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https://timestablesworksheets.com/2-digit-by-1-digit-multiplication-worksheets-without-regrouping/
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# 2 Digit By 1 Digit Multiplication Worksheets Without Regrouping
2 Digit By 1 Digit Multiplication Worksheets Without Regrouping – One of the most challenging and demanding issues you can do with basic school students is purchase them to experience math. Addition worksheets and subtraction worksheets aren’t what most youngsters wish to be performing throughout their day. Math in standard phrases is usually everything very easy to train, but in terms of teaching the better complicated hypotheses of mathematics, all around 3rd level, it could be a little more tough to maintain the students interested. It can be even more difficult to put into practice methods of training that can help the students be aware of the ideas of multiplication without the need of receiving puzzled.
## Download 2 Digit By 1 Digit Multiplication Worksheets Without Regrouping to Learn Multiplication
Multiplication tables are a great way to break into instructing multiplication. The 5 times tables and other little numbers prove much easier to teach most primary school pupils. College students may find it simple to merely continue to keep adding an additional digit on their last final result to find the next number inside the time tables. By educating this primary, you may get the scholars within the disposition and cozy with multiplying. The issues may possibly happen, nonetheless, as soon as you’ve arrived at a higher times tables than 5 various. The key of just adding an additional digit on the final result, to get the following number from the table becomes much more tough.
There is available a stage exactly where pupils will need to start off memorizing the times tables in order to remember them, in contrast to having the capability to utilize a particular technique. In this article you have to have the capacity to put into practice enjoyable types of instructing to create memorizing the times tables a far much less daunting process. This is a crucial part of mathematics. If your college students neglect to come to be good with their times tables soon they’ll have a problem from the afterwards levels of school. So what can you really do to be able to increase their capacity within this discipline?
## 2 Digit By 1 Digit Multiplication Worksheets Without Regrouping is Important to Improve your Math Skills
You can find a number of stuff that both parents and teachers likewise can perform to aid the student become successful. On the part of their moms and dads, they should conduct a number of points. Firstly, a homework system has to be integrated in your own home to make certain that your child is not disregarding their reports in your own home. Dismissing their work at home and trying to perform every thing at school never aids! By sitting down along with your kids throughout the night, turning from the television and supporting them turn out to be knowledgeable about multiplication you will finally be aiding them perform greater in school. I often advise flash cards for multiplication in your house.
If you think your son or daughter will benefit, you can find stuff you can get to boost their reports. You may clearly get textbooks to enable them to, however, many children see looking at like a task nowadays. You can select from a number of pc software program Compact disks and math sites that will make multiplication exciting. These math games enable your little one to obtain fun taking part in games while learning multiplication tables simultaneously.
Educators must naturally established correct due diligence, and ensure that it’s not too significantly. Due diligence is an important part of a student’s reports, but placing excessive can lower morale and create a little one feel overwhelmed. Setting entertaining and sensible amounts of due diligence can seriously help a young child on their own approach to memorizing and knowing the multiplication tables.he most significant issue to remember when trying to help you your child is in order to ensure that is stays enjoyable. Math should be shown to kids in a exciting way. They should not even know they are discovering.
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In this tutorial I want to show how you can use alluvial plots to visualise model response in up to 4 dimensions. `easyalluvial` generates artificial data space using fixed values for unplotted variables or uses the partial dependence plotting method. It is model agnostic but offers some convenient wrappers for `caret` models.
# Introduction
## Taking a peek
When building machine learning model we are usually faced with a trade-off between accurracy and interpretability. However even if we tend to lean towards accuracy and pick a modelling method that results in nearly uninterpretable models we can still make use of a bunch of model agnostic techniques that have been summarized in this excellent ebook Interpretable Machine Learning by Christoph Molnar.
Whithout getting to theoretical I personally always feel the urge to simply take a peek simulate some data and see how the model reacts a method described in Wickham H, Cook D, Hofmann H (2015) Visualizing statistical models: Removing the blindfold. Statistical Analysis and Data Mining 8(4) <doi:10.1002/sam.11271>. In order to simulate data we can generate a a vector with a sequence of values over the entire range of a predictor variable of interest while setting all the others to their median or mode and use this artificial data space to obtain model predictions which we can plot against our variable of interest. An R package that will do this for you is (`plotmo`)[https://cran.r-project.org/web/packages/plotmo/index.html]. Instead of ranging over 1 predictor variable we can create a data grid using 2 predictor variables and plot the response as a third dimension. However this is as far as you can go in a conventional plots. Alluvial plots can line up much more than 3 dimensions on a plane next to each other only limited by the number of flows as it will get too cluttered when there are too many of them.
## Which variables to plot
When using conventional model response plotting beeing limited two 2 variables we can simply resolve this by generating many plots and look at them one by one. Alluvial plots require a bit more attention and cannot easily be screened and compared since visually there is so much going on. Therefore I do not recommend to brute force it by simply creating a lot of random combinations of predictor variables and multiple alluvial plots but instead to focus on those that have the highest calculated feature importance. Feature importance values are natively provided by most modelling packages. So the question is how many can we plot and it turns out 4 features will usually result in well interpretable plot.
## Generating the data space
``````suppressPackageStartupMessages( require(tidyverse) )
suppressPackageStartupMessages( require(easyalluvial) )
suppressPackageStartupMessages( require(mlbench) )
suppressPackageStartupMessages( require(randomForest) )
``````
We start by creating a model
``````data('BostonHousing')
df = as_tibble( BostonHousing )
m = randomForest( lstat ~ ., df )
``````
and looking at the importance features
``````imp = m\$importance %>%
tidy_imp(df)
knitr::kable(imp)
``````
vars imp
medv 8352.4165
rm 3370.7073
indus 2765.8359
age 2716.4687
crim 2134.6400
dis 1819.3780
nox 1696.9545
tax 724.1518
b 682.6062
ptratio 340.1351
rad 222.3688
chas 131.2264
zn 101.5024
When generating the data space we cannot screent and infinite amount of values per variable. We want to create all possible combinations between the values of the 4 variables we want to plot and an alluvial plot we cannot distinguish more than 1000 flows I recommend to go with 5 values which will result in 5 x 5 X 5 X 5 –> 625 combinations. That also leaves some wiggeling room if one of the top 4 variables is a factor with more than 5 levels. `get_data_space()` will split the range of a variable into 3 and picks the median of each split and add the variable minimum and the maximum to the set.
``````dspace = get_data_space(df, imp
, degree = 4 # specifies the number of variables
, bins = 5 # the number of values per variable
)
knitr::kable( head(dspace, 10) )
``````
medv rm indus age crim dis nox tax b ptratio rad chas zn
5 3.561 0.46 2.90 0.25651 3.20745 0.538 330 391.44 19.05 5 0 0
5 3.561 0.46 32.20 0.25651 3.20745 0.538 330 391.44 19.05 5 0 0
5 3.561 0.46 67.95 0.25651 3.20745 0.538 330 391.44 19.05 5 0 0
5 3.561 0.46 94.60 0.25651 3.20745 0.538 330 391.44 19.05 5 0 0
5 3.561 0.46 100.00 0.25651 3.20745 0.538 330 391.44 19.05 5 0 0
5 3.561 4.27 2.90 0.25651 3.20745 0.538 330 391.44 19.05 5 0 0
5 3.561 4.27 32.20 0.25651 3.20745 0.538 330 391.44 19.05 5 0 0
5 3.561 4.27 67.95 0.25651 3.20745 0.538 330 391.44 19.05 5 0 0
5 3.561 4.27 94.60 0.25651 3.20745 0.538 330 391.44 19.05 5 0 0
5 3.561 4.27 100.00 0.25651 3.20745 0.538 330 391.44 19.05 5 0 0
Total rows in dspace: 625
``````dspace %>%
summarise_all( ~ length( unique(.) ) ) %>%
knitr::kable()
``````
medv rm indus age crim dis nox tax b ptratio rad chas zn
5 5 5 5 1 1 1 1 1 1 1 1 1
## Generating model response
``````pred = predict(m, newdata = dspace)
``````
## Plotting
``````alluvial_model_response(pred, dspace, imp, degree = 4, bins = 5)
``````
`easyalluvial` allows you to build exploratory alluvial plots (sankey diagrams) with a single line of code while automatically binning numerical variables. In version `0.2.0` marginal histograms improve the visibility of the numerical variables. Further a method has been added that creates model agnostic 4 dimensional partial dependence plots to visualise model response.
# Introduction
I am happy to announce the release of `easyalluvial 0.2.0` with some exciting new features and some mayor changes compared to version `0.1.8` Some improvements were made on the default plotting options which alter the default plot output thus classify as mayor changes (see below).
``````suppressPackageStartupMessages( require(tidyverse) )
suppressPackageStartupMessages( require(easyalluvial) )
``````
# Marginal Histograms
The automated binning of numerical variables hides their distribution. This new feature alleviates that.
``````p = alluvial_wide(mtcars2, max_variables = 9)
p_grid = add_marginal_histograms(p, mtcars2)
``````
{{< wide-image src=“/images/marg_hits.png” title=”]” >}}
# 4 Dimensional Partial Dependence Plots
Techniques for explaining the predictions of machine learning models have been discussed a lot lately. For an introduction into the topic I can recommend this excellent ebook Interpretable Machine Learning by Christoph Molnar. Inspired by packages like `plotmo` and `pdp` which can make partial dependence plots (PDP) of 2 feature variables against the response variable I tested whether alluvial plots, which can basically line up an unlimited amount of features on a 2 dimensional plane, can be used to make PDPs with more than 2 feature variables. And basically it turns out that you can use 4 features before things get to cluttered. And this is how it looks like:
``````df = select(mtcars2, -ids)
train = caret::train( disp ~ .
, df
, method = 'rf'
, trControl = caret::trainControl( method = 'none' )
, importance = TRUE )
p = alluvial_model_response_caret(train, degree = 4, method = 'pdp')
p_grid = add_marginal_histograms(p, df, plot = F) %>%
add_imp_plot(p, df)
``````
We see that the top 4 important features of the model have been selected and that 5 values have been picked over the range of the numerical variables which together with the levels of the categorical variable have been used to construct a data grid of all possible combinations. This data grid has been combined with the values of the remaining features of each observation in the training data which then has been used to generate model predictions which were consecutively averaged (see the ebook for a better explanation of PDPs).
{{< wide-image src=“/images/pdp.png” title=”]” >}}
#### Step-by-Step
• On the left we see the averaged model predictions that are generated by a specific combination of the 4 most important variables. You can find the combinations by tracing the coloured flows.
• The stratum label of the individual feature variables indicate the variable value and which fraction of the colored flows pass through it.
• On the right you see the feature importance of all variables and the proportion contributed by the plotted variables on the alluvial plot.
• On the top left you see how the distribution of the generated predictions compare to the distribution of the predicted variable (in this case disp) in the training data.
• The marginal histograms indicate the original distributions in the raining data and the lines indicate the location of the values picked for the data grid.
A more in-depth tutorial for this feature can be found on the project’s github page which will also be vailable on this blog in a few days.
If you are as enthusiastic about alluvial plots as me you will appreciate this plot because it can help you to get an intuitive mid-level understanding of how you’re model is making the predictions. Just be aware of a few limitations.
#### Limitations
• There is a loss of information when binning the numerical variables
• The combinations generated when making the grid might be outside the feature distribution space (generate combinations that are impossible)
• We only look at the combination of 4 features and disregard the others
To alleviate this you can reduce the complexity of the model by reducing features (take out correlating variables) or use additional model exploration methods such as classical PDPs, ALE plots, Shapely values, etc, …
#### We do not have to use `caret`
Note: importance is calculated differently when using this implementation of random forest.
``````df = select(mtcars2, -ids)
m = randomForest::randomForest( disp ~ ., df)
imp = m\$importance
dspace = get_data_space(df, imp, degree = 4)
pred = get_pdp_predictions(df, imp
, .f_predict = randomForest:::predict.randomForest
, m
, degree = 4
, bins = 5)
p = alluvial_model_response(pred, dspace, imp, degree = 4, method = 'pdp')
p_grid = add_marginal_histograms(p, df, plot = F) %>%
add_imp_plot(p, df)
``````
# Changes in Default Plotting Settings
• Default colors have been changed. The first 7 colors of `palette_qualitative()` the function that provides the default colors have hand-picked for better contrast.
• The stratum fill color of the variable determining the flow fill color in `alluvial_wide()` hast been set to match the flow fill color.
• label text size can now be modified via `stratum_label_size` parameter. Labels have gotten slightly bigger by default.
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https://www.aqua-calc.com/one-to-one/density/gram-per-cubic-decimeter/pennyweight-per-us-cup/1
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# 1 gram per cubic decimeter [g/dm³] in pennyweights per US cup
## g/dm³ to dwt/US c unit converter of density
1 gram per cubic decimeter [g/dm³] = 0.15 pennyweight per US cup [dwt/c]
### grams per cubic decimeter to pennyweights per US cup density conversion cards
• 1
through
25
grams per cubic decimeter
• 1 g/dm³ to dwt/c = 0.15 dwt/c
• 2 g/dm³ to dwt/c = 0.3 dwt/c
• 3 g/dm³ to dwt/c = 0.46 dwt/c
• 4 g/dm³ to dwt/c = 0.61 dwt/c
• 5 g/dm³ to dwt/c = 0.76 dwt/c
• 6 g/dm³ to dwt/c = 0.91 dwt/c
• 7 g/dm³ to dwt/c = 1.06 dwt/c
• 8 g/dm³ to dwt/c = 1.22 dwt/c
• 9 g/dm³ to dwt/c = 1.37 dwt/c
• 10 g/dm³ to dwt/c = 1.52 dwt/c
• 11 g/dm³ to dwt/c = 1.67 dwt/c
• 12 g/dm³ to dwt/c = 1.83 dwt/c
• 13 g/dm³ to dwt/c = 1.98 dwt/c
• 14 g/dm³ to dwt/c = 2.13 dwt/c
• 15 g/dm³ to dwt/c = 2.28 dwt/c
• 16 g/dm³ to dwt/c = 2.43 dwt/c
• 17 g/dm³ to dwt/c = 2.59 dwt/c
• 18 g/dm³ to dwt/c = 2.74 dwt/c
• 19 g/dm³ to dwt/c = 2.89 dwt/c
• 20 g/dm³ to dwt/c = 3.04 dwt/c
• 21 g/dm³ to dwt/c = 3.19 dwt/c
• 22 g/dm³ to dwt/c = 3.35 dwt/c
• 23 g/dm³ to dwt/c = 3.5 dwt/c
• 24 g/dm³ to dwt/c = 3.65 dwt/c
• 25 g/dm³ to dwt/c = 3.8 dwt/c
• 26
through
50
grams per cubic decimeter
• 26 g/dm³ to dwt/c = 3.96 dwt/c
• 27 g/dm³ to dwt/c = 4.11 dwt/c
• 28 g/dm³ to dwt/c = 4.26 dwt/c
• 29 g/dm³ to dwt/c = 4.41 dwt/c
• 30 g/dm³ to dwt/c = 4.56 dwt/c
• 31 g/dm³ to dwt/c = 4.72 dwt/c
• 32 g/dm³ to dwt/c = 4.87 dwt/c
• 33 g/dm³ to dwt/c = 5.02 dwt/c
• 34 g/dm³ to dwt/c = 5.17 dwt/c
• 35 g/dm³ to dwt/c = 5.32 dwt/c
• 36 g/dm³ to dwt/c = 5.48 dwt/c
• 37 g/dm³ to dwt/c = 5.63 dwt/c
• 38 g/dm³ to dwt/c = 5.78 dwt/c
• 39 g/dm³ to dwt/c = 5.93 dwt/c
• 40 g/dm³ to dwt/c = 6.09 dwt/c
• 41 g/dm³ to dwt/c = 6.24 dwt/c
• 42 g/dm³ to dwt/c = 6.39 dwt/c
• 43 g/dm³ to dwt/c = 6.54 dwt/c
• 44 g/dm³ to dwt/c = 6.69 dwt/c
• 45 g/dm³ to dwt/c = 6.85 dwt/c
• 46 g/dm³ to dwt/c = 7 dwt/c
• 47 g/dm³ to dwt/c = 7.15 dwt/c
• 48 g/dm³ to dwt/c = 7.3 dwt/c
• 49 g/dm³ to dwt/c = 7.45 dwt/c
• 50 g/dm³ to dwt/c = 7.61 dwt/c
• 51
through
75
grams per cubic decimeter
• 51 g/dm³ to dwt/c = 7.76 dwt/c
• 52 g/dm³ to dwt/c = 7.91 dwt/c
• 53 g/dm³ to dwt/c = 8.06 dwt/c
• 54 g/dm³ to dwt/c = 8.22 dwt/c
• 55 g/dm³ to dwt/c = 8.37 dwt/c
• 56 g/dm³ to dwt/c = 8.52 dwt/c
• 57 g/dm³ to dwt/c = 8.67 dwt/c
• 58 g/dm³ to dwt/c = 8.82 dwt/c
• 59 g/dm³ to dwt/c = 8.98 dwt/c
• 60 g/dm³ to dwt/c = 9.13 dwt/c
• 61 g/dm³ to dwt/c = 9.28 dwt/c
• 62 g/dm³ to dwt/c = 9.43 dwt/c
• 63 g/dm³ to dwt/c = 9.58 dwt/c
• 64 g/dm³ to dwt/c = 9.74 dwt/c
• 65 g/dm³ to dwt/c = 9.89 dwt/c
• 66 g/dm³ to dwt/c = 10.04 dwt/c
• 67 g/dm³ to dwt/c = 10.19 dwt/c
• 68 g/dm³ to dwt/c = 10.34 dwt/c
• 69 g/dm³ to dwt/c = 10.5 dwt/c
• 70 g/dm³ to dwt/c = 10.65 dwt/c
• 71 g/dm³ to dwt/c = 10.8 dwt/c
• 72 g/dm³ to dwt/c = 10.95 dwt/c
• 73 g/dm³ to dwt/c = 11.11 dwt/c
• 74 g/dm³ to dwt/c = 11.26 dwt/c
• 75 g/dm³ to dwt/c = 11.41 dwt/c
• 76
through
100
grams per cubic decimeter
• 76 g/dm³ to dwt/c = 11.56 dwt/c
• 77 g/dm³ to dwt/c = 11.71 dwt/c
• 78 g/dm³ to dwt/c = 11.87 dwt/c
• 79 g/dm³ to dwt/c = 12.02 dwt/c
• 80 g/dm³ to dwt/c = 12.17 dwt/c
• 81 g/dm³ to dwt/c = 12.32 dwt/c
• 82 g/dm³ to dwt/c = 12.47 dwt/c
• 83 g/dm³ to dwt/c = 12.63 dwt/c
• 84 g/dm³ to dwt/c = 12.78 dwt/c
• 85 g/dm³ to dwt/c = 12.93 dwt/c
• 86 g/dm³ to dwt/c = 13.08 dwt/c
• 87 g/dm³ to dwt/c = 13.24 dwt/c
• 88 g/dm³ to dwt/c = 13.39 dwt/c
• 89 g/dm³ to dwt/c = 13.54 dwt/c
• 90 g/dm³ to dwt/c = 13.69 dwt/c
• 91 g/dm³ to dwt/c = 13.84 dwt/c
• 92 g/dm³ to dwt/c = 14 dwt/c
• 93 g/dm³ to dwt/c = 14.15 dwt/c
• 94 g/dm³ to dwt/c = 14.3 dwt/c
• 95 g/dm³ to dwt/c = 14.45 dwt/c
• 96 g/dm³ to dwt/c = 14.6 dwt/c
• 97 g/dm³ to dwt/c = 14.76 dwt/c
• 98 g/dm³ to dwt/c = 14.91 dwt/c
• 99 g/dm³ to dwt/c = 15.06 dwt/c
• 100 g/dm³ to dwt/c = 15.21 dwt/c
• dwt/c stands for dwt/US c
#### Foods, Nutrients and Calories
CRACKERS, UPC: 790956404143 contain(s) 433 calories per 100 grams or ≈3.527 ounces [ price ]
BLACKENED SEASONING, UPC: 011110712363 weigh(s) 202.88 gram per (metric cup) or 6.77 ounce per (US cup) [ weight to volume | volume to weight | price | density ]
#### Gravels, Substances and Oils
CaribSea, Freshwater, African Cichlid Mix, Ivory Coast Gravel weighs 1 505.74 kg/m³ (94.00028 lb/ft³) with specific gravity of 1.50574 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Thallium [Tl] weighs 11 850 kg/m³ (739.77133 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-502, liquid (R502) with temperature in the range of -51.12°C (-60.016°F) to 60°C (140°F)
#### Weights and Measurements
A micron per hour squared (µm/h²) is a derived metric SI (System International) measurement unit of acceleration
Power is defined as energy transferred from or to an object by a force on the object per unit of time.
st/ft² to gr/pm² conversion table, st/ft² to gr/pm² unit converter or convert between all units of surface density measurement.
#### Calculators
Volume of a sphere calculator with surface area to volume ratio
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https://uk.mathworks.com/matlabcentral/cody/players/6243205-marian-kind/solved
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Cody
# Marian Kind
Rank
Score
1 – 26 of 26
#### Problem 11. Back and Forth Rows
Created by: Cody Team
Tags matrices
#### Problem 32. Most nonzero elements in row
Created by: Cody Team
Tags matrices
#### Problem 14. Find the numeric mean of the prime numbers in a matrix.
Created by: Cody Team
Tags easy, matrices
#### Problem 25. Remove any row in which a NaN appears
Created by: Cody Team
#### Problem 23. Finding Perfect Squares
Created by: Cody Team
#### Problem 33. Create times-tables
Created by: Cody Team
Tags matrices
#### Problem 12. Fibonacci sequence
Created by: Cody Team
#### Problem 10. Determine whether a vector is monotonically increasing
Created by: Cody Team
#### Problem 19. Swap the first and last columns
Created by: Cody Team
#### Problem 4. Make a checkerboard matrix
Created by: Cody Team
#### Problem 1774. Free passes for everyone!
Created by: Alfonso Nieto-Castanon
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1. ## A few questions.
I just have a few general questions.
1). Lets say I have a joint pdf f(r,h) where r is radius and h is height and the function is in terms of both r and h. How do I transform this into a joint pdf of volume and surface area?
2). I have two independent standard exponential random variables X1 and X2. I want to find the pdf of Y = X1 / (X1 + X2). How do I do this? Y is not a linear combination of X1 and X2.
3). I want to find the linear combination of a standard uniform and a standard exponential random variable. Do I just add them together?
4)> I want to find Z = XY where X and Y are both independent uniform random variables. Do I just multiply them?
I missed the lecture in class and by just looking at the notes, I have no idea how do tackle these problems so any help is appreciated. Thanks.
2. You need to review calculus three change of variables.
Especially the part where these use Jacobians.
You need to make a transformation from $R^2$ to $R^2$
I did the product of uniforms here this past summer.
If you want Y=X1/(X1+X2) then let W=X1 and find the joint density of Y and W.
Then integrate out the random variable W.
3. Could you expand on that a little bit? I'm still not quiet sure what to do
4. Originally Posted by CUEngineering
Could you expand on that a little bit? I'm still not quiet sure what to do
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My Math Forum changing order of integration
Real Analysis Real Analysis Math Forum
November 30th, 2012, 12:37 PM #1 Newbie Joined: Nov 2012 Posts: 1 Thanks: 0 changing order of integration Hi! I'm solving this problem and I'm not sure how to solve it I have definite triple integral of function f(x,y,z). It's domain is set $M = \{0 I would like to get function g(z) = \int\int f(x,y,z) dxdy but I'm quite confused with borders of integration (and od course domain of the function g) -> Could you give me some help? Thanks in advance! --- -> and here goes my idea (but I'm not sure about it): - for each$z$and$y$I'm able to get an interval for$x$:$\max \left(0, \frac{z-2y^2+3ay}{y+a} \right) < x < \min \left( a , \frac{a^2 - 2y^2+2ay + z}{y} \right)$- therefore I can get function \int_max(...)^min(...) f(x,y,z)dxdydz = [h]_{\max(...)}^{\min(...)}$ - for each $z$ I'm able to say which values functions min/max takes - so the function g(z) could be sum of integrals of k(y,z) (the number of integrals and their domains depends on values of min/max function..) -> so, is it good idea or really wrong way of thinking about this problem? -> and what about domain of g(z)? (could it be [min_(x,y)f(x,y,z);max_(x,y)f(x,y,z)] ?) --- Once again thanks in advance! (and sorry for my English.. ). Have a nice day! Doxxik
December 14th, 2012, 05:40 PM #2 Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: changing order of integration All you are told about x and y is that 0< x< a and 0< y< a so that the integrals, with respect to x and y, are from 0 to a. The result will be a function of z and its domain will be the possible values of z. What are the maximum and minimum values of the two functions bounding z on the square 0< x< a and 0< y< a?
Tags changing, integration, order
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# CLRM – Assumption 1: Linear Parameter and correct model specification
Assumption 1 requires that the dependent variable $\textbf{y}$ is a linear combination of the explanatory variables $\textbf{X}$ and the error terms $\epsilon$. Assumption 1 requires the specified model to be linear in parameters, but it does not require the model to be linear in variables. Equation 1 and 2 depict a model which is both, linear in parameter and variables. Note that Equation 1 and 2 show the same model in different notation.
(1) $\textbf{y} = \textbf{X}\boldsymbol\beta + \boldsymbol\epsilon$
(2) $y_{i} = \beta_{0} + \beta_{1}x_{i1} + \beta_{2}x_{i2} + ... + \beta_{K}x_{iK} + \epsilon_{i}$
In order for OLS to work the specified model has to be linear in parameters. Note that if the true relationship between $x_{1}$ and $y$ is non linear it is not possible to estimate the coefficient $\beta$ in any meaningful way. Equation 3 shows an empirical model in which $\beta_{1}$ is of quadratic nature.
(3) $y_{i} = \beta_{0} + (\beta_{1})^{2}x_{i1} + \beta_{2}x_{i2} + ... + \beta_{K}x_{iK} + \epsilon_{i}$
Assumption 1 of CLRM requires the model to be linear in parameters. OLS is not able to estimate Equation 3 in any meaningful way. However, assumption 1 does not require the model to be linear in variables. OLS will produce a meaningful estimation of $\beta_{1}$ in Equation 4.
(4) $y_{i} = \beta_{0} + \beta_{1}(x_{i1})^{2} + \beta_{2}x_{i2} + ... + \beta_{K}x_{iK} + \epsilon_{i}$
Using the method of ordinary least squares (OLS) allows us to estimate models which are linear in parameters, even if the model is non linear in variables. On the contrary it is not possible to estimate models which are non linear in parameters, even if they are linear in variables.
Finally, every model estimated with OLS should contains all relevant explanatory variables and all included explanatory variables should be relevant. Not including all relevant variables gives rise to the omitted variables bias. A very serious issue in regression analysis.
## 9 thoughts on “CLRM – Assumption 1: Linear Parameter and correct model specification”
1. Jane says:
I have a question with regards to this. If for instance, the model if: ln(y) = a+bx^2 +e. do we still say that it satisfied GMA?
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You are Here: Home >< Maths
# Very important question
Announcements Posted on
Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016
1. Are there any other names for the harmonic series????????????????
Plz help
2. Zacken
physicsmaths
TeeEm
3. (Original post by Gome44)
Are there any other names for the harmonic series????????????????
Plz help
1. The -series with .
2.
Any reason why you're looking for alternate names?
4. (Original post by Zacken)
1. The -series with .
2.
Any reason why you're looking for alternate names?
Secret
5. I think TeeEm has left TSR now.
6. (Original post by Gome44)
Secret
I can't think of any other names.
7. (Original post by XxKingSniprxX)
I think TeeEm has left TSR now.
wut. But he was a top lad :'(
8. Yeh sorry mate. Ain't got nothing other then what zacken wrote.
9. Sum of the reciprocals of the positive integers.
10. i think we should call it The Trickster
because it looks like it should converge....
Spoiler:
Show
BUT IT DOESN'T !!!!
11. (Original post by Zacken)
1. The -series with .
2.
Any reason why you're looking for alternate names?
Actually zeta(1) is technically not the harmonic series, since zeta(s) is only defined by the p-series for Re(s)>1.
(But I don't know this for sure.)
12. (Original post by IrrationalRoot)
Actually zeta(1) is technically not the harmonic series, since zeta(s) is only defined by the p-series for Re(s)>1.
(But I don't know this for sure.)
Yeaaah, I was thinking that as well, but meh. I could've sworn I've seen used fairly frequently though. And then there's the whole zeta renormalisation thingy with zeta(1) = -1/12 so idk plus there's the whole analytical continuation thingy.
13. (Original post by Zacken)
Yeaaah, I was thinking that as well, but meh. I could've sworn I've seen used fairly frequently though. And then there's the whole zeta renormalisation thingy with zeta(1) = -1/12 so idk plus there's the whole analytical continuation thingy.
I know, it's all really complicated and technical. I was thinking it was only -1/12 by analytic continuation. Anyway, I think I'll ignore those details until uni .
14. (Original post by IrrationalRoot)
I know, it's all really complicated and technical. I was thinking it was only -1/12 by analytic continuation. Anyway, I think I'll ignore those details until uni .
Same.
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# On the nonlinear Hadamard-type integro-differential equation
## Abstract
This paper studies uniqueness of solutions for a nonlinear Hadamard-type integro-differential equation in the Banach space of absolutely continuous functions based on Babenko’s approach and Banach’s contraction principle. We also include two illustrative examples to demonstrate the use of main theorems.
## Introduction
The Hadamard fractional integration and differentiation are based on the nth integral of the form [1, 2]
\begin{aligned} \bigl(\mathcal{J}^{n}_{ a + , \mu } u\bigr) (x) = & x^{-\mu } \int _{a}^{x} \frac{d t_{1}}{t_{1}} \int _{a}^{t_{1}} \frac{d t_{2}}{t_{2}} \cdots \int _{a}^{t_{n - 1}} t_{n}^{\mu } u(t_{n}) \frac{d t_{n}}{t_{n}} \\ = & \frac{1}{(n - 1)!} \int _{a}^{x} \biggl(\frac{t}{x} \biggr)^{\mu } \biggl(\log \frac{x}{t} \biggr)^{n - 1} u(t) \frac{d t}{t} \end{aligned}
and the corresponding derivative
\begin{aligned}& \bigl(\mathcal{D}^{1}_{a + , \mu } u\bigr) (x) = \bigl(( \delta + \mu ) u\bigr) (x) = x u'(x) + \mu u(x),\quad \delta = x \frac{d}{d x}, \\& \mathcal{D}^{n}_{a + , \mu } u = \mathcal{D}^{1}_{a + , \mu } \bigl(\mathcal{D}^{n - 1}_{a + , \mu } u\bigr),\quad n = 2, 3, \ldots , \end{aligned}
where $$\log ( \cdot ) = \log _{e} (\cdot )$$, $$0 < a < x < b$$, and $$\mu \in R$$.
The fractional version of the Hadamard-type integral and derivative are given by
$$\bigl(\mathcal{J}^{\alpha }_{ a + , \mu } u\bigr) (x) = \frac{1}{\Gamma (\alpha )} \int _{a}^{x} \biggl(\frac{t}{x} \biggr)^{\mu } \biggl(\log \frac{x}{t} \biggr)^{\alpha - 1} u(t) \frac{d t}{t},\quad \alpha > 0$$
and
$$\bigl(\mathcal{D}^{\alpha }_{a + , \mu } u\bigr) (x) = x^{-\mu } \delta ^{n} x^{\mu }\bigl( \mathcal{J}^{n - \alpha }_{ a + , \mu } u\bigr) (x),$$
where $$n = [\alpha ] + 1$$, and $$[\alpha ]$$ being integral part of α.
When $$0 < \alpha < 1$$, the fractional derivative turns out to be
\begin{aligned} \bigl(\mathcal{D}^{\alpha }_{ a + , \mu } u\bigr) (x) =& x^{-\mu } \delta x^{\mu }\bigl(\mathcal{J}^{1 - \alpha }_{ a + , \mu } u\bigr) (x) \\ =& \frac{1}{\Gamma (1 - \alpha )} x^{-\mu + 1} \frac{d}{d x} \int _{a}^{x} t^{\mu - 1} \biggl(\log \frac{x}{t} \biggr)^{- \alpha } u(t) \,dt. \end{aligned}
In particular, for $$\alpha = 1$$,
$$(\mathcal{J}_{ a + , \mu } u) (x) = \bigl(\mathcal{J}^{1}_{ a + , \mu } u\bigr) (x) = \frac{1}{\Gamma (\alpha ) x^{\mu }} \int _{a}^{x} t^{ \mu - 1} u(t) \,dt,$$
which leads to defining the space $${X}_{\mu }[a, b]$$ of those Lebesgue measurable functions u on $$[a, b]$$ for which $$x^{\mu - 1} u(x)$$ is absolutely integrable [2]:
$$X_{\mu }[a, b] = \biggl\{ u : [a, b] \rightarrow C \mbox{ and } \lVert u \rVert _{X_{\mu }} = \int _{a}^{b} x^{\mu - 1} \bigl\vert u(x) \bigr\vert \,dx < \infty \biggr\} .$$
Let $$\operatorname{AC}[a, b]$$ be the set of absolutely continuous functions on $$[a, b]$$. Then it follows from [3] that
$$u \in \operatorname{AC}[a, b] \quad \mbox{if and only if}\quad u(x) = u(a) + \int _{a}^{x} v(t) \,dt, \quad v(t) \in L[a, b].$$
Obviously,
$$\operatorname{AC}[a, b] \subset X_{\mu }[a, b].$$
The latter is a Banach space under its norm. We further define the space
$$\operatorname{AC}_{0}[a, b] = \biggl\{ u : u(x) \in \operatorname{AC}[a, b] \mbox{ with } u(a) = 0 \mbox{ and } \lVert u \rVert _{0} = \int _{a}^{b} \bigl\vert u'(x) \bigr\vert \,dx < \infty \biggr\} .$$
Clearly, $$\lVert u \rVert _{0}$$ is a norm in $$\operatorname{AC}_{0}[a, b]$$. Indeed, if $$\lVert u \rVert _{0} = 0$$ then $$u(x) = u(a) = 0$$. To show that $$\operatorname{AC}_{0}[a, b]$$ is complete, we assume $$\{u_{n} (x) \}$$ is a Cauchy sequence in $$\operatorname{AC}_{0}[a, b]$$, then we need to find a function $$u(x)$$ such that $$u(x)$$ is absolutely continuous and $$u_{n} \rightarrow u$$ under its norm. Since $$\{u_{n} (x) \}$$ is Cauchy in $$\operatorname{AC}_{0}[a, b]$$, we claim that $$u_{n}(a) = 0$$ and $$\{u'_{n} (x) \}$$ is Cauchy in $$L[a, b]$$. Hence, there exists $$g \in L[a, b]$$ such that $$u'_{n} \rightarrow g$$ in $$L[a, b]$$. Define
$$u(x) = \int _{a}^{x} g(\tau ) \,d\tau .$$
Then $$u(a) = 0$$ and $$u(x)$$ is absolutely continuous on $$[a, b]$$, and
$$\lVert u_{n} - u \rVert _{0} \leq \int _{a}^{b} \bigl\vert u_{n}'(x) - g(\tau ) \bigr\vert \,d\tau$$
converges to zero. Therefore, $$\operatorname{AC}_{0}[a, b]$$ is a Banach space.
### Lemma 1.1
If $$\alpha > 0$$, $$\mu \in R$$, and $$0 < a < b < \infty$$, then the operator $$\mathcal{J}^{\alpha }_{ a + , \mu }$$ is bounded in $$\operatorname{AC}_{0}[a, b]$$ and
$$\bigl\lVert \mathcal{J}^{\alpha }_{ a + , \mu } u \bigr\rVert _{0} \leq \frac{C_{\mu }}{\Gamma (\alpha + 1)} \biggl[\log \biggl( \frac{b}{a} \biggr) \biggr]^{\alpha } \lVert u \rVert _{0},$$
where $$C_{\mu }$$ is the maximum value of the function
$$w(t, x) = \biggl(\frac{t}{x} \biggr)^{\mu }$$
on $$[a, b] \times [a, b]$$.
### Proof
Let $$u \in \operatorname{AC}_{0}[a, b]$$. Then
$$u(t) = \int _{a}^{t} v(s) \,ds = \int _{a}^{t} u'(s) \,ds , \quad v(s) = u'(s) \in L[a, b],$$
and
\begin{aligned} \mathcal{J}^{\alpha }_{ a + , \mu } u =& \mathcal{J}^{\alpha }_{ a + , \mu } \int _{a}^{t} v(s) \,ds = \frac{1}{\Gamma (\alpha )} \int _{a}^{x} \biggl(\frac{t}{x} \biggr)^{\mu } \biggl(\log \frac{x}{t} \biggr)^{ \alpha - 1} \int _{a}^{t} v(s) \,ds \frac{d t}{t} \\ =& \frac{1}{\Gamma (\alpha )} \int _{a}^{x} v(s) \,ds \int _{s}^{x} \biggl(\frac{t}{x} \biggr)^{\mu } \biggl(\log \frac{x}{t} \biggr)^{ \alpha - 1} \frac{d t}{t}, \end{aligned}
by changing the order of integration. Using
$$0 \leq \biggl(\frac{t}{x} \biggr)^{\mu }\leq C_{\mu },$$
we imply that
\begin{aligned}& \biggl\vert v(s) \int _{s}^{x} \biggl(\frac{t}{x} \biggr)^{\mu } \biggl( \log \frac{x}{t} \biggr)^{\alpha - 1} \frac{d t}{t} \biggr\vert \leq \frac{C_{\mu }}{\alpha } \bigl\vert v(s) \bigr\vert \biggl[\log \biggl(\frac{b}{a} \biggr) \biggr]^{\alpha }\in L[a, b],\quad \mbox{and} \\& \bigl\lVert \mathcal{J}^{\alpha }_{ a + , \mu } u \bigr\rVert _{0} \leq \frac{C_{\mu }}{\Gamma (\alpha + 1)} \biggl[\log \biggl( \frac{b}{a} \biggr) \biggr]^{\alpha } \lVert u \rVert _{0}. \end{aligned}
This completes the proof of Lemma 1.1. □
Kilbas showed the following lemma in reference [2], which is soon to be used.
### Lemma 1.2
1. (i)
If $$\alpha >0$$, $$\beta > 0$$, $$\mu \in R$$, and $$u \in X_{\mu }[a, b]$$, then the semigroup property holds
$$\mathcal{J}^{\alpha }_{ a + , \mu } \mathcal{J}^{\beta }_{ a + , \mu } u = \mathcal{J}^{\alpha + \beta }_{ a + , \mu }u.$$
2. (ii)
If $$0 < \alpha < 1$$ and $$u \in \operatorname{AC}[a, b]$$, then
$$\mathcal{J}^{\alpha }_{a + , \mu } \mathcal{D}^{\alpha }_{a + , \mu } u = u.$$
Let $$u \in \operatorname{AC}[a, b]$$ and $$0 < \beta < 1$$. It follows from Lemma 1.2 that
$$\mathcal{J}^{\alpha }_{ a + , \mu } \mathcal{D}^{\beta }_{a + , \mu } u = \mathcal{J}^{\alpha - \beta }_{ a + , \mu } u$$
if $$\alpha \geq \beta$$.
Let $$0 < \alpha _{0} < \alpha _{ 1} < \cdots < \alpha _{n} < 1$$ and $$0 \leq \beta _{n + 1} < \cdots < \beta _{m} \in R$$, where $$n = 0, 1, \ldots$$ and $$m > n$$. In this paper, we show the uniqueness of solutions for the following new nonlinear Hadamard-type integro-differential equation for all $$\mu \in R$$ in the space $$\operatorname{AC}_{0}[a, b]$$:
\begin{aligned}& \mathcal{D}^{\alpha _{n}}_{a + , \mu } u + a_{n - 1} \mathcal{D}^{ \alpha _{n - 1}}_{a + , \mu } u + \cdots + a_{0} \mathcal{D}^{ \alpha _{0}}_{a + , \mu } u + b_{n + 1} \mathcal{J}^{\beta _{n + 1}}_{a + , \mu }u + \cdots + b_{m} \mathcal{J}^{\beta _{m}}_{a + , \mu }u \\& \quad = \int _{a}^{x} f\bigl(\tau , u'( \tau )\bigr) \,d\tau \end{aligned}
(1)
by Banach’s contraction principle and Babenko’s approach [4], with two applicable examples presented to illustrate the main results. It seems impossible to obtain these by any existing integral transforms or analytic local model methods. Babenko’s approach treats integral operators like variables in solving differential and integral equations. The method itself is close to the Laplace transform method in the ordinary sense, but it can be used in more cases [5, 6], such as dealing with integral or fractional differential equations with distributions whose Laplace transforms do not exist in the classical sense. Furthermore, it works well on certain differential or integral equations whose solutions cannot be achieved by the local model. Clearly, it is always necessary to show convergence of the series obtained as solutions. Recently, Li studied the generalized Abel’s integral equations of the first [7] and second kind with variable coefficients by Babenko’s technique [810].
It is well known that fractional calculus [3, 11, 12] has been an emergent tool which uses fractional differential and integral equations to develop more sophisticated mathematical models that can accurately describe complex systems. There are many definitions of fractional derivatives available in the literature, such as the Riemann–Liouville derivative which played an important role in the development of the theory of fractional analysis. However, the commonly used is the Hadamard fractional derivative (with $$\mu = 0$$) given by Hadamard in [13]. Butzer et al. [1416] studied various properties of the Hadamard-type derivative which is more generalized than the Hadamard fractional derivative. In particular, Hadamard fractional differential equations with boundary value problems or initial conditions have been investigated by researchers using fixed point theories [17, 18]. In 2014, Thiramanus et al. [19] studied the existence and uniqueness of solutions for a fractional boundary value problem involving Hadamard differential equations of order $$q \in (1, 2]$$ and nonlocal fractional integral boundary conditions by fixed point theories. In 2018, Matar [20] obtained the solution of the linear equations with the initial conditions (three terms on the left-hand side at most and a given function on the right) by the parameter technique, and then investigated the existence problems of the corresponding nonlinear types of Hadamard equations using fixed point theorems. Very recently, Ding et al. [21] applied the fixed point index and nonnegative matrices to study the existence of positive solutions for a system of Hadamard-type fractional differential equations with semipositone nonlinearities. In 1967, Caputo [22] introduced another type of fractional derivative which has an advantage over R-L derivative in differential equations since it does not require to define fractional order initial conditions. Jarad et al. [23] defined the Caputo-type modification of the Hadamard fractional derivatives which preserve physically interpretable initial conditions similar to the ones in Caputo fractional derivatives. Gambo et al. [24] further presented the generalization of the fundamental theorem of fractional calculus (FTFC) in the Caputo–Hadamard setting with several new results. Adjabi et al. [25] studied Cauchy problems for a differential equation with a left Caputo–Hadamard fractional derivative in spaces of continuously differentiable functions.
There are new studies on fixed point theorems for different operators on metric spaces [2628], as well as their applications in differential and integral equations, existence and uniqueness of solutions for equations [2931]. Palve et al. [32] recently constructed the existence and uniqueness of solutions for the fractional implicit differential equation with boundary condition of the form
\begin{aligned}& {{}_{H}}D^{\alpha , \beta }_{1+} u(x) = f\bigl(x, u(x), {{}_{H}}D^{\alpha , \beta }_{1+} u(x) \bigr),\quad 0 < \alpha < 1, 0 \leq \beta \leq 1, x \in [1, b], \\& \mathcal{J}^{1 - \gamma }_{1+, 0} c_{1} u(x) + c_{2} u \bigl(b^{-}\bigr) = c_{3},\quad \alpha \leq \gamma = \alpha + \beta (1 - \alpha ), \end{aligned}
where $${{}_{H}}D^{\alpha , \beta }_{1+}$$ is the Hilfer–Hadamard type fractional derivative of order α and type β given by
$${{}_{H}}D^{\alpha , \beta }_{1+} = \mathcal{J}^{\beta (n - \alpha )}_{1+, 0} D^{n} \mathcal{J}^{(1 - \beta )(n - \alpha )}_{1+, 0}, \quad n - 1 < \alpha < n,$$
and $$c_{1}, c_{2}, c_{3} \in R$$ with $$c_{1} + c_{2} \neq 0$$ and $$c_{2} \neq 0$$. Li [33] obtained uniqueness of solutions for the coupled system of integral equations
$$\textstyle\begin{cases} a_{n} (\mathcal{J}^{\alpha _{n}}_{ a + , \mu } u)(x) + \cdots + a_{1} (\mathcal{J}^{\alpha _{1}}_{ a + , \mu } u)(x) + u(x) = g_{1}(x, u(x), v(x)), \\ b_{n} (\mathcal{J}^{\beta _{n}}_{ a + , \mu } v)(x) + \cdots + b_{1} (\mathcal{J}^{\beta _{1}}_{ a + , \mu } v)(x) + v(x) = g_{2}(x, u(x), v(x)), \end{cases}$$
on the product space $$X_{\mu }[a, b] \times X_{\mu }[a, b]$$ (it is a Banach space), based on Babenko’s approach and Banach’s contraction principle.
## Main results
### Theorem 2.1
Assume that $$a_{i}$$ and $$b_{j}$$ for $$i = 0, 1, \ldots , n -1$$ and $$j = n +1, \ldots , m$$ are arbitrary complex numbers, and $$g \in \operatorname{AC}_{0}[a, b]$$. In addition, we let $$0 < \alpha _{0} < \alpha _{ 1} < \cdots < \alpha _{n} < 1$$ and $$0 \leq \beta _{n + 1} < \cdots < \beta _{m} \in R$$, where $$n = 0, 1, \ldots$$ . Then equation
$$\mathcal{D}^{\alpha _{n}}_{a + , \mu } u + a_{n - 1} \mathcal{D}^{ \alpha _{n - 1}}_{a + , \mu } u + \cdots + a_{0} \mathcal{D}^{ \alpha _{0}}_{a + , \mu } u + b_{n + 1} \mathcal{J}^{\beta _{n + 1}}_{a + , \mu }u + \cdots + b_{m} \mathcal{J}^{\beta _{m}}_{a + , \mu }u = g(x),$$
(2)
has a unique solution
$$u(x) = \sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1}} \cdots b_{m}^{k_{m}} \mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}( \alpha _{n } + \beta _{m}) + \alpha _{n } }_{a + , \mu } g$$
(3)
in the space $$\operatorname{AC}_{0}[a, b]$$.
### Proof
Applying the operator $$\mathcal{J}^{\alpha _{n }}_{a + , \mu }$$ to both sides of equation (2), we get
\begin{aligned}& \mathcal{J}^{\alpha _{n }}_{a + , \mu } \mathcal{D}^{\alpha _{n}}_{a + , \mu } u + a_{n - 1} \mathcal{J}^{\alpha _{n }}_{a + , \mu } \mathcal{D}^{\alpha _{n - 1}}_{a + , \mu } u + \cdots + a_{0} \mathcal{J}^{\alpha _{n }}_{a + , \mu } \mathcal{D}^{\alpha _{0}}_{a + , \mu } u \\& \quad {} + b_{n + 1} \mathcal{J}^{\alpha _{n }}_{a + , \mu } \mathcal{J}^{ \beta _{n + 1}}_{a + , \mu }u + \cdots + b_{m} \mathcal{J}^{ \alpha _{n }}_{a + , \mu } \mathcal{J}^{\beta _{m}}_{a + , \mu }u = \mathcal{J}^{\alpha _{n }}_{a + , \mu } g. \end{aligned}
Using Lemma 1.2,
\begin{aligned}& u + a_{n - 1} \mathcal{J}^{\alpha _{n } - \alpha _{n - 1}}_{a + , \mu } u + \cdots + a_{0} \mathcal{J}^{\alpha _{n } - \alpha _{0}}_{a + , \mu } u \\& \quad {}+ b_{n + 1} \mathcal{J}^{\alpha _{n } + \beta _{n + 1}}_{a + , \mu } u + \cdots + b_{m} \mathcal{J}^{\alpha _{n } + \beta _{m}}_{a + , \mu } u = \mathcal{J}^{\alpha _{n }}_{a + , \mu } g \end{aligned}
by noting that $$0 < \alpha _{0} < \alpha _{ 1} < \cdots < \alpha _{n} < 1$$. Hence,
\begin{aligned}& \bigl(1 + a_{n - 1} \mathcal{J}^{\alpha _{n } - \alpha _{n - 1}}_{a + , \mu } + \cdots + a_{0} \mathcal{J}^{\alpha _{n } - \alpha _{0}}_{a + , \mu } + b_{n + 1} \mathcal{J}^{\alpha _{n } + \beta _{n + 1}}_{a + , \mu } + \cdots + b_{m} \mathcal{J}^{\alpha _{n } + \beta _{m}}_{a + , \mu } \bigr) u \\& \quad = \mathcal{J}^{\alpha _{n }}_{a + , \mu } g. \end{aligned}
By Babenko’s method we come to
\begin{aligned} u(x) = & \bigl(1 + a_{n - 1} \mathcal{J}^{\alpha _{n } - \alpha _{n - 1}}_{a + , \mu } + \cdots + b_{m} \mathcal{J}^{\alpha _{n } + \beta _{m}}_{a + , \mu } \bigr)^{-1} \mathcal{J}^{\alpha _{n }}_{a + , \mu } g \\ = & \sum_{k = 0}^{\infty }(-1)^{k} \bigl(a_{n - 1} \mathcal{J}^{ \alpha _{n } - \alpha _{n - 1}}_{a + , \mu } + \cdots + b_{m} \mathcal{J}^{\alpha _{n } + \beta _{m}}_{a + , \mu } \bigr)^{k} \mathcal{J}^{\alpha _{n }}_{a + , \mu } g \\ = & \sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} \bigl(a_{n - 1} \mathcal{J}^{\alpha _{n } - \alpha _{n - 1}}_{a + , \mu } \bigr)^{k_{1}} \cdots \bigl( b_{m} \mathcal{J}^{\alpha _{n } + \beta _{m}}_{a + , \mu } \bigr)^{k_{m}} \mathcal{J}^{\alpha _{n }}_{a + , \mu } g \\ = & \sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1}} \mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1})}_{a + , \mu } \cdots b_{m}^{k_{m}} \mathcal{J}^{k_{m}(\alpha _{n } + \beta _{m})}_{a + , \mu } \mathcal{J}^{\alpha _{n }}_{a + , \mu } g \\ = & \sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1}} \cdots b_{m}^{k_{m}} \mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}( \alpha _{n } + \beta _{m}) + \alpha _{n } }_{a + , \mu } g, \end{aligned}
using Lemma 1.2 and the multinomial theorem. Clearly, $$u(a) = 0$$ since $$\alpha _{n } > 0$$ and
$$\bigl(\mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}( \alpha _{n } + \beta _{m}) + \alpha _{n } }_{a + , \mu } g \bigr) (a) = 0.$$
It remains to show that the series converges in the space $$\operatorname{AC}_{0}[a, b]$$ and is absolutely continuous on $$[a, b]$$. By Lemma 1.1,
$$\bigl\lVert \mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}(\alpha _{n } + \beta _{m}) + \alpha _{n }}_{ a + , \mu } g \bigr\rVert _{0} \leq K \lVert g \rVert _{0},$$
where
\begin{aligned} K = & \frac{C_{\mu }}{\Gamma (k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}(\alpha _{n } + \beta _{m}) + \alpha _{n } + 1)} \\ &{}\cdot \biggl(\log \frac{b}{a} \biggr)^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}(\alpha _{n } + \beta _{m}) + \alpha _{n }}. \end{aligned}
Therefore,
\begin{aligned} \lVert u \rVert _{0} \leq &C_{\mu }\sum _{k = 0}^{\infty }\sum _{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} \\ &{} \cdot \frac{ ( \vert a_{n - 1} \vert (\log \frac{b}{a} )^{\alpha _{n} - \alpha _{n - 1}} )^{k_{1}} \cdots ( \vert b_{m} \vert (\log \frac{b}{a} )^{\alpha _{n} + \beta _{m}} )^{k_{m}} }{\Gamma ( k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}(\alpha _{n } + \beta _{m}) + \alpha _{n } + 1)} \lVert g \rVert _{0} \\ =& C_{\mu }E_{(\alpha _{n} - \alpha _{n - 1}, \ldots , \alpha _{n} + \beta _{m}, \alpha _{n} + 1)} \biggl( \vert a_{n - 1} \vert \biggl(\log \frac{b}{a} \biggr)^{\alpha _{n} - \alpha _{n - 1}}, \ldots , \vert b_{m} \vert \biggl(\log \frac{b}{a} \biggr)^{\alpha _{n} + \beta _{m}} \biggr) \lVert g \rVert _{0}, \end{aligned}
where
$$E_{(\alpha _{n} - \alpha _{n - 1}, \ldots , \alpha _{n} + \beta _{m}, \alpha _{n} + 1)} \biggl( \vert a_{n - 1} \vert \biggl(\log \frac{b}{a} \biggr)^{ \alpha _{n} - \alpha _{n - 1}}, \ldots , \vert b_{m} \vert \biggl(\log \frac{b}{a} \biggr)^{\alpha _{n} + \beta _{m}} \biggr) < \infty$$
is the value at
$$z_{1} = \vert a_{n - 1} \vert \biggl(\log \frac{b}{a} \biggr)^{\alpha _{n} - \alpha _{n - 1}},\qquad \ldots,\qquad z_{m} = \vert b_{m} \vert \biggl(\log \frac{b}{a} \biggr)^{\alpha _{n} + \beta _{m}}$$
of the multivariate Mittag-Leffler function $$E_{(\alpha _{n} - \alpha _{n - 1}, \ldots , \alpha _{n} + \beta _{m}, \alpha _{n} + 1)}(z_{1}, \ldots , z_{m})$$ given in [12]. Thus, the series on the right-hand side of equation (3) is convergent. To see $$u(x)$$ is absolutely continuous,
\begin{aligned} u(x) = & \sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1}} \cdots b_{m}^{k_{m}} \\ &{}\cdot \mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}( \alpha _{n } + \beta _{m}) + \alpha _{n } }_{a + , \mu } \int _{a}^{t} g'(s) \,ds \\ = & \sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1}} \cdots b_{m}^{k_{m}} \\ &{}\cdot \frac{1}{\Gamma (k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}(\alpha _{n } + \beta _{m}) + \alpha _{n })} \int _{a}^{x} g'(s) \,ds \\ &{}\cdot \int _{s}^{x} \biggl(\frac{t}{x} \biggr)^{\mu } \biggl(\log \frac{x}{t} \biggr)^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}(\alpha _{n } + \beta _{m}) + \alpha _{n } - 1} \frac{d t}{t} \\ = & \int _{a}^{x} \sum _{k = 0}^{\infty }(-1)^{k} \sum _{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1}} \cdots b_{m}^{k_{m}} \\ & {}\cdot\frac{g'(s) }{\Gamma (k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}(\alpha _{n } + \beta _{m}) + \alpha _{n })} \\ &{}\cdot \int _{s}^{x} \biggl(\frac{t}{x} \biggr)^{\mu } \biggl(\log \frac{x}{t} \biggr)^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}(\alpha _{n } + \beta _{m}) + \alpha _{n } - 1} \frac{d t}{t} \,ds, \end{aligned}
as the function inside of the outer integral
\begin{aligned}& \sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1}} \cdots b_{m}^{k_{m}} \\& \quad {}\cdot \frac{g'(s) }{\Gamma (k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}(\alpha _{n } + \beta _{m}) + \alpha _{n })} \\& \quad {}\cdot \int _{s}^{x} \biggl(\frac{t}{x} \biggr)^{\mu } \biggl(\log \frac{x}{t} \biggr)^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}(\alpha _{n } + \beta _{m}) + \alpha _{n } - 1} \frac{d t}{t} \end{aligned}
uniformly converges with respect to t and belongs to $$L[a, b]$$ from Lemma 1.1 and the multivariate Mittag-Leffler function used above. Thus, $$u(x)$$ is absolutely continuous on $$[a, b]$$. To verify that the obtained series is a solution, we substitute it into the left-hand side of equation (2):
\begin{aligned}& \mathcal{D}^{\alpha _{n}}_{a + , \mu } \Biggl(\sum _{k = 0}^{\infty }(-1)^{k} \sum _{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1}} \cdots b_{m}^{k_{m}} \\& \qquad {} \cdot \mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}( \alpha _{n } + \beta _{m}) + \alpha _{n } }_{a + , \mu } g\Biggr) \\& \qquad {}+ \Biggl(\sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1} + 1} \cdots b_{m}^{k_{m}} \\& \qquad {} \cdot \mathcal{J}^{(k_{1} +1) (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}(\alpha _{n } + \beta _{m}) }_{a + , \mu } g\Biggr) \\& \qquad {}+ \cdots + \Biggl(\sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1} } \cdots b_{m}^{k_{m} + 1} \\& \qquad {} \cdot \mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + (k_{m} + 1)(\alpha _{n } + \beta _{m}) }_{a + , \mu } g\Biggr) \\& \quad = \mathcal{D}^{\alpha _{n}}_{a + , \mu } \Biggl(\mathcal{J}^{\alpha _{n}}_{a + , \mu } g + \sum_{k = 1}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1}} \cdots b_{m}^{k_{m}} \\& \qquad {} \cdot \mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}( \alpha _{n } + \beta _{m}) + \alpha _{n } }_{a + , \mu } g\Biggr) \\& \qquad {}+ \Biggl(\sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1} + 1} \cdots b_{m}^{k_{m}} \\& \qquad {} \cdot \mathcal{J}^{(k_{1} +1) (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}(\alpha _{n } + \beta _{m}) }_{a + , \mu } g\Biggr) \\& \qquad {}+ \cdots + \Biggl(\sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1} } \cdots b_{m}^{k_{m} + 1} \\& \qquad {} \cdot \mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + (k_{m} + 1)(\alpha _{n } + \beta _{m}) }_{a + , \mu } g\Biggr) \\& \quad = g + \Biggl(\sum_{k = 1}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1}} \cdots b_{m}^{k_{m}} \\& \qquad {} \cdot \mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}( \alpha _{n } + \beta _{m}) }_{a + , \mu } g\Biggr) \\& \qquad {}+ \Biggl(\sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1} + 1} \cdots b_{m}^{k_{m}} \\& \qquad {} \cdot \mathcal{J}^{(k_{1} +1) (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}(\alpha _{n } + \beta _{m}) }_{a + , \mu } g\Biggr) \\& \qquad {}+ \cdots + \Biggl(\sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1} } \cdots b_{m}^{k_{m} + 1} \\& \qquad {} \cdot \mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + (k_{m} + 1)(\alpha _{n } + \beta _{m}) }_{a + , \mu } g\Biggr) = g \end{aligned}
by the cancelation. Note that all series are absolutely convergent and the term rearrangements are feasible for the cancelation.
Indeed,
\begin{aligned}& -\sum_{k_{1} + \cdots + k_{m} = 1} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1}} \cdots b_{m}^{k_{m}} \mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}( \alpha _{n } + \beta _{m}) }_{a + , \mu } g \\& \qquad {}+ \sum_{k_{1} + \cdots + k_{m} = 0} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1} + 1} \cdots b_{m}^{k_{m}} \mathcal{J}^{(k_{1} +1) (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}( \alpha _{n } + \beta _{m}) }_{a + , \mu } g \\& \qquad {}+ \cdots + \sum_{k_{1} + \cdots + k_{m} = 0} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1} } \cdots b_{m}^{k_{m} + 1} \mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + (k_{m} + 1)(\alpha _{n } + \beta _{m}) }_{a + , \mu } g \\& \quad = 0. \end{aligned}
The rest terms cancel each other similarly.
Clearly, the uniqueness follows immediately from the fact that the integro-differential equation
$$\mathcal{D}^{\alpha _{n}}_{a + , \mu } u + a_{n - 1} \mathcal{D}^{ \alpha _{n - 1}}_{a + , \mu } u + \cdots + a_{0} \mathcal{D}^{ \alpha _{0}}_{a + , \mu } u + b_{n + 1} \mathcal{J}^{\beta _{n + 1}}_{a + , \mu }u + \cdots + b_{m} \mathcal{J}^{\beta _{m}}_{a + , \mu }u = 0$$
only has solution zero by Babenko’s method. This completes the proof of Theorem 2.1. □
### Remark 1
1. (i)
It follows from Theorem 5.3 in [2] that for $$0 < \alpha < 1$$
$$\bigl(\mathcal{D}^{\alpha }_{a + , \mu }u\bigr) (x) = \frac{x^{- \mu }}{\Gamma (1 - \alpha )} \biggl[u_{0}(a) \biggl(\log \frac{x}{a} \biggr)^{- \alpha } + \int _{a}^{x} \biggl(\log \frac{x}{t} \biggr)^{- \alpha } u_{0}'(t) \,dt \biggr],$$
where $$u_{0}(x) = x^{\mu }u(x) \in \operatorname{AC}[a, b]$$. Hence, for $$u \in \operatorname{AC}_{0}[a, b]$$,
\begin{aligned}& \bigl(\mathcal{D}^{\alpha }_{a + , \mu }u\bigr) (x) = \frac{x^{- \mu }}{\Gamma (1 - \alpha )} \int _{a}^{x} \biggl(\log \frac{x}{t} \biggr)^{- \alpha } u_{0}'(t) \,dt,\quad \mbox{and} \\& \bigl(\mathcal{D}^{\alpha }_{a + , \mu }u\bigr) (a) = 0. \end{aligned}
2. (ii)
A solution of equation (2) in the space $$\operatorname{AC}_{0}[a, b]$$ is said to be stable if $$\forall \epsilon > 0$$ $$\exists \delta > 0$$, such that $$\lVert u \rVert _{0} < \epsilon$$ if $$\lVert g \rVert _{0} < \delta$$. Using the inequality
\begin{aligned} \lVert u \rVert _{0} \leq & C_{\mu }E_{(\alpha _{n} - \alpha _{n - 1}, \ldots , \alpha _{n} + \beta _{m}, \alpha _{n} + 1)} \\ &{}\cdot \biggl( \vert a_{n - 1} \vert \biggl(\log \frac{b}{a} \biggr)^{\alpha _{n} - \alpha _{n - 1}}, \ldots , \vert b_{m} \vert \biggl(\log \frac{b}{a} \biggr)^{ \alpha _{n} + \beta _{m}} \biggr) \lVert g \rVert _{0}, \end{aligned}
(4)
we imply that the solution u is stable.
3. (iii)
The multivariate Mittag-Leffler function was initially introduced by Hadid and Luchko [34], who used it for solving linear fractional differential equations with constant coefficients by the operational method. Suthar et al. [35] studied some properties of generalized multivariate Mittag-Leffler function and established two theorems giving the image of this function under certain integral operators. Haubold et al. [36] presented a good survey of the Mittag-Leffler function, generalized Mittag-Leffler functions, Mittag-Leffler type functions, their interesting and useful properties, and applications in certain areas of physical and applied sciences. The Mittag-Leffler function plays an important role in the investigations of the fractional generalization of the kinetic equation, random walks, Lévy flights, superdiffusive transport and in the study of complex models.
Let $$\nu > 0$$ and $$x \geq 0$$. The incomplete gamma function is defined by
$$\gamma (\nu , x) = \int _{0}^{x} t^{\nu - 1} e^{-t} \,dt.$$
From the recurrence relation [37]
$$\gamma ( \nu + 1, x) = \nu \gamma (\nu , x) - x^{\nu }e ^{-x},$$
we get
$$\gamma (\nu , x) = x^{\nu }\Gamma (\nu ) e^{-x} \sum_{j = 0}^{\infty } \frac{x^{j}}{\Gamma (\nu + j + 1)}.$$
(5)
### Example 1
Let $$0< a < x < b$$. Then the Hadamard-type integro-differential equation
$$\bigl(\mathcal{D}^{0.8}_{a + , -1}u\bigr) (x) + \bigl( \mathcal{D}^{0.7}_{a + , -1}u\bigr) (x) + \bigl( \mathcal{D}^{0.1}_{a + , -1}u\bigr) (x) + 2 \bigl( \mathcal{J}^{0.2}_{ a + , -1} u\bigr) (x) - ( \mathcal{J}_{ a + , -1} u) (x) = x^{2},$$
has the solution
\begin{aligned} u(x) = & a x \sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + k_{2} + k_{3} + k_{4} = k}\binom{k}{k_{1}, k_{2}, k_{3}, k_{4}} 2^{k_{3}} (-1)^{k_{4}} \\ &{}\cdot \sum_{j = 0}^{\infty } \frac{ (\log x/a )^{j + 0.1 k_{1} + 0.7 k_{2} + k_{3} + 1.8 k_{4} + 0.8} }{\Gamma (0.1 k_{1} + 0.7 k_{2} + k_{3} + 1.8 k_{4} + 0.8 + j + 1)} \end{aligned}
in the space $$\operatorname{AC}_{0}[a, b]$$. Indeed, it follows from Lemma 2.4 in [2] that
$$\bigl(\mathcal{J}^{\alpha }_{ a + , \mu } t^{w}\bigr) (x) = \frac{\gamma (\alpha , (\mu + w) \log (x/a))}{\Gamma (\alpha )} (\mu + w)^{-\alpha } x^{w},$$
where $$\mu + w > 0$$.
By Theorem 2.1,
\begin{aligned} u(x) = & \sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + k_{2} + k_{3} + k_{4} = k}\binom{k}{k_{1}, k_{2}, k_{3}, k_{4}} 2^{k_{3}} (-1)^{k_{4}} \\ &{}\cdot \bigl(\mathcal{J}^{0.1 k_{1} + 0.7 k_{2} + k_{3} + 1.8 k_{4} + 0.8}_{ a + , - 1} t^{2}\bigr) (x) \\ = & \sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + k_{2} + k_{3} + k_{4} = k}\binom{k}{k_{1}, k_{2}, k_{3}, k_{4}} 2^{k_{3}} (-1)^{k_{4}} \\ &{}\cdot \frac{\gamma (0.1 k_{1} + 0.7 k_{2} + k_{3} + 1.8 k_{4} + 0.8, \log (x/a) )}{\Gamma (0.1 k_{1} + 0.7 k_{2} + k_{3} + 1.8 k_{4} + 0.8)} x^{2}. \end{aligned}
Applying equation (5),
\begin{aligned}& \gamma \bigl(0.1 k_{1} + 0.7 k_{2} + k_{3} + 1.8 k_{4} + 0.8, \log (x/a) \bigr) \\& \quad = (\log x/a )^{0.1 k_{1} + 0.7 k_{2} + k_{3} + 1.8 k_{4} + 0.8} \Gamma (0.1 k_{1} + 0.7 k_{2} + k_{3} + 1.8 k_{4} + 0.8) \\& \qquad {}\cdot \frac{a}{x} \sum_{j = 0}^{\infty } \frac{ (\log x/a )^{j}}{\Gamma (0.1 k_{1} + 0.7 k_{2} + k_{3} + 1.8 k_{4} + 0.8 + j + 1)}. \end{aligned}
Thus,
\begin{aligned} u(x) = & a x \sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + k_{2} + k_{3} + k_{4} = k}\binom{k}{k_{1}, k_{2}, k_{3}, k_{4}} 2^{k_{3}} (-1)^{k_{4}} \\ &{}\cdot \sum_{j = 0}^{\infty } \frac{ (\log x/a )^{j + 0.1 k_{1} + 0.7 k_{2} + k_{3} + 1.8 k_{4} + 0.8} }{\Gamma (0.1 k_{1} + 0.7 k_{2} + k_{3} + 1.8 k_{4} + 0.8 + j + 1)} \end{aligned}
is the solution in the space $$\operatorname{AC}_{0}[a, b]$$.
The following theorem shows the uniqueness of equation (1).
### Theorem 2.2
Assume that $$f: [a, b] \times R \rightarrow R$$ is a continuous function, and there exists a constant C such that
$$\bigl\vert f(x, y_{1}) - f(x, y_{2}) \bigr\vert \leq C \vert y_{1} - y_{2} \vert$$
for all $$x \in [a, b]$$ and $$y_{1}, y_{2} \in R$$. Furthermore,
$$C_{\mu }C E_{(\alpha _{n} - \alpha _{n - 1}, \ldots , \alpha _{n} + \beta _{m}, \alpha _{n} + 1)} \biggl( \vert a_{n - 1} \vert \biggl(\log \frac{b}{a} \biggr)^{\alpha _{n} - \alpha _{n - 1}}, \ldots , \vert b_{m} \vert \biggl(\log \frac{b}{a} \biggr)^{\alpha _{n} + \beta _{m}} \biggr) < 1.$$
Then equation (1) has a unique solution in the space $$\operatorname{AC}_{0}[a, b]$$ for every $$\mu \in R$$.
### Proof
Let $$u \in \operatorname{AC}_{0}[a, b]$$. Then
$$\int _{a}^{x} f\bigl(\tau , u'( \tau )\bigr) \,d\tau \in \operatorname{AC}_{0}[a, b],$$
as $$u'(\tau ) \in L[a, b]$$ and $$f(\tau , u'(\tau )) \in L[a, b]$$. Clearly,
\begin{aligned} \biggl\lVert \int _{a}^{x} f\bigl(\tau , u'( \tau )\bigr) \,d\tau \biggr\rVert _{0} =& \int _{a}^{b} \bigl\vert f\bigl(x, u'(x)\bigr) \bigr\vert \,dx \\ \leq& \int _{a}^{b} \bigl\vert f\bigl(x, u'(x)\bigr) - f(x, 0) \bigr\vert \,dx + \int _{a}^{b} \bigl\vert f(x, 0) \bigr\vert \,dx \\ \leq& C \int _{a}^{b} \bigl\vert u'(x) \bigr\vert \,dx + \int _{a}^{b} \bigl\vert f(x, 0) \bigr\vert \,dx < \infty . \end{aligned}
Define a mapping T on $$\operatorname{AC}_{0}[a, b]$$ by
\begin{aligned} T(u) = & \sum_{k = 0}^{\infty }(-1)^{k} \sum_{k_{1} + \cdots + k_{m} = k} \binom{k}{k_{1}, k_{2}, \ldots , k_{m}} a_{n - 1}^{k_{1}} \cdots b_{m}^{k_{m}} \\ &{}\cdot \mathcal{J}^{k_{1} (\alpha _{n } - \alpha _{n - 1}) + \cdots + k_{m}( \alpha _{n } + \beta _{m}) + \alpha _{n } }_{a + , \mu } \int _{a}^{t} f \bigl( \tau , u'(\tau )\bigr) \,d\tau . \end{aligned}
Using inequality (4), we claim that
$$\bigl\lVert T(u) \bigr\rVert _{0} < \infty\quad \mbox{and}\quad T(u) (a) = 0.$$
Furthermore, $$T(u)$$ is absolutely continuous on $$[a, b]$$ from the proof of Theorem 2.1. Hence, T is a mapping from $$\operatorname{AC}_{0}[a, b]$$ to $$\operatorname{AC}_{0}[a, b]$$. It remains to prove that T is contractive. Indeed,
\begin{aligned}& \bigl\lVert T(u) - T(v) \bigr\rVert _{0} \\& \quad \leq C_{\mu } E_{(\alpha _{n} - \alpha _{n - 1}, \ldots , \alpha _{n} + \beta _{m}, \alpha _{n} + 1)} \biggl( \vert a_{n - 1} \vert \biggl(\log \frac{b}{a} \biggr)^{\alpha _{n} - \alpha _{n - 1}}, \ldots , \vert b_{m} \vert \biggl(\log \frac{b}{a} \biggr)^{\alpha _{n} + \beta _{m}} \biggr) \\& \qquad {} \cdot \biggl\lVert \int _{a}^{t} f \bigl( \tau , u'(\tau )\bigr) \,d\tau - \int _{a}^{t} f \bigl( \tau , v'(\tau )\bigr) \,d\tau \biggr\rVert _{0}. \end{aligned}
Since
\begin{aligned} \biggl\lVert \int _{a}^{t} f \bigl( \tau , u'(\tau )\bigr) \,d\tau - \int _{a}^{t} f \bigl( \tau , v'(\tau )\bigr) \,d\tau \biggr\rVert _{0} =& \int _{a}^{b} \bigl\vert f\bigl(t, u'(t)\bigr) - f\bigl(t, v'(t)\bigr) \bigr\vert \,dt \\ \leq& C \int _{a}^{b} \bigl\vert u'(t) - v'(t) \bigr\vert \,dt = C \lVert u - v \rVert _{0}, \end{aligned}
we derive
\begin{aligned}& \bigl\lVert T(u) - T(v) \bigr\rVert _{0} \\& \quad \leq C_{\mu } C E_{(\alpha _{n} - \alpha _{n - 1}, \ldots , \alpha _{n} + \beta _{m}, \alpha _{n} + 1)} \biggl( \vert a_{n - 1} \vert \biggl(\log \frac{b}{a} \biggr)^{\alpha _{n} - \alpha _{n - 1}}, \ldots , \vert b_{m} \vert \biggl(\log \frac{b}{a} \biggr)^{\alpha _{n} + \beta _{m}} \biggr) \\& \qquad {} \cdot \lVert u - v \rVert _{0}. \end{aligned}
Therefore T is contractive. This completes the proof of Theorem 2.2. □
### Example 2
Let $$a = 1$$, $$b = e$$ and $$\mu = 2$$. Then there is a unique solution for the following nonlinear Hadamard-type integro-differential equation:
\begin{aligned}& \bigl(\mathcal{D}^{0.5}_{ 1 + , 2} u\bigr) (x) + \bigl(\mathcal{J}^{0.5}_{ 1 + , 2} u\bigr) (x) - \bigl(\mathcal{J}^{1.5}_{ 1 + , 2} u\bigr) (x) + \bigl( \mathcal{J}^{2.1}_{ 1 + , 2} u\bigr) (x) \\& \quad = \int _{a}^{x} \biggl(\frac{t^{2}}{C(1 + t^{100})} \sin u'(t) + \cos (\sin t) + e^{t^{2}} \biggr) \,dt, \end{aligned}
(6)
where the constant C is to be determined.
Clearly, $$C_{2} = e^{2}$$ is the maximum value of the function $$(\frac{t}{x} )^{2}$$ over the interval $$[1, e] \times [1, e]$$, and the function
$$f(x , y) = \frac{x^{2}}{C(1 + x^{100})} \sin y + \cos (\sin x) + e^{x^{2}}$$
is a continuous function from $$[1, e] \times R$$ to R and satisfies
$$\bigl\vert f(x, y_{1}) - f(x, y_{2}) \bigr\vert \leq \frac{x^{2}}{C(1 + x^{100})} \vert \sin y_{1} - \sin y_{2} \vert \leq \frac{x^{2}}{C(1 + x^{100})} \vert y_{1} - y_{2} \vert \leq \frac{1}{C} \vert y_{1} - y_{2} \vert .$$
Obviously $$\log b/a = 1$$. By Theorem 2.2, we need to calculate the value
\begin{aligned}& \sum_{k = 0}^{\infty }\sum _{k_{1} + k_{2} + k_{3} = k} \binom{k}{k_{1}, k_{2}, k_{3}} \frac{1}{\Gamma (k_{1} + 2 k_{2} + 2.6 k_{2} + 1.5)} \\& \quad = \sum_{k = 0}^{\infty }\sum _{k_{1} + k_{2} + k_{3} = k} \binom{k}{k_{1}, k_{2}, k_{3}} \frac{1}{\Gamma (k + 1.5 + k_{2} + 1.6k_{3})} \\& \quad = \frac{1}{\Gamma (1.5) } + \sum_{k = 1}^{\infty } \sum_{k_{1} + k_{2} + k_{3} = k} \binom{k}{k_{1}, k_{2}, k_{3}} \frac{1}{\Gamma (k + 1.5 + k_{2} + 1.6k_{3})}. \end{aligned}
For $$k \geq 1$$,
$$\frac{1}{\Gamma (k + 1.5 + k_{2} + 1.6k_{3})} \leq \frac{1}{\Gamma (k + 1) } = \frac{1}{k!}, \quad \mbox{and}\quad \sum_{k_{1} + k_{2} + k_{3} = k} \binom{k}{k_{1}, k_{2}, k_{3}} = 3^{k}.$$
Therefore,
\begin{aligned}& \sum_{k = 0}^{\infty }\sum _{k_{1} + k_{2} + k_{3} = k} \binom{k}{k_{1}, k_{2}, k_{3}} \frac{1}{\Gamma (k_{1} + 2 k_{2} + 2.6 k_{2} + 1.5)} \leq \frac{1}{\Gamma (1.5) } + \sum_{k = 1}^{\infty } \frac{3^{k}}{k!} \\& \quad < \frac{1}{2} + \sum_{k = 0}^{\infty } \frac{3^{k}}{k!}. \end{aligned}
Then, choose a positive C such that
$$C e^{2} \Biggl(\frac{1}{2} + \sum _{k = 0}^{\infty }\frac{3^{k}}{k!} \Biggr) < 1.$$
By Theorem 2.2, equation (6) has a unique solution. We note that the series $$\sum_{k = 0}^{\infty }\frac{3^{k}}{k!}$$ converges.
## Conclusions
Using Babenko’s approach and Banach’s contraction principle, we have derived the uniqueness of solutions for the new nonlinear Hadamard-type integro-differential equation for all $$\mu \in R$$:
\begin{aligned}& \mathcal{D}^{\alpha _{n}}_{a + , \mu } u + a_{n - 1} \mathcal{D}^{ \alpha _{n - 1}}_{a + , \mu } u + \cdots + a_{0} \mathcal{D}^{ \alpha _{0}}_{a + , \mu } u + b_{n + 1} \mathcal{J}^{\beta _{n + 1}}_{a + , \mu }u + \cdots + b_{m} \mathcal{J}^{\beta _{m}}_{a + , \mu }u \\& \quad = \int _{a}^{x} f\bigl(\tau , u'( \tau )\bigr) \,d\tau \end{aligned}
in the Banach space $$\operatorname{AC}_{0}[a, b]$$, with two examples given to illustrate the main theorems. The results obtained are fresh in the present studies, and they cannot be achieved via any existing integral transforms or local model methods to the best knowledge of the author.
Not applicable.
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## Acknowledgements
The author is grateful to the three reviewers for their careful reading of the paper with productive comments and suggestions.
## Funding
This work is supported by NSERC (Canada 2019-03907).
## Author information
Authors
### Contributions
The author prepared, read, and approved the final manuscript.
### Corresponding author
Correspondence to Chenkuan Li.
## Ethics declarations
### Competing interests
The author declares that they have no competing interests.
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Li, C. On the nonlinear Hadamard-type integro-differential equation. Fixed Point Theory Algorithms Sci Eng 2021, 7 (2021). https://doi.org/10.1186/s13663-021-00693-5
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## Video Transcript: MCAT Antilogs Without A Calculator
June 9, 2014 By Leah4sci 1 Comment
[Start Transcript]
Leah here from Leah4sci.com/MCAT and in this video, Iâll show you how to solve MCAT style Antilogs questions without a calculator. Â This video picks up from my last video where I show you how to solve logarithm based questions and you can find this video alongwith my entire series on solving MCAT Math without a calculator by visiting my website https://leah4sci.com/MCATMATH.
In the last video I showed you a trick on how to find a pH, pOH or pKa  value when given a concentration or ka. The trick showed you that when you have a number times ten to a negative power, that power becomes your pka, your pH or your poH.
But what if now you are faced with a question where the actual pH, poH or pka value is given and youâre ask to find the concentration or the ka? For  example you maybe given a question that says:
Find the ka of an acid whose buffer has a pH of 4.19 in a solution containing equal moles of acid and conjugate base.
Iâll cover the Science portion of this question in my Chemistry videos at leah4sci.com/MCATCHEMISTRY but for this video let focus just on the Math. Since weâre dealing with a buffer weâll use the HendersonâHasselbalch equation which says that pH is equal to pka plus the log of Conjugate base over acid.
Youâll also see this written as A minus over HA. Even though we have equal moles of acid and conjugate base or fifty-fifty, whatever number we have for conjugate base is the number for acid and that means we have a ratio of some number over itself or one.
The log of one is zero and that means this entire portion of the equation drops out telling me that the pH is equal to the pka. Knowing that the pH is equal to 4.19 equals the pka we know the pka is also equals to 4.19. But how do we use this to find the ka value of this acid? 4.19 is not a clean and easy number to calculate so letâs break it down:
The first thing you want to do is check how close your answer choices are to each other to see how much you can simplify and how quickly you can come up with the answer. Hereâs the equation weâll use. If pka is equal to negative log of ka (pka= -log ka), since log stands for log base ten, to solve for ka we have to have ten to the power of negative log to cancel out and that means we need ten to the power of negative pka. So the ka value is equal to ten to the minus pka which is equal to ten to the minus four point one nine (ka = 10^pka = 10^-4.19).
A nice and clean number like four would give us a ka value of one times ten to the minus four. But we also have to account for that 4.19 so we donât know weâre looking for the number close to one times ten to the minus four. If this is not enough to isolate your answer, you then want to find the range where your ka will fall out.
Weâll take the number 4.19 and round it down to 4 and up to 5. A pka of  4 has a ka of one times ten to the minus four, a pka of 5 has a ka of one times ten to the minus five. That means the number weâre looking for is somewhere in this range. But if this is still not enough, then you wanna go back to the trick where I showed you how to recognize the different numbers that give you different ranges. In review, if we have a number point one we get an eight time ten to the minus x. And I put  x instead of the number because if we have 4.19 our exponent will be a number times ten to the minus five.  So if we had 4.1 it will be eight times ten to the minus five.
A number point three will be five times ten to the minus that power. In this case, if we have 4.19, letâs round that to 4.2, remember on the MCAT you are allowed to round because it will be close enough. If point one gives me an eight and point five gives me a five then our answer has to be somewhere between eight and five so all you have to recognize is that the number is somewhere between five and eight. So itâll be five times ten to the minus five to eight times ten to the minus five.
Even if we havenât narrowed in on a specific number, for the MCAT this is close enough. In fact, punching ten to the negative four point one nine in the calculator I get an answer of 6.46 times ten to the minus five which on the MCAT is close enough. If you wanna narrow this down a little more, 4.19 is closer to 4.1 than it is to 4.3 and that means weâll be closer to the 8 than to the 5 as is evident by 6.46.
This concludes my video series on MCAT Math Without a Calculator. You can find this entire series on my website at https://leah4sci.com/MCATMath . You can find additional MCAT videos including Physics, Chemistry, Biology and Organic Chemistry on my website https://leah4sci.com/MCAT
Are you stuck on a specific MCAT topic? I offer Private Online Tutoring where I focus on your needs to strengthen your individual weaknesses. Tutoring details can be found using the link below or by visiting my website leah4sci.com/MCATTutor.
Are you overwhelmed by the sheer volume of information required for the MCAT? Are you worried that lack of a proper study plan and low MCAT score will prevent you from getting into Medical School?
My new ebook The MCAT Exam Strategy is 6-Week Guide to Crushing the MCAT will help you formulate a concrete study plan by helping you figure out where you stand now, identify your goals and figure out what it takes to reach them and itâs yours FREE when you sign up for my email newsletter at MCATExamStrategy.com.
By signing up for my email newsletter, youâll also be the first to know when I have new videos, MCAT Study Guide Cheat Sheets Tips and so much more! The link again MCATExamStrategy.com.
[End Transcript]
Watch the Video Here:Â MCAT Math Video 9 – Antilogs
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## Everyone's talking about OpenAI's Q*. Here's what you need to know about the mysterious project.
• A mysterious new OpenAI model known as Q* has got the tech world talking.
• The model is said to have sparked concern at the startup that led to the resulting chaos.
• AI experts say the model could be a big step forward but is unlikely to end the world anytime soon.
As the dust settles on the chaos at OpenAI, we still don't know why CEO Sam Altman was fired â but reports have suggested it could be linked to a mysterious AI model .
The Information reported that a team led by OpenAI chief scientist Ilya Sutskever had made a breakthrough earlier this year, which allowed them to build a new model known as Q* (pronounced "Q star.") The outlet reported that the model could solve basic math problems.
Sources told Reuters that this model provoked an internal firestorm , with several staff members writing a letter to OpenAI's board warning that the new breakthrough could threaten humanity.
This warning was cited as one of the reasons that the board chose to fire Sam Altman, who returned as CEO on Wednesday after days of turmoil at the company, Reuters reported.
The ability to solve basic math problems may not sound that impressive, but AI experts told Business Insider it would represent a huge leap forward from existing models, which struggle to generalize outside the data they are trained on.
"If it has the ability to logically reason and reason about abstract concepts, which right now is what it really struggles with, that's a pretty tremendous leap," said Charles Higgins, a cofounder of the AI-training startup Tromero who's also a Ph.D. candidate in AI safety.
He added, "Maths is about symbolically reasoning â saying, for example, 'If X is bigger than Y and Y is bigger than Z, then X is bigger than Z.' Language models traditionally really struggle at that because they don't logically reason, they just have what are effectively intuitions."
Sophia Kalanovska, a fellow Tromero cofounder and Ph.D. candidate, told BI that Q*'s name implied it was a combination of two well-known AI techniques, Q-learning and A* search.
She said this suggested the new model could combine the deep-learning techniques that power ChatGPT with rules programmed by humans. It's an approach that could help fix the chatbot's hallucination problem .
"I think it's symbolically very important. On a practical level, I don't think it's going to end the world," Kalanovska said.
"I think the reason why people believe that Q* is going to lead to AGI is because, from what we've heard so far, it seems like it will combine the two sides of the brain and be capable of knowing some things out of experience, while still being able to reason about facts," she added, referring to artificial general intelligence.
"That is definitely a step closer to what we consider intelligence, and it is possible that it leads to the model being able to have new ideas, which is not the case with ChatGPT."
The inability to reason and develop new ideas, rather than just regurgitating information from within their training data, is seen as a huge limitation of existing models, even by the people building them .
Andrew Rogoyski, a director at the Surrey Institute for People-Centered AI, told BI that solving unseen problems was a key step toward creating AGI.
"In the case of math, we know existing AIs have been shown to be capable of undergraduate-level math but to struggle with anything more advanced," he said.
"However, if an AI can solve new, unseen problems, not just regurgitate or reshape existing knowledge, then this would be a big deal, even if the math is relatively simple," he added.
Not everyone was so enthused by the reported breakthrough. Gary Marcus, an AI expert and deep-learning critic , expressed doubts about Q*'s reported capabilities in a post on his Substack .
"If I had a nickel for every extrapolation like thatâ'today , it works for grade school students! next year, it will take over the world!'âI'd be Musk-level rich," wrote Marcus.
OpenAI did not immediately respond to a request for comment from Business Insider, made outside normal working hours.
## Watch: Sam Altman moves to Microsoft after OpenAI fires him as CEO
#### IMAGES
1. How to Solve Logs and Natural Logs (ln) Easily : r/Mcat
2. Logarithms and Antilogs Without a Calculator (MCAT Shortcut)
3. Solving Logarithmic Equations
4. EVERYTHING YOU NEED to do MATH w/o Calculator for MCAT (exponents, logs, percent, estimation)
5. Solving log equations 1
6. Question Video: Computing Logarithmic Expressions Using Laws of
#### VIDEO
1. Solve Logarithmic Equations
2. Logarithms
3. Solve Exponential Equations Using Logs
4. Solve the Logarithmic Equation Involving Different Bases
5. 4 Things You Need to Know Before Registering for the MCAT in 2024
6. How to find log and antilog without table
1. Quickly Calculate Logarithms Without a Calculator
The log function is used to solve equations where the variable is an exponent with base 10. For example, 10 X = 100. The log operator allows us to solve for X and write an equivalent expression as log (100) = X. 10 X = 100 is the exponential form of the expression, and log (100) = X is the logarithmic form.
2. How To Tackle MCAT style Logarithm Questions
Therefore, learning a non-calculator trick for solving log questions is a must for every MCAT student. And that's why my newest video shows you how to quickly tackle simple negative log questions, and a trick for solving the more difficult questions, all without a calculator. Logarithms and Negative Logs Without A Calculator
3. Log shortcuts : r/Mcat
Factoring out is the easiest way. For the MCAT, a good approximation will likely lead you to the right answer. A few easy things to remember: log (10 x) = x, likewise -log (10 x) = -x log (0.1<x<1) is something between -1 and 0 log (1<x<10) is something between 0 and 1 always convert to scientific notation, so 16x10 -10 = 1.6x10 -9
4. Logarithms and Antilogs Without a Calculator (MCAT Shortcut)
https://Leah4sci.com/MCATmath presents: Logarithms and Antilog shortcut quick solving without a calculator for the MCAT, GAMSAT, DAT and moređşWatch Next: Vo...
5. Video Transcript: Logarithms and Negative Logs on the MCAT
[Start Transcript] Leah here from Leah4sci.com/MCAT and in this video, I'll show you how to tackle logarithms style tackled questions without a calculator as it may show up on your MCAT.
6. Math for the MCAT: Everything You Need to Know
Part 1: Introduction Part 2: Estimating and rounding numbers a) Multiplication b) Division c) Example problem Part 3: Logarithms a) Logarithm rules b) Common applications of logarithms c) Example problem Part 4: Strategies for eliminating answer choices a) Significant figures b) Exponents c) Example problem Part 5: Fractions
7. How to Solve Logs and Natural Logs (ln) Easily : r/Mcat
log10 = log (5**2) = log5 + log2 = 0.7 + 0.3 = 1 log8 = log (4**2) = log (2**2**2) = log2 + log2 + log2 = 0.3 + 0.3 + 0.3 = 0.9
8. logs without a calculator : r/Mcat
You wont need to do anything but discern between 4 answers for the mcat. So if you're calculating a pH from an [h+] of like, 1.6 x 10 -⾠just select the answer that's between pH 4 and pH 5. There won't be two answers in this range. throwitaway348493 ⢠3 yr. ago
9. MCAT Math Without A Calculator
Yes. Are you allowed to use a calculator on the MCAT? No! But somehow, you still need to work through multi-step calculations, including everything from decimals and exponents to trigonometry and logarithms. And did I mention that there's no calculator on the MCAT?
10. How to Survive MCAT Math Without a Calculator
1. Practice for not having a calculator. You aren't allowed to use a calculator on the MCAT, so you need to practice doing arithmetic calculations by hand. Fortunately, the amount of calculation you'll have to do is small. See how you score on our free MCAT practice test . 2. You only have to be reasonably accurate.
11. EASY WAY TO SOLVE FOR -log : r/Mcat
EASY WAY TO SOLVE FOR -log 55 comments Best Add a Comment WassilyJ ⢠3 yr. ago ⢠Edited 3 yr. ago If anyone is wondering why this works, I'll just attempt to explain it. Hopefully the explanation finds it useful to someone. There's a log property, where: Log (a*b) = log (a) + log (b)
12. MCAT Math Vid 8
http://leah4sci.com/MCAT Presents: MCAT Math Without A Calculator Video 8 - Solving MCAT logarithm calculations without a calculatorđşWatch Next: Logarithms ...
13. MCAT Math Archives
This video picks up from MCAT Math Part 8 - Logs and Negative Logs showing you how to solve questions where the log value is provided. Instead you're asked to solve for the anti-log value in questions relating to ion concentration and Ka values. Anti-logs, just like logarithms, are difficult to calculation without a calculator [âŚ]
14. Solving Antilog MCAT Questions Without A Calculator
Solving Antilog MCAT Questions Without A Calculator This video picks up from MCAT Math Part 8 - Logs and Negative Logs showing you how to solve questions where the log value is provided. Instead you're asked to solve for the anti-log value in questions relating to ion concentration and Ka values.
15. Video Transcript: MCAT Antilogs Without A Calculator
Here's the equation we'll use. If pka is equal to negative log of ka (pka= -log ka), since log stands for log base ten, to solve for ka we have to have ten to the power of negative log to cancel out and that means we need ten to the power of negative pka.
16. Do we have to solve natural logarithms on the test? : r/Mcat
Can't hurt to keep the log trick in your back pocket though: -log (x * 10^-y) = y - 0.x will give you an approximation. Also not important, but if it makes you feel better ln (x)= 2.3* log (x) DatEz 524 (131/130/131/132) ⢠4 yr. ago Know these for sure just in case!
17. how do you calculate logs? : r/Mcat
Lets say you need to calculate this to find pH: -log (1.6 x 10 -4) To calculate that, take the base number (the 1.6, idk if its actually called the base lol), move the decimal one unit left, and subtract it from the exponent 4 (ignore the negative on it) So in this case, you'd go 4 - 0.16 = 3.84 which would be ur pH. TwoSides1 ⢠7 yr. ago ...
18. MCAT: Anti log without a calculator
0:00 / 3:04 MCAT: Anti log without a calculator MVP Series 261 subscribers Subscribe 198 Share 19K views 7 years ago MCAT Prep: In this video we show you how to find the negative log...
19. Tricks for ln and log problems? : r/Mcat
I arrive at the proper formula doing problems with pH, Ka, Kb etc., but miss the answer because I have trouble solving the ln or log portion of the equation. comments sorted by Best Top New Controversial Q&A Add a Comment ksaado ... r/Mcat ⢠497 diagnostic in Jan â> 520 on May 26th đ Thank you to all the community members who helped me. ...
20. Logarithms || MCAT Basic Maths ||
In this Video I have explained about logarithm and how to use them to solve questions..#log #Mcat_basics #MCAT_Physics
21. MCAT Math Vid 9
https://Leah4sci.com/MCATMath presents: MCAT Math Without A Calculator Video 9 - Solving MCAT antilog calculations without a calculatorđşWatch Next: 5 Tips t...
22. Log Collection: Collecting logs of the whole installation process
CN: the logs on compute node: C1 - the installation logs will be saved to /var/log/xcat/xcat.log file on CN. C2 - the debug trace ( set -x or -o xtrace) of bash scripts enabled. C3 - the installer logs will be saved to the CN in /var/log/anaconda for RHEL, /var/log/YaST2 for SLES, /var/log/installer for UBT. N - the logs will not be saved to CN.
23. Everyone's Talking About OpenAI's Q.* Here's What You Need to Know
Here's what you need to know about the mysterious project. Sam Altman is back as OpenAI CEO after several days of chaos at the company. A mysterious new OpenAI model known as Q* has got the tech ...
24. Solid state batteries promise to radically change EVs. But they ...
Electric cars are supposed to be the future, but they still have issues that are keeping away many car buyers. The range is too short. The batteries are too heavy and expensive. They take too long ...
25. Why supply chains are key to helping solve the puzzle of inflation
The New York Fed's Global Supply Chain Pressure Index saw a record-low reading last month, a sign that the worst of the supply jams is behind us. A White House official said there is likely more disinflation to come from the supply channel, citing more capacity for companies to bring prices back down as their own input costs fall.
26. MCAT: negative log without a calculator
MCAT Prep: In this video we show you how to find the negative log without a calculatorVisit our website at http://www.mcat-mvp.com/MCAT2015 to learn more abo...
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https://se.mathworks.com/matlabcentral/cody/problems/2022-find-a-pythagorean-triple/solutions/3336103
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Cody
# Problem 2022. Find a Pythagorean triple
Solution 3336103
Submitted on 22 Oct 2020 by Victor
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
a = 1; b = 2; c = 3; d = 4; flag_correct = false; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
2 Pass
a = 2; b = 3; c = 4; d = 5; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
3 Pass
a = 3; b = 4; c = 5; d = 6; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
4 Pass
a = 3; b = 4; c = 4.5; d = 5; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
5 Pass
a = 3; b = 3.5; c = 4; d = 5; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
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# Practical application of matrices and determinants
I have learned recently about matrices and determinants and also about the geometrical interpretations, i.e, how the matrix is used for linear transformations and how determinants tell us about area/volume changes.
My school textbooks tell me that matrices and determinants can be used to solve a system of equations, but I feel that such a vast concept would have more practical applications. My question is: what are the various ways the concept of matrices and determinants is employed in science or everyday life?
• Matrices are used a lot in machine learning. – Bladewood Mar 17 at 19:00
• With some exaggeration, all of applied mathematics boils down to solving systems of linear equations. – Rodrigo de Azevedo Mar 17 at 19:19
• Solving systems of equations is extremely practical. Every time someone solves a differential equation using the finite element method, or runs a linear regression, or solves an optimization problem using Newton's method, a system of linear equations is solved. There is hardly any engineering or applied math project that doesn't require solving a system of linear equations. – Sasho Nikolov Mar 17 at 19:37
• Matrices are important to computer graphics, but not determinants. – immibis Mar 17 at 22:47
• – user21820 Mar 18 at 7:51
My first brief understanding of matrices is that they offer an elegant way to deal with data (combinatorially, sort of). A classical and really concrete example would be a discrete Markov chain (don't be frightened by its name). Say you are given the following information: if today is rainy, then tomorrow has a 0.9 probability to be rainy; if today is sunny, then tomorrow has a 0.5 probability to be rainy. Then you may organize these data into a matrix:
$$A=\begin{pmatrix} 0.9 & 0.5 \\ 0.1 & 0.5 \end{pmatrix}$$
Now if you compute $$A^2=\begin{pmatrix} 0.86 & 0.7 \\ 0.14 & 0.3 \end{pmatrix}$$, what do you get? 0.86 is the probability that if today is rainy then the day after tomorrow is still rainy and 0.7 is the probability that if today is sunny then the day after tomorrow is rainy. And this pattern holds for $$A^n$$ an arbitrary $$n$$.
That's the simple point: matrices are a way to calculate elegantly. In my understanding, this aligns with the spirit of mathematics. Math occurs when people try to solve practical problems. People find that if they make good definitions and use good notations, things will be a lot easier. Here comes math. And the matrix is such a good notation to make things easier.
Matrices are used widely in computer graphics. If you have the coordinates of an object in 3d space, then scaling, stretching and rotating the object can all be done by considering the coordinates to be vectors and multiplying them by the appropriate matrix. When you want to display that object on-screen, the projection down to a 2D object is also a matrix multiplication.
Determinants are of great theoretical significance in mathematics, since in general "the determinant of something $$= 0$$" means something very special is going on, which may be either good news of bad news depending on the situation.
On the other hand determinants have very little practical use in numerical calculations, since evaluating a determinant of order $$n$$ "from first principles" involves $$n!$$ operations, which is prohibitively expensive unless $$n$$ is very small. Even Cramer's rule, which is often taught in an introductory course on determinants and matrices, is not the cheapest way to solve $$n$$ linear equations in $$n$$ variables numerically if $$n>2$$, which is a pretty serious limitation!
Also, if the typical magnitude of each term in a matrix of of order $$n$$ is $$a$$, the determinant is likely to be of magnitude $$a^n$$, and for large $$n$$ (say $$n > 1000$$) that number will usually be too large or too small to do efficient computer calculations, unless $$|a|$$ is very close to $$1$$.
On the other hand, almost every type of numerical calculation involves the same techniques that are used to solve equations, so the practical applications of matrices are more or less "the whole of applied mathematics, science, and engineering". Most applications involve systems of equations that are much too big to create and solve by hand, so it is hard to give realistic simple examples. In real-world numerical applications, a set of $$n$$ linear equations in $$n$$ variables would still be "small" from a practical point of view if $$n = 100,000,$$ and even $$n = 1,000,000$$ is not usually big enough to cause any real problems - the solution would only take a few seconds on a typical personal computer.
• Why "even Cramer's rule"? That rule is so obviously inefficient that it's hardly worth mentioning, as every introductory course covers Gaussian elimination, which is clearly much more efficient. – Inactive - avoiding CoC Mar 17 at 20:19
• Whilst it doesn't make it more efficient, the determinant calculations in Cramer's rule can be done using Gaussian elimination which means its at least in the same complexity class surely? – jacob1729 Mar 17 at 23:10
Here's an application in calculus. The multivariate generalisation of integration by substitution viz. $$x=f(y)\implies dx=f^\prime(y)dy$$ uses the determinant of a matrix called a Jacobian in place of the $$f^\prime$$ factor. In particular, the chain rule $$dx_i=\sum_j J_{ij}dy_j,\,J_{ij}:=\frac{\partial x_i}{\partial y_j}$$ for $$n$$-dimensional vectors $$\vec{x},\,\vec{y}$$ can be summarised as $$d\vec{x}=Jd\vec{y}$$. Then $$d^n\vec{x}=|\det J|d^n\vec{y}$$.
There are plenty of applications of determinants, but I will just mention one that applies to optimization. A totally unimodular matrix is a matrix (doesn’t have to be square) that every square submatrix has a determinant of 0, 1 or -1. It turns out that (by Cramer’s rule) that if a constraint matrix $$A$$ of a linear program max $$\{c’x:\: Ax \leq b, x \in \mathbb{R}^n_+\}$$ is totally unimodular, it is guaranteed to have an integer solution if a solution exists. In other words, the polyhedron formed by $$P = \{x:\: Ax \leq b\}$$ has integer vertices in $$\mathbb{R}^n$$. This has major implications in integer programming, as we solve an integer program that has a totally unimodular matrix as a linear program. This is advantageous because a linear program can me solved in polynomial time, where there is no polynomial algorithm for integer programs.
Besides the applications already mentioned in the previous answers, just consider that matrices are the fundamental basis for Finite Elements design, today widely used in every sector of engineering.
Actually a truss is a physical representation of a matrix: if its stiffness matrix has null determinant, it means that there can be movements without external forces, i.e. the truss will collapse.
Also, in the continuous analysis of the deformation of bodies, stress and strain each are represented by matrices (tensors).
The inertia of a body to rotation is a matrix.
An electric network is described by a matrix voltages/ currents, and a null determinant denotes a short somewhere.
And so on ...
If the determinant of a matrix is zero, then there are no solutions to a set of equations represented by an nXn matrix set equal to a 1Xn matrix. If it is non-zero, then there are solutions and they can all be found using Cramer's Rule. They are also used in Photoshop for various visual tricks; they are used to cast 3D shapes onto a 2D surface; they are used to analyze seismic waves... and a hundred other applications where data need to be crunched in a simple manner.
In system-theory,
1. systems can be represented by matrices and each column represent the internal-state of the system.
2. If a determinant of one such matrix is zero, then we can say that one of the state associated with certain dynamics is being duplicated.
3. Based on some special matrix operations, we arrive at something called as relative-gain-array (RGA). This will give information on how much each states/output of a system interactes with each other, collectively speaking.
However, top are just a few examples. There are much more.
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http://www.wyzant.com/resources/answers/4763/6_4_t_12t
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Search 73,034 tutors
0 2
## -6(4-t)=12t
pre algebravvvvvvvvvvvvvvvvvbbbbbbbhhhhhgggffff
Hi, Caitlin.
To solve equations like this type, make sure to follow three steps:
1) Simplify both sides of the equation
2) Make sure there is only a variable term (letters) on one side of the equation
3) Think "opposite" - that is, do opposite operations in the opposite order of operations
-6 (4 - t) = 12t
1) The left side is not simplified. We can take care of that by distributing the negative 6 to both members of the ( )'s grouping:
-24 + 6t = 12t
2) We have variable terms on both sides of the equation.
-24 + 6t = 12t
- 6t - 6t
-24 = 6t
3) Now think opposite operation on what's preventing the letter from being alone. The t is being multiplied by 6... so we divide by 6 on both sides:
-24 = 6t
6 6
-4 = t
Or t = -4 to say it "frontwards."
If you are asked to check your answer, put the value of -4 in place of t in the original problem:
-6 (4 - t) = 12t
-6 (4 - (-4)) = 12(-4)
-6 (4 + 4) = 12 • -4
-6 • 8 = 12 • -4
-48 = -48
It worked!
Find t. distribute (-6)=(-6*4) - (-6*t) = 12t leaving (-24)-(-6t) = 12t resulting in (-24) + (6t) = 12t isolate t by subtracting 6t from both sides for -24 = 6t divide both sides by 6 t=-4 Substitute -4 for t in the equation. -6(4-t)=12t -6(4-(-4))=12(-4) -6(4+4)=-48 -6(8)=-48 -48=-48 So, it's a solution
So your overall goal is to get t by itself Right?
So, first distribute the -6 : so mulitply -6*4 and -6*-t -24+6t=12t
Now move +6t to the other side so the t's are all on the same side, do this by subtracting 6t from both sides:
-24+6t-6t = 12t-6t. Then simplify:
-24=6t. now let's switch both sides so the t is on the left
6t = -24
now we need to get the t by itself by dividing both sides by 6:
(6t)/is just leaves t because the 6's cancel out and -24/6 is -4
so t=-4
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https://www.mathsisfun.com/algebra/amplitude-period-frequency-phase-shift.html
Amplitude, Period, Phase Shift and Frequency. Some functions (like Sine and Cosine) repeat forever and are called Periodic Functions. The Period goes from ...
https://mathbitsnotebook.com/Algebra2/TrigGraphs/TGShift.html
The graphs of sine and cosine are the same when sine is shifted left by 90º or d radians. Such a shifting is referred to as a horizontal shift.
https://www.varsitytutors.com/precalculus-help/find-the-phase-shift-of-a-sine-or-cosine-function
Precalculus Help » Graphs and Inverses of Trigonometric Functions » Graphing the Sine and Cosine Functions » Find the Phase Shift of a Sine or Cosine ...
https://www.purplemath.com/modules/grphtrig3.htm
Demonstrates how to adjust a trig graph to account for phase shift, among other ... The regular period for sine waves is 2π, but the variable in this function is ...
Nov 22, 2010 ... Thanks to all of you who support me on Patreon. You da real mvps! \$1 per month helps!! :) https://www.patreon.com/patrickjmt !! In this video, I ...
https://www.purplemath.com/modules/grphtrig.htm
Explains how to relate constants in trig function to amplitudes, vertical shifts, periods, and phase ... graph of sin(2t), showing period is half as long because sine changes ... The formula for sines and cosines says that the regular period is 2π. .... The number C inside with the variable is 2π/3, so this will be the phase shift.
https://calcworkshop.com/graphing-trig-functions/graphing-sine-and-cosine-with-phase-shift/
Graphing Sine and Cosine Functions with a Phase (Horizontal) Shift has never been easier with this easy to follow, step-by-step tricks and tips.
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# I am a 3-digit number. My first digit is thrice my last digit. The sum of my last two digit is one less than my last digit
42,561 results, page 18
1. ## Math
Use the digits below to answer the number riddle. 2 4 9 6 7 3 The difference between two 3-digit numbers is a palidrome between 200 and 300,What are the numbers?
asked by Tina on October 15, 2008
2. ## math
erin is thinking of a 3 digit number.it uses the digits 1,7,and 4.how many numbers can you make that are even?how many can you make that are odd?explain. please answer as soon as possible!!!!!! :)
asked by jimena on September 14, 2011
3. ## math
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4. ## algebra
I am thinking of a three digit number. It is am odd multiple of three, and the product of its digits is 24. It is larger than 275. What are all the numbers I could be thinking of? need help thank you
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5. ## POW#4 Read MY Mind
I am thinking of a three-digit number. It is an odd multiple of three, and the product of its digits is 24. It is larger than 15 squared. What are all the numbers that I could be thinking of?
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6. ## Calculus
What is the largest two digit prime number whose digits are both prime numbers?
asked by Ashley on January 26, 2008
7. ## Math help (emergency)
How many 2-digit numbers can be formed using only the digits 2, 3, 5, and 6. if the digits are not repeated within a number? 11 12 10 2
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8. ## math
i'm a four digit number having all the digits odd except 9 and the digits are in ascending order. who am i?
asked by suman on August 12, 2014
9. ## Math
How many different 3-digit numbers can be formed from the number 1,2 and 3 if the digits do not repeat in the numbers?
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10. ## math
How many integers \$m\$ are there such that \$0
asked by Joy F on October 31, 2018
11. ## Math
i am a 6 digit number 8 ones, 5 hundreds 2 thousands, 19 ten thousands. What is the answer?
asked by CHRIS on June 27, 2017
12. ## math
Given a four digit number XXXX, are there more permutations if repetition is allowed or not allowed?
asked by Anonymous on April 2, 2018
13. ## Chemistry
How many number of significant figures are there in measurement 6.80? I think 2 In measurement 43.52 cm what is digit is the most uncertain? thanks
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14. ## math
1. a door code is made up of six unique digits the last digit is 8. 2.oddd and even digits alternate(o is even) 3.the difference of the adjacent is always bigger than 1. 4.the first 2 digits (reads as one number)as well as he two middle digits(as one
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15. ## maths
A certain nine digit number has only ones in ones period,only two’s in the thousand period and only threes in millions period .Write this number in words in the Indian system
asked by jalaja on September 26, 2012
16. ## Maths
Both of the digits is even. One of the digit is 4. The tens place is twice the amount of the ones place. The number is less than 70. What is the number?
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17. ## algebra
a five digit number that follows the pattern:if you put a 1 after it the number is three times as large as it would be if u had put a 1 before it
asked by Anonymous on November 27, 2012
18. ## Math
Given any random, 3-digit number, what is the probability that a number will be: a) a multiple of 5 b) a multiple of 2 c) a square number How?
asked by Tiara on December 2, 2010
19. ## problem solving
A nine-digit number is formed using each of the digits 1,2,3,...,9 exactly once. For n = 1,2,3,...,9, n divides the first n digits of the number. Find the number.
asked by kakak on December 8, 2014
20. ## math 157
How many three-symbol codes (letter, letter, number) can be made from the letters S, P, Y and one digit fro the set {0, 1, 2,...9} without repitition? What and how do I get started?
asked by Rmz on January 8, 2010
21. ## math
Writing to explain: The digit 5 is usually rounded up, but it can also be rounded down. How would you round the numbers in the equation 9.5 +4.7+3.2+7.5= to the nearest whole number without getting an overestimate or an underestimate?
asked by Jess on January 17, 2014
22. ## math
The number 199,557 contains two sets of digits in which one digit is ten times as great as the other. What are the values of the digits in each set?
asked by liz on October 15, 2014
23. ## Math
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asked by X on September 11, 2007
24. ## Math
How many five digit are numbers there?
asked by Anonymous on May 15, 2017
25. ## Math
What is the value of the underlined digit 0.26. The underlined digit is 2
asked by Anonymous on August 31, 2010
26. ## Fracture Math word
3/5 of digit is something you do with a shovel.
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27. ## math
What is the value of the tens digit in 54?
asked by gt on September 28, 2011
28. ## 12
let X=(3^5^7^9)+1 find the last digit of x
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29. ## math
what is the unit digit in (7^95 - 3^58) ?
asked by medlin on July 2, 2015
write the digit 5's value
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32. ## Math
What are the possible four digit numbers between 8.6 and 8.7
asked by Anonymous on February 11, 2014
33. ## Math
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34. ## Math
A number of two digits is such that twice the ten digits is 8 less than seven times the unit digit. When the digits are reversed, the number is decreased by 9. What is the number?
asked by Zulaiykherh on January 1, 2017
35. ## Maths
A number consists of two digits.THe digit in tens place is twice that in units place.if 18 is subtracted from the number. The digits are reversed.find the number.
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36. ## Computer Science
The digits 0, 1, and 8 look much the same if rotated 180 degrees on the page (turned upside down). Also, the digit 6 looks much like a 9, and vice versa, when rotated 180 degrees on the page. A multi-digit number may also look like itself when rotated on
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37. ## Math
I am a 4-digit even number. I am less than 2500. The digits in my hundreds and the tens place is different. The digits in my hundreds place are less than the digits in the one thousands place. The sum of my last 2-digits equal 9. I am _?
asked by Rico on August 31, 2015
38. ## math
A state's license plate number consists of 3 digits followed by 2 letters. If the digit 0 and the letter X cannot be used, how many distinct possible licence plate numbers are possible?
asked by Anonymous on September 20, 2013
39. ## college math
a license plate is to consist of three digits followed by two uppercase letters. determine the number of different license plates possible. the first digit cannot be zero and repition is not permitted.
asked by ashley on November 22, 2011
40. ## Math
Lockers are numbered with consecutive positive integers beginning with 1, and the digit 2 is used exactly 106 times (including on the last locker). What is the number of the last locker?
asked by Sarah on September 4, 2015
41. ## math
Assume that 3 digits are selected at random from the set { 2, 5, 6, 8, 9 } and are arranged in random order. What is the probability that the resulting 3-digit number is less than 900?
asked by victoria on February 20, 2012
42. ## math
round to the place of the underline digit 16 when 1 is the underline number
asked by tia on September 5, 2012
43. ## math
Using only the numbers 8, 1, 4, 7, & 3, the product of 2 two digit numbers plus a number is 3,355?
asked by Anonymous on September 6, 2011
44. ## math
1A56B is a five digit number. When 1A56B is divided by 36 the remainder is 23. How many possible value can A take? #anyone to help me out
asked by kelvin on April 30, 2017
45. ## Math
How many 3 digit numbers can be made using the following digits only once in each number? Use the digits : 2,3,4
asked by Pietro on December 7, 2013
46. ## math
What will be the digit in the tens place of the sum of the following expression: 7 + 77 + 777 + 7777 + ... + 7777777777777777777 1. (The same as if you added 5x7 and 4x70) The numbere in the 100's and higher places don't contribute to the amount in the
asked by Preston on January 22, 2007
47. ## finite math
suppose that 3 digits are selected at random from the set S={1,2,3,4,5,6} and are arranged in random order. Find the probability that the resulting 3-digit number is less than 300.
asked by allen on September 4, 2012
48. ## Math
Supposed you model numbers with three digits, but you do not use any tens blocks. What digit must be in the standard form of every number you model? Give examples and explain
asked by Diana on August 22, 2013
49. ## math
I am helping my daughter and I need help on this question in the number 57,733 contains two sets of digits in which one digit is ten times as great as the other what are the values of the digits in each set help please!!!!
asked by terry on February 15, 2014
50. ## math
This was on my 5th grader's homework: what is the least number you can make using all of the digits from 0 through 9 exactly once? My answer was 1,023,456,789. The teacher says it's 0,123,456,789. I think that if the zero is the first digit, it's not
asked by stick on August 21, 2012
51. ## Algebra
1)An ice cream store has 31 flavors of ice cream and 10 toppings. A regular sundae has 1 flavor of ice cream, 1 topping, and comes with or without whipped cream. How many different ice cream sundaes can be ordered? A)310 B)372 C)620 D)82 I chose C 2)How
asked by Jon on February 7, 2008
52. ## MATH
IN HOW MANY WHOLE NUMBERS LESS THAN 100 DOES THE DIGIT 9 APPEAR AT LEAST ONCE?
asked by KIRA on February 20, 2008
53. ## maths
if 91z 5 is a multiple of 9 where z is a digit what might be the value of z?
54. ## maths
if 24y5 is a multiple of 3 where y is the digit what might be the value of y?
55. ## math 1321
what is the digit in the ones place of the solution to 3^63
asked by delia :) on September 16, 2012
56. ## math
I am a 3 digit #. All 3 digits are the same . I could not be any greater . What # am I ?
asked by Ridgecrest on October 26, 2011
57. ## math
which digits are always the same in a 4 digit palindrome?
asked by corrie on April 23, 2010
58. ## Math
how to estimate with 1-digit divisors
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59. ## math
Fill in the missing digit: _.4 x 2.4 +3.28 + 16._ _ = 19.6_
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60. ## algebra
find the last digit in 3^9999.
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61. ## math
what is a boldface digit and how do you find its value
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62. ## Math
Using the digit 9 and 1. Write a fraction less than 1/10
asked by Sara on November 4, 2010
63. ## maths
how many of the integers from 1 to 200 contains the digit 1 at least twice?
asked by abraham on May 26, 2016
64. ## random
How many 4 digit combinations can you get using numbers 0-9..?
asked by y912f on February 15, 2012
65. ## Last BrainMath :(
How many whole numbers from 99 to 131 contain the digit 1 exactly once?
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66. ## Math
How do you do 3 digit multiplication? Example 145x312
asked by Neon on April 23, 2012
67. ## Math
I have 3 digits. My middle digit is 4, the first is 8 and the last 2. Who am I?
asked by Noghto on January 15, 2013
68. ## math
please write the mising digit ( ) 6 ( ) x 4 ( ) 5 ----------------- 1 ( ) 4 ( ) ( ) 0 ( ) 1 ( ) 7 ( ) --------------------- ( ) 1 ( ) 0 ( ) 5
asked by aanine on August 12, 2013
69. ## Math
Using the digit 9 and 1. Write a decimal less than 0.3
asked by Sara on November 4, 2010
70. ## Calculus
What is the tens digit of 0! + 1! + 2! + 3! + ... + 2000!?
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71. ## Math
write the missing digit.... [ ] 6 [ ] x 4 [ ] 5 ----------------- 1 [ ] 4 [ ] [ ] 0 [ ] 1[ ]7 [ ] ------------------ [ ]1[ ]0 [ ] 5
asked by Janine on August 12, 2013
72. ## math
how many whole numbers less than 100 does the digit 9 appear at least once??
asked by KIRA on February 20, 2008
73. ## probability
What is the probability that an 8 digit numbers contains "97"?
asked by steph on June 20, 2013
74. ## Math
Using the digit 2 and 1. Write a decimal less than 0.3
asked by Sara on November 4, 2010
75. ## maths
how many 3-digit numbers have exactly 2 digits that are the same?
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76. ## math
how do you round off each factor so that only one digit is not zero
asked by lorenzo on January 29, 2009
77. ## maths
two 2 digit numbers reading across and two 2 digit numbers reading down
asked by Thomas on July 1, 2012
78. ## Hannah
what are the two-digit perfect squares?
asked by Anonymous on September 17, 2008
79. ## Math
Using the digit 5 and 1. Write a fraction less than 1/10
asked by Sara on November 4, 2010
80. ## math
What is the thousands digit of 9^1000?
asked by Anonymous on July 21, 2018
81. ## algebra
In base 5, how many two digit numbers are there?
asked by john on March 6, 2011
82. ## math
How many 3 digit numbers are divisible by 8?
asked by april on May 10, 2011
83. ## Math
Using the digit 5 and 1. Write a decimal less than 0.3
asked by Sara on November 4, 2010
84. ## Algebra
In base 5, how many two digit numbers are there?
asked by Jason on March 5, 2011
85. ## algebra
A pair of emirps consists of two prime numbers such that reversing the digits of one number gives the other. How many pairs of two-digit emirps exist such that each number in the pair is greater than 11?
asked by Sara on April 2, 2011
86. ## math
Computers manufactured by a certain company have a serial number consisting of a letter of the alphabet followed by a six-digit number. If all the serial numbers of this type have been used, how many sets have already been manufactured?
asked by Mike on February 14, 2011
In the multiplication problem at the right different letters stand for different digits, and ABC and DBC each represent a three digit number. What number does DBC represent? Two answer possible; give one
asked by Brooke on March 1, 2014
88. ## math
Computers manufactured by a certain company have a serial number consisting of a letter of the alphabet followed by a five-digit number. If all the serial numbers of this type have been used, how many sets have already been manufactured?
asked by dan on September 11, 2012
89. ## 2nd grade math requestion post
Okay, my child is doing problem solving lesson 5-7 in scott foresman addison wesley book. The question that we are having a problem with is a number line. there are 10 spaces in the number line and the clues are: Clue 1: the number are between 70 and 82
asked by Dawn on January 20, 2009
90. ## statistics
a computer system requires all users to log on using a 6 character password. if each character can be either a digit from 0 to 9 or a letter of the alphabet, what is the number of possible passwords. please show formula used
asked by luckybee on January 4, 2012
91. ## math
I just did this Quiz and I dont understand how I got this wrong. The question is How many 2-digit numbers can be formed using the only digits 2,3,5, and 6, if the digits are not repeated within a number. A.11 B.12 C.10 D.2 I picked 10 but the score sheet
asked by Charlene on May 31, 2018
92. ## Finite Math
Assume that 3 digits are selected at random from the set {1,3,5,6,7,9} and are arranged in random order. What is the probability that the resulting 3-digit number is less than 700? I know the sample space is P(6,3) but that is all I know. Help!
asked by Willow on February 28, 2009
93. ## math
How many ways are there to choose a three-digit lottery number? Note that numbers like 003, 014, 222, 232 or 123 are all allowed. 1000 720 240 None of the above Explain your answer and how you came to that conclusion.
asked by M.A. on August 18, 2010
94. ## computer programming
Write a program that simulates the dial of a phone number. The program will input a phone number and the program will acknowledge the call by either writing an error message or the 8 digit phone number to the console window. The phone number may have
asked by Jeffry on November 21, 2010
95. ## Math
Determine the number of combinations on a bicycle lock that has a 3-digit number combination with only 3 numbers where the numbers can be repeated to open the lock
asked by Pedro on September 8, 2016
96. ## Math
my tenths is twice greater than my thousandths . My hundredths digit is the sum of my tenths and thousandths . Who am I?
asked by anne klein on July 1, 2013
97. ## math
How many different four-digit numbers have the same digits as 1993?
asked by Mike on January 18, 2010
98. ## grd1 math
2+10= i havent learn how to do a dble digit, how do u do this?
asked by just wondering on April 6, 2010
99. ## Math
Using the digit 9 and 1. Write a fraction greather than 1/3
asked by Sara on November 4, 2010
100. ## Math
Using only the digit 5 and 1. Write a fraction greater than 1/3
asked by Sara on November 4, 2010
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http://new.math.uiuc.edu/netgeom/advice/howto-KSEGhelp.html
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Here is a growing collection of useful information on how to use KSEG, and also of some peculiarities. We begin with a an essay by Ilya Baran which is inside the KSEG distribution but hard to find. - Baran’s KSEG Help file
Tricks and Traps
Vanishing button bars
Here is an unexpected trap in some versions of KSEG. The buttons that activate the KSEG tools come in four banks, can be moved around the screen, and even turned off. Which of the bars are visible seems to persist from one session to the next. So, if you turn any of the off, you had better know how to turn them on again.
In short, there is a "secret" dropdown menu on the right mouse button "located" in the extreme top left (or right) corner of the drawing plane. It has four toggles, which some experimentation show, mean the following:
• new has the construction buttons with the line intersection on the left, and ending with the splined curve. Attached to it are four more, starting with the filled polygon, and ending with a filled circle.
• measure has the length measuring buttons, ending with the formula button.
• view deals with pushing the canvas around, and zooming. Especially importantis the 3rd one (2nd magnifying glass) which "zooms-to-fit".
• transform which deals with those transformations KSEG comes with. So far we have used only one of them, namely translation. We needed this to produce the inchworm. The inchworm is a cute way to refer prolonging a given line segment into a line with integer markings on it. Given the segment \$ AB \$ mark the vector \$ B-A \$. There is a yellow arrow over the segment now. Any figure you choose (turns red), for example the original segment itself, now translates by the yellow vector.
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http://www.clear.rice.edu/comp130/12spring/pca/pca_docs.shtml
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COMP 200 & COMP 130
# COMP 130 Elements of Algorithms and ComputationSpring 2012
## matplotlib.mlab.PCA Documentation
The official Python documentation can be found here, but is unfortunately so sparse as to approach completely unusable.
To use PCA, be sure to include the following import statement:
from matplotlib.mlab import PCA
matplotlib.mlab.PCA is the name of a class in Python. Classes are like recipes used to create objects, which can be thought of intelligent entities that hold data and have built-in functions, called "methods", that perform various processes on the data the object holds.
PCA calculates principal component (PC) axes such that the origins of PC axes is at the mean of the distribution along each axis, that is, the origin is at the statistical "center" of the cluster. The PC axes are also scaled such that distances along each axis is in units of the standard deviation of the data point distribution along that axis. It has the advantage of creating a normalized distance notion but it also masks whether or not there is even a significant distribution along that axis.
## Constructor: PCA(dataMatrix) -- construct a new PCA object from a matrix of data points.
A constructor is very much like a function that returns a new object of a particular type, here a PCA class object.
```import numpy as np
dataMatrix = np.array(aListOfLists) # Convert a list-of-lists into a numpy array. aListOfLists is the data points in a regular list-of-lists type matrix.
myPCA = PCA(dataMatrix) # make a new PCA object from a numpy array object
```
## Attributes: numrows, numcols -- the number of data points and the dimension of the measurement space.
numrows is the number of rows in the original data matrix, which corresponds to the number of data points, since each row is a data point.
numcols is the number of columns in the orginal data matrix, which corresponds to the number of axes in the measurement space, i.e. its dimensionality.
```myPCA = PCA(dataMatrix)
myPCA.numrows == len(dataMatrix) #True. The number of rows in the data matrix
myPCA.numcols == len(dataMatrix[i]) #True for all valid index i. The number of columns in any row of the data matrix
```
## Method: project(x) -- the projection of a vector x onto the principal component axes
The input parameter x is a vector (list) of measurements in terms of the original measurement axis, e.g. words for text analysis. The projection of a vector into a different coordinate system is express the same point the vector represents in terms of distances along the coordinate axes of that new system. project(x) returns a vector representing the same point in terms of the PC axes. It is important to remember that the origin of the PC axes is NOT the same as the origin of the original measurement axes (e.g. zero word counts). Instead, the origin is at the mean of the data points along each axis, i.e. at the "center" of the cluster. Also, the scale of the PC axes is not the same as the measurement axes. Distances along the PC axes are measured in terms of standard deviations of the distribution along that axis. Thus project(x) results in a normalized position with respect to the center of the cluster.
```myPCA = PCA(dataMatrix)
pcDataPoint = myPCA.project(aDataPoint) # pcDataPoint is the same point as aDataPoint, but in terms of the PC axes.
```
## Method: center(x) -- the translation and scaling of a vector x to the center of the cluster and scaled as per the standard deviations along the measurement axes
The center() muethod translates the given vector (list), x, by the mean of all the measurements, mu, and then scales the result by the standard deviation of the data along all the measurement axis. The translation puts the new origin at the center of the cluster.
```myPCA = PCA(dataMatrix)
myPCA.center(x) == (x -myPCA.mu)/myPCA.sigma # True, note that subtraction and division are on an element by element basis
myPCA.center(myPCA.mu+myPCA.sigma) == [1, 1, 1, ...] #True. one standard deviation away in all measurement directions.
```
## Attribute: mu -- the vector that points to the origin of the PC axes in terms of the measurement axes.
mu is a vector (list) in the original measurement coordinates that points to the location of the mean values for all PC axes, which is the origin of the PC axes. The value of any element of mu is the average of the corresponding element in every data point in the measurement coordinates.
```import numpy as np
myPCA = PCA(dataMatrix)
myPCA.mu[i] == np.average([v[i] for v in dataMatrix]) # True for any valid index i
myPCA.project(myPCA.mu) == [0, 0, 0,...] # True always because mu is the vector pointing at origin of the PC axes.
myPCA.center(myPCA.mu) == [0, 0, 0,...] # True always because mu is the vector pointing at the center of the cluster.
```
## Attribute: sigma -- the vector that points to 1 standard deviation from the mean along the measurement axes
sigma is a vector (list) in the original measurement coordinates that, from the mean position (mu), points to the location 1 standard deviation along every measurement axis, in terms of the distribution of values along each axis. The value of any element of sigma is the standard deviation of the corresponding element in every data point in the measurement coordinates.
```import numpy as np
myPCA = PCA(dataMatrix)
myPCA.mu[i] == np.std([v[i] for v in dataMatrix]) # True for any valid index i
```
## Attribute: Y -- the original data matrix in terms of the principal component axes
The attribute, Y, of a PCA object is the original data matrix in terms of the principal components basis vectors. That is, if Y is the matrix of data points you would get if you were to express the original data points in terms of the principal component axes instead of the original measurement components (e.g. words for text analysis). The i'th row of Y is equivalent to the projection of the i'th row of the original data matrix onto the principal component coordinates:
```myPCA = PCA(dataMatrix)
myPCA.Y[i] == myPCA.project(dataMatrix[i]) #True for all valid index i
myPCA.Y[i] != myPCA.a[i] #True in general for all valid index i
```
## Attribute: a -- the original data matrix centered on the cluster and scaled by the standard deviation in terms of the measurement axes.
The attribute, a, of a PCA object is the original data matrix in terms of the... The i'th row of a is equivalent to the projection of the i'th row of the original data matrix onto
```myPCA = PCA(dataMatrix)
myPCA.a[i] == myPCA.center(dataMatrix[i]) #True for all valid index i
myPca.a[i] == (dataMatrix[i] - p.mu)/p.sigma # True, note that subtraction and division are on an element by element basis
[np.std(v) for v in np.transpose(myPCA.a)] == [1, 1, 1, ...] # True, the standard deviation of the "centered" data points is 1 along any measurement axis.
myPCA.a[i] != myPCA.Y[i] #True in general for all valid index i
```
## Attribute: fracs -- the proportion of variance of each of the principal component axes.
fracs is a vecotr that measures how much variance is along each of the PC axes. This tells us how much the cluster is aligned along each PC axis. Since the PCA analysis orders the PC axes by descending importance in terms of describing the clustering, we see that fracs is a list of monotonically decreasing values. The fracs vector can be used to give weighting to each axis in terms of its importance in the clustering. Essentially, one wants to ignore the PC axes with low corresponding values in the fracs vector.
## Attribute: Wt -- the PC axes in terms of the measurement axes scaled by the standard deviations
Wt is a matrix whose rows are the PC axes in terms of the measurement axes where the distances along each measurement axis has been scaled by the standard deviation of the data along that axis. In linear algebra terms, Wt is the rotation matrix around the center of the cluster that will transform the measurement axis-expressed vectors whose origins are at the center of the cluster into the PC axis-expressed vectors whose origins are still at the center of the cluster. For instance Wt can transform the data point vectors in a into the data point vectors in Y.
```import numpy as np
myPCA = PCA(dataMatrix)
myPCA.project(p.Wt[i]*p.sigma + p.mu) == [0,0,...,1,...0,0] #True for all valid i where the "1" in the i'th position in the result and rest of the elements are zero.
[np.dot(myPCA.a[i], w) for w in myPCA.Wt] == p.Y[i] # True for all valid index i. This is just the rotation of a[i] into Y[i]```
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http://www.programmingtunes.com/finding-greatest-common-divisor-of-2-numbers-c/
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Monday, January 18, 2021
Home C / C++ Finding Greatest Common Divisor of 2 numbers in C++
# Finding Greatest Common Divisor of 2 numbers in C++
As we know that greatest common divisor of 2 numbers is the largest number that divides both those numbers, How we implement this logic in C++, its given in the below code:
``` #include <iostream>
using namespace std;
void main()
{
int x = 24;
int y = 18;
int temp;
for(int i = 2; i<y; i++)
{
if(x%i == 0 >> y%i == 0)
{
//cout<<i<<endl;
temp = i;
}
}
cout<<temp<<endl;
}
```
First of all we got two numbers, you can get these numbers from the user too, that’s not a big deal. Then we applied a for loop upto any of that numbers and used a condition that if for both the numbers the modulus with loop increment is same, then that is our greatest common divisor of numbers.
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https://www.jiskha.com/questions/1010667/a-projectile-is-launched-from-ground-level-at-37-0-m-s-at-an-angle-of-30-6-above
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# physics
A projectile is launched from ground level at 37.0 m/s at an angle of 30.6 ° above horizontal. Use the launch point as the origin of your coordinate system.
(a) How much time elapses before the projectile is at a point 8.9 m above the ground and heading downwards toward the ground?
(b) How far downrange (the horizontal distance from the origin) was the projectile when it reached the highest point in its flight?
My answer for part a) 3.29 s
How do I get part b?
1. 👍
2. 👎
3. 👁
1. Vo = 37m/s[30.6o]
Xo = 37*cos30.6 = 31.8 m/s.
Yo = 37*sin30.6 = 18.8 m/s.
a. Yo*t + 0.5g*t^2 = 8.9 m.
18.8t - 4.9t^2 = 8.9
-4.9t^2 + 18.8t - 8.9 = 0
t = 3.28 s. and heading downward.
b. Y = Yo - g*Tr = 0 at max ht.
18.8 - 9.8*Tr = 0
9.8Tr = 18.8
Tr = 1.92 s = Rise time or time to reach max ht.
Hor. Distance=Xo * Tr=31.8m/s * 1.92s =
61 m.
1. 👍
2. 👎
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https://gmatclub.com/forum/research-indicates-that-the-health-of-all-multicellular-organisms-189619.html
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# Research indicates that the health of all multicellular organisms
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Re: Research indicates that the health of all multicellular [#permalink]
Finding difficult to eliminate A. Can experts please provide more specific guidance.
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Re: Research indicates that the health of all multicellular [#permalink]
Straight (C). B, D & E are eliminated on the spot for wordiness.
ayushman wrote:
Finding difficult to eliminate A. Can experts please provide more specific guidance.
The uncomfortable "depends on not only" made me eliminate A, C is more elegant.
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Re: Research indicates that the health of all multicellular [#permalink]
Thanks abdul and daagh. Very helpful.
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Research indicates that the health of all multicellular organisms [#permalink]
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Research indicates that the health of all multicellular organisms, including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able to self-destruct when they become superfluous or disordered.
A. including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able
B. including humans, not only depends on the organism's ability to produce new cells but also the ability of individual cells
C. including humans, depends not only on the organism's ability to produce new cells but also on the ability of individual cells
D. which includes humans, depends on not only the organism's ability to produce new cells but also depending on the ability of individual cells
E. which includes humans, not only depends on whether the organism is able to produce new cells but also on the ability of individual cells
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Research indicates that the health of all multicellular organisms, including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able to self-destruct when they become superfluous or disordered.
When looking at this problem it's important that the idiom "not only x but also y" is correct and that the sentence maintains its parallelism.
A. including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able
The word "whether" disrupts the idiom "not only x but also y". Also the sentence is wordy. Should be not only the ability... but also the ability... Not "are able"
B. including humans, not only depends on the organism's ability to produce new cells but also the ability of individual cells
to maintain parallelism it should be not only depends on x... but also depends on y...
C. including humans, depends not only on the organism's ability to produce new cells but also on the ability of individual cells
Like Mary Poppins. Perfect in every way.
D. which includes humans, depends on not only the organism's ability to produce new cells but also depending on the ability of individual cells
"depending" is not necessary. It is also not parallel to "depends"
E. which includes humans, not only depends on whether the organism is able to produce new cells but also on the ability of individual cells
Not parallel.
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Re: Research indicates that the health of all multicellular organisms [#permalink]
Research indicates that the health of all multicellular organisms, including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able to self-destruct when they become superfluous or disordered.
A. including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able--this is not parallel (we're using the correct idiom not only... but also), but the words following not only must be mirrored by what follows but also--this isn't the case--incorrect
B. including humans, not only depends on the organism's ability to produce new cells but also the ability of individual cells--again, same issue, parallelism isn't maintained with respect to the not only, but also idiom
C. including humans, depends not only on the organism's ability to produce new cells but also on the ability of individual cells--correct with respect to grammar, parallelism and meaning
D. which includes humans, depends on not only the organism's ability to produce new cells but also depending on the ability of individual cells--"which includes humans" is modifying the "health of organisms"--incorrect and illogical modifier
E. which includes humans, not only depends on whether the organism is able to produce new cells but also on the ability of individual cells--same error as D
Choice C
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Re: Research indicates that the health of all multicellular organisms [#permalink]
A. including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able (second part lacks "on" after "also")
B. including humans, not only depends on the organism's ability to produce new cells but also the ability of individual cells (the same as in A)
C. including humans, depends not only on the organism's ability to produce new cells but also on the ability of individual cells ("not only on"...is parallel to "but also on")
D. which includes humans, depends on not only the organism's ability to produce new cells but also depending on the ability of individual cells ("Organisms includes" is incorect. Organisms are plural.)
E. which includes humans, not only depends on whether the organism is able to produce new cells but also on the ability of individual cells (the same as in D)
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Re: Research indicates that the health of all multicellular organisms [#permalink]
Research indicates that the health of all multicellular organisms, including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able to self-destruct when they become superfluous or disordered.
A. including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able - 'not only the ability.....but also whether' is not parallel.
B. including humans, not only depends on the organism's ability to produce new cells but also the ability of individual cells - 'not only depends on.....but also the ability of' is not parallel
C. including humans, depends not only on the organism's ability to produce new cells but also on the ability of individual cells - 'not only on the organism's ability.....but also on the ability' is parallel Hence correct.
D. which includes humans, depends on not only the organism's ability to produce new cells but also depending on the ability of individual cells - 'not only the organism's ability....but also depending on'is not parallel. Also verb is repeated.
E. which includes humans, not only depends on whether the organism is able to produce new cells but also on the ability of individual cells - 'not only depends on....but also on the ability' is not parallel. Verb 'Depends' is missing for second sentence.
+1 for Kudos....
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Research indicates that the health of all multicellular organisms [#permalink]
souvik101990 wrote:
New GMAT Club Project of reviving hardest questions! Post answers and explanations to get kudos
Research indicates that the health of all multicellular organisms, including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able to self-destruct when they become superfluous or disordered.
A. including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able
B. including humans, not only depends on the organism's ability to produce new cells but also the ability of individual cells
C. including humans, depends not only on the organism's ability to produce new cells but also on the ability of individual cells
D. which includes humans, depends on not only the organism's ability to produce new cells but also depending on the ability of individual cells
E. which includes humans, not only depends on whether the organism is able to produce new cells but also on the ability of individual cells
Hey Souvik
Can you help me with E. Between C and E, dont you think "including" is vague in modifying for humans in C whereas in E, "which" correctly modifies organisms?
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Re: Research indicates that the health of all multicellular organisms [#permalink]
sinhap07 wrote:
souvik101990 wrote:
New GMAT Club Project of reviving hardest questions! Post answers and explanations to get kudos
Research indicates that the health of all multicellular organisms, including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able to self-destruct when they become superfluous or disordered.
A. including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able
B. including humans, not only depends on the organism's ability to produce new cells but also the ability of individual cells
C. including humans, depends not only on the organism's ability to produce new cells but also on the ability of individual cells
D. which includes humans, depends on not only the organism's ability to produce new cells but also depending on the ability of individual cells
E. which includes humans, not only depends on whether the organism is able to produce new cells but also on the ability of individual cells
Hey Souvik
Can you help me with E. Between C and E, dont you think "including" is vague in modifying for humans in C whereas in E, "which" correctly modifies organisms?
E has two problems. First, incorrect use of "which". Since "which" modifies the closest noun then it should be "which include" since "organisms" are plural. Second, incorrect use of idiom "not only but also". Here "only" modifies "depends", hence after "but" there should be a verb (because of parallelism). Hence E is incorect
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Re: Research indicates that the health of all multicellular organisms [#permalink]
When you have a sentence like : "blablabla, verb-ing blabla", the "ing" modifier refers to the entire preceding clause.
Whereas here, "including" refers to "multicellular organisms".
So I would say this is a wrong modifier.
Can you please explain me ?
Thanks
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Re: Research indicates that the health of all multicellular organisms [#permalink]
Alex75PAris wrote:
Whereas here, "including" refers to "multicellular organisms".
So I would say this is a wrong modifier.
Hi Alex75PAris, that's a good observation. However, including is actually a well known exception to this rule. So, you might just want to remember it.
Our book EducationAisle Sentence Correction Nirvana documents this exception of "including", and provides examples to illustrate. If someone is interested, PM me your email-id, I can mail the corresponding section.
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Research indicates that the health of all multicellular organisms, including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able to self-destruct when they become superfluous or disordered.
A. including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able
B. including humans, not only depends on the organism's ability to produce new cells but also (on) the ability of individual cells
C. including humans, depends not only on the organism's ability to produce new cells but also on the ability of individual cells
D. which includes humans, depends on not only the organism's ability to produce new cells but also depending on the ability of individual cells
E. which includes humans, not only depends on whether the organism is able to produce new cells but also on the ability of individual cells
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Re: Research indicates that the health of all multicellular organisms [#permalink]
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Re: Research indicates that the health of all multicellular organisms [#permalink]
daagh wrote:
A case of not only….. but also parallelism. Only C maintains it. C is the choice
Hi,
I would like to check whether the phrase "ability of" right in the correct choice. Manhattan Sc states that "ability to " is right and not "ability of "
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Re: Research indicates that the health of all multicellular organisms [#permalink]
daagh wrote:
A case of not only….. but also parallelism. Only C maintains it. C is the choice
I agree with daagh sir. I almost fell for option B. However I found that the clause not only depends on but also DEPENDS ON is missing. It's a clear cut parallelism error.
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Research indicates that the health of all multicellular organisms [#permalink]
Easy pickings Issue 1 - SV Agreement
"organisms" = plural "includes" = singular -> Eliminate D and E
D. which includes humans, depends on not only the organism's ability to produce new cells but also depending on the ability of individual cells
E. which includes humans, not only depends on whether the organism is able to produce new cells but also on the ability of individual cells
Issue 2 - Parallelism
A. including humans, depends on not only the ability of the organism to produce new cells but also whether individual cells are able
B. including humans, not only depends on the organism's ability to produce new cells but also (depends on has been ommitted) the ability of individual cells
C. including humans, depends not only on the organism's ability to produce new cells but also on the ability of individual cells
C is correct as it maintains symmetry between parallelism markers "not only...but also"
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Approximately 5 ounces life shelled peanuts = 1 cup. 12 ounces roasted shelled peanuts = 2 cups.
You are watching: How much is 1 oz of peanuts
Similarly, how plenty of ounces is 1/2 cup peanuts? identical values amount, in grams (g) amount, in ounces (oz)
1/4 cup 30 g 1.1 oz
1/3 cup 40 g 1.5 oz
3/8 cup 45 g 1.7 oz
1/2 cup 65 g 2.2 oz
Then, exactly how do ns measure 1 oz the nuts?
A solitary serving that nuts is one ounce or 1/4 come 1/3 cup. 2 finger scoop = One tablespoon.
How lot is one Oz of peanuts?
an ounce that peanuts is about 28 totality peanuts. A handful of peanuts is also around an ounce, which provides it straightforward to measure them out. Peanuts are high in protein, as one-third of their calories come native protein.
Afrae HormannProfessional
How countless peanuts need to you eat a day?
So, including peanuts or peanut butter together a nutritious enhancement to your diet have the right to be done without guilt or break “the diet” bank, when consumed in the best portions. The recommended everyday servings room a grasp of peanuts (1-2 ounces depending on your size) or 2 tablespoons that peanut butter.
How plenty of cups is 100g the peanuts?
How many united state customary cups is 100 grams peanuts? 100 grams that peanuts = 2/3 that a us cup of peanuts.
Raleigh LindenschmidtProfessional
How many ounces is a handful of peanuts?
A serving of peanuts is same to one ounce, or approximately 28 peanuts.
Draguta VerstegeExplainer
How numerous grams room a cup?
How numerous grams in 1 cups? The price is 236.5882375. We assume you room converting in between gram and also cup .
Nasly UhartExplainer
How numerous nuts space in one ounce?
The complying with equal one ounce: 24 almonds, 18 tool cashews, 12 hazelnuts or filberts, 8 medium Brazil nuts, 12 macadamia nuts, 35 peanuts, 15 pecan halves and also 14 English walnut halves (3).
How plenty of nuts is 30g?
A 30g offer of nuts is identical to approximately: 20 almonds. 10 Brazil nuts. 15 cashews.
Hormisdas GuilleminPundit
Are peanuts fattening?
Peanuts room fattening. It"s true the peanuts have fat, but it"s the good-for-you mono- and also polyunsaturated kind that tree nuts also have. They won"t make you fat if you clock your part size-1 oz of peanuts (32 nuts) it is provided 166 calories.
Zuzana AlfonskyPundit
How plenty of grams is a peanut?
One serving of dry-roasted peanuts (30 grams) contains 12 grams that unsaturated fat, only two grams of saturated fat, and no trans-fat.
Shanta MirzaPundit
What walk 1 oz of meat watch like?
3 oz portion is comparable in size to a deck that cards ? 1 oz of cooked meat is similar in size to 3 dice. A 1-inch meatball is around one ounce. 4 oz that raw, lean meat is around 3 ounces after ~ cooking. 3 oz of grilled fish is the size of a checkbook.
Rosina SeiglPundit
How big is an oz of cheese?
One ounce. That"s the proper serving size because that a serving of cheese. You have the right to estimate your portions knowing that one oz of cheese is around the size of a pair the dice.
Erminia LedruPundit
What go 1 oz that almonds look like?
One serving of almonds is 23 almonds, which equates to 1 ounce, ¼ cup or around 1 handful.
Annamae ThomansTeacher
Does 4 ounces equal 1 cup?
4 oz come cups = 0.5 cup in 4 oz.
What does 1 cup of street weigh?
200 grams
Kristofer MichelleSupporter
How countless peanuts walk it require to make a tablespoon that peanut butter?
Quite a few! The national Peanut Board estimates it takes around 540 peanuts come make a 12-ounce seasoned of peanut butter. That"s roughly 45 peanuts per oz of peanut butter.
How many cups space in 16 oz of peanuts?
How many us cups the peanut butter in 16 ounces? Peanut butter counter Chart near 4 ounces.
ounces to us cups of Peanut butter
15 ounces = 1.77 (1 3/4 ) united state cups
16 ounces = 1.89 (1 7/8 ) united state cups
17 ounces = 2.01 (2) us cups
18 ounces = 2.13 (2 1/8 ) us cups
Terina HiscocksBeginner
How plenty of cashews room in a quarter cup?
Use this chart for easy reference, a quarter cup of nuts is approximately equal come 40 almonds, 24 cashews, or 10 walnuts.
See more: Muse Of Lyric Poetry And Music, In Greek Mythology
Neco ZabalaBeginner
How lot does 1 cup that peanut butter weigh?
equivalent values
amount, in grams (g) amount, in ounces (oz)
3/4 cup 190 g 6.6 oz
7/8 cup 220 g 7.7 oz
1 cup 250 g 8.8 oz
2 cups 500 g 17.6 oz
Yosyp IsernhinkeBeginner
How many peanuts space in a gallon?
According to the agricultural Marketing resource Center, typical yield the peanuts is 3000 lb/ acre, with the potential the ~135 gallons of oil per acre. This offers ~22 lb that peanuts (in shell) every gallon that oil.
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Coils of Wire in the Motor's Rotor
PWM Frequency is the count of PWM interval periods per second, expressed in Hertz (Hz). Mathematically, the frequency is equal to the inverse of the interval period's length (PWM_Frequency = 1 / PWM_Interval_Period).
When calculating the PWM Equivalent Voltage, we generally assume that the motor will operate ideally and respond as if it was connected to a non-PWM power source providing the voltage. But that's not the case. For example, a Yellow-TT motor will easily spin if a single 1.5-volt battery is connected, but will not turn until the PWM Equivalent Voltage coming from a Motor FeatherWing reaches 2.0 volts when operating in fast decay mode. And when it does start, it suddenly rotates at 4000 RPM. Why is that?
Since a brushed DC motor’s internal rotor consists of two or more coils of wire wound around laminated magnetic core material, the motor acts like an inductor. Depending on size of the rotor coil, it may take a few milliseconds for the energy to build sufficiently to turn the shaft.
Inductors are electromagnetic components that capture energy from the incremental buildup of the magnetic field created by an electrical current passing through a wire coil.
Rotor coil inductance becomes an important factor to consider when using PWM for motor speed control. The motor coil works best when the applied voltage is relatively steady since it needs time for its magnetic field to reach the needed strength. At higher PWM frequencies, the pulses from the motor controller board change too quickly to provide enough energy to spin the motor until the equivalent voltage reaches 2.0 volts, although switching to using slow decay mode can help.
When the PWM frequency is lowered, the motor’s coils extract more energy from the pulsed PWM signal. That means that the motor will start spinning at a lower equivalent voltage and will operate with improved torque at low speeds. The following graphs compare the Yellow-TT motor's speed response when the default PWM frequency of 1600Hz is changed to 25Hz.
The spin threshold at 25Hz starts at 0.5 volts or less depending on the decay mode selection, increasing the useable motor speed range to as low as 100 RPM. The Yellow-TT gearbox reduces the motor’s RPM by a factor of 48, so the attached wheel will be turning at 2 RPM or about 0.7 cm/sec. A velocity like that will make it much easier for your robot to sneak up on the cat.
So now that we know about current decay mode and PWM frequency, how do we go about choosing the best configuration of the two parameters for our robot’s motors?
This guide was first published on Apr 11, 2021. It was last updated on May 15, 2024.
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It is always said that dealing with floating point numbers is slow. So slow that if precision can be compromised and speed is more important, programmers will use integers instead. Integers ? yes ! But how will you deal with fractions then ? The answer is fixed point arithmetic.
When we deal with decimal numbers, each digit in a number is actually a coefficient of the base, which is 10, to a certain power for the digit. For example:
1234 = 1 \cdot 10^3 + 2 \cdot 10^2 + 3 \cdot 10^1 + 4 \cdot 10^0
We can rewrite the above equation as:
10 \cdot (123.4) = 1234 = 10 \cdot (1 \cdot 10^2 + 2 \cdot 10^1 + 3 \cdot 10^0 + 4 \cdot 10^{-1})
So by expressing with a number multiplied by 10, we can have the decimal point between the third and the fourth digit. By changing the 10 to 100, we can move the decimal point to between the second and the third digit. Basically, we can place the decimal point at any position we want, as long as we know what the multiplier we are using. That is how we express fraction with integers.
The downside is, fixed point numbers do not have fixed effective number of digits. The smaller number it has to express the less effective number of digits it has.
1.234 -> 4 digits 0.123 -> 3 digits 0.012 -> 2 digits 0.001 -> 1 digit
Thus, when dealing with fixed points, the order of computations can really matter. The good point is, we do not need special hardware or software to deal with them since basically all arithmetics now deal with normal integers.
For binary numbers, the base is 2. The basic rules do not change.
10111_2 = 1 \cdot 2^4 + 0 \cdot 2^3 + 1 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0
2 \cdot (1011.1_2) = 10111_2 = 2 \cdot ( 1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0 + 1 \cdot 2^{-1})
If we split an M+N bit number into M bit and N bit numbers with decimal point between them, then the multiplier is 2N. It can be easily shown that 1.0 is then
equals to 2N in the fixed point representation. Then all we have to do to convert a real number to its fixed point representation is to multiply it with 2N (and round, any fraction after multiplication becomes conversion error). And to get the real number from a fixed point representation we simply divide it with 2N.
Addition with fixed point numbers is as simple as addition with integers but of course you need to align the decimal points in case they are different. Multiplication is also as simple as multiplication with integers, but care must be taken as the decimal point position will change. If we multiply M.N with O.P fixed point numbers, the result will be in (M+O).(N+P) format. The O+P can be easily understood as M.N has a multiplier of 2N and O.P has a multiplier of 2P. By multiplying both numbers, we will also have the multipliers be multiplied too, resulting a new multiplier of 2N+P.
Below is a small Perl program to help you convert from real numbers to fixed point numbers and vice versa. Simply execute it with:
fixedpt.pl <qformat> [number]
The qformat should be specified prepended with Q, i.e. to specify M.N format, use QM.N
number, when supplied will be autodetected. If it is detected a real number, the program will output the fixed point counterpart, and vice versa. Fixed point number must be supplied in hexadecimal.
> fixedpt.pl Q2.5 1.61.60000 = 33(16) = 0110011(2)> fixedpt.pl Q2.5 3333(16) = 0110011(2) = 1.59375
If no number is provided, the program will read from standard input.
#!/usr/local/bin/perlif ($#ARGV < 0) { print "usage:$0 [number]\n"; exit;}$QFORMAT =$ARGV[0];if ($QFORMAT =~ /[qQ]([0-9]+)\.([0-9]+)/) {$QM = $1;$QN = $2;}else { print "invalid format specifier. expected Qm.n form\n"; exit;}$QTOTAL = $QM +$QN; # sign bit is included in QM# one = 1.000000..., where the number of 0 is $QN$ONE = 2.0 ** $QN;$MASK = (2.0 ** $QTOTAL) - 1;if ($#ARGV >= 1) { &convert($ARGV[1]);}else { while () { chop; &convert($_); }}sub DecToBin() { local($FPN) = @_; local$FPNBIN, $x,$n; $FPNBIN = ""; while (length($FPNBIN) < $QTOTAL) {$n = int($FPN / 2.0);$x = (abs($FPN -$n * 2.0) > 0.5) ? 1 : 0; $FPNBIN =$x . $FPNBIN;$FPN = $n; } die "overflow !\n" if ($FPN > 0.0); return $FPNBIN;}sub BinToHex() { local($FPNBIN) = @_; local $FPNHEX,$x, $h;$FPNHEX = ""; while (length($FPNBIN) > 0) { if (length($FPNBIN) >= 4) { $x = substr($FPNBIN, -4); $FPNBIN = substr($FPNBIN, 0, length($FPNBIN)-4); } else {$x = $FPNBIN;$h = substr($x, 0, 1);$x = $h .$x while (length($x) < 4);$FPNBIN = ""; } $h = ($x eq "0000") ? "0" : ($x eq "0001") ? "1" : ($x eq "0010") ? "2" : ($x eq "0011") ? "3" : ($x eq "0100") ? "4" : ($x eq "0101") ? "5" : ($x eq "0110") ? "6" : ($x eq "0111") ? "7" : ($x eq "1000") ? "8" : ($x eq "1001") ? "9" : ($x eq "1010") ? "A" : ($x eq "1011") ? "B" : ($x eq "1100") ? "C" : ($x eq "1101") ? "D" : ($x eq "1110") ? "E" : "F"; $FPNHEX =$h . $FPNHEX; } return$FPNHEX;}sub HexToDec() { local($FPN) = @_; local$FPNDEC, $x; local$hexstr = "0123456789ABCDEF"; $FPN =$FPN . ""; # make sure we are processing string $FPNDEC = 0.0; while ($FPN ne "") { $x = uc(substr($FPN, 0, 1)); $x = index($hexstr, $x); die "bad hexadecimal input\n" if ($x == -1); $FPNDEC =$FPNDEC * 16.0 + $x;$FPN = substr($FPN, 1); } die "overflow !\n" if ($FPNDEC > $MASK); return$FPNDEC;}sub convert() { local ($EXPR) = @_; local$NUMBER; if ($EXPR =~ /[^0-9A-Fa-f\.\-]/) {$EXPR = "\$NUMBER =$EXPR"; eval $EXPR; } else {$NUMBER = $EXPR; } if ($NUMBER =~ /\./) { # real number $MAX = 2.0 **$QM; if ($NUMBER >=$MAX) { print "overflow !\n"; exit; } $NEG = ($NUMBER < 0.0) ? 1 : 0; $FPN =$ONE * abs($NUMBER);$FPN = ($MASK -$FPN) + 1.0 if ($NEG); # 2's complement on negative number$FPNBIN = &DecToBin($FPN);$FPNHEX = &BinToHex($FPNBIN); printf "%." .$QN . "f = %s(16) = %s(2)\n", $NUMBER,$FPNHEX, $FPNBIN; } else {$FPN = &HexToDec($NUMBER);$FPNBIN = &DecToBin($FPN); if (substr($FPNBIN, 0, 1) eq "1") { $FPN = ($MASK - $FPN) + 1.0; # 2's complement on negative number$FPN = -$FPN; }$FPN = $FPN /$ONE; printf "%s(16) = %s(2) = %." . $QN . "f\n",$NUMBER, $FPNBIN,$FPN; }}
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### Doing the Math
#### Matt Lowry
The power of numbers, and by extension the whole subject of mathematics, is phenomenal. Math takes us from the realm of guesswork into certainty; math gives us power over our physical world, allowing us to make predictions that are testable. And, when applied in certain ways, can lead us to ponder some of the most compelling questions about our very existence.
Take, for instance, two areas of thought -- the age-old pseudoscience of astrology and the considerably younger search for extra-terrestrial intelligence (SETI). Each is popular, and each has its own peculiar association with numbers.
As much as one-third of the U.S. population still adheres in some way to astrology, outdated and useless as it is. This is, in large part, due to a misunderstanding of basic mathematics. As a way to simply illustrate the incorrectness of astrology to my high school classes, I had them perform a simple test. Each student was to read through a description of each Zodiac symbol (there are 12 in all) and rate which one described them best, on a scale of 1 to 7 for accuracy. It should be noted that each symbol was represented by a number, as opposed to the symbol's name -- this was done to blind the students from knowing ahead of time which Zodiac symbol they were reading about.
The next day, I had the students pick the Zodiac number which they say described them best, and then we looked up on calendars to see which Zodiac symbol actually belonged to each student. Then came the kicker -- I showed my students the key, which matched up the numbers with the actual Zodiac symbols. The whole idea is for the students to see how close, if at all, they came to accurately describing themselves through astrology.
The result? Only 1 out of 42 students actually got it "right" -- that is, one kid picked the description that was their actual Zodiac symbol. When they saw the results, there were a lot of wide eyes and "whoas" throughout the room; the conclusion at that point was obvious. When applied as a tool to separate fact from fiction, numbers and mathematics can be most powerful.
Yet math can also inspire. Take the case of SETI. One of the more quanitifiable concepts associated with SETI is the famous Drake Equation, which is a mathematical way of estimating the number of technological civilizations that may exist in our galaxy. Here's what the equation looks like...
N = R* - fp - ne - fl - fi - fc - L
where N = number of civilizations in the Milky Way Galaxy whose electromagnetic emissions are detectable, R* = rate of formation of stars suitable for the development of intelligent life, fp = fraction of those stars with planetary systems, ne = number of planets, per solar system, with an environment suitable for life, fl = fraction of suitable planets on which life actually appears, fi = the fraction of life bearing planets on which intelligent life emerges, fc = fraction of civilizations that develop a technology that releases detectable signs of their existence into space, and L = length of time such civilizations release detectable signals into space.
At this stage, we have a rough handle on the values of R*, fp, and ne. Frank Drake himself estimates tthe value of R* at about 10/year. Discoveries of numerous (150+) extra-solar planetary systems within the last 15 years have given us an estimate of fp. And in mid-June, the news of the discovery of the first Earth-like planet beyond our solar system was announced -- this now gives more information on the possible value of ne. Of course, we don't truly have any solid idea of the values beyond these first three, and any guesses would be speculation.
But if you're like me, and you cannot resist speculating on such a fascinating question, you could always tinker with the numbers on your own using the Drake Calculator -- www.seti.org/science/drake-calc.html
In time, we'll gather more data and the terms in the Drake Equation will become better known. And perhaps, someday in the far future, our descendants will be able to do the math in a more satisfactory manner. Until then, we'll have to live with the uncertainty... and the inspiration to learn more.
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You are on page 1of 11
# 90
## Olympiads in Informatics, 2013, Vol. 7, 90100
2013 Vilnius University
## Where to Use and How not to Use Polynomial
String Hashing
Faculty of Mathematics, Informatics and Mechanics, University of Warsaw
Banacha 2, 02-097 Warsaw, Poland
## Abstract. We discuss the usefulness of polynomial string hashing in programming competition
tasks. We show why several common choices of parameters of a hash function can easily lead to
a large number of collisions. We particularly concentrate on the case of hashing modulo the size
of the integer type used for computation of fingerprints, that is, modulo a power of two. We also
give examples of tasks in which string hashing yields a solution much simpler than the solutions
obtained using other known algorithms and data structures for string processing.
Key words: programming contests, hashing on strings, task evaluation.
1. Introduction
Hash functions are used to map large data sets of elements of an arbitrary length (the
keys) to smaller data sets of elements of a fixed length (the fingerprints). The basic application of hashing is efficient testing of equality of keys by comparing their fingerprints.
A collision happens when two different keys have the same fingerprint. The way in which
collisions are handled is crucial in most applications of hashing. Hashing is particularly
useful in construction of efficient practical algorithms.
Here we focus on the case of the keys being strings over an integer alphabet =
{0, 1, . . . , A 1}. The elements of are called symbols.
An ideal hash function for strings should obviously depend both on the multiset of the
symbols present in the key and on the order of the symbols. The most common family of
such hash functions treats the symbols of a string as coefficients of a polynomial with an
integer variable p and computes its value modulo an integer constant M :
H(s1 s2 s3 . . . sn ) = (s1 + s2 p + s3 p2 + + sn pn1 ) mod M.
A careful choice of the parameters M , p is important to obtain good properties of the
hash function, i.e., low collision rate.
Fingerprints of strings can be computed in O(n) time using the well-known Horners
method:
hn+1 = 0,
## hi = (si + phi+1 ) mod M
for i = n, n 1, . . . , 1.
(1)
## Where to Use and How not to Use Polynomial String Hashing
91
Polynomial hashing has a rolling property: the fingerprints can be updated efficiently
when symbols are added or removed at the ends of the string (provided that an array of
powers of p modulo M of sufficient length is stored). The popular RabinKarp pattern
matching algorithm is based on this property (Karp and Rabin, 1987). Moreover, if the
fingerprints of all the suffixes of a string are stored as in (1), one can efficiently find the
fingerprint of any substring of the string:
H(si . . . sj ) = (hi pji+1 hj+1 ) mod M.
(2)
## This enables efficient comparison of any pair of substrings of a given string.
When the fingerprints of two strings are equal, we basically have the following options: either we consider the strings equal henceforth, and this way possibly sacrifice
the correctness in the case a collision occurs, or simply check symbol-by-symbol if the
strings are indeed equal, possibly sacrificing the efficiency. The decision should be made
depending on a particular application.
In this article we discuss the usefulness of polynomial string hashing in solutions of
programming competition tasks. We first show why several common ways of choosing
the constants M and p in the hash function can easily lead to a large number of collisions
with a detailed discussion of the case in which M = 2k . Then we consider examples of
tasks in which string hashing applies especially well.
## 2. How to Choose Parameters of Hash Function?
2.1. Basic Constraints
A good requirement for a hash function on strings is that it should be difficult to find a
pair of different strings, preferably of the same length n, that have equal fingerprints. This
excludes the choice of M < n. Indeed, in this case at some point the powers of p corresponding to respective symbols of the string start to repeat. Assume that pi pj mod M
for i < j < n. Then the following two strings of length n have the same fingerprint:
0
. . 0 a1 a2 . . . anj 0
. . 0
.
.
i
ji
and
0 .
. . 0 a1 a2 . . . anj .
j
Similarly, if gcd(M, p) > 1 then powers of p modulo M may repeat for exponents
smaller than n. The safest choice is to set p as one of the generators of the group U (ZM )
the group of all integers relatively prime to M under multiplication modulo M . Such
a generator exists if M equals 2, 4, q a or 2q a where q is an odd prime and a 1 is
integer (Weisstein, on-line). A generator of U (ZM ) can be found by testing a number of
random candidates. We will not get into further details here; it is simply most important
not to choose M and p for which M | pi for any integer i.
A slightly less obvious fact is that it is bad to choose p that is too small. If p < A
(the size of the alphabet) then it is very easy to show two strings of length 2 that cause a
92
collision:
H(01) = H(p0).
2.2. Upper Bound on M
We also need to consider the magnitude of the parameter M . Let us recall that most
programming languages, and especially the languages C, C++, Pascal that are used for
IOI-style competitions, use built-in integer types for integer manipulation. The most popular such types operate on 32-bit or 64-bit numbers which corresponds to computations
modulo 232 and 264 respectively. Thus, to be able to use Horners method for fingerprint
computation (1), the value (M 1) p + (M 1) = (M 1) (p + 1) must fit within
the selected integer type. However, if we wish to compute fingerprints for substrings of a
string using the (2), we need (M 1)2 + (M 1) to fit within the integer type, which
bounds the range of possible values of M significantly. Alternatively, one could choose
a greater constant M and use a fancier integer multiplication algorithm (which is far less
convenient).
2.3. Lower Bound on M
On the other side M is bounded due to
the well-known birthday paradox: if we consider
a collection of m keys with m 1.2 M then the chance of a collision to occur within
this collection is at least 50% (assuming that the distribution of fingerprints is close to
uniform on the set of all strings). Thus if the birthday paradox applies then one needs to
choose M = (m2 ) to have a fair chance to avoid a collision. However, one should note
that not always the birthday paradox applies. As a benchmark consider the following two
problems.
Problem 1: Longest Repeating Substring. Given a string s, compute the longest string
that occurs at least twice as a substring of s.
Problem 2: Lexicographical Comparison. Given a string s and a number of queries
(a, b, c, d) specifying pairs of substrings of s: sa sa+1 . . . sb and sc sc+1 . . . sd , check, for
each query, which of the two substrings is lexicographically smaller.
A solution to Problem 1 uses the fact that if s has a repeating substring of length k
then it has a repeating substring of any length smaller than k. Therefore we can apply binary search to find the maximum length k of a repeating substring. For a candidate value
of k we need to find out if there is any pair of substrings of s of length k with equal fingerprints. In this situation the birthday paradox applies. Here we assume that the distribution
of fingerprints is close to uniform, we also ignore the fact that fingerprints of consecutive
substrings heavily depend on each other both of these simplifying assumptions turn out
not to influence the chance of a collision significantly.
The situation in Problem 2 is different. For a given pair of substrings, we apply binary
search to find the length of their longest common prefix and afterwards we compare the
## Where to Use and How not to Use Polynomial String Hashing
93
symbols that immediately follow the common prefix, provided that they exist. Here we
have a completely different setting, since we only check if specific pairs of substrings are
equal and we do not care about collisions across the pairs. In a uniform model, the chance
1
, and the chance of a collision occurring
of a collision within a single comparison is M
m
. The birthday paradox does not apply
within m substring comparisons does not exceed M
here.
2.4. What if M = 2k ?
A very tempting idea is not to select any value of M at all: simply perform all the computations modulo the size of the integer type, that is, modulo 2k for some positive integer k.
Apart from simplicity we also gain efficiency since the modulo operation is relatively
slow, especially for larger integer types. That is why many contestants often choose this
method when implementing their solutions. However, this might not be the safest choice.
Below we show a known family of strings which causes many collisions for such a hash
function.
This family is the ubiquitous ThueMorse sequence (Allouche and Shallit, 1999). It
is defined recursively as follows:
0 = 0; i = i1 i1
for i > 0
where x
is the sequence resulting by negating the bits of x. We have:
0 = 0, 1 = 01, 2 = 0110, 3 = 01101001, 4 = 0110100110010110, . . .
and clearly the length of i is 2i .
Let W (s) be the fingerprint of a string s without any modulus:
W (s1 s2 s3 . . . sn ) = s1 + s2 p + s3 p2 + + sn pn1 .
We will show the following property of the ThueMorse sequence that uncovers the reason for its aforementioned especially bad behavior when hashing modulo M = 2k .
n ) W (n ).
Theorem 1. For any n 0 and 2 p, 2n(n+1)/2 | W (
Proof. By the recursive definition n = n1 n1 and, similarly, n = n1 n1 we
have:
n1 ) + p2
W (
n ) W (n ) = W (
n1
= W (
n1 )(1 p2
= (1 p2
n1
W (n1 ) W (n1 ) p2
n1
) W (n1 )(1 p2
)(W (
n1 ) W (n1 )).
## Now it is easy to show by induction that:
W (
n ) W (n ) = (1 p2
n1
)(1 p2
n2
) . . . (1 p).
n1
n1
W (
n1 )
94
## To conclude the proof, it suffices to argue that
2i | 1 p2
i1
for any i 1.
(3)
This fact can also be proved by induction. For i = 1 this is a consequence of the fact that
p is odd. For the inductive step (i > 1) we use the following equality:
1 p2
i1
i2
i2
1 + p2
.
= 1 p2
The second factor is even again because p is odd. Due to the inductive hypothesis, the first
factor is divisible by 2i1 . This concludes the inductive proof of (3) and, consequently,
the proof of the whole theorem.
By Theorem 1, the strings n and n certainly yield a collision if n(n + 1)/2 exceeds the number of bits in the integer type. For instance, for 64-bit integers it suffices
to take 11 and 11 which are both of length 2048. Finally, let us note that our example
is really vicious it yields a collision regardless of the parameter p, provided that it is
odd (and choosing an even p is surely bad if M is a power of two). This example can
also be extended by sparsifying the strings (i.e., inserting segments consisting of a large
number of zeros) to eliminate with high probability a heuristic algorithm that additionally
checks equality of a few random symbols at corresponding positions of the strings if their
fingerprints are equal.
The same example was described in a recent blog post related to programming competitions (Akhmedov, 2012).
## 3. Usefulness of String Hashing
In the previous section we have described several conditions that limit the choice of constants for string hashing. In some applications it is difficult to satisfy all these conditions
at the same time. In contrast, in this section we show examples of problems that are more
difficult to solve without using string hashing. Two particularly interesting problems we
describe in detail, and at the end we list several other examples of tasks from Polish programming competitions. In the algorithms we do not describe how to handle collisions,
that is, we make an optimistic assumption that no collisions occur and thus agree to obtain
a heuristic solution.
We have already mentioned that string hashing can be used to check equality of substrings of a given string. For the same purpose one could use the Dictionary of Basic
Factors (O(n log n) time and space preprocessing for a string of length n) or the suffix tree/suffix array (O(n) time and space preprocessing for a constant-sized alphabet in
the basic variant, both data structures are relatively complex). More on these data structures can be found, for instance, in the books Crochemore et al. (2007) and Crochemore
and Rytter (2003). However, in the following problem it is much more difficult to apply any of these data structures instead of string hashing. This is a task from the fi-
95
## nal round of a Polish programming competition Algorithmic Engagements 2011 (see
http://main.edu.pl/en/archive/pa/2011/bio).
Problem 3: Computational Biology. We are given a string s = s1 . . . sn over a constantsized alphabet. A cyclic substring of s is a string t such that all cyclic rotations of t are
substrings of s 1 . For a cyclic substring t of s, we define the number of cyclic occurrences
of t in s as the total number of occurrences of distinct cyclic rotations of t within s. We
would like to find a cyclic substring of s of a given length m that has the largest number
of cyclic occurrences. We are to output this number of occurrences.
For example, consider the string s = BABABBAAB and m = 3. The string AAB is its
cyclic substring with 3 cyclic occurrences: one as AAB, one as ABA and one as BAA. The
string ABB is also a cyclic substring and it has 4 cyclic occurrences: one as ABB, two as
BAB and one as BBA. Thus the result is 4.
On the other hand, consider the string s = ABAABAABAABAAAAA and m = 5. Here
the result is just 1: a single cyclic occurrence of a cyclic substring AAAAA. Note that none
of the strings ABAAB and AABAA is a cyclic substring of the string and therefore they are
not included in the result.
Using string hashing, we solve this problem as follows. We start by computing fingerprints of all m-symbol substrings of s: s1 . . . sm , s2 . . . sm+1 ,. . . this can be done in
O(n) time using the rolling property of the hash function. Using a map-like data structure
storing fingerprints, for each of these substrings we count the number of times it occurs
in s.
Now we treat these substrings as vertices of a directed graph. The vertices are identified by the fingerprints. Each vertex is assigned its multiplicity: the number of occurrences of the corresponding substring in s. The edges of the graph represent cyclic rotations by a single symbol: there is an edge from u1 u2 . . . um to u2 . . . um u1 provided
that the latter string occurs as a substring of s (see Fig. 1). Note that the fingerprint of the
endpoint of any edge can be computed in O(1) time, again due to the rolling property of
the hash function.
Thus our problem reduces to finding the heaviest cycle in a graph in which each vertex
has in-degree and out-degree at most 1. This can be done in linear time, since such a graph
is a collection of cycles and paths. The whole solution has O(n log n) time complexity
due to the application of a map-like data structure.
The part of the above solution that causes difficulties in using the Dictionary of Basic
Factors or the suffix tree/suffix array data structures is the fact that if u1 u2 . . . um is
a substring of s then u2 . . . um u1 is (almost never) a substring of s that could be easily
identified by its position in s.
We proceed to the second task example. Recall that in the pattern matching problem
we are given two strings, a pattern and a text, and we are to find all the occurrences of the
1 A cyclic rotation of a string is constructed by moving its first letter to its end, possibly multiple times. For
example, there are three different cyclic rotations of ABAABA, namely BAABAA, AABAAB and ABAABA.
96
Fig. 1. The graphs of substrings for: s = BABABBAAB and m = 3 (above) and s = ABAABAABAABAAAAA
and m = 5 (below)
pattern in the text. This problem has a number of efficient, linear time solutions, including
the MorrisPratt algorithm, BoyerMoore algorithm, and a family of algorithms working in constant space. Also suffix trees/suffix arrays can be used for this problem. It is,
however, rather difficult to extend any of these algorithms to work for the 2-dimensional
variant of the problem:
Problem 4: 2-Dimensional Pattern Matching. Let the pattern be an array composed of
mm symbols and the text be an array composed of nn symbols. Find all occurrences
of the pattern as a subarray of the text.
The RabinKarp pattern matching algorithm that is based on string hashing can easily
be extended to two dimensions. In each row of the text we compute the fingerprint of each
substring of m symbols, this is done in O(nn ) time using the rolling property of the hash.
We create a new array of size n (n m + 1) that contains these fingerprints. Thus the
problem reduces to 1-dimensional pattern matching in the columns of the new array with
the pattern composed of the fingerprints of the rows of the original pattern (see Fig. 2).
This problem can be solved in linear time using the standard RabinKarp algorithm.
We obtain a linear time algorithm that appears simpler than, e.g., the known
Bird/Baker 2-dimensional pattern matching algorithm which generalizes the MorrisPratt
algorithm to the case of multiple patterns, see Crochemore and Rytter (2003).
Below we list four other examples of tasks in which string hashing can be applied to
obtain efficient solutions. The examples are ordered subjectively from the easiest to the
hardest task. For each task a short comment on its solution is provided.
Problem 5: Quasi-Cyclic-Rotations. We are given two strings s and t of the same length
n over English alphabet. We are to check if we can change exactly one letter in s so that
## Where to Use and How not to Use Polynomial String Hashing
97
Fig. 2. Reduction of 2-dimensional pattern matching to 1-dimensional pattern matching in columns via string
hashing (in this example p = 10, M > 1000)
## it becomes a cyclic rotation of t. This is a task from Algorithmic Engagements 2007:
http://main.edu.pl/en/archive/pa/2007/pra.
To check if s is an exact cyclic rotation of t, one could check if s occurs as a substring
of the string tt this is a pattern matching problem. For quasi-cyclic-rotations we can
modify this approach: for each position i = 1, 2, . . . , n in tt we need to check if the
substring u of length n starting at this position differs from s exactly at one position. For
this it suffices to find the longest common prefix and the longest common suffix of u and
s. The model solution computes these values in linear total time for all us using the PREF
table, also called the table of prefixes (see Section 3.2 of Crochemore and Rytter (2003)).
An alternative solution applies binary search for the longest commom prefix/suffix and
checks a candidate prefix/suffix using string hashing. This solution requires storing of
fingerprints for s and tt and works in O(n log n) time.
Problem 6: Antisymmetry. We are given a string t = t1 . . . tn over {0, 1} alphabet.
we denote the
For a substring u of t, by uR we denote the reversed string u and by u
negation of u obtained by changing all the zeroes to ones and ones to zeroes. A substring
u of t is called antisymmetric if u = u
R . We are to count the number of substrings of
t that are antisymmetric (if the same antisymmetric substring occurs multiple times, we
count each of its occurrences). This is a task from 17th Polish Olympiad in Informatics:
http://main.edu.pl/en/archive/oi/17/ant.
Antisymmetric strings resemble palindromic strings. Recall Manachers algorithm
that finds in linear time, for each position i, the radius R[i] of the longest even-length
palindromic substring centered at this position (see Section 8.1 of Crochemore and Rytter (2003)). The model solution is based on a modification of Manachers algorithm that
finds, for each position i, the radius R [i] of the longest antisymmetric substring centered
98
at this position. However, R [i] could alternatively be computed by applying binary search
and checking if a candidate radius is valid via string hashing. Here string fingerprints for
both t and tR need to be stored. The solution based on string hashing has O(n log n) time
complexity.
Problem 7: Prefixuffix. We are given a string t = t1 . . . tn over
bet. A prefixuffix of t is a pair (p, s) of a prefix and a suffix of t,
at most n/2, such that s is a cyclic rotation of p. The goal of the
the longest prefixuffix of t. This is a task from 19th Polish Olympiad
http://main.edu.pl/en/archive/oi/19/pre.
## English alphaeach of length
in Informatics:
The notion of a prefixuffix generalizes the notion of a border, which is a pair formed
by an equal prefix and suffix of t. The model solution for this task works in linear time and
is really tricky. Assume 2 | n and let t = xy, where x and y have the same length. Consider the word Q(x, y) (a crossover of x and y) defined as x1 yn/2 x2 yn/21 . . . xn/2 y1 .
Then the result is l/2, where l is the length of the longest prefix of Q(x, y) that is a concatenation of two even-length palindromes. The value of l can be found in linear time
using the output of Manachers algorithm (already mentioned in Problem 6).
This solution deserves a short explanation. Note that (p, s) is a prefixuffix if p = uv
and s = vu for some words u and v. Then t = uvwzvu for some words w and z of equal
length. Thus x = uvw, y = zvu and Q(x, y) = Q(u, u)Q(v, v)Q(w, z). Now it suffices
to note that Q(u, u) and Q(v, v) are palindromes. E.g., if t = ababbabbabbaab then
x = ababbab, y = babbaab and
Q(x, y) = abbaaabbbbaabb = abba aabbbbaa bb.
Here l = 12 which yields a prefixuffix of t of length 6: (ababba, abbaab).
An alternative solution using string hashing was much simpler to come up with. Recall
that we need to find a representation t = uvwzvu with uv as long as possible. To find u,
we consider all the borders of t, and to find v, we need to know the longest border of each
substring of t obtained by removing the same number of letters from the front and from
the back of t, that is, a[i] = border (ti ti+1 . . . tni tni+1 ). All the requested borders can
be found using string hashing in linear time. In particular, the latter ones, a[i], can be
computed for i = 1, 2, . . . , n/2 by observing that a[i] a[i 1] 2.
Problem 8: String Similarity. We consider a family of strings, S1 , . . . , Sn , each of
length l. We perform m operations, each of which consists in swapping a pair of letters
across some pair of the strings. The goal is to compute, for each i = 1, . . . , n, how many
strings among {Sj } were equal to Si at some moment in time, at maximum. This is a task
from 15th Polish Olympiad in Informatics: http://main.edu.pl/en/archive
/oi/15/poc.
Here the first idea that comes to mind is to use string hashing to identify which pairs
of strings are equal at respective moments in time. Note that fingerprints of the strings can
be updated easily when the letters are swapped. However, in the case of this task string
hashing is only the beginning of the solution. An efficient, O((nl + m) log (nl + m))
## Where to Use and How not to Use Polynomial String Hashing
99
time solution requires keeping track of groups of equal strings. Roughly speaking, for
each valid fingerprint we compute the number of strings with this fingerprint at each
moment in time and then for each single string we find the maximum size of a group of
equal strings it belongs to at any moment in time.
4. Final Remarks
In general, polynomial string hashing is a useful technique in construction of efficient
string algorithms. One simply needs to remember to carefully select the modulus M and
the variable of the polynomial p depending on the application. A good rule of thumb is to
pick both values as prime numbers with M as large as possible so that no integer overflow
occurs and p being at least the size of the alphabet.
There is a number of string processing problems in which hashing enables to present
solutions that are competitive with the ones obtained using non-randomized algorithms.
This includes pattern matching in one and multiple dimensions and searching for specific
patterns, which includes palindromic substrings, cyclic rotations and common substrings.
The major virtue of hashing is its O(n) time and space consumption. However, one should
always keep in mind that hashing is a heuristic algorithm.
References
Akhmedov, M. (2012). Anti-hash test. Codeforces Blog.
http://codeforces.com/blog/entry/4898.
Allouche, J.-P., Shallit, J. (1999). The ubiquitous Prouhet-Thue-Morse sequence. In: Ding, C., Helleseth, T.,
Niederreiter, H. (Eds.), Sequences and Their Applications, Proceedings SETA98, New York, SpringerVerlag, 116.
Crochemore, M., Hancart, C., Lecroq, T. (2007). Algorithms on Strings, Cambridge University Press.
Crochemore, M., Rytter, W. (2003). Jewels of Stringology, World Scientific.
Karp, R.M., Rabin, M.O. (1987). Efficient randomized pattern-matching algorithms. IBM Journal of Research
and Development Mathematics and Computing, 31(2), 249260.
Weisstein, E.W. Modulo multiplication group. MathWorld a Wolfram Web Resource.
http://mathworld.wolfram.com/ModuloMultiplicationGroup.html.
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The equation below is one of the more commonly understood definitions of the average:
Average = Sum/Count
where the sum is the result of adding all of the given numbers, and the count is the number of values being added. For example, given the 5 numbers, 2, 7, 19, 24, and 25, the average can be calculated as such:
Average = (2 + 7 + 19 + 24 + 25)/5
Average = 77/5
Average = 15.4
IT will Also Tell you
Average Value15.4 Arithmetic: 15.4 Geometric: 10.9801 Harmonic: 6.4337
Sum 77 Count 5 Median 19 Geometric Mean 10.9801 Largest 25 Smallest 2 Range 23
Singha
CEO / Co-Founder
A dynamic force in the world of artificial intelligence (AI) technology. Harbored a vision of AI technology that was accessible to everyone. This vision, fueled by a relentless pursuit of innovation, led to the birth of MGToL. The platform offers a suite of free, user-friendly AI tools and utilities, empowering individuals and businesses with capabilities that were once the domain of experts.
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# Q-The cast of 3 pens and copild is ₹110 represent this sitnation withe a linear equation in two variables also find the cost of one copy if one per cost ₹20 Q- In triangle DEF,A, B and C are mid points of sides EF,DF and DE respectively.If ar (triangleBAF)=24 cm 2means square ,find ar (AFBC) Q- prove that if two chords of a circle bisect each other then the two chords are diameter of the given circle Q-Draw a line segmen SR of length 10 cm divide it into 4 equal parts using compass and ruler Q-construct triangl ABC surch that BC=3.2cm angle B=45° and AC-AB=2.1cm. Q-AB and CD are two parallel chords of a circle such that lengths of AB and CD are 10cm and 24cm resputively.If the chord are on opposite sides of the center and the distance between then is17cm find the radius of the circle.
1
by NeevBava667
2015-04-01T19:03:50+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Price/Cost of a pen = Rs X
Price/Cost of a copy = Rs y
Total cost = 3 X + Y = Rs 110
a pen costs Rs 20. => X = Rs 20
So Y = 110 - 3 * 20 = Rs 50.
cost of a copy = Rs 50.
============================
see diagram.
When we join the mid points of the sides of the triangle DEF, we have the triangle DEF divided into 4 equal triangles. Thus the triangles ABC, BAF, AEC and BCD are all identical congruent triangles. So area of the triangle BAF = 24 cm² = 1/4 th of area of triangle DEF.
Area of parallelogram AFBC = 2 * area of triangle BAF = 48 cm²
==============================
see diagram.
AC and BD are chords bisecting each other at O.
Hence, AO = OC and BO = OD. Also, the angles AOB and COD are same (vertical angles at O). Hence, the two triangles AOB and COD are congruent.
Hence, AB = CD.
Similarly, in triangles AOD and BOC, the angles AOD = angle BOC (vertical angles). As, AO = OC and BO = OD, the two triangles are congruent.
Hence, AD = BC.
The two triangles ACB and ADB are congruent as all the three corresponding sides are equal to one another.
So angle D = angle B.
Similarly, angle C = angle A.
In a cyclic quadrilateral ABCD, the opposite angles are supplementary. Hence, angle B = angle D = 90°,
same way, angle C = angle A = 90°.
In a circle, if a chord AC (or BD) subtends an angle 90° at a point B (or A) on the circumference, then it is the diameter of the circle.
hence, AC and BD are diameters.
=============================================
Draw a line segment SR of length 10 cm horizontally using the ruler. Then using the compass, measure about 6 to 7 cm. With S as the center draw an arc above SR and an arc below SR. With R as the center and with the same radius, draw an arc above SR and an arc below SR , intersecting the earlier arcs drawn.
The points are intersection of arcs can be joined to get the perpendicular bisector of SR. Let it intersect SR at P. So P is the mid point of SR.
Follow the above procedure for the line segments SP and PR. Measure 3 to 4 cm with the compass. Draw arcs with centers S and P above and below SP. Join their points of intersection. Similarly, draw arcs with centers as P and R above and below PR. Join their intersection points.
Join/Draw the perpendicular bisectors for SP and PR. Now the line segment SR is divided into 4 equal parts.
============================================
construct ABC:
draw a line segment horizontally BC of length 3.2 cm.
Now draw a line BD (long one) making the angle CBD = 45°.
We need to find the point A such that AC = AB + 2.1 cm.
==================
see diagram.
AB || CD. AB and CD are chords of the circle. Let R be the radius of the circle.
AB = 10 cm CD = 24 cm. Center of circle is O. Draw perpendicular bisectors for AB and CD. They will join at center O. Given, EF = EO + OF = 17 cm.
we have, AE = EB = 5 cm. CF = FD = 12 cm.
Using Pythagoras theorem in triangle AEO, EO = √(R² - 5²)
Using Pythagoras theorem in triangle BOF, OF = √(R² - 12²)
Hence, EO + OF = √(R² - 25) + √(R² - 144) = 17
√(R² - 144) = 17 - √(R² - 25)
R² - 144 = 17² + (R² - 25) - 2 * 17 * √(R² - 25)
- 408 = - 2 * 17 * √(R² - 25)
12 = √(R² - 25)
R² = 144+25 = 169
R = 13 cm. = radius of the circle.
=========================
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## Thursday, May 5, 2011
### Odd Numbered Integers (1)
As Euler demonstrated it is possible to represent the result of the Riemann Zeta Function for all even values of s (2, 4, 6,...) in terms of an expression involving pi, i.e. (pi^s)/k where k is a rational number.
However - as is well known - a similar sort of expression cannot be provided for the corresponding odd values of s (3, 5, 7,....)
So, from my perspective the first task is to explain in qualitative terms why this situation arises!
From the psychological perspective each value of s represents a "higher" dimension of understanding (which traditionally has been associated with contemplative type development).
In this context a crucial distinction can be made as between the odd and even numbers which is very revealing.
Whereas the odd numbers represent a new level of (rational) differentiation, the even numbers represent, by contrast, new equilibrium states of (intuitive) integration.
Now once again, the conventional (1-dimensional) level is viewed as a means of (rational) differentiation of meaning. Though informally the importance of supporting intuition may be admitted, this is not formally included in interpretation. And as we have seen this leads to the inevitable reduction of qualitative to quantitative type analysis.
In general terms differentiation is based on the separation of poles in experience, whereas integration is based on their complementarity (and ultimate identity).
From another perspective there always remains a certain linear character to the differentiated aspect of understanding (even at higher levels) while the integral aspect is always characterised by complementarity.
As stated previously the qualitative structure of a dimension is inversely related to its corresponding root structure. So for example the two roots of unity are + 1 and - 1 respectively; in corresponding inverse qualitative fashion, 2-dimensional appreciation is characterised by the complementarity of opposite poles of form. So with the quantitative interpretation of roots we employ a (linear) either/or logic; with the corresponding qualitative interpretation of dimensions we employ a (circular) both/and logic.
Now when we look at the root structure for all even numbered roots, they are characterised by the complementarity of opposites where half of the roots can be balanced by the other half (that represents the negative of all roots in the first half).
Now the very nature of such complementary understanding is that it is perfectly circular in nature. However because customary mathematical understanding is based on linear type distinctions we can only attempt to convey the nature of such circular understanding (which ultimately is of a purely intuitive nature) in a reduced linear fashion.
So from a rational perspective the nature or 2-dimensional understanding entails the relationship as between what is circular and linear. In corresponding quantitative fashion the very nature of pi involves the relationship as between circle and line (i.e. as the ratio of circular circumference to line diameter).
Thus the qualitative nature of all even dimensions (as representing the refined rational linear manner of conveying circular meaning) is perfectly replicated in quantitative terms in the numerical expressions of the Zeta Function for even dimensional values that are positive.
When we consider the structure of odd number roots, perfect quantitative complementarity does not exist. This is evidenced by the fact that the odd numbered roots cannot be arranged in a fully complementary manner! In corresponding fashion, qualitative complementarity likewise does not exist. Therefore in both cases, they represent a situation of broken symmetry.
In fact with odd numbered roots, + 1 is always one root obtained that in a sense remains separated from all other roots where a complementary pattern does exist with respect to the imaginary parts of these roots. So the broken symmetry relates here to a distinction as between the behaviour of real and imaginary parts. Likewise in qualitative terms, at the higher odd numbered dimensions a degree of linear understanding is maintained thereby enabling consious differentiation of phenomena that is ultimately inconsistent with the (unconscious) holistic understanding of these dimensions. Thus the imbalance of conscious and unconscious remaining requires progression to the next (even) higher dimension where a new level of integration can be obtained.
So from a psychological perspective, again it is easy to suggest why such broken symmetry might exist.
In development each new dimensional stage of integration (where a temporary equilibrium is reached) follows a corresponding previous dimensional stage of differentiation (where such equilibrium is temporarily broken).
Therefore the (positive) odd-numbered dimensions represent a situation where a new configuration of both (refined) rational and intuitive understanding go hand in hand. In other words experience is literally somewhat uneven whereby the relationship as between both linear and circular type interpretation entails a degree of confusion. When such confusion is then unravelled one moves on to the next integral stage (represented by an even numbered dimension). And then the deepening of such integral awareness requires the further progression through ever "higher" alternating odd and even-numbered dimensions.
Now ultimately with very high numbered dimensions the differentiated rational element of understanding becomes so refined that it is no longer (explicitly) separable from the intuitive. So here both integral and differentiated understanding closely approximate (where the discrete becomes inseparable from the continuous).
Remarkably here we have the qualitative correspondent of e. As is well known both the differentiated and integrated expressions of e^x remain the same.
The same e also plays an enormous role with respect to the distribution of prime numbers. In corresponding qualitative fashion e represents the dynamic state where (integral) intuition cannot be distinguished from (differentiated) reason. This plays an enormous role with respect to the holistic mastery of primitive desires (so that their discrete involuntary nature can be controlled). And it is this same experience that characterises the qualitative appreciation of the holistic nature of prime numbers!
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Anda di halaman 1dari 20
# Cambridge International Examinations
## Cambridge International General Certificate of Secondary Education
* 1 8 4 3 4 6 9 5 1 2 *
0625/33
PHYSICS
Paper 3 Extended
October/November 2014
1 hour 15 minutes
## Candidates answer on the Question Paper.
Write your Centre number, candidate number and name on all the work you hand in.
Write in dark blue or black pen.
You may use a pencil for any diagrams or graphs.
Do not use staples, paper clips, glue or correction fluid.
DO NOT WRITE IN ANY BARCODES.
Electronic calculators may be used.
You may lose marks if you do not show your working or if you do not use appropriate units.
Take the weight of 1 kg to be 10 N (i.e. acceleration of free fall = 10 m / s2).
At the end of the examination, fasten all your work securely together.
The number of marks is given in brackets [ ] at the end of each question or part question.
The syllabus is approved for use in England, Wales and Northern Ireland as a Cambridge International Level 1/Level 2 Certificate.
## This document consists of 19 printed pages and 1 blank page.
DC (SJF/JG) 81773/4
UCLES 2014
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1
A free-fall parachutist jumps from a helium balloon, but does not open his parachute for some
time.
Fig. 1.1 shows the speed-time graph for his fall. Point B indicates when he opens his parachute.
speed
0
time
0
Fig. 1.1
(a) (i)
State the value of the gradient of the graph immediately after time t = 0.
(ii)
## Explain why the gradient has this value.
...........................................................................................................................................
.......................................................................................................................................[1]
(b) State how Fig. 1.1 shows that the acceleration decreased between time t = 0 and the time
to A.
...................................................................................................................................................
...............................................................................................................................................[1]
(c) Explain, in terms of forces, what is happening in section AB of the graph in Fig. 1.1.
...................................................................................................................................................
...................................................................................................................................................
...............................................................................................................................................[2]
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(d) A second parachutist of the same size and mass jumps from the balloon with a larger
parachute. He also opens his parachute at point B.
On Fig. 1.1, sketch a possible speed-time graph for his fall after he opens his parachute.
[3]
[Total: 8]
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2
Fig. 2.1 shows a uniform, rectangular slab of concrete ABCD standing upright on the ground. The
slab has height 0.60 m, width 0.30 m and mass 18 kg. A force of 40 N acts horizontally to the left
at B.
A
B
40 N
0.60 m
C
0.30 m
Fig. 2.1
(a) (i)
## Calculate the weight W of the concrete slab.
W = ........................................................ [1]
(ii)
## The thickness of the slab is 0.040 m.
Calculate the pressure exerted by the slab on the ground.
## pressure = ........................................................ [2]
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(b) (i)
(ii)
On Fig. 2.1, draw and label an arrow to show the weight W of the slab acting at its centre
of mass.
[1]
Calculate
1.
## the moment of the 40 N force about point D,
moment = ........................................................
2.
## the moment of W about point D.
moment = ........................................................
[3]
(iii)
## The ground is rough so that the slab does not slide.
State and explain what happens to the slab as the horizontal force at B is gradually
increased.
...........................................................................................................................................
...........................................................................................................................................
.......................................................................................................................................[2]
[Total: 9]
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3
Fig. 3.1 shows a long, plastic tube, sealed at both ends. The tube contains 0.15 kg of small metal
spheres.
## small metal spheres
Fig. 3.1
A physics teacher turns the tube upside down very quickly and the small metal spheres then fall
through 1.8 m and hit the bottom of the tube.
(a) Calculate
(i)
## decrease in gravitational potential energy = ........................................................ [2]
(ii)
the speed of the spheres as they hit the bottom of the tube.
## speed = ........................................................ [3]
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(b) The gravitational potential energy of the spheres is eventually transformed to thermal energy
in the metal spheres. The physics teacher explains that this procedure can be used to
determine the specific heat capacity of the metal.
(i)
State one other measurement that must be made in order for the specific heat capacity
of the metal to be determined.
...........................................................................................................................................
.......................................................................................................................................[1]
(ii)
Suggest a source of inaccuracy in determining the specific heat capacity using this
experiment.
...........................................................................................................................................
.......................................................................................................................................[1]
(iii)
The teacher turns the tube upside down and lets the spheres fall to the bottom 100 times
within a short period of time.
Explain why turning the tube upside down 100 times, instead of just once, produces a
more accurate value of the specific heat capacity.
...........................................................................................................................................
...........................................................................................................................................
.......................................................................................................................................[2]
[Total: 9]
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(a) Fig. 4.1 shows a syringe containing 100 cm3 of air at atmospheric pressure. Atmospheric
pressure is 1.0 105 Pa.
piston
Fig. 4.1
The open end of the syringe is sealed and the piston is pushed inwards until the air occupies
a volume of 40 cm3. The temperature of the air remains constant.
Calculate the new pressure of the air in the syringe.
## air pressure = ........................................................ [2]
(b) A syringe is used to transfer smokey air from above a flame to a small glass container.
Extremely small solid smoke particles are suspended in the air in the container.
The container is brightly illuminated from the side and viewed through a microscope.
(i)
The movement of the suspended smoke particles is called Brownian motion. Describe
this Brownian motion.
...........................................................................................................................................
...........................................................................................................................................
.......................................................................................................................................[2]
(ii)
## Explain what causes the motion of the smoke particles.
...........................................................................................................................................
...........................................................................................................................................
.......................................................................................................................................[2]
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(c) In the space below, sketch a diagram to represent the molecular structure of a solid. Show the
molecules as small circles of equal sizes.
[2]
[Total: 8]
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5
Light enters a glass fibre from air at an angle of incidence of 62 . The angle of refraction in the
glass is 36 .
(a) The speed of light in air is 3.0 108 m / s.
Determine the speed of light in the glass fibre.
## speed = ........................................................ [4]
(b) Describe how glass fibres are used in communications technology.
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...............................................................................................................................................[3]
[Total: 7]
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6
(i)
## metals are good conductors of electricity,
...........................................................................................................................................
...........................................................................................................................................
(ii)
## insulators do not conduct electricity.
...........................................................................................................................................
...........................................................................................................................................
[3]
(b) The battery of an electric car supplies a current of 96 A at 120 V to the motor which drives
the car.
(i)
State the useful energy change that takes place in the battery.
.......................................................................................................................................[1]
(ii)
(iii)
## The motor operates with an efficiency of 88 %.
Calculate the power output of the motor.
## power = ........................................................ [2]
[Total: 8]
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7
(a) Underline the most appropriate value below for the speed of sound in water.
1.5 m / s
15 m / s
150 m / s
1 500 m / s
[1]
15 000 m / s
## (b) Sound travels in water as a series of compressions and rarefactions.
Describe what is meant by a compression and by a rarefaction.
compression .............................................................................................................................
...................................................................................................................................................
rarefaction .................................................................................................................................
...................................................................................................................................................
[2]
(c) An echo-sounder sends out a pulse of sound to determine the depth of the sea bed. It
measures the time between sending out the pulse and receiving its echo.
Fig. 7.1 shows a boat using an echo-sounder.
echo-sounder
12 m
echo
pulse of sound
sea bed
Fig. 7.1
The sea bed is 12 m below the echo-sounder.
(i)
Use your value for the speed of sound in water from (a) to calculate the time between the
sending out of the pulse and receiving its echo.
## time = ........................................................ [3]
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(ii)
The boat passes over a region of the sea bed of the same depth, where the reflection of
sound waves is weaker.
State whether there is an increase, a decrease or no change in the amplitude and pitch
of the reflected wave.
amplitude ...........................................................................................................................
pitch ...................................................................................................................................
[2]
[Total: 8]
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8
A student sets up a circuit containing three identical cells. Each cell has an e.m.f. (electromotive
force) of 2.0 V.
Fig. 8.1 shows the cells in series with a length of uniform metal wire connected between two
terminals K and L, an ammeter and a resistor X.
uniform
metal
wire
K
A
L
X
Fig. 8.1
(a) State the total e.m.f. of the three cells in series.
## total e.m.f. = ........................................................ [1]
(b) The ammeter reading is 0.25 A.
(i)
## State the name of the unit in which electric charge is measured.
.......................................................................................................................................[1]
(ii)
Calculate the charge that flows through the circuit in twelve minutes.
(iii)
## The metal wire has a resistance of 16 .
Calculate the resistance of resistor X.
## resistance = ........................................................ [2]
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(c) The student removes the 16 wire from the circuit and cuts it into two equal lengths.
He then connects the two lengths in parallel between K and L, as shown in Fig. 8.2.
uniform
metal
wire
A
L
X
Fig. 8.2
Calculate the resistance of the two lengths of wire in parallel.
## resistance = ........................................................ [3]
[Total: 9]
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9
A circuit contains a battery, a variable resistor and a solenoid. Fig. 9.1 shows the magnetic field
pattern produced by the current in the solenoid.
solenoid
## magnetic field line
Fig. 9.1
(a) (i)
State how the magnetic field pattern indicates regions where the magnetic field is
stronger.
...........................................................................................................................................
.......................................................................................................................................[1]
(ii)
State what happens to the magnetic field when the current in the circuit is reversed.
...........................................................................................................................................
.......................................................................................................................................[1]
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(b) A second solenoid is placed next to the first solenoid.
Fig. 9.2 shows the second solenoid connected to a very sensitive ammeter.
very
sensitive
ammeter
A
second solenoid
Fig. 9.2
(i)
## The variable resistor is adjusted so that its resistance changes quickly.
State and explain what is seen to happen in the circuit of the second solenoid.
...........................................................................................................................................
...........................................................................................................................................
...........................................................................................................................................
...........................................................................................................................................
.......................................................................................................................................[3]
(ii)
## The variable resistor is adjusted much more slowly than in (i).
State and explain the difference in what is seen to happen in the circuit of the second
solenoid.
...........................................................................................................................................
...........................................................................................................................................
.......................................................................................................................................[2]
[Total: 7]
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10 A technician sets up a radiation detector in a university laboratory, for use in some experiments.
Even before the radioactive source for the experiments is brought into the laboratory, the detector
registers a small count rate due to background radiation.
(a) Suggest one source of this background radiation.
...................................................................................................................................................
...............................................................................................................................................[1]
(b) The radioactive source emits -rays. It is placed on the laboratory bench close to the detector.
(i)
## State what -rays are.
...........................................................................................................................................
...........................................................................................................................................
.......................................................................................................................................[2]
(ii)
A lead sheet of thickness 10 mm is positioned between the detector and the radioactive
source.
State and explain what happens to the count rate on the detector.
...........................................................................................................................................
...........................................................................................................................................
.......................................................................................................................................[2]
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(c) In a second experiment, -rays pass through air to the detector, as shown in Fig. 10.1.
a-rays
detector
source
Fig. 10.1
One end of a bar magnet is brought close to the path of the -rays.
(i)
Tick one box to indicate the effect on the path of the -rays.
[1]
## deflected into the page
deflected out of the page
deflected downwards
deflected upwards
no deflection
(ii)
...........................................................................................................................................
.......................................................................................................................................[1]
[Total: 7]
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BLANK PAGE
Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every
reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the
publisher will be pleased to make amends at the earliest possible opportunity.
Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local
Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.
UCLES 2014
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TIBCO Software Inc (NAS:TIBX)
Net Cash
\$-3.12 (As of Aug. 2014)
Net cash per share is calculated as Cash and Cash Equivalents minus Total Liabilities and then divided by Shares Outstanding. TIBCO Software Inc's net cash per share for the quarter that ended in Aug. 2014 was \$-3.12.
Definition
In the calculation of a companys net cash, assets other than cash and short term investments are considered to be worth nothing. But the company has to pay its debt and other liabilities in full. This is an extremely conservative way of valuation. Most companies have negative net cash. But sometimes a companys price may be lower than its net-cash.
TIBCO Software Inc's Net Cash Per Share for the fiscal year that ended in Nov. 2013 is calculated as
Net Cash Per Share (A: Nov. 2013 ) = (Cash and Cash Equivalents - Total Liabilities) / Shares Outstanding = (745.951 - 1053.896) / 163.17 = -1.89
TIBCO Software Inc's Net Cash Per Share for the quarter that ended in Aug. 2014 is calculated as
Net Cash Per Share (Q: Aug. 2014 ) = (Cash and Cash Equivalents - Total Liabilities) / Shares Outstanding = (532.454 - 1043.05) / 163.90 = -3.12
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Explanation
Ben Graham invested in situations where the companys stock price was lower than its net-cash. He assigned some value to the companys other current asset. The value is called Net Current Asset Value (NCAV). One research study, covering the years 1970 through 1983 showed that portfolios picked at the beginning of each year, and held for one year, returned 29.4 percent, on average, over the 13-year period, compared to 11.5 percent for the S&P 500 Index. Other studies of Grahams strategy produced similar results.
You can find companies that are traded below their Net Current Asset Value (NCAV) with our Net-Net screener. GuruFocus also publishes a monthly Net-Net newsletter.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
TIBCO Software Inc Annual Data
Nov04 Nov05 Nov06 Nov07 Nov08 Nov09 Nov10 Nov11 Nov12 Nov13 netcash 1.00 1.08 1.23 -0.41 -0.44 -0.46 -0.92 -1.06 -1.71 -1.89
TIBCO Software Inc Quarterly Data
May12 Aug12 Nov12 Feb13 May13 Aug13 Nov13 Feb14 May14 Aug14 netcash -2.37 -2.50 -1.71 -1.21 -1.62 -1.82 -1.89 -1.74 -2.92 -3.12
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19 followers -
19 followers
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## Savings Plan and Future Cost Estimation
Estimated Annual Cost in 2032 \$110,674 Monthly savings required \$875
How much will it cost to send your child to college in 18 years and how much do you need to save? A 4 year degree is estimated to be priced at \$442,697.85 for students enrolling in 2032 if tuition increases average 7% per year until then. Assuming you have no current college savings, monthly deposits of \$874.69 into a 529 or other college savings plan earning and after tax or tax exempt return of 7% will be necessary to achieve this balance.
## Assumptions
Current Annual Cost \$29,500.00 Years until enrollment 18 Annual Tuition Increase 7% Current Savings \$0.00 After Tax Return on Savings 7% Years of enrollment 4
## Current Savings Shortfall
### College Savings Calculator
Discover how much will you need to start saving now to afford college in the future.
## College Savings Schedule
Month Begin Deposit Withdraw Interest End
1 \$0.00 \$874.69 \$0.00 \$0.00 \$874.69
2 \$874.69 \$874.69 \$0.00 \$4.95 \$1,754.33
3 \$1,754.33 \$874.69 \$0.00 \$9.92 \$2,638.94
4 \$2,638.94 \$874.69 \$0.00 \$14.92 \$3,528.55
5 \$3,528.55 \$874.69 \$0.00 \$19.95 \$4,423.19
6 \$4,423.19 \$874.69 \$0.00 \$25.01 \$5,322.89
7 \$5,322.89 \$874.69 \$0.00 \$30.10 \$6,227.68
8 \$6,227.68 \$874.69 \$0.00 \$35.21 \$7,137.58
9 \$7,137.58 \$874.69 \$0.00 \$40.36 \$8,052.63
10 \$8,052.63 \$874.69 \$0.00 \$45.53 \$8,972.85
11 \$8,972.85 \$874.69 \$0.00 \$50.73 \$9,898.27
12 \$9,898.27 \$874.69 \$0.00 \$55.97 \$10,828.93
13 \$10,828.93 \$874.69 \$0.00 \$61.23 \$11,764.85
14 \$11,764.85 \$874.69 \$0.00 \$66.52 \$12,706.06
15 \$12,706.06 \$874.69 \$0.00 \$71.84 \$13,652.59
16 \$13,652.59 \$874.69 \$0.00 \$77.19 \$14,604.47
17 \$14,604.47 \$874.69 \$0.00 \$82.58 \$15,561.74
18 \$15,561.74 \$874.69 \$0.00 \$87.99 \$16,524.42
19 \$16,524.42 \$874.69 \$0.00 \$93.43 \$17,492.54
20 \$17,492.54 \$874.69 \$0.00 \$98.91 \$18,466.14
21 \$18,466.14 \$874.69 \$0.00 \$104.41 \$19,445.24
22 \$19,445.24 \$874.69 \$0.00 \$109.95 \$20,429.88
23 \$20,429.88 \$874.69 \$0.00 \$115.51 \$21,420.08
24 \$21,420.08 \$874.69 \$0.00 \$121.11 \$22,415.88
25 \$22,415.88 \$874.69 \$0.00 \$126.74 \$23,417.31
26 \$23,417.31 \$874.69 \$0.00 \$132.40 \$24,424.40
27 \$24,424.40 \$874.69 \$0.00 \$138.10 \$25,437.19
28 \$25,437.19 \$874.69 \$0.00 \$143.83 \$26,455.71
29 \$26,455.71 \$874.69 \$0.00 \$149.58 \$27,479.98
30 \$27,479.98 \$874.69 \$0.00 \$155.38 \$28,510.05
31 \$28,510.05 \$874.69 \$0.00 \$161.20 \$29,545.94
32 \$29,545.94 \$874.69 \$0.00 \$167.06 \$30,587.69
33 \$30,587.69 \$874.69 \$0.00 \$172.95 \$31,635.33
34 \$31,635.33 \$874.69 \$0.00 \$178.87 \$32,688.89
35 \$32,688.89 \$874.69 \$0.00 \$184.83 \$33,748.41
36 \$33,748.41 \$874.69 \$0.00 \$190.82 \$34,813.92
37 \$34,813.92 \$874.69 \$0.00 \$196.84 \$35,885.45
38 \$35,885.45 \$874.69 \$0.00 \$202.90 \$36,963.04
39 \$36,963.04 \$874.69 \$0.00 \$208.99 \$38,046.72
40 \$38,046.72 \$874.69 \$0.00 \$215.12 \$39,136.53
41 \$39,136.53 \$874.69 \$0.00 \$221.28 \$40,232.50
42 \$40,232.50 \$874.69 \$0.00 \$227.48 \$41,334.67
43 \$41,334.67 \$874.69 \$0.00 \$233.71 \$42,443.07
44 \$42,443.07 \$874.69 \$0.00 \$239.98 \$43,557.74
45 \$43,557.74 \$874.69 \$0.00 \$246.28 \$44,678.71
46 \$44,678.71 \$874.69 \$0.00 \$252.62 \$45,806.02
47 \$45,806.02 \$874.69 \$0.00 \$258.99 \$46,939.70
48 \$46,939.70 \$874.69 \$0.00 \$265.40 \$48,079.79
49 \$48,079.79 \$874.69 \$0.00 \$271.85 \$49,226.33
50 \$49,226.33 \$874.69 \$0.00 \$278.33 \$50,379.35
51 \$50,379.35 \$874.69 \$0.00 \$284.85 \$51,538.89
52 \$51,538.89 \$874.69 \$0.00 \$291.41 \$52,704.99
53 \$52,704.99 \$874.69 \$0.00 \$298.00 \$53,877.68
54 \$53,877.68 \$874.69 \$0.00 \$304.63 \$55,057.00
55 \$55,057.00 \$874.69 \$0.00 \$311.30 \$56,242.99
56 \$56,242.99 \$874.69 \$0.00 \$318.01 \$57,435.69
57 \$57,435.69 \$874.69 \$0.00 \$324.75 \$58,635.13
58 \$58,635.13 \$874.69 \$0.00 \$331.53 \$59,841.35
59 \$59,841.35 \$874.69 \$0.00 \$338.35 \$61,054.39
60 \$61,054.39 \$874.69 \$0.00 \$345.21 \$62,274.29
61 \$62,274.29 \$874.69 \$0.00 \$352.11 \$63,501.09
62 \$63,501.09 \$874.69 \$0.00 \$359.04 \$64,734.82
63 \$64,734.82 \$874.69 \$0.00 \$366.02 \$65,975.53
64 \$65,975.53 \$874.69 \$0.00 \$373.04 \$67,223.26
65 \$67,223.26 \$874.69 \$0.00 \$380.09 \$68,478.04
66 \$68,478.04 \$874.69 \$0.00 \$387.18 \$69,739.91
67 \$69,739.91 \$874.69 \$0.00 \$394.32 \$71,008.92
68 \$71,008.92 \$874.69 \$0.00 \$401.49 \$72,285.10
69 \$72,285.10 \$874.69 \$0.00 \$408.71 \$73,568.50
70 \$73,568.50 \$874.69 \$0.00 \$415.97 \$74,859.16
71 \$74,859.16 \$874.69 \$0.00 \$423.26 \$76,157.11
72 \$76,157.11 \$874.69 \$0.00 \$430.60 \$77,462.40
73 \$77,462.40 \$874.69 \$0.00 \$437.98 \$78,775.07
74 \$78,775.07 \$874.69 \$0.00 \$445.41 \$80,095.17
75 \$80,095.17 \$874.69 \$0.00 \$452.87 \$81,422.73
76 \$81,422.73 \$874.69 \$0.00 \$460.38 \$82,757.80
77 \$82,757.80 \$874.69 \$0.00 \$467.92 \$84,100.41
78 \$84,100.41 \$874.69 \$0.00 \$475.52 \$85,450.62
79 \$85,450.62 \$874.69 \$0.00 \$483.15 \$86,808.46
80 \$86,808.46 \$874.69 \$0.00 \$490.83 \$88,173.98
81 \$88,173.98 \$874.69 \$0.00 \$498.55 \$89,547.22
82 \$89,547.22 \$874.69 \$0.00 \$506.31 \$90,928.22
83 \$90,928.22 \$874.69 \$0.00 \$514.12 \$92,317.03
84 \$92,317.03 \$874.69 \$0.00 \$521.97 \$93,713.69
85 \$93,713.69 \$874.69 \$0.00 \$529.87 \$95,118.25
86 \$95,118.25 \$874.69 \$0.00 \$537.81 \$96,530.75
87 \$96,530.75 \$874.69 \$0.00 \$545.80 \$97,951.24
88 \$97,951.24 \$874.69 \$0.00 \$553.83 \$99,379.76
89 \$99,379.76 \$874.69 \$0.00 \$561.91 \$100,816.36
90 \$100,816.36 \$874.69 \$0.00 \$570.03 \$102,261.08
91 \$102,261.08 \$874.69 \$0.00 \$578.20 \$103,713.97
92 \$103,713.97 \$874.69 \$0.00 \$586.41 \$105,175.07
93 \$105,175.07 \$874.69 \$0.00 \$594.68 \$106,644.44
94 \$106,644.44 \$874.69 \$0.00 \$602.98 \$108,122.11
95 \$108,122.11 \$874.69 \$0.00 \$611.34 \$109,608.14
96 \$109,608.14 \$874.69 \$0.00 \$619.74 \$111,102.57
97 \$111,102.57 \$874.69 \$0.00 \$628.19 \$112,605.45
98 \$112,605.45 \$874.69 \$0.00 \$636.69 \$114,116.83
99 \$114,116.83 \$874.69 \$0.00 \$645.23 \$115,636.75
100 \$115,636.75 \$874.69 \$0.00 \$653.83 \$117,165.27
101 \$117,165.27 \$874.69 \$0.00 \$662.47 \$118,702.43
102 \$118,702.43 \$874.69 \$0.00 \$671.16 \$120,248.28
103 \$120,248.28 \$874.69 \$0.00 \$679.90 \$121,802.87
104 \$121,802.87 \$874.69 \$0.00 \$688.69 \$123,366.25
105 \$123,366.25 \$874.69 \$0.00 \$697.53 \$124,938.47
106 \$124,938.47 \$874.69 \$0.00 \$706.42 \$126,519.58
107 \$126,519.58 \$874.69 \$0.00 \$715.36 \$128,109.63
108 \$128,109.63 \$874.69 \$0.00 \$724.35 \$129,708.67
109 \$129,708.67 \$874.69 \$0.00 \$733.39 \$131,316.75
110 \$131,316.75 \$874.69 \$0.00 \$742.48 \$132,933.92
111 \$132,933.92 \$874.69 \$0.00 \$751.63 \$134,560.24
112 \$134,560.24 \$874.69 \$0.00 \$760.82 \$136,195.75
113 \$136,195.75 \$874.69 \$0.00 \$770.07 \$137,840.51
114 \$137,840.51 \$874.69 \$0.00 \$779.37 \$139,494.57
115 \$139,494.57 \$874.69 \$0.00 \$788.72 \$141,157.98
116 \$141,157.98 \$874.69 \$0.00 \$798.13 \$142,830.80
117 \$142,830.80 \$874.69 \$0.00 \$807.59 \$144,513.08
118 \$144,513.08 \$874.69 \$0.00 \$817.10 \$146,204.87
119 \$146,204.87 \$874.69 \$0.00 \$826.66 \$147,906.22
120 \$147,906.22 \$874.69 \$0.00 \$836.28 \$149,617.19
121 \$149,617.19 \$874.69 \$0.00 \$845.96 \$151,337.84
122 \$151,337.84 \$874.69 \$0.00 \$855.69 \$153,068.22
123 \$153,068.22 \$874.69 \$0.00 \$865.47 \$154,808.38
124 \$154,808.38 \$874.69 \$0.00 \$875.31 \$156,558.38
125 \$156,558.38 \$874.69 \$0.00 \$885.20 \$158,318.27
126 \$158,318.27 \$874.69 \$0.00 \$895.15 \$160,088.11
127 \$160,088.11 \$874.69 \$0.00 \$905.16 \$161,867.96
128 \$161,867.96 \$874.69 \$0.00 \$915.22 \$163,657.87
129 \$163,657.87 \$874.69 \$0.00 \$925.35 \$165,457.91
130 \$165,457.91 \$874.69 \$0.00 \$935.52 \$167,268.12
131 \$167,268.12 \$874.69 \$0.00 \$945.76 \$169,088.57
132 \$169,088.57 \$874.69 \$0.00 \$956.05 \$170,919.31
133 \$170,919.31 \$874.69 \$0.00 \$966.40 \$172,760.40
134 \$172,760.40 \$874.69 \$0.00 \$976.81 \$174,611.90
135 \$174,611.90 \$874.69 \$0.00 \$987.28 \$176,473.87
136 \$176,473.87 \$874.69 \$0.00 \$997.81 \$178,346.37
137 \$178,346.37 \$874.69 \$0.00 \$1,008.40 \$180,229.46
138 \$180,229.46 \$874.69 \$0.00 \$1,019.04 \$182,123.19
139 \$182,123.19 \$874.69 \$0.00 \$1,029.75 \$184,027.63
140 \$184,027.63 \$874.69 \$0.00 \$1,040.52 \$185,942.84
141 \$185,942.84 \$874.69 \$0.00 \$1,051.35 \$187,868.88
142 \$187,868.88 \$874.69 \$0.00 \$1,062.24 \$189,805.81
143 \$189,805.81 \$874.69 \$0.00 \$1,073.19 \$191,753.69
144 \$191,753.69 \$874.69 \$0.00 \$1,084.20 \$193,712.58
145 \$193,712.58 \$874.69 \$0.00 \$1,095.28 \$195,682.55
146 \$195,682.55 \$874.69 \$0.00 \$1,106.42 \$197,663.66
147 \$197,663.66 \$874.69 \$0.00 \$1,117.62 \$199,655.97
148 \$199,655.97 \$874.69 \$0.00 \$1,128.88 \$201,659.54
149 \$201,659.54 \$874.69 \$0.00 \$1,140.21 \$203,674.44
150 \$203,674.44 \$874.69 \$0.00 \$1,151.60 \$205,700.73
151 \$205,700.73 \$874.69 \$0.00 \$1,163.06 \$207,738.48
152 \$207,738.48 \$874.69 \$0.00 \$1,174.58 \$209,787.75
153 \$209,787.75 \$874.69 \$0.00 \$1,186.17 \$211,848.61
154 \$211,848.61 \$874.69 \$0.00 \$1,197.82 \$213,921.12
155 \$213,921.12 \$874.69 \$0.00 \$1,209.54 \$216,005.35
156 \$216,005.35 \$874.69 \$0.00 \$1,221.33 \$218,101.37
157 \$218,101.37 \$874.69 \$0.00 \$1,233.18 \$220,209.24
158 \$220,209.24 \$874.69 \$0.00 \$1,245.10 \$222,329.03
159 \$222,329.03 \$874.69 \$0.00 \$1,257.08 \$224,460.80
160 \$224,460.80 \$874.69 \$0.00 \$1,269.13 \$226,604.62
161 \$226,604.62 \$874.69 \$0.00 \$1,281.26 \$228,760.57
162 \$228,760.57 \$874.69 \$0.00 \$1,293.45 \$230,928.71
163 \$230,928.71 \$874.69 \$0.00 \$1,305.70 \$233,109.10
164 \$233,109.10 \$874.69 \$0.00 \$1,318.03 \$235,301.82
165 \$235,301.82 \$874.69 \$0.00 \$1,330.43 \$237,506.94
166 \$237,506.94 \$874.69 \$0.00 \$1,342.90 \$239,724.53
167 \$239,724.53 \$874.69 \$0.00 \$1,355.44 \$241,954.66
168 \$241,954.66 \$874.69 \$0.00 \$1,368.05 \$244,197.40
169 \$244,197.40 \$874.69 \$0.00 \$1,380.73 \$246,452.82
170 \$246,452.82 \$874.69 \$0.00 \$1,393.48 \$248,720.99
171 \$248,720.99 \$874.69 \$0.00 \$1,406.30 \$251,001.98
172 \$251,001.98 \$874.69 \$0.00 \$1,419.20 \$253,295.87
173 \$253,295.87 \$874.69 \$0.00 \$1,432.17 \$255,602.73
174 \$255,602.73 \$874.69 \$0.00 \$1,445.21 \$257,922.63
175 \$257,922.63 \$874.69 \$0.00 \$1,458.33 \$260,255.65
176 \$260,255.65 \$874.69 \$0.00 \$1,471.52 \$262,601.86
177 \$262,601.86 \$874.69 \$0.00 \$1,484.79 \$264,961.34
178 \$264,961.34 \$874.69 \$0.00 \$1,498.13 \$267,334.16
179 \$267,334.16 \$874.69 \$0.00 \$1,511.55 \$269,720.40
180 \$269,720.40 \$874.69 \$0.00 \$1,525.04 \$272,120.13
181 \$272,120.13 \$874.69 \$0.00 \$1,538.61 \$274,533.43
182 \$274,533.43 \$874.69 \$0.00 \$1,552.25 \$276,960.37
183 \$276,960.37 \$874.69 \$0.00 \$1,565.97 \$279,401.03
184 \$279,401.03 \$874.69 \$0.00 \$1,579.77 \$281,855.49
185 \$281,855.49 \$874.69 \$0.00 \$1,593.65 \$284,323.83
186 \$284,323.83 \$874.69 \$0.00 \$1,607.61 \$286,806.13
187 \$286,806.13 \$874.69 \$0.00 \$1,621.64 \$289,302.46
188 \$289,302.46 \$874.69 \$0.00 \$1,635.76 \$291,812.91
189 \$291,812.91 \$874.69 \$0.00 \$1,649.95 \$294,337.55
190 \$294,337.55 \$874.69 \$0.00 \$1,664.23 \$296,876.47
191 \$296,876.47 \$874.69 \$0.00 \$1,678.58 \$299,429.74
192 \$299,429.74 \$874.69 \$0.00 \$1,693.02 \$301,997.45
193 \$301,997.45 \$874.69 \$0.00 \$1,707.54 \$304,579.68
194 \$304,579.68 \$874.69 \$0.00 \$1,722.14 \$307,176.51
195 \$307,176.51 \$874.69 \$0.00 \$1,736.82 \$309,788.02
196 \$309,788.02 \$874.69 \$0.00 \$1,751.59 \$312,414.30
197 \$312,414.30 \$874.69 \$0.00 \$1,766.44 \$315,055.43
198 \$315,055.43 \$874.69 \$0.00 \$1,781.37 \$317,711.49
199 \$317,711.49 \$874.69 \$0.00 \$1,796.39 \$320,382.57
200 \$320,382.57 \$874.69 \$0.00 \$1,811.49 \$323,068.75
201 \$323,068.75 \$874.69 \$0.00 \$1,826.68 \$325,770.12
202 \$325,770.12 \$874.69 \$0.00 \$1,841.95 \$328,486.76
203 \$328,486.76 \$874.69 \$0.00 \$1,857.31 \$331,218.76
204 \$331,218.76 \$874.69 \$0.00 \$1,872.76 \$333,966.21
205 \$333,966.21 \$874.69 \$0.00 \$1,888.29 \$336,729.19
206 \$336,729.19 \$874.69 \$0.00 \$1,903.92 \$339,507.80
207 \$339,507.80 \$874.69 \$0.00 \$1,919.63 \$342,302.12
208 \$342,302.12 \$874.69 \$0.00 \$1,935.43 \$345,112.24
209 \$345,112.24 \$874.69 \$0.00 \$1,951.31 \$347,938.24
210 \$347,938.24 \$874.69 \$0.00 \$1,967.29 \$350,780.22
211 \$350,780.22 \$874.69 \$0.00 \$1,983.36 \$353,638.27
212 \$353,638.27 \$874.69 \$0.00 \$1,999.52 \$356,512.48
213 \$356,512.48 \$874.69 \$0.00 \$2,015.77 \$359,402.94
214 \$359,402.94 \$874.69 \$0.00 \$2,032.12 \$362,309.75
215 \$362,309.75 \$874.69 \$0.00 \$2,048.55 \$365,232.99
216 \$365,232.99 \$874.69 \$0.00 \$2,065.08 \$368,172.76
217 \$368,172.76 \$874.69 \$99,708.00 \$2,081.70 \$271,421.15
218 \$271,421.15 \$874.69 \$0.00 \$1,534.65 \$273,830.49
219 \$273,830.49 \$874.69 \$0.00 \$1,548.28 \$276,253.46
220 \$276,253.46 \$874.69 \$0.00 \$1,561.98 \$278,690.13
221 \$278,690.13 \$874.69 \$0.00 \$1,575.75 \$281,140.57
222 \$281,140.57 \$874.69 \$0.00 \$1,589.61 \$283,604.87
223 \$283,604.87 \$874.69 \$0.00 \$1,603.54 \$286,083.10
224 \$286,083.10 \$874.69 \$0.00 \$1,617.56 \$288,575.35
225 \$288,575.35 \$874.69 \$0.00 \$1,631.65 \$291,081.69
226 \$291,081.69 \$874.69 \$0.00 \$1,645.82 \$293,602.20
227 \$293,602.20 \$874.69 \$0.00 \$1,660.07 \$296,136.96
228 \$296,136.96 \$874.69 \$0.00 \$1,674.40 \$298,686.05
229 \$298,686.05 \$874.69 \$106,687.56 \$1,688.81 \$194,561.99
230 \$194,561.99 \$874.69 \$0.00 \$1,100.08 \$196,536.76
231 \$196,536.76 \$874.69 \$0.00 \$1,111.25 \$198,522.70
232 \$198,522.70 \$874.69 \$0.00 \$1,122.48 \$200,519.87
233 \$200,519.87 \$874.69 \$0.00 \$1,133.77 \$202,528.33
234 \$202,528.33 \$874.69 \$0.00 \$1,145.12 \$204,548.14
235 \$204,548.14 \$874.69 \$0.00 \$1,156.54 \$206,579.37
236 \$206,579.37 \$874.69 \$0.00 \$1,168.03 \$208,622.09
237 \$208,622.09 \$874.69 \$0.00 \$1,179.58 \$210,676.36
238 \$210,676.36 \$874.69 \$0.00 \$1,191.19 \$212,742.24
239 \$212,742.24 \$874.69 \$0.00 \$1,202.88 \$214,819.81
240 \$214,819.81 \$874.69 \$0.00 \$1,214.62 \$216,909.12
241 \$216,909.12 \$874.69 \$114,155.69 \$1,226.44 \$104,854.56
242 \$104,854.56 \$874.69 \$0.00 \$592.86 \$106,322.11
243 \$106,322.11 \$874.69 \$0.00 \$601.16 \$107,797.96
244 \$107,797.96 \$874.69 \$0.00 \$609.51 \$109,282.16
245 \$109,282.16 \$874.69 \$0.00 \$617.90 \$110,774.75
246 \$110,774.75 \$874.69 \$0.00 \$626.34 \$112,275.78
247 \$112,275.78 \$874.69 \$0.00 \$634.82 \$113,785.29
248 \$113,785.29 \$874.69 \$0.00 \$643.36 \$115,303.34
249 \$115,303.34 \$874.69 \$0.00 \$651.94 \$116,829.97
250 \$116,829.97 \$874.69 \$0.00 \$660.57 \$118,365.23
251 \$118,365.23 \$874.69 \$0.00 \$669.25 \$119,909.17
252 \$119,909.17 \$874.69 \$0.00 \$677.98 \$121,461.84
253 \$121,461.84 \$0.00 \$122,146.59 \$686.76 \$2.01
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Home / Mathematics Mcqs / A Car Owner Buys Petrol At Rs. 7.50, Rs. 8 And Rs. 8.50 Per Liter For Three Successive Years. What Approximately Is The Average Cost Per Liter Of Petrol If He Spends Rs. 4000 Each Year?
# A Car Owner Buys Petrol At Rs. 7.50, Rs. 8 And Rs. 8.50 Per Liter For Three Successive Years. What Approximately Is The Average Cost Per Liter Of Petrol If He Spends Rs. 4000 Each Year?
Question: A Car Owner Buys Petrol At Rs. 7.50, Rs. 8 And Rs. 8.50 Per Liter For Three Successive Years. What Approximately Is The Average Cost Per Liter Of Petrol If He Spends Rs. 4000 Each Year?
a. Rs. 7.98
b. Rs. 10.50
c. Rs. 12.50
d. Rs. 15
[efaccordion id=”01″] [efitems title=”Show Answer” text=”Rs. 10.50″][/efaccordion]
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12:35 AM
@CarlLange Haha, yeah I didn't know what to expect, but worth the cut-and-paste
1:22 AM
I am trying to take intersections of several three-dimensional regions by using regionplot3d command.
Currently thinking how to use looping to take intersections for several 3-D regions.
RegionPlot3D[
x - 2*(A/8)^(i) - p^(i)*(y + z) < 0 &&
x + (A/8)^(i) - p^(i)*(y + z) > 0 &&
x - 2*(A/8)^(i) - p^(i)*(y + z) < 0 &&
x + (A/8)^(i) - p^(i)*(y + z) > 0, {x, -2, 2}, {y, -2, 2}, {z, -2,
2}, Mesh -> None, PlotPoints -> 100]
The region after taking i = 4,5,6,...,20
any idea of looping?
A =3, p =0.5
7 hours later…
7:58 AM
If you ever wanted to get something similar to Matlab's plot interactivity in Mathematica, there's now `ResourceFunction["InteractiveGraphics"]`, with support for zooming, dragging, and data-tip placement:
6
2 hours later…
9:32 AM
@zhk, one thing I forgot to mention: you can simplify the NeumannValues quite a bit by using, for example, NeumannValue[ha*(Tp[t, x, y, z] - Tinf), y == 0 || y==Lap || ...] and the DirichletCondition[{Tp[t, x, y, z] == Tc, Tn[t, x, y, z] == Tc}, x == 0] etc.
5 hours later…
2:47 PM
Quick Dynamic question. It's been a couple of years since I've used more than it's most basic use cases, and I've been deep in the Typescript/React world for UI this year. If I want like a 50x50 grid of squares and clicking on one of them changes the color, can I do that without redrawing the full set of squares? Do I need to make each square depend on a separate symbol instead of the shared array?
I can of course do it with the full redraw, but it takes about a second to update.
Maybe if I switch to ArrayPlot or something instead of Graphics and Rectangles.
3:08 PM
It's interesting to compare React to Dynamic. I remember someone doing a brief comparison to Angular I think. You have to manually specify which data the UI depends on as a whole in React, and it recalculates those dependencies with each change, but it is smart about the redrawing smaller sections. If the recalculated dependency has the same value for a particular section of the UI, it doesn't redraw that section (update the HTML). So that's the part where it is a bit smart/fast/automatic.
The "hooks" syntax is pretty short and clean. Pretty productive and flexible system.
But I don't think people use it for 3D graphics. Use separate libraries that wrap HTML Canvas.
But I have used charting libraries that wrap HTML Canvas and play nicely with React. Including built-in subtle animations and such.
3:37 PM
@MichaelHale Take a look at this for an example of Dynamic and grid updating.
3:49 PM
@RohitNamjoshi Thanks. Unfortunately, it's not any faster than my current solution.
@MichaelHale "current solution" in React or WL Dynamic?
WL. I haven't done this exact thing in React. That's just been influencing my thinking from my day job this year.
3 hours later…
7:15 PM
@MichaelHale The fastest approach is likely to make each square itself a `Dynamic` object. and give it a special `TrackedVariable` that allows you to control when it updates. The whole `Dynamic` system is clunky & I'd try to avoid it were I you
7:29 PM
Mathematica almost mentioned in a great University Challenge episode: youtu.be/mKJLqpQAF-8?t=661
2 hours later…
10:11 PM
@b3m2a1 Yeah. The above is the best I got. It is as fast as I wanted. I don't understand why some slightly simpler variations of that didn't work. But it does give the fast performance and let me read the data back out from the original array.
I thought I could just let "trigger" be some dummy variable, but it seems important to use it meaningfully in the expression. So maybe TrackedSymbols isn't just any symbol, but selecting from those used in the Dynamic.
But it's way better than leaving TrackedSymbols off.
I definitely still think Dynamic is cool. This was just messing around, but I'm certainly one of those people in the group Wolfram talks about about who uses Mathematica behind the scenes for random tasks at my job.
10:39 PM
@MichaelHale It's gotta be in the expr but can be like
```Dynamic[
trigger;
some actual expr,
TrackedSymbols:>{trigger}
]```
Whenever I do UI work I find myself doing stuff like that
10:51 PM
@b3m2a1 Works. Thanks.
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collins_ Aug 3, 2012 at 1:48 PM Edited Aug 3, 2012 at 2:00 PM I'm trying to fitting uncertain mesurements with an user-defined function (FitToFunction method). I'm using C# with VisualStudio11 Beta (Any CPU). To start, i would like to check if the exemple found in this website works : UncertainMeasurementSample data = new UncertainMeasurementSample(); data.Add(new UncertainMeasurement(1.0, new UncertainValue(2.0,3.0))); data.Add(new UncertainMeasurement(2.0, 1.0, 3.0)); data.Add(3.0, 2.0, 1.0); Function model = delegate (double[] a, double x) { double y = a[0] * Math.Sin(2.0 * Math.PI / a[1] + a[2]); return(y); }; FitResult oscilation = data.FitToFunction(model, new double[] { 1.0, 1.0, 0.0 }); First I got a problem with the declaration of Function : I have to replace "Function" by "Func" After compilation and execution of code I got this error (english translation) : System.NullReferenceException: The reference of this object is not defined to one instance of one object from Meta.Numerics.Statistics.UncertainMeasurementSample`1.FitToFunction(Func`3 function, Double[] start) Any ideas ? Collins ichbin Aug 4, 2012 at 9:27 AM Hi Collins, With release 2.0, Meta.Numerics changed from using its own Function delegate to the .NET-standard Func delegate. Clearly we need to update the tutorial documentation. That is, as you pointed out, an easy fix. The NullReferenceException is more troublesome. I can reproduce your result and see what's going on in the code: once we have found the parameter values that minimize the chi-squared function and determined the curvature matrix of the chi-squared function at the minimum, we try to invert the curvature matrix in order to determine the covariance matrix of the fit parameters. That inversion is failing, indicating that the curvature matrix is not positive definite, even though it is supposed to be so at a minimum point. I don't yet know why that is happening (a bug in the inversion code? a bug in the minimum finder? a bug in the curvature determination?) I do know the bug is new since that documentaiton was written, because it did work for us at the time. In any case, thanks for the report. I will reply again when I know more. ichbin Aug 4, 2012 at 10:11 AM Okay, I figured out what's going on here. There is a second problem with the tutorial documentation, and that problem has "infected" your example code. This problem makes the whole fit procedure ill-defined. When this problem is corrected, the fit works fine. Let's just start again with revised example code: ``` // Create a new sample and add some vaguely oscilatory data UncertainMeasurementSample data = new UncertainMeasurementSample(); data.Add(2.0,2.5,0.5); data.Add(3.0,2.0,1.0); data.Add(5.0,-1.0,0.5); data.Add(8.0,-4.0,0.5); data.Add(10.0,-1.5,0.5); data.Add(12.0, 0.0, 1.0); // Define an oscilatory model // a[0] ~ amplitude, a[1] ~ period, a[2] ~ phase Func model = (double[] a, double x) => a[0] * Math.Sin(2.0 * Math.PI * x / a[1] + a[2]); // Fit the model to the data, providing an initial guess for the parameters FitResult oscilation = data.FitToFunction(model, new double[] { 2.0, 10.0, 0.0 }); // Print the results for (int i = 0; i < oscilation.Dimension; i++) { Console.WriteLine(oscilation.Parameter(i)); } ``` This code example is better in lots of ways: it is better commented, fixes the Func issue, has more data points than parameters, and uses modern notation for the model delagate. But the key change is this: the model function actually depends on x. Note that, in the original example, x never appeared in the delegate definition. (I suspect this was simply a transcription typo.) This meant that the function was, from the point of view of the model, just a complicated constant computed from the parameters. The best the fit function could do is to adjust the parameters so that the constant equaled the average value of the data. But since there are multiple ways to do this, there are zero directions at the "minimum" and the curvature matrix is indeed not positive definite and therefore indeed not invertible. In any real application, your model function will actually depend on your independent variable, so in practice you should not encounter this problem. Of course, our code should still behave better when given a non-sensicle model, but that is a considerably less worrying problem than I feared we had when I last wrote. Thanks for pointing out our doc bug. Please let me know if you have any more problems after studying the revised example code. collins_ Aug 6, 2012 at 8:32 AM I copy past your exemple, but it dosn't work. It seems no convergence : "Meta.Numerics.NonconvergenceException' arrived to Meta.Numerics.Functions.FunctionMath.FindMinimum(Func`2 f, Double[] x) for Meta.Numerics.Statistics.UncertainMeasurementSample`1.FitToFunction(Func`3 function, Double[] start)". So i try an easier exemple for fitting " y=2x^2+3x+7 " : var data3 = new Meta.Numerics.Statistics.UncertainMeasurementSample(); data3.Add(1.0, 12.0, 0.01); data3.Add(2.0, 21.0, 0.01); data3.Add(3.0, 34.0, 0.01); data3.Add(4.0, 51.0, 0.01); data3.Add(5.0, 72.0, 0.01); data3.Add(6.0, 97.0, 0.01); data3.Add(7.0, 126.0, 0.01); data3.Add(8.0, 159.0, 0.01); data3.Add(9.0, 196.0, 0.01); data3.Add(10.0, 237.0, 0.01); Func model = (double[] a, double x) => a[0]*x*x + a[1] * x + a[2]; //V1 Meta.Numerics.Statistics.FitResult polyX = data.FitToFunction(model, new double[] { 2.0, 3.0, 7.0 }); //V1 With this exemple, no error but the result is not good : -Param n°0: -3,9657908577218E-18 ± 0,00435194139889244 -Param n°1: 3 ± 0,0491210625687066 -Param n°2: 1,27879034232044E-16 ± 0,117615191762516 Just to confirm, i try with FitToPolynomial : Meta.Numerics.Statistics.FitResult polyX = data3.FitToPolynomial(2); I got good result : ---X^0: 6,99999999999902 ± 0,0194079021706795 ---X^1: 3,00000000000081 ± 0,0145471440193393 ---X^2: 1,99999999999983 ± 0,00300058269399403 ichbin Aug 7, 2012 at 12:48 AM Hi Collins, I think you might have made an error in your comparison of FitToFunction and FitToPolynomial. I notice that you call FitToPolynomial on the data3 object but FitToFunction on the data object (no 3), which presumably contains different data points. When I call FitToFunction on the data3 object with the content as defined in your last post, I get a[0] = 2 ± 0.000435194139889244 a[1] = 3 ± 0.00491210625687065 a[2] = 7 ± 0.0117615191762515 precisely the same coefficents as for FitToPolynomial. (By the way, one of our unit tests, DataSetTest.FitToFunctionPolynomialCompatibilityTest, which you can find in our source, tests precisely this relationship.) With regard to your other problem, the FitToFunction call that gave you a NonconvergenceException, I'm afraid I can't reproduce this. Are you sure you are running precisely the code above, with the same data, model, and starting point? I wonder if perhaps there was a similiar error there, in which a different or changed object that happened to be lying around the same scope was used instead. I am certainly interested in chasing down any nonconvergence behavior you might encounter, but I would need to be able to reproduce it. Cheers, David collins_ Aug 7, 2012 at 10:52 AM So sorry ! Effectively, not a full copy past : data instead data3 ! Now Everything is working fine for polynominal fitting. For your exemple with sin() i got : A[0]: 3,43481931601377 ± 0,36174906992542 A[1]: 14,5075811130185 ± 0,989537309537895 A[2]: 1,33983540139721 ± 0,222553876989726 After i choosed more closer data : data3.Add(2.0, 2, 0.5); data3.Add(3.0, 1.9, 0.5); data3.Add(5.0, 0, 0.5); data3.Add(8.0, -2.0, 0.5); data3.Add(10.0, 0, 0.5); data3.Add(12.0, 2.0, 0.5); And i got : A[0]: 2,07832127074118 ± 0,265143580229382 A[1]: 10,0458902158885 ± 0,893527889071437 A[2]: 0,0274078446223988 ± 0,439562043057793 So as a conclusion, everything is working fine :) Now i can start to play with my own experiments results (i want to fit something like a*sin²(b/x) )... Thanks,
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Relationship between radicals and complex numbers i
So long as we keep that little "i" there to remind us that we still need to multiply by √−1 we are safe to continue with our solution! Using i we can also come up. The imaginary unit or unit imaginary number (i) is a solution to the quadratic equation x2 + 1 = 0 . This is because, although −i and i are not quantitatively equivalent (they are negatives of each other), there is no algebraic difference between i and −i. . The radical sign notation is reserved either for the principal square root. Explain the relationship between radical equations and complex numbers. What are the [Hint: the square root of i will be a complex number, of the form a + bi.
Задняя стенка ангара бесследно исчезла прямо.
Такси все еще двигалось рядом, тоже въехав на газон. Огромный лист гофрированного металла слетел с капота автомобиля и пролетел прямо у него над головой.
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# flintmax
Largest consecutive integer in floating-point format
## Syntax
``f = flintmax``
``f = flintmax(precision)``
``f = flintmax("like",p)``
## Description
example
````f = flintmax` returns the largest consecutive integer in IEEE® double precision, which is `2^53`. Above this value, double-precision format does not have integer precision, and not all integers can be represented exactly.```
example
````f = flintmax(precision)` returns the largest consecutive integer in IEEE single or double precision. `flintmax` returns `single(2^24)` for single precision and `2^53` for double precision.```
example
````f = flintmax("like",p)` returns the largest consecutive integer with the same data type, sparsity, and complexity (real or complex) as the floating-point variable `p`.```
## Examples
collapse all
Return the largest consecutive integer in IEEE® double precision, `2^53`.
```format long e f = flintmax```
```f = 9.007199254740992e+15 ```
Return the largest consecutive integer in IEEE® single precision, `single(2^24)`.
`f = flintmax("single")`
```f = single 16777216 ```
Check the class of `f`.
`class(f)`
```ans = 'single' ```
Above the value returned by `flintmax("single")`, not all integers can be represented exactly with single precision.
Return the largest consecutive integer in IEEE® single precision, `single(2^24)`.
`f = flintmax("single")`
```f = single 16777216 ```
Add `1` to the value returned from `flintmax`.
`f1 = f+1`
```f1 = single 16777216 ```
`f1` is the same as `f`.
`isequal(f,f1)`
```ans = logical 1 ```
Add `2` to the value returned from `flintmax`. The number `16777218` is represented exactly in single precision while `16777217` is not.
`f2 = f+2`
```f2 = single 16777218 ```
Return the largest consecutive integer with the same data type and complexity as an existing array.
First, create a complex vector of `single` data type.
`p = single([0.12+2i -0.5i 3]);`
Return the largest consecutive integer as a scalar that is complex like `p`.
`f = flintmax("like",p)`
```f = single 1.6777e+07 +0.0000e+00i ```
## Input Arguments
collapse all
Floating-point precision type, specified as `"double"` or `"single"`.
Data Types: `char`
Prototype, specified as a floating-point variable.
Data Types: `double` | `single`
Complex Number Support: Yes
## Output Arguments
collapse all
Largest consecutive integer in floating-point format returned as a scalar constant. This constant is `2^53` for double precision and `single(2^24)` for single precision.
## Version History
Introduced in R2013a
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# Absolute Value Footloose Task Cards
\$3.50
Using the Absolute Value Footloose task cards to play this fun math game is a fantastic way to get students moving while they are practicing concepts!
Students solve each of the 30 problems and record their answers in the corresponding boxes on their Footloose grids.
• Identify absolute value
• Compare and order absolute values and integers
• Apply absolute value to real-life situations
Activity includes:
• Footloose grid
• 30 cards (a set WITH background AND a set WITHOUT background – SAME set, just the option of a background for you when you print them)
• Directions
• Coloring notes sheet
⭐️⭐️⭐️⭐️⭐️ My students used these as review of absolute value! I love that they were able to see examples of absolute value in so many different ways!
⭐️⭐️⭐️⭐️⭐️ Loved the resource! I used it as a review for my 7th graders who were not taught this standard last year due to the pandemic. I adapted it to be used with PearDeck and I will definitely be using it again!
⭐️⭐️⭐️⭐️⭐️ We used this as a review. It really helped solidify my students practice with absolute value. Thank you for your effort in this resource!
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https://www.emathhelp.net/drops-metric-to-cups-metric/
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# Drops (metric) to cups (metric)
This free conversion calculator will convert drops (metric) to cups (metric), i.e. gtt (metric) to c. Correct conversion between different measurement scales.
Convert
The formula is $$V_{\text{c}} = \frac{1}{5000} V_{\text{gtt (metric)}}$$$, where $$V_{\text{gtt (metric)}} = 15$$$.
Therefore, $$V_{\text{c}} = \frac{3}{1000}$$$. Answer: $$15 \text{gtt (metric)} = \frac{3}{1000} \text{c} = 0.003 \text{c}$$$.
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https://english.stackexchange.com/questions/458961/numbers-or-number-which-sentence-is-correct?rq=1
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“Numbers” or “number?” Which sentence is correct?
Which sentence is correct? Why?
1. In 2000, the number of Vietnamese students studied in Russia and France was around 3 million and 3.5 million respectively.
2. In 2000, the numbers of Vietnamese students studied in Russia and France were around 3 million and 3.5 million respectively.
• Here's an article that you might find useful: dailywritingtips.com/is-number-singular-or-plural
– VTH
Aug 6, 2018 at 5:43
• Pham Van Nhan, you should wait a bit longer to tick any answer correct. In this case, you chose the wrong one. Also, you should edit the question a bit: the Vietnamese students aren't being studied by Russian and French researchers; the Vietnamese students are studying in Russia and France.
– lly
Aug 6, 2018 at 7:47
• I noted the mistake in the question, it should be 'The number of Vietnamese students who studied in Russia and France'. Thanks for your explanation of the tick, I didn't know that :D Really appreciate!! Aug 6, 2018 at 10:03
The short answer is that it should be in the singular. (I will explain this later.)
However, there is a better solution to this—one that is both syntactically correct and which sounds more natural:
In 2000, Vietnamese students who studied in Russia and France numbered around 3 million and 3.5 million, respectively.
This dispenses with the troublesome noun number, using instead the past tense of the verb number.
To explain why the example sentence is in the singular, consider its expanded version:
In 2000, the number of Vietnamese students who studied in Russia was around 3 million and the number of Vietnamese students who studied in France was around 3.5 million.
In both clauses, number is singular and so is the verb. Shortening it into a single clause with respectively doesn't change this essential fact in this particular case. (Even when X is plural, you can't say the numbers of X.)
In 2000, the number of Vietnamese students who studied in Russia and France was around 3 million and 3.5 million, respectively.
But in addition to the rephrasing I suggested at the start of my answer, there is another possibility:
In 2000, the total number of Vietnamese students who studied in Russia and France was around 6.5 million: 3 million and 3.5 million in the respective countries.
• 'Number' here is a statistic, and like other statistics can be plural-form. See [English for Students] 'The means / averages of the two sets were 12.5 and 13.7 cm respectively.' // 'The numbers of students involved were just under 12 000 and over 43 000 respectively.' //// 'Number' can also be (and often is) used in a singular-form-covers-all way ('The number of patients rose from 1200 to 2700 in a single week'. //// As you say, awkward mismatches with verb forms can arise and should be avoided by reformulation. Jun 18, 2021 at 14:11
The First one is correct in my opinion: According to grammar rule : "The number of" + singular verb.
Moreover, the Object of Prepositions has nothing to do with the Verb of the sentence.
The second seems correct since you are referring to two results of two different counts. Also, I think you mean "studying in," not "studied in."
• Thank you for your answer, noted with "studying". So if we consider "the numbers" as 2 different subjects, one for Russia and another for France, there will be 2 individuals to go with "were". Am I correct? Aug 6, 2018 at 2:33
• Your subject, "numbers," is plural, and so you should use "were."
– Jeh
Aug 6, 2018 at 2:47
• You might want to say "There were about 3 million Vietnamese students in Russia and about 3.5 million in France in 2000." or, alternatively "...students from Vietnam...". Aug 6, 2018 at 4:41
• I can follow your logic but you simply wouldn't use the word "numbers" if there aren't two separate subjects of the counting. ("The numbers of commuting and boarding students at the university were...") Here, it's one group being numbered up in two different contexts. You'd use a singular number or rephrase, like Mr Bassford suggests.
– lly
Aug 6, 2018 at 7:51
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## Saturday, August 29, 2015
### The Famous Sun Dial at Konark, Odisha
The sundial has 8 major spokes that divide 24 hours into 8 equal parts, which means that the time between two major spokes is 3 hours. There are 8 minor spokes as well. Each minor spoke runs exactly in the middle of 2 major spokes. This means that the minor spoke divides the 3 hours in half, so the time between a major spoke and a minor spoke is an hour and half or 90 minutes.
Now, at the edge of the wheel, you can see a lot of beads. If you observe carefully, you can see that there are 30 beads between a minor and a major spoke. So, the 90 minutes are further divided by 30 beads. This means that each bead carries a value of 3 minutes. The beads are large enough, so you can also see if the shadow falls in the center of the bead or on one of the ends of the bead. This way we can further calculate time accurately to the minute.
The sundial shows time in an anti-clockwise fashion. At the top, the major spoke stands for midnight and this spoke stands for 3 A.M and this one for 6 A.M and so on. When I place a finger or a pen at the tail of the animal in the axle, the shadow will fall on the edge of the wheel. Now, I simply note the bead where the shadow falls. Using the math we did before, I can easily tell the current time precisely down to the minute. Imagine how much time and coordination would have happened between the astronomers, engineers and sculptors to create something like this 750 years ago.
Now if you are observing closely, you would have 2 questions in your mind right now. The first question would be, what happens when the sun moves from east to west. Since the wheel is carved on a wall, the sun would not shine on this wheel at all. How can we tell time in the afternoons? Now, the Konark temple has another wheel or sundial, located on the west side of the temple as well. You can just use the other sundial that will work perfectly from afternoon, until sunset.
This is the second and the most interesting question. How do you tell time after sunset? There would be no sun, and hence no shadows from sunset till the next morning’s sunrise. After all, we have 2 sundials in the temple which work only when the sun shines. To this question, I want to point out that the Konark temple does not have just 2 wheels like this. The temple has a total of 24 wheels, all accurately carved just like the sundials. Have you heard of the Moondial? Do you know that the moondials can work just like sun dials during night time? What if the other wheels in the temple could be used as moondials?
Many people think that the other 22 wheels were carved for decorative or religious purposes and do not have an actual use. This is what people thought about the 2 sundials as well. Believe it or not, people thought that all the 24 wheels were just carved for beauty and as Hindu symbols. About 100 years ago, it became known that this was a sundial when an old yogi was seen calculating time secretly. Apparently selected people were using these wheels for generations and for 650 years no one else knew about it. They say that when they asked him about the purpose of the other 22 wheels, the yogi refused to talk and simply walked away.
And our knowledge of just these 2 sundials themselves is actually very limited. You can see how there are multiple circles of beads. You can see carvings and markings all over these sundials, and we don’t the meaning of most of them. For example, this carving on a major spoke has exactly 60 beads. Notice how in some carving you can see leaves and flowers which may mean Spring or Summer. Notice how in some carvings you can see lemurs mating, which only happens during winter. So, these sundials could have even been used as an almanac for a variety of different things. Now you can understand how limited our knowledge is about the rest of the 22 wheels.
## Friday, August 28, 2015
### Hindus Lived 74000 Years Ago Survey Finds
New evidence suggests that Hinduism existed in Indonesia 74,000 ago and it is safe to assume that Hinduism antedates this period.Earlier to this finding a city older that Mohenjo-Daro has been found.
About 76,000 years ago, the volcano Toba – located in what is now Indonesia – erupted to create the largest and most devastating volcanic event of the past 2 million years. Almost 3,000 cubic kilometers of magma was spewed out, while sulfuric acid rained over the earth as far away as Greenland. The world became subject to a volcanic winter, and what followed was one of the most severe ice ages in documented history.
Over in India, the land was showered with 15 centimeters of volcanic ash, which can be seen today, working as a distinct age marker in the earth’s stratigraphy. And yet, contrary to all logic, archaeologists have unearthed assemblages of stone tools both above and below the ash deposit in India’s Jwalapuram Valley.
The tools look remarkably similar to those made by humans in Africa, which indicates that these tools were also human-formed – and yet, if humans were still in India after the depositing of ash (an incredible feat it itself), they would have had an extremely difficult time trying to survive. After all, the sheer magnitude of the eruption suspended both volcanic gas and sulfuric acid in the earth’s atmosphere for years, causing warm sunlight to be redirected away from Earth – and plunging the world into several centuries of temperatures that were at least 3-5 degrees C lower than normal after the event.
Mapping of stone tool artefacts on a Middle Palaeolithic occupation surface under the Toba ash.
Newly discovered archaeological sites in southern and northern India have revealed how people lived before and after the colossal Toba volcanic eruption 74,000 years ago.
The international, multidisciplinary research team, led by Oxford University in collaboration with Indian institutions, unveiled to a conference in Oxford what it calls ‘Pompeii-like excavations’ beneath the Toba ash.
The seven-year project examines the environment that humans lived in, their stone tools, as well as the plants and animal bones of the time. The team has concluded that many forms of life survived the super-eruption, contrary to other research which has suggested significant animal extinctions and genetic bottlenecks.
According to the team, a potentially ground-breaking implication of the new work is that the species responsible for making the stone tools in India was Homo sapiens. Stone tool analysis has revealed that the artefacts consist of cores and flakes, which are classified in India as Middle Palaeolithic and are similar to those made by modern humans in Africa. ‘Though we are still searching for human fossils to definitively prove the case, we are encouraged by the technological similarities. This suggests that human populations were present in India prior to 74,000 years ago, or about 15,000 years earlier than expected based on some genetic clocks,’ said project director Dr Michael Petraglia, Senior Research Fellow in the School of Archaeology at the University of Oxford.
This exciting new information questions the idea that the Toba super-eruption caused a worldwide environmental catastrophe. Dr Michael Petraglia, School of Archaeology
An area of widespread speculation about the Toba super-eruption is that it nearly drove humanity to extinction. The fact that the Middle Palaeolithic tools of similar styles are found right before and after the Toba super-eruption, suggests that the people who survived the eruption were the same populations, using the same kinds of tools, says Dr Petraglia. The research agrees with evidence that other human ancestors, such as the Neanderthals in Europe and the small brained Hobbits in Southeastern Asia, continued to survive well after Toba.
Although some scholars have speculated that the Toba volcano led to severe and wholesale environmental destruction, the Oxford-led research in India suggests that a mosaic of ecological settings was present, and some areas experienced a relatively rapid recovery after the volcanic event.
The team has not discovered much bone in Toba ash sites, but in the Billasurgam cave complex in Kurnool, Andhra Pradesh, the researchers have found deposits which they believe range from at least 100,000 years ago to the present. They contain a wealth of animal bones such as wild cattle, carnivores and monkeys. They have also identified plant materials in the Toba ash sites and caves, yielding important information about the impact of the Toba super-eruption on the ecological settings.
Dr Petraglia said: ‘This exciting new information questions the idea that the Toba super-eruption caused a worldwide environmental catastrophe. That is not to say that there were no ecological effects. We do have evidence that the ash temporarily disrupted vegetative communities and it certainly choked and polluted some fresh water sources, probably causing harm to wildlife and maybe even humans.’
Older Than Harappa.
--------------------------
“A team of archaeologists from the Deccan College Post Graduate and Research Institute is back from Haryana where they stumbled upon a record 70 Harappan graves at a site in Farmana, discovering the largest burial site of this civilization in India so far. It is an extraordinary archaeological finding. A big housing complex that matured during the Harappan era was discovered by these archaeologists who have been working in this little known village for the past three years. The archaeological team here uncovered an entire town plan. The skeletal remains belong to an era between 2500 BC to 2000 BC.
http://ancientstandard.com/2007/08/18/%E2%80%9Cvolcanic-mega-eruption-no-problem-how%E2%80%99ve-you-been%E2%80%9D-ca-74000-bc/
## Wednesday, August 26, 2015
### Do the Vedas prohibit idol worship?
It is a very controversial thing about this verses along the religious websites, especially non-Hindu people/websites, they are claiming that idol worship is prohibited in Hinduism.
"andhaṁ tamaḥ praviśanti ye 'sambhūtim upāsate
tato bhūya iva te tamo ya u sambhūtyām ratāḥ ||"
This is present in the Isha Upanishad or the Shukla Yajur Veda chapter forty. But to say that this verse prohibits idol worship would be incorrect. All that it says is, those who worship only the asambhuta (which has not originated) and those who worship only the sambhuta (which has originated) enter into darkness.
But the thing is, because Sanskrit words can mean multiple things, people give it different meanings. But simply speaking, asambhuta here means the unmanifested absolute formless mode of supreme Brahman and sambhuta means the different manifested forms like the devas or demigods. It is because, the absolute neither comes to exist nor ceases to exist (without origin), but the various gods come to exist and also after their time is over cease to exist (with origin). Similarly, there is another verse which forbids both knowledge and ignorance:
"andhaṁ tamaḥ praviśanti ye 'vidyām upāsate
tato bhūya iva te tamo ya u vidyāyām ratāḥ [Isha Up. - 9] ||"
Now tell me, what kind of logic is this! It is understandable if we say one will enter darkness if he worships or follows ignorance, but why would any one enter darkness if he follows knowledge?
So these verses only encourage to have a complete knowledge of the both the aspects instead of following or knowing only one partially. It is because both knowledge and ignorance are part of God:
"vidyāvidye mama tanū [SB - 11.11.3]"
- Both knowledge and ignorance are my body (energy potencies)
And God is both with and without forms:
"dve vāva brahmaṇo rūpe, mūrtaṃ caivāmūrtaṃ ca [Brh. Up - 2.3.1] ||"
- God (Brahman) has two modes, formless (nirakara, asambhuta) as well as form (sakar, sambhuta).
The Vedas and scriptures are full of contradicting statements for a certain reason. If one takes up only one statement and tries to define everything else as per it, then he will only reach biased and wrong conclusions. Complete knowledge is always required. So another verse of that same Upanishad explicitly mentions to known both knowledge and ignorance. Because only by knowing both the modes of God that one will be able to have the complete and absolute knowledge:
"vidyāṁ cāvidyāṁ ca yas tad vedobhayaṁ saha
avidyayā mṛtyuṁ tīrtvā vidyayāmṛtam aśnute [Isha Up. - 11] ||"
Meaning
Only one who can learn the process of nescience (avidya) and that of transcendental knowledge (vidya) side by side can transcend the influence of repeated birth and death and enjoy the full blessings of immortality.
So don't take any such arguments seriously. Idol worship is neither mandatory nor prohibited in Hinduism. In fact puranas like Shrimad Bhagavatam itself mention the process of deity form worship of the God and what the idols can be made of:
"śailī dāru-mayī lauhī lepyā lekhyā ca saikatī
mano-mayī maṇi-mayī pratimāṣṭa-vidhā smṛtā [SB - 11.27.12] ||"
Meaning
The Deity form of the Lord is said to appear in eight varieties — stone, wood, metal, earth, paint, sand, the mind or jewels.
## Tuesday, August 25, 2015
### Mind Blowing Facts about Sanskrit
• Sanskrit has the highest number of vocabularies than any other language in the world.
• 102 arab 78 crore 50 lakh words have been used till now in Sanskrit. If it will be used in computers & technology, then more these number of words will be used in next 100 years.
• Sanskrit has the power to say a sentence in a minimum number of words than any other language.
• America has a University dedicated to Sanskrit and the NASA too has a department in it to research on Sanskrit manuscripts.
• Sanskrit is the best computer friendly language.(Ref: Forbes Magazine July 1987).
• Sanskrit is a highly regularized language. In fact, NASA declared it to be the “only unambiguous spoken language on the planet” – and very suitable for computer comprehension.
• Sanskrit is an official language of the Indian state of Uttarakhand.
• There is a report by a NASA scientist that America is creating 6th and 7th generation super computers based on Sanskrit language. Project deadline is 2025 for 6th generation and 2034 for 7th generation computer. After this there will be a revolution all over the world to learn Sanskrit.
• The language is rich in most advanced science, contained in their books called Vedas, Upanishads, Shruti, Smriti, Puranas, Mahabharata, Ramayana etc. (Ref: Russian State University, NASA etc. NASA possesses 60,000 palm leaf manuscripts, which they are studying.)
• Learning of Sanskrit improves brain functioning. Students start getting better marks in other subjects like Mathematics, Science etc., which some people find difficult. It enhances the memory power. James Junior School, London, has made Sanskrit compulsory. Students of this school are among the toppers year after year. This has been followed by some schools in Ireland also.
• Research has shown that the phonetics of this language has roots in various energy points of the body and reading, speaking or reciting Sanskrit stimulates these points and raises the energy levels, whereby resistance against illnesses, relaxation to mind and reduction of stress are achieved.
• Sanskrit is the only language, which uses all the nerves of the tongue. By its pronunciation, energy points in the body are activated that causes the blood circulation to improve. This, coupled with the enhanced brain functioning and higher energy levels, ensures better health. Blood Pressure, diabetes, cholesterol etc. are controlled. (Ref: American Hindu University after constant study)
• There are reports that Russians, Germans and Americans are actively doing research on Hindu’s sacred books and are producing them back to the world in their name. Seventeen countries around the world have a University or two to study Sanskrit to gain technological advantages.
• Surprisingly, it is not just a language. Sanskrit is the primordial conduit between Human Thought and the Soul; Physics and Metaphysics; Subtle and Gross; Culture and Art; Nature and its Author; Created and the Creator.
• Sanskrit is the scholarly language of 3 major World religions – Hinduism, Buddhism (along with Pali) and Jainism (second to Prakrit).
• Today, there are a handful of Indian villages (in Rajasthan, Madhya Pradesh, Orissa, Karnataka and Uttar Pradesh) where Sanskrit is still spoken as the main language. For example in the village of Mathur in Karnataka, more than 90% of the population knows Sanskrit. Mathur/Mattur is a village 10 kms from Shimoga speaks Sanskrit on daily basis (day-to-day communication).
• Even a Sanskrit daily newspaper exists! Sudharma, published out of Mysore, has been running since 1970 and is now available online as an e-paper (sudharma.epapertoday.com)!
• The best type of calendar being used is hindu calendar(as the new year starts with the geological change of the solar system)
ref: german state university
• The UK is presently researching on a defence system based on Hindu’s shri chakra.
• Another interesting fact about Sanskrit language was that the process of introducing new words into the language continued for a long period until it was stopped by the great grammarian Panini who wrote an entire grammar for the language laying down rules for the derivation of each and every word in Sanskrit and disallowed the introducing of new words by giving a full list of Roots and Nouns. Even after Panini, some changes occur which were regularised by Vararuchi and finally by Patanjali. Any infringement of the rules as laid down by Patanjali was regarded as a grammatical error and hence the Sanskrit Language has remained in same without any change from the date of Patanjali (about 250 B.C.) up to this day.
• Sanskrit is the only language in the world that exists since millions of years. Millions of languages that emerged from Sanskrit are dead and millions will come but Sanskrit will remain eternal. It is truly language of Bhagwan.
## Friday, August 21, 2015
### Harvard University Says ... Kumbh mela is better organized than Fifa WC !!!
A book produced by scholars and students of Harvard University along with architects and town planners of international repute, declared that Maha Kumbh 2013 to be better organised than FIFA World Cup in Brazil and Commonwealth Games in New Delhi.
----------------------------
– The tent township is much larger than the size of Manhattan in terms of population.
– More than 100 million come to a small place and stay there for 55 days.
– Nearly 5 million people bath in the holy rivers Ganga, Yamuna and Saraswati.
The Largest hub of communication in one place
-----------------------------------
– There are 390 million communication events (calls, messages, etc).
– 146 million (145,736,764) text messages
– 245 million (245,252,102) calls
## Sunday, August 9, 2015
### Vyuhs(व्यूह) described in Mahabharat
Vyuh (Sanskrit: व्यूह) means - 'to arrange troops in a battle array', 'to arrange, put or place in order, to dispose, separate, divide, alter, transpose, disarrange, resolve.
At various times during battle, the supreme commander of either army ordered special formations(vyuhas). Each formation had a specific purpose; some were defensive while others were offensive. Each formation had its specific strengths and weaknesses.
The Mahabharat lists the following:
1. Krauncha vyuha(heron formation)
2. Makara vyuha(crocodile formation)
3. Kurma vyuha(tortoise or turtle formation)
4. Trishula vyuha(trident formation)
5. Chakrvyuha(wheel or discus formation)
6. Kamala vyuha or Padma vyuha(lotus formation)
7. Garud vyuha(eagle formation)
8. Oormi vyuha(ocean formation)
9. Mandala vyuha(galaxy formation)
10. Vajra vyuha(diamond or thunderbolt formation)
11. Shakata vyuha(box or cart formation)
12. Asura vyuha(demon formation)
13. Deva vyuha(divine formation)
14. Soochi vyuha(needle formation)
15. Sringataka vyuha(horned formation)
16. Chandrakala vyuha(crescent or curved blade formation)
17. Mala vyuha(garland formation)
Each formation was met bya counter formation by the other side.
For instance , the Sarpa Vyuha was met with Garuda Vyuha(Serpent formation against Eagle.Formation was usually met with Garuda or eagle Formation’Eagle is a Natural Enemy of Heron.Note the Swoop of the Garuda, nullifying and swamping the posture of the Heron.
There were also courses to get into and out of each formation.
## Thursday, August 6, 2015
### Raghuvanshi: The Most Respected Lineage Among Hindu Kings
King Raghu was son of King Dilīp. The practice of donating amass wealth, protecting cows and sharing prosperity with people and Sages was done more passionately by King Raghu in his Kingdom.
King Raghu was one of the greatest donors. Demands of common people, whoever visited the palace were met, while giving them dignity and respect. Sages, Sadhus were regarded as protectors of the land and held with high esteem in the palace. There was common saying in the kingdom that “no one returned from the palace of King Raghu empty-handed.”
Such was the passion to keep people happy that all promises and demands of the praja was fulfilled by King Raghu.
This also led to the another famous saying:
Raghukul Reet Sada Chali aayi, Prann Jaaye Par Vachan Na Jaaye
Meaning: The legacy of Raghukul continues; words and promises are kept even to the extent of giving life to fulfill it.
Clash of King Dilip with Indra on Completion of Yagna
----------------------------------
Performing 100 yagnas with proper Vedic principles under guidance of Sages was done before by Indra, King of Gods (Indra is a position acquired by people who are eligible to become one).
King Raghu’s father King Dilīp was a very pious king and devotee of Brahma, Vishnu and Mahesh, so he performed as many as 100 yajnas. It required immense devotion, penance and focus to perform 100 yagnas and only Indra was successful to do so. King Dilip was nearing the completion of his 100th yagna, Indra felt jealous and so he placed many hurdles in the path of the ongoing 100th yagna, but King Raghu with his piousness, bhakti and devotion was able to continue the 100th yagna and successfully complete it. The most celebrated of all, Raghukul Dynasty was thus born. It was series of pious karmas of forefathers of King Raghu and then his future sons that made it possible for Vishnu to take Avatar as Bhagwan Ram so as to eventaully make the end of Treta Yug more dharmic filled with prosperity, piousness and free from Asurs (Rakshas).
King Raghu was soft spoken, kind hearted and intelligent ruler. He took blessings of Sages and their guidance to rule the Kingdom. His foresightedness was responsible that kept his Kingdom very happy; free from wrath, distress and sorrow. He was ardent devotee of Vedic gods, very brave King and knew rightful ways to treat enemies.
To avoid invasion, establish dharma and make this world peaceful place – pious and free from adharmis, Anti-Vedic people. King Raghu showcased his great warrior skills when he marched towards central Asia (now Uzbekistan, Tajikistan, southern Kyrgyzstan and Kazakhstan). King Raghu made them aware of the Vedic science, principles and peaceful ways of leading life.
The kingdom of Raghu extend from his capital Ayodhya (Awadh) to the Bay of Bengal, then south along the eastern shore of India to Cape Comorin, then north along the western shore until the region surrounded by the Indus (Sindhu river), finally east through the tremendous Himalayan range into Assam
## Wednesday, August 5, 2015
### Hindu legacy in Quanzhou,China
Evidence of Hinduism in China have been found in and around Quanzhou in Fujian province, suggesting a Hindu community and particularly Tamil Hindu traders in medieval China. The evidence consists of a Tamil-Chinese bilingual inscription dated April 1281 AD devoted to deity Śiva, as well as over 300 artifacts, idols and Chola-style temple structures discovered in Fujian rovince since 1933.Archeological studies suggest at least Vaishnavism and Shaivism schools of Hinduism had arrived in China in its history.
At present, there are no Hindus in Quanzhou. However, there previously existed a Tamil Hindu community in the city who, in the late 13th century, built the Kaiyuan Temple dedicated to Lord Shiva. The temple is now in ruins, but over 300 carvings are still within the city. Many are currently on display in the Quanzhou museum, and some have become a part of Buddhist temple—Kaiyuan Temple. Behind its main hall "Mahavira Hall”, there are some columns decorated by some Hinduism carvings. The carvings are dispersed across five primary sites in Quanzhou and the neighboring areas. They were made in the South Indian style, and share close similarities with 13th-century temples constructed in the Kaveri Delta region in Tamil Nadu. Nearly all the carvings were carved with greenish-gray granite, which was widely available in the nearby hills and used in the region's local architecture.Shiva-related themes depicted in Quanzhou temple include the story of Gajaranya Kshetra - an elephant worshipping a Shiva linga, the story of Saivite saint Thirumular depicted through a cow anointing a linga and two Hindu wrestler story from Indian region now called as Andhra Pradesh.
In addition to Shiva, Vishnu sculpture has been discovered in Nanjiaochang area.Two pillars on the Kaiyuan temple have seven images dedicated to Vishnu - one with Garuda, one in the man-lion Narasimha avatar, one depicting the legend of Gajendra moksha, one with Lakshmi, one as Krishna stealing milkmaids clothing to tease them, one depicting the story of Vishnu as Krishna subduing serpent Kaliya, and another of Krishna in Mahabharat
## Monday, August 3, 2015
### Sculpting The Entire Rock to Realize Divine Imagination
The Kailash temple is not built. All is cut and carved from one gigantic piece of rock, hewn out of the Charanandri hills of the Sahyadri range of the Deccan Plateau at a village, which once was called Elapura, (later renamed as Ellora by british), 30 km northwest of Sambhajinagar (aurangabad). The Kailash temple was cut from the top down in a U-shape form, about 50 Meters deep in the back and sliding to lower levels on the sides to the front where there is an entry gate. From where the cutting was started is not known, definitely it could be from the Ganesh Dwar (every Hindu temple has a place that is dedicated to Bhagwan Ganesh, Son of Bhagwan Shiv), but later were the de-pilings done simultaneously or step by step – nothing is clear even after deep research conducted by several experts.
After deeply analyzing the amount of man-hours and efforts involved, experts summarized that the scale at which the work was undertaken is enormous. It covers twice the area of the Parthenon in Athens and is 1.5 times high (and bigger than all ancient temples of the world), and it entailed removing 200,000 tonnes of rock. It is believed to have taken 7,000 labourers and 150 years to complete the project.
Practically speaking with modern technology, given the space and plan around the Kailash temple, it is almost impossible to implement Ten 10-ton JCB machines to remove the rock pieces, as movement of each machine would require space and such huge machines can muffle up lot of space around them, their function is very complicated and when the ground is hollow the machine cannot operate and pose problem in itself.
Hypothetically, If the scale of work was carried today, it would have required at least 10 largest ever 10-ton JCB machines to excavate the pieces of rock of 20,000 tonnes by each machine. Each machine is able to excavate 100 tons of rock pieces in phased manner so even removal of 1000 tonnes each day would have taken more than 6 months i.e. 200 days of continous work !
Is It Possible to Replicate the Divine Feat by Humans Today
------------------------------------
Today building a massive structure like Kailash temple would require pre-design and 3D conceptions using latest CAD softwares and high-tech computers. Imagining how the structure will look at which area we should have distance properly maintained to start carving, where we should pause, which side we should start building entrace, how the internal design be formed. There are hundreds of questions which require answers and only getting resolutions around these queries alone would require several months of hundreds of dedicated designs, 3D graphic artistes and designers who have knowledge of construction and civil work.
The precision with which the sculpts can be cut is still not invented by engineers in the world today. So we will need manual labors to minutely carve the temple. It will require at least 10,000 skilled workers to carry out the digging, carving, sculpting and materializing the entire structure as envisioned. The total duration required to complete the task cannot be estimated because till date in modern times, no single mid-size rock is cut to create a temple.
Repeating such a feat even by using modern technology is almost impossible but ancient Hindu Sages made it possible simply with their spiritual powers, astute direction to skilled and dedicated workers with endless divine blessings of Bhagwan.
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# Bias of Sample Variance
## Theorem
Let $X_1, X_2, \ldots, X_n$ form a random sample from a population with mean $\mu$ and variance $\sigma^2$.
Let:
$\ds \bar X = \frac 1 n \sum_{i \mathop = 1}^n X_i$
Then:
$\ds \hat {\sigma^2} = \frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2$
is a biased estimator of $\sigma^2$, with:
$\operatorname{bias} \paren {\hat {\sigma ^2} } = -\dfrac {\sigma^2} n$
## Proof
If $\hat {\sigma^2}$ is a biased estimator of $\sigma^2$, then:
$\expect {\hat {\sigma^2} } \ne \sigma^2$
We have:
$\ds \expect {\hat {\sigma^2} }$ $=$ $\ds \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}$ $\ds$ $=$ $\ds \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren{\paren {X_i - \mu} - \paren {\bar X - \mu} }^2}$ writing $X_i - \bar X = X_i - \bar X - \mu + \mu$ $\ds$ $=$ $\ds \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren{\paren {X_i - \mu}^2 - 2 \paren {\bar X - \mu} \paren {X_i -\mu} + \paren {\bar X - \mu}^2} }$ Square of Difference $\ds$ $=$ $\ds \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \mu}^2 - \frac 2 n \paren {\bar X - \mu} \sum_{i \mathop = 1}^n \paren {X_i -\mu} + \frac 1 n \paren {\bar X - \mu}^2 \sum_{i \mathop = 1}^n 1}$ Summation is Linear
We have that:
$\ds \frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \mu}$ $=$ $\ds \frac 1 n \sum_{i \mathop = 1}^n X_i - \frac n n \mu$ from $\ds \sum_{i \mathop = 1}^n 1 = n$, noting that $\mu$ is independent of $i$. $\ds$ $=$ $\ds \bar X - \mu$ Definition of $\bar X$
So:
$\ds \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \mu}^2 - \frac 2 n \paren {\bar X - \mu} \sum_{i \mathop = 1}^n \paren {X_i -\mu} + \frac 1 n \paren {\bar X - \mu}^2 \sum_{i \mathop = 1}^n 1}$ $=$ $\ds \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \mu}^2 - 2 \paren {\bar X - \mu}^2 + \paren {\bar X - \mu}^2}$ $\ds$ $=$ $\ds \frac 1 n \expect {\sum_{i \mathop = 1}^n \paren {X_i - \mu}^2} - \expect {\paren {\bar X - \mu}^2}$ Expectation is Linear $\ds$ $=$ $\ds \frac 1 n \sum_{i \mathop = 1}^n \expect {\paren {X_i - \mu}^2} - \var {\bar X}$ Definition of Variance, Expectation is Linear $\ds$ $=$ $\ds \frac 1 n \sum_{i \mathop = 1}^n \var {X_i} - \frac {\sigma^2} n$ Definition of Variance, Variance of Sample Mean $\ds$ $=$ $\ds \frac n n \sigma^2 - \frac {\sigma^2} n$ $\var {X_i} = \sigma^2$, $\ds \sum_{i \mathop = 1}^n 1 = n$ $\ds$ $=$ $\ds \sigma^2 - \frac {\sigma^2} n$ $\ds$ $\ne$ $\ds \sigma^2$
So $\hat {\sigma^2}$ is a biased estimator of $\sigma^2$.
Further, we have:
$\operatorname{bias} \paren {\hat {\sigma ^2}} = \sigma^2 - \dfrac {\sigma^2} n - \sigma^2 = -\dfrac {\sigma^2} n$
$\blacksquare$
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HL Deb 18 February 1954 vol 185 cc1019-38
4.5 p.m.
My Lords, this is a small but rather important Bill to the shipping industry. It is to amend the Merchant Shipping Acts in respect of the method of calculating the net tonnage of shipping, which is, of course, derived from the gross tonnage. As the existing calculations are set out by Statute, we need a new Act of Parliament to change them. The size of a merchant ship is commonly measured in gross tonnage. This is not a measurement of weight, but of space contained in the ship converted at the rate of 100 cubic feet to one ton. To calculate the gross tonnage of a merchant ship, one delves among the 748 sections and twenty-two Schedules of the Merchant Shipping Act, 1894. There, Section 77, together with the Second Schedule, sets out the extremely complicated method of calculation.
But one also requires another measurement, namely, the net tonnage, to give an idea of the earning capacity of the ship, and for other reasons. This is obtained by deducting from the gross tonnage the allowances for the various spaces in the ship not available for freight, and Sections 78 and 79 of that Act tell one how to do this, the result being called the net tonnage. The crew space et cetera, is deducted in a straightforward way, set out in Section 79; but when one comes to deducting the space for propelling machinery the procedure is more complicated, and is set out in Section 79. In future, for brevity, I shall speak of this space as the engine room, for it represents the engine room, the boilers and the shaft tunnel of the ship—in fact, the space required to propel the ship. The standard deduction from the gross tonnage for the engine room is one and three quarter times the actual space used; but there is a complication in that, if the engine room represents between 13 per cent. and 20 per cent. of the whole of the gross tonnage, then instead of the one and three quarter formula one makes a flat rate deduction of 32 per cent. of the gross tonnage.
Now this flat rate of 32 per cent. produces certain absurdities which it is the object of the Bill to set right. For instance, a 12.9 per cent. engine room attracts an allowance of one and three-quarter times 12.9—say, 22½per cent.; whereas, if the size of the engine room jumps by a fraction to 13.1 per cent., then the allowance immediately jumps to 32 per cent. That is the full measure of the anomaly. The amount of this allowance—hence the net registered tonnage resulting—is extremely vital to shipowners because harbour dues are paid on the net tonnage, and on the existing formula the difference between a 12.9 per cent. engine room and a 13.1 per cent. engine room on a tanker of 18,000 gross tons represents an allowance of approximately 1,700 tons. If one were to assume harbour dues at 6s. per ton (as I believe they are at some of the terminal ports) that would mean that the 12.9 per cent vessel would pay over £500 more than the 13.1 per cent. vessel when it went into harbour. Clearly, this is a penal difference, and naturally, shipbuilders have done their best to see that the engine room occupies more than 13 per cent. of the whole ship.
In the days of reciprocating engines, 13 per cent. and over may have been a normal engine room, but to-day, with the advance in marine engineering, this is no longer so, particularly in the slower vessels. In the circumstances, designers have in some cases been forced to devote a little more space than is really necessary to the engine room and sometimes even to add superstructure connected with the engine room to bring the whole space over 13 per cent. This, of course, does not have any marked effect upon the ship, but it is slightly undesirable from either one or another point of view—either the engine room itself may be unnecessarily large, in which case the sub-division of the hull tends to suffer, or else, if the addition is by way of superstructure, the ship is carrying some additional unnecessary weight on top.
In calculating the net tonnage in this Bill we are proposing to abolish the jump in allowance which occurs at 13 per cent, and instead to allow the proportion of 32 per cent. which the actual size of the engine room bears to 13 per cent. For instance, a 12 per cent. vessel will get 12/13ths of 32 per cent, an 11 per cent. vessel will get 11/13ths of 32 per cent., a 10 per cent. vessel will get 10/13ths of 32 per cent., and so on, instead of one and three-quarter times the actual space, as at present. In fact, the new allowance will be substantially better than the old in every case where engine rooms are less than 13 per cent., and it will be unchanged above that size. I have dealt only with screw steamers, but precisely the same applies to paddle steamers, except that the figures are different, to allow for the normally larger size of paddle steamer engine rooms.
Now in this proposed change in the law there seem to be four different types of parties who may be interested, and all have agreed to the change. First, there are the owners. They want ships to be as safe as possible and as efficient as possible, without sacrificing any comfort in operation. Under this Bill, they will be able to go ahead and build ships incorporating the latest engineering developments, without having to aim at a 13 per cent. engine room the whole time to avoid extra heavy port dues. I am advised that, in some cases, it is expected they will be able to design 11 per cent. or 12 per cent. engine rooms. Furthermore, any saving on engine room may, in certain cases, be available for cargo. The owners welcome the Bill. Second, there are the officers and crew. They want ships to be safe, comfortable to operate, and as efficient as possible. As regards safety, as I have already said, any changes that the Bill makes will make for superior safety, owing to the possibility of better sub-division. I may say here that there is no question at all of the load line, the Plimsoll line, being affected. As regards comfort in the engine room, we have particularly provided that if the Minister's surveyors are not satisfied with the light and air conditions in the engine room the ship will not get the benefit of the new allowance, and, if it has already got it, it will lose it. The officers and crew are, of course, greatly interested, too, in the efficiency of the ship, because upon the efficiency of our merchant marine their livelihood depends. This Bill definitely allows designers to design more efficiently, and the unions representing officers and men have approved the Bill.
The third party are the harbour authorities, and as they stand to lose some revenue from harbour dues they might have objected, but, in the interests of progress, they have approved the proposals. The fourth body are the overseas countries with whom we have reciprocal arrangements for the measurement of tonnage. These arrangements extend to the Commonwealth and eighteen foreign countries, and, of these, all have agreed in principle except the Soviet, who have not replied, and Italy with whom we are in some explanatory correspondence at the moment. We hope that, in due course, the whole world will come into line. It is obviously to the advantage of all other countries to do so because until they have agreed the new arrangements with us, their ships cannot obtain the benefit of them when they come to pay harbour dues in British harbours. Other bodies who have been consulted and who have approved are the Institute of Naval Architects and the Shipbuilding Conference.
The clauses of the Bill are only three in number and they speak for themselves, so that I need not detail them to your Lordships. I think one can, however, congratulate the draftsmen on having produced a technical Bill which is reasonably simple for the layman to understand. If your Lordships pass the Bill, the Minister hopes to bring the new legislation into operation by statutory instrument some time this year. I claim that the Bill is progressive and uncontroversial, and I commend it to your Lordships as one likely to promote the efficiency of our Merchant Navy. I beg to move that the Bill he now read a second time.
Moved, That the Bill be now read 2".—(Lord Hawke.)
4.17 p.m.
I have little criticism to make. The noble Lord, Lord Hawke, has rightly said that the Bill removes a limitation which ship architects and designers have felt for many years. I think it makes for a more sensible arrangement of the propulsive power in ships. In the past, any designer who attempted to put his propulsion machinery aft has been at a great disadvantage. I have been interested in the latest example, the new Shaw Savill liner, which has departed from the practice in large ocean-going liners by having her engines aft. This is claimed, by those who know far more about it than I do, to be the best practice to-day. The noble Lord pointed out that it will enable all ships affected by this Bill to expand their cargo space and, therefore, their earning capacity.
It must be plain to all of us that the British Mercantile Marine is up against a gigantic problem. The cost of building ships to-day is prodigious. I note with concern that the new P. & O. liner "Arcadia," which has been built for the company with which the noble Viscount, Lord Runciman, is so intimately connected, and which starts on her maiden voyage next week, cost approximately the same amount as the "Queen Mary" and the "Queen Elizabeth" and it is less than half the size. How British shipping companies are going to get a return on huge capital outlays like that puzzles me somewhat. When there was a change of Government, I thought it was a good time to have a holiday and went to Australia. The cost of going to Australia two years ago was heavy, but I looked up the cost of going to Australia to-day and, at those rates, I do not think Australia will see me again, not because I have no desire to go there but because of high fares and the activities of the Chancellor of the Exchequer. The shipping industry is up against a big problem and any measure such as this which will help them to save money in design or enable them to make a better contribution to their costs is to be welcomed.
The scope of the Bill is limited to the space occupied by the propelling power, and does not go any further. Speaking as a layman, in face of many experts, I should have thought that common sense would have dictated that the whole question of tonnage measurement should be reconsidered internationally at the earliest opportunity. The noble Lord has said that the agreement of twenty-two countries has been obtained for this modest Bill, and with the example in another sphere of the difficulty of getting international agreement between four countries, I think the Minister of Transport is to be congratulated on obtaining this agreement amongst twenty-two countries—though I am sorry to see that Russia and Italy have not thought fit to come into the arrangement at the present time. There is in existence an Inter-Governmental Maritime Consultative Organisation—a high-sounding title, which is shortened to I.M.C.O.—one of whose tasks is to see whether tonnage measurement cannot be internationally agreed. On the trip to Australia that I have just mentioned, I was amused to see the crew taking down the deck screens at one end of the Suez Canal and putting them up when they came to the other end; and they went through these operations on several other occasions. I was interested to know the reason why, and on inquiry I discovered that it was because covered deck space was charged at one rate in the Canal and at another rate somewhere else; so we altered the design of the ship as we went along, to suit the dues payable. I should have thought it advisable to have an international tonnage measurement. We usually look at the gross tonnage of a ship and then wonder why the registered tonnage is so much less.
We on this side of the House have no serious complaint about the Bill, but I should like the noble Lord to place this suggestion before his right honourable friend the Minister of Transport. Is the Minister quite satisfied that the surveyors have a standard to which to work? Clause 1 (1) (b) of the Bill uses the words: …unless the surveyor of ships is satisfied that the space provided for the working of the boilers and machinery and the ventilation and lighting of that space are adequate. There is no definition in the Bill of the word "adequate," and it appears that the question of what is adequate in the case of lighting and ventilating the engine room is to be left to the judgment of the surveyor concerned. I think I am right in saying that no such wide discrimination is allowed when it comes to sleeping quarters: there the surveyor of a ship has no latitude. The space which a member of the crew is entitled to have is already laid down. Does the noble Lord not think that the powers given by this Bill to surveyors are too wide? I should like him to say something upon that when he winds up. Perhaps he will be able to tell your Lordships whether the Ministry lay down a standard, because, if so, perhaps it is not necessary for it to be included in this Bill. But, at least, I feel that we should have some information about it.
Other than that, we welcome the Bill, and hope that it will be the first of many to remedy these anomalies, and thus make it easier for the ship designer and shipbuilder to compete in world trade. I believe that the noble Lord, Lord Hawke, would agree with me, also, in saying that probably the time has now arrived when the Merchant Shipping Acts should be consolidated. One has only to turn to the last page of this Bill to see that those Acts go right back to 1894. From that year up to 1954 there have been many of them. Does not the noble Lord think it would be as well if consolidation took place now? With those few remarks, and again welcoming the Bill on behalf of noble Lords on this side of the House, I will content myself by congratulating the noble Lord, Lord Hawke, on an excellent representation to your Lordships' House of a Bill which I am sure he could not have known a great deal about.
4.32 p.m.
My Lords, I should like to join with the noble Lord who has just sat down in congratulating my noble friend Lord Hawke or the extreme clarity with, which he moved the Second Reading of a Bill on a remarkably complicated and technical subject. I should also like to confirm, as I think I can fairly do, that the content of this Bill is welcome, not only to the naval architects but also to those whose business is to build ships and to run them. We are happy to feel that noble Lords opposite are in favour of this measure, too. I should like to touch, on two points only, both of which have already been mentioned by the noble Lord, Lord Lucas of Chilworth, though I am afraid that what I may have to say upon them is not, precisely in the same sense as the sentiments to which he has just given utterance.
The first is the point—a new one, so far as I am aware, in legislation of this kind—which leaves it to the discretion of a surveyor to decide whether certain spaces are, or are not, adequate for their purpose. If I remember rightly, the provision in the existing Merchant Shipping Acts is merely that the surveyor shall satisfy himself that the spaces are reasonable in extent, and that they cannot be used for anything else. He is now asked, to go rather further than that, and, while I do not think there is any other satisfactory way round it, one cannot help regretting faintly, as a matter of principle, another piece of legislation the effect of which is to depend upon the judgment of an individual not directly answerable to Parliament, or to anybody else other than his own masters.
On the other hand, I very much doubt, whether it would be a satisfactory way of attempting to get over that point to introduce regulations, by which the surveyor would have to be governed. It is not, if I may say so with respect, the same thing as producing regulations for the amount of air and space which members of the crew should have in their own accommodation. After all, from that point of view, human beings are all very much the same: they may look different in other respects, but, through and by, they breathe about the same amount of air; they require beds of approximately the same length, and will be satisfied by approximately the same amenities. Merchant engines, however, are now, and are likely to be in the future, exceedingly different in their characteristics.
If you seek to compare the appearance of a triple expansion engine, with or without an auxiliary exhaust turbine, a diesel engine, a turbine installation—to say nothing of the more modern proposals of gas turbines, and, for all we know, in the future, steam raised by atomic power, which we may find in ships' engine rooms—I doubt whether any regulations could be devised which would appropriately meet the conditions of engineers tending so wide a range of engines. They give off differing amounts of heat: some are larger than others, and some are smaller than others. As a purely practical point, I am inclined to think that the only successful thing to do—as I believe the shipping industry confidently does—is to place faith in the Ministry of Transport surveyors. We know them to be skilled and impartial people, and to be reasonable and far-sighted. I feel that, on the whole, the interests of both those who will have to work in the engine rooms and those responsible for building the ships—and, indeed, those like the Minister himself, who are responsible for the public safety—will be best served if these things are worked out as they go along, rather than by any attempt to draft in advance regulations on a subject of this kind.
It is perhaps slightly in the same spirit that I should like to speak on one other subject raised by the noble Lord, Lord Lucas of Chilworth. Whilst welcoming the extent of the Bill, he went on to ask whether it would not be a good thing if the whole system of tonnage measurement could be revised. It is perfectly true that the system of measuring a ship's tonnage, as can be gathered, to some extent, from what my noble friend Lord Hawke said originally, is neither a particularly simple, nor in some ways, a particularly logical thing. It may well be that a philosopher from another planet, coming and inquiring how ships were measured, would not only be perplexed to understand it, but would also think it great nonsense. And, speaking as one who has had a good deal to do with the measurement of tonnage, I should be inclined to agree with him. It is a curiously illogical thing which has just grown up from time to time. If your Lordships will pardon the simile, I feel that it is rather like the case of a man who has lived for a great many years with a wife who, perhaps, is neither particularly attractive nor even particularly intelligent; who is slightly given, at times, it may be, to nagging and forcing him to a number of rather peculiar shifts and devices in order to preserve domestic harmony. Nevertheless, if somebody comes along from outside and says, "My good fellow, your wife is a really most unsatisfactory woman. Why do you not get rid of her and have a nice new, reasonable, good-tempered, pretty, sweet, straightforward woman to live with from now on?", curiously enough, though it might be greatly in his interests to do so, the husband is apt to suggest that it is not such a good idea as it might at first sight appear.
The noble Viscount will no doubt agree that that would depend on the extent of the nagging the man had to put up with.
That is true; but I venture to think that if the woman had been really intolerable the husband might have found some way of getting rid of her before. The truth of the matter is that this system of tonnage measurement has grown up over a long time. It contains a number of odd things, but we have got used to it. Except in regard to the particular point which this Bill, happily, is designed to put right, I do not believe there are any particulars in which it so seriously hampers the design and operation of efficient ships that it is worth while going to an immense amount of trouble to change it just for the sake of being tidy.
The noble Lord, Lord Lucas of Chilworth, commented upon the difficulty of getting a great many countries to agree upon anything. I have myself had some slight experience of trying to negotiate amendments to tonnage regulations, and, so far as I can remember, the last one which we did negotiate, which was one connected with the proper allowances for ballast tanks, was finally compromised after something like nineteen years of fairly continuous discussion. If we have to embark upon an international discussion for the complete revision of all this, Heaven knows whether or not our, great-grandchildren will see the answer. Meanwhile, a great deal of trouble will have been gone through. It is true that a number of highly deserving experts may have been provided with an interesting livelihood in the process, but I doubt whether that particular game is worth the candle. I venture to hope that we shall try in the future, as is being attempted with this Bill now, to improve on things as we go along and as the necessity for it becomes apparent, rather than, as I fear will be the case, let the better be the enemy of the good.
Finally, although this may appear a comparatively small matter to-day, and although the number of engine rooms which will be affected is not at the moment particularly large—though as the noble Lord, Lord Lucas said, a growing tendency to place engine rooms aft is likely to be encouraged, and I think desirably encouraged, by this measure—nevertheless, with the developments which are in sight for marine engines, all of which tend to make them more corn-pact and lighter, I think that the knowledge that designers can go ahead without the rather ridiculous handicap of having to try to inflate their engine rooms to an unnecessarily large size will be a stimulus to design and can be nothing but beneficial to designers, builders, and shipowners alike.
4.43 p.m.
My Lords, in welcoming this Bill, perhaps I should declare my interest as a director of a ship-owning company. There is no doubt that the present rules of measurement of net tonnage have had an adverse effect upon modern ship design. With smaller and efficient engines it has become all too apparent, through the years, that some kind of revision was necessary, aid I congratulate Her Majesty's Government on bringing this Bill before your Lordships' House to-day. I understand that this Bill, when it becomes an Act, will come into operation at a date to be approved by the Minister—which will be, I suppose, when the other countries have been able to pass their own legislation on this matter.
As has been mentioned by the noble Viscount, Lord Runciman, the requirement in the Bill that the Ministry's surveyor must be satisfied as to adequate space as regards propelling machinery is a new feature. I, too, think that it would be a mistake to have regulations, because the factors are so variable. Perhaps the noble Lord who is to reply for Her Majesty's Government: will confirm that each case will be determined by the Ministry of Transport from plans supplied of the ship, and that the examination, which of course I agree is necessary, will not be unduly delayed. It has been suggested in some quarters—and, I believe, in another place—that the adoption of the new measurement for engine rooms would adversely affect: the revenue from harbour, pilotage, and light dues, through a reduction in the registered net tonnage, the basis, of course, on which they are paid. The noble Lord, Lord Hawke, rather suggested that there might be a reduction in harbour dues. In cases where the propulsion space is under 13 per cent., I venture to point out that quite the contrary is the case. Shipowners will prefer to have smaller engine rooms in their ships, which will in fact, increase the net tonnage. On the other hand, the increase in the dues payable an the increase in net tonnage will be more than offset by the increased carrying capacity, lower construction costs, and, last but not least, greater safety.
I was rather astonished at some views put forward in another place. It was said that marine underwriters were concerned about the provisions of this Bill because new ships would be able to carry more cargo in view of the smaller engine room. In fact, the whole question of how much cargo and what weight a ship is allowed to carry is governed by the load line convention and the rules made under it: and this Bill in no way affects these rules. There is, of course, no question of reducing the margin of safety in that respect—indeed, the safety factor will be increased, because it will be possible to have a better sub-division of water-tight compartments in these new ships. I, too, support the views put for. ward to-day by the noble Viscount, Lord Runciman, that we should proceed with some caution as regards amendment of the tonnage rules generally. They are of a complex nature: international agreement might be somewhat difficult to obtain; and in any case the rules are not of a very serious nature. I consider this Bill to be an excellent one, and I support it in every way.
4.46 p.m.
My Lords, the noble Lord, Lord Hawke, in commending this Bill, mentioned two points. One was the reduction in harbour dues which will follow. Shipowners to-day have to watch their overheads very carefully, and a Bill which reduces harbour dues and, at the same time, gives a little additional cargo space must be welcome news indeed to shipowners.
The noble Lord will forgive me for interrupting. There will be no reduction in harbour dues, except in certain cases.
At any rate, in those cases there will be. The noble Lord, Lord Hawke, mentioned that fact, and I am sure that it will be welcomed. The noble Lord, Lord Hawke, also mentioned the question of superior safety. I believe it to be true that, as a result of ships having been designed and built with larger engine rooms than were necessary, safety and efficiency may to some extent have been affected; at any rate, the possibility of their being affected was there. A welcome feature of the Bill is that if in future engine rooms are less than 13 per cent. of the gross tonnage, the deduction from the 32 per cent. will now be tapered off, to some extent, instead of being rather abrupt, as was the case before. In reducing the size of the engine rooms, it will be essential that ample working space is provided and there, I take it, the responsibility will be that of the surveyor. The noble Lord mentioned that the Bill is welcome to the unions concerned. I am associated with one union, the Navigators' and Engineer Officers' Union. I have consulted the officials of that union about this Bill, and I am happy to be able to say that they welcome it.
The noble Lord, Lord Lucas of Chilworth, said something about consolidation of the Merchant Shipping Acts. That would be a formidable task upon which to embark. But considering the great technical developments that have taken place in propelling plants in the latter part of this half-century, and as so much greater power is now obtainable, and has for some time been obtainable, from much smaller engines, it is rather remarkable that a reform in the manner envisaged by this Bill has not been carried out long before. After all, the Act which this is to replace is dated, I think, 1894, and I feel that this matter might have been looked at before now. There is another point which I think has not been mentioned this afternoon. Not only is the propelling plant more compact, and takes up less room, but oil fuel has replaced coal and requires less space for stowage. I believe that oil fuel is frequently carried in the double bottom, which is, as I understand it, exempted space.
It may have come as a surprise to many people to know that there are no regulations affecting ventilation and lighting in the engine-room spaces. I noticed what the noble Viscount, Lord Runciman of Doxford, said about the difficulty, in his view, of drawing up regulations affecting the ventilation of engine-room spaces because of the different characteristics and behaviour of various types of plant. I should have thought, however, that it would have been possible to establish some formula relating ventilation to engine-room temperatures, although they vary. This is a technical point, and I am not fully qualified to speak upon it, but on the question of lighting I should have thought that there was no difficulty whatsoever in providing regulations. Ventilation and lighting are very important matters. Great attention has been paid from time to time to ventilation of accommodation for the officers and crews, and I should have thought it was equally important to impose regulations affecting the ventilation and lighting of these engine-room spaces. May I call your Lordships' attention to the fact that engineer officers and men spend a third of their lives on duty in these engine-room spaces and that the ship may be practically continuously employed in tropical or semi-tropical waters? Certainly it is of the utmost importance that they should not have to spend so much of their time in what may be ill-ventilated engine rooms.
In this Bill a great deal of reliance, as has been mentioned by other noble Lords, is placed upon the surveyor. The surveyor must be satisfied that the ventilation and lighting of boiler and machinery spaces are—the word employed is "adequate." It seems to me that the surveyor is given very wide powers indeed if he is the authority to decide whether ventilation and lighting are really adequate. Perhaps we might hear a little more about the surveyor. The Minister may feel able to tell us—as has been suggested to me, I think very pertinently, by the noble Lord, Lord Balfour of Inchrye—a little more about the functions of the surveyor: to whom he is responsible, and so on, so that we may know whether the conditions of his work are such as to make him a reliable judge of what is adequate in these respects. With those few words I should like, on my own behalf and on behalf of the union which I have mentioned, to say once again that we welcome this Bill and that we are grateful to the Government for having brought forward a measure which will remedy some of the anomalies which have perhaps existed for too long.
4.54 p.m.
My Lords, I speak a second time by leave of the House. I should first like to extend my thanks to noble Lords for their kind reception of the Bill and for the remarks they have made indicating that I have managed to explain it with a reasonable degree of clarity. It is always welcome when a Bill meets with approval on all sides of this House. To take the various points that were made, I should like to say first that I think my noble friend Lord Runciman of Doxford answered the noble Lord, Lord Lucas of Chilworth, to some extent on several points, particularly on this question of getting a universal tonnage throughout the world. It presents the most formidable difficulties, because the interested parties are so many and so various that the possibilities of disagreement go up in geometrical (or should it be arithmetical?) progression according to the parties concerned. However, it is an aim; and this Inter-governmental Maritime Consultative Organisation—I.M.C.O.—is a body which will come into operation at some future date, in which case we shall have gone some way towards standard tonnage for the world. I should next deal with the question of consolidation and I do this with considerable trepidation because my noble and learned friend on the Woolsack is the master of consolidation. No doubt he will correct me if I am wrong. I think I can say straight away that the consolidator's time is just not available in the near future for this monumental task—for it is a monumental task. The 1894 Act, as I have said, is one of the longest on the Statute Book.
And there are a good many others knocking at the door.
Before the noble Lord leaves that point, may I say that I appreciate what he has said; but can he give the House an assurance that even if this matter is not within arm's length to knock at the door, it is at least in the queue somewhere?
It is certainly in the mind of my right honourable friend's advisers, but do not think they aspire to a very high place in the queue.
The main point raised by several speakers is this question of the surveyors. In my first speech I made one slight slip which will come out on Lord Winster's side, and that is that should a new vessel of under 13 per cent. not measure up to the standards of light and air which appear proper to the surveyors, the engine room of that vessel will get no allowance at all until it is put right, while an existing vessel would net get the benefit of the new regulations but would, of course, not be deprived of the existing ones which it has had all the time.
Will it be the surveyor who will decide whether the lighting and ventilation are correct to qualify under this Bill?
I will come to that in a moment. In the case of an existing vessel the surveyors are the authorities. They go on board these vessels at frequent intervals—for fire inspection alone once a year is the minimum, and they are out and about on these vessels. They are mostly men of seafaring experience and are highly qualified to judge whether the light and air in an engine room is sufficient to enable the machinery to be worked in comfort by the crew. I was greatly impressed by my noble friend's argument that it is possible to lay down regulations for human beings, who are roughly the same. I do not altogether agree with his statement that they all require the same length of bed; but they certainly require roughly the same amount of air. It is extremely difficult to give it in vessels which are of different sizes and shapes. Her Majesty's Government set great importance on this matter. My right honourable friend has given it a great deal of thought; but he is firmly convinced, and I am sure he is right, that he can do these things better by human judgment than by putting everything down in regulations. Perhaps I may be allowed to remind your Lordships that regulations for a minimum for any purpose nearly always become in the end regulations for a maximum as well. That, I think, would have unfortunate consequences. We are unshaken in our belief that it is better to trust to our surveyors.
On the point made by the noble Lord, Lord Teynham, on the subject of the new vessels and their plans, may I say that they will be given provisional approval by the Ministry, but the final approval will have to be given by the surveyors on the spot. I think one can promise the minimum of delay, as the noble Lord asked.
I hope the noble Lord will forgive my interrupting: I am grateful to him for giving way. On this question of discretion, there could be somewhat of a difference between the views of one surveyor and those of another. I agree with the noble Lord that regulations in this matter can be irksome, but one surveyor may suffer from claustrophobia while another may not. The Minister of Transport is the arbiter. May we have some assurance from the noble Lord that the Minister will give to his surveyors some kind of direction which will be a yardstick to which they may work? If the Minister still thinks that it is far better to leave it to the unfettered discretion of the surveyor, is there machinery for either the representatives of the crew or the representatives of the shipowner to appeal against the surveyor's decision?
The crew or the owner can complain to the Minister. The procedure then would presumably be for the Minister to send down, perhaps two surveyors, or a more senior surveyor, to examine the substance of the complaint. But I do not think that a surveyor who suffered from claustrophobia would long remain on the pay-roll of my right honourable friend the Minister of Transport. These people are practical men, and must remain so.
I am sorry to press the noble Lord again on this point, but surely engine-room temperatures and boiler-room temperatures are recorded in the ship's log, and over a period of a year it is quite possible to determine what is the average temperature in those machinery spaces. Cannot this be related to same question of adequacy of air space for working in those temperatures? It is not a matter of guesswork: the temperatures are recorded.
I am not a seafaring engineer but, having read my Kipling, I should have said that it is by no means certain that the temperature in an engine room is uniform the whole time. There obviously must be wide fluctuations, even in comparison with the temperature of the outer air. If the noble Lord thinks that there are engine rooms which are not up to standard, it is open for the union to complain to the Minister that the engine room of such-and-such a ship is not fit to work in; and a surveyor will then be sent along to deal with the matter. To my mind, that is a far more practical method than sending along a surveyor with a notebook full of figures which might be quite inapplicable in the case in question.
The only other point which I record as having been raised is this question of harbour dues. I think there is an opening for difference of opinion as to whether in fact the total of harbour dues in the future will be more or less than at present. My own speculative estimate of the future—this is not Her Majesty's Government's estimate; it is a purely personal one—is that there should be a slight decrease in the incidence of harbour dues per ton on cargo brought into the ports of this country. Whether those cargoes will be carried in more or in fewer vessels I do not know. It is obviously not a matter on which one can dogmatise. I repeat, I am most grateful for the reception given to this Bill by your Lordships. I feel sure that you, will accord it a Second Reading.
5.5 p.m.
My Lords, before the Second Reading is given, may I ask the noble Lord, Lord Hawke, a question (I have listened to the debate, as have other noble Lords) on the powers given to the surveyor? It was not quite clear from the noble Lord's two speeches that the survey ors are, in fact, employees of the Ministry of Transport—I understand that that is the case. Therefore, when the Minister, if there is some conflict between the owners, or the crew, and the surveyors, says that further surveyors will be sent down, it means, in fact, that he is sending down again officials from exactly the same Department. So the executive is given great powers. I think the Minister made a most convincing case that this is a matter that cannot be defined by regulation, but if we are to give the executive —that is to say, the surveyors employed by the Minister—the power to decide whether this concession should or should not be granted—and presumably that would be decided only after the ship had been constructed, not at the stage of the plans—
May I interrupt? I said that the approval in principle was given at the planning stage, but the final approval only on the ground, so to speak.
The final approval must be given on the ground. Would the Minister ask his right honourable friend, between now and the further stages of the Bill, whether he would consider the provision in the Bill of some administrative machinery whereby there could be an appeal to some impartial tribunal, or someone other than the Minister's own employees, in respect of any fundamental difference as to what is satisfactory ventilation and lighting between, on the one side, those who own the ship or travel in it and, on the other, those employees of the Government who do the surveying? I merely ask him to consider that suggestion between now and the next stage.
I will certainly look into the suggestion of my noble friend, but I think he is making rather a mountain out of a molehill, because, so far as I am aware, these matters have never failed to be settled amicably. In the last resort, the Minister is responsible to Parliament, and if, in the case of some ship or other, there is a complete inability to reconcile the two sides, the question could certainly be raised in Parliament. I feel certain that an agreement would result. Nevertheless, I will certainly bring my noble friend's suggestion to the attention of my right honourable friend.
On Question, Bill read 2ª; and committed to a Committee of the Whole House.
House adjourned at nine minutes past five o'clock.
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# gridDistance
0th
Percentile
##### Distance on a grid
The function calculates the distance to cells of a RasterLayer when the path has to go through the centers of neighboring raster cells (currently only implemented as a 'queen' case in which cells have 8 neighbors).
The distance is in meters if the coordinate reference system (CRS) of the RasterLayer is longitude/latitude (+proj=longlat) and in the units of the CRS (typically meters) in other cases.
Distances are computed by summing local distances between cells, which are connected with their neighbours in 8 directions.
Keywords
spatial
##### Usage
# S4 method for RasterLayer
gridDistance(x, origin, omit=NULL, filename="", ...)
##### Arguments
x
RasterLayer
origin
value(s) of the cells from which the distance is calculated
omit
value(s) of the cells which cannot be traversed (optional)
filename
character. output filename (optional)
...
additional arguments as for writeRaster
##### Details
If the RasterLayer to be processed is big, it will be processed in chunks. This may lead to errors in the case of complex objects spread over different chunks (meandering rivers, for instance). You can try to solve these issues by varying the chunk size, see function setOptions().
##### Value
RasterLayer
See distance for 'as the crow flies' distance. Additional distance measures and options (directions, cost-distance) are available in the 'gdistance' package.
##### Aliases
• gridDistance
• gridDistance,RasterLayer-method
##### Examples
# NOT RUN {
#world lon/lat raster
r <- raster(ncol=10,nrow=10, vals=1)
r[48] <- 2
r[66:68] <- 3
d <- gridDistance(r,origin=2,omit=3)
plot(d)
#UTM small area
crs(r) <- "+proj=utm +zone=15 +ellps=GRS80 +datum=NAD83 +units=m +no_defs"
d <- gridDistance(r,origin=2,omit=3)
plot(d)
# }
Documentation reproduced from package raster, version 3.0-12, License: GPL (>= 3)
### Community examples
Looks like there are no examples yet.
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# Geometric interpretation of the exact sequence for the cotangent bundle of the projective space
Edit: As Dan Petersen pointed out, this question is a duplicate of a previous one. I would leave it for the moderators to decide if this should be closed. On the other hand, may be this should be left open on the merit of the excellent answers and comments (@Emerton: Thanks!).
I was trying to understand the following exact sequence (for $X := \mathbb{P}^n_k$, where $k$ is an algebraically closed field): $$0 \to \Omega_X \to \mathcal{O}_X(-1)^{n+1} \to \mathcal{O}_X \to 0$$ The explanation (as in the proof of Theorem II.8.13 of Hartshorne) is given by some algebraic formulae, which I am having trouble to digest. I was trying to see in more geometric terms what is going on, and was somewhat successful in the case of the surjection $\mathcal{O}_X(-1)^{n+1} \to \mathcal{O}_X$, namely: we can regard $\mathcal{O}_X(1)$ (respectively $\mathcal{O}_X(-1)$) as the normal bundle $\mathcal{N}$ of (respectively conormal bundle) of $X$ in $Z := \mathbb{P}^{n+1}_k$. Any global section of $\mathcal{O}_X(1)$ therefore induces a map (via evaluation) from $\mathcal{O}_X(-1)$ to $\mathcal{O}_X$. The above surjection comes from taking $n+1$-linearly independent global sections of $\mathcal{O}_X(1)$.
But I do not understand how to interpret the injection $\Omega_X \to \mathcal{O}_X(-1)^{n+1}$. How would someone 'naturally' come up with the algebraic formula?
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I am definitely not an expert, but what happens if you realize the conormal bundle as a sub bundle of the cotangent bundle? Don't you find that the cotangent bundle of the embedded variety gets realized as a piece of the conormal? – Filippo Alberto Edoardo Jun 2 '12 at 4:53
There is a quotient map $\pi:\mathbb A^{n+1} \setminus \{0\} \to \mathbb P^n$, and so we can pull back a differential form $\omega$ on $U \subset \mathbb P^n$ to a differential form $\pi^* \omega$ on $\pi^{-1}(U)$. The sheaf of differential forms on $\mathbb A^{n+1} \setminus \{0\}$ is globally free, with a basis $dx_0, \ldots, dx_n$. If you think about it, you can then interpret the pull-back map as an embedding $\Omega_{\mathbb P^n} \hookrightarrow \mathcal O(-1)^{n+1}$. Not every section of $\mathcal O(-1)^{n+1}$ will give rise to a differential form, though --- when you interpret such ... – Emerton Jun 2 '12 at 4:58
... a section as a differential form on $\mathbb A^{n+1}\setminus \{0\}$, to come from $\Omega_{\mathbb P^n}$ is has to be invariant under the scaling action of $\mathbb G_m$. This can be tested by pairing with the Euler class $x_0\partial_{x_0} + \cdots + x_n\partial_{x_n}$, and this gives the map $\mathcal O(-1)^{n+1} \to \mathcal O$ which appears in the exact sequence. – Emerton Jun 2 '12 at 5:00
You might be interested to know that Ravi Vakil has a short but sweet exposition of this exact sequence and its various generalizations in section 23.3 of his notes (math.stanford.edu/~vakil/216blog/FOAGmay1612public.pdf) – Sam Lichtenstein Jun 2 '12 at 5:58
This is an exact duplicate of mathoverflow.net/questions/5211 – Dan Petersen Jun 2 '12 at 12:17
By dualizing and twisting we obtain the equivalent exact sequence of vector bundles
$$0\to \tau\to \mathbb P^n_k\times k^{n+1} \to T_{\mathbb P^n}(-1)\to 0 \quad (*)$$ The first morphism is just the inclusion of the tautological vector bundle $\tau$ into the trivial bundle and is geometrically transparent.
To understand the second morphism geometrically, fix a point $p\in \mathbb P^n_k$ and the corresponding line $l\subset \mathbb P^n_k$ (I forgot to say I'm using the pre-Grothendieck definition of projective space as a set of lines) .
At $p$ the exact sequence $(*)$ becomes the exact sequence of vector spaces$$0\to l\to k^{n+1} \to T_{\mathbb P^n}[p]\otimes l\to 0$$
Exactness then translates into the canonical isomorphism $$T_{\mathbb P^n}[p] = \mathcal L(l,k^{n+1}/l) \quad (**)$$
So the whole problem boils down to understanding $(**)$, i.e.understanding in a canonical way the fiber of the tangent bundle to $\mathbb P^n$ at a point $p=(a_0....:a_n)\in \mathbb P^n$.
Here is the idea inspired by differential geometry.
The "curve" $\epsilon \mapsto (a_0+\epsilon t_0,....,a_n+\epsilon t_n)\; (\epsilon^2=0)$ [algebraic geometers consider very short curves!] gives rise to a tangent vector $t\in T_{\mathbb P^n}[p]$.
The canonically associated linear map $\lambda _t:l\to k^{n+1}/l$ is then characterized by the condition $$\lambda _t(a_0,...,a_n)=\overline {(t_0,...,t_n)}$$
[Be careful that if you change the vector $(a_0,...,a_n)$ representing $p$ to a colinear vector $(a_0',...,a_n')$, you also have to change $(t_0,...,t_n)$ to another $(t_0',...,t_n')$]
The details are in Dolgachev's online notes , Example 13.2
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Thanks! This is pretty close to what I wanted. – auniket Jun 2 '12 at 19:38
Here is another (unknown?) way of optaining the Euler sequence (though not really geometric): Since $\Omega^1_{\mathbb{P}}$ is a coherent sheaf; by Serre it has a "twisted presentation". For that one has to find some $k > 0$ such that $\Omega^1_{\mathbb{P}}(k)$ is generated by global sections. You will find that $k=2$ suffices, namely there is an epimorphism $\bigoplus_{u<v} \mathcal{O}(-2) \twoheadrightarrow \Omega^1$, which is given on $D_+(x_i)$ by mapping
$$x_i^{-2} e_{uv} \mapsto \dfrac{x_u}{x_i} \cdot d\left(\dfrac{x_v}{x_i}\right)- \dfrac{x_v}{x_i} \cdot d\left(\dfrac{x_u}{x_i}\right).$$
You can also compute the relations between these elements and arrive at the exact sequence
$$\bigoplus_{u<v<w} \mathcal{O}(-3) \to \bigoplus_{u<v} \mathcal{O}(-2) \to \Omega^1 \to 0.$$
But now the (graded) Koszul resolution of $R[x_0,\dotsc,x_n]/(x_0,\dotsc,x_n)$ (here $R$ is an arbitrary base ring; it doesn't have to be an algebraically closed field) yields the long exact sequence
$$\dotsc \to \bigoplus_{u<v<w} \mathcal{O}(-3) \to \bigoplus_{u<v} \mathcal{O}(-2) \to \bigoplus_{u} \mathcal{O}(-1) \to \mathcal{O} \to 0.$$
These combine to the Euler sequence $0 \to \Omega^1 \to \bigoplus_{u} \mathcal{O}(-1) \to \mathcal{O} \to 0$.
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# Thread: Find The Y Intercept - Urgent
1. ## Find The Y Intercept - Urgent
I need serious help.
I need to find the Y intercept of the graph
$y= -0.02(x^3 - 38x^2 + 441x - 1620)$
I also need to X when y=3
I also need to find the equation of the inclined path if it hasa gradient of 4/5
2. for the y-intercept, just plug in x = 0
3. Awesome. Any help with the others?
4. Originally Posted by mibamars
$y= -0.02(x^3 - 38x^2 + 441x - 1620)$
I also need to X when y=3
$y= -0.02(x^3 - 38x^2 + 441x - 1620)$
when $y = 3$ we have
$3= -0.02(x^3 - 38x^2 + 441x - 1620)$
$\Rightarrow x^3 - 38x^2 + 441x - 1470 = 0$
Trying the factors of 1470, we realize that x = 14 is a root, so (x - 14) is a factor. Dividing the cubic by (x - 14) using long division, we obtain:
$x^3 - 38x^2 + 441x - 1470 = (x - 14) \left( x^2 - 24 x + 105 \right) = 0$
$\Rightarrow x - 14 = 0 \mbox { or } x^2 - 24x + 105 = 0$
$\Rightarrow x = 14 \mbox { or } x = \frac {24 \pm \sqrt {156}}{2}$
Originally Posted by mibamars
I also need to find the equation of the inclined path if it hasa gradient of 4/5
when you say inclined path you mean tangent line right?
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# Do Adjustable Hosels Really Change Loft?
## Recommended Posts
I've read and listened to all the talk about adjustable necks on drivers and other clubs being able to change the loft of the club, some by a number of degrees. However, for the life of me I can not visualize how a straight shaft into an adjustable neck can change the loft of the club face. I have Ping G, I changed the adjustment but don't really see a change to loft. I can understand fade or draw bias but not loft. Can someone in laymen's terms explain?
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It's actually very simple. Picture in your mind a traditional bonded hosel driver from 15 years ago let's say. Now look at the club from the butt end staring down the shaft with the head sitting flat in address position. So you are looking at the butt of the grip in the foreground and you can see the clubhead in the background. Now imagine that the butt end of the grip is the center of a clock face (ie where the hands are connected to the clock face in the center).
All the adjustable hosel is doing is moving the direction the hosel points. So you take a modern adjustable driver and place it into your imagined clock setup. Head in the same position on the ground. However now in the neutral position the grip butt is now pointing towards 12 o'clock (ie more upright lie angle). Modern drivers are more upright in general. If we rotate the hosel 180 degrees, the grip butt will now be pointing towards 6 o'clock and the lie angle will be flatter.
Now if we rotate the hosel to change the loft, all we are doing is rotating the hosel so that the shaft is now leaning either towards or away from your target. On our clock diagram this means the hosel is pointing towards 9 o'clock if we are wanting to increase loft and 3 o'clock if we are wanting to lower loft. Of course this is assuming a right handed golfer and club. You may have heard that the adjustable hosels don't actually change the loft, and this is why some people say that. After you rotate the hosel and have the shaft pointing more in this 9 o'clock/ 3 o'clock plane, the head remains sitting flat on the ground until you manually move the shaft back so that it points at the center of the face again (like with our initial bonded hosel driver). So if the shaft was pointing at 9 and we move it back to center and keep the face square to the target, we have just increased the loft. Likewise if we take the shaft that had been pointing at 3 o'clock and again move it back to center keeping the face aligned to the target, we have now lowered the static loft.
Of course the other result of this is when we loft up and hold the shaft back in that neutral position, most drivers will want to fall closed because of the weighting and sole design. Likewise, a delofted driver held in that neutral shaft position will fall open. Not all drivers do this depending on design, and as most golfers align the head manually towards the target, it not usually an issue.
This is the basis of how all these adjustable hosels work. Moving the hosel to point more towards 9 or 3 will give the greatest loft change. Moving it in the 12/6 o'clock direction will change the lie angle. Moving it somewhere in between will give lesser loft change along with a lie angle change.
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7 hours ago, cooke119 said:
However, for the life of me I can not visualize how a straight shaft into an adjustable neck can change the loft of the club face
It doesn't go straight in.
If you want the more complicated version, read the @Adam C post. 😄
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Here is my understanding, hopefully correct and maybe simpler.
In reality an adjustable hosel doesn’t change the loft of a driver.
What it really does is open or close the face. It is still the same loft. As Adam C noted above, when you turn the driver to face the target, you will with increase the face angle, or loft, (clockwise), or decrease the face angle, or loft, (counter-clockwise).
Because of how the shaft comes out of the head, this allows you to square the head to change the face angle.
John
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36 minutes ago, billchao said:
Bill,
This is the video I watched years ago that explained it. Thanks for posting it.
John
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8 hours ago, billchao said:
Thanks for posting. My only issue is I rarely have a tee box that is as level as a golf shop floor. So soling the club is not as easy as he states. Also, isn’t bending the driver hosel doing the same thing? The sole of the head isn’t changing, so its relationship to the ground will change.
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Gaaahhhhhhh those pleated khakis!!!
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9 hours ago, billchao said:
I like a lot of what Wishon discusses in videos and articles but not this one. This is just a backhanded attempt to push his Wishon clubs bendable hosels while make arguments that just don't make logical sense.
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1 hour ago, boogielicious said:
Thanks for posting. My only issue is I rarely have a tee box that is as level as a golf shop floor. So soling the club is not as easy as he states. Also, isn’t bending the driver hosel doing the same thing? The sole of the head isn’t changing, so its relationship to the ground will change.
That’s true of any club you sole, though. Rarely will you have a perfectly level lie. It makes you wonder about whether one or two degrees of adjustment on your clubs actually makes a significant difference. We’re not robots.
23 minutes ago, Adam C said:
I like a lot of what Wishon discusses in videos and articles but not this one. This is just a backhanded attempt to push his Wishon clubs bendable hosels while make arguments that just don't make logical sense.
I think he does a good job of explaining how adjustable hostels work, but I agree he muddles things up a bit trying to promote his own products.
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Thanks everyone, I think it is getting through my thick head, the video helped but confused me a little about having to adjust the face angle at address, and you're correct I have never been in a tee box where the ground is perfectly flat. However, I try not to ground the club right before my swing anyway. I'll read the posts and watch the video again to help clear it up.
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22 hours ago, colin007 said:
Gaaahhhhhhh those pleated khakis!!!
Jim Harbaugh loves them!
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I’ve recently bought a Ping G410 driver, kept it on neutral first few weeks , changed it to flat.
Hard to see much difference ( and on shot shape /direction etc)
Compared to other Ping drivers I’ve had, at address it always ‘looks’ a bit open as opposed to previous models.
Its definitely more upright , the toe sticks way up in the air, mind you standard irons stick toe up for me too, I was measured as an Orange dot 2* flat many Ping Eye 2 moons ago
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I don't know when people got so pissed off at knowledge! How many club designs have you made? How many pro golfers have you designed clubs for? I've been hearing for years that when you up the loft on these adjustable drivers, you close the face! So, if you want to realize more loft, you have to open up the face from where it naturally soles.
That being said, I've never been a naturally soled Driver hitter! I want that face to look square. But when you do that with these modern clubs, you are actually closing the face! That is, if you have the loft adjusted upward. And getting the ball more up in the air is something that many of us could use.
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12 hours ago, Buckeyebowman said:
That being said, I've never been a naturally soled Driver hitter! I want that face to look square. But when you do that with these modern clubs, you are actually closing the face!
I think you’re misunderstanding how adjustable drivers work. The change in loft only presents itself if you square the face. And if you square the face, it’s not closed...
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4 minutes ago, billchao said:
I think you’re misunderstanding how adjustable drivers work. The change in loft only presents itself if you square the face. And if you square the face, it’s not closed...
Additionally, not all drivers worked that way. Some independently adjusted loft and face angle.
I think the Nike Covert was the first to independently adjust loft/face angle, but others have done so recently IIRC.
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I did a little photo exercise with two identical Titleist 917 D2 drivers with 10.5 loft. It shows the face angle change at address with the club soled lying on a flat floor.
Here is standard. The lie slightly open to my eye.
Here is same loft but 2 degrees upright and one flat. The upright is slightly closed to my eye and the B1 a bit more open than A1.
Extremes: 12.5, 2 degrees upright and 9.5 1 degree flat. Left is closed, right is the most open.
The opposite, 9.5 2 degrees upright, 12.5 1 degree flat. I didn’t see much of a difference but right was slightly more open. The photo is off a bit on angle.
But most importantly, I aim the face and don’t rely on the way it looks soled. The tee box is never flat and I open the face at address because my swing path is in to out.
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This is a copy of my original post on Kudelski Security’s Research Blog for my archives.
Have you ever heard of Functional Encryption (FE)? If so, you may be associating it with some sort of homomorphic encryption, which is not wrong, but not exactly right neither.
Let us see today what FE is along with a few examples, roughly how it differs from Fully Homomorphic Encryption, and how the FENTEC project is working on it!
Now, let us define what we mean when we are talking about FE. It is only recently, in 2010, that Dan Boneh, Amit Sahai and Brent Waters formalized the notion of functional encryption. We can roughly describe FE by saying it is a public-key encryption scheme with different decryption keys allowing a user to learn specific functions of the encrypted data.
So, in an FE scheme for function $F(\cdot, \cdot)$, an authority holding a master key generates a key $s_k$ that enables the computation of the function $F(k, \cdot)$ on encrypted data, so that an evaluator knowing a ciphertext $c$ of the data $x$ and the key $s_k$ is able to compute $F(k,x)$, but without being able to learn anything more than the result of the function evaluation about $x$.
### FE vs FHE, why the ache H?
If you are familiar with the concept of (Fully) Homomorphic Encryption, (F)HE, it is interesting to draw the following parallel:
With FHE one can compute arbitrary functions on encrypted data without decrypting that data. This is very interesting for delegating computation to untrusted third parties. The downside (depending on your use case) of FHE being that the result is still encrypted and thus needs to be sent back to the entity holding the private key in order to decrypt the result of the computation. We can somehow represent the FHE process as follows, with $E$ being an encryption scheme and $F$ being the function we want to compute on the encrypted data:
$$E(x_1), E(x_2), ..., E(x_n) \rightarrow E(F(x_1,x_2, \cdots, x_n))$$
On the other hand, with FE, the result is directly accessible in plaintext after computation, which means we could see it as follows:
$$E(x_1), E(x_2), \cdots , E(x_n) \rightarrow F(x_1,x_2, \cdots, x_n)$$
In some sense, we can see FE as being a scheme that does both evaluation and decryption of the result of a function at the same time, without leaking the private key allowing to decrypt the data and without leaking information about the plaintexts $x_1, x_2, \cdots, x_n$ other than the result of their evaluation by the function $F$.
Obviously, we don't want everyone to be able to evaluate any function they might want to, as otherwise it would be easy to leak information about individual plaintexts. (Look! I'd like to evaluate the identity function, pretty please.) Hence, only the private key holder is actually able to decrypt $E(x_i)$, and only they can generate the evaluation keys for specific functions of their choice. This means that functional encryption requires a "central authority" to issue the "evaluation key" to the parties in charge of the functional evaluation.
So, functional encryption per se is a generalization of the notion of public-key encryption, which allows users to delegate to third parties the computation of certain classes of functions of the encrypted data by generating specific secret keys for these functions. Unlike standard encryption schemes, it allows for a more fine-grained control of the decryption capabilities of third parties.
Functional Encryption is extremely useful since it basically allows us to intentionally leak some information about encrypted data to chosen users. For example, we could get the average value of a set of encrypted data without revealing the data itself, or we could go further and obtain even more statistical data about an encrypted dataset. With the current privacy concerns and requirements raised by new laws such as GDPR, the need for efficient FE schemes appears even more clearly as it would allow some processing on encrypted data to be done by certain third parties without leaking the actual plaintexts to anyone. This means we could go further than pseudonymization of personal data, to ensure some kind of stricter confidentiality! The FENTEC blog also has an entry about cryptography and laws if that's interesting to you.
### The FENTEC project
Now, you may wonder: what is this FENTEC project he's talking about? Well, it is a project that received funding from the European Union’s Horizon 2020 research and innovation programme under grant agreement No 780108, and whose goal is to further develop Functional Encryption Technologies as an alternative to the all-or-nothing approach of traditional encryption systems. (That is: either you have encryption, or you don't, there were no middle ground until Functional Encryption.)
The project brings together multiple universities and industrial partners in order to design and implement efficient, innovative FE systems which are application oriented and can be used in a wide range of scenarios. Kudelski Security's research team is glad to be one of the industrial partners taking part in the FENTEC project, which will last 36 months. The project officially started in January 2018 and will thus end in December 2020.
### Back to FE
Due to its generality, functional encryption encompasses and unifies many other advanced encryption schemes that used to be studied independently, such as identity-based encryption, searchable public-key encryption, hidden-vector encryption, identity-based encryption with wildcards, attribute-based encryption, and inner-product functional encryption.
While FE schemes are still very young, a lot of things happened since 2010 and there are now quite a few interesting schemes allowing for functionalities that seemed hard to achieve 8 years ago. It has now come to a point where certain cryptography conferences even have a session named "Functional Encryption"!
Let us take a look at several different types of functional encryption schemes. For example:
• Inner product functional encryption (IPFE) schemes, where the plaintext is a vector and the encrypted data can be used along with an evaluation key to compute the inner product of the said vector with another vector. There are multiple variants of IPFE actually: multi-clients, multi-inputs, decentralized, function hiding, etc.
• Attribute-based encryption (ABE) schemes, where the encrypted data is linked with a set of attributes and secret keys along with certain policies that allow to control which ciphertexts can be decrypted depending on the attributes we possess.
• "General-purpose" FE schemes, that allow to evaluate any kind of function $f$ (or circuit, depending on the scheme) on the encrypted data $\mathrm{Enc}(x)$. (I wish we'd call these "Fully Functional Encryption schemes", even if they don't really work in practice)
However, it is also important to notice here that while there has been a lot of work focused on the theoretical aspects of FE in order to go as far as we can, all the general-purpose FE schemes are currently too inefficient for practical usages. This is also one of the focuses of the FENTEC project: making FE usable in practice by designing and implementing practical schemes that can be used in industrial use-cases. This objective is supported by the design and implementation of new schemes, allowing for richer functionalities and practical instantiations, but also by designing dedicated co-processors that can further accelerate the required computations, in order to be able to bridge the theory with the practice. You can read more about the hardware components of the project in this FENTEC blog post.
## Wanna use FE today? Please, do.
But what if you want to use FE tomorrow? Well, that shouldn't be a problem! Indeed, in the frame of the FENTEC project, the team at XLAB has been implementing many of the schemes designed by the partner universities in a C library "CiFEr" and a Go library "GoFE", that are ready to go!
You can read more about the FENTEC libraries on the FENTEC blog, or you can go directly to Github to start playing with CiFEr & GoFE. By the way, guess what? We tested it, and it even runs in your browser using WASM, shall you so wish!
There even are a couple examples available on the FENTEC's Github repositories, namely:
### The gory details: the guts of a FE scheme
Among currently implemented schemes you can find a lot of "Inner Product Functional Encryption" ones. But what does it mean? Well, exactly what it says: these are schemes that allow a third party to compute the inner product of two vectors using FE.
Let's say that you would like to encrypt a given vector $a$ and obtain its inner product with a vector $y$. First, we need to have a central authority in order to implement functional encryption.
In this case, the central authority would issue a "master public key" $mpk$, as well as an evaluation key $z_y$ for a given vector $y$. Then, anybody knowing the public key could encrypt a vector $a$, allowing any third party in possession of the evaluation key $z_y$ would then be able to compute the values of $\langle a, y \rangle$ when provided with $E_{mpk}(a)$, without learning anything about the vector $a$.
However, note that in this FE scheme, the vector $y$ corresponding to the evaluation key $z_y$ has to be known by the third party in order to evaluate the inner product. That is: only the vector $a$ is remaining secret.
Now, what if you want both vectors $a$ and $y$ to remain secret while you still want that third party to evaluate their inner product?
Thankfully, this is a field of research that has also known tremendous improvement over the past years and is named "FE with function hiding". Basically, an inner product encryption scheme is "function-hiding" if the keys and ciphertexts reveal no additional information about both vectors $a$ and $y$ beyond their inner product $\langle a,y\rangle$. Newer inner product FE schemes are more and more featuring function hiding.
## To wind-up
While it is still young, Functional Encryption has many interesting usages we can foresee. Among them, one that seems particularly interesting to us is the possibility to move the decision making process based on end-to-end encrypted data from the back-end systems to some gateways in complex networks. We call this "local decision making", and being able to enable gateways to perform such local decision making is a big step forward to secure IoT networks and other highly decentralized networks that might want to implement end-to-end encryption, without losing too much decision capabilities at the gateways level.
At Kudelski Security, in the frame of the FENTEC project, we are currently working on a prototype leveraging an inner product FE scheme in order to detect motion in a video stream between a camera and a backend system at the gateway level, using the so-called "motion vectors" that are present in the H.264/MPEG-4 AVC standard.
Notice how motion vectors are good candidates for application of an Inner Product Functional Encryption schemes, as Inner Products are defined on vectors spaces! We are still exploring the best movement detection methods, but are looking forward to having a fully working prototype leveraging FE to detect motion by the end of next year!
In case you'd like to stay tuned for more about FE, FENTEC and motion detection, don't hesitate to follow us on Twitter, or to follow the FENTEC project!
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# How does regular exercise help make us more alert and energetic?
###### Question:
How does regular exercise help make us more alert and energetic?
### How many sixteenths are in 15/16?
How many sixteenths are in 15/16?...
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• # question_answer Each stands for 1 unit. The weight of is ______ units. A) 5 B) 3 C) 1 D) 2
Solution :
We have, But weight of = 1 unit Hence, weight of = 2 units
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## G = D20⋊21D4order 320 = 26·5
### 9th semidirect product of D20 and D4 acting via D4/C22=C2
Series: Derived Chief Lower central Upper central
Derived series C1 — C2×C10 — D20⋊21D4
Chief series C1 — C5 — C10 — C2×C10 — C22×D5 — C23×D5 — D5×C22⋊C4 — D20⋊21D4
Lower central C5 — C2×C10 — D20⋊21D4
Upper central C1 — C22 — C22⋊Q8
Generators and relations for D2021D4
G = < a,b,c,d | a20=b2=c4=d2=1, bab=a-1, cac-1=dad=a9, cbc-1=dbd=a18b, dcd=c-1 >
Subgroups: 1510 in 334 conjugacy classes, 107 normal (43 characteristic)
C1, C2 [×3], C2 [×9], C4 [×2], C4 [×8], C22, C22 [×2], C22 [×27], C5, C2×C4 [×2], C2×C4 [×4], C2×C4 [×13], D4 [×18], Q8 [×2], C23, C23 [×15], D5 [×7], C10 [×3], C10 [×2], C42, C22⋊C4 [×2], C22⋊C4 [×10], C4⋊C4, C4⋊C4 [×2], C4⋊C4, C22×C4, C22×C4 [×5], C2×D4 [×13], C2×Q8, C4○D4 [×4], C24 [×2], Dic5 [×3], C20 [×2], C20 [×5], D10 [×4], D10 [×21], C2×C10, C2×C10 [×2], C2×C10 [×2], C2×C22⋊C4 [×2], C4×D4 [×2], C22≀C2 [×2], C4⋊D4 [×3], C22⋊Q8, C22.D4 [×2], C4.4D4, C22×D4, C2×C4○D4, C4×D5 [×8], D20 [×4], D20 [×12], C2×Dic5 [×3], C5⋊D4 [×2], C2×C20 [×2], C2×C20 [×4], C2×C20 [×2], C5×Q8 [×2], C22×D5, C22×D5 [×4], C22×D5 [×10], C22×C10, D45D4, C4×Dic5, C10.D4, D10⋊C4, D10⋊C4 [×8], C23.D5, C5×C22⋊C4 [×2], C5×C4⋊C4, C5×C4⋊C4 [×2], C2×C4×D5, C2×C4×D5 [×4], C2×D20 [×2], C2×D20 [×6], C2×D20 [×4], Q82D5 [×4], C2×C5⋊D4, C22×C20, Q8×C10, C23×D5 [×2], D5×C22⋊C4 [×2], C22⋊D20 [×2], D208C4, D10.13D4 [×2], C4⋊D20, C4⋊D20 [×2], C4×C5⋊D4, C20.23D4, C5×C22⋊Q8, C22×D20, C2×Q82D5, D2021D4
Quotients: C1, C2 [×15], C22 [×35], D4 [×4], C23 [×15], D5, C2×D4 [×6], C4○D4 [×2], C24, D10 [×7], C22×D4, C2×C4○D4, 2+ 1+4, C22×D5 [×7], D45D4, D4×D5 [×2], Q82D5 [×2], C23×D5, C2×D4×D5, C2×Q82D5, D48D10, D2021D4
Smallest permutation representation of D2021D4
On 80 points
Generators in S80
```(1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20)(21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40)(41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60)(61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80)
(1 15)(2 14)(3 13)(4 12)(5 11)(6 10)(7 9)(16 20)(17 19)(22 40)(23 39)(24 38)(25 37)(26 36)(27 35)(28 34)(29 33)(30 32)(41 57)(42 56)(43 55)(44 54)(45 53)(46 52)(47 51)(48 50)(58 60)(62 80)(63 79)(64 78)(65 77)(66 76)(67 75)(68 74)(69 73)(70 72)
(1 39 52 79)(2 28 53 68)(3 37 54 77)(4 26 55 66)(5 35 56 75)(6 24 57 64)(7 33 58 73)(8 22 59 62)(9 31 60 71)(10 40 41 80)(11 29 42 69)(12 38 43 78)(13 27 44 67)(14 36 45 76)(15 25 46 65)(16 34 47 74)(17 23 48 63)(18 32 49 72)(19 21 50 61)(20 30 51 70)
(1 69)(2 78)(3 67)(4 76)(5 65)(6 74)(7 63)(8 72)(9 61)(10 70)(11 79)(12 68)(13 77)(14 66)(15 75)(16 64)(17 73)(18 62)(19 71)(20 80)(21 60)(22 49)(23 58)(24 47)(25 56)(26 45)(27 54)(28 43)(29 52)(30 41)(31 50)(32 59)(33 48)(34 57)(35 46)(36 55)(37 44)(38 53)(39 42)(40 51)```
`G:=sub<Sym(80)| (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20)(21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80), (1,15)(2,14)(3,13)(4,12)(5,11)(6,10)(7,9)(16,20)(17,19)(22,40)(23,39)(24,38)(25,37)(26,36)(27,35)(28,34)(29,33)(30,32)(41,57)(42,56)(43,55)(44,54)(45,53)(46,52)(47,51)(48,50)(58,60)(62,80)(63,79)(64,78)(65,77)(66,76)(67,75)(68,74)(69,73)(70,72), (1,39,52,79)(2,28,53,68)(3,37,54,77)(4,26,55,66)(5,35,56,75)(6,24,57,64)(7,33,58,73)(8,22,59,62)(9,31,60,71)(10,40,41,80)(11,29,42,69)(12,38,43,78)(13,27,44,67)(14,36,45,76)(15,25,46,65)(16,34,47,74)(17,23,48,63)(18,32,49,72)(19,21,50,61)(20,30,51,70), (1,69)(2,78)(3,67)(4,76)(5,65)(6,74)(7,63)(8,72)(9,61)(10,70)(11,79)(12,68)(13,77)(14,66)(15,75)(16,64)(17,73)(18,62)(19,71)(20,80)(21,60)(22,49)(23,58)(24,47)(25,56)(26,45)(27,54)(28,43)(29,52)(30,41)(31,50)(32,59)(33,48)(34,57)(35,46)(36,55)(37,44)(38,53)(39,42)(40,51)>;`
`G:=Group( (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20)(21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80), (1,15)(2,14)(3,13)(4,12)(5,11)(6,10)(7,9)(16,20)(17,19)(22,40)(23,39)(24,38)(25,37)(26,36)(27,35)(28,34)(29,33)(30,32)(41,57)(42,56)(43,55)(44,54)(45,53)(46,52)(47,51)(48,50)(58,60)(62,80)(63,79)(64,78)(65,77)(66,76)(67,75)(68,74)(69,73)(70,72), (1,39,52,79)(2,28,53,68)(3,37,54,77)(4,26,55,66)(5,35,56,75)(6,24,57,64)(7,33,58,73)(8,22,59,62)(9,31,60,71)(10,40,41,80)(11,29,42,69)(12,38,43,78)(13,27,44,67)(14,36,45,76)(15,25,46,65)(16,34,47,74)(17,23,48,63)(18,32,49,72)(19,21,50,61)(20,30,51,70), (1,69)(2,78)(3,67)(4,76)(5,65)(6,74)(7,63)(8,72)(9,61)(10,70)(11,79)(12,68)(13,77)(14,66)(15,75)(16,64)(17,73)(18,62)(19,71)(20,80)(21,60)(22,49)(23,58)(24,47)(25,56)(26,45)(27,54)(28,43)(29,52)(30,41)(31,50)(32,59)(33,48)(34,57)(35,46)(36,55)(37,44)(38,53)(39,42)(40,51) );`
`G=PermutationGroup([(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20),(21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40),(41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60),(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80)], [(1,15),(2,14),(3,13),(4,12),(5,11),(6,10),(7,9),(16,20),(17,19),(22,40),(23,39),(24,38),(25,37),(26,36),(27,35),(28,34),(29,33),(30,32),(41,57),(42,56),(43,55),(44,54),(45,53),(46,52),(47,51),(48,50),(58,60),(62,80),(63,79),(64,78),(65,77),(66,76),(67,75),(68,74),(69,73),(70,72)], [(1,39,52,79),(2,28,53,68),(3,37,54,77),(4,26,55,66),(5,35,56,75),(6,24,57,64),(7,33,58,73),(8,22,59,62),(9,31,60,71),(10,40,41,80),(11,29,42,69),(12,38,43,78),(13,27,44,67),(14,36,45,76),(15,25,46,65),(16,34,47,74),(17,23,48,63),(18,32,49,72),(19,21,50,61),(20,30,51,70)], [(1,69),(2,78),(3,67),(4,76),(5,65),(6,74),(7,63),(8,72),(9,61),(10,70),(11,79),(12,68),(13,77),(14,66),(15,75),(16,64),(17,73),(18,62),(19,71),(20,80),(21,60),(22,49),(23,58),(24,47),(25,56),(26,45),(27,54),(28,43),(29,52),(30,41),(31,50),(32,59),(33,48),(34,57),(35,46),(36,55),(37,44),(38,53),(39,42),(40,51)])`
53 conjugacy classes
class 1 2A 2B 2C 2D 2E 2F 2G 2H 2I 2J 2K 2L 4A 4B 4C ··· 4G 4H 4I 4J 4K 4L 5A 5B 10A ··· 10F 10G 10H 10I 10J 20A ··· 20H 20I ··· 20P order 1 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 ··· 4 4 4 4 4 4 5 5 10 ··· 10 10 10 10 10 20 ··· 20 20 ··· 20 size 1 1 1 1 2 2 10 10 10 10 20 20 20 2 2 4 ··· 4 10 10 10 10 20 2 2 2 ··· 2 4 4 4 4 4 ··· 4 8 ··· 8
53 irreducible representations
dim 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 4 4 4 4 type + + + + + + + + + + + + + + + + + + + + + image C1 C2 C2 C2 C2 C2 C2 C2 C2 C2 C2 D4 D5 C4○D4 D10 D10 D10 D10 2+ 1+4 D4×D5 Q8⋊2D5 D4⋊8D10 kernel D20⋊21D4 D5×C22⋊C4 C22⋊D20 D20⋊8C4 D10.13D4 C4⋊D20 C4×C5⋊D4 C20.23D4 C5×C22⋊Q8 C22×D20 C2×Q8⋊2D5 D20 C22⋊Q8 C2×C10 C22⋊C4 C4⋊C4 C22×C4 C2×Q8 C10 C4 C22 C2 # reps 1 2 2 1 2 3 1 1 1 1 1 4 2 4 4 6 2 2 1 4 4 4
Matrix representation of D2021D4 in GL6(𝔽41)
0 1 0 0 0 0 40 6 0 0 0 0 0 0 40 0 0 0 0 0 0 40 0 0 0 0 0 0 40 36 0 0 0 0 25 1
,
0 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 40 0 0 0 0 0 25 1
,
40 0 0 0 0 0 35 1 0 0 0 0 0 0 0 1 0 0 0 0 40 0 0 0 0 0 0 0 32 37 0 0 0 0 20 9
,
40 0 0 0 0 0 35 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 9 4 0 0 0 0 21 32
`G:=sub<GL(6,GF(41))| [0,40,0,0,0,0,1,6,0,0,0,0,0,0,40,0,0,0,0,0,0,40,0,0,0,0,0,0,40,25,0,0,0,0,36,1],[0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,40,25,0,0,0,0,0,1],[40,35,0,0,0,0,0,1,0,0,0,0,0,0,0,40,0,0,0,0,1,0,0,0,0,0,0,0,32,20,0,0,0,0,37,9],[40,35,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,9,21,0,0,0,0,4,32] >;`
D2021D4 in GAP, Magma, Sage, TeX
`D_{20}\rtimes_{21}D_4`
`% in TeX`
`G:=Group("D20:21D4");`
`// GroupNames label`
`G:=SmallGroup(320,1302);`
`// by ID`
`G=gap.SmallGroup(320,1302);`
`# by ID`
`G:=PCGroup([7,-2,-2,-2,-2,-2,-2,-5,219,184,1571,297,192,12550]);`
`// Polycyclic`
`G:=Group<a,b,c,d|a^20=b^2=c^4=d^2=1,b*a*b=a^-1,c*a*c^-1=d*a*d=a^9,c*b*c^-1=d*b*d=a^18*b,d*c*d=c^-1>;`
`// generators/relations`
×
𝔽
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https://oeis.org/A110094
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crawl-data/CC-MAIN-2021-25/segments/1623488550571.96/warc/CC-MAIN-20210624015641-20210624045641-00627.warc.gz
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A110094 Startorial primes. 0
2, 3, 5, 7, 23, 719, 5039, 1451521, 2903041, 5806081, 46448639, 92897281, 371589121, 10032906239, 30098718719, 270888468479, 812665405439, 7313988648961, 21941965946881, 89874292518420479 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS These are primes of the form A109834 startorials (base 10) +1 or -1. This is by analogy to factorial primes (A002981), superfactorial primes (A073828), hyperfactorial primes, ultrafactorial primes (comment in A046882), subfactorial primes (A100015), double factorial primes (A080778), multifactorial primes (A037083). LINKS FORMULA {a(n)} = {A109834(k)+1 an element of A000040, or A109834(k)-1 an element of A000040, for some k}. CROSSREFS Cf. A000040, A002981, A073828, A080778, A037083, A100015, A103317, A046882, A109834. Sequence in context: A343834 A070029 A262339 * A088054 A249509 A085907 Adjacent sequences: A110091 A110092 A110093 * A110095 A110096 A110097 KEYWORD base,easy,nonn AUTHOR Jonathan Vos Post, Sep 04 2005 STATUS approved
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Last modified June 24 00:10 EDT 2021. Contains 345403 sequences. (Running on oeis4.)
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crawl-data/CC-MAIN-2016-30/segments/1469257831770.41/warc/CC-MAIN-20160723071031-00108-ip-10-185-27-174.ec2.internal.warc.gz
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #51 2011-01-02 20:02:43
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 104,353
### Re: The Prophecy
What? The idiotic end that water kills aliens? What no rain where they come from?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
A number by itself is useful, but it is far more useful to know how accurate or certain that number is.
Online
## #52 2011-01-03 04:41:35
LQ
Real Member
Registered: 2006-12-04
Posts: 1,285
### Re: The Prophecy
Hey bobbym, check out this episode of star trek, I want your full evaluation.
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
Offline
## #53 2011-01-03 07:04:19
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 104,353
### Re: The Prophecy
Hi LQ;
I know that episode. I have seen it many times. What about it?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
A number by itself is useful, but it is far more useful to know how accurate or certain that number is.
Online
## #54 2011-01-03 20:50:44
LQ
Real Member
Registered: 2006-12-04
Posts: 1,285
### Re: The Prophecy
Is it happening, allthough in a smaller scale?
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
Offline
## #55 2011-01-04 01:29:10
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 104,353
### Re: The Prophecy
Hi LQ;
What part? The Q represent a higher dimensional being. It is all covered in Edwin Abbott's book "Flatland." It was recreated for the television series, "Nova" by Ann Druyan. A creature like that would be more than a match for their contrived anti matter technology.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
A number by itself is useful, but it is far more useful to know how accurate or certain that number is.
Online
## #56 2011-01-04 20:24:15
Tigeree
Member
Registered: 2005-11-19
Posts: 13,883
### Re: The Prophecy
bobbym wrote:
What? The idiotic end that water kills aliens? What no rain where they come from?
No not Signs! Unbreakable! Although water killing aliens is pretty funny. Maybe they were spawned from the Wicked Witch of the West.
Although now that it's mentioned... I believe water was Brucey's weakness in Unbreakable too. How dare M. Night use the same idea twice! Can he not think of something interesting to kill aliens with, like dogs or some kinda virus??
People don't notice whether it's winter or summer when they're happy.
~ Anton Chekhov
Cheer up, emo kid.
Offline
## #57 2011-01-04 20:38:49
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 104,353
### Re: The Prophecy
Well he certainly knows how to kill the audience.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
A number by itself is useful, but it is far more useful to know how accurate or certain that number is.
Online
## #58 2011-01-04 21:30:41
Tigeree
Member
Registered: 2005-11-19
Posts: 13,883
### Re: The Prophecy
Yeah, that's a shame. He was going so well. And then he just lost it...
But I think he got it back on the Last Airbender. That was a great movie. You gotta see it!
People don't notice whether it's winter or summer when they're happy.
~ Anton Chekhov
Cheer up, emo kid.
Offline
## #59 2011-01-04 21:32:31
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 104,353
### Re: The Prophecy
No way! He joins Michael Crichton, Dan Brown and the guy who wrote "The Firm."
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
A number by itself is useful, but it is far more useful to know how accurate or certain that number is.
Online
## #60 2011-01-04 21:38:23
Tigeree
Member
Registered: 2005-11-19
Posts: 13,883
### Re: The Prophecy
Okay, I'm not even gunna ask what "the firm" is.
Dan Brown...
Michael Crichton....
No, he's cooler than that.
People don't notice whether it's winter or summer when they're happy.
~ Anton Chekhov
Cheer up, emo kid.
Offline
## #61 2011-01-04 21:40:40
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 104,353
### Re: The Prophecy
He is more "Valley Girl" than them but he is obviously fixated on water. Probably hates taking a bath.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
A number by itself is useful, but it is far more useful to know how accurate or certain that number is.
Online
## #62 2011-01-04 21:46:52
Tigeree
Member
Registered: 2005-11-19
Posts: 13,883
### Re: The Prophecy
And that's a problem because?...
People don't notice whether it's winter or summer when they're happy.
~ Anton Chekhov
Cheer up, emo kid.
Offline
## #63 2011-01-04 21:50:53
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 104,353
### Re: The Prophecy
Well he has to mingle with people.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
A number by itself is useful, but it is far more useful to know how accurate or certain that number is.
Online
## #64 2011-01-04 21:53:36
Tigeree
Member
Registered: 2005-11-19
Posts: 13,883
### Re: The Prophecy
Not necessarily....
People don't notice whether it's winter or summer when they're happy.
~ Anton Chekhov
Cheer up, emo kid.
Offline
## #65 2011-01-04 21:55:26
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 104,353
### Re: The Prophecy
I think he does. See, he thinks he is a swinger. Actually he is just a bad writer, bathed or not.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
A number by itself is useful, but it is far more useful to know how accurate or certain that number is.
Online
## #66 2011-01-04 22:19:54
Tigeree
Member
Registered: 2005-11-19
Posts: 13,883
### Re: The Prophecy
well, then stop watching his movies! See? That's not so hard. That way you don't have to complain about him, & I don't have to listen to you complaining.
People don't notice whether it's winter or summer when they're happy.
~ Anton Chekhov
Cheer up, emo kid.
Offline
## #67 2011-01-04 22:21:32
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 104,353
### Re: The Prophecy
Whose movies?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
A number by itself is useful, but it is far more useful to know how accurate or certain that number is.
Online
## #68 2011-01-04 22:32:03
Tigeree
Member
Registered: 2005-11-19
Posts: 13,883
### Re: The Prophecy
M. Night's movies.
Not Hitchcock's movies, he'll always be awesome!
People don't notice whether it's winter or summer when they're happy.
~ Anton Chekhov
Cheer up, emo kid.
Offline
## #69 2011-01-04 22:36:23
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 104,353
### Re: The Prophecy
Yes, Hitchcock was good. Night is kaboobly doo.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
A number by itself is useful, but it is far more useful to know how accurate or certain that number is.
Online
## #70 2011-01-04 22:41:33
Tigeree
Member
Registered: 2005-11-19
Posts: 13,883
### Re: The Prophecy
People don't notice whether it's winter or summer when they're happy.
~ Anton Chekhov
Cheer up, emo kid.
Offline
## #71 2011-01-04 22:42:52
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 104,353
### Re: The Prophecy
Thank you. It is a good word. It fits him too.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
A number by itself is useful, but it is far more useful to know how accurate or certain that number is.
Online
## #72 2011-01-04 22:47:28
Tigeree
Member
Registered: 2005-11-19
Posts: 13,883
### Re: The Prophecy
I'm really trying not to laugh, but it's not working that well.
Ah, bobby, you do use strange and wonderful words.
People don't notice whether it's winter or summer when they're happy.
~ Anton Chekhov
Cheer up, emo kid.
Offline
## #73 2011-01-04 22:49:26
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 104,353
### Re: The Prophecy
Mister m to you.
Got to go, bye for now.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
A number by itself is useful, but it is far more useful to know how accurate or certain that number is.
Online
## #74 2011-01-04 23:01:22
Tigeree
Member
Registered: 2005-11-19
Posts: 13,883
### Re: The Prophecy
I did start calling you that for a while but it felt somehow uncomfortable. And I felt too much like Harley Quinn. Which is weird because she's my idol.
People don't notice whether it's winter or summer when they're happy.
~ Anton Chekhov
Cheer up, emo kid.
Offline
## #75 2011-01-05 04:14:16
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 104,353
### Re: The Prophecy
Since this thread is called The Prophecy, here is mine.
Get used to it!!
Who is Harley Quinn, Anthony Quinn's daughter?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
A number by itself is useful, but it is far more useful to know how accurate or certain that number is.
Online
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Streetsblog.net
# It’s True: The Typical Car Is Parked 95 Percent of the Time
11:56 AM EST on March 10, 2016
Cars are a very inefficient transportation technology for too many reasons to count. They take up huge amounts of space but get driven around mostly empty -- the average private car in the U.S. carries only 1.6 people. A lot of the time, people drive distances that are short enough to easily walk or bike -- 28 percent of car trips are a mile or less, according to the Governors Highway Safety Association.
But perhaps the most profound inefficiency is that cars mostly just sit there. Early on in Donald Shoup's influential tome, The High Cost of Free Parking, he points out that cars are parked 95 percent of the time.
Paul Barter at Network blog Reinventing Parking wanted to show the basis for this assertion. Turns out, it's a surprisingly bullet-proof figure that holds up across different methods of calculation and in different countries. He crunched the numbers three ways:
Option #1: based on the number of cars, the number of car trips and the average time duration of car trips:
A UK report on parking put out last year by the RAC Foundation (and well worth a read by the way!) uses this method based on data from the UK National Travel Survey (NTS) (see p.23):
"... there are about 25 billion car trips per year, and with some 27 million cars, this suggests an average of just under 18 trips per car every week. Since the duration of the average car trip is about 20 minutes, the typical car is only on the move for 6 hours in the week: for the remaining 162 hours it is stationary – parked."
Since there are 168 hours in a week, the typical UK car is parked 96.5% of the time - even higher than Shoup's US estimate!
With this method, be careful to use car trips and not just trips by car. In other words, you want trips by car as driver so that you don't count trips as a car passenger. If you only have a total for both drivers and passengers then you will need to find an estimate of overall average car occupancy and divide by that to get the vehicle trips (driver trips) number.
Option #2: Based on time drivers spend driving (from transportation surveys) and assuming one car per driver.
If your local travel survey spits out a number for time spent driving then you can use that directly for a rough estimate.
This is how Shoup gets his 95% number. See Appendix B (p.624) of his epic tome "The High Cost of Free Parking". He cites the 1995 Nationwide Personal Transportaton Survey (NPTS) of the US Department of Transportation as finding that the average time drivers spent driving was 73 minutes (1.2 hours). Assuming one car per driver (which is roughly OK probably for the US context), this gives 5% as the time each car is in motion.
Option #3: Using car kilometres per car and overall average speeds
Transportation studies for metropolitan areas often provide data on the average yearly distance driven per car and the overall (24 hour, 7 days) traffic speed. If so, you can use this easy method (although note that those 24/7 speed numbers may not be the most solid of transport statistics I suspect). If you don't have car km per car you may be able to get it by dividing total car km by the number of cars, being careful that the two numbers are for the same study area.
The average time each car is in motion is the car km per year divided by the average speed.
I took a look at the 1995 UITP Millennium Cities Database numbers using this method. The average percentage of time that cars were parked for the 84 cities in that study was 95.8%. They were typically in motion for 1.02 hours per day.
Barter says the data underscores the massive savings that could be realized if cars were used more efficiently via car-sharing or on-demand services like taxis or Uber.
Elsewhere on the Network today: The Dallas Morning News' Transportation reports that a survey of who uses toll lanes doesn't quite fit the "Lexus Lanes" stereotype. The Washington Post has again published an editorial in which the writer fantasizes about killing cyclists, notes the FABB Blog. And Bike Portland has a map of where the city's new Biketown bike-share stations will go.
## More from Streetsblog USA
### Wednesday’s Headlines Wonder What If?
If New York City doesn't go through with congestion pricing, history is unlikely to look kindly on the decision, according to a NY Times urban policy writer.
June 19, 2024
### Three-Quarters of Black Motorists Are Struggling With the Cost of Car Ownership
Forcing everyone to rely on cars hurts millions of U.S. families financially — but for Black families, the burden is far worse.
June 19, 2024
### National Green Groups Condemn Hochul’s Congestion Pricing ‘Pause’
Had New York engaged congestion pricing, the state would have "played a nation-leading role." Alas.
June 18, 2024
### All Aboard for Tuesday’s Headlines
Amtrak is on pace to break 2019's record of 32 million riders. And The Guardian goes in-depth on plans for high-speed rail in the U.S.
June 18, 2024
See all posts
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Individuals or companies that invest money have every right to ask, “What’s in it for me?” Whether it involves funding a retirement account for a millennial or a piece of capital equipment purchased by a business owner, knowing what you’ll get in return for your dollars is paramount. We’ll help you break down the return on investment (ROI) ratio — what it is and how it’s calculated.
## What Is the Return on Investment Ratio?
ROI analytics is a vital tool for businesses and individuals. Whether you’re a business owner looking to turbocharge revenue or an investor in search of the next explosive stock, the ROI ratio holds a lot of value. The calculation is a straightforward method that indicates if a past investment was a worthy one, or if that decision needs to be re-evaluated.
There are two separate methods to figure out ROI. In one method, you’ll look at your beginning investment of dollars spent and dividing your net return by that initial cost. As an alternate method, subtract the initial value of the investment from the final value of the investment and divide that difference by the starting amount of the investment.
Either method will give you the same result, expressed as a percentage. In the first case, you will have already established your net return, and in the second example, you’ll need to factor in a beginning and ending value to arrive at net return. Which method you use will likely depend on the tools and data you have at your disposal. Some software programs may yield a net return while manual calculations may require another step.
## Return on Investment Ratio Formula
Let’s take a look at determining ROI in practice. If you want to calculate return, simply plug some prescribed variables into either of the two equations below:
### First formula
ROI = Net Return on Investment / Cost of Investment × 100%
As an example, let’s say \$20,000 in revenue was generated from an enhanced marketing effort and expenditure of \$100,000. Therefore, we can plug in the variables as such:
ROI = \$20,000/ \$100,000 × 100%
And arrive at our ROI:
ROI = 0.20 × 100% or 20%
### Second formula
ROI = Final Value of Investment − Initial Value of Investment
/ Initial Cost of Investment × 100%
Sticking with the advertising campaign scenario, we need to add the additional revenue to the initial investment to get our first value, and then we’re off to the races:
ROI = (\$100,000 + \$20,000) − (\$100,000) / \$100,000 × 100%
ROI = \$20,000 / \$100,000 × 100%
There’s an extra step but we see the same result:
ROI = 0.20 × 100% or 20%
## What Is a Good Return on Investment Ratio?
If you want to know what constitutes a good return on investment, the answer is usually this: It depends.
While arriving at ROI is fairly simple for straight-dollar dump-ins on personal securities investments, assessing all the factors on the business side of the equation is a bit more complex, especially with regard to assessing return.
### Investors
If you’re a growth investor placing that same \$100,000 investment in a stock or other security that earns 12% per year, there’s probably a job for you on Wall Street. All kidding aside, that ROI, were it achieved year after year, should please any investor. Yet, should you have expected 15% or even 25% in return?
For individual investors, a “good” ROI is relative, and it’s usually compared to a benchmark.
Each year, all professional money managers actively seek to best the percentage return of the S&P 500 Index. This index reflects the aggregate stock price performance of the largest 500 publicly traded companies in the United States. To put the S&P 500 in terms of ROI, the index historically returns about 7% annually on average. If the market mavens on “The Street ” consider beating the index a job well done (and they do), then so should you.
Of course, you might want more or even settle for less. Growth investing is often looked at as a strategy aimed at making your dollars outpace the rate of inflation, which runs about 3-3.5% per year on average, depending on whom you consult. So, if that’s your objective, a 6% annual average return might meet your expectations.
### Corporations
So, what about the business owner or executive who wants to gauge performance? For large corporations that must answer to shareholders and maintain a team of internal analysts, there’s a wealth of financial data to compare ROI on past projects. Stock analysts outside the company confine ROI comparisons to organizations in similar industries: a bank with a bank or a retailer with a retailer. Thus, if your ROI leads the pack among your peers, you can certainly define that as “good.”
With small business, the benchmarks may be tougher to establish. Data and competitor information is less plentiful at that level but you could consider ROI in terms of how quickly the initial investment is recovered. If you could coax an 18% ROI out of that \$100,000 ad campaign, you will have recovered your upfront costs in 4 years.
For a business that’s spending \$100,000 on a new advertising campaign, the upfront investment amount is easy enough to calculate. However, the actual return in dollars requires more careful analysis. If your sales rose by \$50,000 since you launched this new ad initiative, you might reasonably conclude that your idea was responsible for the total increase. However, you shouldn’t assume that your marketing efforts alone led to the boost.
There could be any number of reasons for that good news. Those variables might have included a price increase, a reduction in the cost of goods or services or an extraordinary effort on the part of your sales team.
You can get closer to the true ROI for individual assets or pursuits such as advertising, however.
Let’s assume your sales grew \$30,000 in the time period before the marketing plan was implemented. You can then deduce that the additional \$20,000 in revenue was due to the new initiative when measured over a similar time frame— perhaps a quarter or a fiscal year.
This also illustrates a shortcoming with ROI measurement: There’s no allowance for time. A 20% ROI in a single year might look fantastic, but spread over 5 years — not so much. That’s an easy fix, though. Simply measure your investment dollars and the return in dollars over any timeframe you choose, and compare ROI to the previous period, such as year-over-year or month-over-month.
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# Constrained Linear Regression in Python
You mention you would find Lasso Regression or Ridge Regression acceptable. These and many other constrained linear models are available in the scikit-learn package. Check out the section on generalized linear models.,Are there any other Python options for performing constrained least squares fits?,Or are there python routines for performing Lasso Regression or Ridge Regression or some other regression method which penalizes large b coefficient values?,Please be sure to answer the question. Provide details and share your research!
I recently prepared some tutorials on Linear Regression in Python. Here is one of the options (Gekko) that includes constraints on the coefficients.
# Constrained Multiple Linear Regression
import numpy as np
nd = 100 # number of data sets
nc = 5 # number of inputs
x = np.random.rand(nd,nc)
y = np.random.rand(nd)
from gekko import GEKKO
m = GEKKO(remote=False); m.options.IMODE=2
c = m.Array(m.FV,nc+1)
for ci in c:
ci.STATUS=1
ci.LOWER = -10
ci.UPPER = 10
xd = m.Array(m.Param,nc)
for i in range(nc):
xd[i].value = x[:,i]
yd = m.Param(y); yp = m.Var()
s = m.sum([c[i]*xd[i] for i in range(nc)])
m.Equation(yp==s+c[-1])
m.Minimize((yd-yp)**2)
m.solve(disp=True)
a = [c[i].value[0] for i in range(nc+1)]
print('Solve time: ' + str(m.options.SOLVETIME))
print('Coefficients: ' + str(a))
Source: https://stackoverflow.com/questions/10154922/constrained-linear-regression-in-python
Scikit-learn does not allow such constraints on the coefficients. ,I am currently running multiple linear regression on a dataset. At first, I didn't realize I needed to put constraints over my weights; as a matter of fact, I need to have specific positive & negative weights.,Any idea of how I could assign a constraint on the coefficient of a specific variable ? Probably going to be burdensome to define each constraint but I have no idea how to do otherwise.,I've written a class that imposes upper and lower bounds on LinearRegression coefficients. You can extend it to use Ridge or evel Lasso penalty if you want:
As an example, let's suppose my model is :
y = W0*x0 + W1*x1 + W2*x2
I am working on Python using the ScikitLearn packages. This is how I get my best model :
def ridge(Xtrain, Xtest, Ytrain, Ytest, position):
param_grid={'alpha':[0.01 , 0.1, 1, 10, 50, 100, 1000]}
gs = grid_search.GridSearchCV(Ridge(), param_grid=param_grid, n_jobs=-1, cv=3)
gs.fit(Xtrain, Ytrain)
hatytrain = gs.predict(Xtrain)
hatytest = gs.predict(Xtest)
Source: https://www.py4u.net/discuss/193366
Please be sure to answer the question. Provide details and share your research!,There are is a constrained least squares method scipy.optimize.lsq_linear. Another option is to use an optimizing solver for Python. Here is one of the options (Gekko) that I maintain that includes coefficient constraints.,Having said that, there is no standard implementation of Non-negative least squares in Scikit-Learn. The pull request is still open.,Software Recommendations
I use a workaround with Lasso on Scikit Learn (It is definitely not the best way to do things but it works well). Lasso has a parameter positive which can be set to True and force the coefficients to be positive. Further, setting the Regularization coefficient alpha to lie close to 0 makes the Lasso mimic Linear Regression with no regularization. Here's the code:
from sklearn.linear_model import Lasso
lin = Lasso(alpha=0.0001,precompute=True,max_iter=1000,
positive=True, random_state=9999, selection='random')
lin.fit(X,y)
Source: https://datascience.stackexchange.com/questions/18258/how-to-force-weights-to-be-non-negative-in-linear-regression
In [1]:
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# IBPS Reasoning Quiz for 2019 Exams: 5th March 2019
Dear Aspirants,
### Reasoning Questions for IBPS Exam 2019:
Reasoning Ability is an onerous section. With the increasing complexity of questions, it becomes hard for one to give it the cold shoulder. The only way to make the grade in this particular section in the forthcoming banking exams. And, to let you practice with the best of the latest pattern questions, here is the Adda247 Reasoning Quiz based on the exact same pattern of questions that are being asked in the exams.
Watch Video Solution Here
Directions (1-5): Read the given information carefully and answer the given questions.
Two F1 racing drivers A and B are participating in a racing competition. A and B starts their journey from same point but not necessary in the same order. Driver A starts moving 6 km towards north direction to point P then take a right turn and move 18km to reach point R, then take a left turn and move 12km to reach point S. Point P, Q and R lies in a same line. Distance between P and Q is half of the distance between point Q and R. Driver B starts to move toward west 2km more than the distance between point Q and R and reach point W and then take a right turn and walk 18km to reach point U. U, V and W lies in the same line as distance of V and W is double of the distance between U and V. From point U, Driver B take a right turn and reach point S. Point T lies between U and S in exact midway. Q is to the east of P and U is to the north of V.
Q1. What is the distance between P and Q?
14km
12km
16km
18km
6km
Solution:
Q2. What is the distance between T and S?
14km
12km
16km
18km
6km
Solution:
Q3. Driver A is moving in which direction currently?
West
North-east
South
North
East
Solution:
Q4. What is the total distance between U and S?
29km
32km
36km
30km
28km
Solution:
Q5. In which direction is point T with respect to point W?
North
South
North West
North East
South West.
Solution:
Directions (6-10): Study the following information carefully and answer the given questions.
Seven Persons - A, B, C, D, E, F and G live on seven different floors of a building but not necessarily in the same order. Lowermost floor is numbered 1. Each of them went the following destinations for official meeting - P, Q, R, S, T, U and V. Each left on a different day of the week starting from Monday.
· B lives on an odd numbered floor but not on floor number three and left on Monday
· The one who went to R lives immediately above G who left on some day after Thursday
· The one who went to P lives on an odd numbered floor above F, who went before C.
· E went to V and lives on the 4th floor and does not went on Friday or Sunday.
· A lives on the floor above E and left for T on Saturday
· The one who lives on top floor went to S and the person living on the bottom floor went to U but not on Monday. C lives on an even numbered floor but not above A. Atleast One day gap is between the days on which C left and the person who went to Q left.
· D left on Tuesday but does not live immediate above E.
The one who left on Wednesday does not live immediate below B. F does not live in bottom two floors.
Q6. Who lives on the topmost floor?
D
A
B
C
None of the above
Solution:
Q7. If C left for his destination on Friday, then G left for his destination on?
Sunday
Wednesday
Thursday
Either Sunday or Wednesday
Can’t be determined
Solution:
Q8. Who went to Q?
B
D
F
C
None of these
Solution:
Q9. Who lives between the persons who went to R and V?
The one who went on Monday
The one who went on Thursday
F
The person on the seventh floor
Either (b) or (c)
Solution:
Q10. For which destination did the person who went on Monday left?
S
P
Q
R
U
Solution:
Directions (11-15): Study the following information carefully to answer the given questions.
Five students J, K, L, M and N appeared in a test that comprised of five different subjects namely English, Hindi, Math, Physics and Chemistry. Each student got a different rank in each of these five subjects and the rank of no two students were the same in the same subject. It is known that
· J got 1st rank in English and the lowest rank in Chemistry.
· M got 5th rank in Physics which was also N's rank in Math.
· L got 2nd rank in Math and his rank in Hindi was better than his rank in Math.
· The person who got 4th rank in Math also got 3rd rank in English but he was not M.
· M got 3rd rank in chemistry and J got a better rank in Hindi compared to his rank in Math.
· N didn't get 1st rank in Physics and got 3rd rank in hindi.
Q11. L got the lowest rank in which of the following subject?
Hindi
Physics
English
Chemistry
Can't be determined
Solution:
Q12. Who got 1st rank in Physics?
M
N
J
K
None of These
Solution:
Q13. What was K's rank in chemistry?
2nd
4th
1st
5th
Can't be determined
Solution:
Q14. Who got 2nd rank in English?
K
L
M
N
Can't be determined
Solution:
Q15. Who got 3rd Rank in Hindi?
N
J
K
M
Can’t be determined
Solution:
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https://www.numbersaplenty.com/23313312333011
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Search a number
23313312333011 = 107217881423673
BaseRepresentation
bin1010100110100000011011…
…01100110011110011010011
310001112201201211011101122202
411103100031230303303103
511023431130424124021
6121325553212153415
74624221420123011
oct523201554636323
9101481654141582
1023313312333011
117479136042918
12274633916886b
13100158893c75c
145a85282c58b1
152a6675833a0b
hex15340db33cd3
23313312333011 has 4 divisors (see below), whose sum is σ = 23531193756792. Its totient is φ = 23095430909232.
The previous prime is 23313312332971. The next prime is 23313312333029. The reversal of 23313312333011 is 11033321331332.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 23313312333011 - 238 = 23038434426067 is a prime.
It is a super-2 number, since 2×233133123330112 (a number of 28 digits) contains 22 as substring.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (23313312333061) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 108940711730 + ... + 108940711943.
It is an arithmetic number, because the mean of its divisors is an integer number (5882798439198).
Almost surely, 223313312333011 is an apocalyptic number.
23313312333011 is a deficient number, since it is larger than the sum of its proper divisors (217881423781).
23313312333011 is a wasteful number, since it uses less digits than its factorization.
23313312333011 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 217881423780.
The product of its (nonzero) digits is 8748, while the sum is 29.
Adding to 23313312333011 its reverse (11033321331332), we get a palindrome (34346633664343).
The spelling of 23313312333011 in words is "twenty-three trillion, three hundred thirteen billion, three hundred twelve million, three hundred thirty-three thousand, eleven".
Divisors: 1 107 217881423673 23313312333011
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https://www.physicsforums.com/threads/mass-reconstruction-what-does-it-mean.435687/
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# Mass Reconstruction: What Does It Mean?
• godtripp
In summary, the concept of "reconstructing" mass involves using energy and momentum conservation to compute the 4-vector of a particle and its Lorentz square. By analyzing the decay products and summing up their total energies and momentums, we can trace back the energy and mass of the parent particle. This is done by equating the invariant mass of the particle to the difference between the square of the total energy and the square of the total momentum of the decay products.
godtripp
What is it? I always hear it and never know what exactly it means to "reconstruct" mass.
Using energy and momentum conservation, compute the 4-vector of the particle, and it's Lorentz square is the square of the mass.
Say for instance you detect the decay products of the particle, add up all 4-vectors, the square of the mass is the difference between (1) the square of the total energy of the decay products and (2) the square of the total momentum carried by the decay products.
You could also do it if by any chance you can assume the particle is the only one missing in a certain balance, and computing the missing 4-vector.
sorry, I'm going to have to verify in layman's terms.. this is way above my current education in physics. (sophmore year undergrad)
So what you're doing is analyzing the decay products, summing up their total energies to trace back the energy , and thus the masses, of their parent particles?
godtripp said:
summing up their total energies to trace back the energy
And the momentum.
For any particle:
$$(mc^2)^2 = E^2 - ({\vec p }c)^2$$
If the particle then decays into a collection of other particles, the total energy and the total momentum are both conserved, so
$$(mc^2)^2 = (\Sigma E_i)^2 - (\Sigma {\vec p_i} c)^2$$
In these equations, m is the "invariant mass", also known as "rest mass"; not the "relativistic mass" which you find in many introductory treatments of relativity.
Last edited:
## 1. What is mass reconstruction?
Mass reconstruction is the process of calculating the distribution of matter in the universe, based on observed distortions in the light from distant galaxies. It is an important technique used in cosmology to understand the large-scale structure of the universe.
## 2. How is mass reconstruction done?
Mass reconstruction is done using a method called gravitational lensing, which occurs when light from a distant galaxy is bent by the gravitational pull of a massive object, such as a galaxy cluster. By measuring the distortion of the light, scientists can infer the distribution of matter in the universe.
## 3. Why is mass reconstruction important?
Mass reconstruction is important because it provides valuable insights into the composition and evolution of the universe. By understanding the distribution of matter, scientists can better understand the processes that shaped the universe and the role of dark matter in its formation.
## 4. What are the challenges in mass reconstruction?
One of the main challenges in mass reconstruction is separating the signal from the noise. Gravitational lensing effects can be very subtle and can be influenced by various factors, such as the properties of the intervening matter and the shape of the galaxies being observed. This makes it important to use advanced statistical techniques to accurately reconstruct the mass distribution.
## 5. How is mass reconstruction used in research?
Mass reconstruction is used in a variety of research areas, including the study of dark matter, the large-scale structure of the universe, and the formation of galaxies. It is also used in conjunction with other observational and theoretical techniques to test and refine our understanding of cosmology and the laws of physics.
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https://kr.mathworks.com/matlabcentral/answers/315378-calculate-shortest-distance-between-2-lines
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# Calculate shortest distance between 2 lines.
조회 수: 62(최근 30일)
Ema 5 Dec 2016
I got 2 lines:
Line 1 = (x,y,z) ∈ (Real numbers/3d), x=10+3t, y=−1+2t, z=−1.5−0.01t, t ∈ Real numbers
Line 2 = (x,y,z) ∈ (Real numbers/3d), x=−1−0.1t, y=2t, z=−2.5+0.02t, t ∈ Real numbers
unit of lenght (meters)
First i don't seem to be able to plot it since i have not stated the t free variable.
Since i get this error
Error using plot3
Data must be numeric, datetime, duration or an array convertible to double.
Error in Euklidisk_geometri_1 (line 5)
x = 10+3*t ; y = -1+2*t ; z = -1.5+0.01*t; plot3(x,y,z)
Second i wonder if my calulculations are correct in matlab: I get D(distance) to be 1.0001 meters.
clf
syms t
x = 10+3*t ; y = -1+2*t ; z = -1.5+0.01*t; plot3(x,y,z)
hold on
x = -1-0.1*t ; y = 2*t ; z = -2.5 +0.02*t; plot3(x,y,z)
L1 = (([10,-1,-1.5])+t*([3,2,-0.01]))
L2 = (([-1,2,-2.5])+t*([-0.1,2,0.02]))
v1 = [3,2,-0.01]
v2 = [-0.1,2,0.02]
N = (cross(v1,v2))
P1 = [10,-1,-1.5]
P2 = [-1,2,-2.5]
P = P2-P1
C = P.*N
d = (P.*N/N.^2).*N
D = norm(sqrt(d.^2))
#### 댓글 수: 1
Tamir Suliman 5 Dec 2016
Data must be numeric, datetime, duration or an array convertible to double.
instead of syms t you should have some thing like
t=-10:0.001:10;
로그인 to comment.
### 답변(2개)
Tamir Suliman 5 Dec 2016
Data must be numeric, datetime, duration or an array convertible to double.
instead of syms t you should have some thing like
t=-10:0.001:10;
#### 댓글 수: 0
로그인 to comment.
Walter Roberson 5 Dec 2016
#### 댓글 수: 0
로그인 to comment.
이 질문에 답변하려면 로그인을(를) 수행하십시오.
Translated by
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https://www.airmilescalculator.com/distance/ord-to-yxc/
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Distance between Chicago, IL (ORD) and Cranbrook (YXC)
Flight distance from Chicago to Cranbrook (Chicago O'Hare International Airport – Cranbrook/Canadian Rockies International Airport) is 1436 miles / 2311 kilometers / 1248 nautical miles. Estimated flight time is 3 hours 13 minutes.
Driving distance from Chicago (ORD) to Cranbrook (YXC) is 1709 miles / 2751 kilometers and travel time by car is about 29 hours 51 minutes.
Map of flight path and driving directions from Chicago to Cranbrook.
Shortest flight path between Chicago O'Hare International Airport (ORD) and Cranbrook/Canadian Rockies International Airport (YXC).
How far is Cranbrook from Chicago?
There are several ways to calculate distances between Chicago and Cranbrook. Here are two common methods:
Vincenty's formula (applied above)
• 1435.947 miles
• 2310.933 kilometers
• 1247.804 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
• 1432.513 miles
• 2305.406 kilometers
• 1244.820 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 Chicago O'Hare International Airport
City: Chicago, IL
Country: United States
IATA Code: ORD
ICAO Code: KORD
Coordinates: 41°58′42″N, 87°54′17″W
City: Cranbrook
IATA Code: YXC
ICAO Code: CYXC
Coordinates: 49°36′38″N, 115°46′55″W
Time difference and current local times
The time difference between Chicago and Cranbrook is 1 hour. Cranbrook is 1 hour behind Chicago.
CST
MST
Carbon dioxide emissions
Estimated CO2 emissions per passenger is 175 kg (387 pounds).
Frequent Flyer Miles Calculator
Chicago (ORD) → Cranbrook (YXC).
Distance:
1436
Elite level bonus:
0
Booking class bonus:
0
In total
Total frequent flyer miles:
1436
Round trip?
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https://rdrr.io/cran/EVchargcost/src/R/plot_functions.R
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# R/plot_functions.R In EVchargcost: Computes and Plot the Optimal Charging Strategy for Electric Vehicles
#### Documented in plot_functions
```plot_functions <- function(a, alpha, beta, delta, gamma, tau, R, x_values, y_values){
#' @title Plots the charging function, the electricity price function and the optimal cost function
#' @description Function that plots the charging function, the electricity price function and the optimal cost function
#' @param a vector with the breaking points of charging function in the x-axis
#' @param alpha vector with the slopes of the charging function on each segment
#' @param beta vector with the y-intercepts of the charging function on each segment
#' @param delta vector with the times duration of each segment of electricity price function
#' @param gamma vector with the prices of the electricity on each segment of electricity price function
#' @param tau consumption of the vehicle (numerical value)
#' @param R range of the vehicle (numerical value)
#' @param x_values vector with the x-values of the breaking points of the charging cost function
#' @param y_values vector with the y-values of the breaking points of the charging cost function
#' @return A plot with the charging function, the electricity price function and the optimal cost function
#' @examples
#' a <- c(0,3.3,6.6,10)
#' alpha <- c(0.1757576, 0.07272727, 0.05294118)
#' beta <- c(0, 0.34, 0.4705882)
#' delta <- c(4, 3, 5)
#' gamma <- c(0.45, 0.25, 0.5)
#' tau <- 0.15
#' R <- 250
#' opt_cost_function = minimum_cost(a, alpha, beta, delta, gamma, tau, R)
#' xvalues <- opt_cost_function[["xvalues"]]
#' yvalues <- opt_cost_function[["yvalues"]]
#' plot_functions(a, alpha, beta, delta, gamma, tau, R, xvalues, yvalues)
B <- length(alpha)
P <- length(gamma)
x <- a
y <- c(0)
for (i in 1:B){
y <- c(y, alpha[i]*a[i+1]+beta[i])
}
df1 <- data.frame(x = x, y = y)
plot1 <- ggplot2::ggplot(df1, ggplot2::aes(x, y)) + ggplot2::geom_line(color = "darkblue") + labs(x = "Duration", y = "State of Charge", title = "Charging function") + ggplot2::geom_point(size=1, color="darkblue")
xinit <- c(0)
xend <- c()
for (i in 1:(P-1)){
xinit <- c(xinit, sum(delta[1:i]))
xend <- c(xend, sum(delta[1:i]))
}
xend <- c(xend, sum(delta))
y <- gamma
df2 <- data.frame(xinit = xinit, y = y, xend = xend)
plot2 <- ggplot2::ggplot(df2, ggplot2::aes(x, y)) + ggplot2::geom_segment(ggplot2::aes(x = xinit, y = y, xend = xend, yend = y), linetype = "solid", color = "darkorange") + labs(x = "Time", y = "Price", title = "Electricity price function")
x <- x_values
y <- y_values
df3 <- data.frame(x = x, y = y)
plot3 <- ggplot2::ggplot(df3, ggplot2::aes(x, y)) + ggplot2::geom_line(color = "darkgreen") + labs(x = "State of Charge", y = "Cost", title = "Charging cost function") + ggplot2::geom_point(size=1, color="darkgreen")
p <- cowplot::plot_grid(plot1, plot2, plot3, ncol = 3)
return(p)
}
```
## Try the EVchargcost package in your browser
Any scripts or data that you put into this service are public.
EVchargcost documentation built on May 29, 2024, 8:29 a.m.
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http://www.elliottelectric.com/StaticPages/ElectricalReferences/Guides/calculate-fuse-ratings-for-motor-protection-using-full-load-amps-current.aspx
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# How to Calculate Fuse Ratings for AC Motor Protection Based on Motor Full Load Amps Current
Using AC Motor Protection Tables to Select Fuse Ratings
Fuse ratings selected in accordance with the following recommendations also meet NEC® requirements for Motor Branch Circuit and Short Circuit Protection.
Selecting Fuses for Motor Running Protection Based on Motor Actual Full Load Currents
Better protection is achieved when fuse ratings are based on motor actual full load amps (FLA) as obtained from motor nameplates. Locate motor nameplate FLA in the column appropriate for the type of motor and type of protection required. Read to the left to obtain the recommended fuse ampere rating.
#### Full Load Amps: 0 - 2.5
Fuse
Amp
Rating
Service Factor
≥ 1.15 or with
Rise ≤ 40 ℃
Service Factor
< 1.15 or with
Rise > 40 ℃
1/10 0.08 - 0.09 0.09 - 0.10
1/8 0.10 - 0.11 0.11 - 0.125
15/100 0.12 - 0.15 0.14 - 0.15
2/10 0.16 - 0.19 0.18 - 0.20
1/4 0.20 - 0.23 0.22 - 0.25
3/10 0.24 - 0.30 0.27 - 0.30
4/10 0.32 - 0.39 0.35 - 0.40
1/2 0.40 - 0.47 0.44 - 0.50
6/10 0.48 - 0.60 0.53 - 0.60
8/10 0.64 - 0.79 0.70 - 0.80
1 0.80 - 0.89 0.87 - 0.97
1 1/8 0.90 - 0.99 0.98 - 1.08
1 1/4 1.00 - 1.11 1.09 - 1.21
1 4/10 1.12 - 1.19 1.22 - 1.30
1 1/2 1.20 - 1.27 1.31 - 1.39
1 6/10 1.28 - 1.43 1.40 - 1.56
1 8/10 1.44 - 1.59 1.57 - 1.73
2 1.60 - 1.79 1.74 - 1.95
2 1/4 1.80 - 1.99 1.96 - 2.17
2 1/2 2.00 - 2.23 2.18 - 2.43
2 8/10 2.24 - 2.39 2.44 - 2.60
3 2.40 - 2.55 2.61 - 2.78
#### Full Load Amps: 0 - 2.5
Fuse
Amp
Rating
Service Factor
≥ 1.15 or with
Rise ≤ 40 ℃
Service Factor
< 1.15 or with
Rise > ℃
1/10 0.00 - 0.08 0.00 - 0.09
1/8 0.09 - 0.10 0.10 - 0.11
15/100 0.11 - 0.12 0.12 - 0.13
2/10 0.13 - 0.16 0.14 - 0.17
1/4 0.17 - 0.20 0.18 - 0.22
3/10 0.21 - 0.24 0.23 - 0.26
4/10 0.25 - 0.32 0.27 - 0.35
1/2 0.33 - 0.40 0.36 - 0.43
6/10 0.41 - 0.48 0.44 - 0.52
8/10 0.49 - 0.64 0.53 - 0.70
1 0.65 - 0.80 0.71 - 0.87
1 1/8 0.81 - 0.90 0.88 - 0.98
1 1/4 0.91 - 1.00 0.99 - 1.09
1 4/10 1.01 - 1.12 1.10 - 1.22
1 1/2 1.13 - 1.20 1.23 - 1.30
1 6/10 1.21 - 1.28 1.31 - 1.39
1 8/10 1.29 - 1.44 1.40 - 1.57
2 1.45 - 1.60 1.58 - 1.74
2 1/4 1.61 - 1.80 1.75 - 1.96
2 1/2 1.81 - 2.00 1.97 - 2.17
2 8/10 2.01 - 2.24 2.18 - 2.43
3 2.25 - 2.40 2.44 - 2.60
#### Full Load Amps: 2.5 - 35
Fuse
Amp
Rating
Service Factor
≥ 1.15 or with
Rise ≤ 40 ℃
Service Factor
< 1.15 or with
Rise > 40 ℃
2 8/10 2.24 - 2.39 2.44 - 2.60
3 2.40 - 2.55 2.61 - 2.78
3 2/10 2.56 - 2.79 2.79 - 3.04
3 1/2 2.80 - 3.19 3.05 - 3.47
4 3.20 - 3.59 3.48 - 3.91
4 1/2 3.60 - 3.99 3.92 - 4.34
5 4.00 - 4.47 4.35 - 4.86
5 6/10 4.48 - 4.79 4.87 - 5.21
6 4.80 - 4.99 5.22 - 5.43
6 1/4 5.00 - 5.59 5.44 - 6.08
7 5.60 - 5.99 6.09 - 6.52
7 1/2 6.00 - 6.39 6.53 - 6.95
8 6.40 - 7.19 6.96 - 7.82
9 7.20 - 7.99 7.83 - 8.69
10 8.00 - 9.59 8.70 - 10.00
12 9.60 - 11.99 10.44 - 12.00
15 12.00 - 13.99 13.05 - 15.00
17 1/2 14.00 - 15.99 15.22 - 17.39
20 16.00 - 19.99 17.40 - 20.00
25 20.00 - 23.99 21.74 - 25.00
30 24.00 - 27.99 26.09 - 30.00
35 28.00 - 31.99 30.44 - 34.78
40 32.00 - 35.99 34.79 - 39.12
#### Full Load Amperes: 2.5 - 35
Fuse
Ampere
Rating
Service Factor
≥ 1.15 or with
Rise ≤ 40 ℃
Service Factor
< 1.15 or with
Rise > 40 ℃
3 2.25 - 2.40 2.44 - 2.60
3 2/10 2.41 - 2.56 2.61 - 2.78
3 1/2 2.57 - 2.80 2.79 - 3.04
4 2.81 - 3.20 3.05 - 3.48
4 1/2 3.21 - 3.60 3.49 - 3.91
5 3.61 - 4.00 3.92 - 4.35
5 6/10 4.01 - 4.48 4.36 - 4.87
6 4.49 - 4.80 4.88 - 5.22
6 1/4 4.81 - 5.00 5.23 - 5.43
7 5.01 - 5.60 5.44 - 6.09
7 1/2 5.61 - 6.00 6.10 - 6.52
8 6.01 - 6.40 6.53 - 6.96
9 6.41 - 7.20 6.97 - 7.83
10 7.21 - 8.00 7.84 - 8.70
12 8.01 - 9.60 8.71 - 10.43
15 9.61 - 12.00 10.44 - 13.04
17 1/2 12.01 - 14.00 13.05 - 15.21
20 14.01 - 16.00 15.22 - 17.39
25 16.01 - 20.00 17.40 - 21.74
30 20.01 - 24.00 21.75 - 26.09
35 24.01 - 28.00 26.01 - 30.43
40 28.01 - 32.00 30.44 - 37.78
45 32.01 - 36.00 37.79 - 39.13
#### Full Load Amps: 35 - 600
Fuse
Amp
Rating
Service Factor
≥ 1.15 or with
Rise ≤ 40 ℃
Service Factor
< 1.15 or with
Rise > 40 ℃
40 32.00 - 35.99 34.79 - 39.12
45 36.00 - 39.99 39.13 - 43.47
50 40.00 - 47.99 43.48 - 50.00
60 48.00 - 55.99 52.17 - 60.00
70 56.00 - 59.99 60.87 - 65.21
75 60.00 - 63.99 65.22 - 69.56
80 64.00 - 71.99 69.57 - 78.25
90 72.00 - 79.99 78.26 - 86.95
100 80.00 - 87.99 86.96 - 95.64
110 88.00 - 99.99 95.65 - 108.69
125 100.00 - 119.99 108.70 - 125.00
150 120.00 - 139.99 131.30 - 150.00
175 140.00 - 159.99 152.17 - 173.90
200 160.00 - 179.99 173.91 - 195.64
225 180.00 - 199.99 195.65 - 217.38
250 200.00 - 239.99 217.39 - 250.00
300 240.00 - 279.99 260.87 - 300.00
350 280.00 - 319.99 304.35 - 347.82
400 320.00 - 359.99 347.83 - 391.29
450 360.00 - 399.99 391.30 - 434.77
500 400.00 - 479.99 434.78 - 500.00
600 480.00 - 600.00 521.74 - 600.00
#### Full Load Amps: 35 - 600
Fuse
Amp
Rating
Service Factor
≥ 1.15 or with
Rise ≤ 40 ℃
Service Factor
< 1.15 or with
Rise > 40 ℃
40 28.01 - 32.00 30.44 - 37.78
45 32.01 - 36.00 37.79 - 39.13
50 36.01 - 40.00 39.14 - 43.48
60 40.01 - 48.00 43.49 - 52.17
70 48.01 - 56.00 52.18 - 60.87
75 56.01 - 60.00 60.88 - 65.22
80 60.01 - 64.00 65.23 - 69.57
90 64.01 - 72.00 69.58 - 78.26
100 72.01 - 80.00 78.27 - 86.96
110 80.01 - 88.00 86.97 - 95.65
125 88.01 - 100.00 95.66 - 108.70
150 100.01 - 120.00 108.71 - 130.43
175 120.01 - 140.00 130.44 - 152.17
200 140.01 - 160.00 152.18 - 173.91
225 160.01 - 180.00 173.92 - 195.62
250 180.01 - 200.00 195.63 - 217.39
300 200.01 - 240.00 217.40 - 260.87
350 240.01 - 280.00 260.88 - 304.35
400 280.01 - 320.00 304.36 - 347.83
450 320.01 - 360.00 347.84 - 391.30
500 360.01 - 400.00 391.31 - 434.78
600 400.01 - 480.00 434.79 - 521.74
Warning: When using this information to perform electrical work, call a licensed electrician and consult the NEC® for safety. All licensed electricians have passed examinations covering the National Electric Code®, know state and local building codes, and may carry insurance to cover damages.
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https://sonichours.com/how-many-days-until-july-18-2020/
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General
# How Many Days Until July 18 2020
The question of how many days until July 18 2020 can be an easy one to answer. There are 156 days until July 18th of next year. That’s a Saturday, and it is in week 29 of the year. This date is day number 200 of the year, so it has a duration of approximately 167 days. The year 2021 is a leap year, so it will have 365.9 calendars days.
There are 165 days until July 18, 2020, which is five months and four days. If you count the weeks and days of each month, you’ll find that there are 366 days between January 18 and July 18. However, the number of weekends is the most important thing. In order to have a good count of how many weeks and months are left before July 18, the first step is to know the number of weekends.
The end of the month on June 9 2020 is also a Saturday. That means there are 165 days between that date and July 18. That’s about five weeks and four days, or three months and four weeks. In all, this is about nine hundred and sixty-six days. This time period includes 117 weekdays, 56.160 minutes and 3,369,600 seconds. Obviously, this is quite a bit longer than it seems.
In case you were wondering how many days until July 18 2020, remember that the year is a leap year. The year is actually a half year longer than it is a regular calendar year. So if you’re wondering, “How many days until July 18 2020?” just go ahead and look up the date! It’s easy to get confused. So just remember to be patient! So many days are left to count until you reach the goal of reaching your dream holiday. So start celebrating! It’s just around the corner!
As you can see, the end date of July 18 2020 is a Saturday. That means that there are 39 days between the end date of June 9 and July 17 of 2020. That’s a five-week and a half year and a quarter. Then, there are seven months and a half weeks until July 18. That means there are three hundred and six months until July 18 of 2020.
The end date of July 18 2020 is a Saturday. That’s the end of the year. The remaining 165 days are 5 weeks and four days. So, from January 16 to July 18, the total length of time between the two dates is 936 hours, 237 minutes, and 14,256,000 seconds. That’s a lot of time! So if you’re a planner, now is the time to make plans.
When asked how many days are left until July 18, 2020, the answer is 165. This is a leap year. That’s an extra day to celebrate Independence Day. Then, there’s a leap year. If you’re a Cancer, you’ll be glad to know that there are 166 days between July 18. So, there’s only a little more time left in the world.
The number of days until July 18 2020 is 165. It is equivalent to 23 weeks and four days. In other words, there are 167 days from today to July 18 on the date of the leap year. There are 157 days between January 1 and July 17 and one day until July 18. In other words, there’s just under one year and two years. The next year, in 2020, will be a leap year.
July 18 will be the third full week of a leap year. It is a leap year, meaning it has three extra weeks. In other words, it has 164 days before July 18! As a result, you have 169 days until July 18. If you want to be precise, use a calendar or timer. There’s no need to worry. Just remember to celebrate the leap year!
Visit the rest of the site for more useful articles!
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## Modelling Changenumber of dvds and blu-rays sold per year
I found the number of DVDs and Blu-Rays sold in millions on this website . As shown on the following graph, the number of DVDs sold since they were created in 1995 (although my data only starts in 1998) kept increasing until 2008, when the amount sold suddenly dropped despite the invention of the Blu-Ray in 2007. This was probably due to the fact that it became possible to watch movies without buying a physical copy of them, like on Netflix. The number of DVDs sold keeps dropping until the end of my data in 2015.
My graph: Number of DVDs and Blu-Rays sold per year (in millions)
I split my graph into three sections according to the data: first section, the rapid increase. Second section, the increase slows and third section, the data drops. Here are my data tables for each section.
Data Table Section 1
Data Table Section 2
Data Table Section 3
Here are my lines of best fit for each section of the graph
After splitting the graph into sections, I found the line of best fit for each. To do that, I put a ruler on my graph and found the place where the line should be approximately. Then I had to find the more accurate slope of the lines and their y intercepts. Only then could I graph the lines.
### here are my calculations:
#### Interval/section 1
Slope ≈ 28.6 and y intercept ≈ -24.6
#### Interval/Section 2
(a few steps are skipped since they were already shown in interval 1)
Slope ≈ 17.5 and y intercept ≈ 91.5
#### Interval/section 3
Slope ≈ -19.6 and y intercept ≈ 452.4
### Line of best fit
The long green curve is the line of best fit for the entire data set
I used a quadratic equation to find my line of best fit. At first, I wanted to use a linear equation because it was a better fit for section 3 of the graph and would therefore have given me more accurate predictions, however, it did not at all fit with sections 1 and 2, which have a positive correlation, so overall the curve is better. The data on my graph shows that in 2067, -6 700 million DVDs will be sold and in 2112, - 21 350 DVDs will be sold. Of course, this is not physically possible, so I've come to the conclusion that by 2067 there will be no more DVDs or Blu-Rays sold.
by Clara
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# Lambda Calculus, a Suprisingly Powerful Language
I’m reading the book Types and Programming Languages from Benjamin C. Pierce from time to time at the moment. The chapter explaining the untyped lambda calculus gave me more than one occasion to pause and think. The lambda calculus, invented in the 30s, is a formal system for expressing computation with functions and can also been seen as a programming language. A very simple one, yet a surprisingly very powerful one. A language where all computations are done with functions, thus the first functional programming language in the history of mankind. It plays an important role in the field of research but also as a theoretical foundation for implementing functional languages such as Haskell, OCaml, Scheme, Clojure etc.
The lambda calculus is composed of three elements (or in programming language theory jargon, of three terms):
• variables, of the form `x`
• abstractions, of the form `λx.t`
• applications, of the form `t s`
An abstraction can almost be seen like a function in “classic” programming languages. An application “applies” its arguments to an abstraction. There is nothing more. There are no booleans, numbers, strings, conditional statements etc. Functions accept only one argument. Yet, everything can be computed with the lambda calculus.
A variable `x` is bound if it occurs in the abstraction of the body `t` of an abstraction `λx.t`. It is is free if it appears if it is not bound by an enclosing abstraction on `x`. For example in ```(λx.x) x``` the first `x` is bound, the second one is free. A term with no free variables is called a combinator. 1
The simplest combinator is the identity function. It returns its argument unchanged:
`id = λx. x`
There are different strategies when choosing on how arguments are evaluated before applying them to functions. That is, there are different strategies to reduce the terms. The most common one is call-by-value. Under this strategy a function can be applied only when its right-hand side has been reduce to a value. A value is a term that cannot be further reduced.
Thus `(λx.x x) (λx.x) y` reduces to `(λx.x) (λx.x) y` in the first evaluation step then to `(λx.x) y` then to `y`. Functions with multiple arguments can be simulated by having a function returning another function: `λx.λy.x y`, for example. This is familiar to functional programmers as currying.
What really impress me is that every computation are possible with such a simple language. I will give a few examples from the book, some are answers from exercises, so I you indent to read it I would suggest to stop your reading here.
Boolean values can be encoded as functions like this:
`tru = λt. λf. t`
`fls = λt. λf. f`
Tru is a function that discards the value of its second argument and returns the value of its first argument. Fls is a function that discards the value of its first arguments and returns the value of its second argument. They are used as a representation of the True and False booleans.
The test combinator can be defined like this:
`test = λl. λm. λn. l m n`
If the first argument of test is `tru` the expression will reduce to `m`, if it is `fls` it will reduce to `n`. For example ```test tru x y``` evaluates to `tru x y` which evaluates to `x`. Awesome, it is almost like a if statement!
Numbers can be encoded too. The Church numerals 2 are defined like this:
`c0 = λs. λz. z`
`c1 = λs. λz. s z`
`c2 = λs. λz. s (s z)`
and so on and so on. Functions such as `scc` (successor) and `plus` can also be defined.
`scc = λn. λs. λz. s (n s z)`
`plus = λm. λn. λs. λz. m s (n s z)`
Data structures such as pairs can also be encoded:
`pair = λf. λs. λb. b f s`
`fst = λp. p tru`
`snd = λp. p fls`
Take a paper and a pen and evaluate `fst (pair v w)`. It evaluates to `v`, as expected.
Lists can also be encoded. The list `[x, y, z]` can be encoded as the abstraction `λc. λn. c x (c y (c z n))`. That is a fold (also known as a reduce) is used to encode the list.
The usual functions to manipulate lists can be defined:
`nil = λc. λn. n`
`isnil = λl. l (λh. λt. fls) tru`
`cons = λh. λt. λc. λn. c h (t c n)`
`head = λl. l (λh.λt.h) fls`
`tail = λl. fst (l (λx. λp. pair (snd p) (cons x (snd p))) (pair nil nil))`
It is also possible to encode lists with pairs.
A mechanism similar to recursion can be defined by having functions reducing their arguments to a form similar to the function itself. For example this combinator:
`(λx. x x) (λx. x x)` reduces to himself.
More advanced tricks allow to define the equivalent of recursive functions such as the canonical definition of factorial.
Lambda calculus is a very simple but amazingly powerful language. It is also striking to make the comparison with some programming languages such as Java that could not even pass simply a function around before the most recent versions.
Congratulation for reading until there! Given all the shortcuts I took to explain, if you understood everything, well done.
## Footnotes:
1
I’m paraphrasing the book here and also a lot below!
2
Alonzo Church invended the lambda calculus.
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https://github.com/sympy/sympy/issues/12019
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Matrix.inv_mod gives multiplicative inverse in python2.7 OSX but not in python3.5 WIN #12019
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opened this Issue Jan 4, 2017 · 0 comments
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commented Jan 4, 2017 edited
Note: The sympy versions are different as well, with the newer version demonstrating the (apparent?) error ``````import numpy as np from sympy import Matrix # works F = np.array([[2, 7, 1, 4], [7, 2, 7, 1], [4, 7, 2, 7], [1, 4, 7, 2]], dtype=int) G = np.array([[int(c) for c in r] for r in Matrix(F).inv_mod(11).tolist()], dtype=int) np.dot(G, F) % 11 # doesn't X = np.array( [[2, 8, 17, 2, 20, 12, 5, 4, 17, 4, 13, 2], [4, 2, 8, 17, 2, 20, 12, 5, 4, 17, 4, 13], [2, 4, 2, 8, 17, 2, 20, 12, 5, 4, 17, 4], [13, 2, 4, 2, 8, 17, 2, 20, 12, 5, 4, 17], [4, 13, 2, 4, 2, 8, 17, 2, 20, 12, 5, 4], [17, 4, 13, 2, 4, 2, 8, 17, 2, 20, 12, 5], [4, 17, 4, 13, 2, 4, 2, 8, 17, 2, 20, 12], [5, 4, 17, 4, 13, 2, 4, 2, 8, 17, 2, 20], [12, 5, 4, 17, 4, 13, 2, 4, 2, 8, 17, 2], [20, 12, 5, 4, 17, 4, 13, 2, 4, 2, 8, 17], [2, 20, 12, 5, 4, 17, 4, 13, 2, 4, 2, 8], [17, 2, 20, 12, 5, 4, 17, 4, 13, 2, 4, 2]], dtype=int) Y = np.array([[int(c) for c in r] for r in Matrix(X).inv_mod(19).tolist()], dtype=int) np.dot(Y, X) % 19 `````` With OSX: ``````\$ pip freeze | egrep "numpy|scipy|sympy" numpy==1.10.4 scipy==0.16.0 sympy==0.7.6.1 Python 2.7.11 |Anaconda 2.4.0 (x86_64)| (default, Dec 6 2015, 18:57:58) [GCC 4.2.1 (Apple Inc. build 5577)] on darwin >>> np.dot(G, F) % 11 array([[1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1]]) >>> np.dot(Y, X) % 19 array([[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]]) `````` but on Windows: ``````Python 3.5.2 |Anaconda 4.2.0 (64-bit)| (default, Jul 5 2016, 11:41:13) [MSC v.1 900 64 bit (AMD64)] on win32 >>> numpy.__version__ '1.11.1' >>> scipy.__version__ '0.18.1' >>> sympy.__version__ '1.0' >>> np.dot(G, F) % 11 array([[1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1]], dtype=int32) >>> np.dot(Y, X) % 19 array([[ 7, 1, 8, 4, 5, 1, 2, 12, 10, 10, 12, 7], [ 7, 8, 1, 7, 15, 7, 18, 8, 16, 13, 11, 12], [ 8, 6, 6, 1, 18, 18, 12, 7, 4, 6, 2, 2], [11, 15, 7, 4, 6, 10, 13, 12, 14, 18, 5, 6], [11, 0, 10, 11, 3, 2, 16, 15, 5, 12, 0, 12], [12, 17, 3, 1, 6, 6, 16, 14, 14, 12, 17, 10], [14, 15, 4, 13, 16, 13, 16, 2, 10, 7, 10, 2], [18, 2, 11, 8, 0, 16, 13, 17, 9, 10, 7, 0], [16, 5, 11, 6, 12, 9, 6, 6, 10, 4, 0, 10], [18, 10, 12, 9, 13, 15, 13, 0, 15, 17, 3, 11], [ 6, 8, 2, 13, 18, 6, 11, 14, 7, 4, 10, 18], [12, 4, 9, 3, 17, 15, 7, 15, 2, 6, 16, 3]], dtype=int32) ``````
changed the title from mod_inv gives multiplicative inverse in python2.7 OSX but not in python3.5 WIN to Matrix.inv_mod gives multiplicative inverse in python2.7 OSX but not in python3.5 WIN Jan 4, 2017
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https://cn.maplesoft.com/support/help/errors/view.aspx?path=Fractals/LSystem/Iterate&L=C
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Iterate - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
Fractals[LSystem]
Iterate
Formal Language Iterator
Calling Sequence Iterate(state, rules, iterations)
Parameters
state - string or Vector(integer) ; the current state of the LSystem rules - list(character=string) or table(integer=list(integer)) ; the iteration rules for the given state iterations - nonnegint ; (optional) number of iterations to be applied
Options
• returnVector : keyword option of the form returnVector=value where value is true or false. If returnVector=true then a Vector corresponding to the ascii representation of the new state will be returned, otherwise the new state will be returned in the same format as state.
Description
• The Iterate command takes in a state and applies an iteration rule to each element, which is typically used to form an LSystem state.
• The state parameter is a sequence of characters. It can be input as a string, or as Vector of integers in the range [1,255], whose values correspond to the ascii values of the characters of the sequence.
• The iteration rules are defined by the rules parameter. An iteration will replace each element in the state with its corresponding iteration rule. If rules is input as list(character=string), the right-hand side of each equation in the list will replace the corresponding left hand side character at each occurrence of the character in the state. An element with no iteration rule defined will map to itself. The rules parameter may also be input as table(integer=list(integer)) (that is, a table where integers in the range [1,255] index to lists of integers in the range [1,255]). Each integer corresponds to an ASCII value and each occurrence of an ASCII value in state will get replaced by the corresponding list of integers.
• The iterations parameter defines the number of iterations that will be performed. This value is 1 by default.
• Iterate returns the iterated state in the same format as the input of state, unless otherwise specified by the returnVector option.
• The Iterate command can be used with LSystem[LSystemPlot] to create LSystem plots.
Examples
> $\mathrm{with}\left(\mathrm{Fractals}:-\mathrm{LSystem}\right):$
> $\mathrm{state},\mathrm{rules}≔"F",\left["F"="FF-\left[-F+F+F\right]+\left[+F-F-F\right]"\right]$
${\mathrm{state}}{,}{\mathrm{rules}}{≔}{"F"}{,}\left[{"F"}{=}{"FF-\left[-F+F+F\right]+\left[+F-F-F\right]"}\right]$ (1)
> $\mathrm{newstate}≔\mathrm{Iterate}\left(\mathrm{state},\mathrm{rules}\right)$
${\mathrm{newstate}}{≔}{"FF-\left[-F+F+F\right]+\left[+F-F-F\right]"}$ (2)
> $\mathrm{state},\mathrm{rules}≔"A",\left["A"="ABBA","B"="BAAB"\right]$
${\mathrm{state}}{,}{\mathrm{rules}}{≔}{"A"}{,}\left[{"A"}{=}{"ABBA"}{,}{"B"}{=}{"BAAB"}\right]$ (3)
> $\mathrm{newstate}≔\mathrm{Iterate}\left(\mathrm{state},\mathrm{rules},3\right)$
${\mathrm{newstate}}{≔}{"ABBABAABBAABABBABAABABBAABBABAABBAABABBAABBABAABABBABAABBAABABBA"}$ (4)
> $\mathrm{state},\mathrm{rules}≔"ABC",\left["A"="B","B"="C","C"="A"\right]$
${\mathrm{state}}{,}{\mathrm{rules}}{≔}{"ABC"}{,}\left[{"A"}{=}{"B"}{,}{"B"}{=}{"C"}{,}{"C"}{=}{"A"}\right]$ (5)
> $\mathrm{newstate}≔\mathrm{Iterate}\left(\mathrm{state},\mathrm{rules},3\right)$
${\mathrm{newstate}}{≔}{"ABC"}$ (6)
Compatibility
• The Fractals:-LSystem:-Iterate command was introduced in Maple 2015.
• For more information on Maple 2015 changes, see Updates in Maple 2015.
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https://www.aqua-calc.com/calculate/volume-to-weight/substance/lanthanum-blank-oxybromide
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# Weight of Lanthanum oxybromide
## lanthanum oxybromide: convert volume to weight
### Weight of 1 cubic centimeter of Lanthanum oxybromide
carat 31.4 ounce 0.22 gram 6.28 pound 0.01 kilogram 0.01 tonne 6.28 × 10-6 milligram 6 280
### The entered volume of Lanthanum oxybromide in various units of volume
centimeter³ 1 milliliter 1 foot³ 3.53 × 10-5 oil barrel 6.29 × 10-6 Imperial gallon 0 US cup 0 inch³ 0.06 US fluid ounce 0.03 liter 0 US gallon 0 meter³ 1 × 10-6 US pint 0 metric cup 0 US quart 0 metric tablespoon 0.07 US tablespoon 0.07 metric teaspoon 0.2 US teaspoon 0.2
• For instance, calculate how many ounces, pounds, milligrams, grams, kilograms or tonnes of a selected substance in a liter, gallon, fluid ounce, cubic centimeter or in a cubic inch. This page computes weight of the substance per given volume, and answers the question: How much the substance weighs per volume.
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BUTTERMILK PANCAKE and WAFFLE COMPLETE MIX, BUTTERMILK, UPC: 032251122903 weigh(s) 129 grams per metric cup or 4.3 ounces per US cup, and contain(s) 361 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
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CaribSea, Marine, Arag-Alive, Indo-Pacific Black weighs 1 441.7 kg/m³ (90.00239 lb/ft³) with specific gravity of 1.4417 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Chloroperoxyl, gas [ClO2] weighs 9.99 kg/m³ (0.00577459 oz/in³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-410A, liquid (R410A) with temperature in the range of -40°C (-40°F) to 60°C (140°F)
#### Weights and Measurements
A foot-pound (ft-lb) is a non-SI (non-System International) measurement unit of energy or heat
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lb/US c to g/ml conversion table, lb/US c to g/ml unit converter or convert between all units of density measurement.
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# Remainder Theorem: Definition & Examples
Instructions:
question 1 of 3
### For the function g(x) = -2x^4 - 3x^3 + 5x^2 - 4x + 1, find g(-2) by using the remainder theorem.
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### 2. For the function f(x) = x^3 + 2x^2 - 7x + 4, which of the following (if any) are zeroes?
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Quizzes, practice exams & worksheets
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Mathematics is filled with efficient methods for performing complex operations. The remainder theorem is one such method that can be utilized when solving particular problems and together. The quiz and worksheet for this lesson will help you to understand and recall the important information regarding this mathematical theorem.
## Quiz & Worksheet Goals
The quiz and worksheet will test you on:
• Answer problems using the remainder theorem
• Find the zeroes in a particular equation
• Mathematical equations & processes
• Example problems
## Skills Practiced
These tools will assess your skills of:
• Reading comprehension - ensure that you draw the most important information from the related remainder theorem lesson
• Problem solving - use acquired knowledge to solve function practice problems
• Interpreting information - verify that you can read information regarding mathematical processes and interpret it correctly
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https://www.jiskha.com/questions/76137/f-t-sqrtt-2-1-f-9-sqrt-0-2-1
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# algebra
f(t)=sqrtt^2+1
f(-9)=sqrt-0^2+1
1. 👍
2. 👎
3. 👁
1. Since sqrt t^2 = t, I will assume you mean
f(t) = sqrt (t^2+1)
and not f(t) = sqrt(t^2)+1 = t + 1
I also will assume that the positive square root is to be used.
I have no idea what your second line is supposed to mean. You have not asked a question.
If you are trying to calculate f(0) or f(-9),then
f(0) = sqrt (1) = 1
f(-9) = sqrt(81 + 1) = 9.055
1. 👍
2. 👎
2. p=2l+2w forl
1. 👍
2. 👎
## Similar Questions
1. ### Math
1. The length of the hypotenuse of a 30-60-90 triangle is 7. Find the perimeter. A) 7/2+21/2 sqrt 3 B) 21+7 sqrt 3 C) 7+21 sqrt 3 D) 21/2 + 7/2 sqrt 3 Could someone please help me, I don't know how to do this. Thank you!
2. ### algebra
am I right? 1. Simplify radical expression sqrt 50 5 sqrt ^2*** 2 sqrt ^5 5 sqrt ^10 5 2. Simplify the radical expression sqrt 56x^2 28x 2x sqrt 14*** 2x sqrt 7 sqrt 14x2 3. Simplify the radical expression. sqrt 490y^5w^6 2 sqrt
3. ### algebra
Simplify: 2 sqrt (3) + 6 sqrt(2) - 4 sqrt(3) + sqrt (2) a) 8 sqrt(2) - 3 sqrt(3) b) 6 sqrt(2) - 8 sqrt(3) c) 5 sqrt(6) d) 7 sqrt(2) - 2 sqrt(3) the answer i picked was d
4. ### math;)
Find the unit vector in the direction of u=(-3,2). Write your answer as a linear combination of the standard unit vectors i and j. a. u=-3[sqrt(13)/13]i+2[sqrt(13)/13]j b. u=-3[sqrt(5)/5]i+2[sqrt(5)/5]j c.
1. ### algebra
Simplify: sqrt (21) (sqrt(7) + sqrt(3)) a) 7 sqrt (3) + 3 sqrt(7) b) 7 sqrt(3) + sqrt (3) c) sqrt (210) d) sqrt (147) + sqrt (63)''. a
2. ### Algebra
Evaluate sqrt7x (sqrt x-7 sqrt7) Show your work. sqrt(7)*sqrt(x)-sqrt(7)*7*sqrt(7) sqrt(7*x)-7*sqrt(7*7) sqrt(7x)-7*sqrt(7^2) x*sqrt 7x-49*x ^^^ would this be my final answer?
3. ### algebra
what is the simplified form of 3 sqrt (5c) x sqrt (15c^3) choices are a. 15c^2 sqrt (3) b.6c^2 sqrt (5) c. 5c^2 sqrt(3) d. 12c^4 sqrt(5) please help i don't understand how to do the question i would really appreciate it
4. ### ap calculus
Which of the following definite integrals gives the length of y = e^(e^x) between x=0 and x=1? All the answers are preceded by the integral sign from 0 to 1. (a) sqrt[1 + e^(2*(x+e^x))] dx (b) sqrt[1 + e^(4x)] dx (c) sqrt[1 +
1. ### Algebra
Which of these expressions is in simplified form? A. a ^3 sqrt 4 - b ^3 sqrt 2 / 2 B. sqrt 1/2x + sqrt 1/2z C. x^2 - 3x sqrt y / sqrt 3 D. sqrt 125x - x^2
2. ### Math:)
A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise around a point that is centered at the origin. What is the exact value of the position of the rider after the carousel rotates
3. ### is this correct?- math
find the domain of the real valued function; f(x) = sqrt(5 - (sqrtx)) my solution: 5 - (sqrt x) >=0 -(sqrt x) >= -5 (sqrt x)
4. ### Calculus
Graph the curve and find its exact length. x = e^t + e^-t, y = 5 - 2t, from 0 to 3 Length = Integral from 0 to 3 of: Sqrt[(dx/dt)^2 + (dy/dt)^2] dx/dt = e^t - e^-t, correct? dy/dt = -t^2 - 5t, correct? So: Integral from 0 to 3 of
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## Astronomy 161: Review Guide for Second Quiz
### Mechanics
Same as for the first quiz. The quiz will have 10 multiple choice questions, and it will occupy the first 20 minutes of Friday's class. If you're late, that's your problem ... in accordance with the makeup policy, you can only take a makeup quiz if you have notified me in advance that you will be absent. The quiz is closed book, but you may bring a single sheet (both sides) of handwritten notes. You may bring a calculator, but you will not need it. Please bring a number 2 pencil.
### Topics
The quiz will cover the material in Lectures 5-9 (i.e., through Wednesday's lecture). The reading for all of these lectures is chapter 4 of the textbook.
Names to know: Ptolemy (2nd Century AD), Copernicus (1473-1543), Tycho (1546-1601), Kepler (1571-1630), Galileo (1564-1642), Newton (1642-1727)
I don't expect you to know dates, but you should know the general order in which things happened, which is evident from the birth and death dates above: Copernicus developed his model first, Tycho carried out his observations later in the same century, Kepler and Galileo were contemporaries with Kepler building on Tycho's observations, and Newton worked after Kepler and Galileo had made and published their discoveries.
While there will not be any complicated calculations on the quiz or the midterm, we have by now covered several important equations. I expect you to understand the physical content of these equations and the ways that they can be applied, and to be able to do simple calculations with them (e.g., if I double the distance between two objects, what happens to the gravitational force between them?). The important equations through lecture 9 are:
• a = d x (theta / 57.3 degrees)
• P2 = a3
• F = m a
• F = G M1 M2 / d2
• (r / 1 AU)3 = (P / 1 year)2 x (M / Msun)
### Review Advice
As before, my recommendation is to make sure that you are up to date with the reading and to spend a couple of hours going over the lecture notes, concentrating on any of the topics you found confusing. Think particularly about how to answer the "Key Questions" mentioned at the start of each lecture --- if you can run confidently through the answers to these questions in your head, then you will be in excellent shape for the quiz. You should also review the in-class questions from these lectures.
Go to the A161 home page
Go to David Weinberg's Home Page
Updated: 2005 April 18[dhw]
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# hi
More Options: Make a Folding Card
#### Storyboard Description
This storyboard does not have a description.
#### Storyboard Text
• I wonder how many floors are in my apartment altogether. Since there are 61 floors above ground level and 1 floor below ground level you would need to find the difference, 61 - (-1)= 62. The difference between the highest and lowest floor in my apartment is 62.
• Today I walked -2 x 5= -10 floors from the 20th floor. My final location would be 40 floors above my apartment. Since I live on the 30th floor, 10 floors below the 20th floor would equal 40 floors above me. Therefore, my location is +40.
• Yesterday, I walked down 90 feet in 9 minutes. -90/9= -10 feet per minute. Today, I walked down 240 feet in 12 minutes. -240/12= -20 feet per minute. Therefore, I walked faster today.
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# Search results
Found 521 matches
Damping factor (Series RLC circuit)
Damping is caused by the resistance in the circuit. It determines whether or not the circuit will resonate naturally. Circuits which will resonate in this ... more
Quality Factor
In physics and engineering the quality factor or Q factor is a dimensionless parameter that describes how under-damped an oscillator or resonator is, or ... more
Resonance frequency RLC
Resonance frequency is the ability of a circuit to resonate at a specific frequency, the resonance frequency related to the angular frequency
... more
Damping ratio (related to Quality factor)
Formula first contributed by:
trooper
In engineering, the damping ratio is a dimensionless measure describing how ... more
Resonance frequency in LC circuits
An LC circuit, also called a resonant circuit, tank circuit, or tuned circuit, is an electric circuit consisting of an inductor, represented by the letter ... more
Q factor for a series resonant circuit (RC circuits)
It is defined as the peak energy stored in the circuit divided by the average energy dissipated in it per cycle at resonance; Q factor is directly ... more
Resonance frequency in LC circuits (angular)
An LC circuit, also called a resonant circuit, tank circuit, or tuned circuit, is an electric circuit consisting of an inductor, represented by the letter ... more
Q factor for a series resonant circuit (RL circuits)
It is defined as the peak energy stored in the circuit divided by the average energy dissipated in it per cycle at resonance; Q factor is directly ... more
Miles Equation
In 1954, Miles developed his version of this equation for GRMS as he was researching fatigue failure of aircraft structural ... more
Vacuum wavelength
When an electromagnetic wave travels through a medium in which it gets attenuated (this is called an “opaque” or “attenuating” ... more
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Go4Expert (http://www.go4expert.com/)
- C++ (http://www.go4expert.com/forums/cpp/)
- - convert fortran to C++ (http://www.go4expert.com/forums/convert-fortran-cpp-t11975/)
norainon 9Jul2008 14:02
convert fortran to C++
i have a question regarding to convert from fortran language to C++ programming especially in diognal array
for example;
DO 10 I=1,3*(X+2)
10 ID(I)=(I-1)*m+1
how do i convert into C++ programming
xpi0t0s 15Jul2008 13:27
Re: convert fortran to C++
I'm not clear what the above code has to do with diagonal arrays; ID looks like a simple array to me.
int ID[SUITABLE_LIMIT];
for (int i=1; i<=3*(x+2); i++) ID[i]=(i-1)*m+1;
Don't forget C++ arrays are zero based so SUITABLE_LIMIT must be at least the maximum value of 3*(x+2) PLUS ONE. Also the above loop doesn't initialise ID[0]; watch out for that one, maybe use an array name that indicates it's 1-based rather than zero based, something along the lines of Hungarian Notation but simpler, e.g int ID_nz[..], nz meaning "no zero element". I always end up confusing myself if I use some arrays that are zero based and some that are not, in the same program.
abdunabi 7Mar2012 18:56
Re: convert fortran to C++
I have a question regarding to convert from fortran language to C++ programming. How I convert into C++ programming , Please someone help me!!!
Code:
``` DIMENSION TK(15), PK(15), TC(15),DT(15),DP(15),H(15),PAR(15),Q(15), M(15), PO(15), PO1(15), PO2(15), PI(15), 1 FORMAT (213) 5 FORMAT (2F8.4,I3) 3 FORMAT (3F10.4,I3) 15 FORMAT (///,8F10.5,/) READ (5,1) NG, M1 READ (5,3) A,B,P,M2 READ (5,5) DTF,DPF,Z 25 FORMAT (F10.3) 98 READ (5,25) T IF(T.EQ.0) GO TO 99 DO 100 I=1, NG M(I)=M1+I-1 TK(I)=A*(8*M(I)+M2)**P+8 DT(I)=(M(I)-1)*0.02+DTF TC(I)=TK(I)/(0.567+DT(I)-DT(I)**2) DP(I)=(M(I)-1)*0.227+DPF PK(I)=(14*M(I)+Z)/(DP(I)+0.34)**2 H(I)=(TK(I)/TC(I))*(ALOG10(PK(I))/(1-TK(I)/TC(I))) PAR(I)=ALOG10(PK(I))+H(I)*(1-TC(I)/T) PO2(I)=(14*M(I)+Z)/(82.06*TK(I)) PI(I)=5*(T/TC(I)-1) PO1(I)=PO2(I)*10**PI(I) PO3=1 PO(I)=PO1(I)+PO3*(1-T/TC(I))**0.3 100 Q(I)=10**PAR(I) 20 FORMAT (//,10F10.5,/) WRITE(6,20) T WRITE(6,15) (Q(I),I=1,NG) WRITE(6,6) (PO(I),I=1,NG) 6 FORMAT (///,8F10.6,/) GO TO 98 99 CONTINUE STOP END```
All times are GMT +5.5. The time now is 14:13.
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$(3-x)\sqrt[3]{2x}$
Simplifying each term and then combining like terms, the given expression, $\sqrt[3]{54x}-\sqrt[3]{2x^4} ,$ is equivalent to \begin{array}{l}\require{cancel} \sqrt[3]{27\cdot2x}-\sqrt[3]{x^3\cdot2x} \\\\= \sqrt[3]{(3)^3\cdot2x}-\sqrt[3]{(x)^3\cdot2x} \\\\= 3\sqrt[3]{2x}-x\sqrt[3]{2x} \\\\= (3-x)\sqrt[3]{2x} .\end{array} * Note that it is assumed that all variables represent positive numbers.
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Quaternions. Problem with understanding essence of.
I can't calm down when I can't realized of principles of working something. Quaternion is that place where I spend а few days! Unfortunately, not one lectures or books what I saw or read don't explain principe of working of quaternion in 4d, as it understood sir Hamilton. Yes we works in 4d and it's so hard to explain, but if quaternions is a expand of complex numbers, it's must works simillary. For example, why we use in Euler formula for quaternions sin for any three coordinates of vector, if they perpendicular to each other in imagine space? At the same time we use cos only for real part. $$ql = \left[cos (\frac{1}{2} \theta) , sin (\frac{1}{2} \theta) (i+j+k)\right]$$
I explain it for myself as three Cartesian plane with the one common real component which defines a vector position in 4d. This is my attempt to somehow imagine a quaternion. Is it far from reality?
Why we disregard of real part of coordinate in first part of Euler formula?
$$ql = \left[cos (\frac{1}{2} \theta) (w), sin (\frac{1}{2} \theta) (i+j+k)\right]$$
Why we need multiply quaternions by involving sandwich if we don't need that in complex number multipliction?
$$p^. = qpq^-1$$
May be I'm on wrong way to understanding of quaternion by equating it to 2d complex number, let me know.
• It is often misleading to compare a non-commutative ring to a commutative one. Although quaternions can seem at first sight only an "enhanced version" of complex numbers, the non-commutativity of the product give to them some properties that have just no counterpart in $\mathbb{C}$ (non-trivial conjugation, for instance, that you call for some strange reason "sandwich"). – Francesco Polizzi Jun 20 '17 at 13:13
• Sorri, I met this slang in couple books. :) But it's theory of algebras and this only makes it difficult to understand the essence of quaternions. – hardCode Jun 20 '17 at 13:38
Well it is always tricky trying to answer the question "why do things work they way they do" because the answer is usually a moving target.
Unfortunately, not one lectures or books what I saw or read don't explain principe of working of quaternion in 4d,
Well, Hamilton was not really 'thinking in $4$ dimensions' AFAIK... he was very interested in doing three dimensional geometry with quaternions. If you still think this is a sticking point you will have to explain what exactly you want to understand. As with most mathematics, you can get a get a long way by simply "getting used" to how things work, and then forming your own pictures as you go. Expecting a clear picture from the outset is often too ambitious.
I explain it for myself as three Cartesian plane with the one common real component which defines a vector position in 4d.
I do not think it is useful to think of quaternions as a vector position in $4$-D (that is going no further than thinking of $\mathbb R^4$.) Complex numbers are certainly a lot more than vector positions in $2$-D. You could think of $3$-space as two orthogonal Cartesian planes meeting on a real line, and you can think of $4$-space as three mutually orthogonal planes meeting on a real line. But this has more to do with $\mathbb R^3$ and $\mathbb R^4$ and not really anything to do with the quaternions.
This is my attempt to somehow imagine a quaternion.
Why is imagining quaternions any more challenging than imagining integers, rational numbers, or real numbers? It's just another, albeit different, number system that you can add, multiply and divide in. IMO more strenuous attempts to "imagine" ("imagine as an object in reality"?) do not yield anything useful compared to the amount of thinking that goes into it.
Is it far from reality?
I think we see this question sometimes, but it doesn't have an answer. I don't know what reality you're talking about. The usefulness of whatever picture a person has is relative to their own understanding of the subject, and stands on the merits of its own appeal. There is no standard of reality to measure a description against.
Why we disregard of real part of coordinate in first part of Euler formula?
I don't know what you're talking about. As I understand it, we do not disregard that part. I will give you my heuristic for rationalizing quaternion rotation below. I'm excerpting a couple slides from a talk I gave on quaternions:
1. If the coefficients of $q$ have Euclidean length $1$, then $q^{-1}=\bar q$.
2. If $v$ and $w$ have real part zero, then
1. The real part of $vw$ is $-1(v\bullet w)$.
2. The pure quaternion part of $vw$ is $v\times w$.
3. If $u^2=-1$, $(\cos(\theta)+u\sin(\theta))(\cos(\theta)-u\sin(\theta))=\cos(\theta)^2+\sin(\theta)^2=1$ (basic trigonometry)
4. If $u^2=-1$, $(\cos(\theta)+u\sin(\theta))(\cos(\theta)+u\sin(\theta))=\cos(2\theta)+u\sin(2\theta)$ (De Moivre's formula)
Rationalizing quaternions' multiplication action on the model of $3$-space
The model of $3$-space I'm referring to, of course, is the space of quaternions with real part zero. As usual, we take $q = \cos(\theta/2)+u\sin(\theta/2)$ as the rotation quaternion, where $u$ is a unit vector pointing along the axis of rotation and $\theta$ is the angle of rotation around the axis measured using the right-hand rule. We aim to make the 'sandwich' action look more like what happens in complex arithmetic
1. $u$ is unmoved by $q$: $$quq^{-1}= (\cos(\theta/2)+u\sin(\theta/2))u(\cos(\theta/2)-u\sin(\theta/2))=\\ (\cos(\theta/2)+u\sin(\theta/2))(\cos(\theta/2)-u\sin(\theta/2))u=\\ (\cos(\theta/2)^2-(u\sin(\theta/2)^2))u=\\ (\cos(\theta/2)^2+\sin(\theta/2)^2)u=u$$
2. if $v$ is a unit length pure quaternion orthogonal to $u$: $$qvq^{-1}= (\cos(\theta/2)+u\sin(\theta/2))v(\cos(\theta/2)-u\sin(\theta/2))=\\ (\cos(\theta/2)+u\sin(\theta/2))(\cos(\theta/2)+u\sin(\theta/2))v=\\ (\cos(\theta/2)+u\sin(\theta/2))^2v=\\ (\cos(\theta)+u\sin(\theta))v\leftarrow\text{looks like a rotation in the complex plane}$$
3. $q$ leaves $u$ unchanged and rotates its normal plane by $\theta$. Everything else follows rigidly, so we have the rotation explained in terms that look like complex arithmetic.
There is one critical thing to notice here, though: in the last expression, $u$ and $v$ would both be $i$ in complex arithmetic. Let me try to explain that. The circle of quaternions that cause rotations around $u$ live in the plane $P$ spanned by $1$ and $u$. They are acting on the set $P^\perp$, the orthogonal complement of the first plane. You can see we need at least $4$ dimensions to fit these together in the same space.
I don't know the right way to explain how this can be aligned with complex multiplication, but I believe there is a concrete rationalization. In simple terms, I think it has to do with shifting perspective that the things you are operating on live in $P^\perp$ to operating on things in $P$. In harder terms, I believe it has to do with a duality between $P$ and $P^\perp$, which I've seen explained in some texts on Clifford/geometric algebra, but I do not properly know.
Final word
I believe also there is another good explanation of how the 'sandwich' action arises, an explanation that relies on exponential maps and Lie algebra, which again I have not fully absorbed. I leave it to someone who is more familiar with that field to provide a complementary answer along those lines.
• Related but not quite a duplicate. – rschwieb Jun 20 '17 at 13:36
• Even more related but has quite a few elements that are different from this question. – rschwieb Jun 20 '17 at 13:38
I found this helpful article for basic understanding of quaternions.
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1. Order of Kilopi
Join Date
Nov 2002
Posts
6,238
in all mainstream versions of time dilation that i've read about, the operation revolves around determining gamma.
Instead of reading about it, how about following the actual equations used in Einstein's paper and show us precisely, in the equations, where, what and why you specifically disagree with the conclusions.
2. Tensor
i recently reopened this thread with a post detailing some simulated experiments.
the reasoning for the experiments was to check the validity of the pythagorean geometry involved in the generation of gamma.
the results showed that the required hypotenuse line cannot be formed in the unit time allowed.
therefore there is no base for the maths to apply to.
3. Established Member
Join Date
Nov 2004
Posts
1,541
This paradox is observed in the Astronomy. A Quasar ejects a jet, This jet emits a light. For us a distant observers seems it like velocity faster then "c".
It is only illusion because a time dilation. There is just one second longer and light can travel a longer distance in the same time.
In a Madman example a photon will reach both mirrors for an inner and outside moving observers because a time dilation and space deformation. The outside observer will see an additionally redshift of the photon. We do not exactly know who is moving - the 2 mirrors or an outside observer. Effect is the same.
i recently reopened this thread with a post detailing some simulated experiments.
the reasoning for the experiments was to check the validity of the pythagorean geometry involved in the generation of gamma.
the results showed that the required hypotenuse line cannot be formed in the unit time allowed.
therefore there is no base for the maths to apply to.
Your reasoning was flawed. You said:
but einstein then assumes (without proof) that the light will travel from one mirror to the other in the same amount of time regardless of whether the clock is still or in motion.
What he assumed was that light would travel at the same speed in a frame inertially moving relative to the light clock, which results in the time taken for the light to travel from one mirror to the other to be greater in that moving frame, not the same.
and if your version of time dilation requires a change in relative motion only...then how does a body sustain a time dilation effect for any appreciable time?
Ah, this old chesnut had me confused for ages and is what initially brought me to bad astronomy. Time dialation is symmetrical all the time both twins are moving apart (or together) intertially -- they both see each others clock running slow [ after taking into account the time taken for the light signals from each other's clock to reach them - this is what is usually meant by "what they see" in these sort of thought experiments ]. This is not paradoxical because they cannot directly compare clocks without coming back together, which would require at least one of them to accelerate.
It is only when one turns around that the symmetry is broken. A sudden turn around causes a sudden forward leap of the other's clock. It is caused by the Lorentz transformation from one intertial frame to the other, the transformation changes what is now over there. And the further away "over there" is, the bigger the leap.
A more realistic turn around over some time causes the other's clock to run fast, according to the turning twin, throughout the turn around. Interestingly, if you integrate over many small Lorentz transformations to work out the effect of accelerating the twin, rather than just suddenly changing his direction of motion, then the effect comes out the same as gravitational time dilation due to a unform g field equivalent to the acceleartion -- the stay-at-home twin is very high up in this pseudo g-field of the turning-around twin and has a subsequently faster running clock (according to the turning-twin) throughout the turn around.
5. worzel (and Tim too)
yes it was wrong of me to use the term "time".
velocity or speed would have been better.
look at the 2nd gif again.
the vertical speed of the light pulse is the same as shown in the first gif (and since it is the unit measurement for light-speed i called it "time" by mistake).
worzel
i can't believe in sudden alterations or "symmetry breaking" concepts as being required for time dilation.
supposedly the basic operation relies upon the strict formation of a right angled triangle.
from which gamma may be derived.
this requires a set unit of velocity for eg: a spaceship.
in the gif examples i posted (taken from and based on many physics websites) the exercise is organised to allow the transmission of the light signal over a physical distance (the hypotenuse) which is always longer than the vertical distance (hence the difference, and supposed acceptable input value for time dilation).
this transmission completes the function by creating the required physical parameter known as "the observer".
without the observer no data is transferred and no result of "time dilation" may be returned.
for example...no one can see the "triangle" or "the zig-zag"....because only the mirror/observer contact points receive data (if the triangle were a polygon..then the observers see only the vertices).
we would not see the light clock "tick-tock" in a meadow near a train because it is not sending that data to us.
if the observer/mirror is kept as part of the exercise then he/it will observe either a tick or a tock...but not both.
6. Ok, slow down a bit Madman. Start from the beginning.
Do you agree that experiments before the formulation of SR suggested that the speed of light was always the same (against expectations at the time)?
Do you agree that Maxwell's equations imply that the speed of light is always c in a vacuum?
PS I don't understand Maxwell's equations, but I'm happy to accept the word of those that do until such time as I can understand them myself.
7. yes...we are dealing at all times (in these exercises) with light travelling at speed C.
relative motion of frames would be seen as doppler effects.
8. @czeslaw
************************************************** *******************************
how the extreme value of Gamma is determined by decimal values when velocity is brought asymptotically closer to C.
************************************************** *******************************
first, the formula for Gamma.
**********************************
gamma = 1/sqrt(1-(v*v)/(c*c))
**********************************
and a few examples of deriving Gamma for various velocities...
velocity = 0.9C
gamma = 1/sqrt(1-(90*90)/(100*100))
gamma = 1/sqrt(1-(8100/10000))
gamma = 1/sqrt(1-0.81)
gamma = 1/sqrt(0.19)
gamma = 1/0.4359
gamma = 2.2941
*********************************
velocity = 0.99C
gamma = 1/sqrt(1-(99*99)/(100*100))
gamma = 1/sqrt(1-(9801/10000))
gamma = 1/sqrt(1-0.9801)
gamma = 1/sqrt(0.0199)
gamma = 1/0.1411
gamma = 7.0872
*********************************
velocity = 0.999C
gamma = 1/sqrt(1-(999*999)/(1000*1000))
gamma = 1/sqrt(1-(998001/1000000))
gamma = 1/sqrt(1-0.998001)
gamma = 1/sqrt(0.001999)
gamma = 1/0.0447
gamma = 22.3714
*********************************
note the 3rd last lines in each block of calculations (above)...which i've listed below.
gamma = 1/sqrt(0.19)
gamma = 1/sqrt(0.0199)
gamma = 1/sqrt(0.001999)
to get those values...a subtraction is performed.
gamma = 1/sqrt(1-0.81)
gamma = 1/sqrt(1-0.9801)
gamma = 1/sqrt(1-0.998001)
if the square and square root conversions are stripped out?...then we are left with the core operation.
C - velocity = time dilation (seed value)
*****************************************
the speed of light (C) is given the value 1.
velocity ranges from 0 to 1 (a fraction of light speed C).
velocity is subtracted from C (and the remainder becomes a seed value for Gamma).
so basically (in this slimmed-down exercise) we are dealing with the operational range of 1 - 0 up to 1 - 1.
ie(roughly):
1 - 0 = 1 (no time dilation)
1 - 0.1 = 0.9
1 - 0.2 = 0.8
1 - 0.3 = 0.7
1 - 0.4 = 0.6
1 - 0.5 = 0.5
1 - 0.6 = 0.4
1 - 0.7 = 0.3
1 - 0.8 = 0.2
1 - 0.9 = 0.1
1 - 1 = 0 (time stops?)
************************************************** ****
so what happens when velocity is brought asymptotically closer (nearer and nearer) to C....ie: >0.9C?
******************************************
C - velocity = time dilation seed value
******************************************
1 - 0.9 = 0.1
1 - 0.99 = 0.01
1 - 0.999 = 0.001
1 - 0.9999 = 0.0001
1 - 0.99999 = 0.00001
1 - 0.999999 = 0.000001
1 - 0.9999999 = 0.0000001
1 - 0.99999999 = 0.00000001
*****************************************
as the velocity increases towards 1, the product of "C - velocity" will become a smaller fraction.
accordingly with each decimal increase in velocity, the gamma seed value will be pushed a decimal place smaller as a fraction.
this is solely a product of the mathematical model chosen...it in no way relates to any supposed "special" qualities of "relativistic" speed.
but instead shows the real reason why Gamma attains "significant" values at speeds "very close to C".
**************************************************
here's another version
http://www.fourmilab.ch/cship/timedial.html
and a slightly tidied up copy of the proper results (from that page) to show the numerical progression.
************************************************** **
velocity.................................days..... .....years
************************************************** **
0.9.........................................2.2... .......0.006
0.99.......................................7...... .......0.019
0.999....................................22....... ......0.061
0.9999..................................70........ ......0.19
0.99999...............................220......... .....0.61
0.999999.............................700.......... ....1.9
0.9999999..........................2200........... ...6.1
0.99999999........................7000............ .19
0.999999999.....................22000............. 61
0.9999999999...................70000 ...........190
0.99999999999 ...............220000............610
0.999999999999..............700000...........1900
0.9999999999999...........2200000...........6100
0.99999999999999.........7000000..........19000
0.999999999999999......22000000..........61000
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so..are you and Grey now (instead of Sam5) arguing that this is not required?
I'm arguing that you're saying Einstein was wrong in assuming the light could travel along the diagonal in unit time, but since Einstein never said that, he can't have been wrong about it.
10. Grey
my impression is that the light is supposed to travel the length of the hypotenuse in the unit time.
this is supposed to create the quandary that einstein resolves by formulating the following.
and this is to explain how the light could travel the larger distance of the hypotenuse in the same amount of time as if it had travelled along the eg: vertical distance.
the light travels a longer distance than it should be able to...therefore the distance must contract.
it does it in less time than it should be able...therefore time is contracted too.
otherwise, how do your muons supposedly get to earth?
my impression is that the light is supposed to travel the length of the hypotenuse in the unit time.
this is supposed to create the quandary that einstein resolves by formulating the following.
No, "light speed must not alter" was the starting point, the quandary if you like. That time and space are interwoven in a 4d lorentzian geometry rather than a 3d Euclidean space and 1d absolute time was the resolution to that quandary. The problem is we're all hard-wired to think in Newtonianesque absolute space and time so the first hurdle to understanding relativity is to realize that this is just an assumption that happens to be very accurate on the scale of relative motions that we're used to, but can't be correct if light speed is constant for all intertial observers. I'm not sure, but I think you may still be stuck at the first hurdle.
and this is to explain how the light could travel the larger distance of the hypotenuse in the same amount of time as if it had travelled along the eg: vertical distance.
the light travels a longer distance than it should be able to...therefore the distance must contract.
it does it in less time than it should be able...therefore time is contracted too.
The light clock example demonstrates only time contraction, there is no length in the direction of motion to contract.
yes...we are dealing at all times (in these exercises) with light travelling at speed C.
relative motion of frames would be seen as doppler effects.
Right, stop there. Forget about doppler effects for now. Just focus on this. If a photon travels between two points at c relative to me, and the very same photon travels at c relative to you as well, even though you are moving relative to me at 0.5c then what's going on?
This implies that if I shoot a laser at your rapidly retreating back, but miss, the laser beam will leave me at c relative to me and then overtake you at c relative to you, even if you are running away from me and the gun at 0.99999c. Think about that, it is very strange.
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A.You say this 2 mirrors are moving and a distant observer stay at rest.
B.I say this 2 mirrors stay at rest but the distant observer is moving.
We both are right and the photons reach both mirrors in A and B.
Speed of light is always "c" - time dilation explains it.
13. worzel
in effect the light pulse travels the length of the hypotenuse in the unit time.
since (according to the gamma formula) the hypotenuse doesn't actually exist (only the 2 sides "C and V").
the pythagorean relationship that would be afforded by the hypotenuse is used in the maths to generate gamma...but the discrete time that would be associated with it's generation is not.
otherwise if we waited for the light pulse to complete it's journey along the hypotenuse it would take a time longer than 1 (aka: C)
gamma = 1/sqrt(1-(v*v)/(1.?c*1.?c))
this cannot occur if we are to form a triangle with sides C and V.
in other words...
the completion of the lines C and V require time...the hypotenuse must be completed at the same time as both of these.
14. What is the problem, really?
If the observer is at rest wrt the mirrors, the pulse reaches the other mirror in one time unit.
If another observer is moving wrt the mirrors, the pulse takes longer, gamma times the time unit.
This is implicitly understood to mean that both observers have also taken into consideration the time it took for them to "see" the events and calculated backwards to get the actual time between events.
According to the second observer, time passes slower for the mirrors. The first one would disagree. Both are correct, but since they have no way of comparing measurements directly, there isn't a problem.
15. Madman, suppose there is a light beam clock sitting next to me. I see the beam of light bouncing back and forth, right? Now, suppose that I go driving past the stationary light clock at high speed. What path do I see the light beam take?
16. I'll clarify. This is the situation as seen by the first observer.
These two describe the situation as seen by the second:
Both depictions describe the same situation, but from different perspectives. Both are equally correct, given their frame of reference.
17. Grey
when the experiment is arranged in the following way, i can accept that everything works out right.
but not as arranged and shown in the last 2 gifs (re-posted by AstroSmurf).
worzel
in effect the light pulse travels the length of the hypotenuse in the unit time.
No it doesn't. It completes the journey from one mirror to the other in the unit time when we are at rest relative to the light clock. If we move relative to the light clock then the path traced by the photon in our moving frame is longer and hence the time taken to complete the journey from one to the other is greater than one. You can use pythagorus' theorem to figure out by how much, gamma.
since (according to the gamma formula) the hypotenuse doesn't actually exist (only the 2 sides "C and V").
If we move distance x relative to the light clock while a photon moves distance y between the two mirrors, and x and y are at right angles then they form the two right sides of a right angled triangle and the hypotenuse is the path travelled by the photon in our frame of reference. I don't understand what your problem is with that.
the pythagorean relationship that would be afforded by the hypotenuse is used in the maths to generate gamma...but the discrete time that would be associated with it's generation is not.
What discrete time? The path between the mirrors is longer in our moving frame so the photon must take more time because it always goes at the same speed.
otherwise if we waited for the light pulse to complete it's journey along the hypotenuse it would take a time longer than 1 (aka: C)
It does in our moving frame, that's called time dilation.
the completion of the lines C and V require time...the hypotenuse must be completed at the same time as both of these.
I don't understand. y is the distance between the mirrors, x is the distance we move while the photon travels from on mirror to the other. So the photon travels along this hypotenuse of x and y in our moving frame. Where's the problem?
19. worzel
it is the maximum value in the operation and is the "time limit" for the completion of the triangle.
if the hypotenuse is not formed in that time...the triangle is not formed with side C.
i posted the last gif to illustrate a format in which the completion of the hypotenuse occurs with the correct time and coordinates.
in the 2 gifs reposted by AstroSmurf it does not.
i pointed out that regardless of this anomaly the gamma formula operates as if the hypotenuse is formed according to the unit time.
there is no anomaly when just using the formula, because it does not rely on the hypotenuse as an input....it only uses the y and x (C and V).
20. Ach, this is getting confusing!
Madman, let's start at the beginning.
You've got a photon gun pointing straight at a receiver. The distance between them is d. The whole apparatus is stationary. The photon gun fires a photon, the photon travels through the intervening space at velocity c for time d/c, and then hits the receiver. Right?
Now, let's say the whole apparatus is moving perpendicular to the gun-receiver line. It's moving at a significant velocity, say 0.6c. The photon gun fires a photon. Some questions:
1) Will the photon still hit the receiver?
2) Will the photon still be traveling at velocity c, and thus by necessity require more time than d/c to make the journey?
3) Is it true that there is no way to differentiate between
a) a stationary observer observing a moving gun-receiver apparatus; and
b) a moving observer observing a stationary gun-receiver apparatus?
If you answered "no" to any of those questions, then your problem is not with Relativity, it's with the experiments that have shown that the correct answer to those questions is "yes."
If you answered "yes" to all three questions, how do you propose to resolve the inherent conflicts?
21. worzel
look at the blue wave gif.
the movement of the top mirror produces an internal transmission delay.....the light must travel further to reach the mirror and take more time to do it.
the speed of light remains the same and it's delay in reaching the 2nd mirror is basically a doppler shift.
***************************************
these exercises based on 2d are too limited.
imagine a 3d object with light clocks pointed in every direction...give it a vector and you'll find that the internal clocks pointed in one direction are time dilated...and those pointed in the opposite direction are time contracted.
both are caused by doppler shifting due to relative motion.
we can only achieve einsteinian gamma formulations by using a very abstract and limited 2d model.
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the reasoning for the experiments was to check the validity of the pythagorean geometry involved in the generation of gamma.
the results showed that the required hypotenuse line cannot be formed in the unit time allowed.
therefore there is no base for the maths to apply to.
But, that was my whole point. You claim the line cannot form in the time allowed. SR claims otherwise. Now, we have two mutually exclusive claims (yours and SRs). You should be able to show exactly where SR is wrong. If you can't, I would say it's a fair assumption that you can't because there is nothing wrong and it's your claim is a simple misunderstanding on your part.
I would also point out that SR has predicts several other effects and has been incorporated QM, and is part of all the QM predictions. Observations, for both SR and QM match predictions to a high degree of accuracy. Why, if SR is so wrong, do those predictions match observations so closely.
when the experiment is arranged in the following way, i can accept that everything works out right.
It might be better to describe this verbally, or at least explain what the various parts of your animated gif represent. In this case, though, it seems clear that an observer at rest with respect to the red mirror sees the light travel in a straight line, while an observer with respect to the blue mirror sees the light travel along a diagonal path. Is this correct?
but not as arranged and shown in the last 2 gifs (re-posted by AstroSmurf).
The only important difference I see between these two is that in the one you just posted, the source is not moving. Are you claiming that makes a difference? You're also emphasizing a different reference frame, but of course either should be equally valid, right?
But mostly, I'm not sure that you've really answered my question. Say there are two parallel mirrors with a light beam bouncing between them, the whole thing in an inertial reference frame that I'm considering to be "at rest", at least for the moment. If I go zipping past these mirrors, perpendicular to the light path, do we agree that I will see the light pulse travelling in a zig-zag pattern?
worzel
No, y was the distance between the mirrors, as I said.
it is the maximum value in the operation and is the "time limit" for the completion of the triangle.
What does that mean? It is just the distance between the mirrors, nothing more.
if the hypotenuse is not formed in that time...the triangle is not formed with side C.
i posted the last gif to illustrate a format in which the completion of the hypotenuse occurs with the correct time and coordinates.
If you mean that the light can't travel along this diagonal path (i.e. in the frame of the moving observer) in the same amount of time that it does the vertical path (i.e. in the frame of the stationary observer) then you're right. It takes longer to traverse the diagonal path, so the time taken for the photon to traverse the distance between the mirrors is longer in the moving frame -- da da -- time dilation.
i pointed out that regardless of this anomaly the gamma formula operates as if the hypotenuse is formed according to the unit time.
I think you're just having difficulty letting go of your intuitive notion of absolute time. The 1 second tick of the light clock is longer in the moving frame precisely because it can't travel the diagonal path in the same time that it could travel the vertical path if it always travelled at c in all (inertial) frames.
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"In a paper published in 1918 Einstein corrected the “relative motion” error of his 1905 paper and he added “forces” and “atomic clocks” to his thought experiments. He changed the reason for the single clock slow-down from “relative motion” to “forces” exerted on the oscillating atoms in the single atomic clock that slowed down, and thus he basically returned to Lorentz’s basic electrodynamics concept of 1895, and Einstein's own 1911 gravitational redshift theory."
So time dilation is a is not reality, but clocks slowing down is.
So there is no such thing as time dilation.
26. Originally Posted by upriver
So time dilation is a is not reality, but clocks slowing down is.
So there is no such thing as time dilation.
So Sam5 claims.
27. Originally Posted by upriver
So time dilation is a is not reality, but clocks slowing down is.
So there is no such thing as time dilation.
That would be incompatible with SR because the amount by which the clocks differ in the twin paradox is related to how far apart the twins are when one turns around. If it was just the acceleration felt that caused the turning twin's clock to slow then it shouldn't matter how far away the other twin is.
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There isn't an absolutely velocity or absolutely time.
The mirrors move relatively to Earth, Earth relatively to Sun and so on. There is only relative velocity.
A faster velocity means a higher relative energy and processes (atom clock) are running slower for distant observer.
If the observers are in different reference frames (velocity, gravity) they observe different reality (time, redshift-blueshift). If they come together everyting comes back too.
According to Lorentz transformation , you see - there is not a Black Hole in a center of a galaxy but if you come to the galaxy center you fall in a Black Hole. It seems to be a paradox but it is a reality in different reference frames.
Some people call it time dilation, may be it is not correct English word.
29. I think time dilation and length contraction are good terms for these phenomena. It is important to stress to newbies that clocks do run slower and the rods are shorter for a moving observer. It isn't just some sort of illusion. Most books on SR start out with someting like an some imaginary grid of rods with light receptors and clocks at each intersection and a database recording all the events that happen as they happen on the grid and then say that whenever they talking about Bod "seeing" this or that what they mean is that this is what Bob's database would have recorded on his grid of clocks and receptors which is stationary with respect to him.
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Originally Posted by upriver
"In a paper published in 1918 Einstein corrected the “relative motion” error of his 1905 paper and he added “forces” and “atomic clocks” to his thought experiments. He changed the reason for the single clock slow-down from “relative motion” to “forces” exerted on the oscillating atoms in the single atomic clock that slowed down, and thus he basically returned to Lorentz’s basic electrodynamics concept of 1895, and Einstein's own 1911 gravitational redshift theory."
So time dilation is a is not reality, but clocks slowing down is.
So there is no such thing as time dilation.
It’s not quite that simple. There is no “time dilation” caused by “relative motion” alone, so the 1905 SR theory is wrong about that. Get yourself a copy of Einstein’s 1918 paper and see what he says about it.
However, atomic clocks (and perhaps other atomic processes too) can slow down if, for example, the atoms of the clock are near the surface of a massive planet or star and if their internal harmonic oscillation rates slow down. However, the old fable that you will live longer at sea level and you will age faster on a mountain top (because your atoms are oscillating more rapidly on the mountain top) is wrong. An atomic clock will “tick” more rapidly on a mountain top, but it’s such a slight amount that the atomic oscillation rate change has nothing to do with our aging rates. Our human aging rates are determined more by our body and cell temperatures (which control thermodynamic time) than by any very slight slowdown or speed up in our atomic oscillation rates when we move from sea level to a mountain top.
Also, different types of clocks slow down and speed up at different rates, depending on which laws of physics are governing the tick rates of the clocks. For example, an atomic clock will “tick” faster on a mountain top, while a pendulum clock will tick slower on a mountain top. An atomic clock will “tick” slower at sea level, while a pendulum clock will tick faster at sea level.
Frozen human embryos can be kept from aging for many years, 5, 10, etc, and then they can be thawed out and the embryo will grow into a baby. This is an example of thermodynamic time that is not related to atomic time. You can have two twin frozen embryos, and thaw one out now, and thaw the other out 10 years from now, and you will have one twin being 10 years older than the other. One will be in the 4th grade while the other is still a baby. This is thermodynamic time. It’s actually performed by doctors every day, although they usually thaw out only one embryo of each group they collect.
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Question Video: Finding the Equation of a Curve Given the Slope of the Tangent and a Point that Lies on the Curve | Nagwa Question Video: Finding the Equation of a Curve Given the Slope of the Tangent and a Point that Lies on the Curve | Nagwa
# Question Video: Finding the Equation of a Curve Given the Slope of the Tangent and a Point that Lies on the Curve Mathematics • Third Year of Secondary School
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Find the equation of a curve that passes through the point (0, 0) and is such that for each point (π₯, π¦) on the curve the slope of the tangent at that point is β3π₯β΅ the ninth root of π₯βΈ.
04:10
### Video Transcript
Find the equation of a curve that passes through the point zero, zero and is such that for each point π₯, π¦ on the curve the slope of the tangent at that point is negative three π₯ to the power of five multiplied by the ninth root of π₯ to the power of eight.
In this question, weβre given the slope function or dπ¦ by dπ₯, which is equal to negative three π₯ to the power of five multiplied by the ninth root of π₯ to the power of eight. And we know that in order to find the equation of the curve, weβll need to integrate this function. However, before doing this, we will try to simplify the expression for dπ¦ by dπ₯. We begin by recalling one of our exponent laws. This state that the πth root of π₯ is equal to π₯ to the power of one over π. We can therefore rewrite the second term in our expression as π₯ to the power of eight raised to the power of one-ninth.
Next, we recall the power rule of exponents. This states that π₯ to the power of π raised to the power of π is equal to π₯ to the power of π multiplied by π. Multiplying eight by one-ninth, our expression simplifies to negative three π₯ to the power of five multiplied by π₯ to the power of eight-ninths. Finally, we recall that π₯ to the power of π multiplied by π₯ to the power of π is equal to π₯ to the power of π plus π. We need to add the exponents five and eight-ninths. Since five can be rewritten as forty-five ninths, we have forty-five ninths plus eight-ninths. This is equal to fifty-three ninths.
Our expression for the slope function dπ¦ by dπ₯ simplifies to negative three π₯ to the power of fifty-three ninths. We are now in a position to integrate this expression to find the equation of the curve. π¦ is equal to the integral of negative three π₯ to the power of fifty-three ninths with respect to π₯. The power rule of integration states that the integral of π₯ to the power of π with respect to π₯ is equal to π₯ to the power of π plus one divided by π plus one plus our constant of integration πΆ. Adding one or nine-ninths to our power gives us sixty-two ninths. And π¦ is therefore equal to negative three π₯ to the power of 62 over nine divided by 62 over nine plus πΆ. This in turn simplifies to π¦ is equal to negative 27π₯ to the power of 62 over nine all divided by 62 plus the constant of integration πΆ.
Our next step is to find the value of the constant of integration πΆ. We are told that the curve passes through the origin, the point with coordinates zero, zero. Substituting π₯ is equal to zero and π¦ is equal to zero into our equation, we see that πΆ is also equal to zero. We can therefore conclude that the equation of the curve is π¦ is equal to negative 27π₯ to the power of 62 over nine divided by 62.
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# Multiples of a number
This tool calculates the first 150 positive multiples of an integer number.
## How to find the multiples of a number?
Simply multiply this number by 1, 2, 3, 4, 5, etc.
Example: the multiples of 3 are: 3x1, 3x2, 3x3, 3x4, etc, these are equal to,
3, 6, 9, 12, 15, 18, 21, etc
The list of multiples of a number is infinite (we can't list all of them!).
## Easily identify multiples of 2, 3, 5...
- A multiple of 2 always ends with an even digit: 0, 2, 4, 6 or 8.
- A mutliple of 3: the sum of its digits is a multiple of 3.
- A multiple of 4: its last 2 digits form a number which is a multiple of 4.
- A multiple of 5: always ends with 0 or 5.
- A mutliple of 9: the sum of its digits is a multiple of 9.
- A multiple of 10: always ends with 0.
## Special cases
- A number is always multiple of itself.
- 0 is a multiple of all numbers (0 = 0 x n).
## Programming
### Python
This python program finds the first c multiples of a given integer n.
def multiples(n, c):
m = []
for i in range (c):
m.append((i+1) *n)
return m
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# Solve for x in the equation 2x^2+3x-7=x^2+5x+39
Solve for x in the equation 2x^2+3x-7=x^2+5x+39
## This Post Has 10 Comments
1. Akira8889 says:
Hey there, hope I can help!
$\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}$
$2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)$
Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
$x^2-2x-46=0$
Lets use the quadratic formula now
$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$
$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}$
$\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$
Multiply the numbers 2 * 1 = 2
$\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$
$2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}$
$\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \ \left(-2\right)^2=2^2, 2^2 = 4$
$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \ \sqrt{4+184} \ \textgreater \ \sqrt{188} \ \textgreater \ 2 + \sqrt{188}$
$\frac{2+\sqrt{188}}{2} \ \textgreater \ Prime\;factorize\;188 \ \textgreater \ 2^2\cdot \:47 \ \textgreater \ \sqrt{2^2\cdot \:47}$
$\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ \sqrt{47}\sqrt{2^2}$
$\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ \sqrt{2^2}=2 \ \textgreater \ 2\sqrt{47} \ \textgreater \ \frac{2+2\sqrt{47}}{2}$
$Factor\;2+2\sqrt{47} \ \textgreater \ Rewrite\;as\;1\cdot \:2+2\sqrt{47}$
$\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \ 2\left(1+\sqrt{47}\right) \ \textgreater \ \frac{2\left(1+\sqrt{47}\right)}{2}$
$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \ 1+\sqrt{47}$
Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.
$\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$
$\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$
$2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ 2-\sqrt{188} \ \textgreater \ \frac{2-\sqrt{188}}{2}$
$\sqrt{188} = 2\sqrt{47} \ \textgreater \ \frac{2-2\sqrt{47}}{2}$
$2-2\sqrt{47} \ \textgreater \ 2\left(1-\sqrt{47}\right) \ \textgreater \ \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \ 1-\sqrt{47}$
Therefore our final solutions are
$x=1+\sqrt{47},\:x=1-\sqrt{47}$
Hope this helps!
2. markeledwards699 says:
$2x^2 - 3x - 4 = 0 \\ x= \frac{-b \pm \sqrt{ b^{2} -4ac} }{2a} \\ = \frac{-(-3) \pm \sqrt{ (-3)^{2} -4 \times2 \times (-4)} }{2 \times 2} \\ = \frac{3 \pm \sqrt{ 9 +32} }{4} \\ = \frac{3 \pm \sqrt{41} }{4} \\ = \frac{3 + \sqrt{41} }{4} \ or \ \frac{3 - \sqrt{41} }{4} \\ =2.35 \ or \ -0.85$
3. lagarde says:
D
hope this helps :o)
4. misaki2002 says:
Sorry I can't explain this one much but it is x = 1+- square root of 47
5. bsheepicornozj0gc says:
There is a combination of both answers x=7.8556546,−5.8556546
6. areanna02 says:
$X=1+\sqrt{47}$
Step-by-step explanation:
Hope this helps
Mark me as brainiest
7. Baby010391 says:
x=1±√47
Step-by-step explanation:
it's up above.
$Solve for x in the equation 2x^2+3x-7=x^2+5x+39$
8. twalters88 says:
Subtract x^2 from both sides
x^2 + 3x - 7 = 5x + 39
Subtract 5x from both sides
x^2 - 2x - 7 = 39
x^2 - 2x = 46
Complete the square by adding (b/2)^2 to both sides, b = ( -2)
(-2/2) = -1, then square that (-1)^2 = 1
x^2 - 2x + 1 = 46 + 1
Simplify the expression by factoring
(x - 1)^2 = 47
Take square root on each side
x - 1 = (sqrt (47))
Solve for x
x = 1 + (sqrt (47))
Since 47 is prime, 47 cannot be broken down by the square root and this is the answer to your problem.
9. isabellemaine says:
x1= 1 + √47
x2= 1 - √47
I use a handheld calculator to solve the problem
10. jackieanguiano4758 says:
$x=1\pm\sqrt{47}$
Step-by-step explanation:
We have been given an equation $2x^2+3x-7=x^2+5x+39$. We are asked to find the solution for our given equation.
$2x^2+3x-7=x^2+5x+39$
$2x^2-x^2+3x-7=x^2-x^2+5x+39$
$x^2+3x-7=5x+39$
$x^2+3x-5x-7-39=5x-5x+39-39$
$x^2-2x-46=0$
Using quadratic formula, we will get:
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-46)}}{2(1)}$
$x=\frac{2\pm\sqrt{4+184}}{2}$
$x=\frac{2\pm\sqrt{188}}{2}$
$x=\frac{2\pm2\sqrt{47}}{2}$
$x=1\pm\sqrt{47}$
Therefore, the solutions for our given equation are $x=1\pm\sqrt{47}$.
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# POW: A Digital Proof
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Essay, Pages 2 (438 words)
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Essay, Pages 2 (438 words)
This problem of the week has a main gain goal set upon boxes. There are five boxes numbered one through zero. Underneath the boxes have the numbers written under them. In the boxes, there are numbers that should be entered in the boxes that all evenly works out. For instance, the number that you put in box zero must be the same as the number of zeros that were used. The same procedures apply when using other boxes. The same number cannot be used in the box, for example, a four cannot be placed in box four or the number two cannot be enlisted in box two.
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The same number is tolerable to be used more than once. The only exception is that no number higher than four can be expended. My goal for this POW is to corroborate and demonstrate that I have found all solutions and that all solutions work and is credible.
Process:
This POW was very challenging to achieve at first.
A few methods that I used to help me with this POW was using when Mr. Kohnen first showed us how to play a game of Sudoku. Sudoku is number game where you have to have numbers one through nine in a box going across and up and down. There are nine boxes so only one number can be in the box. This game helped me get a generalization of how to complete the task of this POW. I also had the opportunity to work with other classmates.
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Working with other classmates helped me a lot because we were able to come up with solutions faster and other solutions of how to solve the problem. Coming up with different ways to find the answer was useful because as you’re getting close to the answer, all that needs to be done is switching around numbers that will facilitate and lead you to the answer.
This is he incorrect to do this because there is a one in box two and more than one two is used in the boxes. Likewise, there is a two in box four, but the number of fours is not two. I also used the process of elimination to find the correct answer.
Solution:
After doing all of the work and figuring out the solutions, here is how I found a way to find the correct solution. I am assured that there are no other ways of how to find any other solutions. The zeros generate a larger amount of numbers. Whichever number that is placed in the zero boxes that is how may zeros that need to be revealed.
## Cite this essay
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1. Inequality help
Wasn't sure where to put this...profound apologies if this is in the wrong forum...
Suppose there are $n$ distinct odd, positive integers $\{a_i\}_{i=1}^n$ where the absolute value of the difference between each pair of numbers is distinct i.e. each $|a_k - a_i|, \ 1\leq k < i \leq n$ is distinct.
Prove that $$\sum_{i=1}^n a_i \geq \tfrac{1}{3} n(n^2 + 2)$$
I tried doing this by induction and have gotten to the stage of my induction hypothesis but am not sure how to proceed. I have tried to prove that $a_{k+1} \geq 1 + k + k^2$ (as this would help yield the result for my induction) but, again, no success....
Can anyone give me a hint as to how I would go about the induction (or even if induction is the correct way and I should try a different approach)?
Thank you
2. Re: Inequality help
Hi,
You're definitely on the right track. The attachment shows an outline of a proof. If you have any questions, post again.
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## Calculus, 10th Edition (Anton)
$\lim\limits_{y \to 6^{-}}\frac{y+6}{y²-36}$ = $-\infty$
y² - 36 = (y + 6)(y - 6) $\frac{y+6}{y²-36}$ = $\frac{1}{y-6}$ $\lim\limits_{y \to 6^{+}}\frac{y+6}{y²-36}$ = $\lim\limits_{y \to 6^{-}}\frac{1}{y-6}$ = $\frac{1}{0^{-}}$ = $-\infty$
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https://eng.libretexts.org/Bookshelves/Computer_Science/Applied_Programming/Book%3A_Introduction_to_Computer_Graphics_(Eck)/02%3A_Two-Dimensional_Graphics/2.02%3A_Shapes
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# 2.2: Shapes
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( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
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We have been talking about low-level graphics concepts like pixels and coordinates, but fortunately we don’t usually have to work on the lowest levels. Most graphics systems let you work with higher-level shapes, such as triangles and circles, rather than individual pixels. And a lot of the hard work with coordinates is done using transforms rather than by working with coordinates directly. In this section and the next, we will look at some of the higher-level capabilities that are typically provided by 2D graphics APIs.
## Basic Shapes
In a graphics API, there will be certain basic shapes that can be drawn with one command, whereas more complex shapes will require multiple commands. Exactly what qualifies as a basic shape varies from one API to another. In the JavaScript API for drawing on an HTML canvas, for example, the only basic shapes are lines and rectangles. In this subsection, I consider lines, rectangles, and ovals to be basic.
By “line,” I really mean line segment, that is a straight line segment connecting two given points in the plane. A simple one-pixel-wide line segment, without antialiasing, is the most basic shape. It can be drawn by coloring pixels that lie along the infinitely thin geometric line segment. An algorithm for drawing the line has to decide exactly which pixels to color. One of the first computer graphics algorithms, Bresenham’s algorithm for line drawing, implements a very efficient procedure for doing so. I won’t discuss such low-level details here, but it’s worth looking them up if you want to start learning about what graphics hardware actually has to do. In any case, lines are typically more complicated. Antialiasing is one complication. Line width is another. A wide line might actually be drawn as a rectangle.
Lines can have other attributes, or properties, that affect their appearance. One question is, what should happen at the end of a wide line? Appearance might be improved by adding a rounded “cap” on the ends of the line. A square cap—that is, extending the line by half of the line width—might also make sense. Another question is, when two lines meet as part of a larger shape, how should the lines be joined? And many graphics systems support lines that are patterns of dashes and dots. This illustration shows some of the possibilities:
On the left are three wide lines with no cap, a round cap, and a square cap. The geometric line segment is shown as a dotted line. (The no-cap style is called “butt.”) To the right are four lines with different patters of dots and dashes. In the middle are three different styles of line joins: mitered, rounded, and beveled.
The basic rectangular shape has sides that are vertical and horizontal. (A tilted rectangle generally has to be made by applying a rotation.) Such a rectangle can be specified with two points, (x1,y1) and (x2,y2), that give the endpoints of one of the diagonals of the rectangle. Alternatively, the width and the height can be given, along with a single base point, (x,y). In that case, the width and height have to be positive, or the rectangle is empty. The base point (x,y) will be the upper left corner of the rectangle if y increases from top to bottom, and it will be the lower left corner of the rectangle if y increases from bottom to top.
Suppose that you are given points (x1,y1) and (x2,y2), and that you want to draw the rectangle that they determine. And suppose that the only rectangle-drawing command that you have available is one that requires a point (x,y), a width, and a height. For that command, x must be the smaller of x1 and x2, and the width can be computed as the absolute value of x1 minus x2. And similarly for y and the height. In pseudocode,
DrawRectangle from points (x1,y1) and (x2,y2): x = min( x1, x2 ) y = min( y1, y2 ) width = abs( x1 - x2 ) height = abs( y1 - y2 ) DrawRectangle( x, y, width, height )
A common variation on rectangles is to allow rounded corners. For a “round rect,” the corners are replaced by elliptical arcs. The degree of rounding can be specified by giving the horizontal radius and vertical radius of the ellipse. Here are some examples of round rects. For the shape at the right, the two radii of the ellipse are shown:
My final basic shape is the oval. (An oval is also called an ellipse.) An oval is a closed curve that has two radii. For a basic oval, we assume that the radii are vertical and horizontal. An oval with this property can be specified by giving the rectangle that just contains it. Or it can be specified by giving its center point and the lengths of its vertical radius and its horizontal radius. In this illustration, the oval on the left is shown with its containing rectangle and with its center point and radii:
The oval on the right is a circle. A circle is just an oval in which the two radii have the same length.
If ovals are not available as basic shapes, they can be approximated by drawing a large number of line segments. The number of lines that is needed for a good approximation depends on the size of the oval. It’s useful to know how to do this. Suppose that an oval has center point (x,y), horizontal radius r1, and vertical radius r2. Mathematically, the points on the oval are given by
( x + r1*cos(angle), y + r2*sin(angle) )
where angle takes on values from 0 to 360 if angles are measured in degrees or from 0 to $$2\pi$$ if they are measured in radians. Here sin and cos are the standard sine and cosine functions. To get an approximation for an oval, we can use this formula to generate some number of points and then connect those points with line segments. In pseudocode, assuming that angles are measured in radians and that pi represents the mathematical constant $$\pi$$,
Draw Oval with center (x,y), horizontal radius r1, and vertical radius r2: for i = 0 to numberOfLines: angle1 = i * (2*pi/numberOfLines) angle2 = (i+1) * (2*pi/numberOfLines) a1 = x + r1*cos(angle1) b1 = y + r2*sin(angle1) a2 = x + r1*cos(angle2) b2 = y + r2*sin(angle2) Draw Line from (x1,y1) to (x2,y2)
For a circle, of course, you would just have r1 = r2. This is the first time we have used the sine and cosine functions, but it won’t be the last. These functions play an important role in computer graphics because of their association with circles, circular motion, and rotation. We will meet them again when we talk about transforms in the next section.
## Stroke and Fill
There are two ways to make a shape visible in a drawing. You can stroke it. Or, if it is a closed shape such as a rectangle or an oval, you can fill it. Stroking a line is like dragging a pen along the line. Stroking a rectangle or oval is like dragging a pen along its boundary. Filling a shape means coloring all the points that are contained inside that shape. It’s possible to both stroke and fill the same shape; in that case, the interior of the shape and the outline of the shape can have a different appearance.
When a shape intersects itself, like the two shapes in the illustration below, it’s not entirely clear what should count as the interior of the shape. In fact, there are at least two different rules for filling such a shape. Both are based on something called the winding number. The winding number of a shape about a point is, roughly, how many times the shape winds around the point in the positive direction, which I take here to be counterclockwise. Winding number can be negative when the winding is in the opposite direction. In the illustration, the shapes on the left are traced in the direction shown, and the winding number about each region is shown as a number inside the region.
The shapes are also shown filled using the two fill rules. For the shapes in the center, the fill rule is to color any region that has a non-zero winding number. For the shapes shown on the right, the rule is to color any region whose winding number is odd; regions with even winding number are not filled.
There is still the question of what a shape should be filled with. Of course, it can be filled with a color, but other types of fill are possible, including patterns and gradients. A pattern is an image, usually a small image. When used to fill a shape, a pattern can be repeated horizontally and vertically as necessary to cover the entire shape. A gradient is similar in that it is a way for color to vary from point to point, but instead of taking the colors from an image, they are computed. There are a lot of variations to the basic idea, but there is always a line segment along which the color varies. The color is specified at the endpoints of the line segment, and possibly at additional points; between those points, the color is interpolated. For other points on the line that contains the line segment, the pattern on the line segment can be repeated, or the color of the endpoint can simply be extended. For a linear gradient, the color is constant along lines perpendicular to the basic line segment, so you get lines of solid color going in that direction. In a radial gradient, the color is constant along circles centered at one of the endpoints of the line segment. And that doesn’t exhaust the possibilities. To give you an idea what patterns and gradients can look like, here is a shape, filled with two gradients and two patterns:
The first shape is filled with a simple linear gradient defined by just two colors, while the second shape uses a radial gradient.
Patterns and gradients are not necessarily restricted to filling shapes. Stroking a shape is, after all, the same as filling a band of pixels along the boundary of the shape, and that can be done with a gradient or a pattern, instead of with a solid color.
Finally, I will mention that a string of text can be considered to be a shape for the purpose of drawing it. The boundary of the shape is the outline of the characters. The text is drawn by filling that shape. In some graphics systems, it is also possible to stroke the outline of the shape that defines the text. In the following illustration, the string “Graphics” is shown, on top, filled with a pattern and, below that, filled with a gradient and stroked with solid black:
## Polygons, Curves, and Paths
It is impossible for a graphics API to include every possible shape as a basic shape, but there is usually some way to create more complex shapes. For example, consider polygons. A polygon is a closed shape consisting of a sequence of line segments. Each line segment is joined to the next at its endpoint, and the last line segment connects back to the first. The endpoints are called the vertices of the polygon, and a polygon can be defined by listing its vertices.
In a regular polygon, all the sides are the same length and all the angles between sides are equal. Squares and equilateral triangles are examples of regular polygons. A convex polygon has the property that whenever two points are inside or on the polygon, then the entire line segment between those points is also inside or on the polygon. Intuitively, a convex polygon has no “indentations” along its boundary. (Concavity can be a property of any shape, not just of polygons.)
Sometimes, polygons are required to be “simple,” meaning that the polygon has no self-intersections. That is, all the vertices are different, and a side can only intersect another side at its endpoints. And polygons are usually required to be “planar,” meaning that all the vertices lie in the same plane. (Of course, in 2D graphics, everything lies in the same plane, so this is not an issue. However, it does become an issue in 3D.)
How then should we draw polygons? That is, what capabilities would we like to have in a graphics API for drawing them. One possibility is to have commands for stroking and for filling polygons, where the vertices of the polygon are given as an array of points or as an array of x-coordinates plus an array of y-coordinates. In fact, that is sometimes done; for example, the Java graphics API includes such commands. Another, more flexible, approach is to introduce the idea of a “path.” Java, SVG, and the HTML canvas API all support this idea. A path is a general shape that can include both line segments and curved segments. Segments can, but don’t have to be, connected to other segments at their endpoints. A path is created by giving a series of commands that tell, essentially, how a pen would be moved to draw the path. While a path is being created, there is a point that represents the pen’s current location. There will be a command for moving the pen without drawing, and commands for drawing various kinds of segments. For drawing polygons, we need commands such as
• createPath() — start a new, empty path
• moveTo(x,y) — move the pen to the point (x,y), without adding a segment to the the path; that is, without drawing anything
• lineTo(x,y) — add a line segment to the path that starts at the current pen location and ends at the point (x,y), and move the pen to (x,y)
• closePath() — add a line segment from the current pen location back to the starting point, unless the pen is already there, producing a closed path.
(For closePath, I need to define “starting point.” A path can be made up of “subpaths” A subpath consists of a series of connected segments. A moveTo always starts a new subpath. A closePath ends the current segment and implicitly starts a new one. So “starting point” means the position of the pen after the most recent moveTo or closePath.)
Suppose that we want a path that represents the triangle with vertices at (100,100), (300,100), and (200, 200). We can do that with the commands
createPath() moveTo( 100, 100 ) lineTo( 300, 100 ) lineTo( 200, 200 ) closePath()
The closePath command at the end could be replaced by lineTo(100,100), to move the pen back to the first vertex.
A path represents an abstract geometric object. Creating one does not make it visible on the screen. Once we have a path, to make it visible we need additional commands for stroking and filling the path.
Earlier in this section, we saw how to approximate an oval by drawing, in effect, a regular polygon with a large number of sides. In that example, I drew each side as a separate line segment, so we really had a bunch of separate lines rather than a polygon. There is no way to fill such a thing. It would be better to approximate the oval with a polygonal path. For an oval with center (x,y) and radii r1 and r2:
createPath() moveTo( x + r1, y ) for i = 1 to numberOfPoints-1 angle = i * (2*pi/numberOfLines) lineTo( x + r1*cos(angle), y + r2*sin(angle) ) closePath()
Using this path, we could draw a filled oval as well as stroke it. Even if we just want to draw the outline of a polygon, it’s still better to create the polygon as a path rather than to draw the line segments as separate sides. With a path, the computer knows that the sides are part of single shape. This makes it possible to control the appearance of the “join” between consecutive sides, as noted earlier in this section.
I noted above that a path can contain other kinds of segments besides lines. For example, it might be possible to include an arc of a circle as a segment. Another type of curve is a Bezier curve. Bezier curves can be used to create very general curved shapes. They are fairly intuitive, so that they are often used in programs that allow users to design curves interactively. Mathematically, Bezier curves are defined by parametric polynomial equations, but you don’t need to understand what that means to use them. There are two kinds of Bezier curve in common use, cubic Bezier curves and quadratic Bezier curves; they are defined by cubic and quadratic polynomials respectively. When the general term “Bezier curve” is used, it usually refers to cubic Bezier curves.
A cubic Bezier curve segment is defined by the two endpoints of the segment together with two control points. To understand how it works, it’s best to think about how a pen would draw the curve segment. The pen starts at the first endpoint, headed in the direction of the first control point. The distance of the control point from the endpoint controls the speed of the pen as it starts drawing the curve. The second control point controls the direction and speed of the pen as it gets to the second endpoint of the curve. There is a unique cubic curve that satisfies these conditions.
The illustration above shows three cubic Bezier curve segments. The two curve segments on the right are connected at an endpoint to form a longer curve. The curves are drawn as thick black lines. The endpoints are shown as black dots and the control points as blue squares, with a thin red line connecting each control point to the corresponding endpoint. (Ordinarily, only the curve would be drawn, except in an interface that lets the user edit the curve by hand.) Note that at an endpoint, the curve segment is tangent to the line that connects the endpoint to the control point. Note also that there can be a sharp point or corner where two curve segments meet. However, one segment will merge smoothly into the next if control points are properly chosen.
This will all be easier to understand with some hands-on experience. The interactive demo c2/cubic-bezier.html lets you edit cubic Bezier curve segments by dragging their endpoints and control points.
When a cubic Bezier curve segment is added to a path, the path’s current pen location acts as the first endpoint of the segment. The command for adding the segment to the path must specify the two control points and the second endpoint. A typical command might look like
cubicCurveTo( cx1, cy1, cx2, cy2, x, y )
This would add a curve from the current location to point (x,y), using (cx1,cy1) and (cx2,cy2) as the control points. That is, the pen leaves the current location heading towards (cx1,cy1), and it ends at the point (x,y), arriving there from the direction of (cx2,cy2).
Quadratic Bezier curve segments are similar to the cubic version, but in the quadratic case, there is only one control point for the segment. The curve leaves the first endpoint heading in the direction of the control point, and it arrives at the second endpoint coming from the direction of the control point. The curve in this case will be an arc of a parabola.
Again, this is easier to understand this with some hands-on experience. Try the interactive demo c2/quadratic-bezier.html.
This page titled 2.2: Shapes is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David J. Eck via source content that was edited to the style and standards of the LibreTexts platform.
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# Chapter 21 Electric Current and Direct- Current Circuits
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1 Chapter 21 Electric Current and Direct- Current Circuits 1
2 Overview of Chapter 21 Electric Current and Resistance Energy and Power in Electric Circuits Resistors in Series and Parallel Kirchhoff s Rules Circuits Containing Capacitors RC Circuits 2
3 21-1 Electric Current Electric current is the flow of electric charge from one place to another. Electric circuit: A closed path through which charge can flow, returning to its starting point. 3
4 Battery has a potential difference between terminals 21-1 Electric Current Chemical reactions separate electrons from atoms Analogy 4 - Current to flows through the flashlight bulb similar to someone lifting water which flows through the paddle wheel.
5 21-1 Electric Current A battery that is disconnected from any circuit has an electric potential difference between its terminals called the electromotive force (emf) Remember despite its name, the emf is an electric potential, not a force. Amount of work it takes to move a charge ΔQ from one terminal to the other is: 5
6 21-1 Electric Current Current flows from the positive terminal to the negative Decided before it was realized that electrons are negatively charged. The electrons move the other way 6
7 21-1 Electric Current Actual motion of electrons along a wire is quite slow Electrons spend most of their time bouncing around randomly with small velocity component - Opposite to the direction of the current. The electric signal propagates much more quickly 7
8 21-2 Resistance and Ohm s Law Wires present some resistance to the motion of electrons. Ohm s law relates the voltage to the current: Be careful Ohm s law is not a universal law and is only useful for certain materials (which include most metallic conductors). 8
9 21-2 Resistance and Ohm s Law Solving for the resistance, we find: The units of resistance, volts per ampere, are called ohms: 9
10 21-2 Resistance and Ohm s Law Two wires of the same length and diameter will have different resistances if they are made of different materials. Property of a material is called the resistivity. 10
11 21-2 Resistance and Ohm s Law The difference between insulators, semiconductors, is quantified by resistivity.. 11
12 21-2 Resistance and Ohm s Law In general.. Resistance of materials goes up as the temperature goes up, due to thermal effects. This property can be used in thermometers. Resistivity decreases as the temperature decreases Class of materials called superconductors Resistivity drops suddenly to zero at a finite temperature, called the critical temperature T C. 12
13 21-2 Resistance and Ohm s Law Superconductors in action. The ATLAS experiment at the LHC. MRI machines. 13
14 21-3 Energy and Power in Electric Circuits When a charge moves across a potential difference, its potential energy changes: Therefore, the power it takes to do this is Remember 14
15 21-3 Energy and Power in Electric Circuits In materials for which Ohm s law holds, the power can also be written: This power mostly dissipates as heat inside a resistive material 15
16 21-3 Energy and Power in Electric Circuits When the electric company sends you a bill, your usage is quoted in kilowatt-hours (kwh). They are charging you for energy use, and kwh are a measure of energy. 16
17 21-4 Resistors in Series and Parallel Resistors connected end to end are said to be in series. They can be replaced by a single equivalent resistance without changing the current in the circuit. 17
18 21-4 Resistors in Series and Parallel Relation due to current through the series resistors being the same for each.. 18
19 21-4 Resistors in Series and Parallel Resistors are in parallel when they are across the same potential difference They can again be replaced by a single equivalent resistance.. 19
20 21-4 Resistors in Series and Parallel Relation due to potential difference across parallel resistors being the same for each.. 20
21 21-4 Resistors in Series and Parallel If a circuit is more complex: Start with combinations of resistors that are either purely in series or in parallel. Replace these with their equivalent resistances; as you go on you will be able to replace more and more of them. 21
22 21-5 Kirchhoff s Rules Even more complex circuits cannot be broken down into series and parallel pieces. Kirchhoff s rules are needed. The junction rule is a consequence of charge conservation; the loop rule is a consequence of energy conservation. 22
23 21-5 Kirchhoff s Rules The junction rule: At any junction, the current entering the junction must equal the current leaving it. 23
24 21-5 Kirchhoff s Rules The loop rule: The algebraic sum of the potential differences around a closed loop must be zero (it must return to its original value at the original point). 24
25 Using Kirchhoff s rules: 21-5 Kirchhoff s Rules The variables for which you are solving are the currents through the resistors. You need as many independent equations as you have variables to solve for. You will need both loop and junction rules. 25
26 21-6 Circuits Containing Capacitors Capacitors can also be connected in series or in parallel. When capacitors are connected in parallel, the potential difference across each one is the same. 26
27 21-6 Circuits Containing Capacitors Therefore, the equivalent capacitance is the sum of the individual capacitances: 27
28 21-6 Circuits Containing Capacitors Capacitors connected in series do not have the same potential difference across them. They do all carry the same charge. The total potential difference is the sum of the potential differences across each one. 28
29 21-6 Circuits Containing Capacitors Therefore, the equivalent capacitance is Capacitors in series combine like resistors in parallel, and vice versa. 29
30 21-7 RC Circuits Circuit containing only batteries and capacitors Charge appears almost instantaneously on the capacitors when the circuit is connected. If the circuit contains resistors as well Not the case Known as a RC circuit 30
31 21-7 RC Circuits It can be shown that the charge on the capacitor increases as: Here, τ is the time constant of the circuit: And is the final charge on the capacitor, Q. 31
32 21-7 RC Circuits Here is the charge vs. time for an RC circuit: 32
33 21-7 RC Circuits It can be shown that the current in the circuit has a related behavior: 33
35 Hint: Resistivity of silver is 1.59 x 10-8 Ω m Answer: 0.5 Ω 35
37 Note on Kirchoff s laws Loop rule: -Sum of the potential differences around a closed loop must be zero - ε = +15 V + I R 3 R 1 What does this mean? R 2 -EMF adds a potential difference to current -Resistors take it away ε + V 1 +V 2 +V 3 = 0 ε -IR 1 -IR 2 -IR 3 = 0 37
38 Note on Kirchoff s laws ε T ε T I I ε ε R 1 R 1 R 2 R 2 ε -IR 1 -IR 2 +ε 1 = 0 ε -IR 1 -IR 2 -ε 2 = 0 38
40 40
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# Class 11 RD Sharma Solutions – Chapter 15 Linear Inequations – Exercise 15.6
• Last Updated : 24 Nov, 2022
### Question 1(i). Solve the following systems of linear inequation graphically:
2x + 3y ≤ 6
3x + 2y ≤ 6
x ≥ 0, y ≥ 0
Solution:
Let us convert the given inequations to equations.
2x + 3y = 6 -(1)
3x + 2y = 6 -(2)
x = 0 -(3)
y = 0 -(4)
Let us consider eq(1)
The line 2x + 3y = 6 meets the x-axis at (3, 0) and the y-axis at (0, 2).
And Origin(0, 0) satisfies the inequation 2x + 3y ≤ 6.
So, the section containing the origin will represent the solution set of inequation 2x + 3y ≤ 6.
Let us consider eq(2)
The line 3x + 2y = 6 meets the x-axis at (2, 0) and the y-axis at (0, 3).
And Origin(0, 0) satisfies the inequation 3x + 2y ≤ 6.
So, the section containing the origin will represent the solution set of inequation 3x + 2y ≤ 6.
x ≥ 0 and y ≥ 0 represent the first quadrant.
Thus, the shaded region represents the solution of the given inequations.
### Question 1(ii). Solve the following systems of linear inequation graphically:
2x + 3y ≤ 6
x + 4y ≤ 4
x ≥ 0, y ≥ 0
Solution:
Let us convert the given inequations to equations.
2x + 3y = 6 -(1)
x + 4y = 4 -(2)
x = 0 -(3)
y = 0 -(4)
Let us consider eq(1)
The line 2x + 3y = 6 meets the x-axis at (3, 0) and the y-axis at (0, 2).
And Origin(0, 0) satisfies the inequation 2x + 3y ≤ 6.
So, the section containing the origin will represent the solution set of inequation 2x + 3y ≤ 6.
Let us consider eq(2)
The line x + 4y = 4 meets the x-axis at (4, 0) and the y-axis at (0, 1).
And Origin(0, 0) satisfies the inequation x + 4y ≤ 4.
So, the section containing the origin will represent the solution set of inequation x + 4y ≤ 4.
x ≥ 0 and y ≥ 0 represent the first quadrant.
Thus, the shaded region represents the solution set of given inequations.
### Question 1(iii). Solve the following systems of linear equation graphically:
x – y ≤ 1
x + 2y ≤ 8
2x + y ≥ 2
x ≥ 0, y ≥ 0
Solution:
Let us convert the given inequations to equations.
x – y = 1 – (1)
x + 2y = 8 -(2)
2x + y = 2 -(3)
x = 0, y = 0 -(4)
Let us consider eq(1)
The line x – y = 1 meets the x-axis at (1, 0) and the y-axis at (0, -1).
And Origin(0, 0) satisfies the inequation x – y ≤ 1
So, the section containing the origin will represent the solution set of inequation x – y ≤ 1.
Let us consider eq(2)
The line x + 2y = 8 meets the x-axis at (8, 0) and the y-axis at (0, 4).
And Origin(0, 0) satisfies the inequation x + 2y ≤ 8.
So, the section containing the origin will represent the solution set of inequation x + 2y ≤ 8.
Let us consider eq(3)
The line 2x + y = 2 meets the x-axis at (1, 0) and the y-axis at (0, 2).
And Origin(0, 0) doesnot satisfy the inequation 2x + y ≥ 2.
So, the section containing the origin will represent the solution set of inequation 2x + y ≥ 2.
x ≥ 0 and y ≥ 0 represent the first quadrant.
Thus the shaded region represents the solution set of given inequations.
### Question 1(iv). Solve the following systems of linear equation graphically:
x + y ≥ 1
7x + 9y ≤ 63
x ≤ 6
y ≤ 5
x ≥ 0, y ≥ 0
Solution:
Let us convert the given inequations to equations.
x + y = 1 -(1)
7x + 9y = 63 -(2)
x = 6 -(3)
y = 5 -(4)
x = 0, y = 0
Let us consider eq(1)
The line x + y= 1 meets the x-axis at (1, 0) and the y-axis at (0, 1).
And Origin(0, 0) doesnot satisfy the inequation x + y ≥ 1.
So, the section containing the origin will represent the solution set of inequation x + y ≥ 1.
Let us consider eq(2)
The line 7x + 9y = 63 meets the x-axis at (9, 0) and the y-axis at (0, 7).
And Origin(0, 0) satisfies the inequation 7x + 9y ≤ 63.
So, the section containing the origin will represent the solution set of inequation 7x + 9y ≤ 63.
Let us consider eq(3)
The line x = 6 is parallel to y-axis and it is at a distance of 6 units from it.
And Origin(0, 0) satisfies the inequation x ≤ 6.
So, the section containing the origin will represent the solution set of inequation x ≤ 6.
Let us consider eq(4)
The line y = 5 is parallel to x-axis and it is at a distance of 5 units from it.
And Origin(0, 0) satisfies the inequation y ≤ 5.
So, the section containing the origin will represent the solution set of inequation y ≤ 5.
x ≥ 0 and y ≥ 0 represent the first quadrant.
Thus, the shaded region represents the solution sets of given inequations.
### Question 1(v). Solve the following systems of linear equation graphically:
2x + 3y ≤ 35
y ≥ 3
x ≥ 2
x ≥ 0, y ≥ 0
Solution:
Let us convert the given inequations to equations.
2x + 3y = 35 -(1)
y = 3 -(2)
x = 2 -(3)
x = 0, y = 0
Let us consider eq(1)
The line 2x + 3y = 35 meets the x-axis at (35/2, 0) and the y-axis at (0, 35/3).
And Origin(0, 0) satisfies the inequation 2x + 3y ≤ 35.
So, the section containing the origin will represent the solution set of inequation 2x + 3y ≤ 35.
Let us consider eq(2)
The line y = 3 is parallel to x-axis and it is at a distance of 3 units from it.
And Origin(0, 0) doesnot satisfy the inequation y ≥ 3.
So, the section containing the origin will represent the solution set of inequation y ≥ 3.
Let us consider eq(3)
The line x = 2 is parallel to y-axis and it is at a distance of 2 units from it.
And Origin(0, 0) doesnot satisfy the inequation x ≥ 2.
So, the section containing the origin will represent the solution set of inequation x ≥ 2.
x ≥ 0 and y ≥ 0 represent the first quadrant.
Thus, the shaded region represents the solution sets of given inequations.
### Question 2(i). Show that the solution set of the following linear equations is empty set:
x – 2y ≥ 0
2x – y ≤ -2
x ≥ 0
y ≥ 0
Solution:
Let us convert the given inequations to equations.
x – 2y = 0 (1)
2x – y = -2 -(2)
x = 0 -(3)
y = 0 -(4)
Let us consider eq(1)
The line x – 2y = 0 meets the x-axis at (0, 0) and the y-axis at (0, 0).
If x = 1 then y = 1/2
And Origin(0, 0) satisfies the inequation x – 2y ≥ 0.
So, the section containing the origin will represent the solution set of inequation x – 2y ≥ 0.
Let us consider eq(2)
The line 2x – y = -2 meets the x-axis at (-1, 0) and y-axis at (0, 2).
And Origin(0, 0) doesn’t satisfy the inequation 2x – y ≤ -2.
So, the section containing the origin will represent the solution set of inequation 2x – y ≤ -2.
Clearly, x ≥ 0 and y ≥ 0 represent the first quadrant.
From the given graph we conclude that there is no common region in all the lines.
So, the solution set to given inequations is empty.
Hence Proved
### Question 2(ii). Show that the solution set of the following linear equations is empty set:
x + 2y ≤ 3
3x + 4y ≥ 12
y ≥ 1
x ≥ 0
y ≥ 0
Solution:
Let us convert the given inequations to equations.
x + 2y = 3 -(1)
3x + 4y = 12 -(2)
y = 1 -(3)
x = 0 -(4)
y = 0 -(5)
Let us consider eq(1)
The line x + 2y = 3 meets the x-axis at (3, 0) and the y-axis at (0, 3/2).
And Origin(0, 0) satisfies the inequation x + 2y ≤ 3.
So, the section containing the origin will represent the solution set of inequation x + 2y ≤ 3.
Let us consider eq(2)
The line 3x + 4y = 12 meets the x-axis at (4, 0) and y-axis at (0, 3).
And Origin(0, 0) doesnot satisfy the inequation 3x + 4y ≥ 12.
So, the section containing the origin will represent the solution set of inequation 3x + 4y ≥ 12.
Let us consider eq(3)
The line y = 1 is parallel to x-axis and it is at a distance of 1 unit from it.
And Origin(0, 0) doesnot satisfy the inequation y ≥ 1.
So, the section containing the origin will represent the solution set of inequation y ≥ 1.
x ≥ 0 and y ≥ 0 represent the first quadrant.
From the given graph we conclude that there is no common region in all the lines.
So, the solution set to given inequations is empty.
Hence Proved
### Question 3. Find the linear inequations for which shaded region in the given below figure is the solution set. Draw the diagram for the solution set of linear inequations.
Solution:
Let us consider the line 2x + 3y = 6
The shaded region and origin(0, 0) are on the opposite side of this line.
And Origin(0, 0) doesnot satisfy the inequation 2x + 3y ≥ 6
Let us consider the line 4x + 6y = 24
The shaded region and origin (0, 0) are on the same side of this line.
And Origin(0, 0) satisfies the inequation 4x + 6y ≤ 24
Let us consider the line x – 2y = 2
The shaded region and origin (0, 0) are on the same side of this line.
And Origin(0, 0) satisfies the inequation x – 2y ≤ 2
Let us consider the line -3x + 2y = 3
The shaded region and origin (0, 0) are on the same side of this line.
And Origin(0, 0) satisfies the inequation -3x + 2y ≤ 3
We also have to take x ≥ 0 and y ≥ 0.
Thus, the linear equations comprising the given solution set are:
2x + 3y ≥ 6, 4x + 6y ≤ 24, x – 2y ≤ 2, -3x + 2y ≤ 3, x ≥ 0 and y≥. 0.
### Question 4. Find the linear inequations for which the solution set is the shaded region given below.
Solution:
Let us consider line x + y = 4
The shaded region and origin (0, 0) are on the same side of this line.
And Origin(0, 0) doesnot satisfy the inequation x + y ≤ 4
Let us consider line y = 3
The shaded region and origin (0, 0) are on the same side of this line.
And Origin(0, 0) satisfies the inequation y ≤ 3.
Let us consider line x = 3
The shaded region and origin (0, 0) are on the same side of this line.
And Origin(0, 0) satisfies the inequation x ≤ 3.
Let us consider line x + 5y = 4
The shaded region and origin (0, 0) are on the opposite side of this line.
And Origin(0, 0) doesnot satisfy the inequation x + 5y ≥ 4.
Let us consider line 6x + 2y = 8
The shaded region and origin (0, 0) are on the opposite side of this line.
And Origin(0, 0) doesnot satisfy the inequation 6x + 2y ≥ 8.
We also have to take x ≥ 0 and y ≥ 0.
Thus, the linear equations comprising the given solution set are:
x + y ≤ 4, x ≤ 3, y ≤ 3 x + 5y ≥ 4, 6x + 2y ≥ 8, x ≥ 0 and y ≥ 0.
### Question 5. Show that the solution set of the following linear inequations is an unbounded set:
x + y ≥ 9
3x + y ≥ 12
x ≥ 0 and y ≥ 0.
Solution:
Let us convert the given inequations to equations.
x + y = 9 -(1)
3x + y = 12 -(2)
x = 0 and y = 0 -(3)
Let us consider eq(1)
The line x + y = 9 meets the x-axis at (9, 0) and the y-axis at (0, 9).
And Origin(0, 0) doesnot satisfy the inequation x + y ≥ 9.
So, the section containing the origin will represent the solution set of inequation x + y ≥ 9.
Let us consider eq(2)
The line 3x + y = 12 meets the x-axis at (4, 0) and y-axis at (0, 12).
And Origin(0, 0) doesnot satisfy the inequation 3x + y ≥ 12.
So, the section containing the origin will represent the solution set of inequation 3x + y ≥ 12.
Clearly, x ≥ 0 and y ≥ 0 represent the first quadrant.
Hence, the solution set lies in the first quadrant.
From the given graph we conclude that the shaded region is unbounded.
Hence, the solution set to the given set of inequalities is an unbounded set.
### Question 6(i). Solve the following systems of inequations graphically:
2x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6
Solution:
Let us convert the given inequations to equations.
2x + y = 8 -(1)
x + 2y = 8 -(2)
x + y = 6 -(3)
Let us consider eq(1)
The line 2x + y = 8 meets the x-axis at (4, 0) and the y-axis at (0, 8).
And Origin(0, 0) doesnot satisfy the inequation 2x + y ≥ 8.
So, the section containing the origin will represent the solution set of inequation 2x + y ≥ 8.
Let us consider eq(2)
The line x + 2y ≥ 8 meets the x-axis at (8, 0) and y-axis at (0, 4).
And Origin(0, 0) doesnot satisfy the inequation x + 2y ≥ 8.
So, the section containing the origin will represent the solution set of inequation x + 2y ≥ 8.
Let us consider eq(3)
The line x + y = 6 meets the x-axis at (6, 0) and the y-axis at (0, 6).
And Origin(0, 0) satisfies the inequation x + y ≤ 6.
So, the section containing the origin will represent the solution set of inequation x + y ≤ 6.
Hence, the solution set to the inequalities is the intersection of the above three solutions.
Thus, the shaded region represents the solution set of the given set of inequalities.
### Question 6(ii). Solve the following systems of inequations graphically:
12x + 12y ≤ 840, 3x + 6y ≤ 300, 8x + 4y ≤ 480, x ≥ 0 and y ≥ 0.
Solution:
Let us convert the given inequations to equations.
12x + 12y = 840 -(1)
3x + 6y = 300 -(2)
8x + 4y = 480 -(3)
x = 0 and y = 0 -(4)
Let us consider eq(1)
The line 12x + 12y = 840 meets the x-axis at (70, 0) and the y-axis at (0, 70).
And Origin(0, 0) satisfies the inequation 12x + 12y ≤ 840.
So, the section containing the origin will represent the solution set of inequation 12x + 12y ≤ 840.
Let us consider eq(2)
The line 3x + 6y = 300 meets the x-axis at (100, 0) and the y-axis at (0, 50).
And Origin(0, 0) satisfies the inequation 3x + 6y ≤ 300.
So, the section containing the origin will represent the solution set of inequation 3x + 6y ≤ 300.
Let us consider eq(3)
The line 8x + 4y = 480 meets the x-axis at (60, 0) and the y-axis at (0, 120).
And Origin(0, 0) satisfies the inequation 8x + 4y ≤ 480.
So, the section containing the origin will represent the solution set of inequation 8x + 4y ≤ 480.
Clearly, x ≥ 0 and y ≥ 0 represent the first quadrant.
Hence, the solution set to the inequalities is the intersection of the above three solutions.
Thus, the shaded region represents the solution set of the given set of inequalities.
### Question 6(iii). Solve the following systems of inequations graphically:
x + 2y ≤ 40, 3x + y ≥ 30, 4x + 3y ≥ 60, x ≥ 0 and y ≥ 0.
Solution:
Let us convert the given inequations to equations.
x + 2y = 40 -(1)
3x + y = 30 -(2)
4x + 3y = 60 -(3)
x = 0 and y = 0 -(4)
Let us consider eq(1)
The line x + 2y = 40 meets the x-axis at (40, 0) and the y-axis at (0, 20).
And Origin(0, 0) satisfies the inequation x + 2y ≤ 40.
So, the section containing the origin will represent the solution set of inequation x + 2y ≤ 40.
Let us consider eq(2)
The line 3x + y = 30 meets the x-axis at (10, 0) and y-axis at (0, 30).
And Origin(0, 0) doesnot satisfy the inequation 3x + y ≥ 30.
So, the section containing the origin will represent the solution set of inequation 3x + y ≥ 30.
Let us consider eq(3)
The line 4x + 3y = 60 meets the x-axis at (15, 0) and y-axis at (0, 20).
And Origin(0, 0) doesnot satisfy the inequation 4x + 3y ≥ 60.
So, the section containing the origin will represent the solution set of inequation4x + 3y ≥ 60.
Clearly, x ≥ 0 and y ≥ 0 represent the first quadrant.
Hence, the solution set to the inequalities is the intersection of the above three solutions.
Thus, the shaded region represents the solution set of the given set of inequalities.
### Question 6(iv) Solve the following systems of inequations graphically:
5x + y ≥ 10, 2x + 2y ≥ 12, x + 4y ≥ 12, x ≥ 0 and y ≥ 0.
Solution:
Let us convert the given inequations to equations.
5x + y = 10 -(1)
2x + 2y = 12 -(2)
x + 4y = 12 -(3)
x = 0 and y = 0 -(4)
Let us consider eq(1)
The line 5x + y = 10 meets the x-axis at (2, 0) and y-axis at (0, 10).
And Origin(0, 0) doesnot satisfy the inequation 5x + y ≥ 10.
So, the section containing the origin will represent the solution set of inequation 5x + y ≥ 10.
Let us consider eq(2)
The line 2x + 2y = 12 meets the x-axis at (6, 0) and y-axis at (0, 6).
And Origin(0, 0) doesnot satisfy the inequation 2x + 2y ≥ 12.
So, the section containing the origin will represent the solution set of inequation 2x + 2y ≥ 12.
Let us consider eq(3)
The line x + 4y = 12 meets the x-axis at (12, 0) and y-axis at (0, 3).
And Origin(0, 0) doesnot satisfy the inequation x + 4y ≥ 12.
So, the section containing the origin will represent the solution set of inequationx + 4y ≥ 12.
Clearly, x ≥ 0 and y ≥ 0 represent the first quadrant.
Hence, the solution set to the inequalities is the intersection of the above three solutions.
Thus, the shaded region represents the solution set of the given set of inequalities.
### Question 7. Show that the following system of linear equations has no solutions
x + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
Solution:
Let us convert the given inequations to equations.
x + 2y = 3 -(1)
3x + 4y = 12 -(2)
x = 0 and y = 1 -(3)
The points satisfying the eq(1) are
The points satisfying the eq(2) are
Now, let us draw the graph representing the given inequations
So, from the above graph we can observe that there is no common region.
Hence we can say that, the solution set of these inequalities has no solution.
### Question 8. Show that the solution set of the following system of linear inequations is an unbounded region:
2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0
Solution:
Given that
2x + y ≥ 8 -(1)
x + 2y ≥ 10 -(2)
x ≥ 0 -(3)
y ≥ 0 -(4)
The points satisfying the line 2x + y = 8 are
The points satisfying the x + 2y = 10 are
Clearly, x ≥ 0 and y ≥ 0 represent the first quadrant.
Let us draw the graph representing the given inequations
Since the common shaded region is the solution set of given set of inequalities.
Hence we proved that solution set of given linear inequations is an unbounded region.
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Mathbox for Scott Fenton < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > inffzOLD Structured version Visualization version GIF version
Theorem inffzOLD 31947
Description: The infimum of a finite sequence of integers. (Contributed by Scott Fenton, 8-Aug-2013.) Obsolete version of inffz 31946 as of 10-Oct-2021. (New usage is discouraged.) (Proof modification is discouraged.)
Assertion
Ref Expression
inffzOLD (𝑁 ∈ (ℤ𝑀) → sup((𝑀...𝑁), ℤ, < ) = 𝑀)
Proof of Theorem inffzOLD
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 zssre 11585 . . . . 5 ℤ ⊆ ℝ
2 ltso 10319 . . . . 5 < Or ℝ
3 soss 5188 . . . . 5 (ℤ ⊆ ℝ → ( < Or ℝ → < Or ℤ))
41, 2, 3mp2 9 . . . 4 < Or ℤ
5 cnvso 5818 . . . 4 ( < Or ℤ ↔ < Or ℤ)
64, 5mpbi 220 . . 3 < Or ℤ
76a1i 11 . 2 (𝑁 ∈ (ℤ𝑀) → < Or ℤ)
8 eluzel2 11892 . 2 (𝑁 ∈ (ℤ𝑀) → 𝑀 ∈ ℤ)
9 eluzfz1 12554 . 2 (𝑁 ∈ (ℤ𝑀) → 𝑀 ∈ (𝑀...𝑁))
10 elfzle1 12550 . . . . 5 (𝑥 ∈ (𝑀...𝑁) → 𝑀𝑥)
1110adantl 467 . . . 4 ((𝑁 ∈ (ℤ𝑀) ∧ 𝑥 ∈ (𝑀...𝑁)) → 𝑀𝑥)
128zred 11683 . . . . 5 (𝑁 ∈ (ℤ𝑀) → 𝑀 ∈ ℝ)
13 elfzelz 12548 . . . . . 6 (𝑥 ∈ (𝑀...𝑁) → 𝑥 ∈ ℤ)
1413zred 11683 . . . . 5 (𝑥 ∈ (𝑀...𝑁) → 𝑥 ∈ ℝ)
15 lenlt 10317 . . . . 5 ((𝑀 ∈ ℝ ∧ 𝑥 ∈ ℝ) → (𝑀𝑥 ↔ ¬ 𝑥 < 𝑀))
1612, 14, 15syl2an 575 . . . 4 ((𝑁 ∈ (ℤ𝑀) ∧ 𝑥 ∈ (𝑀...𝑁)) → (𝑀𝑥 ↔ ¬ 𝑥 < 𝑀))
1711, 16mpbid 222 . . 3 ((𝑁 ∈ (ℤ𝑀) ∧ 𝑥 ∈ (𝑀...𝑁)) → ¬ 𝑥 < 𝑀)
18 brcnvg 5441 . . . . 5 ((𝑀 ∈ ℤ ∧ 𝑥 ∈ (𝑀...𝑁)) → (𝑀 < 𝑥𝑥 < 𝑀))
1918notbid 307 . . . 4 ((𝑀 ∈ ℤ ∧ 𝑥 ∈ (𝑀...𝑁)) → (¬ 𝑀 < 𝑥 ↔ ¬ 𝑥 < 𝑀))
208, 19sylan 561 . . 3 ((𝑁 ∈ (ℤ𝑀) ∧ 𝑥 ∈ (𝑀...𝑁)) → (¬ 𝑀 < 𝑥 ↔ ¬ 𝑥 < 𝑀))
2117, 20mpbird 247 . 2 ((𝑁 ∈ (ℤ𝑀) ∧ 𝑥 ∈ (𝑀...𝑁)) → ¬ 𝑀 < 𝑥)
227, 8, 9, 21supmax 8528 1 (𝑁 ∈ (ℤ𝑀) → sup((𝑀...𝑁), ℤ, < ) = 𝑀)
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 196 ∧ wa 382 = wceq 1630 ∈ wcel 2144 ⊆ wss 3721 class class class wbr 4784 Or wor 5169 ◡ccnv 5248 ‘cfv 6031 (class class class)co 6792 supcsup 8501 ℝcr 10136 < clt 10275 ≤ cle 10276 ℤcz 11578 ℤ≥cuz 11887 ...cfz 12532 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1869 ax-4 1884 ax-5 1990 ax-6 2056 ax-7 2092 ax-8 2146 ax-9 2153 ax-10 2173 ax-11 2189 ax-12 2202 ax-13 2407 ax-ext 2750 ax-sep 4912 ax-nul 4920 ax-pow 4971 ax-pr 5034 ax-un 7095 ax-cnex 10193 ax-resscn 10194 ax-pre-lttri 10211 ax-pre-lttrn 10212 This theorem depends on definitions: df-bi 197 df-an 383 df-or 827 df-3or 1071 df-3an 1072 df-tru 1633 df-ex 1852 df-nf 1857 df-sb 2049 df-eu 2621 df-mo 2622 df-clab 2757 df-cleq 2763 df-clel 2766 df-nfc 2901 df-ne 2943 df-nel 3046 df-ral 3065 df-rex 3066 df-reu 3067 df-rmo 3068 df-rab 3069 df-v 3351 df-sbc 3586 df-csb 3681 df-dif 3724 df-un 3726 df-in 3728 df-ss 3735 df-nul 4062 df-if 4224 df-pw 4297 df-sn 4315 df-pr 4317 df-op 4321 df-uni 4573 df-iun 4654 df-br 4785 df-opab 4845 df-mpt 4862 df-id 5157 df-po 5170 df-so 5171 df-xp 5255 df-rel 5256 df-cnv 5257 df-co 5258 df-dm 5259 df-rn 5260 df-res 5261 df-ima 5262 df-iota 5994 df-fun 6033 df-fn 6034 df-f 6035 df-f1 6036 df-fo 6037 df-f1o 6038 df-fv 6039 df-riota 6753 df-ov 6795 df-oprab 6796 df-mpt2 6797 df-1st 7314 df-2nd 7315 df-er 7895 df-en 8109 df-dom 8110 df-sdom 8111 df-sup 8503 df-pnf 10277 df-mnf 10278 df-xr 10279 df-ltxr 10280 df-le 10281 df-neg 10470 df-z 11579 df-uz 11888 df-fz 12533 This theorem is referenced by: (None)
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Finance 3 Pages
# Wacc
##### Question
Calculate the weighted average cost of capital (WACC) using the market value of equity to determine a more realistic cost of capital. Visit a Web site to get the current value of the common stock price per share and multiply this value times the most recent number of shares of common stock outstanding. In this exercise, you will be ignoring preferred stock because its weight value will most likely be too low to impact the final result. You may use the following table to complete this portion of the task:
Company Common Stock, price per share Number of Common Shares Outstanding Market Value of Common Equity
Now, you can estimate the total market value of the company by adding the book value of total liabilities to the market value of the firm’s common equity and determine their market value weights.
Company Total Liabilities Market Value of Common Equity Market Value of the Firm Values Weights
Using the cost of each component as determined in the Phase 4 IP, calculate the firm's market value WACC.
After-Tax Cost of Debt Cost of Common Equity WACC Unweighted Cost Market Weight of Component Market Weighted Cost of Component
The firm is considering investing in a capital project that will have an initial cost of \$12 million. The project is expected to have a productive life of 5 years, and at the end of this period, it is expected to have a salvage value of \$2 million. The net value of the project will be depreciated using the straight-line method for the full 5 years. The project is expected to increase the firm’s revenue by \$10 million per year, and related expenses (not including depreciation) are expected to increase by about \$6.5 million per year.
The first thing you need to do is calculate the annual depreciation. Feel free to use the following table:
Initial Investment less Salvage Value Depreciable Value Life of Project (years) Depreciation/year
Now, you need to calculate the relevant cash flows for the project. You can use the average tax rate that was calculated in Phase 4 to determine the additional taxes the firm will have to pay. The following template may be of some assistance to you:
Years 0 1 2 3 4 5 Initial Investment (negative) Increase in Revenue Less Increase in Operating Expenses Increase in Operating Income Less Depreciation/year Taxable Income Less Taxes at Average Rate Net Income Plus Depreciation/year Operating Cash Flow Plus Salvage Value (Year 5) Relevant Cash Flows (0–5)
At this point, you are ready to apply the capital budgeting techniques of net present value (NPV) and the internal rate of return (IRR). To calculate the NPV, use the market-value WACC as your discount rate and the required rate of return for IRR.
• After completing the required calculations, explain your results in a Word document, and attach the spreadsheet showing your work. Be sure to explain the following:
• Based on your calculations, would you recommend approving or rejecting the project, and why?
• Why would you expect both NPV and IRR to support the same conclusion to accept or reject the project?
• If you employed a cost of capital of 20%, would you have made the same accept or reject decision, and why?
• What are some of the advantages and disadvantages of the 2 methods (NPV and IRR), and under what circumstances is one more reliable over the other?
Why is operating cash flow used in capital budgeting and not net income?
##### Solution
Title: Wacc
Length: 3 pages (902 Words)
Style: APA
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Determination of the market value of the firm
The analysis was based on Tesco Plc. a supermarket chain whose home market is in the United Kingdom. The share price as per the end of the year 2014 on which the study is based is £230. The amount of shares outstanding as at December 2014 are 80,680,000. The market value as of the same date is determined to be £8,556,400,000.
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The OEIS is supported by the many generous donors to the OEIS Foundation.
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A304156 T(n,k)=Number of nXk 0..1 arrays with every element unequal to 2, 3, 4 or 6 king-move adjacent elements, with upper left element zero. 7
%I #4 May 07 2018 09:04:46
%S 0,0,0,0,3,0,0,5,5,0,0,18,4,18,0,0,61,42,42,61,0,0,209,130,346,130,
%T 209,0,0,702,464,1767,1767,464,702,0,0,2381,1722,10182,13279,10182,
%U 1722,2381,0,0,8069,6378,60352,100942,100942,60352,6378,8069,0,0,27330,22939,350540
%N T(n,k)=Number of nXk 0..1 arrays with every element unequal to 2, 3, 4 or 6 king-move adjacent elements, with upper left element zero.
%C Table starts
%C .0....0.....0.......0........0..........0...........0.............0
%C .0....3.....5......18.......61........209.........702..........2381
%C .0....5.....4......42......130........464........1722..........6378
%C .0...18....42.....346.....1767......10182.......60352........350540
%C .0...61...130....1767....13279.....100942......839935.......6803634
%C .0..209...464...10182...100942....1100030....12806826.....145611915
%C .0..702..1722...60352...839935...12806826...217640633....3520683600
%C .0.2381..6378..350540..6803634..145611915..3520683600...80851458613
%C .0.8069.22939.2034140.54261600.1634366796.56360283031.1830588080314
%H R. H. Hardin, <a href="/A304156/b304156.txt">Table of n, a(n) for n = 1..180</a>
%F Empirical for column k:
%F k=1: a(n) = a(n-1)
%F k=2: a(n) = 3*a(n-1) +a(n-2) +2*a(n-3) -2*a(n-4) -4*a(n-5) for n>6
%F k=3: [order 18] for n>19
%F k=4: [order 66] for n>67
%e Some solutions for n=5 k=4
%e ..0..1..0..1. .0..1..0..0. .0..1..0..1. .0..1..1..0. .0..1..0..1
%e ..1..0..1..0. .1..0..1..1. .0..1..1..0. .1..0..0..1. .1..1..0..0
%e ..1..1..1..1. .1..0..0..0. .1..0..1..0. .1..0..0..1. .0..1..1..1
%e ..0..1..0..0. .1..0..0..1. .0..1..1..1. .1..0..0..1. .0..1..1..1
%e ..1..0..1..1. .0..1..0..1. .1..0..1..0. .1..0..0..1. .0..1..0..0
%Y Column 2 is A303684.
%K nonn,tabl
%O 1,5
%A _R. H. Hardin_, May 07 2018
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Analysis of Free Fall Motion
* Please refer to the text below for how to make a graph.
Free fall motion
If you hold the ball in your hand and release it, it will fall.
In the absence of air resistance, the only force acting on the ball while it is falling is 'gravity.'
If you take pictures of free-falling balls at equal time intervals, the distance between the balls gradually increases. The graph of distance traveled over time can be expressed as follows.
The distance between the balls is the distance traveled by the balls in a certain amount of time, meaning the speed. Therefore, it can be seen that a ball moving in free fall has a constant increase in speed with time. The time-speed graph can be expressed as:
Analysis of the time-speed graph of free-fall motion reveals that the speed increases by '9.8 m/s' every second, where '9.8' is called the 'gravity acceleration constant' of the Earth.
Some textbooks calculate 9.8 by rounding it to 10. (Convenience of calculation process)
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# How To Use Contraharmonic Mean Filter – C# Guide
We can use contraharmonic mean filter to process image data in spatial domain. It's most effective against salt and pepper noise.
Andraz Krzisnik
How To Use Contraharmonic Mean Filter...
We use contraharmonic mean filter to process image data in spatial domain. Therefore we will need to use kernel and convolution to apply it. We could also say that it’s the most useful for removing impulse or salt and pepper noise.
Mean filters are in general most useful to deal with images that only have noise degradation. In other words, image alterations of any other kind might complicate restoration process.
Kernel and convolution are essential for filtering in spatial domain. I’ll try to keep it short and simple about it, but I also recommend you get to know how it works in detail.
Kernel is basically a small matrix we put on top of our image. It works as a window to take values in a local area of the image and compute one output value from each position. This is also the reason why spatial filtering outputs images with slightly smaller dimensions.
Convolution is the process of calculating the values from kernel and image to compute new output value. We also move kernel along the entire image to compute all output values to form the image.
## Contraharmonic mean filter
As we mentioned before, contraharmonic mean filter is most effective to use for images with salt and pepper noise. While harmonic mean filter deals with salt noise and fails for the pepper noise, this one works for both.
However, we can’t use it to remove both at the same time. Let’s take a look at the following formula to understand better how it works.
The g(s, t) function represents pixel value from corrupted image, while s and t variables represent the coordinates inside the kernel.
There is also Q variable in this formula, which is something extra and only this mean filter has it. Furthermore, it represents the order of the filter. If we set it to positive number, it’s going to remove pepper noise and if we set it to negative, it’s going to remove salt noise.
If we set Q to 0, it’s going to reduce the filter into arithmetic mean filter and if we turn it to -1, it’s going to become harmonic mean filter.
Therefore, to remove salt and pepper noise, we need to use this filter twice to remove each individually.
## C# code
``````public static Bitmap ContraharmonicMean(this Bitmap image, double order)
{
int w = image.Width;
int h = image.Height;
BitmapData image_data = image.LockBits(
new Rectangle(0, 0, w, h),
PixelFormat.Format24bppRgb);
int bytes = image_data.Stride * image_data.Height;
byte[] buffer = new byte[bytes];
Marshal.Copy(image_data.Scan0, buffer, 0, bytes);
image.UnlockBits(image_data);
int r = 1;
int wres = w - 2 * r;
int hres = h - 2 * r;
Bitmap result_image = new Bitmap(wres, hres);
BitmapData result_data = result_image.LockBits(
new Rectangle(0, 0, wres, hres),
ImageLockMode.WriteOnly,
PixelFormat.Format24bppRgb);
int res_bytes = result_data.Stride * result_data.Height;
byte[] result = new byte[res_bytes];
for (int x = r; x < w - r; x++)
{
for (int y = r; y < h - r; y++)
{
int pixel_location = x * 3 + y * image_data.Stride;
int res_pixel_loc = (x - r) * 3 + (y - r) * result_data.Stride;
double[] sum1 = new double[3];
double[] sum2 = new double[3];
for (int kx = -r; kx <= r; kx++)
{
for (int ky = -r; ky <= r; ky++)
{
int kernel_pixel = pixel_location + kx * 3 + ky * image_data.Stride;
for (int c = 0; c < 3; c++)
{
sum1[c] += Math.Pow(buffer[kernel_pixel + c], order + 1);
sum2[c] += Math.Pow(buffer[kernel_pixel + c], order);
}
}
}
for (int c = 0; c < 3; c++)
{
result[res_pixel_loc + c] = (byte)(sum1[c] / sum2[c]);
}
}
}
Marshal.Copy(result, 0, result_data.Scan0, res_bytes);
result_image.UnlockBits(result_data);
return result_image;
}``````
## Conclusion
Contraharmonic mean filter is one of the filters we can use to turn into others if we need it. Therefore, it’s a kind of universal filter for processing salt and pepper noise. However, it has some shortcomings, such as we need to know if the noise is light or dark to set the order of it right.
### Related Articles
#### How To Adjust Image Tone – C# Guide
This guide shows how to apply image tone corrections for flat, dark and light images. The purpose of it is to adjust brightness and contrast.
Posted on by Andraz Krzisnik
#### How To Use Gaussian Highpass Filter – C# Guide
Gaussian highpass filter processes images in frequency domain. It attenuates low frequencies without creating ringing artifacts.
Posted on by Andraz Krzisnik
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Cody
# Problem 411. Back to basics 21 - Matrix replicating
Solution 1971641
Submitted on 11 Oct 2019 by Cao Van Cuong
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
x = [1]; y_correct = [1 1;1 1]; assert(isequal(matrix_replication(x),y_correct))
2 Pass
x = [1 2;3 4]; y_correct = [1 2 1 2; 3 4 3 4; 1 2 1 2; 3 4 3 4]; assert(isequal(matrix_replication(x),y_correct))
3 Pass
x = [1 2]; y_correct = [1 2 1 2; 1 2 1 2]; assert(isequal(matrix_replication(x),y_correct))
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### CAT 2018 Question Paper (Slot 2) - LRDI Question 21
Instructions
Each of the 23 boxes in the picture below represents a product manufactured by one of the following three companies: Alfa, Bravo and Charlie. The area of a box is proportional to the revenue from the corresponding product, while its centre represents the Product popularity and Market potential scores of the product (out of 20). The shadings of some of the boxes have got erased.
The companies classified their products into four categories based on a combination of scores (out of 20) on the two parameters - Product popularity and Market potential as given below:
Question 21
# Considering all companies' products, which product category had the highest revenue?
Solution
Let us divide the given figure in four quadrants (Q1, Q2, Q3, Q4).
Let us solve this problem by considering only one category at a time.
(A) Blockbuster category: We have two information regarding Blockbuster category.
1. Alfa and Bravo had the same number of products in the Blockbuster category.
There are a total of 7 products in Blockbuster category. Alfa already has two products in blockbuster category. If Alfa has 3 products in blockbuster category then Bravo will also have 3 products in Blockbuster category which is not possible as there are 2 products of Charlie. Hence, we can say that Alfa and Bravo have 2 products each in Blockbuster category whereas Charlie has 3 products in Blockbuster category.
2. It is also given that Charlie had a higher revenue than Bravo from products in the Blockbuster category.
We know that one of the product 1 and 2 is from Charlie and the other is from Bravo.
If product 1 is from Charlie, then we can say that products 1, 7 and 5 are from Charlie. Therefore, revenue generated by products in Charlie category = 2 + 4 + 6 = 12 units. (Assuming area of a square to be 1 unit)
In this case product 2 and product n are from Bravo.Therefore, revenue generated by products in Bravo category = 6 + 9 = 15 units.
We can see that products from Charlie have a higher revenue than Bravo. Hence, this case in not possible.
Therefore, we can say that Product 1 is from Bravo and Product 2 is from Charlie. We have determined each product's company name in Blockbuster category.
(B) No-hope category: We have two information regarding No-hope category.
(1) Charlie had more products than Bravo but fewer products than Alfa in the No-hope category. Bravo and Charlie had the same revenue from products in the No-hope category.There are a total of 6 products in no-hope category.
Therefore, we can say that Alfa, Charlie and Bravo have 3, 2 and 1 products in No-hope category in that order.
Bravo and Charlie had the same revenue from products in the No-hope category.
Revenue generated for Bravo in the No-hope category = 4 units. We need same revenue for Charlie which ha s 2 products in this category. Hence, we can say that Product 10 and one of product 8 and 9 is from Charlie and other is from Alfa. Let's assume that product 8 is from Charlie and product 9 is from Alfa.
(C) Doubtful category: We have two information regarding Doubtful category.
(1). Charlie did not have any product in the Doubtful category, while Alfa had one product more than Bravo in this category .
(2). Bravo had a higher revenue than Alfa from products in the Doubtful category.
We can see that there are a total of 7 products in this category. Hence, we can say that 4 products are from Alfa and 3 products are from Bravo.
We can see that one of product 14, 15 and 16 is from Bravo company and others are from Alfa company. Bravo will have higher revenue than Alfa only when product no. 14 is from Bravo and others (15 and 16) are from Alfa category.
In this case total revenue by Bravo = Product 14 + Product 19 + Product 20 = 9 + 6 + 2 = 17
Similarly, total revenue by Charlie = Product 15 + Product 16 + Product 17 + Product 18 = 6 + 1 + 1 + 4 = 12
(D) Promising category: We have only 1 direct information regarding Promising category.
1. Each company had an equal number of products in the Promising category.
There are a total of 3 products in promising category with different revenue. Therefore, we can say that each company had 1 product in promising category. We are given that Alfa and Charlie had the same total revenue considering all products. We can calculate the revenue generated by Alfa and Charlie from the products in categories.
Revenue generated by Charlie from all categories except Promising = From Blockbuster + From No-hope + From Doubtful
$$\Rightarrow$$ (9+6+2) + (3+1) + (0) = 21 units
Revenue generated by Alfa from all categories except Promising = From Blockbuster + From No-hope + From Doubtful
$$\Rightarrow$$ (6+3) + (4+2+1) + (1+6+4+1) = 28 units
We can see the difference between revenue generated by Charlie and Alfa from remaining categories is 7 units. Hence, we can say that Charlie's product's revenue should be 7 units more than Alfa's product's revenue in Promising category. That is possible only in one case where product 22 is from Charlie and product 21 is form Alfa. Consequently, we can say that product 23 is from Bravo. Now we have identified each product's company name we can answer all the questions.
Revenue generated by the products in Promising category = 2 + 9 + 3 = 14 units.
Revenue generated by the products in Doubtful category = 1 + 9 + 4 + 6 + 2 + 1 + 6 = 29 units.
Revenue generated by the products in No-hope category = 4 + 4 + 3 + 2 + 1 + 1 = 15 units.
Revenue generated by the products in Blockbuster category = 6 + 3 + 6 + 2 + 4 + 6 + 9 = 36 units.
We can see that the revenue generated is the highest for Blockbuster category. Hence, option B is the correct answer.
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# norm of orthogonal projection of some vector in Hilbert space
Let $H$ be Hilbert space and $u_1,u_2,...u_n \in H$ (vectors dont have to be orthogonal)
$V=span\{u_1,u_2,...u_n\}\subset H$
and $S$ is unit sphere in $V$. $P_V$ is orthogonal projection on V.
Now lets take some $h\in H$ $$\sup_{v\in S}|\langle h,v \rangle| = \|P_Vh \|$$
How can I show this property? I know that $\|P_Vh-h \|= \inf_{v\in V}\|v-h\|$ for $V$ closed and convex , $(P_Vh-h) \perp V$ , uniqueness , $P^2=P$ ... but supremum confused me. Is there some short elegant way to show this?
P.S. give hint if its easy and short
## 1 Answer
Hint: From $(P_Vh-h) \perp V$ you know $\langle h,v \rangle=\langle P_Vh,v \rangle$. Which $v\in S$ maximizes this expression?
• now cauchy-schwartz $|\langle P_V h,v \rangle| \le \|P_V h\| \| v \|$ equality of $v=\alpha P_V h$ , $\|v\|=1$ so $\alpha = 1/ \| P_V h \|$. correct? – jack Feb 21 '16 at 23:21
• @jack: Yes, except that's $\alpha^{-1}$ (and you need to treat $h=0$ separately). – joriki Feb 21 '16 at 23:24
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# is a generalization of Classical and Interval Statistics
While the Classical Statistics deals with determinate data and determinate inference methods only, the Neutrosophic Statistics deals with indeterminate data, i.e. data that has some degree of indeterminacy (unclear, vague, partially unknown, contradictory, incomplete, etc.), and indeterminate inference methods that contain degrees of indeterminacy as well (for example, instead of crisp arguments and values for the probability distributions, charts, diagrams, algorithms, functions etc. one may have inexact or ambiguous arguments and values).
For example, the population or sample sizes might not be exactly known because of some individuals that partially belong to the population or sample, and partially they do not belong, or individuals whose membership is completely unknown. Also, there are population or sample individuals whose data could be indeterminate.
The Neutrosophic Statistics was founded by Prof. Dr. Florentin Smarandache, from the University of New Mexico, United States, in 1998, who developed it in 2014 by introducing the Neutrosophic Descriptive Statistics (NDS). Further on, Prof. Dr. Muhammad Aslam, from the King Abdulaziz University, Saudi Arabia, introduced in 2018 the Neutrosophic Inferential Statistics (NIS), Neutrosophic Applied Statistics (NAS), and Neutrosophic Statistical Quality Control (NSQC).
The Neutrosophic Statistics is also a generalization of Interval Statistics, because of, among others, while Interval Statistics is based on Interval Analysis, Neutrosophic Statistics is based on Set Analysis (meaning all kinds of sets, not only intervals, for example finite discrete sets). Also, when computing the mean, variance, standard deviation, probability distributions etc. in classical and interval statistics it is automatically assumed that all individuals belong 100% to the respective sample or population, but in our world one often meet individuals that only partially belong, partially do not belong, and partially their belong-ness is indeterminate. The neutrosophic statistics results are more accurate than the classical and interval statistics, since for example the individuals who belong only partially do not have to be considered at the same level as one those that fully belong.
The Neutrosophic Probability Distributions may be represented by three curves: one representing the chance of the event to occur, other the chance of the event not to occur, and a third one the indeterminate chance of the event to occur or not.
Neutrosophic Statistics is more elastic than Classical Statistics.
If all data and inference methods are determinate, then the Neutrosophic Statistics coincides with the Classical Statistics.
If all sets that are used are intervals, and all individuals belong 100% to the sample and population, and there is only one probability distribution curve, then the Neutrosophic Statistics coincides with the Interval Statistics.
But, since in our world we have more indeterminate data than determinate data, therefore more neutrosophic statistical procedures are needed than classical ones.
Of course, the Neutrosophic DataSets (where the data have some degree of indeterminacy) are used in Neutrosophic Statistics.
The Neutrosophic Numbers of the form N = a+bI have been defined by W. B. Vasantha Kandasamy and F. Smarandache in 2003 [see B2], and they were interpreted as: "a" is the determinate part of the number N, and "bI" is the indeterminate part of the number N by F. Smarandache in 2014 [see B3]. For the neutrosophic statistics "I" is a subset.
Neutrosophic Statistics is the analysis of events described by the Neutrosophic Probability.
Neutrosophic Probability is a generalization of the classical probability and imprecise probability in which the chance that an event A occurs is t% true - where t varies in the subset T, i% indeterminate - where i varies in the subset I, and f% false - where f varies in the subset F. In classical probability the sum of all space probabilities is equal to 1, while in Neutrosophic Probability it is equal to 3.
In Imprecise Probability: the probability of an event is a subset T in [0, 1], not a number p in [0, 1], what’s left is supposed to be the opposite, subset F (also from the unit interval [0, 1]); there is no indeterminate subset I in imprecise probability [see B9].
The function that models the Neutrosophic Probability of a random variable x is called Neutrosophic distribution: NP(x) = ( T(x), I(x), F(x) ), where T(x) represents the probability that value x occurs, F(x) represents the probability that value x does not occur, and I(x) represents the indeterminate / unknown probability of value x [see B3].
Comparison between Neutrosophic Statistics (NS) and Interval Statistics (IS)
We show that NS and IS are different, and in many applications the NS is more general than IS.
NS is not reduced to only using neutrosophic numbers in statistical applications, as some people asserted, but it is much broader. NS deals with all types of indeterminacy, while IS deals only with indeterminacy that can be represented by intervals. However, not all indeterminacies (uncertainties) may be represented by intervals.
Below we present several advantages of applying NS over IS:
- Neutrosophic Statistics is based on Set Analysis, while Interval Statistics on Interval Analysis, therefore the Interval Statistics is a particular case of the Neutrosophic Statistics that uses all types of sets, not only intervals.
- The numerical neutrosophic numbers permit the reduction of indeterminacy through operations, while the intervals increase the indeterminacy (see a simple example: let N1 = 4+3I, N2 = 5-2I, where indeterminacy I = [0,1]; using NS one has: N1 + N2 = 9 + I = [9, 10];
using IS one has: N1 + N2 = [4, 7] + [3, 5] = [7, 12]; clearly the indeterminacy or the real data point being in [7, 12], is bigger than the real data point being in [9, 10]).
- Instead of intervals, for specific applications NS uses hesitant sets {discrete finite sets of numbers}, which make the calculations easier and reduce the indeterminacy (for example, if the real values may be 0.4, 7.9, or 41.5 (not sure which ones), instead of taking the interval [0.4, 41.5] as in IS, it is easier in NS to take just the hesitant set {0.4, 7.9, 41.5} of cardinal 3).
- NS deals with sample or population whose size is not well-known.
- NS deals with sample or population which contain individuals that only partially belong to the sample/population and others whose appurtenance is unknown.
- NS deals with sample or population individuals whose degree of appurtenance to the sample or population may be outside of the interval [0, 1], as in neutrosophic overset (degree > 1), underset (degree < 0), and in general neutrosophic offset (both appurtenance degrees, > 1 and < 0, for various individuals) [see B4].
- Neutrosophic (or Indeterminate) Data is a vague, unclear, incomplete, partially unknown, conflicting indeterminate data.
- NS also deals with refined neutrosophic data used in the Big Data.
- NS may employ partially indeterminate curves.
- NS also uses Thick Functions (as intersections of curves, that may not be represented by intervals) as probability distributions [see B3].
- Neutrosophic Probability Distribution (NPD) of an event (x) to occur is represented by three curves: NPD(x) = (T(x), I(x), F(x)), where T(x) represent the chance that the event E occurs, I(x) the indeterminate-chance that the event E occurs or not, and F(x) the chance that the event x does not occur. With T(x), I(x), F(x) being classical or neutrosophic (unclear, approximate, thick) functions – depending on each application, and T(x) + I(x) + F(x) in [0, 3] {see B9}.
- Diagrams, histograms, pictographs, line/bar/cylinder graphs, plots with neutrosophic data (not represented by intervals) [see B9].
- Not well-known (or completely unknown): the mean, variance, standard deviation, probability distribution function, and other statistic
- The Qualitative Data is represented by a finite discrete neutrosophic label set, instead of a label interval.
- You cannot use Interval Statistics or Interval (Imprecise) Probability to compute the probability of a die on a cracked surface, or coin on a crack surface, or s defect die or coin [see B9].
We deal with indeterminacy with respect to the probability or statistics space or space's elements, indeterminacy with respect to the observer that evaluates the event, indeterminacy with respect to the event [4].
You cannot approximate the indeterminacy from these examples by using some interval, so you need neutrosophic probability and statistics that deal with all types of indeterminacies.
- In conclusion: we cannot represent all types of indeterminacies by intervals.
More than 100 papers, nine books, one PhD thesis, and five international scientific seminars have been published or presented on neutrosophic statistics, including many journals by Elsevier and Springer of high impact factor.
References
Books
B (sixth edition). InfoLearnQuest, 1998 - 2007, 156 p. http://fs.unm.edu/eBook-Neutrosophics6.pdf
B2. W. B. Vasantha Kandasamy, Florentin Smarandache, Fuzzy Cognitive Maps and Neutrosophic Cognitive Maps, Xiquan, Phoenix, 211 p., 2003, http://fs.unm.edu/NCMs.pdf
B3. Florentin Smarandache: Introduction to Neutrosophic Statistics. Sitech & Education Publishing, 2014, 124 p. http://fs.unm.edu/NeutrosophicStatistics.pdf
B4. Florentin Smarandache: Neutrosophic Overset, Neutrosophic Underset, and Neutrosophic Offset. Similarly for Neutrosophic Over-/Under-/Off- Logic, Probability, and Statistics. Pons Editions, Brussels, 2016, 168 p. http://fs.unm.edu/NeutrosophicOversetUndersetOffset.pdf
B5. Maikel Leyva Vázquez, Florentin Smarandache: Neutrosofía: Nuevos avances en el tratamiento de la incertidumbre. Pons Editions, Bruselas, 2018, 74 p. http://fs.unm.edu/NeutrosofiaNuevosAvances.pdf
B6. Tatiana Veronica Gutierrez Quinonez, Fabian Andres Espinoza, Ingrid Kathyuska Giraldo, Angel Steven Asanza, Mauricio Daniel Montenegro: Estadistica y Probabilidades: Una Vision Neutrosofica desde el Aprendizaje Basado en Problemas en la Construccion del Conocimiento. Pons Editions, Bruselas, 2020, 131 p. http://fs.unm.edu/EstadisticaYProbabilidadNeutrosofica.pdf
B7. F. Smarandache, Neutrosophic Statistics vs. Classical Statistics, section in Nidus Idearum / Superluminal Physics, Vol. 7, third edition, p. 117, 2019, http://fs.unm.edu/NidusIdearum7-ed3.pdf .
B8. F. Smarandache, Nidus Idearum de Neutrosophia (Book Series), Editions Pons, Brussels, Belgium, Vols. 1-7, 2016-2019; http://fs.unm.edu/ScienceLibrary.htm
B9. F. Smarandache, Introduction to Neutrosophic Measure, Neutrosophic Integral, and Neutrosophic Probability, Sitech Publishing House, Craiova, 2013, http://fs.unm.edu/NeutrosophicMeasureIntegralProbability.pdf
PhD Thesis
PhD1. Rafif Alhabib: Formulation of the classical probability and some probability distributions due to neutrosophic logic and its impact on Decision Making. PhD Thesis in Arabic, held under the supervision of Dr. M. M. Ranna, Dr. H. Farah, Dr. A. A. Salama, Faculty of Science, Department of Mathematical Statistics, University of Aleppo, Syrian Arab Republic, 2019. http://fs.unm.edu/NS/FormulationOfTheClassicalProbability-PhDThesis.pdf
Scientific Presentations
SP1. Muhammad Aslam, Testing wind speed using Neutrosophic Weibull distribution, Bejing Jiaotong University, P. R. China, 14 November 2022.
Articles
1. Florentin Smarandache: Operators on Single-Valued Neutrosophic Oversets, Neutrosophic Undersets, and Neutrosophic OffsetsJournal of Mathematics and Informatics, Vol. 5, 2016, 63-67.
2. Florentin Smarandache: Interval-Valued Neutrosophic Oversets, Neutrosophic Undersets, and Neutrosophic OffsetsInternational Journal of Science and Engineering Investigations, Vol. 5, issue 54, 2016, Paper ID: 55416-01, 4 p.
3. Nouran M. Radwan, M. Badr Senousy, Alaa El Din M. Riad: Approaches for Managing Uncertainty in Learning Management SystemsEgyptian Computer Science Journal, vol. 40, no. 2, May 2016, 10 p.
4. Muhammad Aslam: A Variable Acceptance Sampling Plan under Neutrosophic Statistical Interval MethodSymmetry 2019, 11, 114, DOI: 10.3390/sym11010114.
5. Soumyadip Dhar, Malay K. KunduAccurate segmentation of complex document image using digital shearlet transform with neutrosophic set as uncertainty handing toolApplied Soft Computing, vol. 61, 2017, 412–426.
6. B. Kavitha, S. Karthikeyan, P. Sheeba MaybellAn ensemble design of intrusion system for handling uncertainty using Neutrosophic Logic ClassifierKnowlwdge-Based Systems, vol. 28, 2012, 88-96.
7. Muhammad AslamA new attribute sampling plan using neutrosophic statistical interval methodComplex & Intelligent Systems, 6 p. DOI: 10.1007/s40747-018-0088-6
8. Muhammad Aslam, Nasrullah Khan, Mohammed AlbassamControl Chart for Failure-Censored Reliability Tests under Uncertainty EnvironmentSymmetry 2018, 10, 690, DOI: 10.3390/sym10120690.
9. Muhammad Aslam, Nasrullah Khan, Ali Hussein AL-MarshadiDesign of Variable Sampling Plan for Pareto Distribution Using Neutrosophic Statistical Interval MethodSymmetry 2019, 11, 80, DOI: 10.3390/sym11010080.
10. Jun Ye, Jiqian Chen, Rui Yong, Shigui DuExpression and Analysis of Joint Roughness Coefficient Using Neutrosophic Number FunctionsInformation, Volume 8, 2017, 13 pages.
11. Jiqian Chen, Jun Ye, Shigui Du, Rui YongExpressions of Rock Joint Roughness Coefficient Using Neutrosophic Interval Statistical NumbersSymmetry, Volume 9, 2017, 7 pages.
12. Adrian Rubio-Solis, George PanoutsosFuzzy Uncertainty Assessment in RBF Neural Networks using neutrosophic sets for Multiclass Classification. Presented at 2014 IEEE International Conference on Fuzzy Systems (FUZZ-IEEE) July 6-11, 2014, Beijing, China, 8 pages.
13. Pierpaolo D’UrsoInformational Paradigm, management of uncertainty and theoretical formalisms in the clustering framework: A reviewInformation Sciences, 400–401 (2017), pp. 30-62, 33 pages.
14. Muhammad Aslam, Mohammed AlbassamInspection Plan Based on the Process Capability Index Using the Neutrosophic Statistical MethodMathematics 2019, 7, 631, DOI: 10.3390/math7070631.
15. Mirela Teodorescu, Florentin Smarandache, Daniela GifuMaintenance Operating System Uncertainties Approached through Neutrosophic Theory. 8 p.
16. Muhammad Aslam, Rashad A. R. Bantan, Nasrullah KhanMonitoring the Process Based on Belief Statistic for Neutrosophic Gamma Distributed ProductProcesses 2019, 7, 209, DOI: 10.3390/pr7040209.
17. Rafael Rojas-Gualdron, Florentin Smarandache, Carlos Diaz-BohorquezApplication of The Neutrosophical Theory to Deal with Uncertainty in Supply Chain Risk Management. AGLALA 2019; 10 (2): 1-19.
18. Florentin Smarandache, Gheorghe SavoiuNeutrosophic Index Numbers: Neutrosophic Logic Applied In The Statistical Indicators TheoryCritical Review, Vol. XI, 2015, pp. 67-100.
19. Murat Kirisci, Necip SimsekNeutrosophic normed spaces and statistical convergenceJournal of Analysis, 11 April 2020, DOI: 10.1007/s41478-020-00234-0.
20. S.K. PatroThe Neutrosophic Statistical Distribution: More Problems, More Solutions. 17 p.
21. Deepesh Kunwar, Jayant Singh, Florentin SmarandacheNeutrosophic statistical evaluation of migration with particular reference to JaipurOctogon Mathematical Magazine, vol. 26, no. 2, October 2018, 560-568.
22. Deepesh Kunwar, Jayant Singh, Florentin SmarandacheNeutrosophic statistical techniques to find migration pattern in JaipurOctogon Mathematical Magazine, vol. 26, no. 2, October 2018, 583-592.
23. Muhammad Aslam, Osama H. Arif, Rehan Ahmad Khan SherwaniNew Diagnosis Test under the Neutrosophic Statistics: An Application to Diabetic Patients. Hindawi, BioMed Research International, Volume 2020, Article ID 2086185, 7 pages; DOI: 10.1155/2020/2086185.
24. Jose L. Salmeron, Florentin SmarandacheProcessing Uncertainty and Indeterminacy in Information Systems success mapping. 13 p., arXiv:cs/0512047v2.
25. Wenzhong Jiang, Jun Ye, Wenhua Cui: Scale Effect and Anisotropic Analysis of Rock Joint Roughness Coefficient Neutrosophic Interval Statistical Numbers Based on Neutrosophic StatisticsJournal of Soft Computing in Civil Engineering, 2-4 / 2018, 62-71; DOI: 10.5281/zenodo.3130240.
26. Muhammad Aslam, P. Jeyadurga, Saminathan Balamurali, Ali Hussein Al-MarshadiTime-Truncated Group Plan under aWeibull Distribution based on Neutrosophic StatisticsMathematics 2019, 7, 905; DOI: 10.3390/math7100905
27. A.A. Salama, M. Elsayed Wahed, Eman YousifA Multi-objective Transportation Data Problems and their Based on Fuzzy Random VariablesNeutrosophic Knowledge, vol. 1, 2020, 41-53; DOI: 10.5281/zenodo.4269558.
28. Philippe SchweizerUncertainty: two probabilities for the three states of neutrosophy. International Journal of Neutrosophic Science (IJNS), Volume 2, Issue 1, 2020, 18-26; DOI: 10.5281/zenodo.3989350.
29. Carlos N. Bouza-Herrera, Mir SubzarEstimating the Ratio of a Crisp Variable and a Neutrosophic VariableInternational Journal of Neutrosophic Science (IJNS), Volume 11, Issue 1, 2020, 9-21; DOI: 10.5281/zenodo.4275712
30. Angel Carlos Yumar Carralero, Darvin Manuel Ramirez Guerra, Giorver Perez IribarAnalisis estadistico neutrosofico en la aplicacion de ejercicios fisicos en la rehabilitacion del adulto mayor con gonartrosisNeutrosophic Computing and Machine Learning, Vol. 13, 1-9, 2020; DOI: https://zenodo.org/record/3901770.
31. Alexandra Dolores Molina Manzo, Rosa Leonor Maldonado Manzano, Blanca Esmeralda Brito Herrera, Johanna Irene Escobar JaraAnalisis estadistico neutrosofico de la incidencia del voto facultativo de los jovenes entre 16 y 18 anos en el proceso electoral del EcuadorNeutrosophic Computing and Machine Learning, Vol. 11, 9-14, 2020; DOI: https://zenodo.org/record/3474439.
32. Johana Cristina Sierra Morán, Jenny Fernanda Enríquez Chuga, Wilmer Medardo Arias Collaguazo And Carlos Wilman Maldonado Gudiño:Neutrosophic statistics applied to the analysis of socially responsible participation in the community , Neutrosophic Sets and Systems, vol. 26, 2019, pp. 19 -28. DOI: 10.5281/zenodo.3244232
33. Paúl Alejandro Centeno Maldonado, Yusmany Puertas Martinez, Gabriela Stephanie Escobar Valverde, and Juan Danilo Inca Erazo: Neutrosophic statistics methods applied to demonstrate the extra-contractual liability of the state from the Administrative Organic Code, Neutrosophic Sets and Systems, vol. 26, 2019, pp. 29-34. DOI: 10.5281/zenodo.3244262
35. Lilia Esther Valencia Cruzaty, Mariela Reyes Tomalá, Carlos Manuel Castillo Gallo and Florentin Smarandache, Neutrosophic Sets and Systems, vol. 34, 2020, pp. 33-39. DOI: 10.5281/zenodo.3843289; http://fs.unm.edu/NSS/NeutrosophicStatisticMethod.pdf
Journal of Fuzzy Extension & Applications (JFEA), Volume 2, Issue 1, Winter 2021, 33-40; DOI: 10.22105/JFEA.2021.272508.1073.
37. Muhammad Aslam, Rashad A.R. Bantan, Nasrullah Khan: Design of tests for mean and variance under complexity-an application to rock measurement data. Elsevier: Measurement, Volume 177, June 2021, 109312; DOI: 10.1016/j.measurement.2021.109312.
38. O.H. Arif, Muhammad Aslam: Springer: Complex & Intelligent Systems (2021); DOI: .
39. Nasrullah Khan, Muhammad Aslam, Asma Arshad, Ambreen Shafqat: Springer: Journal of Metrology Society of India (2021); DOI: 10.1007/s12647-021-00436-2.
40. Muhammad Aslam: Springer: Soft Computing (2021); DOI: 10.1007/s00500-021-05661-0.
41. Muhammad Aslam: Springer: Journal of Metrology Society of India (2021); DOI: 10.1007/s12647-021-00429-1.
42. Muhammad Aslam, Nasrullah Khan: Springer: Journal of Metrology Society of India (2021); DOI: 10.1007/s12647-020-00428-8.
43. Muhammad Aslam, Gadde Srinivasa Rao, Nasrullah Khan, Liaquat Ahmad: Taylor&Francis: Communications in Statistics - Theory and Methods (2020); DOI: 10.1080/03610918.2019.1702212.
44. Ali Hussein Al-Marshadi, Ambreen Shafqat, Muhammad Aslam, Abdullah Alharbey: Performance of a New Time-Truncated Control Chart for Weibull Distribution Under Uncertainty. Atlantis Press: International Journal of Computational Intelligence Systems, Volume 14, Issue 1, 2021, 1256 - 1262; DOI: .
45. Muhammad Aslam: Testing average wind speed using sampling plan for Weibull distribution under indeterminacy. Nature: Scientific Reports, 11, (2021); DOI: 10.1038/s41598-021-87136-8.
46. Muhammad Aslam, G. Srinivasa Rao, Nasrullah Khan:. Springer: Complex & Intelligent Systems, 7, 891–900 (2021); DOI: .
47. Muhammad Aslam, G. Srinivasa Rao, Ambreen Shafqat, Liaquat Ahmad, Rehan Ahmad Khan Sherwani: Monitoring circuit boards products in the presence of indeterminacy. Elsevier: Measurement, Volume 168, 15 January 2021, 108404; DOI: 10.1016/j.measurement.2020.108404.
48. Mohammed Albassam, Nasrullah Khan, Muhammad Aslam: Neutrosophic D’Agostino Test of Normality: An Application to Water Data. Hindawi: Journal of Mathematics - Theory, Algorithms, and Applications within Neutrosophic Modelling and Optimisation, 2021, , 5 pages; DOI: 10.1155/2021/5582102.
49. Mohammed Albassam: Radar data analysis in the presence of uncertainty. Taylor&Francis: European Journal of Remote Sensing, 54:1, 140-144, 2021; DOI: .
50. Muhammad Aslam: Springer: Complex & Intelligent Systems, 7, 359–365, 2021; DOI: .
51. Abdullah M. Almarashi, Muhammad Aslam: Process Monitoring for Gamma Distributed Product under Neutrosophic Statistics Using Resampling Scheme. Hindawi: Journal of Mathematics: Soft Computing Algorithms Based on Fuzzy Extensions, Volume 2021, , 12 pages; DOI: 10.1155/2021/6635846.
52. Muhammad Aslam: Springer: Theoretical and Applied Climatology, 143, 1227–1234, 2021; DOI: 10.1007/s00704-020-03509-5.
53. Muhammad Aslam, Ali Algarni: Hindawi: International Journal of Photoenergy, Volume 2020, , 6 pages; DOI: 10.1155/2020/6662389.
54. Muhammad Aslam: Nature: Sc. Rep., Volume 10 (2020).
55. Azhar Ali Janjua, Muhammad Aslam, Naheed Sultana: Springer: Theoretical and Applied Climatology, Volume 142, pages 1641–1648 (2020); DOI: 10.1007/s00704-020-03398-8.
56. Rehan Ahmad Khan Sherwan, Mishal Naeem, Muhammad Aslam, Muhammad Ali Raza, Muhammad Abid, Shumaila Abbas: University of New Mexico: Neutrosophic Sets and Systems, Vol. 41, 209-214, 2021; DOI: .
57. Muhammad Aslam, Ambreen Shafqat, Mohammed Albassam, Jean-Claude Malela-Majika, Sandile C. Shongwe: PLoS ONE 16(2): e0246185, 2021; DOI: .
58. Muhammad Aslam: Taylor&Francis: International Journal of Injury Control and Safety Promotion, Volume 28, 2021 - Issue 1, 39-45; DOI: 10.1080/17457300.2020.1835990.
59. Muhammad Aslam: Taylor&Francis: International Journal of Cast Metals Research, Volume 34, 2021 - Issue 1, 1-5; DOI: 10.1080/13640461.2020.1846959.
60. Ishmal Shahzadi, Muhammad Aslam, Hussain Aslam: University of New Mexico: Neutrosophic Sets and Systems, Vol. 39, 101-106, 2020.
61. Nasrullah Khan, Muhammad Aslam, P. Jeyadurga, S. Balamurali: Nature: Sc. Rep., volume 11 (2021).
62. Muhammad Aslam, Rashad A.R. Bantan: Elsevier: Measurement, Volume 166, December 2020, 108201; DOI: 10.1016/j.measurement.2020.108201.
63. Muhammad Aslam, Rashad A. R. Bantan, Nasrullah Khan: Springer: Soft Computing, Volume 24, 16617–16626 (2020); DOI: 10.1007/s00500-020-04964-y.
64. M. Albassam, Muhammad Aslam: IEEE Access, vol. 8, pp. 172379-172386, 2020; DOI: .
65. Ahmed Ibrahim Shawky , Muhammad Aslam, Khushnoor Khan: Hindawi: Journal of Mathematics, Volume 2020, , 14 pages; DOI: 10.1155/2020/7680286.
66. Muhammad Aslam: Science Direct: Journal of King Saud University - Science, Volume 32, Issue 6, September 2020, 2696-2700; DOI: .
67. Muhammad Aslam: MDPI: Symmetry, 2018, 10 (5), 132; DOI: .
68. Muhammad Aslam, Osama H. Arif: MDPI: Symmetry, 2018, 10 (9), 403; DOI: .
69. Muhammad Aslam, Nasrullah Khan, Muhammad Zahir Khan: MDPI: Symmetry, 2018, 10 (11), 562; DOI: .
70. Muhammad Zahir Khan, Muhammad Farid Khan, Muhammad Aslam, Abdur Razzaque Mughal: MDPI: Mathematics, 2019, 7 (1), 9; DOI: .
71. Muhammad Aslam, Ali Hussein Al-Marshadi: MDPI: Symmetry, 2018, 10 (12), 754; DOI: .
72. Muhammad Zahir Khan, Muhammad Farid Khan, Muhammad Aslam, Seyed Taghi Akhavan Niaki, Abdur Razzaque Mughal: MDPI: Information, 2018, 9 (12), 312; DOI: .
73. Muhammad Aslam, Nasrullah Khan, Mohammed Albassam: MDPI: Symmetry, 2018, 10 (12), 690; DOI: .
74. Muhammad Aslam, Mohammed Albassam: MDPI: Symmetry, 2019, 11 (3), 330; DOI: .
75. Muhammad Aslam, Mansour Sattam Aldosari: MDPI: Symmetry, 2019, 11 (2), 193; DOI: .
76. Muhammad Aslam: MDPI: Symmetry, 2019, 11 (1), 114; DOI: .
77. Muhammad Aslam, Nasrullah Khan, Ali Hussein Al-Marshadi: MDPI: Symmetry, 2019, 11 (1), 80; DOI: .
78. Muhammad Aslam, Rashad A. R. Bantan, Nasrullah Khan: MDPI: Processes, 2019, 7 (10), 742; DOI: .
79. Muhammad Aslam, Ali Hussein Al-Marshadi, Nasrullah Khan: MDPI: Mathematics, 2019, 7 (10), 957; DOI: .
80. Muhammad Aslam, P. Jeyadurga, Saminathan Balamurali, Ali Hussein Al-Marshadi: MDPI: Mathematics, 2019, 7 (10), 905; DOI: .
81. Muhammad Aslam, Osama Hasan Arif: MDPI: Mathematics, 2019, 7 (9), 870; DOI: .
82. Muhammad Aslam, Mohammed Albassam: MDPI: Mathematics, 2019, 7 (7), 631; DOI: .
83. Muhammad Aslam, Rashad A. R. Bantan, Nasrullah Khan: MDPI: Processes, 2019, 7 (4), 209; DOI: .
84. Muhammad Aslam: Hindawi: Advances in Fuzzy Systems, Volume 2019, , 8 pages; DOI: 10.1155/2019/8953051.
85. Muhammad Aslam, Rashad A. R. Bantan, Nasrullah Khan: Springer: International Journal of Fuzzy Systems, Volume 21, 433–440 (2019); DOI: 10.1007/s40815-018-0577-1.
86. Muhammad Aslam, Osama H. Arif: Hindawi: Complexity, Volume 2020, , 6 pages; DOI: 10.1155/2020/2935435.
87. Mohammed Albassam, Nasrullah Khan,Muhammad Aslam: Hindawi: Complexity, Volume 2020, , 8 pages; DOI: 10.1155/2020/3690879.
88. Muhammad Aslam, Osama H. Arif, Rehan Ahmad Khan Sherwani: Hindawi: BioMed Research International, Volume 2020, , 7 pages; DOI: 10.1155/2020/2086185.
88. Muhammad Aslam, Ali Hussein Al-Marshadi: Hindawi: Complexity, Volume 2019, , 7 pages; DOI: 10.1155/2019/8178067.
89. Muhammad Aslam, Osama H. Arif: Hindawi: Journal of Analytical Methods in Chemistry, Volume 2020, Article ID 1406028, 6 pages; DOI: .
90. Muhammad Aslam, Abdulmohsen Al-Shareef, Khushnoor Khan: Nature: Sc. Rep., Volume 10, Article number: 12182 (2020).
91. Muhammad Aslam: IEEE Access, vol. 6, pp. 64153-64158, 2018; DOI: .
92. Muhammad Aslam: IEEE Access, vol. 7, pp. 25253-25262, 2019; DOI: .
93. Muhammad Aslam, M. Azam, M. Albassam: IEEE Access, vol. 7, pp. 38568-38576, 2019; DOI: .
94. Naeem Jan, Muhammad Aslam, Kifayat Ullah, Tahir Mahmood, Jun Wang: An approach towards decision making and shortest path problems using the concepts of interval-valued Pythagorean fuzzy information. Wiley: International Journal of Intelligent Systems, Volume 34, Issue 10, October 2019, 2403-2428.
95. Muhammad Aslam: IEEE Access, vol. 7, 2019, 2163-3536; DOI: .
96. Muhammad Aslam, R. A. R. Bantan, N. Khan: IEEE Access, vol. 7, pp. 8858-8864, 2019; DOI: .
97. Muhammad Aslam, Muhammad Ali Raza: Springer: International Journal of Fuzzy Systems, volume 21, 978–992 (2019); DOI: 10.1007/s40815-018-0560-x.
98. Muhammad Aslam: Springer: International Journal of Fuzzy Systems, volume 21, 1214–1220 (2019); DOI: 10.1007/s40815-018-0588-y.
99. Muhammad Aslam: Springer: Complex & Intelligent Systems, volume 5, 403–407 (2019); DOI: .
100. Muhammad Aslam, Mohammed Albassam: Elsevier: Journal of King Saud University - Science, Volume 32, Issue 6, September 2020, 2728-2732; DOI: .
101. Muhammad Aslam, Mansour Sattam Aldosari: Elsevier: Journal of King Saud University - Science, Volume 32, Issue 6, September 2020, 2831-2834; DOI: .
102. Muhammad Aslam: Elsevier: Journal of King Saud University - Science, Volume 32, Issue 3, April 2020, 2005-2008; DOI: .
103. Muhammad Aslam: Springer: Complex & Intelligent Systems, 5, 365–370 (2019); DOI: .
104. Muhammad Aslam, Saminathan Balamurali, Jeyadurga Periyasamypandian, Ali Hussein Al-Marshadi: IEEE Access, vol. 7, 164186-164193, 2019; DOI: .
105. Muhammad Aslam, R. A. R. Bantan, N. Khan: IEEE Access, vol. 7, pp. 152233-152242, 2019; DOI: .
106. Muhammad Aslam: American Chemical Society: ACS Omega 2020, 5, 1, 914-917; DOI: .
107. Muhammad Aslam: Taylor&Francis: Journal of Taibah University for Science, Volume 14, 2020, Issue 1; DOI: .
108. Muhammad Aslam, Muhammad Ali Raza, Liaquat Ahmad: IOS Press: Journal of Intelligent & Fuzzy Systems, vol. 37, no. 6, pp. 7839-7850, 2019; DOI: 10.3233/JIFS-182849.
109. Muhammad Kashif, Hafiza Nida, Muhammad Imran Khan, Muhammad Aslam: University of New Mexico: Neutrosophic Sets and Systems, vol. 30, 143-148, 2019.
110. Muhammad Aslam, Nasrullah Khan: University of New Mexico: Journal of Intelligent & Fuzzy Systems, vol. 36, no. 3, pp. 2615-2623, 2019; DOI: 10.3233/JIFS-181767.
111. N Khan, L Ahmad, M Azam, M Aslam, F Smarandache, Control Chart for Monitoring Variation Using Multiple Dependent State Sampling Under Neutrosophic Statistics, in the book Neutrosophic Operational Research (eds. F. Smarandache, M. Abdel-Basset), Springer, pp 55-70, 10 September 2021, https://link.springer.com/chapter/10.1007/978-3-030-57197-9_4.
112. Rehan Ahmad Khan Sherwani, Muhammad Aslam, Muhammad Ali, Raza Muhammad, Farooq Muhammad, Abid Muhammad Tahir, Neutrosophic Normal Probability Distribution—A Spine of Parametric Neutrosophic Statistical Tests: Properties and Applications, in the book Neutrosophic Operational Research (eds. F. Smarandache, M. Abdel-Basset), Springer, pp 153-169, 10 September 2021, https://link.springer.com/chapter/10.1007/978-3-030-57197-9_8.
113. Rehan Ahmad Khan Sherwani, Muhammad Aslam, Huma Shakeel, Kamran Abbas, Farrukh Jamal, Neutrosophic Statistics for Grouped Data: Theory and Applications, in the book Neutrosophic Operational Research (eds. F. Smarandache, M. Abdel-Basset), Springer, pp 263-289, 10 September 2021, https://link.springer.com/chapter/10.1007/978-3-030-57197-9_14.
114. Arif, O.H., Aslam, M. A new sudden death chart for the Weibull distribution under complexity. Complex Intell. Syst. 7, 2093–2101 (2021), Springer, https://doi.org/10.1007/s40747-021-00316-x, https://link.springer.com/article/10.1007/s40747-021-00316-x
115. Wen-Qi Duan, Zahid Khan, Muhammad Gulistan, Adnan KhurshidNeutrosophic Exponential Distribution: Modeling and Applications for Complex Data AnalysisComplexity, vol. 2021, Article ID 5970613, 8 pages, 2021. https://doi.org/10.1155/2021/5970613
116. Nasrullah Khan, Liaquat Ahmad, G. Srinivasa Rao, Muhammad Aslam, Ali Hussein AL Marshadi, A New X bar Control Chart for Multiple Dependent State Sampling Using Neutrosophic Exponentially Weighted Moving Average Statistics with Application to Monitoring Road Accidents and Road Injuries, International Journal of Computational Intelligence Systems, Springer, (2021) 14:182 https://doi.org/10.1007/s44196-021-00033-w, 30 September 2021, https://link.springer.com/content/pdf/10.1007/s44196-021-00033-w.pdf
117. F. Smarandache, Neutrosophic Statistics vs. Interval Statistics, and Plithogenic Statistics as the most general form of statistics (second edition), International Journal of Neutrosophic Science (IJNS), Vol. 19, No. 01, PP. 148-165, 2022, http://fs.unm.edu/NS/NeutrosophicStatistics-vs-IntervalStatistics.pdf
Project
Pr1. F. Smarandache, Neutrosophic Statistics is a generalization of Classical and Interval Statistics, research project, ResearchGate (Germany), https://www.researchgate.net/project/Neutrosophic-Statistics-is-a-generalization-of-Classical-and-Interval-Statistics
Seminars
S1. History of Neutrosophic Set, Logic, Probability and Statistics and their Applications, Mathematics and Statistics Departments, King Abdulaziz University, Jeddah, Saudi Arabia, 19 December 2019.
S2. Neutrosophic Set and Logic / Interval Neutrosophic Set and Logic / Neutrosophic Probability and Neutrosophic Statistics / Neutrosophic Precalculus and Calculus / Symbolic Neutrosophic Theory / Open Challenges of Neutrosophic Set, lecture series, by F. Smarandache, Nguyen Tat Thanh University, Ho Chi Minh City, Vietnam, 31st May - 3th June 2016.
S3. Neutrosophic Set and Logic / Interval Neutrosophic Set and Logic / Neutrosophic Probability and Neutrosophic Statistics / Neutrosophic Precalculus and Calculus / Symbolic Neutrosophic Theory / Open Challenges of Neutrosophic Setby F. Smarandache, Ho Chi Minh City University of Technology (HUTECH), Ho Chi Minh City, Vietnam, 30th May 2016.
S4. Neutrosophic Set and Logic / Interval Neutrosophic Set and Logic / Neutrosophic Probability and Neutrosophic Statistics / Neutrosophic Precalculus and Calculus / Symbolic Neutrosophic Theory / Open Challenges of Neutrosophic Set, lecture series, by F. Smarandache, Vietnam national University, Vietnam Institute for Advanced Study in Mathematics, Hanoi, Vietnam, lecture series, 14th May – 26th May 2016.
S5. Foundations of Neutrosophic Logic, Set, Probability and Statistics and their Applications in Science. n-Valued Refined Neutrosophic Set, Logic, Probability and Statistics, by F. Smarandache, Universidad Complutense de Madrid, Facultad de Ciencia Matematicas, Departamento de Geometria y Topologia, Instituto Matematico Interdisciplinar (IMI), Madrid, Spain, 9th July 2014.
MATHEMATICS Algebra Geometries Multispace Neutrosophic Environment Number Theory Statistics Plithogenic Set / Logic / Probability / Statistics MATHEMATICS Algebra Geometries Multispace Neutrosophic Environment Number Theory Statistics Neutrosophy, a new branch of philosophy PHYSICS Absolute Theory of Relativity Quantum Paradoxes Unmatter Neutrosophic Physics Superluminal and Instantaneous Physics BIOLOGY Neutrosophic Theory of Evolution Syndrome ECONOMICS Poly-Emporium Theory LINGUISTICS Linguistic Paradoxes Linguistic Tautologies PSYCHOLOGY Neutropsychic Personality Illusion Law on Sensations and Stimuli Synonymity Test Complex SOCIOLOGY Social Paradox Sociological Theory LITERATURE pArAdOXisM oUTER-aRT Theatre Literatura Romana Bancuri Science Library Literature Library GlobeTrekker | Videos Biography
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https://writerbayessays.com/02/solvedto-make-concrete-rita-dumped-two-bags-of-gravel-each-weighing-3-3-4-kg-into-a-wheelbarrow-then-she-added-a-1-7-8-kg-bag-of-cement-what-percent-of-the-weight-of-the-mixture-is-the-weight-of/
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# (solved)To make concrete, Rita dumped two bags of gravel each weighing 3 3/4 kg into a wheelbarrow. Then she added a 1 7/8 kg bag of cement. What percent of the weight of the mixture is the weight of the grav. . . . .
To make concrete, Rita dumped two bags of gravel each weighing 3 3/4 kg into a wheelbarrow. Then she added a 1 7/8 kg bag of cement. What percent of the weight of the mixture is the weight of the gravel?
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https://gamedev.stackexchange.com/questions/824/have-any-video-game-designs-used-non-uniform-random-numbers-in-interesting-ways
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# Have any video game designs used non-uniform random numbers in interesting ways?
A variety of video games use uniformly distributed numbers to decide the outcome of an event, such as a "50% chance to hit" almost always means to check if a random floating point number from 0-1 is greater than 0.5. Many games will layer a few of these uniform percentages on top of each other, for instance a D&D hit roll is a uniformly distributed number from 1-20, except that 1 and 20 have special outcomes. To my mind it seems like things like critical hits are added by designers to try and emulate the fact that in reality hitting/missing or winning/losing is not actually a binary outcome.
In many cases the real life amount of "damage" done by an attack would likely be closer to a gaussian/bell curve distribution, which many results in the middle but the occasional very exciting outlier and smooth curve connecting them. Dice games like Settlers of Catan emulate gaussian distributions by adding together multiple independent rolls, but I feel like I've almost never seen this mechanic in video games.
It seems like games like Civilization (Sid Meier talked extensively at GDC about player perception not matching the actual math used in the game) would benefit from results that matched how things work in the real world. Have any video games used a gaussian or otherwise non-uniform distribution of random numbers in interesting ways?
• lol @ Civ example. What an archer beating a Tank isn't fun? Haha! I suppose it depends if you are the one with the archer. :-) Jul 20, 2010 at 21:35
• I can't find the link right now, but the gist was that players were very upset when a "70% chance to win" as shown in the UI lead to occasionally losing. He blamed the fact that humans are bad at doing probability but I blame the fact that a "70% chance to win" doesn't even make sense in a world of analogue outcomes to battles. Jul 20, 2010 at 21:58
• Actually I think the problem was that the probability was calculated wrong, as it was only a heuristic prediction. You'd get 90% predicted when your chance was 98% or something to that effect. By the way, DnD also uses non uniform numbers- damage and HP are multiple dice rolls. Feb 11, 2012 at 13:35
Shooters often use Gaussian random distribution for weapon accuracy. (If you use linear random numbers and you have bullet decals, it's very easy for a player to see that the accuracy distribution is square, which "feels wrong.")
One interesting random selection method that you don't mention, but which shows up fairly often in games, is "random without replacement." This is analogous to drawing cards from a deck; the game runs in random order through a set of possible outcomes (put together with the desired distribution) and then "reshuffles." This is done to reduce the chance of lucky or unlucky streaks.
• The concept in your second paragraph is also known as a "Shuffle Bag". Jul 20, 2010 at 22:01
• It's also common for shooters to randomly choose a point on a circle (ie an angle and radius) for bullet hits. This eliminates the square as well and ensures that more bullets hit near the center than not. Jul 21, 2010 at 11:42
I use Poisson distributions quite a bit - mainly when trying to work out the number of times that a random event should occur within a period of time.
This has the nice property that the distribution of events if you have two 1 second timesteps is the same as one 2 second timestep, so great for simulations with variable-length time steps.
Something that I have done in the past and that World of Warcraft currently does is to increase the chance of a random event based upon the number of tries since the last successful event. That is for instance, a quest item may have a 20% drop rate on the first NPC kill, a 22% on the 2nd, a 25% on the 3rd, resetting back to 20% when it does drop. I implemented a tech tree where the chance to discover a tech increased each turn until it was on the order of a 99.9% chance of occurring.
• I think this probability boosting is commonly called a pity timer. Mar 4 at 23:20
Random encounters in RPG-s are often non-uniform.
Set X to a random integer between 64 and 255.
For each step in plains, decrement X by 4.
For each step in forest, swamp, or desert, decrement X by 8.
When X < 0, a fight ensues. Go to step 1.
This makes encounters more realistic because animals/trees etc. like to have some space between them.
• I never really though about that, but that's a pretty slick approach to a lot of count-down randomizations. Nov 9, 2015 at 13:55
I've seen critical hits in some action-RPGs before, but usually for action-based games the numbers are kept intentionally simple for a reason: the player is busy running and jumping and dodging and shooting, and they really don't have time to do probability calculations in their head.
So, you're more likely to see non-uniform numbers in things like turn-based strategy games. A typical example are "rogue-like" games (Nethack, Angband) which have non-uniform weapon damage -- one weapon might do 3d5 damage for example, and another does 4d4, and the game will tell you these numbers and it's up to you to decide which one is better (factoring in other variables like weapon weight, character proficiency, etc.)
• even warcraft 3 uses dice-based damage, most of the damages in the scenario editor use the formula "X+YdZ"
– Henk
Jul 22, 2010 at 0:59
League of Legends has a hero named Gangplank with an attack that used to have pretty much uniform distribution which they changed to a gaussian. There is a circle area of effect where multiple hits may occur and they decided to make it cluster towards the center.
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https://www.convertunits.com/from/millimetre/to/arms-length
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## ››Convert millimetre to arms-length
millimetre arms-length
How many millimetre in 1 arms-length? The answer is 700.
We assume you are converting between millimetre and arms-length.
You can view more details on each measurement unit:
millimetre or arms-length
The SI base unit for length is the metre.
1 metre is equal to 1000 millimetre, or 1.4285714285714 arms-length.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between millimetres and arms-length.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of millimetre to arms-length
1 millimetre to arms-length = 0.00143 arms-length
10 millimetre to arms-length = 0.01429 arms-length
50 millimetre to arms-length = 0.07143 arms-length
100 millimetre to arms-length = 0.14286 arms-length
200 millimetre to arms-length = 0.28571 arms-length
500 millimetre to arms-length = 0.71429 arms-length
1000 millimetre to arms-length = 1.42857 arms-length
## ››Want other units?
You can do the reverse unit conversion from arms-length to millimetre, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Millimeter
A millimetre (American spelling: millimeter, symbol mm) is one thousandth of a metre, which is the International System of Units (SI) base unit of length. The millimetre is part of a metric system. A corresponding unit of area is the square millimetre and a corresponding unit of volume is the cubic millimetre.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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