm Volume (Delayed 15m) : 266,254 AFTER HOURS \$104.07 +0.40 +0.38% March 14, 2014 4:11 pm Volume (Delayed 15m): P/E Ratio 22.44 Market Cap \$4.43 Billion Dividend Yield 1.35% Rev. per Employee \$139,023 a financial information company, says it uses patterns and labels in its graphs, and avoids using colors that most colorblind people can't discern.
Some popular videogames, such as PopCap's Peggle and Zuma Blitz, now let gamers switch to colorblind modes, where color codes turn into shapes.
Advocates for the colorblind say the world could be more accommodating to the nearly 10% of people who have trouble seeing shades of red and green. Battery chargers blink orange for empty and green for full. Hotel keycards flash green for entry and red for stop.
Vision and design experts say the best maps, charts and presentations use words and icons in addition to color. "Color deficiency is way low on the totem pole of accessibility problems, but when doing the right thing isn't hard, you should at least take that into account," says Mr. Baio.
Write to Melinda Beck at HealthJournal@wsj.com | 2,255 | 10,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2014-10 | longest | en | 0.914146 |
https://www.airmilescalculator.com/distance/aoi-to-goa/ | 1,611,687,277,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704803308.89/warc/CC-MAIN-20210126170854-20210126200854-00571.warc.gz | 652,806,828 | 17,558 | # Distance between Ancona (AOI) and Genova (GOA)
Flight distance from Ancona to Genova (Marche Airport – Genoa Cristoforo Colombo Airport) is 232 miles / 373 kilometers / 202 nautical miles. Estimated flight time is 56 minutes.
Driving distance from Ancona (AOI) to Genova (GOA) is 318 miles / 511 kilometers and travel time by car is about 4 hours 57 minutes.
## Map of flight path and driving directions from Ancona to Genova.
Shortest flight path between Marche Airport (AOI) and Genoa Cristoforo Colombo Airport (GOA).
## How far is Genova from Ancona?
There are several ways to calculate distances between Ancona and Genova. Here are two common methods:
Vincenty's formula (applied above)
• 232.034 miles
• 373.423 kilometers
• 201.632 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 231.445 miles
• 372.475 kilometers
• 201.120 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Marche Airport
City: Ancona
Country: Italy
IATA Code: AOI
ICAO Code: LIPY
Coordinates: 43°36′58″N, 13°21′44″E
B Genoa Cristoforo Colombo Airport
City: Genova
Country: Italy
IATA Code: GOA
ICAO Code: LIMJ
Coordinates: 44°24′47″N, 8°50′14″E
## Time difference and current local times
There is no time difference between Ancona and Genova.
CET
CET
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 59 kg (130 pounds).
## Frequent Flyer Miles Calculator
Ancona (AOI) → Genova (GOA).
Distance:
232
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
232
Round trip? | 488 | 1,815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-04 | latest | en | 0.801203 |
http://www.stradanove.net/schein/82396067d70315fb1d6e221765e9b42762-how-to-calculate-homogenization-efficiency | 1,675,535,145,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500151.93/warc/CC-MAIN-20230204173912-20230204203912-00665.warc.gz | 89,426,385 | 11,601 | The release study in vitro of puerarin . Consider the second equation (thermal energy group). 11 - Reading the pressure value (resolution) The scale in analog pressure gauges have limited readability. A lot of efforts have been devoted to developing mathematical and/or numerical approaches for calculating the effective/homogenized material properties. Westpac Consumer Confidence Index: An index measuring the level of consumer confidence in Australia. Now, let's look at Sigma Level: it is a high-level baseline metric to understand process capability to meet customer requirements. When the inflow is higher than the average flow (or ETP design flow) in the tank, flow accumulation of excess water occurs (level in the tank . The proposed method uses a derivative-free formulation of a training loss, which incorporates Brownian walkers to find the macroscopic description of a multiscale . 3 Homogenization efficiency for composite materials 147 The sum in relation (3) was calculated for all n points generated by the software FEMM (300 points for each curve); () Step 4: Verify that the equalized flow meets desired discharge limits. For a graph where log (DNA copy#) is on the x-axis and Ct on the . Step 3: Use the diverted volume to calculate the surge basin's volume so continuous flow to the treatment facility can be maintained. Sets of procedures and calculations are presented in order to obtain equivalent properties in all levels, lamina and laminate. MBBR tanks allow for a higher concentration of . Sets of procedures and calculations are presented in order to obtain equivalent properties in all levels, lamina and laminate. Still, modern methods of the homogenization equipment calculations are based on determining other factors: cavitation, turbulence, velocity gradient, creating areas of subcavitation dispersion, etc. For this reason a means of accurately calculating the amount of recycle time needed to complete the process must be established. Numerical comparison was made between the magnetic properties of non-homogeneous materials and. Dhanalakshmi et al. All the resources and contents you need to know about economics and finance!
\$ per KwHr How much the electric company charges. High-pressure homogenization is suitable for fluid materials Sets of procedures and calculations are presented in order to obtain equivalent properties in all levels, lamina and laminate. Non uniform distribution of production cause under utilization of furnace in a particular shift, metal held up etc Instruments were evaluated for repeatability by calculating the sample standard deviation and the . qPCR Efficiency Calculator. When the inflow is higher than the average flow (or ETP design flow) in the tank, flow accumulation of excess water occurs (level in the tank . The efficiency of homogenization is apparently based on the number of fat globules larger than 2 (if any), as measured under specified conditions. To improve modeling efficiency, users are advised to always implement support structure homogenization. You then can calculate S 2mix and followed by CV mix (%). Judging the efficiency of homogenization One of the best methods is to examine a sample under a high power microscope and to note and analyze the size of the fat globules. Energy density can be measured in energy per volume or per mass. case of homogenization, the ends do justify the means, thus it is valuable to dissect the methods available and to assess both their strengths and limitations. The homogenization index was reduced with increasing time of homogenization with ultrasonic treatments. The value of 0.7 was based on the fact that fat globules with smaller diameter do not form clumps because they break up . In the Spanish "MULTITEST" project, the efficiencies of automatic homogenization methods were . Let us analyze the main factors of dispersion in modern homogenizers of dairy industry (Table 1 ). That standard is perpetuated in the field of homogenization and high pressure pumps. Homogenization Pressure - Gap Size Counter Pressure 22 HPH Efficiency Requirements Homogenization valves are therefore designed with the following criteria in mind - Ensure sufficient time for growth of cavitational bubbles - "Explosive" increase of pressure towards the exit - Maximum height of the "Dirac-like" pressure increase Once you've calculated CV mix (%), you can't just compare it to the specification. (Comput Mech 54(6):1497-1514, 2014) reduces the memory requirements of a strain based implementation of the CG algorithm following Zeman et al. The high efficiency available with Gaulin and Rannie homogenizers is . New analytical models have been developed for predicting equivalent Young's and shear moduli of laminate composites. The generated beam profile is generally not perfectly flat, but exhibits some level of random fluctuations, sometimes also relatively regular interference patterns. It's a tradition that dates back to our invention of the homogenizer in the early 1900's. That desire to excel in saving our customers
Therefore, the impact of high-pressure homogenization (30 MPa, 50M Pa) or . Homogenization efficiency Homogenization must always be sufficiently efficient to prevent creaming. The part must be seated directly on the platform for proper creation of support structures. | 1,054 | 5,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-06 | latest | en | 0.918209 |
https://www.convert-measurement-units.com/convert+Kilobyte+SI+to+Yottabyte+SI.php | 1,669,805,024,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710734.75/warc/CC-MAIN-20221130092453-20221130122453-00542.warc.gz | 758,029,797 | 12,983 | Convert Kilobyte SI to Yottabyte SI (Bytes / Bits)
## Kilobyte SI into Yottabyte SI
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Kilobyte+SI+to+Yottabyte+SI.php
## How many Yottabyte SI make 1 Kilobyte SI?
1 Kilobyte SI = 0.000 000 000 000 000 000 001 Yottabyte SI - Measurement calculator that can be used to convert Kilobyte SI to Yottabyte SI, among others.
# Convert Kilobyte SI to Yottabyte SI:
1. Choose the right category from the selection list, in this case 'Bytes / Bits'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Kilobyte SI'.
4. Finally choose the unit you want the value to be converted to, in this case 'Yottabyte SI'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '665 Kilobyte SI'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Bytes / Bits'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '20 Kilobyte SI to Yottabyte SI' or '7 Kilobyte SI into Yottabyte SI' or '60 Kilobyte SI -> Yottabyte SI' or '74 Kilobyte SI = Yottabyte SI'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(3 * 95) Kilobyte SI'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '665 Kilobyte SI + 1995 Yottabyte SI' or '57mm x 48cm x 50dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 4.095 999 962 726 4×1024. For this form of presentation, the number will be segmented into an exponent, here 24, and the actual number, here 4.095 999 962 726 4. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 4.095 999 962 726 4E+24. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 4 095 999 962 726 400 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 943 | 3,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2022-49 | latest | en | 0.789535 |
https://www.fcc.gov/media/radio/fm-frequencies-end-odd-decimal | 1,718,767,498,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00479.warc.gz | 686,321,091 | 13,171 | An official website of the United States government
Official websites use .gov
A .gov website belongs to an official government organization in the United States.
Secure .gov websites use HTTPS
A lock ( ) or https:// means you’ve safely connected to the .gov website. Share sensitive information only on official, secure websites.
The FM broadcast in the United States starts at 88.0 MHz and ends at 108.0 MHz. The band is divided into 100 channels, each 200 kHz (0.2 MHz) wide. The center frequency is located at 1/2 the bandwidth of the FM Channel, or 100 kHz (0.1 MHz) up from the lower end of the channel. For example, the center frequency for Channel 201 (the first FM channel) is 88.0 MHz + 0.1 MHz = 88.1 MHz.
So the FM center frequencies are determined as follows:
``` 88.0 MHz + 0.1 MHz = 88.1 MHz
88.2 MHz + 0.1 MHz = 88.3 MHz
88.4 MHz + 0.1 MHz = 88.5 MHz
up to
107.8 MHz + 0.1 MHz = 107.9 MHz
```
Every FM center frequency ends with a decimal extension of .1, .3, .5, .7, or .9.
To convert FM channel numbers to/from a corresponding frequency, use the table in Section 73.201 of the FCC's Rules, or a conversion tool.
In the AM band, each AM station has a maximum bandwidth of 10 kHz, extending 5 kHz above and 5 kHz below the assigned center frequency. The AM band in the United States covers frequencies from 540 kHz up to 1700 kHz, in 10 kHz steps (540, 550, 560 ... 1680, 1690, 1700). 530 kHz in the United States is not available for broadcast use, but is reserved for the use of very low powered Travelers' Information Stations. AM band stations do not have assigned channel numbers, they are referenced by the frequency in kilohertz only.
AM and FM station assignments in other countries may not be made according these procedures. In some countries, an FM station may be assigned a frequency with an even decimal such as 106.2 MHz. In many places, AM broadcast stations are assigned on frequencies with a 9 kHz bandwidth (531 kHz, 540 kHz, 549 kHz, etc.). There are a few AM stations in the United States assigned in this manner, in Guam. | 552 | 2,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-26 | latest | en | 0.900731 |
https://groups.yahoo.com/neo/groups/synoptic-l/conversations/topics/10244?unwrap=1&var=0 | 1,432,302,419,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207925201.39/warc/CC-MAIN-20150521113205-00344-ip-10-180-206-219.ec2.internal.warc.gz | 867,243,159 | 16,472 | Re: [Synoptic-L] Four and the riddle of the loaves and leftover baskets
Expand Messages
• ... [snip] ... [snip] ... Rick, If you are sticking to the method that the PENTE- root appears five times, please reconsider. Its is enough that HEPTA appears
Message 1 of 2 , Jul 7, 2005
• 0 Attachment
--- Richard Richmond <rickr2889@...> wrote:
[snip]
>
> tessares; these other uses of the root have caused
> me
> to reconsider Mark's formula, and for that I am
> grateful. I had assumed that all the numbers were
> relevant to the formula and it appears that four
> thousand is not relevant as a control number.
>
>
[snip]
> The word four does not actually appear in the
> formula
> text and it is not a word that directly involves the
> loaves or the leftovers; therefore it makes sense
> that
> it would not be a control number for counting. On
> the
> other hand, 5, 7 and 12 all appear here as cardinal
> numbers while four does not. There may be a separate
> significance to the words that appear four times but
> I
> think you are correct and that it is not part of
> Mark's test formula. Technically five thousand is
> not
> either except that it contains the root of five
> which
> does appear in its indeclinable form aa a cardinal
> number within the context of the riddle. So then
> John,
> let me say thank you for your challenging insight.
>
Rick,
If you are sticking to the method that the PENTE- root
appears five times, please reconsider. Its is enough
that HEPTA appears 9 times, not 7, and DODEKA 15, not
12. Perhaps, you wish to either rethink your formula
or rephrase it.
Marks 6 uses of PENTE-
MK 6:38
Mk 6:40 PENTEKONTA
MK 6:41
MK 6:44
MK 8:19 (bis)
Marks 9 uses of HEPTA
Mk 8:5
Mk 8:6
Mk 8:8
Mk 8:20 (bis) not a dittographic error
Mk 12:20
Mk 12:22
Mk 12:23
Mk 16:9 original
Marks 15 uses of DODEKA
Mk 3:14
Mk 3:16
Mk 4:10
Mk 5:25
Mk 5:42
Mk 6:7
Mk 6:43
Mk 8:19
Mk 9:35
Mk 10:32
Mk 11:11
Mk 14:10
Mk 14:17
Mk 14:20
Mk 14:43
Best regards,
John N. Lupia, III
John N. Lupia, III
Beachwood, New Jersey 08722 USA
Fax: (732) 349-3910
http://groups.yahoo.com/group/Roman-Catholic-News/
God Bless America
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Your message has been successfully submitted and would be delivered to recipients shortly. | 797 | 2,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2015-22 | longest | en | 0.915743 |
https://testbook.com/question-answer/the-bar-graph-given-below-shows-the-production-of--615d987e4e47647e4ffb5a7c | 1,643,229,255,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304961.89/warc/CC-MAIN-20220126192506-20220126222506-00249.warc.gz | 607,315,602 | 31,871 | The bar graph given below shows the production of paper (in lakh tones) by two different companies L and M for the given years.J1 = Average production by company L from 1996 to 2000J2 = Average production by company M from 1996 to 1999 What is the value of J1 - J2?
This question was previously asked in
DFCCIL Executive (Ops & BD) 28th Sept 2021 Paper (Shift 1)
View all DFCCIL Executive Papers >
1. 2.75
2. 2.25
3. 1.75
4. 1.5
Answer (Detailed Solution Below)
Option 3 : 1.75
Free
CT 1: Current Affairs (Government Policies and Schemes)
55555
10 Questions 10 Marks 10 Mins
Detailed Solution
Given:
The production of company L from 1996 to 2000 = 45, 75, 95, 120, and 105 respectively
The production of company M from 1996 to 1999 = 60, 105, 70, and 110 respectively
Formula used:
$$Average\ =\ {The\ sum\ of\ all\ items \over The\ total\ number\ of\ items}$$
Calculation:
⇒ J1 = $${45\ +\ 75\ +\ 95\ +\ 120\ +\ 105 \over 5}\ =\ {440 \over 5}\ =\ 88$$
⇒ J2 = $${60\ +\ 105\ +\ 70\ +\ 110 \over 4}\ =\ {345\over 4}\ =\ 86.25$$
⇒ J1 - J2 = 88 - 86.25 = 1.75 (In lakh tones)
∴ The required result will be 1.75. | 409 | 1,122 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2022-05 | latest | en | 0.778508 |
https://practice.geeksforgeeks.org/problems/sum-of-all-subsets-formed-by-first-n-natural-numbers/0 | 1,582,044,932,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143784.14/warc/CC-MAIN-20200218150621-20200218180621-00551.warc.gz | 538,623,740 | 23,008 | Sum of all subsets formed by first N natural numbers
##### Submissions: 399 Accuracy: 36.13% Difficulty: Basic Marks: 1
Given a number N, the task is to find the sum of all the elements from all possible subsets of a set formed by first N natural numbers.
Input:
The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case contains a number N as input.
Output:
For each test case, print the sum in new line.
Constraints:
1 <= T <= 50
1 <= N <= 50
Example:
Input:
2
2
3
Output:
6
24
Explanation:
```Input : n = 2
Output : 6
Possible subsets are {{1}, {2},
{1, 2}}. Sum of elements in subsets
is 1 + 2 + 1 + 2 = 6.```
#### ** For More Input/Output Examples Use 'Expected Output' option **
Author: arun03
If you have purchased any course from GeeksforGeeks then please ask your doubt on course discussion forum. You will get quick replies from GFG Moderators there. | 262 | 943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-10 | longest | en | 0.834097 |
https://inches.guru/19-1/1-inch-to-cm | 1,723,676,665,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641137585.92/warc/CC-MAIN-20240814221537-20240815011537-00036.warc.gz | 244,113,712 | 4,335 | # Convert 19 1/1 inches to cm
19 1/1 inches = 50.8 cm
Result: 19 1/1 inches is equivalent to 50.8 centimeters.
/
## How many cm is 19 1/1 inches?
The answer is 50.8 centimeters, which is also the answer to the question "What is 19 1/1 inches in cm?".
## How to convert 19 1/1 inches to cm?
To convert 19 1/1 inches to centimeters, follow these two steps: first convert the 19 1/1 inch fraction to decimal inches, and then convert the decimal inches to decimal centimeters.
To convert 19 1/1 inches from a fraction to a decimal, divide the numerator 1 by the denominator 1 and add the whole number 19 to get a decimal inch of 20 inches.
Since there are 2.54 cm in an inch, multiplying 20 inches by 2.54 gives us the answer in decimal inches to millimeters.
The answer and calculation process are listed below to show how to convert 19 1/1 inches to cm.
19 1/1 inches = 1 ÷ 1 + 19 = 20 in
20 in × 2.54 = 50.8 cm
Therefore, 19 1/1 inches is equal to 50.8 centimeters.
## Inch fraction to cm conversion table
Inches (fraction)Inches (decimal)Centimeters
9 1/1 in10 in25.4 cm
10 1/1 in11 in27.94 cm
11 1/1 in12 in30.48 cm
12 1/1 in13 in33.02 cm
13 1/1 in14 in35.56 cm
14 1/1 in15 in38.1 cm
15 1/1 in16 in40.64 cm
16 1/1 in17 in43.18 cm
17 1/1 in18 in45.72 cm
18 1/1 in19 in48.26 cm
19 1/1 in20 in50.8 cm
20 1/1 in21 in53.34 cm
21 1/1 in22 in55.88 cm
22 1/1 in23 in58.42 cm
23 1/1 in24 in60.96 cm
24 1/1 in25 in63.5 cm
25 1/1 in26 in66.04 cm
26 1/1 in27 in68.58 cm
27 1/1 in28 in71.12 cm
28 1/1 in29 in73.66 cm | 581 | 1,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-33 | latest | en | 0.671864 |
http://www.cfd-online.com/Forums/main/1101-calculation-rotating-coordinates-print.html | 1,477,211,728,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719192.24/warc/CC-MAIN-20161020183839-00232-ip-10-171-6-4.ec2.internal.warc.gz | 361,927,465 | 2,678 | CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- Main CFD Forum (http://www.cfd-online.com/Forums/main/)
- - calculation at rotating coordinates (http://www.cfd-online.com/Forums/main/1101-calculation-rotating-coordinates.html)
JYPark July 29, 1999 06:16
calculation at rotating coordinates
I am converting the code used at fixed coordinates to the code at rotating coordinates to analysis flow in the turbomachinery. I use the absolute velocity components at the rotating coordinates and Van Leer's flux vector splitting and implicit euler backward time marching method. I am searching the references which explain the procedure in detail. If you know the references, please let me know. Thanks very much in advance.
Hongjun Li July 29, 1999 10:08
Re: calculation at rotating coordinates
The difference b/t fixed and rotating code is the governing equations. The numerical schemes can be exactly the same. So, all you have to do is to rewrite the governing equations in rotating frame of coordinate in terms of relative velocities (not absolute velocity). In fact there are only two additional terms in the momentum equations: the centrifugal force and the Coriolis force terms; the continuity equation has the same form; and in the energy equation, you may need to use rothalpy rather than enthalpy. As soon as you have the correct equations, you may insert the additional terms into the existing code. Sorry, I do not have any public reference to give you. I wrote some detailed reports but they have proprietray information inside so they are for internal use only.
For general information on equations in rotation frame of reference, you may read "Numerical Computation of Internal and External Flows: by Hirsch, Wiley-Intersciency, 1988. (Ch. 1, Sec. 1.4).
Timothy Cowan August 5, 1999 00:28
Re: calculation at rotating coordinates
Check out the following paper:
Kandil and Chuang, "Unsteady Transonic Airfoil Computation Using Implicit Euler Scheme on Body-Fixed Grid," AIAA Journal, Vol. 27, No. 8.
These authors have written several papers (in addition to the above) on this subject. It should at least get you started...
-Tim
Apurva Shukla August 14, 1999 18:55
Re: calculation at rotating coordinates
Hello,
I am also working on turbomachinery flow simulation. I will suggest a paper by B. Lakshminarayana in ASME Journal of Fluid Mechanics, 1991, Frreman Schoolar lecture, I donot remember the month, but U can easilt locate them. You will get a good information & philosophy to start with.
Good Luck
Apurv
All times are GMT -4. The time now is 04:35. | 618 | 2,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2016-44 | latest | en | 0.857276 |
https://www.topperlearning.com/answer/find-all-the-prime-factor-of-935-and-arrange-them-in-ascending-order-now-state-the-relation-if-any-between-two-consecutive-prime-factors/duyplnrr | 1,718,254,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.11/warc/CC-MAIN-20240613025523-20240613055523-00687.warc.gz | 947,713,687 | 55,631 | Request a call back
Find all the prime factor of 935 and arrange them in ascending order now state the relation if any between two consecutive prime factors
Asked by piyushgariya65.6cpsb | 04 Jun, 2021, 12:52: PM
The prime factors of 935 are 5,11 and 17.
Ascending order: 5, 11 , 17
11 - 5 = 6 and 17 - 11 = 6
The difference between two consecutive prime factors is a constant which is 6.
Answered by Yasmeen Khan | 04 Jun, 2021, 01:16: PM
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Asked by Ashishshukla.dfs1 | 09 Jun, 2020, 10:48: AM | 346 | 866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-26 | latest | en | 0.901817 |
http://lbartman.com/worksheet/adding-and-subtracting-linear-expressions-worksheet.php | 1,611,196,792,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522150.18/warc/CC-MAIN-20210121004224-20210121034224-00672.warc.gz | 58,010,869 | 12,291 | ## lbartman.com - the pro math teacher
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# Adding And Subtracting Linear Expressions Worksheet
Public on 02 Oct, 2016 by Cyun Lee
### homework rationals review algebra ii trigonometry
Name : __________________
Seat Num. : __________________
Date : __________________
2543 + 9311 = ...
4975 + 4053 = ...
7138 + 4558 = ...
8512 + 9726 = ...
7815 + 2896 = ...
3341 + 3714 = ...
6579 + 5941 = ...
7897 + 4594 = ...
8957 + 8173 = ...
9691 + 5246 = ...
4605 + 2572 = ...
2997 + 9951 = ...
5744 + 3172 = ...
1054 + 2080 = ...
3496 + 4785 = ...
3819 + 5763 = ...
3761 + 9999 = ...
8282 + 7715 = ...
9750 + 3026 = ...
2847 + 2748 = ...
8244 + 1596 = ...
9872 + 8869 = ...
2762 + 5924 = ...
1795 + 3760 = ...
5502 + 4316 = ...
4003 + 1991 = ...
3450 + 1667 = ...
7463 + 4437 = ...
5232 + 8166 = ...
5357 + 3266 = ...
8521 + 9515 = ...
5107 + 3592 = ...
4867 + 7551 = ...
1897 + 1020 = ...
7177 + 3942 = ...
8408 + 9014 = ...
4690 + 2415 = ...
3396 + 2155 = ...
3726 + 7196 = ...
3122 + 8607 = ...
6418 + 9549 = ...
3334 + 9654 = ...
1751 + 4624 = ...
3347 + 9824 = ...
6571 + 1138 = ...
8883 + 6598 = ...
9061 + 6498 = ...
6869 + 2016 = ...
6006 + 2071 = ...
4799 + 7080 = ...
2718 + 5135 = ...
3554 + 1996 = ...
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6918 + 5664 = ...
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3596 + 2006 = ...
1847 + 8026 = ...
4595 + 5131 = ...
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2830 + 7401 = ...
4020 + 3072 = ...
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8379 + 1996 = ...
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2746 + 4030 = ...
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1911 + 2186 = ...
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3318 + 7805 = ...
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3115 + 3815 = ...
4222 + 6035 = ...
4764 + 4507 = ...
7891 + 7026 = ...
8981 + 7192 = ...
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# Each employee of Company Z is an employee of either division
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Each employee of Company Z is an employee of either division [#permalink]
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23 Apr 2008, 20:29
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Each employee of Company Z is an employee of either division X or division Y, but not both. If each division has some part-time employees, is the ratio of the number of full-time employees to the number of part-time employees greater for division X than for Company Z?
I. The ratio of the number of full-time employees to the number of part-time employees is less for division Y than for Company Z
II. More than half of the full-time employees of Company Z are employees of division X, and more than half of the part-time employees of Company Z are employees of division Y.
Please share your explanation
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Re: DS: tricky [#permalink]
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23 Apr 2008, 22:00
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Lets say number of part time employees in Y = Py
Lets say number of part time employees in X = Px
Lets say number of full time employees in Y = Fy
Lets say number of full time employees in X = Fx
Total number of Part time employees = Px + Py
Total number of Full time employees = Fx + Fy
Question is asking Fx/Px > Fx+Fy/Px+Py
Statement 1:
Tells us that Fy/Py < Fx+Fy/Px+Py. This statement does not talks about division X so it is insufficient.
Statement 2:
Assume Number of Part Time employee 100x where x is any positive number, and number of full time employees 100y, where y is any positive number. As per this statement at least 51x are working in department y and at most 49x is working in department x. Similarly at least 51y are working for department x and at most 49y is working for department y.
Ratio for Department x= 51y/49x
Ratio for whole company = 100y/100x = y/x
So question is answered. Answer B.
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Re: DS: tricky [#permalink]
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28 Apr 2008, 00:52
abhijit_sen wrote:
Lets say number of part time employees in Y = Py
Lets say number of part time employees in X = Px
Lets say number of full time employees in Y = Fy
Lets say number of full time employees in X = Fx
Total number of Part time employees = Px + Py
Total number of Full time employees = Fx + Fy
Question is asking Fx/Px > Fx+Fy/Px+Py
Statement 1:
Tells us that Fy/Py < Fx+Fy/Px+Py. This statement does not talks about division X so it is insufficient.
Statement 2:
Assume Number of Part Time employee 100x where x is any positive number, and number of full time employees 100y, where y is any positive number. As per this statement at least 51x are working in department y and at most 49x is working in department x. Similarly at least 51y are working for department x and at most 49y is working for department y.
Ratio for Department x= 51y/49x
Ratio for whole company = 100y/100x = y/x
So question is answered. Answer B.
Quote:
I think its D.
If Rz>Ry the x ratio (Rx) must be greater than Rz. Rz isn't the weighted average of Rx and Ry, but it must be between them (or equal to them).
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Re: DS: tricky [#permalink]
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28 Apr 2008, 01:36
eschn3am wrote:
Answer D
Company average is the weighted average of the 2 divisions. Statement 1 tells us that Y is below the company average, so X must be above it to compensate.
Statement 2 tells us that X is more than half the full time employees and less than half of the part time, so it has a greater ratio than Y, which means it has a greater ratio than the company (since one must be greater and the other smaller, unless they're both equal to the co. average)
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Re: DS: tricky [#permalink]
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29 Apr 2008, 01:25
Let me, a Vietnamese Engineer, solve it
Let Xf: Full-time employees of X
Let Xp: Part-time employees of X
Let Yf: Full-time employees of Y
Let Yp: Part-time employees of Y
Since Full-time employees of company Z: Xf+Yf
Part-time employees of company Z: Xp+Yp
The question is: Xf/Xp>(Xf+Yf)/(Xp+Yp)? {*}
We must be sure that all Xi and Yi are both greater than zero. So, simplifying {*} then we have
Xf/Xp>Yf/Yp (necessary and sufficient conditions of {*} to be true)
1. Yf/Yp<(Xf+Yf)/(Xp+Yp) ==> Xf/Xp>Yf/Yp ==> Sufficient
2. Xf>(Xf+Yf)/2 ==> Xf>Yf (a)
Yp>(Xp+Yp)/2 ==> Yp>Xp (b)
Since a and b are true we have Xf.Yp> Yf.Yp>Yf.Xp ==> Xf/Xp>Yf/Yp ==> Sufficient
So the answer is D.
If you have any troubles pls send your email to duongminhtan@hotmail.com
Last edited by lexis on 30 Apr 2008, 00:37, edited 2 times in total.
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Re: DS: tricky [#permalink]
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29 Apr 2008, 17:56
lexis wrote:
Let me, an Vietnamese Engineer, solve it
Let Xf: Full-time employees of X
Let Xp: Part-time employees of X
Let Yf: Full-time employees of Y
Let Yp: Part-time employees of Y
Since Full-time employees of company Z: Xf+Yf
Part-time employees of company Z: Xp+Yp
The question is: Xf/Xp>(Xf+Yf)/(Xp+Yp)? *
We must be sure that all Xi and Yi are both greater than zero. So, simplifying * then we have
Xf/Xp>Yf/Yp (necessary and sufficient conditions of * to be true)
1. Yf/Yp<(Xf+Yf)/(Xp+Yp) ==> Xf/Xp>Yf/Yp ==> Sufficient
2. Xf>(Xf+Yf)/2 ==> Xf>Yf (a)
Yp>(Xp+Yp)/2 ==> Yp>Xp (b)
Since a and b are true we have Xf.Yp> Yf.Yp>Yf.Xp ==> Xf/Xp>Yf/Yp ==> Sufficient
So the answer is D.
If you have any troubles pls send your email to duongminhtan@hotmail.com
I still dint get how A...
I went for B..can someone explain in a simpler way please!!
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Re: DS: tricky [#permalink]
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29 Apr 2008, 21:49
Yep, I got tricked and picked B too.
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Re: DS: tricky [#permalink]
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30 Apr 2008, 03:32
Keep in mind that if we prove Xf/Xp>Yf/Yp from given information in one case, then the case must be true.
Hope you could get it.
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Re: DS: tricky [#permalink] 30 Apr 2008, 03:32
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# Each employee of Company Z is an employee of either division
post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,354 | 7,867 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-47 | latest | en | 0.900903 |
http://mathhelpforum.com/calculus/114857-inverse-trigonometry-print.html | 1,529,901,778,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867424.77/warc/CC-MAIN-20180625033646-20180625053646-00630.warc.gz | 205,062,207 | 2,802 | Inverse Trigonometry.
• Nov 15th 2009, 08:11 PM
Sunyata
Inverse Trigonometry.
A curve with gradient function (1-x)/(1+x^2) goes through the origin. Show that the curve cuts the X-axis when arctan(x) = 0.5ln(1+x^2)
What exactly am I meant to do for these types of questions?
• Nov 15th 2009, 08:20 PM
redsoxfan325
Quote:
Originally Posted by Sunyata
A curve with gradient function (1-x)/(1+x^2) goes through the origin. Show that the curve cuts the X-axis when arctan(x) = 0.5ln(1+x^2)
What exactly am I meant to do for these types of questions?
To find the original function you need to integrate
$\displaystyle \int\frac{1-x}{1+x^2}\,dx=\int\frac{1}{1+x^2}\,dx-\int\frac{x}{1+x^2}\,dx$
The first integral is $\displaystyle \arctan(x)$ (you should find this on any table of integrals) and the second integral can be solved using a simple u-sub ($\displaystyle u=1+x^2$), so you end up with $\displaystyle f(x)=\arctan(x)-\frac{1}{2}\ln(1+x^2)$. Then you set $\displaystyle f(x)=0$ to get the answer.
It seems a little strange that the question would ignore the presence of the integration constant though. | 347 | 1,113 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-26 | latest | en | 0.817384 |
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Enthusiast
## Why does <> in my query not work?
Hello,
Could someone please help me understand why my WHERE statement is not working? My query below works if I use a case statement,
but I don't understand WHY the parenthesis does not combine to one logical operation and is instead seen as two seperate ones.
- Why does not (Year_num <> 2005 AND Month_num<> 8) not get seen as one combination ?
Christine
```SELECT CASE WHEN (s.msa_high >= 50 AND s.msa_high <= 60) THEN 'low'
WHEN (s.msa_high > 60 AND s.msa_high <= 70) THEN 'medium'
WHEN s.msa_high >70 THEN 'high'
END AS EduCat, SUM(CleanedData.SalesDays) AS TotSalesDays, SUM(CleanedData.Revenue) AS TotRev
FROM (SELECT EXTRACT(YEAR from saledate) AS Year_num, EXTRACT(MONTH from saledate) AS Month_num, store AS Store, COUNT(DISTINCT saledate) AS SalesDays,
SUM(AMT) AS Revenue, Revenue/SalesDays AS Avg_Daily_Revenue
FROM trnsact
WHERE stype = 'P' AND (Year_num <> 2005 AND Month_num<> 8)
GROUP BY Year_num, Month_num, Store
HAVING SalesDays >= 20 ) AS CleanedData
INNER JOIN store_msa s ON s.store = CleanedData.store
GROUP BY EduCat;```
Accepted Solutions
## Re: Why does <> in my query not work?
Christine,
This is happenigns due to De Morgan's law of boolean logic / set theory (https://en.wikipedia.org/wiki/De_Morgan%27s_laws). Quickly, when you "distribute" a NOT your ANDs switch to ORs.
Your case statement is saying what you want: I want all rows that are not in August of 2005
With logical opperators we can write: WHERE NOT (year_num = 2005 AND month_num = 8)
To distribute the NOT with logical opperators we end up with: WHERE (year_num <> 2005 OR month_num <> 8)
This is confusing to us becase we are looking for rows that are not something and also not something else so we want to use an AND, but check out the Ven Diagrams in the De Morgan's Law link and see if it makes sence why we want OR instead.
Good luck!
Michelle
1 ACCEPTED SOLUTION
10 REPLIES
## Re: Why does <> in my query not work?
Please explain what "doesn't work" means. And please provide the case expression used instead that gets the desired result.
Enthusiast
## Re: Why does <> in my query not work?
Hi Todd,
Thank you. I mean that it is not giving me the desired result which is the one below.
```SELECT CASE WHEN (s.msa_high > 50 AND s.msa_high <= 60) THEN 'low'
WHEN (s.msa_high > 60 AND s.msa_high <= 70) THEN 'medium'
WHEN s.msa_high >70 THEN 'high'
END AS EduCat, SUM(CleanedData.SalesDays) AS TotSalesDays, SUM(CleanedData.Revenue) AS TotRev
FROM (SELECT EXTRACT(YEAR from saledate) AS Year_num, EXTRACT(MONTH from saledate) AS Month_num, store AS Store, COUNT(DISTINCT saledate) AS SalesDays,
SUM(AMT) AS Revenue, Revenue/SalesDays AS Avg_Daily_Revenue, (CASE
WHEN Year_num= '2005' AND Month_num= '8' THEN 'exclude' ELSE 'include' END) AS Filter FROM trnsact WHERE stype='p' AND Filter='include' GROUP BY store, Year_num, Month_num HAVING SalesDays >= 20 )AS CleanedData
INNER JOIN store_msa s ON s.store = CleanedData.store
GROUP BY EduCat;```
## Re: Why does <> in my query not work?
What are the data types and data representation in those types? In the case expression character compares are being done. In the <> expressions numeric compares are being done.
Enthusiast
## Re: Why does <> in my query not work?
Hello,
I want to exclude the year 2005 and month of August from sales. Stypye =p( purchase.)
Please see a screenshot of the results obtained.
Without case clause and <> (wrong result)
With case clause (right result)
Thanks Todd
Highlighted
Enthusiast
## Re: Why does <> in my query not work?
Pardon *stype = purchase
## Re: Why does <> in my query not work?
Perhaps I was not clear enough with my question:
WHEN Year_num= '2005' AND Month_num= '8' is performing character string comparisons due to the numbers being placed inside quotes.
AND (Year_num <> 2005 AND Month_num<> 8) is performing numeric comparisons - no quotes.
Implicit data type casting has to be done when they types do not match.
What data types are Year_num and Month_num? How is the data stored in those columns?
Enthusiast
## Re: Why does <> in my query not work?
Ahh I am sorry, I am new to SQL so I did not correctly understand. Please bear with me, its this what your looking for?
Enthusiast
## Re: Why does <> in my query not work?
Hello Todd,
Could you please kindly clarify what you need from me? I want to learn why my query is not giving me the right result.
## Re: Why does <> in my query not work?
Christine,
This is happenigns due to De Morgan's law of boolean logic / set theory (https://en.wikipedia.org/wiki/De_Morgan%27s_laws). Quickly, when you "distribute" a NOT your ANDs switch to ORs.
Your case statement is saying what you want: I want all rows that are not in August of 2005
With logical opperators we can write: WHERE NOT (year_num = 2005 AND month_num = 8)
To distribute the NOT with logical opperators we end up with: WHERE (year_num <> 2005 OR month_num <> 8)
This is confusing to us becase we are looking for rows that are not something and also not something else so we want to use an AND, but check out the Ven Diagrams in the De Morgan's Law link and see if it makes sence why we want OR instead.
Good luck!
Michelle | 1,420 | 5,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-51 | latest | en | 0.829761 |
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In this video, we learn the definition of rational numbers, rational expressions, and rational functions . Watch the Tagalog math tutorial video above and then answer the exercises below. The complete video playlist of the Sipnayan tutorials on Rational Expressions can be found here.
Practice Exercise
True or False
1.) The fraction $\frac{\pi}{3}$ is a rational number.
2.) The fraction $\frac{2x - 1}{5x + 1}$ is a rational expression.
3.) The fraction $\frac{\sqrt{x} + 3}{x^2 + 3}$ is a rational expression.
4.) The function $f(x) = \frac{1}{x}$ is a rational function.
5.) The fraction $\frac{3}{8}$ is a rational number.
1.False. The numerator and denominator of a rational number must be integers.
2. True
3. False. $\sqrt{x} + 3$ is not a polynomial expression.
4. True
5. True
# Paano gawin ang addition ng integers gamit ang number line?
Pag tayo ay nag-add ng number, ang representation nito sa number line ay ang paggalaw papuntang kanan. Pag tayo ay nagsubtract (o nag-add ng negative number), ang represenation nito ay ang paggalawa pakaliwa.
Ang number line ay isa lamang sa mga representations na pwede nating tingnan kung tayo ay naga-add ng integers. Panoorin ang video sa itaas upang maintindihan ang nangyayari sa addition ng integers.
Pwede mapanood ang mga tagalog math video tutorials tungkol sa integers (kasama addition of integers) ng tuloy tuloy sa Youtube (25 videos).
# Paano i-prove ang Pythagorean Theorem?
This Taglish (mixed Tagalog and English) math tutorial video discusses the proof of the Pythagorean Theorem.
The Pythagorean Theorem states the sum of the squares of the length of the shorter side of a right triangle is equal to the square of its hypotenuse. That is, in a right triangle with sides $a$ and $b$ and hypotenuse $c$, then $c^2 = a^2 + b^2$.
# Paano mag-subtract ng fractions?
This video tutorial discusses subtraction of fractions. The examples discussed in this video include similar fractions, dissimilar fractions, and mixed fractions.
Subtracting fractions is very similar to adding fractions. For similar fractions, the numerator are subtracted and the numerator copied in the answer. For dissimilar fractions, least common multiple is needed in order to perform the operation. | 577 | 2,302 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 20, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2021-31 | latest | en | 0.463487 |
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posted by .
A plane on a search mission flew east from an airport, turned, and flew west back to the airport. The plane crised at 300km/h when flying east, and 400km/h when flying west. The plane was in the air for 7 hours. How far from the airport did the plane travel?
• Math -
Vi = 300km/h, East.
V2 = 400km/h,West.
X hours,East.
(7-X) hours,West.
De = Dw.
300x = 400(7-x)
300x = 2800-400x
300x+400x = 2800
700x = 2800
X = 4 h,East.
De = V*t = 300km/h * 4h = 1200 km =
Dist. travel from airport. | 180 | 512 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-26 | latest | en | 0.940231 |
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Turning data into measurable objectives and goals can be a difficult task. Learn about gathering information for IEP goals and find sample objectives and goals in the areas of reading and math for primary level students.
## Collecting Data for Reading and Math Goals
When establishing reading and math goals, it is important to collect data to determine a baseline and monitor progress. Data can be collected using fluency probes, MAZE probes for comprehension, word lists, timed math computation probes (such as mad minutes), and math concepts and applications assessments. Data can also be gathered using standardized assessments, such as the Woodcock-Johnson III (W-J III) or Wechsler Individual Achievement Test (WIAT III); but since these assessments should not be given more than once per year, they should be used in conjunction with standards and curriculum based assessments, not in lieu of.
Using data obtained from reading probes, curriculum based assessments, standardized assessments, and other data collection tools, the following measurable examples of goals and objectives can be used with students who have learning disabilities in the areas of reading fluency and reading comprehension.
• Given a timed (one minute) unrehearsed reading probe at the _____ grade level, the student will move from improve his oral reading fluency from a baseline of (insert current number of words read accurately per minute) to (insert anticipated number of words read correctly) on two out of three consecutive probes.
• When presented with a list of Dolch sight words (or insert list student's school uses), the student will be able to read independently (insert percentage anticipated) of the words. His current baseline is (insert current percentage correctly read).
• Given a series of curriculum based reading probes measuring the ability to decode words in text fluency, the student will increase his oral reading fluency from (current number of words per minute) at the (instructional grade level) to (anticipated number of words per minute) at the (anticipated grade level).
• Given a (insert grade level child reads at) grade level reading passage, the student will correctly respond to (anticipated percentage correct) of comprehension questions from a current baseline of (current percentage correct) on two out of three assessments.
• Given five grade level reading passages, the student will correctly identify the main idea for four out of five from a current baseline of (current percentage correct).
• Given specially designed instruction and reading material at his instructional level (provide instructional level), the student will demonstrate understanding of the materials by obtaining at least a 75% (or rubric score of 3-Proficient), on three out of four reading comprehension probes.
## Math Computation
Data should be regularly collected from timed probes, curriculum based and standardized tests, and other formative and summative assessments to establish baselines and set achievable goals.
1. Math Computation
• Given weekly timed math probes at the (current grade level), the student will improve from (current number of digits correct) to (anticipated number of digits correct) on two out of three opportunities.
• On a series of probes representing (grade level) mathematics computation problems, the student will correctly compute (anticipated number) digits correct, increasing from (current number) digits correct on three consecutive weekly probes.
2. Math Concepts and Application
• Given weekly timed math concepts and application probes at the (current grade level), the student will improve from (baseline) problems to correct to (anticipated) problems correct.
• Given various combinations of coins, the students will count and identify amounts up to (predicted amount) with an accuracy of (predicted percentage correct). | 735 | 3,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-34 | latest | en | 0.930786 |
https://www.javatpoint.com/types-of-operands-in-computer-organization | 1,726,818,549,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00097.warc.gz | 748,688,935 | 18,202 | # Types of Operands in Computer Organization
Computer instruction is a binary code that determines the micro-operations in a sequence for a computer. They are saved in the memory along with the information. Each computer has its specific group of instructions. They can be categorized into two elements as Operation codes (Opcodes) and Address. Opcodes specify the operation for specific instructions, and an address determines the registers or the areas used for that operation.
Operands are definite elements of computer instruction that show what information is to be operated on. The most important general categories of data are
2. Numbers
3. Characters
4. Logical data
In many cases, some calculation must be performed on the operand reference to determine the main or virtual memory address.
In this context, addresses can be considered to be unsigned integers. Other common data types are numbers, characters, and logical data, and each of these is briefly described below. Some machines define specialized data types or data structures. For example, machine operations may operate directly on a list or a string of characters.
Addresses are nothing but a form of data. Here some calculations must be performed on the operand reference in an instruction, which is to determine the physical address of an instruction.
### Numbers
All machine languages include numeric data types. Even in non-numeric data processing, numbers are needed to act as counters, field widths, etc. An important difference between numbers used in ordinary mathematics and numbers stored in a computer is that the latter is limited. Thus, the programmer is faced with understanding the consequences of rounding, overflow and underflow.
Here are the three types of numerical data in computers, such as:
1. Integer or fixed point: Fixed point representation is used to store integers, the positive and negative whole numbers (… -3, -2, -1, 0, 1, 2, 3, …). However, the programmer assigns a radix point location to each number and tracks the radix point through every operation. High-level programs, such as C and BASIC usually allocate 16 bits to store each integer. Each fixed point binary number has three important parameters that describe it:
• Whether the number is signed or unsigned,
• The position of the radix point to the right side of the sign bit (for signed numbers), or the position of the radix point to the most significant bit (for unsigned numbers).
• And the number of fractional bits stored.
2. Floating point: A Floating Point number usually has a decimal point, which means 0, 3.14, 6.5, and-125.5 are Floating Point
The term floating point is derived from the fact that there is no fixed number of digits before and after the decimal point, which means the decimal point can float. There are also representations in which the number of digits before and after the decimal point is set, called fixed-point representations. In general, floating-point representations are slower and less accurate than fixed-point representations, but they can handle a larger range of numbers.
3. Decimal number: The decimals are an extension of our number system. We also know that decimals can be considered fractions with 10, 100, 1000, etc. The numbers expressed in the decimal form are called decimal numbersor decimals. For example:1, 4.09, 13.83, etc. A decimal number has two parts, and a dot separates these parts (.) called the decimal point.
• Whole number part: The digits lying to the left of the decimal point form the whole number part. The places begin with ones, tens, hundreds, thousands and so on.
• Decimal part: The decimal point and the digits laying on the right of the decimal point form the decimal part. The places begin with tenths, hundredths, thousandths and so on.
### Characters
A common form of data is text or character strings. While textual data are most convenient for humans. But computers work in binary. So, all characters, whether letters, punctuation or digits, are stored as binary numbers. All of the characters that a computer can use are called character sets. Here are the two common standards, such as:
1. American Standard Code for Information Interchange (ASCII)
2. Unicode
ASCII uses seven bits, giving a character set of 128 characters. The characters are represented in a table called the ASCII table. The 128 characters include:
• 32 control codes (mainly to do with printing)
• 32 punctuation codes, symbols, and space
• 26 upper-case letters
• 26 lower-case letters
• numeric digits 0-9
We can say that the letter 'A' is the first letter of the alphabet; 'B' is the second, and so on, all the way up to 'Z', which is the 26th letter. In ASCII, each character has its own assigned number. Denary, binary and hexadecimal representations of ASCII characters are shown in the below table.
A65100000141
Z9010110105A
a97110000161
z12211110107A
048011000030
957011100139
Space32010000020
!33010000121
A is represented by the denary number 65 (binary 1000001, hexadecimal 41), B by 66 (binary 1000010, hexadecimal 42) and so on up to Z, which is represented by the denary number 90 (binary 1011010, hexadecimal 5A).
Similarly, lower-case letters start at denary 97 (binary 1100001, hexadecimal 61) and end at denary 122 (binary 1111010, hexadecimal 7A). When data is stored or transmitted, its ASCII or Unicode number is used, not the character itself.
For example, the word "Computer" would be represented as:
1000011 1101111 1101101 1110000 1110101 1110100 1100101 1110010
On the other hand, IRA is also widely used outside the United States. A unique 7-bit pattern represents each character in this code. Thus, 128 different characters can be represented, and more than necessary to represent printable characters, and some of the patterns represent control characters. Some control characters control the printing of characters on a page, and others are concerned with communications procedures.
IRA-encoded characters are always stored and transmitted using 8 bits per character. The 8 bit may be set to 0 or used as a parity bit for error detection. In the latter case, the bit is set such that the total number of binary 1s in each octet is always odd (odd parity) or always even (even parity).
### Logical data
Normally, each word or other addressable unit (byte, half-word, and so on) is treated as a single unit of data. Sometimes, it is useful to consider an n-bit unit consisting of 1-bit items of data, each item having the value 0 or 1. When data are viewed this way, they are considered to be logical data.
The Boolean data can only represent two values: true or false. Although only two values are possible, they are rarely implemented as a single binary digit for efficiency reasons. Many programming languages do not have an explicit Boolean type, instead of interpreting 0 as false and other values as true. Boolean data refers to the logical structure of how the language is interpreted to the machine language. In this case, a Boolean 0 refers to the logic False, and true is always a non zero, especially one known as Boolean 1.
There are two advantages to the bit-oriented view:
• We may want to store an array of Boolean or binary data items, in which each item can take on only the values 0 and 1. With logical data, memory can be used most efficiently for this storage.
• There are occasions when we want to manipulate the bits of a data item. | 1,630 | 7,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.909396 |
https://www.mrexcel.com/board/threads/sum-of-alternate-columns-with-subtotals.1128370/ | 1,600,496,686,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400190270.10/warc/CC-MAIN-20200919044311-20200919074311-00024.warc.gz | 994,894,365 | 19,473 | # Sum of alternate columns with subtotals
#### HJimHB
##### New Member
Hi
I have a report which pulls values from 2 sources (13 from each) for each month of the year, so 312 columns in total.
For each line, I need to subtotal each source per month to compare these totals, and also get the annual totals for comparison.
How can I get this?
### Excel Facts
Return population for a City
If you have a list of cities in A2:A100, use Data, Geography. Then =A2.Population and copy down.
#### HighAndWilder
##### Board Regular
Are you able to provide a screenshot of the source worksheets?
Also an idea as to how you would like the report to be structured.
#### HJimHB
##### New Member
Hi - see (edited) image above. I need to add all the "Cog" and "SFA" columns to reconcile the totals for each month, and also to reconcile the annual totals, for each row (around 200 in total).
The column headings on row 2 are the same each month; I had to do in to add the months to row 1 in order to easily identify when one month's figures end and the next month's start.
In terms of what the final table would look like, I was trying to have 3 columns per month (Cog total, SFA total, difference) and then 3 for the year.
Thank you
#### HighAndWilder
##### Board Regular
The ease at which one can tackle a task such as this is often determined by how the raw data is structured.
Whilst not wanting to go into the concept of normalisation (see link below) I suggest that you restructure how you save your raw data.
1NF, 2NF, 3NF and BCNF in Database Normalization | Studytonight
Each column should represent 'something' and the single row column header should identify what the column represents. Each colum header must be unique.
Each row should represent 'something' and ideally the first column value should represent what the row represents. I'm not sure what your rows represent.
What is in column A?
Date
CogSFACode
RowIdentificationID
then the structure would be long and thin, something which you should nearly always aim for in raw data.
It's well worth putting the effort in. It can then be filtered and sorted easily
You could then use some of the standard Excel functions such as SUMIF, SUMIFS and COUNTIF, COUNTIFS and the
Aiming for a long and thin report table is also a good idea. Having to scroll horizontally is always awkward.
Can you post the raw data in a workbook and include some data that identifies what the rows represent and I'll have a look at it tomorrow whilst I wait for
Corona Virus to go away!!! It may be that I can write some code to restructure the data.
Take care and be safe.
#### HJimHB
##### New Member
Hi
Thank you for the response. Unfortunately the raw data is a report that we run from a third-party system and we have no control over the formatting. Is there an easy way to reformat the reports to be in the order you suggest?
We need to run the reports at least every month and often more frequently than that so ideally I was hoping to build something where I could paste the report into a sheet and the formulae would extract the necessary information. Is that possible?
I've gone through the file and redacted any confidential info; how can I upload this (sorry, new to the forum)?
Thank you, and hope you're safe as well.
#### HighAndWilder
##### Board Regular
I am safe and well Thanks. Plenty of food and facilities to self isolate.
I trust that all is ok with you.
Apparently one cannot upload a file but thats not the end of the world.
I just need some more information to enable me to think how to convert the data into a better structure.
1. Does each sheet of data only cover one year?
2. Are there only ever 12 columns per month? Never any less?
3. What is in column A?
4. What is in column B?
Thanks
#### Fluff
##### MrExcel MVP, Moderator
Maybe something like
+Fluff.xlsm
ABCDEFGHIJKLMNOPQRST
1JulyJulyJulyJulyJulyJulyAugAugAugAugAugAugJulyJulyAugAug
2Cog on ProgSFA on ProgCog BalSFA BalCog AimSFA AimCog on ProgSFA on ProgCog BalSFA BalCog AimSFA AimCogSFACogSFA
3123456789101112912-32730-3
4121110987654321302731293
Data
Cell Formulas
RangeFormula
O3:P4,R3:S4O3=SUMIFS(\$B3:\$M3,\$B\$2:\$M\$2,O\$2&"*",\$B\$1:\$M\$1,O\$1)
Q3:Q4,T3:T4Q3=O3-P3
#### HighAndWilder
##### Board Regular
Thanks
So what does the value in cell B3, for example represent?
An instance of 'something' that happened in July for which the code 'Cog on Prog' applies?
The idea is to have the raw data once it has been restructured in one sheet and the analysis in another.
#### Fluff
##### MrExcel MVP, Moderator
I'm not the OP
I was just suggesting a possible solution, with some simplified data
#### HJimHB
##### New Member
Hi both
This is financial information that pulls from two sources (Cog and SFA) that we need to reconcile. For each month, there are 26 columns, 13 from each source, that we need to subtotal for each month to reconcile, and then get a total for the year.
Thanks | 1,227 | 4,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-40 | latest | en | 0.9238 |
http://mhtg.blogspot.com/2013/03/wonderful-wednesday.html | 1,527,456,815,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870470.67/warc/CC-MAIN-20180527205925-20180527225925-00340.warc.gz | 184,530,685 | 13,621 | Wednesday, March 20, 2013
Wonderful Wednesday
The children began their day in the media center with Mrs. Rose.
When they returned to class, we went over a few odds and ends. First, I shared the link that I emailed to all parents this morning. I told the children they did not have to download the software (not a class requirement) and they were not allowed to do so without their parent's permission. Then I showed the students how to use the software with homework assignments posted on Edmodo.
After that, I reminded students that the realistic fiction picture book is due no later than Friday. My grades are due over Spring Break. So, whatever is completed by Friday afternoon will be scored.
Finally, I shared the news that we will be getting kindergarten book buddies on Friday. We will be meeting with our kindergarten buddies on Friday to read a story and work on comprehension skills. I've never done this before but I am sure it will be a fun and rewarding experience for all involved!
Next, I reminded the students of my personal goal (well, one of them) for the next marking period...I would like to incorporate science and social studies instruction into our independent work. With that in mind, we defined economics and defined the idea of income, expense and financial decision making.
During independent work time today the students will work on completing their realistic fiction picture book. Then they will read the Healthful Foods article and write 3 things they learned and 1 question they still have. After that, they will complete an activity called Money Goal where they have to make and explain a financial decision.
In math we continued working towards the memorization of the basic multiplication facts. First, though, we learned how to quickly multiply tens, hundreds and thousands, by counting zeroes and adding them to our answer...let me give you an example. To multiply 3 x 400, you would calculate 3x4 and add 2 zeroes to the end.
While Mrs. Head and I met with small groups, the students chose between 4 different ways to practice the multiplication facts. One choice was Ken Ken. Another was the multiplication app through Edmodo. The last two choices were the games we played, in class, on Monday and Tuesday. These were games that are printed on worksheets.
One group I met with practiced identifying equivalent fractions. The second group drew representations of improper fractions.
Mrs. Head worked with students to correctly partition grids and identify fractions. She also met with a group to identify fractions on a number line.
There is a multiplication worksheet fro homework.
Following lunch and recess the students continued working on their realistic fiction picture book, reading and responding to the Healthful Foods article, and writing about a financial decision.
There is an EDMODO homework assignment. The students need to read an article and write 1-2 sentences explaining the issue.
Towards the end of the afternoon reading/writing block, some students shared their realistic fiction books with the class. These will be sent home on Friday, after we share them with our kindergarten buddies.
We ended the day with science. We began watching a Power Point presentation about nutrition labels. The class had great questions and comments to add, so we didn't finish. We will revisit the Power Point tomorrow afternoon.
Students are encouraged to check out this nutrition web site. | 697 | 3,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-22 | latest | en | 0.955059 |
http://www.ehow.co.uk/how_7170780_measure-miter-cuts-tape-measure.html | 1,532,020,809,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591150.71/warc/CC-MAIN-20180719164439-20180719184439-00005.warc.gz | 447,970,651 | 12,695 | DISCOVER
How to Calculate Miter Length
Updated February 21, 2017
In nearly all situations, measuring trim boards for mitre cuts can be accomplished with just a tape measure. However, some understanding of how to make the cut is also necessary, along with an accurate linear dimension for the piece of trim or moulding. Forty-five degree angled cuts are the order of the day for most mitre cuts, and once you understand how this little detail works, calculating the length can be completed with one measurement.
Pull a board from the stock and set the board face down on the sawhorses. Be sure to check the board for quality and straightness.
Take your speed square and make a square line across the width of the board at one end.
Set the angle of the circular saw blade to 45 degrees and make the crosscut with the angled cut pointing towards the midpoint of the board.
Measure the distance across the wall from inside corner to corner. Do this to the nearest 1/16th of an inch.
Starting from the angled cut, extend the tape measure and make a mark at the distance that was calculated in the previous step.
Take the speed square and extend the mark across the width of the board.
With your speed square, make a perpendicular line across the width of the board. Do this within a few inches of the end of the board.
Set the angle of the circular saw blade to 45 degrees and make the crosscut with the angled cut going away from the end of the board.
Make a measurement from the inside corner to the edge of the wall, where the outside mitre will occur.
Add the thickness of the board to the overall length of your measurement. For example, if your measurement is 24 inches and the board is 3/4 of an inch thick, your final numerical value will be 24 3/4 inches.
Measure the correct length of linear dimension and make a mark. Now your mark is at the right length, but on the wrong side of the board.
Extend the mark to the edge of the board with a speed square.
Flip the board over and extend the mark across the surface of the board. You can now make your final cut.
Tip
Cutting one end before transferring the measurement to the piece of trim is a good strategy. All measurements should be made to 1/16th of an inch. If you are using an electric or non-electric mitre box, you need to turn the board at a right angle to the bottom plane of the mitre box and set the saw at 45 degrees.
Warning
Using an assistant to hold one end of the tape measure will help ensure an accurate measurement. A large framing square can be used to check the squareness of each corner. If the corner is not square, you might have to adjust the angle of the cut.
Things You'll Need
• Tape measure
• Stock of trim boards
• Pencil
• Framing square
• Speed square
• Saw horses
• Circular saw | 616 | 2,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2018-30 | latest | en | 0.938686 |
https://www.mathworks.com/matlabcentral/cody/problems/1813-06-matrix-equations-1/solutions/304692 | 1,508,571,046,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824618.72/warc/CC-MAIN-20171021062002-20171021082002-00745.warc.gz | 961,565,506 | 11,607 | Cody
# Problem 1813. 06 - Matrix Equations 1
Solution 304692
Submitted on 15 Aug 2013 by Mehmet OZC
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% aVec = [3.14 15 8 26]; bVec = [2.71;8;28;182]; aMat = ones(9,9)*2; ref = (aVec*bVec)*aMat^2; user = MyFunc(); assert(isequal(user,ref))
2 Pass
%% aRef = [3.14 15 8 26]; [xMat aVec] = MyFunc(); assert(isequal(aRef,aVec))
3 Pass
%% bRef = [2.71;8;28;182]; [xMat aVec bVec] = MyFunc(); assert(isequal(bRef,bVec))
4 Pass
%% aRef = ones(9,9)*2; [xMat aVec bVec aMat] = MyFunc(); assert(isequal(aRef,aMat)) | 259 | 683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-43 | latest | en | 0.628003 |
https://logic-masters.de/Raetselportal/Raetsel/zeigen.php?id=0003K6 | 1,596,589,605,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735906.77/warc/CC-MAIN-20200805010001-20200805040001-00209.warc.gz | 336,638,692 | 3,757 | ## Hidden Skyscrapers Sudoku
(Eingestellt am 31. Mai 2020, 12:00 Uhr von Carrick22)
Hidden Skyscapers Sudoku
• A tribute to the two (at the time of writing this) "Inner Sky" puzzles by @glum_hippo.
Apply classic Sudoku rules : Each row, column and bold-outlined region must contain the digits 1-9 once.
• Furthermore, each digit represents a skyscraper of the given height. A cell contains one or more arrows if and only if it correctly indicates the numbers of visible skyscrapers in the direction of the given arrow(s). Higher skyscrapers block the view of lower ones. NB : All possible arrows are given.
You can use Penpa-edit for solving it on your browser.
Lösungscode: Row 2, followed by row 7.
Zuletzt geändert am 31. Mai 2020, 12:00 Uhr
Gelöst von Uhu, dm_litv, zorant, ManuH, Julianl, japoorva, SirWoezel, marcmees, ch1983, Statistica, r45, MartinR, kishy72, KlausRG, moss, flaemmchen, Mody, ProwlingTiger, NikolaZ, sf2l, rimodech, lutzreimer, skywalker, Hasenvogel, Workout9, bob, Richard, geronimo92, amitsowani, marsigel, anu_chakravarti, ffricke, ArchonE
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### Kommentare
am 11. Juni 2020, 18:45 Uhr von Workout9
It looked easy... But it was complicated! Never give up!
am 2. Juni 2020, 14:30 Uhr von glum_hippo
Tried, failed. Fail again, fail better. But I do like the puzzle so far, and I appreciate the tribute! So easy to make a mistake!
am 31. Mai 2020, 12:00 Uhr von Carrick22
Some typos fixing and link provided.
Schwierigkeit: Bewertung: 92 % Gelöst: 33 mal Beobachtet: 0 mal ID: 0003K6
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Apr 7, 2017
$\theta = 4 \text{ radians}$
Explanation:
Given: $s = 20 \text{ radian cm}$ $d = 10 \text{ cm}$
$r = \frac{d}{2}$
$r = \frac{10 \text{ cm}}{2}$
$r = 5 \text{ cm}$
Using the formula:
$s = \theta r$
$\theta = \frac{s}{r}$
theta = (20" radian cm")/(5" cm"#
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A.1 INTRODUCTION
This appendix contains the inlet control equations used to develop the design charts of this publication (HDS-5). Section A.2 contains the equations for the unsubmerged and submerged inlet control equations. Section A.3 demonstrates how the Section A.2 equations are used to create dimensionless design curves for culvert shapes with coefficients (Section A.3.1) and without (Section A.3.2). Section A.4 discusses how the dimensionless design curves are used to develop the nomographs in Appendix C. Section A.5 discusses how the dimensionless design curves are used to develop the polynomial equations used in FHWA software.
A.2 INLET CONTROL EQUATIONS
The equations used to develop the inlet control nomographs in Appendix C are based on the research conducted by the National Bureau of Standards (NBS) under the sponsorship of the Bureau of Public Roads (now the Federal Highway Administration). John L. French of the NBS produced seven progress reports as a result of this research. Of these, the first (NBS 1955) and fourth (NBS 1961) through seventh reports (NBS 1966a, 1966b, and 1967) dealt with the hydraulics of pipe and box culvert entrances, with and without tapered inlets. Herbert G. Bossy of the FHWA provides an excellent synthesis of the research in his paper, “Hydraulic of conventional Highway Culverts” (Bossy 1961). Additional background on the development of the equations is found in HEC-13 (FHWA 1972a) and unpublished notebooks and notes (Bossy 1963 and Normann 1974).
The two basic conditions of inlet control depend upon whether the inlet end of the culvert is or is not submerged by the upstream headwater. If the inlet is not submerged by the headwater, the inlet performs as a weir and the unsubmerged equations are used (Section A.2.1). If the inlet is submerged by the headwater, the inlet performs as an orifice and the submerged equations are used (Section A.2.2).
Between the unsubmerged and the submerged conditions, there is a transition zone for which the NBS research provided only limited information. The transition zone is defined empirically by drawing a curve between and tangent to the curves defined by the unsubmerged and submerged equations. In most cases, the transition zone is short and the curve is easily constructed.
A.2.1 Unsubmerged Inlet Control Equations
The unsubmerged equation has two forms. Form (1) is based on the specific head at critical depth, adjusted with two correction factors. Form (2) is an exponential equation similar to a weir equation. Form (1) is preferable from a theoretical standpoint, but Form (2) is easier to apply and is the only documented form of equation for some of the inlet control equations. Equations (A.1) and (A.2) apply up to about Q/(A*D0.5) = 3.5 (1.93 SI).
Form (1) HWi/D = Hc/D + K*[(Ku*Q)/(A*D0.5)]M + Ks*S (A.1) Form (2) HWi/D = K*[(Ku*Q)/(A*D0.5)]M (A.2)
Where: HWi Headwater depth above inlet control section invert, m (ft) D Interior height of culvert barrel, ft (m) Hc Specific head at critical depth (dc + Vc2/(2*g)), ft (m) Q Discharge, ft3/s (m3/s) A Full cross sectional area of culvert barrel, ft2 (m2) S Culvert barrel slope, ft/ft (m/m) K, M, c, Y Constants from Tables A.1, A.2, A.3 Ku Unit conversion 1.0 (1.811 SI) Ks Slope correction, -0.5 (mitered inlets +0.7)
A.2.2 Submerged Inlet Control Equations
The submerged equation (A.3) applies above about Q/(A*D0.5) = 4.0 (2.21 SI). The terms are defined in Section A.2.1.
HWi/D = c*[(Ku*Q)/(A*D0.5)]2 + Y + Ks*S (A.3)
A.3 INLET CONTROL DIMENSIONLESS DESIGN CURVES
The equations in Section A.2 may be used to develop design curves for any conduit shape or size. Careful examination of the equation constants for a given form of equation reveals that there is very little difference between the constants for a given inlet configuration. Therefore, given the necessary conduit geometry for a new shape from the manufacturer, a similar shape is chosen and the constants are used to develop new design curves. The curves may be quasi- dimensionless, in terms of Q/(A*D0.5) and HWi/D, or dimensional, in terms of Q and HWi for a particular conduit size. To make the curves truly dimensionless, Q/(A*D0.5) must be divided by g0.5, but this produces small decimal numbers. Note that coefficients for rectangular (box) shapes should not be used for nonrectangular (circular, arch, pipe-arch, etc.) shapes and vice- versa. A constant slope value of 2 percent (0.02) is usually selected for the development of design curves. This is because the slope effect is small and the resultant headwater is conservatively high for sites with slopes exceeding 2 percent (except for mitered inlets). The procedure is illustrated in Section A.3.1.
A.3.1 Elliptical Structural Plate Example
Develop a dimensionless design curve for elliptical structural plate corrugated metal culverts, with the long axis horizontal. Assume a thin wall projecting inlet. Use the coefficients and exponents for a corrugated metal pipe-arch, a shape similar to an ellipse.
From Table A.1, Chart 34, Scale 3:
• Unsubmerged: equation Form (1) with K = 0.0340 M = 1.5
• Submerged: c = 0.0496 and Y = 0.53
Unsubmerged, equation Form (1) (Equation A.1):
HWi/D = Hc/D + 0.0340*(Q/(A*D0.5))1.5 – 0.5*0.02
Submerged (Equation A.3):
HWi/D = 0.0496*(Q/(A*D0.5))2 + 0.53 – 0.5*0.02
A direct relationship between HWi/D and Q/(A*D0.5) may be obtained for the submerged condition. For the unsubmerged condition, it is necessary to obtain the flow rate and equivalent specific head at critical depth. At critical depth, the critical velocity head is equal to one-half the hydraulic depth.
Vc2/(2*g) = yh/2 = Ap/2*Tp
Therefore, specific head at critical depth divided by D is:
Hc/D = dc/D + yh/(2*D) (A.4)
Since the Froude number equals 1.0 at critical depth, Vc can be determuned from the Froude number equation and set equal to Vc in the continuity equation to solve for Qc.
Fr = Vc/(g*yh)0.5 = 1, and Vc = Qc/Ap = (g*yh)0.5, from which Qc = Ap*(g*yh)0.5, or
Qc/(A*D0.5) = Ap/A*(g•(yh/D))0.5 (A.5)
From geometric data supplied by the manufacturer for a horizontal ellipse (Kaiser 1984), the necessary geometry is obtained to calculate Hc/D and Qc/(A*D0.5).
dc/D yh/D (Equation A.4)
Hc/D
Ap/A (Equation A.5)
Qc/(A*D0.5)
0.1 0.04 0.12 0.04 0.05
0.2 0.14 0.27 0.14 0.30
0.4 0.30 0.55 0.38 1.18
0.6 0.49 0.84 0.64 2.54
0.8 0.85 1.22 0.88 4.60
0.9 1.27 1.53 0.97 6.20
1.0 1.00
From unsubmerged equation and the above table:
Qc/(A*D0.5) 0.0340*
(Qc/(A*D0.5)1.5
+Hc/D -0.5*S = HWi/D
0.05 0.0004 0.12 0.01 0.11
0.30 0.0054 0.27 0.01 0.27
1.18 0.044 0.55 0.01 0.58
2.54 0.138 0.84 0.01 0.97
4.60 0.336 1.22 0.01 1.54
6.20 0.525 1.53 0.01 2.05
For the submerged equation, any value of Q/(A*D0.5) may be selected, since critical depth is not involved:
Qc/(A*D0.5) 0.0496*
(Qc/(A*D0.5))2
+Y -0.5*S = HWi/D
1 0.05 0.53 0.01 *0.57
2 0.20 0.53 0.01 *0.72
4 0.79 0.53 0.01 1.31
6 1.79 0.53 0.01 2.31
8 3.17 0.53 0.01 3.69
*Obviously unsubmerged
Note that overlapping values of HWi/D were calculated in order to define the transition zone between the unsubmerged and the submerged states of flow. The results of the above calculations are plotted in Figure A.1. A transition line is drawn between the unsubmerged and the submerged curves. The scales are dimensionless in Figure A.1, but the figures could be used to develop dimensional curves for any selected size of elliptical conduit by multiplying: Q/(A*D0.5) by A*D0.5 and HWi/D by D.
x
Figure A.1. Dimensionless performance curve for structural plate elliptical conduit, long axis horizontal, thin wall projecting entrance.
A.3.2 Dimensionless Design Charts for Culverts without Coefficientc
The dimensionless inlet control design charts provided for long span arches (Chart 52) and for circular and elliptical pipes (Chart 51) were derived using the inlet control equations in Section A.2, selected constants from Table A.1, conduit geometry obtained from various tables and manufacturer’s information (FHWA 1972b, Kaiser 1984, AISI 1983).
Some inlet configurations have no hydraulic tests. In lieu of such tests, the selected edge conditions should approximate the untested configurations and lead to a good estimate of culvert performance. In some cases, it will be necessary to evaluate the inlet edge configuration at a specific flow depth. For example, some inlets may behave as mitered inlets at low headwaters and as thin wall projecting inlets at high headwaters. The designer must apply engineering judgment in selection of the proper relationships for these major structures.
Unsubmerged Conditions. Equation (A.1) was used to calculate HWi/D for selected inlet edge configurations. The following constants were taken from Table A.1, Chart 34 for pipe- arches, except for the 45 degree beveled edge inlet. These constants were taken from Chart 3, Scale A, for circular pipe. No constants were available from tests on pipe-arch models with beveled edges.
Inlet Edge K M Ku*S
Thin Wall Projecting 0.0340 1.5 -0.01
Mitered to Embankment 0.0300 1.0 +0.01
Square Edge in Headwall 0.0083 2.0 -0.01
Beveled Edge (45° Bevels) 0.0018 2.5 -0.01
Geometric relationships for the circular and elliptical (long axis horizontal) conduits were obtained from Tables 4 and 7 (FHWA 1972b), respectively. Geometric relationships for the high and low profile long span arches were obtained from DP-131 (Kaiser 1984) and the results were checked against tables in AISI handbook (AISI 1983).
Submerged Conditions. Equation (A.3) was used to calculate HWi/D for the same inlet configurations using the following constants:
Inlet Edge c Y Ku*S
Thin Wall Projecting 0.0496 0.53 -0.01
Mitered to Embankment 0.0463 0.75 +0.01
Square Edge in Headwall 0.0496 0.57 – 0.01
Beveled Edge (45° Bevels) 0.0300 0.74 – 0.01
In terms of Q/(A*D0.5), all non-rectangular shapes have practically the same dimensionless curves for submerged, inlet control flow. This is not true if Q/(B*D1.5) is used as the dimensionless flow parameter.
To convert Q/(B*D1.5) to Q/(A*D0.5), divide by A/(B*D) for the particular shape of interest as shown in Equation (A.6). This assumes that the shape is geometrically similar, so that A/(B*D) is nearly constant for a range of sizes.
(Q/(B*D1.5))/(A/(B*D)) = (Q/(B*D1.5))*(B*D/A) = Q/(A*D0.5) (A.6)
Dimensionless Curves. By plotting the results of the unsubmerged and submerged calculations and connecting the resultant curves with transition lines, the dimensionless design curves shown in Charts 51 and 52 of Appendix C were developed. All high and low profile arches can be represented by a single curve for each inlet edge configuration. A similar set of curves was developed for circular and elliptical shapes. It is recommended that the high and low profile arch curves in Chart 52 be used for all true arch shapes (those with a flat bottom) and that the curves in Chart 51 be used for curved shapes including circles, ellipses, pipe-arches, and pear shapes.
A.3.3 Dimensionless Critical Depth Charts
Some of the long span culverts and special culvert shapes have no critical depth charts. These special shapes are available in numerous sizes, making it impractical to produce individual critical depth curves for each culvert size and shape. Therefore, dimensionless critical depth curves were developed for the shapes which have adequate geometric relationships in the manufacturer’s literature. It should be noted that these special shapes are not truly geometrically similar, and any generalized set of geometric relationships will involve some degree of error. The amount of error is unknown since the geometric relationships were developed by the manufacturers.
The manufacturers’ literature contains geometric relationships which include the hydraulic depth divided by the rise (inside height) of the conduit (yh/D) and area of the flow prism divided by the barrel area (Ap/A) for various partial depth ratios, y/D. From Equation (A.5):
Q/(A*D0.5) = Ap/A*(g•yh/D)0.5 (A.7)
Setting y/D equal to dc/D, it is possible to determine Ap/A and yh/D at a given relative depth and then to calculate Qc/(A*D0.5). Dimensionless plots of dc/D versus Qc/(A*D0.5) have been developed for the following culvert materials and shapes:
• Chart 20, corrugated metal box culverts (see Second edition HDS 5)
• Chart 44, corrugated metal arches (see Second edition HDS 5)
• Chart 53, Structural plate corrugated metal ellipses, long axis horizontal
• Chart 54, Structural plate corrugated metal arches, low and high profile
A.4 INLET CONTROL NOMOGRAPHS
The nomographs in Appendix C were developed using the equations in Section A.2 and the constants shown in Table A.1. The unsubmerged and submerged equations for a given shape, material and edge configuration were plotted using the dimensionless design curve procedures shown in Section A.3.1. A constant slope value of 2 percent (0.02) was used for the development of these design curves. This is because the slope effect is small and the resultant headwater is conservatively high for sites with slopes exceeding 2 percent (except for mitered inlets). A smooth transition was drawn by hand. This curve was the data used for constructing a nomograph. Dr. F. T. Mavis describes the process of making nomographs in “The Construction of Nomographic Charts” (Mavis 1939). Nomographs were used extensively in engineering prior to the introduction of microcomputers in the early 1980s.
In formulating inlet and outlet control design nomographs, a certain degree of error is introduced into the design process. This error is due to the fact that the nomograph construction involves graphical fitting techniques resulting in scales which do not exactly match the equations. Checks by the authors of the first edition and others indicate that all of the nomographs from HEC-5 have precisions of + 10 percent of the equation values in terms of headwater (inlet control) or head loss (outlet control). The nomographs for tapered inlets have errors of less than 5%, again in terms of headwater.
A.5 INLET CONTROL POLYNOMIAL EQUATIONS
The polynomial equations were developed to be used in software. The equations in Section A.2 with the constants shown in the tables of constants for a given shape, material and edge configuration were plotted using the dimensionless design curve procedures shown in Section A.3.1. The coordinates of selected points can be read from the curve and a best fit statistical analysis performed. A polynomial equation of the following form has been found to provide an adequate fit.
HWi/D = A + B*[Q/(B*D1.5)] + C*[Q/(B*D1.5)]2 + … + X*[Q/(B*D1.5)]n + Ks*S
For fitting the polynomial equations, Ks = 0 was used for most equations so that the slope correction could be applied by the software. The flow factor can be based on A*D0.5 rather than B*D1.5. The constants for the best fit equations are found in the HY-8 User Manual provided with HY-8. For equations that included Ks*S, the A term is adjusted so that Ku*S = 0. HY-8 uses the polynomial equations for all shapes that have constants determined in the laboratory or by FHWA. These include:
• Table A.1 – circles, boxes and tapered inlets (NBS, Bossy 1961)
• Table A.2 – pipe-arches, ellipses, metal boxes and arches (Bossy 1961)
• Table A.3 – South Dakota DOT RCB (FHWA 2006c)
• Table A.4 – open bottom concrete boxes (Chase 1999)
• Table A.5 – embedded circular shapes (NCHRP 2011)
• Table A.6 – embedded elliptical shapes (NCHRP 2011)
For shapes without constants, HY-8 uses Chart 52 developed using the procedures of Section A.3.2.
Note From HDS-5 Section 3.1.3 Inlet Control
The original equations for computer software were generally 5th order polynomial curve-fitted equations that were developed to be as accurate as the nomograph solution (plus or minus 10%) within the headwater range of 0.5*D to 3.0*D. These equations are still being used in HY-8, but have been supplemented with a weir equation from 0.0*D to 0.5*D and an orifice equation above 3.0*D.
Table A.1 Constants for Inlet Control Equations for Charts in Appendix G.
Chart Nomograph Equation Unsubmerged Submerged
No. Shape and Material Scale Inlet Configuration Form K M c Y References
1 Circular Concrete 1 Square edge w/ headwall 1 0.0098 2.0 0.0398 0.67 1, 2
1 Circular Concrete 2 Groove end w/ headwall 1 0.0018 2.0 0.0292 0.74 1, 2
1 Circular Concrete 3 Groove end projecting 1 0.0045 2.0 0.0317 0.69 1, 2
2 Circular CMP 1 Headwall 1 0.0078 2.0 0.0379 0.69 1, 2
2 Circular CMP 2 Mitered to slope 1 0.0210 1.33 0.0463 0.75 1, 2
2 Circular CMP 3 Projecting 1 0.0340 1.50 0.0553 0.54 1, 2
3 Circular A Beveled ring, 45° bevels 1 0.0018 2.50 0.0300 0.74 2
3 Circular B Beveled ring, 33.7° bevels* 1 0.0018 2.50 0.0243 0.83 2
8 Rect. Box Concrete 1 30° to 75° wingwall flares 1 0.026 1.0 0.0347 0.81 1, 3
8 Rect. Box Concrete 2 90° and 15° wingwall flares 1 0.061 0.75 0.0400 0.80 1, 3
8 Rect. Box Concrete 3 0° wingwall flares 1 0.061 0.75 0.0423 0.82 1, 3
9 Rect. Box Concrete 1 45° wingwall flare d = .043D 2 0.510 0.667 0.0309 0.80 3
9 Rect. Box Concrete 2 18° to 33.7° wingwall flare d = .083D 2 0.486 0.667 0.0249 0.83 3
10 Rect. Box Concrete 1 90° headwall w/3/4″ chamfers 2 0.515 0.667 0.0375 0.79 3
10 Rect. Box Concrete 2 90° headwall w/45° bevels 2 0.495 0.667 0.0314 0.82 3
10 Rect. Box Concrete 3 90° headwall w/33.7° bevels 2 0.486 0.667 0.0252 0.865 3
11 Rect. Box Concrete 1 3/4″ chamfers; 45° skewed headwall 2 0.545 0.667 .04505 0.73 3
11 Rect. Box Concrete 2 3/4″ chamfers; 30° skewed headwall 2 0.533 0.667 0.0425 0.705 3
11 Rect. Box Concrete 3 3/4″ chamfers; 15° skewed headwall 2 0.522 0.667 0.0402 0.68 3
11 Rect. Box Concrete 4 45° bevels; 10°-45° skewed headwall 2 0.498 0.667 0.0327 0.75 3
12 Rect. Box 3/4″ chamf. Conc. 1 45° non-offset wingwall flares 2 0.497 0.667 0.0339 0.803 3
12 Rect. Box 3/4″ chamf. Conc. 2 18.4° non-offset wingwall flares 2 0.493 0.667 0.0361 0.806 3
12 Rect. Box 3/4″ chamf. Conc. 3 18.4° non-offset wingwall flares 30° skewed barrel 2 0.495 0.667 0.0386 0.71 3
13 Rec. Box Top Bev. Conc. 1 45° wingwall flares – offset 2 0.497 0.667 0.0302 0.835 3
13 Rec. Box Top Bev. Conc. 2 33.7° wingwall flares – offset 2 0.495 0.667 0.0252 0.881 3
13 Rec. Box Top Bev. Conc. 3 18.4° wingwall flares – offset 2 0.493 0.667 0.0227 0.887 3
55 Circular 1 Smooth tapered inlet throat 2 0.534 0.555 0.0196 0.90 4
55 Circular 2 Rough tapered inlet throat 2 0.519 0.64 0.0210 0.90 4
56 Elliptical Face 1 Tapered inlet – beveled edges 2 0.536 0.622 0.0368 0.83 4
56 Elliptical Face 2 Tapered inlet – square edges 2 0.5035 0.719 0.0478 0.80 4
56 Elliptical Face 3 Tapered inlet – thin edge projecting 2 0.547 0.80 0.0598 0.75 4
57 Rectangular Concrete 1 Tapered inlet throat 2 0.475 0.667 0.0179 0.97 4
58 Rectangular Concrete 1 Side tapered – less favorable edges 2 0.56 0.667 0.0446 0.85 4
58 Rectangular Concrete 2 Side tapered – more favorable edges 2 0.56 0.667 0.0378 0.87 4
59 Rectangular Concrete 1 Slope tapered – less favorable edges 2 0.50 0.667 0.0446 0.65 4
59 Rectangular Concrete 2 Slope tapered – more favorable edges 2 0.50 0.667 0.0378 0.71 4
1Bossy 1963, 2FHWA 1974, 3NBS 5th, 4HEC-13
Table A.2 Constants for Inlet Control Equations for Discontinued Charts (see HDS-5).
Chart Nomograph Equation Unsubmerged Submerged
No. Shape and Material Scale Inlet Configuration Form K M c Y References
16-19 Boxes CM 2 90° headwall 1 0.0083 2.0 0.0379 0.69 1
16-19 Boxes CM 3 Thick wall projecting 1 0.0145 1.75 0.0419 0.64 1
16-19 Boxes CM 5 Thin wall projecting 1 0.0340 1.5 0.0496 0.57 1
29 Horizontal Ellipse Concrete 1 Square edge w/ headwall 1 0.0100 2.0 0.0398 0.67 1
29 Horizontal Ellipse Concrete 2 Groove end w/ headwall 1 0.0018 2.5 0.0292 0.74 1
29 Horizontal Ellipse Concrete 3 Groove end projecting 1 0.0045 2.0 0.0317 0.69 1
30 Vertical Ellipse Concrete 1 Square edge w/ headwall 1 0.0100 2.0 0.0398 0.67 1
30 Vertical Ellipse Concrete 2 Groove end w/ headwall 1 0.0018 2.5 0.0292 0.74 1
30 Vertical Ellipse Concrete 3 Groove end projecting 1 0.0095 2.0 0.0317 0.69 1
34 Pipe Arch 18″ Corner radius CM 1 90° headwall 1 0.0083 2.0 0.0379 0.69 1
34 Pipe Arch 18″ Corner radius CM 2 Mitered to slope 1 0.0300 1.0 0.0463 0.75 1
34 Pipe Arch 18″ Corner radius CM 3 Projecting 1 0.0340 1.5 0.0496 0.57 1
35 Pipe Arch 18″ Corner radius CM 1 Projecting 1 0.0300 1.5 0.0496 0.57 2
35 Pipe Arch 18″ Corner radius CM 2 No Bevels 1 0.0088 2.0 0.0368 0.68 2
35 Pipe Arch 18″ Corner radius CM 3 33.7° Bevels 1 0.0030 2.0 0.0269 0.77 2
36 Pipe Arch 31″ Corner radius CM 1 Projecting 1 0.0300 1.5 0.0496 0.57 2
36 Pipe Arch 31″ Corner radius CM 2 No Bevels 1 0.0088 2.0 0.0368 0.68 2
36 Pipe Arch 31″ Corner radius CM 3 33.7° Bevels 1 0.0030 2.0 0.0269 0.77 2
41-43 Arch CM 1 90° headwall 1 0.0083 2.0 0.0379 0.69 1
41-43 Arch CM 2 Mitered to slope 1 0.0300 1.0 0.0463 0.75 1
41-43 Arch CM 3 Thin wall projecting 1 0.0340 1.5 0.0496 0.57 1
1FHWA 1974, 2Bossy 1963
Table A.3. Constants for Inlet Control Equations for South Dakota Concrete Box (HY-8 User Manual and Table 11 of FHWA 2006).
Sketch Wingwall Flare Top Bevel Top Radius Corner Fillet RCB Inlet Configuration Equation Form Unsub- merged K Unsub- merged M Sub- merged c Sub- merged Y
1 30° 45° Single barrel 2 0.44 0.74 0.040 0.48
2 30° 45° 6″ Multiple barrel (2, 3, and 4 cells) 2 0.47 0.68 0.04 0.62
3 30° 45° Single barrel (2:1 to 4:1 span-to-rise ratio) 2 0.48 0.65 0.041 0.57
4 30° 45° Multiple barrels (15°skewed headwall) 2 0.69 0.49 0.029 0.95
5 30° 45° Multiple barrels (30° to 45° skewed headwall) 2 0.69 0.49 0.027 1.02
6 none Single barrel, top edge 90° 2 0.55 0.64 0.047 0.55
7 45° 6″ Single barrel, (0 and 6-inch corner fillets) 2 0.56 0.62 0.045 0.55
8 45° 6″ Multiple barrels (2, 3, and 4 cells) 2 0.55 0.59 0.038 0.69
9 45° Single barrels 2:1 to 4:1 span-to-rise ratio) 2 0.61 0.57 0.041 0.67
10 8″ 6″ Single barrel (0 and 6-inch fillets) 2 0.56 0.62 0.038 0.67
11 8″ 12″ Single barrel (12-inch corner fillets) 2 0.56 0.62 0.038 0.67
12 8″ 12″ Multiple barrels (2, 3, and 4 cells) 2 0.55 0.6 0.023 0.96
13 8″ 12″ Single barrel (2:1 to 4:1 span-to-rise ratio) 2 0.61 0.57 0.033 0.79
Sketches are shown in the HY-8 documentation and research report. Since sketches 2 and 8 show fillets, a 6-inch fillet is assumed.
Sketches 1 through 5 have the first configuration. Sketches 7 through 13 have the second.
Table A.4. Constants for Inlet Control Equations for Concrete Open-Bottom Arch (Chase 1999).
Span to Rise1 Wingwall Flare Top Edge Inlet Configuration Equation Form Unsub- merged K Unsub- merged M Sub- merged c Sub- merged Y
2:1 90° Mitered to conform to slope 2 0.44 0.74 0.040 0.48
2:1 45° 90° Headwall with wingwalls 2 0.47 0.68 0.04 0.62
2:1 90° 90° Headwall 2 0.48 0.65 0.041 0.57
4:1 90° Mitered to conform to slope 2 0.69 0.49 0.029 0.95
4:1 45° 90° Headwall with wingwalls 2 0.69 0.49 0.027 1.02
4:1 90° 90° Headwall 2 0.56 0.62 0.045 0.55
1The 2:1 constants are used for ratios less than or equal to 3:1 and the 4:1 constants for ratios greater than 3:1.
Table A.5. Constants for Inlet Control Equations for Embedded Circular Shapes (NCHRP 15-24).
Embedded Top Edge Inlet Configuration Unsub- merged K Form 1 Unsub- merged M Form 1 Unsub- merged K Form 2 Unsub- merged M Form 2 Sub- merged c Sub- merged Y
0.2D thin Projecting End, Ponded 0.0860 0.58 0.4293 0.64 0.0303 0.58
0.2D thin Projecting End, Channelized 0.0737 0.45 0.4175 0.62 0.0250 0.63
0.2D Mitered End 1.5H:1V 0.0431 0.58 0.4002 0.63 0.0235 0.61
0.2D 90° Square Headwall 0.0566 0.44 0.4001 0.63 0.0198 0.69
0.2D 45° Beveled End 0.0292 0.57 0.3869 0.63 0.0161 0.73
0.4D thin Projecting End, Ponded 0.0840 0.76 0.4706 0.69 0.0453 0.69
0.4D thin Projecting End, Channelized 0.0927 0.59 0.4789 0.66 0.0441 0.52
0.4D Mitered End 1.5H:1V 0.0317 0.77 0.4185 0.68 0.0363 0.65
0.4D 90° Square Headwall 0.0490 0.71 0.4354 0.68 0.0332 0.67
0.4D 45° Beveled End 0.0358 0.62 0.4223 0.67 0.0245 0.75
0.5D thin Projecting End, Ponded 0.1057 0.69 0.4955 0.71 0.0606 0.54
0.5D thin Projecting End, Channelized 0.1055 0.59 0.4955 0.69 0.0570 0.48
0.5D Mitered End 1.5H:1V 0.0351 0.59 0.4419 0.68 0.0504 0.44
0.5D 90° Square Headwall 0.0595 0.59 0.0595 0.59 0.0402 0.65
0.5D 45° Beveled End 0.0464 0.46 0.4364 0.69 0.0324 0.67
Table A.6. Constants for Inlet Control Equations for Embedded Elliptical Shape (NCHRP 15-24).
Embedded Top Edge Inlet Configuration Unsubmerged K Form1 Unsubmerged M Form 1 Unsubmerged K Form 2 Unsubmerged M Form 2 Submerged c Sub- merged Y
0.5D thin Projecting End, Ponded 0.1231 0.51 0.5261 0.65 0.0643 0.50
0.5D thin Projecting End, Channelized 0.0928 0.54 0.4937 0.67 0.0649 0.12
0.5D Mitered End 1.5H:1V 0.0599 0.60 0.4820 0.67 0.0541 0.50
0.5D 90° Square Headwall 0.0819 0.45 0.4867 0.66 0.0431 0.61
0.5D 45° Beveled End 0.0551 0.52 0.4663 0.63 0.0318 0.68
8,647 total views, 3 views today | 8,580 | 24,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-23 | latest | en | 0.913225 |
https://oeis.org/A029069 | 1,675,064,469,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00548.warc.gz | 454,260,004 | 3,713 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A029069 Expansion of 1/((1-x)*(1-x^4)*(1-x^5)*(1-x^12)). 1
1, 1, 1, 1, 2, 3, 3, 3, 4, 5, 6, 6, 8, 9, 10, 11, 13, 15, 16, 17, 20, 22, 24, 25, 29, 32, 34, 36, 40, 44, 47, 49, 54, 58, 62, 65, 71, 76, 80, 84, 91, 97, 102, 106, 114, 121, 127, 132, 141, 149, 156, 162, 172, 181, 189 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,5 COMMENTS Number of partitions of n into parts 1, 4, 5 and 12. - Ilya Gutkovskiy, May 17 2017 LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,0,-1,0,0,-1,1,0,1,-1,0,0, -1,0,1,0,0,1,-1). MATHEMATICA CoefficientList[Series[1/((1-x)(1-x^4)(1-x^5)(1-x^12)), {x, 0, 60}], x] (* Harvey P. Dale, Aug 11 2011 *) PROG (PARI) x='x+O('x^50); Vec(1/((1 - x)*(1 - x^4)*(1 - x^5)*(1 - x^12))) \\ G. C. Greubel, May 17 2017 CROSSREFS Sequence in context: A120504 A025772 A211535 * A266341 A279521 A076896 Adjacent sequences: A029066 A029067 A029068 * A029070 A029071 A029072 KEYWORD nonn AUTHOR STATUS approved
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Last modified January 30 02:40 EST 2023. Contains 359939 sequences. (Running on oeis4.) | 622 | 1,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-06 | latest | en | 0.629988 |
https://studyqas.com/acomputer-valued-at-6500-depreciates-at-the-rate-of-14-3/ | 1,669,511,704,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710155.67/warc/CC-MAIN-20221127005113-20221127035113-00278.warc.gz | 582,684,403 | 40,454 | # Acomputer valued at $6500 depreciates at the rate of 14.3% per year. find the value (round Acomputer valued at$6500 depreciates at the rate of 14.3% per year. find the value (round to the nearest dollar) of the computer after three years.
## This Post Has 5 Comments
1. NeverEndingCycle says:
4091.25
You can get this number by multiplying by .857, which is the amount remaining after taking off 14.3%. Do this for all 3 years to get the answer above.
2. Expert says:
d.
121π mi^2
hope you got it chief
3. Expert says:
answer: (c) m∠qpo + (2x + 16)° = 180°
step-by-step explanation:
the opposite angles of a quadrilateral are supplementary.
so, ∠o + ∠q = 180° and ∠p + ∠r = 180°
since ∠r = 2x + 16°, we can use substitution as follows:
∠p + ∠r = 180°
∠p + (2x + 16)° = 180°
4. solikhalifeoy3j1r says:
After 3 years the computer will be valued at 3711.5. I'm sure you can round it up to 3712.
Hope this helps 🙂
5. yarrito20011307 says:
Given that,
Value of a computer= $6500 Depreciation rate= 14.3% Now, depreciation after 1st year= 6500 x 14.3% Depreciation after 1st year= 929.5 Value of a computer after 1 year = 6500 – 929.5= 5570.5 Depreciation after 2nd year= 5570.5 x 14.3% Depreciation after 2nd year= 796.58 Value of the computer after 2 year=5570.5-796.58=4773.92 Depreciation after 3rd year= 4773.92 x 14.3% Depreciation after 3rd year= 682.67 Value of the computer after 3 year= 4773.92-682.67= 4091.25 Therefore, the value of the computer after 3 year is$4091 approximately. | 519 | 1,515 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-49 | latest | en | 0.883349 |
https://optionegdugl.netlify.app/kirchner70059ni/how-to-find-the-future-value-of-cash-flows-fen | 1,713,662,733,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817699.6/warc/CC-MAIN-20240421005612-20240421035612-00051.warc.gz | 399,360,146 | 12,045 | ## The future value, FV , of a series of cash flows is the future value, at future time N (total periods in the future), of the sum of the future values of all cash flows, CF.
Determining the appropriate discount rate is the key to valuing future cash flows properly, whether
How to Determine Future Value of Cash Flows. Cash flows are one-time or periodic inflows of money, such as dividends, or outflows, such as tuition expenses. 21 Jun 2019 Determining the appropriate discount rate is the key to properly valuing future cash flows, whether they be earnings or obligations. Calculate the future value (FV) of an investment of \$500 for a period of 3 years that pays an interest rate of 6% compounded semi-annually. FV = 500*(1+6%/2)^ (2* Concept 1: Calculating PV and FV of Different Cash Flows. Present value is the current value of a future cash flow. Longer the time period till the future amount is Define the present value of a series of cash flows. Define an annuity. Identify the factors you need to know to calculate the value of an annuity. Discuss the | 253 | 1,070 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-18 | latest | en | 0.889978 |
https://mathoverflow.net/questions/368067/smallest-ordinal-alpha-such-that-l-cap-pl-alpha-is-uncountable | 1,675,265,724,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499946.80/warc/CC-MAIN-20230201144459-20230201174459-00563.warc.gz | 402,648,999 | 26,801 | # smallest ordinal $\alpha$ such that $L \cap P(L_\alpha)$ is uncountable
Let $$V$$ denote the von Neumann universe and $$L$$ Gödel's constructible universe. For any set $$X$$, let $$P(X)$$ denote the power set of $$X$$.
Assume that $$0^\sharp$$ exists (and ZFC).
What is the smallest ordinal $$\alpha$$ such that $$L \cap P(L_{\alpha})$$ is uncountable? (If $$V = L$$, then $$\alpha = \omega$$, but if $$0^\sharp$$ exists, then $$\alpha > \omega$$.)
• It’s $\omega_1$. Aug 1, 2020 at 5:46
• I suspected so. Do you have a reference for a theorem that implies this? Aug 1, 2020 at 5:55
• If $0^\sharp$ exists, then every cardinal is inaccessible in L. Aug 1, 2020 at 6:04
• Ah, so $L \cap V_\alpha$ is countable for all countable $\alpha$, too, since $L \cap V_\alpha$ is the $V_\alpha$ of $L$. Thank you. Aug 1, 2020 at 6:08
• In ZFC alone, the $\alpha$ in the title can be described as: (1) If genuine $\omega_1$ is a successor cardinal of $L$, then $\alpha$ is its immediate predecessor cardinal of $L$. (2) If genuine $\omega_1$ is a limit (and therefore inaccessible) cardinal of $L$, then it is equal to $\alpha$. The additional hypothesis that $0^\#$ exists implies that case (2) occurs. Aug 1, 2020 at 14:22
Really, this was answered in the comments; I'm putting this answer down to move this off the unanswered queue. I've made this CW and will delete it if one of the original commenters adds their own answer.
We have in $$L$$, for each (infinite) $$\alpha$$, the following bijections:
• $$f_\alpha:\alpha\rightarrow L_\alpha$$.
• $$g_\alpha: \mathcal{P}(L_\alpha)^L=\mathcal{P}(L_\alpha)\cap L\rightarrow L_{(\vert\alpha\vert^+)^L}$$.
Hence $$\vert\mathcal{P}(L_\alpha)^L\vert=\vert(\vert\alpha\vert^+)^L\vert$$. Now assuming $$0^\sharp$$ we have that $$\omega_1^V$$ is a limit cardinal in $$L$$, so for each $$\alpha<\omega_1^V$$ we have $$\vert\mathcal{P}(L_\alpha)^L\vert=\aleph_0$$.
So the answer to your question is $$\omega_1^V$$.
Note that all this requires is that $$\omega_1^V$$ be a limit cardinal in $$L$$. More generally, let $$\kappa$$ be the supremum of the $$L$$-cardinals whose $$L$$-successor is $$<\omega_1^V$$; then the $$\kappa$$th level of $$L$$ is the first whose $$L$$-powerset is truly uncountable.
• OK, except I think you meant $P(L_\alpha)^L = L \cap P(L_\alpha)$, not $P(L_\alpha)^L = P(\alpha) \cap L$. Aug 3, 2020 at 0:30
• @JesseElliott Whoops, quite right - fixed! Aug 3, 2020 at 0:36 | 810 | 2,439 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 32, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-06 | latest | en | 0.828779 |
https://mormonconferences.org/aa/1-2-in-cedar-siding-r-value.html | 1,621,122,372,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991488.53/warc/CC-MAIN-20210515223209-20210516013209-00498.warc.gz | 415,505,946 | 10,259 | # 1 2 In Cedar Siding R Value
### R value table common building materials siding materials hardboard 1 2 0 34 plywood 5 8 0 77 3 4 0 93 wood bevel lapped 0 80 aluminum steel vinyl.
1 2 in cedar siding r value.
R value measures how well certain building insulation materials can resist heat. Most siding materials have an r value of less than 1. Red cedar finger joint primed bevel siding grip rite 13 x 2 in. The rigid board insulation does add significantly to the insulation of the exterior wall.
The edges of the studs are about 25 of the surface of the exterior wall. In simple terms r values measure the effectiveness of insulation and a higher number represents more effective insulation. White cedar has an r value of about r 1 4 per inch so it isn t too hard to calculate the r value of white cedar siding. 6d stainless steel ring shank siding nails 1 lb.
R value see table 2. I suggest using a 1 inch thick panel of rigid foam. This adds an r value of about 5 or 7 and insulates over the edges of all the wall framing studs. For instance if you have a material with an r value of 12 attached to another material with an r value of 3 then both materials combined have an r value of 15.
Vinyl composite sill moulding. If we re talking about cedar shingles there are usually a maximum of three layers of shingles at any one point in the wall. 30 divide this per inch numberby 2 to obtain the r value for 1 2 medium density fiberboard r 1 22. Hardiepanel siding cempanel siding.
The trickiest part of the calculation is determining the siding thickness. That includes standard vinyl steel or aluminum siding without added foam insulation. 2 inch nominal 1 inch 1 88 2 x 4 3 inch 4 38 2 x 6 5 inch 6 88 cedar logs and lumber 1 33 sheathing materials plywood 1 25 inch 0 31 inch 0 47 inch 0 63 inch 0 77 inch 0 94 fiberboard 2 64 inch 1 32 inch 2 06 r value of building materials material r inch r thick ness. When determining how much insulation you need don t guesstimate calculate.
An r 15 or r 18 can be achieved by using 2 6 construction instead of 2 4 construction. Esr 1844 noncombustibility see page 1 section 3 1 flame spread and smoke developed see page 1 section 3 1 1 hour fire walls see page 2 section 4 3 design wind loads see page 4 5 table 4 r values see page 3 table 2. Fiberboard insulating sheathing regular density per inch.
### Dp 2420 Weathered Wood Siding Wood Siding Wall Exterior Wood Panel Siding
Source : pinterest.com | 642 | 2,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-21 | latest | en | 0.811405 |
https://watch-stores.com/clock/where-is-the-point-a-3-4-after-rotation-90-degrees-clockwise.html | 1,656,785,503,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104189587.61/warc/CC-MAIN-20220702162147-20220702192147-00413.warc.gz | 653,743,487 | 17,453 | # Where is the point A 3 4 after rotation 90 degrees clockwise?
Contents
Because (3,4) is in quadrant I and will end up in quadrant IV with a 90º clockwise rotation, the x-value must be positive and the y-value negative.
## How do you rotate a point 90 degrees clockwise about a point?
Answer: To rotate the figure 90 degrees clockwise about a point, every point(x,y) will rotate to (y, -x).
## What is the rule after a 90 degree clockwise rotation?
Rule : When we rotate a figure of 90 degrees clockwise, each point of the given figure has to be changed from (x, y) to (y, -x) and graph the rotated figure.
## What are the rules for 90 degrees clockwise and counterclockwise?
Here are the rotation rules: 90° clockwise rotation: (x,y) becomes (y,-x) 90° counterclockwise rotation: (x,y) becomes (y,x) 180° clockwise and counterclockwise rotation: (x, y) becomes (-x,-y)
## What is 90 degrees counterclockwise?
When we rotate a figure of 90 degrees counterclockwise, each point of the given figure has to be changed from (x, y) to (-y, x) and graph the rotated figure. Example 1 : If this triangle is rotated 90° counterclockwise, find the vertices of the rotated figure and graph. …
## What is 90 degrees in an angle?
right angle-an 90 degree angle. obtuse angle-an angle between 90 and 180 degrees. straight angle-a 180 degree angle.
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## What are the coordinates of the figure after a 90 clockwise rotation about the origin?
When the point is rotated through 90° clockwise about the origin, the point M (h, k) takes the image M’ (k, -h). Therefore, the new position of point M (-2, 3) will become M’ (3, 2). 2. Find the co-ordinates of the points obtained on rotating the point given below through 90° about the origin in clockwise direction. | 475 | 1,820 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2022-27 | latest | en | 0.875531 |
https://calculator.academy/adoption-capacity-calculator/ | 1,686,307,563,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00496.warc.gz | 171,629,354 | 54,153 | Enter the target length of stay (days) and the monthly daily average number of adoptions into the Calculator. The calculator will evaluate the Adoption Capacity.
AC = TLS * MDAA
Variables:
• AC is the Adoption Capacity ()
• TLS is the target length of stay (days)
• MDAA is the monthly daily average number of adoptions
To calculate Adoption Capacity, multiply the length of stay by the monthly daily average number of adoptions.
## How to Calculate Adoption Capacity?
The following steps outline how to calculate the Adoption Capacity.
1. First, determine the target length of stay (days).
2. Next, determine the monthly daily average number of adoptions.
3. Next, gather the formula from above = AC = TLS * MDAA.
4. Finally, calculate the Adoption Capacity.
5. After inserting the variables and calculating the result, check your answer with the calculator above.
Example Problem :
Use the following variables as an example problem to test your knowledge.
target length of stay (days) = 4
monthly daily average number of adoptions = 20 | 218 | 1,048 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-23 | latest | en | 0.851336 |
http://www.docstoc.com/docs/69198620/Continuity-at-a-Point-Worksheets | 1,386,830,644,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164561235/warc/CC-MAIN-20131204134241-00074-ip-10-33-133-15.ec2.internal.warc.gz | 416,346,317 | 15,915 | Continuity at a Point Worksheets
Document Sample
``` 2.3 Continuity AP Calculus
2.3 CONTINUITY
On page 25 of your notes from §2.1 we referred to “well behaved” functions. “Well behaved” functions allowed us to find
the limit by direct substitution. "Well behaved" functions turn out to be continuous functions. In this section we will
discuss continuity at a point, continuity on an interval, and the different types of discontinuities.
In non – technical terms, a function is continuous if you can draw the function “without ever lifting your pencil”. The
following graphs demonstrate three types of discontinuous graphs.
y y y
x x x
Let’s go back to the example we used in §2.1 when we discussed one – sided limits.
y
x
Example: Find the points (intervals) at which the function above is continuous, and the points at which it is discontinuous.
Example: For each point of discontinuity, c, find f (c), lim+ f ( x ) , lim− f ( x ) , and lim f ( x ) if they exist.
x→c x→c x →c
36
2.3 Continuity AP Calculus
Definition: Continuity at a Point
Interior Point: A function y = f (x) is continuous at an interior point c of its domain if
lim f ( x ) = f ( c )
x →c
: This last statement implies that lim f ( x ) exists. This limit only exists if the limit from the left and right of c are equal!
x →c
It also implies that the function value at c … f (c) exists.
Endpoint: A function y = f (x) is continuous at a left endpoint a or is continuous at a right endpoint b of its domain if
lim+ f ( x ) = f ( a ) or lim− f ( x ) = f ( b ) , respectively.
x→a x →b
Discontinuities: Removable versus Non-Removable
To say a function is discontinuous is not sufficient. We would like to know what type of discontinuity exists. If the
function is not continuous, but I could make it continuous by appropriately defining or redefining f(c), then we say that f
has a removable discontinuity. Otherwise, we say f has a non-removable discontinuity.
Once again, informally we say that f has a removable discontinuity if there is a “hole” in the function, but f has a non-
removable discontinuity if there is a “jump” or a vertical asymptote.
All polynomials are continuous. For rational functions, we try to algebraically “remove” the discontinuity by canceling
factors found in both the denominator and the numerator if possible.
Example: Which (if any) of the three graphs at the top of the other side of the paper have a removable discontinuity?
Example: Discuss the continuity of each function
1
(a) f ( x ) =
x −1
2x2 + x − 6
(b) g ( x ) =
x+2
⎧ −2 x + 3 ; x < 1
⎪
(c) h ( x ) = ⎨ 2
⎪x
⎩ ;x ≥1
37
2.3 Continuity AP Calculus
Example: Determine the value of c such that the function is continuous on the entire real line.
⎧ x + 3, x ≤ 2
⎪
f ( x) = ⎪
⎨
⎪cx + 6, x > 2
⎪
⎩
Properties of Continuity
Since continuity is defined using limits, the properties of limits carry over into continuity.
Properties of Continuity
If b is a real number and f and g are continuous at x = c, then the following functions are also continuous at c.
1. Constant multiple: bf 2. Sum and difference: f ± g
3. Product: fg f
4. Quotient: ; g (c) ≠ 0
g
The Intermediate Value Theorem
If f is continuous on the closed interval [a, b] then f takes on every value between f (a) and f (b). Suppose k is any number
between f(a) and f(b), then there is at least one number c in [a, b] such that f(c) = k.
: The Intermediate value theorem tells you that at least one c exists, but it does not give you a method for finding c. This
theorem is an example of an existence theorem.
Example: In the Intermediate Value Theorem, which axis is k on? What about c?
Example: Consider the function f below.
y
Is f continuous on [a, b] ?
f(a) Is f ( b ) < k < f ( a ) ?
k In this example, if a < c < b ,
f(b) then there are ________ c’s
such that f (c) = k.
x
a b Label the c's on the graph as
c1 , c2 , …
38
2.3 Continuity AP Calculus
Example: Is there any real number exactly 2 more than its cube? Give any such values accurate to 3 decimal places.
x2 + x ⎡5 ⎤
Example: Let f ( x) = . Verify that the Intermediate Value Theorem applies to the interval ⎢ , 4⎥ and find the
x −1 ⎢⎣ 2 ⎦⎥
value of c guaranteed by the theorem if f (c) = 6 .
Notecards from Section 2.3: Definition of continuity at x = c, Types of Discontinuities, Intermediate Value Theorem
39
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How are you planning on using Docstoc? | 1,329 | 5,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2013-48 | longest | en | 0.865628 |
https://www.amesweb.info/CompressionMemberDesign/Examples/ColumnEccentricCompression.aspx | 1,532,081,383,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591578.1/warc/CC-MAIN-20180720100114-20180720120114-00419.warc.gz | 808,760,341 | 5,879 | ##### Sample Problem : I Beam Buckling Calculations
A structure with mass of 110 kg will be carried by two I-beams as shown in the figure. I-beams are so located that center of mass of structure remain just at the center between two beams. The height of the I-beam is 1 meter and cross sectional dimensions of I-beam are also given in the figure. If the design factor of the I beam is taken as 3, what is the factor of safety of I- beams ? (Assume that all the mass of structure is concentrated at Cog). The material of the I-beam is ASTM A572 steel with yield strength of 290 MPa and elastic modulus of 140 GPa. I beams are fixed to ground and structure so end conditions are fixed-fixed.
## Step 1 : Write down input parameters (including material properties) which are defined in the sample example.
The force due to the mass of the structure can be calculated by F=m*a. So F=110*9.81=1079.1 N. Since there are two columns, the force per column is 1079.1/2=539.55 N
INPUT PROPERTIES SUMMARY Parameter Symbol Value Unit Width of Column B 76 mm Flange Thickness h 7.6 mm Web Thickness b 4 mm Flange-flange inner face height H 119 mm Length of Column L 1 m Design factor nd 3 --- Eccentricity e 2 m Applied Force P 539.55 N End Condition Fixed-Fixed Yield Strength (A572 Steel) Sy 290 MPa Elastic modulus(A588 Steel) E 140 GPa Density(A204 Steel) ρ 7.85 g/cm3
## Step 5 : Go back to "Compression Member Design" page and calculate factor of safeties for x-x and y-y direction by using parameters given in step 1 and step 4.
According to calculations, factor of safety is calculated as 3.92 (minimum value) and this value is larger than design factor(3). According to input parameters, the columns can carry the structure.
##### Summary
The problem is completely solved with calculators which are summarized as follows.
Calculator Usage Column Buckling and Compression Member Design To calculate column factor of safety for given parameters. In this example, column is classified as strut or short column with eccentric loading and stress equations have been used. Sectional Properties Calculator To calculate sectional properties of standard profiles. In this example, I-beam has been calculated. | 518 | 2,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-30 | latest | en | 0.896187 |
https://www.vorlesungen.ethz.ch/Vorlesungsverzeichnis/sucheLehrangebot.view?abschnittId=87684&semkez=2020W&ansicht=2&lang=de&seite=1 | 1,718,583,684,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861674.39/warc/CC-MAIN-20240616233956-20240617023956-00133.warc.gz | 942,117,283 | 7,086 | # Suchergebnis: Katalogdaten im Herbstsemester 2020
Nummer Titel Typ ECTS Umfang Dozierende Elektrotechnik und Informationstechnologie Bachelor 3. Semester: Prüfungsblöcke Prüfungsblock 1 401-0353-00L Analysis 3 O 4 KP 2V + 2U M. Iacobelli Kurzbeschreibung In this lecture we treat problems in applied analysis. The focus lies on the solution of quasilinear first order PDEs with the method of characteristics, and on the study of three fundamental types of partial differential equations of second order: the Laplace equation, the heat equation, and the wave equation. Lernziel The aim of this class is to provide students with a general overview of first and second order PDEs, and teach them how to solve some of these equations using characteristics and/or separation of variables. Inhalt 1.) General introduction to PDEs and their classification (linear, quasilinear, semilinear, nonlinear / elliptic, parabolic, hyperbolic)2.) Quasilinear first order PDEs- Solution with the method of characteristics- COnservation laws3.) Hyperbolic PDEs- wave equation- d'Alembert formula in (1+1)-dimensions- method of separation of variables4.) Parabolic PDEs- heat equation- maximum principle- method of separation of variables5.) Elliptic PDEs- Laplace equation- maximum principle- method of separation of variables- variational method Literatur Y. Pinchover, J. Rubinstein, "An Introduction to Partial Differential Equations", Cambridge University Press (12. Mai 2005) Voraussetzungen / Besonderes Prerequisites: Analysis I and II, Fourier series (Complex Analysis) 402-0053-00L Physics II O 8 KP 4V + 2U A. Imamoglu Kurzbeschreibung The goal of the Physics II class is an introduction to quantum mechanics Lernziel To work effectively in many areas of modern engineering, such as renewable energy and nanotechnology, students must possess a basic understanding of quantum mechanics. The aim of this course is to provide this knowledge while making connections to applications of relevancy to engineers. After completing this course, students will understand the basic postulates of quantum mechanics and be able to apply mathematical methods for solving various problems including atoms, molecules, and solids. Additional examples from engineering disciplines will also be integrated. Inhalt Content:- Wave mechanics: the old quantum theory- Postulates and formalism of Quantum Mechanics- First application: the quantum well and the harmonic Oscillator- QM in three dimension: the Hydrogen atom- Identical particles: Pauli's principle- Crystalline Systems and band structures- Quantum statistics- Approximation Methods- Applications in Engineering- Entanglement and superposition Skript Lecture notes (hand-written) will be distributed via the Moodle interface Literatur David J. Griffiths, "Introduction to quantum mechanics" Second edition, Cambridge University Press. Link Voraussetzungen / Besonderes Prerequisites: Physics I. 227-0045-00L Signal- und Systemtheorie I O 4 KP 2V + 2U H. Bölcskei Kurzbeschreibung Signaltheorie und Systemtheorie (zeitkontinuierlich und zeitdiskret): Signalanalyse im Zeit- und Frequenzbereich, Signalräume, Hilberträume, verallgemeinerte Funktionen, lineare zeitinvariante Systeme, Abtasttheoreme, zeitdiskrete Signale und Systeme, digitale Filterstrukturen, diskrete Fourier-Transformation (DFT), endlich-dimensionale Signale und Systeme, schnelle Fouriertransformation (FFT). Lernziel Einführung in die mathematische Signaltheorie und Systemtheorie. Inhalt Signaltheorie und Systemtheorie (zeitkontinuierlich und zeitdiskret): Signalanalyse im Zeit- und Frequenzbereich, Signalräume, Hilberträume, verallgemeinerte Funktionen, lineare zeitinvariante Systeme, Abtasttheoreme, zeitdiskrete Signale und Systeme, digitale Filterstrukturen, diskrete Fourier-Transformation (DFT), endlich-dimensionale Signale und Systeme, schnelle Fouriertransformation (FFT). Skript Vorlesungsskriptum, Übungsskriptum mit Lösungen. 252-0836-00L Informatik II O 4 KP 2V + 1U F. Mattern Kurzbeschreibung Die Vorlesung vermittelt die gebräuchlichsten Problemlösungsverfahren, Algorithmen und Datenstrukturen. Themen sind u.a.: Divide and Conquer-Prinzip, Rekursion, Sortieralgorithmen, Backtracking, Suche in Spielbäumen, Datenstrukturen (Listen, Stacks, Binärbäume etc.) zeitdiskrete Simulation, Nebenläufigkeit, Komplexität, Verifikation. Bei den Übungen wird die Programmiersprache Java verwendet. Lernziel Einführung in die Methoden der Informatik sowie Vermittlung von Grundlagen zur selbständigen Bewältigung von anspruchsvolleren Übungen und Studienarbeiten mit Informatikkomponente im nachfolgenden Bachelor- und Masterstudium. Inhalt Teil II der Vorlesung vermittelt die gebräuchlichsten Problemlösungsverfahren, Algorithmen und Datenstrukturen. Der Stoff umfasst auch grundlegende Konzepte und Mechanismen der Programmstrukturierung. Darüber hinaus wird generell das Denken in formalen Systemen, die Notwendigkeit zur Abstraktion, sowie die Bedeutung geeigneter Modellbildungen für die Informatik motiviert. Der Schwerpunkt der Vorlesung liegt auf der praktischen Informatik; konkrete Themen sind u.a.: Komplexität und Korrektheit von Algorithmen, Divide and Conquer-Prinzip, Rekursion, Sortieralgorithmen, Backtracking, Suche in Spielbäumen, Datenstrukturen (Listen, Stacks, binäre Bäume etc.), zeitdiskrete Simulation, Nebenläufigkeit, Verifikation. Bei den praktischen Übungen wird die Programmiersprache Java verwendet, dabei werden auch Aspekte wie Modularisierung, Abstraktion und Objektkapselung behandelt. Gelegentlich werden auch kurze Hinweise zum geschichtlichen Kontext der jeweiligen Konzepte gegeben. In den Übungen wird u.a. in Gruppen ein Spielprogramm für "Reversi" programmiert, am Ende des Semesters findet dazu ein Turnier statt. Skript Folienkopien, erweitert um "bonus slides" mit weiterführenden Anmerkungen sowie Darstellungen des historischen Kontextes. Literatur Lehrbuch von Mark Allan Weiss: Data Structures and Problem Solving Using Java, Addison Wesley. Voraussetzungen / Besonderes Voraussetzungen: Informatik I. Prüfungsblock 2 227-0077-10L Halbleiter-Schaltungstechnik O 4 KP 2V + 2U Q. Huang Kurzbeschreibung Einführungsvorlesung in die Halbleiter-Schaltungstechnik. Der Transistor als aktives Bauelement. Analyse und Entwurf transistorbasierter elektronischer Schaltungen wie Verstärker und Filter; Operationsverstärker und darauf basierende Schaltungen. Lernziel Moderne elektronische Schaltungen auf Transistorbasis haben unser Leben verändert und spielen in unserer Wirtschaft seit einem halben Jahrhundert eine Schlüsselrolle. Das Hauptziel dieser Vorlesung ist es, den Studenten das Konzept des aktiven Bauteils näher zu bringen. Dies beinhaltet Operationsverstärker und deren Anwendung für Verstärkerschaltungen, für Signalaufbereitung, Schaltfunktionen und Filter. Zusätzlich zur Behandlung von typischen elektonischen Schaltungen, welche in üblichen Anwendungen einschliesslich Gruppenarbeiten und Fachpraktika anzutreffen sind, können die Studenten ihre Kenntnisse linearer Schaltungen, welche auf nicht-linearen Bauteilen basieren, vertiefen. Auch auf Nichtidealitäten elektronischer Schaltungen und auf Entwurfskonzepte (als Gegenteil der Analyse) wird eingegangen. Die Veranstaltung stellt eine Voraussetzung für Themengebiete wie analoge, integrierte Schaltungen, HF-Schaltungen für drahtlose Kommunikation, A/D und D/A-Wandler und Optoelektronik dar, welche in höheren Semestern angeboten werden. Inhalt Rekapitulation des Transistors als Bauteil (bipolar und MOSFET), Gross- und Kleinsignalverhalten, Arbeitspunkt und Arbeitspunkteinstellung. Eintransistorverstärker, einfache Rückkopplung zur Arbeitspunkteinstellung. Frequenzgang von einfachen Verstärkern. Methoden zur Bandbreitenerweiterung. Differenzverstärker, Operationsverstärker, Verstärker mit variabler Bandbreite. Instrumentierungsverstärker: Gleichtaktunterdrückung, Rauschen, Störsignale, Chopper-Technik. Transimpedanzverstärker. Aktive Filter: einfache aktive Filter, Filter mit biquadratischen Stufen. Filter höherer Ordnung, Realisierung mit biquadratischen Stufen und mit Leiterstruktur. Switched-Cap-Filter. Literatur Göbel, H.: Einführung in die Halbleiter-Schaltungstechnik. Springer-Verlag Berlin Heidelberg, 6th edition, 2019.Pederson, D.O. and Mayaram, K.: Analog Integrated Circuits for Communication. Springer US, 2nd edition, 2008.Sansen, W.M.C.: Analog Design Essentials. Springer US, 1st edition, 2006.Su, K.L.: Analog Filters. Springer US, 2nd edition, 2002. 401-0053-00L Diskrete Mathematik O 4 KP 2V + 1U D. Adjiashvili Kurzbeschreibung Einführung in Grundlagen der diskreten Mathematik: Kombinatorik (elementare Zählprobleme), Graphentheorie, Algebra, und Anwendungen davon. Lernziel Entwicklung eines guten Verständnisses von einigen der prominentesten Gebiete der diskreten Mathematik.
• Seite 1 von 1 | 2,381 | 8,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-26 | latest | en | 0.705398 |
http://mathhelpforum.com/calculus/66807-curves-tangent-lines-etc-print.html | 1,508,774,623,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826114.69/warc/CC-MAIN-20171023145244-20171023165244-00458.warc.gz | 208,798,987 | 3,177 | # Curves. tangent lines, etc.
• Jan 4th 2009, 11:11 AM
Curves. tangent lines, etc.
Consider the curve defined by $-8x^2+5xy+y^3=-149$
I know that dy/dx = (16x-5y)/(5x+3y^2)
I also know that the equation of the line tangent to the curve at point (4,-1) is y+1=3(x-4).
However, now I'm asked the following..
1. There's a number k so that the point (4.2,k) is on the curve. Using the tangent line, approximate the value of k.
2. Write an equation that can be solved to find the actual value of k so that point (4.2,k) is on the curve.
3. Solve the equation found in part (d) for the actual value of k.
I don't even know where to BEGIN. I'm not looking for the answer, I'm looking for someone patient enough to teach me and guide me to the solution.
• Jan 4th 2009, 11:44 AM
skeeter
Quote:
There's a number k so that the point (4.2,k) is on the curve. Using the tangent line, approximate the value of k.
$y + 1 = 3(x - 4)$
$k + 1 = 3(4.2 - 4)$
$k \approx \, ?$
Quote:
Write an equation that can be solved to find the actual value of k so that point (4.2,k) is on the curve
$-8(4.2)^2 + 5(4.2)k + k^3 = -149$
Quote:
Solve the equation found in part (d) for the actual value of k.
get out your calculator and solve for $k$ ... you should get a value close to your estimate.
• Jan 5th 2009, 01:30 PM
Quote:
Originally Posted by skeeter
$y + 1 = 3(x - 4)$
$k + 1 = 3(4.2 - 4)$
$k \approx \, ?$
$-8(4.2)^2 + 5(4.2)k + k^3 = -149$
get out your calculator and solve for $k$ ... you should get a value close to your estimate.
Thank you SO much. I mean, you pretty much worked out the entire problem for me, but I understand what you did perfectly. | 548 | 1,653 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-43 | longest | en | 0.866678 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/37/5/i/a/ | 1,680,388,108,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950363.89/warc/CC-MAIN-20230401221921-20230402011921-00100.warc.gz | 949,684,026 | 59,976 | # Properties
Label 37.5.i.a Level $37$ Weight $5$ Character orbit 37.i Analytic conductor $3.825$ Analytic rank $0$ Dimension $144$ CM no Inner twists $2$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [37,5,Mod(2,37)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(37, base_ring=CyclotomicField(36))
chi = DirichletCharacter(H, H._module([1]))
N = Newforms(chi, 5, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("37.2");
S:= CuspForms(chi, 5);
N := Newforms(S);
Level: $$N$$ $$=$$ $$37$$ Weight: $$k$$ $$=$$ $$5$$ Character orbit: $$[\chi]$$ $$=$$ 37.i (of order $$36$$, degree $$12$$, minimal)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: no Analytic conductor: $$3.82468863410$$ Analytic rank: $$0$$ Dimension: $$144$$ Relative dimension: $$12$$ over $$\Q(\zeta_{36})$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{36}]$
## $q$-expansion
The dimension is sufficiently large that we do not compute an algebraic $$q$$-expansion, but we have computed the trace expansion.
$$\operatorname{Tr}(f)(q) =$$ $$144 q - 12 q^{2} - 12 q^{3} + 6 q^{4} - 120 q^{5} - 12 q^{6} - 12 q^{7} - 282 q^{8} + 198 q^{9}+O(q^{10})$$ 144 * q - 12 * q^2 - 12 * q^3 + 6 * q^4 - 120 * q^5 - 12 * q^6 - 12 * q^7 - 282 * q^8 + 198 * q^9 $$\operatorname{Tr}(f)(q) =$$ $$144 q - 12 q^{2} - 12 q^{3} + 6 q^{4} - 120 q^{5} - 12 q^{6} - 12 q^{7} - 282 q^{8} + 198 q^{9} - 6 q^{10} - 18 q^{11} + 948 q^{12} - 12 q^{13} - 588 q^{14} - 810 q^{15} - 942 q^{16} + 708 q^{17} + 2418 q^{18} + 132 q^{19} - 1932 q^{20} - 750 q^{21} - 108 q^{22} + 1986 q^{23} - 4632 q^{24} + 1752 q^{25} + 1434 q^{26} + 5814 q^{27} + 9396 q^{28} - 768 q^{29} - 3204 q^{30} - 11364 q^{31} - 13818 q^{32} - 4602 q^{33} - 4044 q^{34} - 7140 q^{35} + 3096 q^{37} + 9732 q^{38} - 174 q^{39} + 24564 q^{40} + 7152 q^{41} + 11892 q^{42} + 14556 q^{43} - 522 q^{44} + 24912 q^{45} - 23952 q^{46} - 1788 q^{47} - 34578 q^{48} - 10050 q^{49} - 11238 q^{50} - 20652 q^{51} + 29394 q^{52} - 1902 q^{53} - 5916 q^{54} + 1692 q^{55} + 13236 q^{56} - 5484 q^{57} + 28128 q^{58} + 26538 q^{59} + 85248 q^{60} - 15996 q^{61} + 17016 q^{62} - 9366 q^{63} - 53316 q^{64} - 27984 q^{65} - 128484 q^{66} + 8346 q^{67} - 76026 q^{68} - 27294 q^{69} - 56298 q^{70} - 31692 q^{71} - 16314 q^{72} + 13614 q^{74} + 49320 q^{75} + 17598 q^{76} + 26190 q^{77} + 119382 q^{78} + 72852 q^{79} + 115608 q^{80} + 34026 q^{81} + 109680 q^{82} - 12792 q^{83} + 84462 q^{84} + 25254 q^{85} + 34890 q^{86} - 86928 q^{87} - 82800 q^{88} + 4992 q^{89} - 175584 q^{90} - 118062 q^{91} + 112794 q^{92} + 119424 q^{93} + 4692 q^{94} + 23802 q^{95} + 109008 q^{96} + 18186 q^{97} - 155478 q^{98} - 4032 q^{99}+O(q^{100})$$ 144 * q - 12 * q^2 - 12 * q^3 + 6 * q^4 - 120 * q^5 - 12 * q^6 - 12 * q^7 - 282 * q^8 + 198 * q^9 - 6 * q^10 - 18 * q^11 + 948 * q^12 - 12 * q^13 - 588 * q^14 - 810 * q^15 - 942 * q^16 + 708 * q^17 + 2418 * q^18 + 132 * q^19 - 1932 * q^20 - 750 * q^21 - 108 * q^22 + 1986 * q^23 - 4632 * q^24 + 1752 * q^25 + 1434 * q^26 + 5814 * q^27 + 9396 * q^28 - 768 * q^29 - 3204 * q^30 - 11364 * q^31 - 13818 * q^32 - 4602 * q^33 - 4044 * q^34 - 7140 * q^35 + 3096 * q^37 + 9732 * q^38 - 174 * q^39 + 24564 * q^40 + 7152 * q^41 + 11892 * q^42 + 14556 * q^43 - 522 * q^44 + 24912 * q^45 - 23952 * q^46 - 1788 * q^47 - 34578 * q^48 - 10050 * q^49 - 11238 * q^50 - 20652 * q^51 + 29394 * q^52 - 1902 * q^53 - 5916 * q^54 + 1692 * q^55 + 13236 * q^56 - 5484 * q^57 + 28128 * q^58 + 26538 * q^59 + 85248 * q^60 - 15996 * q^61 + 17016 * q^62 - 9366 * q^63 - 53316 * q^64 - 27984 * q^65 - 128484 * q^66 + 8346 * q^67 - 76026 * q^68 - 27294 * q^69 - 56298 * q^70 - 31692 * q^71 - 16314 * q^72 + 13614 * q^74 + 49320 * q^75 + 17598 * q^76 + 26190 * q^77 + 119382 * q^78 + 72852 * q^79 + 115608 * q^80 + 34026 * q^81 + 109680 * q^82 - 12792 * q^83 + 84462 * q^84 + 25254 * q^85 + 34890 * q^86 - 86928 * q^87 - 82800 * q^88 + 4992 * q^89 - 175584 * q^90 - 118062 * q^91 + 112794 * q^92 + 119424 * q^93 + 4692 * q^94 + 23802 * q^95 + 109008 * q^96 + 18186 * q^97 - 155478 * q^98 - 4032 * q^99
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
2.1 −7.43299 0.650302i −8.90517 + 10.6128i 39.0695 + 6.88901i −18.8246 + 40.3695i 73.0935 73.0935i −26.2057 + 9.53809i −170.609 45.7146i −19.2633 109.247i 166.176 287.824i
2.2 −7.07168 0.618692i 10.4867 12.4976i 33.8689 + 5.97201i −4.92870 + 10.5696i −81.8911 + 81.8911i 75.0961 27.3328i −126.106 33.7901i −32.1531 182.349i 41.3935 71.6956i
2.3 −5.82370 0.509508i 0.698255 0.832148i 17.8989 + 3.15607i 10.8573 23.2835i −4.49041 + 4.49041i −65.7606 + 23.9349i −12.2821 3.29097i 13.8606 + 78.6073i −75.0928 + 130.064i
2.4 −4.38998 0.384074i 0.247322 0.294747i 3.36752 + 0.593785i −2.39998 + 5.14676i −1.19894 + 1.19894i 28.5706 10.3989i 53.5502 + 14.3487i 14.0398 + 79.6236i 12.5126 21.6724i
2.5 −2.93366 0.256662i −9.35942 + 11.1541i −7.21645 1.27246i 9.43269 20.2285i 30.3202 30.3202i 39.1303 14.2423i 66.3563 + 17.7801i −22.7502 129.023i −32.8641 + 56.9224i
2.6 −1.13246 0.0990776i 9.68265 11.5393i −14.4843 2.55397i 6.08797 13.0557i −12.1085 + 12.1085i −66.6341 + 24.2528i 33.7187 + 9.03489i −25.3370 143.693i −8.18792 + 14.1819i
2.7 −0.818396 0.0716004i 2.52396 3.00793i −15.0923 2.66118i −18.6068 + 39.9023i −2.28096 + 2.28096i 8.45114 3.07597i 24.8574 + 6.66052i 11.3882 + 64.5856i 18.0847 31.3237i
2.8 2.03026 + 0.177625i −6.00370 + 7.15494i −11.6665 2.05712i −1.02327 + 2.19440i −13.4600 + 13.4600i −47.7624 + 17.3841i −54.8178 14.6884i −1.08314 6.14280i −2.46728 + 4.27345i
2.9 2.08231 + 0.182179i 2.85756 3.40550i −11.4541 2.01967i 15.1773 32.5477i 6.57073 6.57073i 66.1524 24.0775i −55.7877 14.9483i 10.6337 + 60.3066i 37.5333 65.0095i
2.10 5.32165 + 0.465584i 8.08788 9.63876i 12.3463 + 2.17699i −8.12880 + 17.4323i 47.5286 47.5286i 2.51614 0.915801i −17.8701 4.78828i −13.4264 76.1449i −51.3748 + 88.9838i
2.11 5.93109 + 0.518903i −7.67181 + 9.14291i 19.1516 + 3.37695i −13.8164 + 29.6295i −50.2465 + 50.2465i 80.3866 29.2583i 19.8237 + 5.31175i −10.6706 60.5159i −97.3214 + 168.566i
2.12 6.83559 + 0.598037i −0.770402 + 0.918130i 30.6107 + 5.39749i 9.09446 19.5031i −5.81523 + 5.81523i −42.6753 + 15.5325i 99.9681 + 26.7864i 13.8161 + 78.3548i 73.8296 127.877i
5.1 −3.29658 + 7.06954i −3.57062 + 9.81021i −28.8263 34.3539i 0.215509 0.307778i −57.5828 57.5828i 11.8673 + 67.3028i 217.341 58.2363i −21.4413 17.9914i 1.46541 + 2.53816i
5.2 −2.62972 + 5.63946i 2.63021 7.22644i −14.6034 17.4037i 17.6934 25.2688i 33.8365 + 33.8365i −11.0584 62.7152i 40.3836 10.8208i 16.7462 + 14.0517i 95.9735 + 166.231i
5.3 −2.37818 + 5.10001i 4.82693 13.2619i −10.0698 12.0008i −25.5003 + 36.4182i 56.1566 + 56.1566i 13.1184 + 74.3980i −1.81618 + 0.486643i −90.5289 75.9628i −125.089 216.661i
5.4 −1.86456 + 3.99856i −1.37794 + 3.78585i −2.22731 2.65440i −13.4352 + 19.1875i −12.5687 12.5687i −8.06935 45.7636i −53.4187 + 14.3135i 49.6157 + 41.6325i −51.6715 89.4976i
5.5 −1.15872 + 2.48488i −5.54437 + 15.2330i 5.45259 + 6.49814i 6.68567 9.54813i −31.4280 31.4280i −2.18344 12.3829i −64.8386 + 17.3735i −139.256 116.849i 15.9792 + 27.6767i
5.6 −1.02938 + 2.20752i 0.471556 1.29559i 6.47109 + 7.71195i 15.7198 22.4501i 2.37463 + 2.37463i 13.1089 + 74.3445i −61.3292 + 16.4331i 60.5934 + 50.8439i 33.3774 + 57.8114i
5.7 0.240642 0.516058i 4.70675 12.9317i 10.0762 + 12.0083i 1.54727 2.20973i −5.54086 5.54086i −8.44071 47.8697i 17.4218 4.66816i −83.0255 69.6667i −0.768009 1.33023i
5.8 0.479059 1.02734i −0.611056 + 1.67886i 9.45866 + 11.2724i −16.2436 + 23.1983i 1.43204 + 1.43204i 0.371208 + 2.10523i 33.6307 9.01132i 59.6044 + 50.0140i 16.0510 + 27.8012i
See next 80 embeddings (of 144 total)
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 35.12 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
37.i odd 36 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 37.5.i.a 144
37.i odd 36 1 inner 37.5.i.a 144
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
37.5.i.a 144 1.a even 1 1 trivial
37.5.i.a 144 37.i odd 36 1 inner
## Hecke kernels
This newform subspace is the entire newspace $$S_{5}^{\mathrm{new}}(37, [\chi])$$. | 4,582 | 8,866 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-14 | latest | en | 0.416721 |
https://encyclios.org/threshold-of-hearing/ | 1,659,962,740,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570827.41/warc/CC-MAIN-20220808122331-20220808152331-00779.warc.gz | 241,062,509 | 23,718 | # Threshold of hearing
The hearing threshold is the sound level below which a person’s ear is unable to detect any sound. For adults, 0 dB is the reference level. Sound level measurements in decibels are generally referenced to a standard threshold of hearing at 1000 Hz for the human ear which can be stated in terms of sound intensity:
$I_0=10^{-12}\;\dfrac{\textrm{W}}{\textrm{m}^2}=10^{-16}\;\dfrac{\textrm{W}}{\textrm{cm}^2}$
or in terms of sound pressure:
$P_0=2\times 10^{-5}\;\dfrac{\textrm{N}}{\textrm{m}^2}$
This value has wide acceptance as a nominal standard threshold and corresponds to 0 decibels. It represents a pressure change of less than one billionth of standard atmospheric pressure. This is indicative of the incredible sensitivity of human hearing.
The actual average threshold of hearing at 1000 Hz is more like 2.5×10-12 W/m2 or about 4 decibels, but zero decibels is a convenient reference. The threshold of hearing varies with frequency, as illustrated by the measured hearing curves below.
## Threshold shift
threshold shift is an increase in the hearing threshold for a particular sound frequency. It means that the hearing sensitivity decreases and that it becomes harder for the listener to detect soft sounds. Threshold shifts can be temporary or permanent.
A temporary threshold shift (TTS) is a temporary shift in the auditory threshold. It may occur suddenly after exposure to a high level of noise, a situation in which most people experience reduced hearing. A temporary threshold shift results in temporary hearing loss.
The opposite of a temporary threshold shift is a permanent threshold shift. A permanent threshold shift (PTS) is when the ability to hear is reduced permanently, which causes a permanent hearing loss. A professional hearing test can measure the level of the threshold shift and the results can be seen on an audiogram. PTS is sensorineural and varies across frequencies, depending on characteristics of the exposure, the transmission characteristics of the external and middle ears, and the innate sensitivity of different regions of the cochlea to damage.
Exposure to intense sound can produce TTS, acute changes in hearing sensitivity that recover over time, or PTS, a loss that does not recover to pre-exposure levels. In general, a threshold shift ≥ 10 dB at 2, 3 and 4 kHz is required for reporting purposes in human studies. The high-frequency regions of the cochlea are most sensitive to noise damage. Resonance of the ear canal also results in a frequency region of high noise sensitivity at 4–6 kHz. A primary noise target is the cochlear hair cell. While the mechanisms that underlie such hair cell damage remain unclear, there is evidence to support a role for reactive oxygen species, stress pathway signaling and apoptosis.
Another target is the synapse between the hair cell and the primary afferent neurons. Large numbers of these synapses and their neurons can be lost after noise, even though hearing thresholds may return to normal. This affects auditory processing and detection of signals in noise. The consequences of TTS and PTS include significant deficits in communication that can impact performance of military duties or obtaining/retaining civilian employment. Tinnitus and exacerbation of post-traumatic stress disorder are also potential sequelae.
### Characteristics of temporary threshold shift
TTS is a change in hearing threshold that recovers to pre-exposure levels (baseline) over time. The amount of time to recover to baseline may be relatively fast (minutes to hours) or slow (days to weeks). The severity of the initial insult, as well as the time course of the recovery, are dependent on a number of factors including: the type of insult or trauma, the intensity and duration of the insult (single vs repeated, short vs long exposures), and the stimulus type (impulse/impact sound or continuous noise including wide or narrow-band noise).
Individual susceptibility is dependent on the use of hearing protective devices, the quiet time or rest between exposures, and the level of hearing loss prior to exposure. Individual susceptibility to TTS may also be influenced by age, sex, prior history of noise exposure, diabetes, genotype and other personal or environmental factors such as smoking and diet. While these factors are at play for PTS as well, unlike PTS, TTS is a change in hearing sensitivity which recovers to baseline or within test/retest criteria in minutes, hours, days or weeks with the upper limit being 30 days post exposure. TTS and PTS outcomes will vary as a function of the insult and individual factors.
#### Mechanisms of temporary threshold shift
Historically, TTS was largely thought to be a mechanical process that involved structures within the outer and middle ear including the ear drum, ossicular chain and middle ear muscles through the acoustic reflex. Extremely intense noise exposure is also known to mechanically damage the cochlea, disrupting the connections between the tectorial membrane and outer hair cell stereocilia, damaging the stereocilia themselves, breaching the integrity of the reticular lamina or even disrupting the basilar membrane.
However, recent work in several preclinical studies has demonstrated a significant involvement of several sensorineural inner ear structures including hair cells and their stereocilia, supporting cells within the organ of Corti, endothelial cells and fibrocytes within the stria vascularis and spiral ligament, and dendritic processes of the auditory nerve. Molecular and biochemical changes have been identified that include pro-inflammatory and pro-apoptotic processes. These changes have been shown to alter the normal function of several critical processes within the cochlea including the endolymphatic potential that drives hair cell depolarization, cellular membranes and mitochondria responsible for hair cell and supporting cell activity, and neural innervation of the inner hair cell that conduct impulses to the auditory brainstem. In addition, changes in the activity or metabolism of neurons in the cochlear nucleus, superior-olivary complex and inferior colliculus have been observed. In support of this noise-induced change in inner ear biology and pharmacology and its relevance in establishing the TTS, several preclinical studies have demonstrated a significant reduction in TTS when the animals were administered otoprotective compounds or drugs immediately prior to noise exposure.
### Characteristics of permanent threshold shift
Noise damage is typically most extensive at frequencies above those of the exposure, a phenomenon well explained by nonlinearities in the cochlear mechanical response to sound. This is most apparent for TTS and for low levels of PTS. However, noises to which human ears are exposed often are broadband in frequency composition.
These signals are shaped (some frequencies amplified, others reduced by filtering) by passage through the external and middle ears. Resonance in the ear canal produces amplification of acoustic frequencies whose wavelengths are approximately 4 times the length of the canal, which for humans results in enhancement of frequencies around 4 kHz. This contributes to an enhanced “notch” of PTS at 4–6 kHz for exposure to broad-band stimuli.
Finally, as with many other forms of damage, the basal cochlea appears to be most vulnerable to noise. While the reason for this is not entirely clear, it may be related to higher levels of antioxidants in apical hair cells as well as higher rates of metabolic activity in basal hair cells. This basal sensitivity results in a tendency for TTS and PTS to be more extensive at high frequencies.
#### Mechanisms of permanent threshold shift
While intense sounds such as blast can damage the conductive apparatus of the outer and middle ears, producing permanent hearing loss through tympanic membrane rupture or ossicular dislocation, PTS is generally considered to be a sensorineural phenomenon restricted to the cells of the cochlea. The most recognized cause of PTS is damage to and loss of cochlear hair cells. The mechanisms by which this damage can occur are not known with certainty. However, there is extensive evidence implicating the generation of reactive oxygen species (ROS) within hair cells during and after overexposure. This leads to the activation of stress signaling pathways such as the JNK MAP kinase cascade, which can in turn lead to cell damage, apoptosis and/or necrosis. The biochemical pathways leading to hair cell damage/death are undoubtedly complex, and also appear to include competing survival pathways that attempt to rescue hair cells and restore their function. It is the balance of these competing pathways that determine the fate of the cell. The outer hair cells, responsible for the exquisite sensitivity and frequency and selectivity of the cochlea, are the most sensitive to damage.
Noise also can target hair cell synapses and neurons directly, even when the hair cells themselves remain and recover normal function. The insult is seen acutely as a glutamate-like ‘excitotoxicity’ that includes swelling and retraction of afferent terminals from beneath inner hair cells. Recent work in animal models shows that noise-induced loss of synapses and afferent terminals is rapid and permanent. Loss of spiral ganglion neurons is comparatively slow, and can be ‘primary,’ that is, occurring without noise-induced hair cell loss or ‘secondary’ to the loss of their inner hair-cell targets. Such synaptic and neural loss can exacerbate the functional consequences of noise exposure by reducing the ability of the VIIIth nerve to encode auditory signals with fidelity, with or without loss of threshold sensitivity. Thus, lack of PTS does not imply that auditory function is normal.
## References
1. Ryan, A. F., Kujawa, S. G., Hammill, T., Le Prell, C., & Kil, J. (2016). Temporary and Permanent Noise-induced Threshold Shifts: A Review of Basic and Clinical Observations. Otology & neurotology : official publication of the American Otological Society, American Neurotology Society [and] European Academy of Otology and Neurotology37(8), e271–e275. https://doi.org/10.1097/MAO.0000000000001071
2. Thurston FE. The worker’s ear: a history of noise-induced hearing loss. Am J Ind Med. 2013;56:367–377. [PubMed] [Google Scholar]
3. Eldredge DH, Mills JH, Bohne BA. Anatomical, behavioral, and electrophysiological observations on chinchillas after long exposures to noise. Adv Otorhinolaryngol. 1973;20:64–81. [PubMed] [Google Scholar]
4. Ryan A, Bone RC. Noise-induced threshold shift and cochlear pathology in the Mongolian gerbil. J Acoust Soc Am. 1978;63:1145–1151. [PubMed] [Google Scholar]
5. Clark WW, Bohne BA. Effects of noise on hearing. JAMA. 1999 May 5;281(17):1658–1659. [PubMed] [Google Scholar]
6. Lonsbury-Martin BL, Martin GK, Bohne BA. Repeated TTS exposures in monkeys: alterations in hearing, cochlear structure, and single-unit thresholds. J Acoust Soc Am. 1987;81:1507–1518. [PubMed] [Google Scholar]
7. NIOSH. National Institute of Occupational Safety and Health. publication 98-126: http://www.cdc.gov/niosh/docs/98-126/pdfs/98-126.pdf.
8. Cody AR, Johnstone BM. Acoustic trauma: single neuron basis for the “half-octave shift” J Acoust Soc Am. 1981;70:707–711. [PubMed] [Google Scholar]
9. Robles L, Ruggero MA. Mechanics of the mammalian cochlea. Physiol Rev. 2001;81:1305–1352. [PMC free article] [PubMed] [Google Scholar]
10. Rosowski JJ. The effects of external- and middle-ear filtering on auditory threshold and noise-induced hearing loss. J Acoust Soc Am. 1991;90:124–135. [PubMed] [Google Scholar]
11. Sha S-H, Taylor R, Forge A, Schacht J. Differential vulnerability of basal and apical hair cells is based on intrinsic susceptibility to free radicals. Hear Res. 2001;155:1–8. [PubMed] [Google Scholar]
12. Mulroy MJ, Fromm RF, Curtis S. Changes in the synaptic region of auditory hair cells during noise-induced temporary threshold shift. Hear Res. 1990;49:79–87. [PubMed] [Google Scholar]
13. Kujawa SG, Liberman MC. Adding insult to injury: cochlear nerve degeneration after “temporary” noise-induced hearing loss. J Neurosci. 2009;29:14077–14085. [PMC free article] [PubMed] [Google Scholar]
14. Henderson D, Bielefeld EC, Harris KC, Hu BH. The role of oxidative stress in noise-induced hearing loss. Ear Hear. 2006;27:1–19. [PubMed] [Google Scholar]
15. Yan D, Zhu Y, Walsh T, et al. Mutation of the ATP-gated P2X2 receptor leads to progressive hearing loss and increased susceptibility to noise. Proc Natl Acad Sci U S A. 2013;110:2228–2233. [PMC free article] [PubMed] [Google Scholar]
16. Ryan AF, Axelsson GA, Woolf NK. Central auditory metabolic activity induced by intense noise exposure. Hear Res. 1992;61:24–30. [PubMed] [Google Scholar]
17. Seidman MD, Shivapuja BG, Quirk WS. The protective effects of allopurinol and superoxide dismutase on noise-induced cochlear damage. Otolaryngol Head Neck Surg. 1993;109:1052–1056. [PubMed] [Google Scholar]
18. Attias J, Sapir S, Bresloff I, Reshef-Haran I, Ising H. Reduction in noise-induced temporary threshold shift in humans following oral magnesium intake. Clin Otolaryngol. 2004;29:635–641. [PubMed] [Google Scholar]18. Yamasoba T, Pourbakht A, Sakamoto T, Suzuki M. Ebselen prevents noise-induced excitotoxicity and temporary threshold shift. Neurosci Lett. 2005;380:234–238. [PubMed] [Google Scholar]
19. Lynch ED, Kil J. Compounds for the prevention and treatment of noise-induced hearing loss. Drug Discovery Today. 2005;10:1291–1298. [PubMed] [Google Scholar]
20. Kil J, Pierce C, Tran H, Gu R, Lynch ED. Ebselen treatment reduces noise induced hearing loss via the mimicry and induction of glutathione peroxidase. Hear Res. 2007;226:44–51. [PubMed] [Google Scholar]
21. Pirvola U, Xing-Qun L, Virkkala J, et al. Rescue of hearing, auditory hair cells, and neurons by CEP-1347/KT7515, an inhibitor of c-Jun N-terminal kinase activation. J Neurosci. 2000;20:43–50. [PMC free article] [PubMed] [Google Scholar]
22. Bohne BA, Harding GW, Lee SC. Death pathways in noise-damaged outer hair cells. Hear Res. 2007;223:61–70. [PubMed] [Google Scholar]
23. Robertson D. Functional significance of dendritic swelling after loud sounds in the guinea pig cochlea. Hear Res. 1983;9:263–278. [PubMed] [Google Scholar]24. Lin HW, Furman AC, Kujawa SG, Liberman MC. Primary neural degeneration in the Guinea pig cochlea after reversible noise-induced threshold shift. J Assoc Res Otolaryngol. 2011;12:605–616. [PMC free article] [PubMed] [Google Scholar]25. Kujawa SG, Liberman MC. Acceleration of age-related hearing loss by early noise exposure: evidence of a misspent youth. J Neurosci. 2006;26:2115–2123. [PMC free article] [PubMed] [Google Scholar]
24. Bohne BA. Time course of nerve-fiber regeneration in the noise-damaged mammalian cochlea. Int J Dev Neurosci. 1997;15:601–617. [PubMed] [Google Scholar]
25. Puel JL, Ruel J, Gervais d’Aldin C, Pujol R. Excitotoxicity and repair of cochlear synapses after noise-trauma induced hearing loss. Neuroreport. 1998;9:2109–2114. [PubMed] [Google Scholar]
26. Bharadwaj HM, Verhulst S, Shaheen L, Liberman MC, Shinn-Cunningham BG. Cochlear neuropathy and the coding of supra-threshold sound. Front Syst Neurosci. 2014;8:26. [PMC free article] [PubMed] [Google Scholar]
27. Housley GD, Morton-Jones R, Vlajkovic SM, et al. ATP-gated ion channels mediate adaptation to elevated sound levels. Proc Natl Acad Sci U S A. 2013;110:7494–7499. [PMC free article] [PubMed] [Google Scholar]
28. Kirchner DB, Evenson E, Dobie RA, et al. Occupational Noise-Induced Hearing Loss. JOEM. 2012;54:106–108. [PubMed] [Google Scholar]
29. AAO Committee on Medical Aspects of Noise and American Council of Otolaryngology. Guide for the evaluation of hearing handicap. JAMA. 1979;241(19):2055–2059. [PubMed] [Google Scholar]
30. ASHA. American Speech-Language Hearing Association: Degree of Hearing Loss.http://www.asha.org/public/hearing/Degree-of-Hearing-Loss/ [Google Scholar]
31. Arlinger S. Negative consequences of uncorrected hearing loss–a review. Int J Audiol. 2003;42(Suppl 2):2S17–2S20. [PubMed] [Google Scholar]34. Lew HL, Garvert DW, Pogoda TK, et al. Auditory and visual impairments in patients with blast-related traumatic brain injury: Effect of dual sensory impairment on Functional Independence Measure. J Rehabil Res Dev. 2009;46:819–826. [PubMed] [Google Scholar]
32. Bone RC, Ryan AF. Audiometric and histologic correlates of the interaction between kanamycin and subtraumatic levels of noise in the chinchilla. Otolaryngology. 1978;86 ORL400-4. [PubMed] [Google Scholar]
33. Campo P, Venet T, Rumeau C, et al. Impact of noise or styrene exposure on the kinetics of presbycusis. Hear Res. 2011;280:122–132. [PubMed] [Google Scholar]
34. Mazurek B, Olze H, Haupt H, Szczepek AJ. The more the worse: the grade of noise-induced hearing loss associates with the severity of tinnitus. Int J Environ Res Public Health. 2010;7:3071–3079. [PMC free article] [PubMed] [Google Scholar]
35. Gomaa MA, Elmagd MH, Elbadry MM, Kader RM. Depression, Anxiety and Stress Scale in patients with tinnitus and hearing loss. Eur Arch Otorhinolaryngol. 2014;271:2177–2184. [PubMed] [Google Scholar]
36. Jansen EJ, Helleman HW, Dreschler WA, de Laat JA. Noise induced hearing loss and other hearing complaints among musicians of symphony orchestras. Int Arch Occup Environ Health. 2009;82:153–164. [PubMed] [Google Scholar] | 4,344 | 17,495 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-33 | latest | en | 0.875596 |
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Ms. Sue could you please have a moment to look at all my other previous posts, because in come of my posts, I have made corrections, and in some I have added more information, I just want you to check them over and see if they're | 684 | 2,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2021-17 | latest | en | 0.967493 |
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## Geometry and Mensuration: Level 3 Test 7
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Question 1
The figure shows the rectangle ABCD with a semicircle and a circle inscribed inside in it as shown. What is the ratio of the area of the circle to that of the semicircle?
A $\displaystyle {{\left( \sqrt{2}-1 \right)}^{2}}:1$ B $\displaystyle 2{{\left( \sqrt{2}-1 \right)}^{2}}:1$ C $\displaystyle {{\left( \sqrt{2}-1 \right)}^{2}}:2$ D None of these
Question 1 Explanation:
$\displaystyle \begin{array}{l}Let\,\,R\,\,be\,\,the\,\,radius\,\,of\,\,the\,\,semicircle\,\\Therefore\,\,OC=R=CB\\\sin ce\,\,OCB\,\,is\,a\,\,right\,\,angle\,\,triangle\,\,\\so\,\,OB=R\sqrt{2}\\Now\,\,area\,\,of\,semicircle=\frac{\pi {{r}^{2}}}{2}=\frac{\pi {{R}^{2}}}{2}\\Now\,\,FB=OB-OF=R\sqrt{2}-R=R\left( \sqrt{2}-1 \right)\\therfore\,\,the\,\,area\,\,of\,\,circle=\frac{\pi {{R}^{2}}{{\left( \sqrt{2}-1 \right)}^{2}}}{{{2}^{2}}}\\therfore\,\,the\,\,required\,\,ratio=\frac{\frac{\pi {{R}^{2}}{{\left( \sqrt{2}-1 \right)}^{2}}}{{{2}^{2}}}}{\frac{\pi {{R}^{2}}}{2}}={{\left( \sqrt{2}-1 \right)}^{2}}:2\end{array}$
Question 2
A wooden box (open at the top) of thickness 0•5 cm, length 21 cm, width 11 cm and height 6 cm is painted on the inside. The expenses of painting are Rs.70. What is the rate of painting per square centimeter?
A Re. 0•7 B Re 0•5 C Re 0•1 D Re 0•2
Question 2 Explanation:
The actual surface area painted = 2{(11-1)(21-1)+20X5+10X5} = 700 sq. units.
Thus cost per sq. cm =70/700 =0.10 Rs
Correct option is (c)
Question 3
Based on the figure below, what is the value of x, if Y= 10.
A 10 B 11 C 12 D None of these
Question 3 Explanation:
$\begin{array}{l}Now\,\,AB=\sqrt{{{\left( x+3 \right)}^{2}}+{{(x-3)}^{2}}}=\sqrt{2({{x}^{2}}+9)}\\BD=\sqrt{100-{{(x-3)}^{2}}}=\sqrt{2\left( {{x}^{2}}+9 \right)}\\BD=\sqrt{100-{{(x-3)}^{2}}}\\thus\\\sqrt{100-(x-3){}^{2}}+x=\sqrt{2({{x}^{2}}+9)}\\Putting\text{ }the\text{ }values\text{ }of\text{ }x\text{ }none\text{ }of\text{ }them\text{ }satisfy\text{ }the\text{ }equation.\\Thus\text{ }the\text{ }correct\text{ }option\text{ }is\text{ }\left( d \right)\end{array}$
Question 4
A certain city has a circular wall around it, and this wall has four gates pointing north south, east and west. A house stands outside the city, three kms north of the north gate, and it can just be seen from a point nine kms east of the south gate. What is the diameter of the wall that surrounds the city?
A 6 km B 9 km C 12 km D None of these
Question 4 Explanation:
$\displaystyle \begin{array}{l}Let\text{ }the\text{ }radius\text{ }be\text{ }r\text{ }and\text{ }the\text{ }center\text{ }be\text{ }0,0\\Now\text{ }since\text{ }between\text{ }the\text{ }point\text{ }from\text{ }where\text{ }it\text{ }can\text{ }be\text{ }just\text{ }seen\text{ }\left( -9,-r \right)\\\sqrt{6r+9}+9=\sqrt{{{(2r+3)}^{2}}+{{9}^{2}}}\\By\,\,putting\,the\,\,values\,\,only\,\,b\,\,satisfies\end{array}$
Question 5
Consider a cylinder of height h cm and radius r = r/2cm as shown in the figure (not drawn to scale).
A string of a certain length, when wound on its cylindrical surface, starting at point A and ending at
point B, gives a maximum of n turns (in other words, the string's length is the minimum length required
to wind n turns.
What is the vertical spacing in ems between two consecutive turns
A h/n B h/√n C h/h2 D Cannot be determined with given information
Question 5 Explanation:
Opening up the cylinder into a rectangle we can see that the
string is wound to a height of h with equal vertical spacing for n times.
Thus the distance = h/n
Correct option is (a)
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
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