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https://www.electricalonline4u.com/2016/02/solar-panels-in-parallel.html | 1,659,995,027,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.1/warc/CC-MAIN-20220808213349-20220809003349-00226.warc.gz | 667,442,943 | 40,469 | # Wiring Solar Panels In Parallel & Solar Parallel Calculation
Nowadays the use of solar panels are getting very high number and people get a huge advantages form solar panels. In my last article i discus the series connection of panels, and this post you learn how to connect solar panels in parallel and parallel calculation.
## How To Connect Solar Panels In Parallel ?
The wiring of solar panels in parallel connection is too easy and we use in many places for different requirements. The calculation of solar panels Voltage and Ampere is opposite from series connection, In series solar cell we plus the voltage and Ampere will be same but in connecting solar panels in parallel connection our voltage V will be same and the current Ampere will plus of all solar panels.
E.g example " if we connect 12 volts 8 amps 4 solar in parallel then our complete output voltage will 12 volts 32 amps (amperes).
Now lets how to do wiring connection of solar in parallel with on another. For Doing the connection follow below steps.
• First of all connect all solar panels negative connection with one point.
• Same connect all panels Positive terminals with one point.
• Now you have one negative - and one Positive + wire.
• Get supply form these to wires.
I know that a diagram is very important for completely understanding, SO lets a look to the diagram.
Same follow the method of doing connection as i shown in above solar panels wiring diagram. I shown complete calculation of all 4 solar.
And INS SHA ALLAH after this diagram & post you will easily wire solar panels in parallel connection.
### Sikandar Haidar
Electricalonline4u.com is a hub of electrical and electronics tutorials and easy diagrams. | 350 | 1,712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-33 | latest | en | 0.853553 |
https://www.teachoo.com/teach-others/7529/Let-set-A-%3D--x-is-a-multiple-of-3---B-%3D--x-is-a-multiple-of-5---then/ | 1,674,806,075,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00318.warc.gz | 1,027,370,902 | 29,220 | ## Let set A = {x: x is a multiple of 3}, set B = {x: x is a multiple of 5}. Then A ∩ B is given by
1. {3, 6, 9, ...}
2. {5, 10, 15, 20, ...}
3. {15, 30, 45, ....}
4. None of the above
For set A
Multiples of 3 are those numbers which come in table of 3
So, Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, ....
Set A = {3, 6, 9, 12, 15, 18, 21, ....}
For set B
Multiples of 5 are those numbers which come in table of 5
So, Multiples of 5 = 5, 10, 15, 20, 25, 30 ....
Set B = {5, 10, 15, 20, 25, 30 ....}
Now,
A ∩ B = {15, 30, ... }
So, C is the correct answer. | 257 | 565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-06 | latest | en | 0.91188 |
https://www.clutchprep.com/chemistry/practice-problems/69433/the-ka-of-propanoic-acid-c2h5cooh-is-1-34-x-10-5-calculate-the-ph-of-the-solutio | 1,618,383,207,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077336.28/warc/CC-MAIN-20210414064832-20210414094832-00457.warc.gz | 795,212,076 | 33,684 | # Problem: The Ka of propanoic acid (C2H5COOH) is 1.34 × 10-5. Calculate the pH of the solution and the concentrations of C2H5COOH and C2H5COO– in a 0.643 M propanoic acid solution at equilibrium.pH =[C2H5COOH] =[C2H5COO-] =
80% (81 ratings)
###### Problem Details
The Ka of propanoic acid (C2H5COOH) is 1.34 × 10-5. Calculate the pH of the solution and the concentrations of C2H5COOH and C2H5COO in a 0.643 M propanoic acid solution at equilibrium.
pH =
[C2H5COOH] =
[C2H5COO-] =
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Weak Acids concept. If you need more Weak Acids practice, you can also practice Weak Acids practice problems.
What is the difficulty of this problem?
Our tutors rated the difficulty ofThe Ka of propanoic acid (C2H5COOH) is 1.34 × 10-5. Calculat...as medium difficulty.
How long does this problem take to solve?
Our expert Chemistry tutor, Dasha took 6 minutes and 26 seconds to solve this problem. You can follow their steps in the video explanation above.
What professor is this problem relevant for?
Based on our data, we think this problem is relevant for Professor Kobko-Litskevitch's class at HUNTER. | 363 | 1,255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-17 | latest | en | 0.913673 |
https://justaaa.com/statistics-and-probability/629742-because-many-passengers-who-make-reservations-do | 1,702,223,893,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102612.80/warc/CC-MAIN-20231210155147-20231210185147-00004.warc.gz | 369,720,300 | 9,451 | Question
# Because many passengers who make reservations do not show up, airlines often overbook flights (sell more...
Because many passengers who make reservations do not show up, airlines often overbook flights (sell more tickets than there are seats). A certain airplane holds 296 passengers. If the airline believes the rate of passenger no-shows is 5% and sells 308 tickets, is it likely they will not have enough seats and someone will get bumped?
Bold a right parenthesis font size decreased by 1
Use the normal model to approximate the binomial to determine the probability of at least 297 passengers showing up.
Bold b right parenthesis font size decreased by 1
Should the airline change the number of tickets they sell for this flight? Explain.
Bold a right parenthesis font size decreased by 1
The probability of at least 297 passengers showing up is
nothing.
(Round to three decimal places as needed.)
b) Should the airline change the number of tickets they sell for this flight? Explain.
A.
The proportion is fairly high, so it is likely that they should sell less. However, the decision also depends on the relative costs of not selling seats and bumping passengers.
B.
Since the proportion is so low, they should change the number of tickets they sell.
C.
Since the proportion is so high, they should not change the number of tickets they sell.
D.
The proportion is fairly low, so it is likely that they should not change the number of tickets they sell. However, the decision also depends on the relative costs of not selling seats and bumping passengers.
This is an example of binomial distribution
the airline believes the probability of passenger no-shows is 0.05
airplane capacity is 296. If more than 296people show someone will get bumped.
the airline sold 308 tickets
The probability that someone will be bumped( if 11 or fewer people do not show up) is ( use excel formula BINOM.DIST(11,308,0.05,TRUE))
The probability of at least 297 passengers showing up is 0.1530
b) the probability is greater than 15%. Hence it likely they will not have enough seats and someone will get bumped.
option A is right
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 495 | 2,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2023-50 | latest | en | 0.936216 |
https://calconcalculator.com/construction/square-yards-calculator/ | 1,702,219,391,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102469.83/warc/CC-MAIN-20231210123756-20231210153756-00693.warc.gz | 173,833,406 | 19,840 | If you came here, you’re probably not too sure what square yards are, or how you can measure them. Fear not, for we have the guide you need. In the following text, you’re going to learn what square yards are, and how you can measure them for different shapes of area.
Take a look other related calculators, such as:
## What are square yards?
In the construction industry, square yards are used to measure the amount of area on a given structure. This measurement is usually used with regards to buildings and rooms, but can also be used to measure yards as well.
These units may be used in various applications including residential and commercial building plans. The number of square yards needed for any given application depends upon its intended purpose, size and shape – all things that must be determined before calculating surface area using this method.
## How to calculate square yards?
There are several ways to calculate square yards. The method you are going to use depends on the shape of the area you’re measuring. So, let’s go over the most common shapes.
## Rectangle
A rectangle is a quadrilateral with four sides and four angles of equal measure. A rectangle has two pairs of parallel and opposite sides. The length of a rectangle is the distance between opposite sides, and its width is the distance between opposite sides. For a rectangle, the surface area is equal to:
A = a \times b
A square is also a rectangle with all four sides having equal length, but unlike rectangles, squares have right angles at every corner (where four lines intersect). Rectangles are most easily constructed by first constructing parallel lines that will be used as boundaries for both length and width measurements on each side. For a square, the surface area is equal to:
A = a^2
## Circle
A circle is a shape that is round. It’s also called an “annulus” or an if it has a small hole in the middle. A circle has no end and no beginning, like infinity. If you travel around a circle, you’ll always get back to where you started from!
The formula for calculating the surface area of a circle:
A = r^2 \times \pi
## Triangles
A triangle is a polygon with three sides, three angles and three vertices. To calculate its surface area, you can use the formula:
A = b \times h
b is the length of the base, and h is the height of the triangle.
## How to use the square yards calculator
With this tool, all you need to do is enter the length and width of the area, and the calculator will give you the surface area.
Furthermore, it can also give you the full price, if you enter the price per square yard.
## FAQ
### Is a yard and a square yard the same?
A yard is a unit of length, while a square yards is a unit of surface area.
### How many square feet are in a square yard?
One square yard contains 9 square feet.
### How many square yards are in a football field?
A football field is 6400 square yards. | 632 | 2,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2023-50 | longest | en | 0.962221 |
https://blog.finxter.com/5-best-ways-to-count-the-number-of-items-in-a-python-list/ | 1,723,434,556,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641028735.71/warc/CC-MAIN-20240812030550-20240812060550-00469.warc.gz | 109,238,915 | 17,780 | 5 Best Ways to Count the Number of Items in a Python List
π‘ Problem Formulation: When working with lists in Python, a common task is to determine the number of items they contain. For instance, given the list `[1, 2, 3, 4, 5]`, we want an output of `5` to indicate the list contains five items. This task underpins operations of counting, aggregation, and data manipulation, making it a fundamental skill for python developers.
Method 1: Using the len() Function
One of the most straightforward ways to count the number of items in a list is to use Python’s built-in `len()` function. This function returns the number of items in an object. When the object is a list, `len()` provides the total count of elements in the list.
Here’s an example:
```my_list = [1, 2, 3, 4, 5]
print(len(my_list))```
Output: `5`
This example showcases the simplicity of the `len()` function β a single line of code returns the count of items in the list `my_list`. The key advantage is its directness and constant time complexity, which is ideal for any list size.
Method 2: Using a For Loop to Manually Count
For scenarios where you may want to customize the count operation, such as counting only specific items in a list, you might use a for loop to iterate through the list and count items based on a condition.
Here’s an example:
```my_list = ['apple', 'banana', 'cherry', 'apple']
count = 0
for item in my_list:
if item == 'apple':
count += 1
print(count)```
Output: `2`
The example counts how many times ‘apple’ appears in the list by manually incrementing the variable `count`. While this method provides more control, it is less efficient than `len()` for simply counting all items.
Method 3: Using the collections.Counter Class
If you need to count how many times each value appears in the list, the `collections.Counter` class is a specialized dictionary for this purpose. It can be particularly useful when working with large datasets.
Here’s an example:
```from collections import Counter
my_list = ['apple', 'banana', 'cherry', 'apple']
count = Counter(my_list)
print(count['apple'])```
Output: `2`
The `Counter` class from the `collections` module provides a tally of all items in the list. It’s a powerful method for frequency counting but may be overkill for simple total counts.
Method 4: Using the sum() Function with a Generator Expression
A combination of the `sum()` function and a generator expression can also be used to count items in a list. This is versatile and memory-efficient, particularly for large lists.
Here’s an example:
```my_list = [True, False, True, True, False]
print(sum(1 for item in my_list if item))```
Output: `3`
The example counts the number of `True` values in a list of boolean elements. The generator expression iterates over each item, and the `sum()` function tallies up the count of `True` items. It’s both compact and efficient.
Bonus One-Liner Method 5: Using the list.count() Method
For counting the number of occurrences of a specific item in a list, you can use the `list.count()` method which returns the count directly.
Here’s an example:
```my_list = [1, 2, 2, 3, 4, 2, 5]
print(my_list.count(2))```
Output: `3`
This one-liner is straight to the point when you only want to know the number of occurrences of a specific element in the list, making it very readable and easy to implement.
Summary/Discussion
• Method 1: Using the len() Function. Strengths: Simple and efficient for counting the total number of items in a list. Weaknesses: Cannot be used to count specific conditions within a list.
• Method 2: Using a For Loop to Manually Count. Strengths: Offers control over what gets counted via conditions. Weaknesses: Less efficient than other methods for simple counts.
• Method 3: Using the collections.Counter Class. Strengths: Ideal for counting the frequency of each unique item. Weaknesses: Overly complex for counting the total number of items.
• Method 4: Using the sum() Function with a Generator Expression. Strengths: Memory efficient and adaptable for counting with conditions. Weaknesses: Slightly less straightforward than the `len()` function.
• Bonus Method 5: Using the list.count() Method. Strengths: Directly counts the occurrences of a particular item. Weaknesses: Inefficient for counting multiple distinct items. | 1,012 | 4,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-33 | latest | en | 0.816006 |
https://www.physicsforums.com/threads/finding-the-variable-s-using-the-intercept-in-simple-harmonic-motion-of-spring.435872/ | 1,718,504,603,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00350.warc.gz | 835,455,449 | 15,642 | # Finding the variable S using the intercept in simple harmonic motion of spring.
• Finaid
In summary, the conversation discusses an experiment to determine the value of g using simple harmonic motion and a harmonic oscillator. The individual has all the necessary variables and equations, but is struggling with finding the value of S in the T^2 equation. They have graphs and data for T^2 and x' and have calculated the value of k and g, but are unsure how to use the intercept in their calculations. They also mention that S should theoretically be about a third of the mass of the spring but are getting large values. They are seeking help with this issue.
Finaid
In an experiment to determine g using simple harmonic motion using a harmonic oscillator, a motion sensor and data logging equipment. I have all the variables but I can't figure out how to get S in the following equation:
T^2 = ((4pi^2)/k)M + ((4pi^2)/k)S ...(1)
I have a graph of T^2 (the period) vs Mass of the hanging load.
The slope is 1.863 +/- 0.016 s^2 kg^-1
The intercept is 0.006 +/- 0.001
From the above equation: k= 4pi^2/slope
=> k=21.19 +/- 0.18 kg s^-2
I also have a graph of x' (extension of spring) vs. Mass of load.
The slope is 0.421 +/- 0.034 m kg^-1
The intercept is -0.009 +/- 0.002
From the equation:
x' = (g/k)M - x1 ...(2)
g/k = slope
=> g = k(slope) = 8.92 m s^-2
and the uncertainty is +/- 0.73 m s^-2
I also know that S should theoretically be about a third of the mass of the spring (which is 0.0139kg) but i keep getting huge values ranging from 200 to 6000. I don't understand what I'm supposed to do except that it has something to do with the intercept. Any help would be much appreciated! This is just for a practical write up and isn't very important in the experiment but it's the only thing I haven't been able to work out and it's really annoying me...
And let me know if I've left out any information...
Welcome to Physics Forums.
Finaid said:
In an experiment to determine g using simple harmonic motion using a harmonic oscillator, a motion sensor and data logging equipment. I have all the variables but I can't figure out how to get S in the following equation:
T^2 = ((4pi^2)/k)M + ((4pi^2)/k)S ...(1)
If T2 and M are the variables here, then the y-intercept in this equation is _____?
I have a graph of T^2 (the period) vs Mass of the hanging load.
The slope is 1.863 +/- 0.016 s^2 kg^-1
The intercept is 0.006 +/- 0.001
From the above equation: k= 4pi^2/slope
=> k=21.19 +/- 0.18 kg s^-2
I also have a graph of x' (extension of spring) vs. Mass of load.
The slope is 0.421 +/- 0.034 m kg^-1
The intercept is -0.009 +/- 0.002
From the equation:
x' = (g/k)M - x1 ...(2)
g/k = slope
=> g = k(slope) = 8.92 m s^-2
and the uncertainty is +/- 0.73 m s^-2
I also know that S should theoretically be about a third of the mass of the spring (which is 0.0139kg) but i keep getting huge values ranging from 200 to 6000. I don't understand what I'm supposed to do except that it has something to do with the intercept. Any help would be much appreciated! This is just for a practical write up and isn't very important in the experiment but it's the only thing I haven't been able to work out and it's really annoying me...
And let me know if I've left out any information...
## What is the variable S in simple harmonic motion of a spring?
The variable S represents the displacement of the spring from its equilibrium position.
## Why is the intercept used to find the variable S?
The intercept is used because it is the point where the displacement of the spring is equal to zero, making it the equilibrium position. This allows for an accurate measurement of the displacement from this point.
## How is the intercept determined in simple harmonic motion of a spring?
The intercept is determined by measuring the distance from the equilibrium point to where the spring crosses the x-axis on a displacement vs. time graph. This point represents the maximum displacement of the spring.
## What is simple harmonic motion?
Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the opposite direction of the displacement. This results in a back-and-forth motion around an equilibrium position.
## Why is the variable S important in simple harmonic motion of a spring?
The variable S is important because it represents the amplitude or maximum displacement of the spring, which is a key factor in determining the motion and energy of the system. It is also used to calculate other important quantities such as the period and frequency of the motion.
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1K | 1,328 | 5,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-26 | latest | en | 0.94629 |
https://online.stat.psu.edu/stat506/book/export/html/666 | 1,721,290,099,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00531.warc.gz | 393,459,213 | 4,860 | # 8.2 - Variance and Cost in Cluster and Systematic Sampling versus S.R.S.
8.2 - Variance and Cost in Cluster and Systematic Sampling versus S.R.S.
For simplicity, suppose that each of N primary units has an equal number $$\overline{M}$$ of secondary units. To simplify the variance computations and to explore the relationship between cluster and simple random sampling, we note the identity:
$$\sum\limits_{i=1}^N \sum\limits_{j=1}^{\overline{M}}(y_{ij}-\mu)^2= \sum\limits_{i=1}^N \sum\limits_{j=1}^{\overline{M}}(y_{ij}-\bar{y}_i)^2+\overline{M}\sum\limits_{i=1}^N (\bar{y}_i-\mu)^2$$
$$\text{where } \bar{y}_i=\sum\limits_{j=1}^{\overline{M}}\dfrac{y_{ij}}{\overline{M}}$$
SST = SSW + SSB
SST: the total sum of square
SSW: within-cluster sum of squares (within-primary units)
SSB: between-cluster sum of squares (between-primary units)
The within-primary-unit variance is:
$$\sigma^2_w=\left\{\sum\limits_{i=1}^N \sum\limits_{j=1}^{\overline{M}}(y_{ij}-\bar{y}_i)^2\right\}/[N(\overline{M}-1)]$$
The between-primary-unit variance is:
$$\sigma^2_b=\left\{\sum\limits_{i=1}^N (\bar{y}_i-\mu)^2\right\}/(N-1)$$
The identity can be rewritten as:
$$(N\overline{M}-1)\sigma^2=N(\overline{M}-1)\sigma^2_w+(N-1)\overline{M}\sigma^2_b$$
Thus, an unbiased estimator of $$\sigma^2$$ from a simple random cluster sample is:
$$\hat{\sigma}^2=\dfrac{N(\overline{M}-1)S^2_w+(N-1)\overline{M}S^2_b}{N\overline{M}-1}$$
Since the data was obtained by cluster sampling, we cannot use $$s^2$$ to estimate $$\sigma^2$$ but we can use $$\hat{\sigma}^2$$ to estimate $$\sigma^2$$.
The relative efficiency of simple random sampling versus simple random cluster sampling is:
$$\dfrac{Var(\bar{y}_{srs})}{Var(\hat{\mu})}=\dfrac{\overline{M}\sigma^2}{\sigma^2_u}$$
It can be estimated by:
$$\dfrac{\hat{V}ar(\bar{y}_{srs})}{\hat{V}ar(\hat{\mu})}=\dfrac{\overline{M}\hat{\sigma}^2}{s^2_u}$$
Note!
$$s^2_u=\dfrac{1}{n-1}\sum\limits_{i=1}^n(y_i-\bar{y})^2=\dfrac{1}{n-1}\sum\limits_{i=1}^n (\overline{M}\bar{y}_i-\overline{M}\hat{\mu})^2={\overline{M}}^2 \dfrac{\sum\limits_{i=1}^n(\bar{y}_i-\hat{\mu})^2}{n-1}={\overline{M}}^2 s^2_b$$
Recall: $$Var(\bar{y}_{srs})=\dfrac{N-n}{N}\cdot \dfrac{\sigma^2}{n\overline{M}}$$ and $$Var(\hat{\mu})=\dfrac{N-n}{N}\cdot \dfrac{\sigma^2_u}{n{\overline{M}}^2}$$
where $$\sigma^2_u$$ is the finite population variance of $$y_i$$ .
## Example 8-2: Number of cell phones per household
The marketing research department of a communication company wishes to estimate the average number of cell phones purchased per household in a given community. Therefore, the 4,000 households in the community are listed in 400 geographical clusters of 10 households each, and a simple random sample of 4 clusters is selected to reduce the traveling cost for interviewing each household. The data are given in the following table:
Cluster Number of cell phones Total
1 3 5 6 4 5 6 3 2 4 5 43
2 2 0 2 1 1 0 1 1 0 1 9
3 3 2 3 2 4 2 2 1 2 2 23
4 5 2 3 2 1 1 2 2 4 1 23
### Using Minitab
Stat > ANOVA > One-way
The data should be entered in two columns, the response contains all 40 responses and the column for factor indicates whether it is cluster 1, cluster 2, cluster 3, or cluster 4.
#### Minitab output
##### One-Way ANOVA: cluster 1, cluster 2, cluster 3, cluster 4
Source DF SS MS F P
Factor 3 58.70 19.57 16.31 0.000
Error 36 43.20 1.20
Total 39 101.90
Let's find the relative efficiency of simple random sampling versus cluster sampling for the data in this example.
In this example, N = 400, n = 4, and $$\overline{M}=10$$.
We need to find $$s_b^2, s_w^2$$.
Note the identity for the population: $$(N\overline{M}-1)\sigma^2=N(\overline{M}-1)\sigma^2_w+(N-1)\overline{M}\sigma^2_b$$
The identity for the sample is: $$(n\overline{M}-1)s^2=n(\overline{M}-1)s^2_w+(n-1)\overline{M}s^2_b$$
SS total = SS error + SS factor
From the ANOVA table of the example, we can find sb2 by:
$$\text{SS factor}=(4-1)10s^2_b=58.70$$
$$s^2_b=\dfrac{58.70}{30}=1.957$$
We can find $$s_w^2$$ by:
$$\text{SS error}=4(10-1)s^2_w=43.2$$
$$s^2_w=1.20$$
### Try it!
Compute $$\hat{\sigma}^2$$.
$$\hat{\sigma}^2=\dfrac{N(\overline{M}-1)\sigma^2_w+(N-1)\overline{M}\sigma^2_b}{N\overline{M}-1} =\dfrac{(400\times 9 \times 1.2)+[(400-1)\times 10 \times 1.957]}{400\times 10 -1}=3.03$$
And now we can determine the relative efficiency of simple random sampling versus cluster sampling by plugging the values into the formula:
$$s^2_u={\overline{M}}^2 s^2_b=100 \times 1.957=195.7$$
### Try it!
Compute the relative efficiency of simple random sampling versus cluster sampling. What does that tell us?
$$\dfrac{\hat{V}ar(\bar{y}_{srs})} {\hat{V}ar(\hat{\mu})}=\dfrac{\overline{M}\hat{\sigma}^2}{s^2_u}=\dfrac{10 \times 3.03}{195.7}=0.155$$
What is this telling us?
Thus, the variance of simple random sampling is just 15.5% of that of cluster sampling if the same sample size is used. We can see that in this example simple random sampling is more efficient if the only variance is considered.
Note! It is a BIG mistake to analyze a cluster sample as if it were a simple random sample, (often with the reported standard error much less than it should be). You will end up being much too optimistic and not conservative regarding your results as you should be.
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# Can you simplyify both sides of the equal sign when working out intersection points?
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1. Would multiplying and dividing both side of an equation when working out intersection point of 2 curves change the points of intersection?
2. Never divide - factorise
Make sure you don't add solutions when you multiply
Other than that it's ok
3. Also if it's by a constant it's ok.
4. (Original post by TheAdviser101)
Would multiplying and dividing both side of an equation when working out intersection point of 2 curves change the points of intersection?
Do you have an example? Though as above, never divide by variables, factor them out instead.
5. (Original post by RDKGames)
Do you have an example? Though as above, never divide by variables, factor them out instead.
9x^2 + 6x + 12 = 3x - 6
So in this case if divide both graphs by 3 (take out a factor of 3), the points of intersection will be identical as the original? btw what do you mean by dividing by variables? Can you give an example. Thanks
6. (Original post by TheAdviser101)
9x^2 + 6x + 12 = 3x - 6
So in this case if divide both graphs by 3 (take out a factor of 3), the points of intersection will be identical as the original? btw what do you mean by dividing by variables? Can you give an example. Thanks
Dividing and taking out a factor is not the same thing. Since 3 is a constant non-zero term you can freely divide by it and the quantity's will still remain equal, so yes the answer's will be the same.
Dividing by variables would be something like dividing both sides by x like so: (as x varies - hence a variable) which is wrong as you are losing a solution.
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16-Pearl
## Mathcad Community Challenge September 2022 - Resistors in Parallel
This month’s challenge is related to electrical engineering. We have a simple circuit with an electrical potential of 220 Volts. Initially we have a single 10 Ohm resistor. Then we add a second resistor in parallel, with 10% higher resistance. Then we add a third resistor in parallel, with 10% higher resistance than the previous resistor. And so on.
• Calculate the current in the circuit for the single resistor case.
• Calculate the resistance of each additional resistor and current through each resistor for 2, 3, 4, 5, and 10 resistors in parallel.
• Can you write a function or program that calculates the resistance of each resistor and current through each resistor for n resistors in parallel?
These calculations are fairly straightforward, so it will be interesting to see what tools – vectors, matrices, loops, plots, charts, etc. – that you use to solve the problem. As always, how you document your calculations is important as your worksheet will be visible to the community.
Here is an example of three resistors in parallel, as drawn in Creo Schematics:
Dave Martin - dmartin@creowindchill.com - https://www.mcaeconsulting.com
33 REPLIES 33
18-Opal
(To:DaveMartin)
Hello,
My first try (Mathcad Prime 7):
23-Emerald III
(To:DaveMartin)
Here's a solution in Prime 4 express.
Luc
23-Emerald I
(To:DaveMartin)
Calculating resistance and current for each resistor is simple.
Calculating resistance and current for the network is more pertinent:
Prime Express 4 attached
12-Amethyst
(To:DaveMartin)
I like to use Mathcad symbolics for insights into how to derive solutions. My final and generalized answer for the total resistance equation is:
The complete solution is attached.
20-Turquoise
(To:RantEng)
In your Mathcad sheet, you write,
When then the resistance approaches zero.
However, Fred, ppal and others show the total parallel resistance is
12-Amethyst
(To:ttokoro)
Good catch. I stand corrected. I was not considering that the the resistors are increasing in value. Before stating so, I should have taken the extra step to derive the limit as you did,
Bob
17-Peridot
(To:DaveMartin)
20-Turquoise
(To:DaveMartin)
20-Turquoise
(To:ttokoro)
Prime 8 with plot.
20-Turquoise
(To:ttokoro)
20-Turquoise
(To:ttokoro)
17-Peridot
(To:ttokoro)
Hi @ttokoro , would you care to please attach your worksheet? Not just screenshots?
I manage the Creo and PTC Mathcad YouTube channels for PTC, as well as all PTC Mathcad marketing in general.
20-Turquoise
(To:ttokoro)
23-Emerald III
(To:DaveMartin)
Here's another incarnation, using recursion to generate the resistor values:
Prime 4 attached, works also in Prime Express.
Luc
17-Peridot
(To:DaveMartin)
Just another one.
11-Garnet
(To:DaveMartin)
Some nice analyses, indeed.
But... don't do this at home!
Even just the first 10 ohm resistor is dissipating 4840 Watts, and the others not much less! Gonna get toasty in there!
23-Emerald I
(To:Strib)
Yeah, don't try this as home, Or anywhere else but as an exercise.
(We could argue that you deserve the burnt fingers and having to sweep up the charred remains, but . . .)
Here's your sign!
1-Newbie
(To:DaveMartin)
1-Newbie
(To:SC_2975838)
11-Garnet
(To:DaveMartin)
11-Garnet
(To:Strib)
17-Peridot
(To:Strib)
Please attach the file.
11-Garnet
(To:ppal)
Sure, happy to attach the file. But NOTE: there's no Mathcad work here! The solution is so simple that's not needed. I simply used Prime as a typing pad.
The only thing clever is realizing that it's trivial if viewed from the standpoint of the conductances, and then representing the summation of the conductances and resolving that series with an expression we all remember from high school. I marked that line with *** in the attached mcdx file.
17-Peridot
(To:Strib)
Something like this?
11-Garnet
(To:ppal)
Yep, that's good with the OP's original parameters.
4-Participant
(To:DaveMartin)
23-Emerald I
(To:ATschida)
Very nice calculation! Well done!
Not criticisms, just comments:
1) Not sure why you changed the default ORIGIN. If that's your default choice, okay; but understanding somebody else's sheet may be confusing,
2) (And, IMHO opinion, more important) Don't divide out units in functions unless you absolutely must! Your equation for total resistance stripped the units. (I know why,) But you didn't need to, and the equivalent equation is simpler and easy to visualize:
Welcome to Mathcad!
4-Participant
(To:Fred_Kohlhepp)
Fred:
Thanks for the comments.... I should have simplified the equation a bit!
With regards to the ORIGIN change... the thought was that I wanted there to be no R0... R1 was to be the first resistor. In playing around with the code though, it appears it works a bit differently that I had understood. Working on that...
thanks
Adam
3-Visitor
(To:DaveMartin)
Opting to make a work sheet that uses "for" loops, is easy for a student to follow, and is scalable.
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Top Tags | 1,346 | 5,307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-18 | latest | en | 0.879405 |
http://newtoyslist.com/rubik/rubiks-cube-building-rubix-design-and-build.html | 1,566,130,128,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313803.9/warc/CC-MAIN-20190818104019-20190818130019-00221.warc.gz | 143,436,200 | 7,437 | ```Begin with your cube solved. Once again, you want to start off with your Rubik’s cube in its solved position. Making a fish pattern on your Rubik’s cube is simple. The end result will have swapped two edge pieces so that the remaining ones look like a fish with fins. If you hold the cube diagonally it will look like a fish swimming away from you.[7]
```
If you're still reading, congratulations on not being put off by the time requirements! The first thing you are going to need to know about solving the cube is how the turns you make can be represented by letters. Later on in this guide, you're going to need a few algorithms. These are combinations of moves that rotate pieces or just move them around to get them where you want them. These algorithms are written using this notation, so you can always come back to this section if you've forgotten by the time we need them.
The standard Rubik's cube has sides of about 2.2 in (5.7 cm) per square. Various other sizes have also been produced such as a 1.5 in (3.8 cm) mini cube, a 0.8 in (2 cm) key chain micro cube, and a 3.5 in (9 cm) giant cube. While the standard cube is a 3 × 3 × 3 segmentation other types have also been introduced. Some of the more interesting ones include the 2 × 2 × 2 cube, the 4 × 4 × 4 cube (called Rubik's Revenge) and the 5 × 5 × 5 cube. The shape has also been varied and puzzles in the form of a tetrahedral, a pyramid, and an octahedral are among types that were produced. The Rubik's cube also led to the development of game derivatives like the Rubik's cube puzzle and the Rub it cube eraser. Rubix Building Solutions
Sport, Outdoor & Furniture Kids Furniture Sand Toys, Pools & Inflatables Kids Sports & Balls Trampolines & Playgyms Blasters & Water Pistols Vehicles Hobbies & Radio Controlled Remote Control Collectible Vehicles Slot Cars & accessories Model Trains & accessories Model Kits Vehicles & Play Sets STEM STEM - Science STEM - Technology STEM - Engineering STEM - Mathematics
Since the center pieces cannot be moved relatively to each other it's important to solve the edge pieces correctly in relation to each other. For example, when solving the white in our case- the green center piece is to the left of the red center piece, therefore the green-white edge piece should to be solved to the left of the red-white edge piece (see image).
The project uses the Pi to directly solve the Rubik’s cube. The BrickPi3 takes the unsolved Rubik’s cube and the Raspberry Pi takes a picture of each side of the Rubik’s cube with the Raspberry Pi Camera. The Pi creates a text map of the color squares that shows where they are located on the cube. When it has fully mapped the cube, the Pi uses the “kociemba” python library to map out the moves needed to solve the Rubik’s cube. This information is taken by the Pi and BrickPi3 to solve the Rubik’s cube using the LEGO motors. The result: a solved Rubik’s cube. Rubiks Build It Solve It
If you were around in the 1980’s and did not live under a rock, you had a Rubik’s cube. It was the 3D combination puzzle that had children and adults mesmerized trying to solve the impossible puzzle. We would spend hours twisting and turning the cube to figure the solution. The cube was invented by an Hungarian professor of architecture, Erno Rubik, in 1974. Although it took Erno over one month to solve his very own puzzle, it became a fad and everyone had one. Consequently, it became the world’s best selling toy ever at that time.
If you're reading this, you're probably holding a cube in your hand and already feeling bad about yourself for needing to look up the solution. But don't worry! In fact, most of the “super-human-intelligence beings” (a common misconception) who have solved the cube thousands of times in their lifetimes were sitting as you are now. Whether you want to learn it to impress a girl, because your friends bet you couldn't, or just to close the book on the biggest time waste of your childhood by finally defeating it, this guide will take you through the simplest way to conquer the puzzle.
If it comes to constructing the Rubik’s Cube, it’s not as difficult as it seems. In reality, it is going to take approximately fifteen minutes and the directions are simple to follow. If it comes to putting the coloured tiles, be sure to look closely at where you’re supposed to put them since in the event that you snap them in the incorrect location, you won’t have the ability to eliminate them. Yes, you will continue to have the ability to use this Rubik’s Cube, however you won’t be able to follow along with the documentation manual on solving the mystery.
The Rubik's cube (sometimes misspelled rubix cube) is a mechanical 3D puzzle, invented more than 30 years ago and still considered as the best-selling toy of all times! Yet, solving the Rubik's Cube is considered a nearly-impossible task, which requires an IQ of 160... Is that really so hard? Definitely not!! Just follow this simple step by step solving guide and you'll shortly find out that you can solve the Rubik's cube as well… Let's get to work!
Since the center pieces cannot be moved relatively to each other it's important to solve the edge pieces correctly in relation to each other. For example, when solving the white in our case- the green center piece is to the left of the red center piece, therefore the green-white edge piece should to be solved to the left of the red-white edge piece (see image).
For decorative purposes, a colorant is typically added to the plastic. The pieces of a Rubik's cube are typically black. During production, colored stickers are put on the outside of the cube to denote the color of a side. The plastics that are used during production are supplied to the manufacturer in a pellet form complete with the filler and colorants. These pellets can then be loaded into the molding machines directly.
Repeat the process. Turn back to your blue side and repeat the turns on opposite sides. Then, return once more to the red side and turn the opposite sides in opposite directions. And last, return once more to the blue side and turn the opposite sides in opposite directions. When you finish, you should have a staircase-like zig-zag across four sides of your Rubik’s cube.[4] Rubiks Build It Solve It Review
###### The Rubik's "Build It Solve It" kit comes with all the parts necessary to build your own Rubik's cube. Easy "how to" instructions, plastic cube pieces and tiles (including some spares), center core pieces, metal screws, springs and screwdriver - all organized in a handy storage tray. Once you've built your cube, turn to the "you can do the Rubik's cube" booklet and learn how to solve it! By learning how to build your own Rubik's cube, you will greatly improve your understanding of how this fascinating puzzle functions, literally from the inside out.
3 The Rubik's cube parts are taken to an assembly line. In this phase of production, the individual cube pieces are put together. Starting with the nylon core, each ABS center cube is riveted to the core with a spring spacer. The rivet is carefully controlled with a depth stop to ensure the spring is compressed just the right amount. Each center cube has a plastic cover that is glued on to hide the rivet. One of the six center cubes is left until the last part of the assembly. The ABS edges and corner pieces are individually stacked around the core. The cube is built from the bottom up and the last piece to be assembled is the final center cube which is again riveted into the core with a spring spacer and the final cap is glued on.
Important! The center pieces are part of the core and subsequently cannot move relatively to each other. For that reason they are already "solved". The solving process is actually bringing all corner and edge pieces to the "already solved" center pieces (meaning there are only 20 pieces to solve out of the 26). For example, the blue center piece will always be opposite to the green center piece (on a standard color-scheme cube). It doesn't matter how hard you will try scrambling the cube, it will just stay that way.
Repeat the process. Turn back to your blue side and repeat the turns on opposite sides. Then, return once more to the red side and turn the opposite sides in opposite directions. And last, return once more to the blue side and turn the opposite sides in opposite directions. When you finish, you should have a staircase-like zig-zag across four sides of your Rubik’s cube.[4] Rubiks Build It Solve It Review
There are 5 different positions your cube can be in now, one of which could be solved. The rest of them have all four corners solved, so do the required amount of U moves so that every corner is in its right place. 2 of the 4 remaining possibilities have a solved bar (as mentioned above, where all three colours on that side are the same), and the other 2 have no solved bars.
```This Rubik’s Build It Solve It building kit is just for one player, and is one of the new toys for 2017. Winning Moves recommend that it will suit children of ages 8 and up. We have mentioned before, that this building kit is ideal for children and adults who like to figure things out. Anyone who likes to put puzzles together will love this. The Rubik’s Build It Solve It building kit will give you an inside look on how the cube works. You will also see how it is put together, and get some tips from the instruction manual on how you can solve it.
```
```The Rubik's cube (sometimes misspelled rubix cube) is a mechanical 3D puzzle, invented more than 30 years ago and still considered as the best-selling toy of all times! Yet, solving the Rubik's Cube is considered a nearly-impossible task, which requires an IQ of 160... Is that really so hard? Definitely not!! Just follow this simple step by step solving guide and you'll shortly find out that you can solve the Rubik's cube as well… Let's get to work!
```
```Simply put the 1x1x3 is a pseudo puzzle, It fills a gap in the collection but its not exactly complicated to solve. The way this puzzle was made was by using two centres and a core of a QiYi Sail. As these parts already spin like a 1x1x3 should all I had to do was make these parts into cubies by adding some apoxie sculpt and sanding them smooth. This puzzle was made in an afternoon and stickered the following morning while I was also building my 'Mefferts bandage cube'.
```
Here, we're looking at the colours that aren't solved. There are 21 different cases for the top layer, but we only need a couple of algorithms to solve them all. The first thing we want to find is headlights. Only 2 of the cases don't have any headlights (one of them is if you skip this step, and the cube is already solved). For the one case without headlights, just perform the algorithm below from any angle. This is a better case because when you do the next step, the cube will be solved already.
The Rubik's cube appears to be made up of 26 smaller cubes. In its solved state, it has six faces, each made up of nine small square faces of the same color. While it appears that all of the small faces can be moved, only the corners and edges can actually move. The center cubes are each fixed and only rotate in place. When the cube is taken apart it can be seen that the center cubes are each connected by axles to an inner core. The corners and edges are not fixed to anything. This allows them to move around the center cubes. The cube maintains its shape because the corners and edges hold each other in place and are retained by the center cubes. Each piece has an internal tab that is retained by the center cubes and trapped by the surrounding pieces. These tabs are shaped to fit along a curved track that is created by the backs of the other pieces. The central cubes are fixed with a spring and rivet and retain all the surrounding pieces. The spring exerts just the right pressure to hold all the pieces in place while giving enough flexibility for a smooth and forgiving function.
Your goal is to have all four edges matching their centres. If you have this, then the cross is solved. If you have only two, then you could have one of two cases. Either the two matching edges are adjacent (next to each other) or opposite each other. If they are adjacent, hold the cube so that the two solved pieces are facing the front and left of the cube (shown in the left picture), then perform:
The Rubik's "Build It Solve It" kit comes with all the parts necessary to build your own Rubik's cube. Easy "how to" instructions, plastic cube pieces and tiles (including some spares), center core pieces, metal screws, springs and screwdriver - all organized in a handy storage tray. Once you've built your cube, turn to the "you can do the Rubik's cube" booklet and learn how to solve it! By learning how to build your own Rubik's cube, you will greatly improve your understanding of how this fascinating puzzle functions, literally from the inside out.
### Rubiks Build It Solve It Review
This Rubik’s Build It, Solve It kit is for one player – it is recommended for ages 8 and up. As we said before, it is great for children and adults that don’t mind trying to figure out how the cube works – it’s great for those that enjoy putting puzzles together. This kit right here is going to give an inside look on how the cube works and how it’s put together. Plus, you’ll receive some tips in the instruction manual on how to solve it. Rubix Build | 2,996 | 13,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-35 | latest | en | 0.948702 |
https://brainly.in/question/201163 | 1,484,786,648,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280410.21/warc/CC-MAIN-20170116095120-00407-ip-10-171-10-70.ec2.internal.warc.gz | 798,774,353 | 9,872 | Sin Sq A.tanA+cossqA.cotA+2sinA.cosA=TanA+cotA=SecA.cosecA. solve this question plzz
1
by HARSH2001
2015-10-08T15:52:50+05:30
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To solve:
Sin² A tan A + Cos²A Cot A + 2 Sin A Cos A
= Tan A + Cot A = Sec A Cosec A
Sin³ A / Cos A + Cos³ A / Sin A + 2 Sin A Cos A
= [ Sin⁴ A + Cos⁴ A + 2 Sin² A Cos² A ] / [Cos A Sin A]
= ( sin² A + Cos² A)² / (Cos A Sin A)
= Sec A Cosec A
=================
Tan A + Cos A
= sin A/ cos A + Cos A / Sin A
= (sin² A + Cos²A)/ (sin A Cos A)
= Sec A Cosec A
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कक्षा 14 अध्याय SIMPLIFICATION से प्रश्न
# का मूल्य है (ख) (सी) (घ)
हल कीजिए : <br> (i) <br> (v)
6:51
भाव का का बराबर (ए) 1 (बी) (सी) (घ)
1:52
हल करें (i) (II) (Iii) (Iv) (वी) (Vi) (Vii)
10:37
1 3 5 3 4 5 9 2 8 7 2 3 6 5 2 7 3 8 1 2 1 8 4 9 8 1 2 4 7 3 5 2 4 8 9 8 2 4 <br> Which of the following element is 5 th to the right of 10 th from the right end?
का (ख) (सी) (घ)
2:30
1 3 5 3 4 5 9 2 8 7 2 3 6 5 2 7 3 8 1 2 1 8 4 9 8 1 2 4 7 3 5 2 4 8 9 8 2 4 <br> How many such odd digits are there in the given arrangement, each of which is immediately followed and preceded by an odd number?
1 3 5 3 4 5 9 2 8 7 2 3 6 5 2 7 3 8 1 2 1 8 4 9 8 1 2 4 7 3 5 2 4 8 9 8 2 4 <br> Which element is exactly mideway between the seventh element form the left end and sixteenth from the right end?
1 3 5 3 4 5 9 2 8 7 2 3 6 5 2 7 3 8 1 2 1 8 4 9 8 1 2 4 7 3 5 2 4 8 9 8 2 4 <br> How many perfect squares are there to the right of the fouteenth element form the right end?
1 3 5 3 4 5 9 2 8 7 2 3 6 5 2 7 3 8 1 2 1 8 4 9 8 1 2 4 7 3 5 2 4 8 9 8 2 4 <br> How many perfect cubes are there in the above arrangement, each is immediately preceded by an odd number and immediately followed by an even number?
का समाधान (ख) (सी) (घ) (ई)
0:54
सेट लिखें सेट बिल्डर फॉर्म में।
2:41
निम्नलिखित आँकड़ों का बहुलक ज्ञात कीजिए : <br> (i) 1, 3, 5, 5, 5, 4, 6 <br> (ii) 2, 4, 6, 3, 4, 3, 4, 4, 7, 2 <br> (iii) 5, 7, 9, 8, 7, 9, 7, 7, 3, 5, 2
2:17
3:14
सरल बनाएं:
1:46
नवीनतम ब्लॉग पोस्ट
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सूक्ष्म अवधारणाएँ | 1,213 | 2,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-24 | latest | en | 0.584425 |
http://www.expertsmind.com/library/calculate-the-option-premium-for-each-contract-51415012.aspx | 1,642,972,967,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00674.warc.gz | 83,400,760 | 12,979 | ### Calculate the option premium for each contract
Assignment Help Finance Basics
##### Reference no: EM131415012
Case- SLEEPLESS IN L.A.
1. Go to www.cboe.com to find out the information about the stock options on Apple
2. list the contract specification of the options using "Terms" worksheet.
3. Use "quotes" worksheet to list the delayed quoates for your options.
4.Take a screen shot of all put/call options quotes for different months (3 months if available) and paste them into a word doc.
5 Write down the meaning behind each symbol you put to the list.
6. calculate the option premium for each contract.
7. calcuate the instrinsic value, time value for each contract
8. state the pattern of the option prices, volumes, open interest ,if any, in each table on the "quotes" worksheet.
Attachment:- case.rar
Verified Expert
The solution is on the Black-Scholes model on investment options.The solution has been done using Microsoft word,.
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Develop and explain strategic goals for finance, product, and expansion, providing an explanation for each. Use the "SMART" methodology as a guide for setting your goals
### Reviews
#### inf1415012
3/22/2017 4:51:09 AM
Hello, thanks alot guys. I cherished it, and i am so grateful for your assistance. i welcome it, and the paper is awesome. much obliged.
#### inf1415012
3/22/2017 4:50:53 AM
One of the attachments is wrong, but the pdf is right, except the excel part, in fact forget about the excel part because I mistakenly sent it. In short, the assignment I am sending has two parts of the attachments, which are pdf and word document. The following attachment is the word document containing questions to the pdf case. 21499036_1assignment questions case 2.docx
#### len1415012
3/6/2017 12:43:43 AM
This class is called BADM 418 financial future and options. list the contract specification of the options using "Terms" worksheet Use "quotes" worksheet to list the delayed quoates for your options. Take a screen shot of all put/call options quotes for different months (3 months if available) and paste them into a word doc. Write down the meaning behind each symbol you put to the list. calculate the option premium for each contract. calcuate the instrinsic value, time value for each contract | 883 | 3,910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-05 | latest | en | 0.884891 |
http://www.slideserve.com/amadahy-mitchell/conversion-of-number-system | 1,493,403,052,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123046.75/warc/CC-MAIN-20170423031203-00396-ip-10-145-167-34.ec2.internal.warc.gz | 667,964,007 | 19,107 | # Conversion of Number System - PowerPoint PPT Presentation
1 / 54
Conversion of Number System. Conversion Among Bases. The possibilities:. Decimal. Octal. Binary. Hexadecimal. Binary to Decimal. Decimal. Octal. Binary. Hexadecimal. Binary to Decimal. Technique Multiply each bit by 2 n , where n is the “weight” of the bit
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Conversion of Number System
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### Conversion of Number System
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Conversion Among Bases
• The possibilities:
Decimal
Octal
Binary
Hexadecimal
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Binary to Decimal
Decimal
Octal
Binary
Hexadecimal
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Binary to Decimal
• Technique
• Multiply each bit by 2n, where n is the “weight” of the bit
• The weight is the position of the bit, starting from 0 on the right
• Add the results
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Example 1
Bit “0”
1010112 => 1 x 20 = 11 x 21 = 20 x 22 = 01 x 23 = 80 x 24 = 01 x 25 = 32
4310
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### Example2
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Octal to Decimal
Decimal
Octal
Binary
Hexadecimal
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Octal to Decimal
• Technique
• Multiply each bit by 8n, where n is the “weight” of the bit
• The weight is the position of the bit, starting from 0 on the right
• Add the results
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Example 1
7248 => 4 x 80 = 42 x 81 = 167 x 82 = 44846810
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### Example 2
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Hexadecimal to Decimal
Decimal
Octal
Binary
Hexadecimal
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Hexadecimal to Decimal
• Technique
• Multiply each bit by 16n, where n is the “weight” of the bit
• The weight is the position of the bit, starting from 0 on the right
• Add the results
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Example 1
ABC16 =>C x 160 = 12 x 1 = 12 B x 161 = 11 x 16 = 176 A x 162 = 10 x 256 = 2560
274810
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### Example 2
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Decimal to Binary
Decimal
Octal
Binary
Hexadecimal
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Decimal to Binary
• Technique
• Divide by two, keep track of the remainder
• First remainder is bit 0 (LSB, least-significant bit)
• Second remainder is bit 1
• Etc.
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2 125 62 1
2 31 0
2 15 1
2 3 1
2 7 1
2 0 1
2 1 1
Example 1
12510 = ?2
12510 = 11111012
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### Example 2
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Octal to Binary
Decimal
Octal
Binary
Hexadecimal
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Octal to Binary
• Technique
• Convert each octal digit to a 3-bit equivalent binary representation
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7 0 5
111 000 101
Example 1
7058 = ?2
7058 = 1110001012
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### Example 2
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Hexadecimal to Binary
Decimal
Octal
Binary
Hexadecimal
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Hexadecimal to Binary
• Technique
• Convert each hexadecimal digit to a 4-bit equivalent binary representation
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1 0 A F
0001 0000 1010 1111
Example 1
10AF16 = ?2
10AF16 = 00010000101011112
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### Example 2
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Decimal to Octal
Decimal
Octal
Binary
Hexadecimal
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Decimal to Octal
• Technique
• Divide by 8
• Keep track of the remainder
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8
19 2
8
2 3
8
0 2
Example 1
123410 = ?8
8 1234
154 2
123410 = 23228
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### Example 2
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Decimal to Hexadecimal
Decimal
Octal
Binary
Hexadecimal
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Decimal to Hexadecimal
• Technique
• Divide by 16
• Keep track of the remainder
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16 1234
77 2
16
4 13 = D
16
0 4
Example 1
123410 = ?16
123410 = 4D216
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### Example 2
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Binary to Octal
Decimal
Octal
Binary
Hexadecimal
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Binary to Octal
• Technique
• Group bits in threes, starting on right
• Convert to octal digits
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1 011 010 111
1 3 2 7
Example 1
10110101112 = ?8
10110101112 = 13278
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### Example 2
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Binary to Hexadecimal
Decimal
Octal
Binary
Hexadecimal
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Binary to Hexadecimal
• Technique
• Group bits in fours, starting on right
• Convert to hexadecimal digits
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Example 1
10101110112 = ?16
• 10 1011 1011
• B B
10101110112 = 2BB16
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### Example 2
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Octal to Hexadecimal
Decimal
Octal
Binary
Hexadecimal
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Octal to Hexadecimal
• Technique
• Use binary as an intermediary
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• 1 0 7 6
• 001 000 111 110
2 3 E
Example 1
10768 = ?16
10768 = 23E16
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### Example 2
• Octal8 -> hexadecimal16
• 278 -> hexadecimal16
First convert the octal number to binary.
2 7
421 421
010 111
278 -> 010 111 2
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### Example2 Cont.,
• Convert to hexadecimal.
0001 0111
8421 8421
0+0+0+1 = 1 0+4+2+1 = 7
278 -> 1716
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Hexadecimal to Octal
Decimal
Octal
Binary
Hexadecimal
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Hexadecimal to Octal
• Technique
• Use binary as an intermediary
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• 1 F 0 C
• 0001 1111 0000 1100
1 7 4 1 4
Example 1
1F0C16 = ?8
1F0C16 = 174148
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### Example 2
• We do not convert directly from hexadecimal to octal but instead first convert to binary and then to octal.
• 4516 -> octal8
• First convert the hexadecimal number to binary.
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### Example2 Cont.,
• Hexadecimal to Binary
4 5
8421 8421
0100 0101
4516 -> 010001012
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### Example2 Cont.,
• Then Convert to Octal
001 000 101
421 421 421
0+0+1 = 1 0+0+0 = 0 4+0+1 = 5
4516 -> 1058
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…..Thank you….
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# How to Solve Right Triangles?
Nov 26, 2022
Triangles are made up of three line segments. These three segments meet to form three angles. The lengths of the sides and sizes of the angles are related to one another. If you know the size (length) of three out of the six parts of the triangle (at least one side must be included), you can find the sizes of the remaining sides and angles.
And if the triangle is right-angled, you can use simple trigonometric ratios to find the missing parts. In a general triangle that is acute or obtuse, you will need to use other techniques like the sine and cosine laws.
## How to Solve a Right Triangle?
You should know the methodology to solve right triangles. Let the three angles of a triangle ABC be labeled as ∠ A, ∠ B, and ∠ C in capital letters, and the three sides of the triangle should be labeled as a, B, and C in small letters. The diagram below shows the labeled figure. If any three of these six measurements are known (other than knowing the measures of the three angles), you can calculate the values of the other three measurements
Solving right triangles here means finding the missing values in the triangle. One of the angles is 90° if the triangle is a right triangle. Therefore, you can solve the right triangle if you are given the measures of two of the three sides or even when you are given the measure of one side and one of the other two angles.
A right triangle is a triangle in which one angle is a right angle that is an angle with a value of 90 degrees (90°). The relation between the sides and angles of a right triangle is the basis for trigonometry.
The side opposite the right angle is called the hypotenuse. It is side c in the above figure. The sides adjacent to the right angle are called the legs of the triangle. Here sides A and B are the legs. Side A is the side adjacent to angle B and opposite to angle A. Side B is the side adjacent to angle A and opposite to angle B.
In several applications, specific angles are given unique names. The Angle of elevation and the Angle of depression are two of these unique terms.
### The Pythagorean Theorem
The Pythagorean Theorem, also known as Pythagoras’ Theorem, helps to define the relationship among the three sides of a right triangle. Pythagoras’ theorem states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The theorem can be written as the below-given equation where a, b, and c are the three sides of the triangle ABC.
a²+b²=c²
In this equation, c represents the length of the hypotenuse, and a and b are the lengths of the triangle’s other two sides.
Example:
A right triangle has side lengths of 3 cm and 4 cm. Find the length of the hypotenuse.
Substitute
a=3 and b=4 into the Pythagorean Theorem and solve for c
a²+b²=c²
3²+4²=c²
9+16=c²
25=c²
c²=25
√c²=√25
c=5 cm
### Solving for a Side in Right Triangles with the Help of Trigonometry Functions
When we say to solve a triangle, we mean to know all three sides and all three angles. Let us look at finding the value of a missing side length in a right triangle by choosing the right trigonometric ratio for a given angle.
When working with right triangles and circles, the sine, cosine, and tangent ratios are three of the most fundamental tools. Let us understand this further.
The sum of internal angles in a triangle is 180°. If we have a right triangle with a non-right angle of 𝜃, the remaining angle will measure 90°−𝜃.
Then, we recall the AAA criterion for similarity. This tells us that two triangles are similar if they have the same internal angles. If two triangles are similar, then the lengths of their sides are scalar multiples of each other.
These two facts allow us to notice an interesting property of right triangles: the value of the ratio of any two lengths of the sides of a right triangle is only dependent on the angle and the choice of the two sides.
We can use trigonometric ratios to find unknown sides in right triangles.
Let’s look at an example.
Given △ABC, A, B, and C, find AC
Solution
Step 1: Determine which trigonometric ratio to use.
Let’s focus on angle B since that is the angle that is explicitly given in the diagram.
Note that we are given the length of the hypotenuse, and we are asked to find the length of the side opposite angle B. The trigonometric ratio that contains both of those sides is the sine.
Step 2: Create an equation using the trig ratio sine and solve for the unknown side.
sin(B) = opposite / hypotenuse (Define sine)
sin(50°) = AC/6 (Substitute)
6sin(50°) = AC
4.60 ≈ AC
### Finding an Angle in a Right-Angled Triangle
If we know the lengths of two of its sides, we can find an unknown angle in a right-angled triangle.
This can be easily done by Sine, Cosine, or Tangent!
A special phrase to remember which one to use is ”OLD HARRY AND HIS OLD AUNT”
O – Opposite
H – Hypotenuse
Sine: sin(θ) = Opposite / Hypotenuse.
Cosine: cos(θ) = Adjacent / Hypotenuse
Tangent: tan(θ) = Opposite / Adjacent
Example:
Find AB
Tan x° = opposite/adjacent = 300/400 = 0.75
Therefore, tan-1 of 0.75 = 36.9°
### How do Solve Special Right Triangles?
To solve for a side in right triangles, first, you should recognize a right-angled triangle. A special right triangle is a right triangle whose sides are in a particular ratio, called the Pythagorean triplets. You can also use the Pythagoras’ theorem”, but if you can see that it is a special triangle, it can save you some calculations.
### What is a 45°-45°-90° Triangle?
A 45°-45°-90° triangle is a special right triangle whose angles are 45°, 45°, and 90°. The lengths of the sides of a 45°-45°-90° triangle are in the ratio of 1: 1: √2.
A right triangle with two sides of equal lengths must be a 45°-45°-90° triangle.
You can also recognize a 45°-45°-90° triangle by the angles. A right triangle with a 45° angle must be a 45°-45°-90° special right triangle.
Side1 : Side2 : Hypotenuse = x : x : x√2
Example 1:
Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are both 3 inches.
Solution:
Step 1: This is a right triangle with two equal sides so it must be a 45°-45°-90° triangle.
Step 2: You are given that both sides are 3. If the first and second value of the ratio x:x:x√2 is 3 then the length of the third side is 3√2.
Answer: The length of the hypotenuse is 3√2 inches.
### What is a 30°-60°-90° Triangle?
Another special right triangle is the 30°-60°-90° triangle. This is a right triangle whose angles are 30°-60°-90°. The lengths of the sides of a 30°-60°-90° triangle are in the ratio of 1 : √3: 2.
You can also recognize a 30°-60°-90° triangle by the angles. As long as you know that one of the angles in the right-angle triangle is either 30° or 60° then it must be a 30°-60°-90° special right triangle. A right triangle with a 30° angle or 60° angle must be a 30°-60°-90° special right triangle.
Side1 : Side2 : Hypotenuse = x : x√3 : 2x
Example 1:
Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 4 inches and 3 inches.
Solution:
Step 1: Test the ratio of the lengths to see if it fits the n:n√2:2n ratio.
4:4√3:? = x:x√3:2x
Step 2: Yes, it is a 30°-60°-90° triangle for x = 4
Step 3: Calculate the third side.
2x = 2 × 4 = 8
Answer: The length of the hypotenuse is 8 inches.
### What is a 3-4-5 Triangle?
A 3-4-5 triangle is a right triangle whose lengths are in the ratio of 3:4:5. When you are given the lengths of two sides of a right triangle, check the ratio of the lengths to see if it fits the 3:4:5 ratio.
Side1 : Side2 : Hypotenuse = 3n : 4n : 5n
Example 1:
Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 6 inches and 8 inches.
Solution:
Step 1: Test the ratio of the lengths to see if it fits the 3n: 4n: 5n ratio.
6 : 8 : ? = 3(2) : 4(2) : ?
Step 2: It is a 3-4-5 triangle for n = 2.
Step 3: Calculate the third side.
5n = 5 × 2 = 10
Answer: The length of the hypotenuse is 10 inches.
1. How do you solve right triangles?
Ans. To solve right triangles, you have to use the Pythagorean theorem. This states that:
a2 + b2 = c2
That means that if you know the lengths of two sides of a right triangle (a and b), you can find the length of the third side (c).
2. How do you solve a right triangle with one side?
Ans. To solve a right triangle with one side:
Choose the given side of the triangle as the hypotenuse, and label it “a”.
Let the other two sides of your triangle be labeled “b” and “c”, respectively.
Use the Pythagorean theorem to find the length of c: a2 + b2 = c2
3. How do I find the angles of a right triangle?
Ans. The angles of a right triangle are found using the Pythagorean Theorem. This equation is a2 + b2 = c2, where a, b, and c are the lengths of the sides of the triangle.
In order to find an angle’s measure in a right triangle, you can use this formula:
sin(a) = opposite side / hypotenuse
cos(a) = adjacent side / hypotenuse
4. How do you find a right triangle with 3 sides?
Ans. To find a right triangle with three sides:
Draw a line from point A to point D.
Draw a line from point B to point C.
Draw a line from point C to point E.
The triangle formed by points A, B, and C is a right triangle if the sum of the lengths of its legs is equal to the length of its hypotenuse.
5. How to solve right triangle trigonometry?
Ans. Right triangle trigonometry can be tricky. Here’s how to do it:
1. Find the sine, cosine, or tangent of each angle in the triangle by using a calculator or plugging them into a chart.
2. Use these values to solve for unknown sides or angles of the triangle, using a calculator or plugging them into a chart.
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […] | 2,897 | 11,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-33 | latest | en | 0.934202 |
https://www.simscale.com/forum/t/continuity-problem-with-a-multiphase-simulation/81697 | 1,607,072,212,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141735395.99/warc/CC-MAIN-20201204071014-20201204101014-00011.warc.gz | 853,134,003 | 11,867 | # Continuity problem with a multiphase simulation
Hey,
i have again a time continuity problem with my multiphase (Water, Air) simualtion. It should be a rain overflow in the sewerage simulatet.
Here is the Link to the projekt.
The picture shows you the geometrie. The BC1(Inlet) is a Velocity inlet with 2 m/s, the BC2(Outlet) ist a pressure out, but the Water should not reach it.
At the first try i used the set numbers for the mesh in x,y, and z direction, furthemore i had problem to see and select the inner walls for the mesh refinements. The walls of the rectangular basin are very thin.
Mesh 1 Run 2:
I used a cartesian Box at the inlet, filled with Alpha phase 1. In order to run over quick.
But the sim stops at 54%.
Mesh 1 Run 3:
I used a second cartesian box an the bassin, also filled with AP1 (Water).
It stops at 53%.
Mesh 1 Run 4:
The second CB is 0.2 m higher at Run 3.
It stops at 69%.
Now i make at second Mesh with thicker walls, so that i can easly see and select it, to make the mesh refinements, i also use approximatley the double count of cells in x,y and z direction.
The boundary conditions are the same like Mesh 1.
Mesh 2 Run 1: I used only the inlet pipe cartesian box 1.
Mesh 2 Run 2: Add a second cartesian box at the basin with the high of 2.5.
Mesh 2 Run 3: Change the high of CB2 to 3.0.
Mesh 2 Run 4: Change the high of CB2 to, 3.4.
Als Runs are broke up at app. 40%-60%.
Here is a screenshot from the last second of Mesh 2 Run 1.
Maybe it helps.
Why is the calculation stops at around 60% every time ?
Kai
Hi @Kai_himself!
Again an awesome case you are trying to solve here! What I am wondering about is the boundary conditions. You defined only one inlet and a single face as an outlet. Where is are your walls defined and what happens to the other inlet (I suppose it should also be an inlet) ? Also tagging the @PowerUsers_CFD here to see if there is anything else you might have missed.
All the best!
Jousef
1 Like
Hey @jousefm,
oh, it is so bad ?
No, the rain overflow is used to seperate the rainwater in the sewerage. The sewage plant is not able to handle so much water. So you have do seperate the dilute rainwater into a river.
The first idea was to let the pipe waterlevel rise, to get over the sill into the basin. Im still trying with the right boundary condition for the pipe outlet, with a “normal” pressure outlet, the waterlevel are dont move up or down. So i decide for this try, the pipe output is negligible. The other effect is, i became a nice wave over the sill :).
When i will start the simulation i got a report “for all missing boundary condiotions are assumed to be a no-slip wall boundary condition”.
Is a manual definded of walls “better” for the stability of the simulation ? Then i will do and try it again.
Here is a picture of a example Rainoverflow from Wikipedia.
Greetz Kai
Hey @jousefm,
mybe i found something, i think, i have read somewhere, that you look for big velocity gradient on walls. The inlet velocity is 2m/s, i got a area with a velocity of roundabout 1000m/s.
Is this why my simulation breaks up ?
Greetz Kai
Hi @Kai_himself!
That makes no sense and there is something weird going on - having a look at it in the evening if that’s okay.
All the best!
Jousef
1 Like
Hey @jousefm,
i check the other runs, the have all at the last sim step a little spot with a very high velocity.
Did you find anything out ?
Greetz Kai
Hi @Kai_himself!
Could not find out so far what the issue might be. I had an idea what it could be but that did not work out unfortunately. Will test another run later on and let you know. @Get_Barried, did you find out anything new?
Best,
Jousef
1 Like
Hi @Kai_himself,
Sorry Kai, I saw this post when you posted it awhile back but was caught up by my work.
So I’ve a quick preliminary look and yes your guess of the sudden acceleration of flow to unrealistic speeds is causing your simulation to crash. This brings me to suspect this is because you’ve assigned the “square outflow” to be a pressure outlet which would be fine but you still have air within the geometry for a inlet that is completely filled with water. This causes the air to create a kind of “vacuum” when eventually all the air flows out of the geometry and the solver just completely crashes as it runs into some sort of continuity error.
Based on what your test case is, I would suggest somehow matching the actual “overflow” case by having a phase fraction inlet that accounts for the amount of air entering the geometry as well as an actual outlet for the end of the pipe rather than forcing all the flow to exit through the “overflow” area. Let the simulation run and hopefully we’re able to achieve a high enough mass flow rate that causes the overflow by fixing the outlet flow rate such that the inlet flow rate is larger than that of the original pipe outflow rate.
As for the specifics of how to do this, I’m not too sure myself. I suggest we work step by step to solve it issue by issue while keeping in mind that we need to “replicate” the actual overflow scenario.
Cheers!
Regards,
Barry
2 Likes
Hey @jousefm and @Get_Barried,
Greetz Kai
1 Like
Hey @Get_Barried,
so i make a first try, with the opposit pipe outlet.
It has the half velocity like the inlet, but the sim broke up at ca. 50%, again with a high velocity spot at the top.
Run 7
Greetz Kai
Hi @Kai_himself,
The Cartesian box still doesn’t cover the area where flow is supposed to overflow and the solver is unable to allow flow to overflow in the specified area. Try extending the box to cover the ledge and gap between the two boxes and see if it works, do refer to the picture below.
Cheers!
Regards,
Barry
1 Like
Hey @Get_Barried,
so i ran a few simulations with different cartesian boxes.
Have i understood that correctly, do you mean this gap between the two boxes ?
This are the runs and the expension of the boxes.
Now i try a inflow with a Phase fraction value of 0.99.
I will get back.
Kai
Hi @Kai_himself,
Yup that is the gap I am referring to.
Do let me know how it goes.
Cheers.
Regards,
Barry
1 Like
Hi @Kai_himself,
Any progress so far?
Cheers.
Regards,
Barry
1 Like
Hey @Get_Barried,
Thanks for the question. No, not as desire :(, i got 20 Runs with different conditions, but every run got an continuity issue, and at the last simulatet frames i got a little spot with high magnitudes.
To fill the Gap between the two cartesian boxes i deside to use only one Box, is it ok or did you have other experiences ?
I make a Excel chart with some specification about every run. I cant upload it here, so i got it to google.
Greetz Kai
Hi @Kai_himself,
Well that is unfortunate. I’ve requested permission for access to the file, can you grant it?
Cheers.
Regards,
Barry
1 Like
Hi @Kai_himself,
May I suggest making a much simpler overflow case to see if that can work? Once we have the settings for that correct we can translate them over and deduce where the issues lies.
Do let me know.
Cheers.
Regards,
Barry
1 Like
Hey @Get_Barried,
yes of course, its ok.
Greets Kai
Hey @Get_Barried,
i build a simpler overflow, do you think its simple enough ? I creat e it on onshape here is the link.
Greetz Kai
Hi @Kai_himself,
It looks good. When you import it do ensure it says it is watertight and go ahead to see if it runs. Be mindful about the need for air to escape the geometry as well besides water.
Cheers.
Regards,
Barry
1 Like | 1,914 | 7,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-50 | latest | en | 0.897793 |
https://wiki.sagemath.org/graph_plotting?action=diff&rev1=3&rev2=4 | 1,627,097,452,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150129.50/warc/CC-MAIN-20210724032221-20210724062221-00038.warc.gz | 617,269,964 | 5,685 | Differences between revisions 3 and 4
⇤ ← Revision 3 as of 2007-01-16 20:10:34 → Size: 1427 Editor: anonymous Comment: ← Revision 4 as of 2007-02-16 05:15:07 → ⇥ Size: 1210 Editor: anonymous Comment: Deletions are marked like this. Additions are marked like this. Line 5: Line 5: The SAGE [http://sage.math.washington.edu:9001/graph Graph Theory Project] aims to implement Graph objects and algorithms in ["SAGE"]. Line 9: Line 7: I'm integrating graph plotting fucntionality in ["SAGE"] one piece at a time. == 2D Plotting == Line 11: Line 9: So far, after small revisions to the graphics class, I've written a NetworkX primitive that takes a NetworkX graph on initialization, and renders that graph using NetworkX's native spring layout routine. In this routine, each edge is treated as a spring; after each node is randomly placed on the plot screen, fifty iterations allow the "springs" to align themselves in equilibrium, often revealing geometric symmetries of the graph (try plotting a Platonic solid...). * matplotlib plotting is supported (albeit awkwardly) by NetworkX. Smooth interfacing of this functionality with SAGE (especially the notebook) is almost complete. Part of this implementation was writing a NetworkX graphics primitive. In this routine, each edge is treated as a spring; after each node is randomly placed on the plot screen, fifty iterations allow the "springs" to align themselves in equilibrium, often revealing geometric symmetries of the graph (try plotting a Platonic solid...). There are also options to pre-specify vertex positions: the graph class now comes with an optional positioning variable, so that if a user likes to think of a graph in a certain visual layout, that layout can be made part of the graph information. Boundary nodes default to plot a different color, and edge labels will soon be displayed. Pending another graphics primitive, graphics objects can be associated with nodes so that the plots show up in place of the nodes when a graph is displayed. Line 13: Line 11: The ["SAGE"] graph class now comes with an optional positioning variable, so that if a user likes to think of a graph in a certain visual layout, that layout can be made part of the graph information. The graph class supports plotting via matplotlib, and soon will also support Tachyon plotting. In addition, the migration of NetworkX 0.32 -> 0.33 must be tested. Boundary nodes (soon to be implemented) will default to plot a different color, and edge labels will be supported. Pending another graphics primitive, soon graphics objects can be associated with nodes so that in the plots show up in place of the nodes when a graph is displayed. * Examples:{{{graphs.CubeGraph(5).show(vertex_labels=False, node_size=100)}}}
## Introduction
Robert Miller is working on this project.
## 2D Plotting
• matplotlib plotting is supported (albeit awkwardly) by NetworkX. Smooth interfacing of this functionality with SAGE (especially the notebook) is almost complete. Part of this implementation was writing a NetworkX graphics primitive. In this routine, each edge is treated as a spring; after each node is randomly placed on the plot screen, fifty iterations allow the "springs" to align themselves in equilibrium, often revealing geometric symmetries of the graph (try plotting a Platonic solid...). There are also options to pre-specify vertex positions: the graph class now comes with an optional positioning variable, so that if a user likes to think of a graph in a certain visual layout, that layout can be made part of the graph information. Boundary nodes default to plot a different color, and edge labels will soon be displayed. Pending another graphics primitive, graphics objects can be associated with nodes so that the plots show up in place of the nodes when a graph is displayed.
• Examples:
graphs.CubeGraph(5).show(vertex_labels=False, node_size=100)
graph_plotting (last edited 2008-11-14 13:42:15 by anonymous) | 886 | 3,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-31 | latest | en | 0.91097 |
https://oeis.org/A212271 | 1,624,555,978,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556482.89/warc/CC-MAIN-20210624171713-20210624201713-00017.warc.gz | 403,636,373 | 3,885 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A212271 Number of ways to place k non-attacking ferses on an n x n cylindrical chessboard, summed over all k >= 0. 4
2, 9, 80, 1600, 79033, 8156736, 2055960192, 1108756350625, 1411080429618656, 3943472747846953216, 25425527581172360096017, 365481944233773616212640000, 11980566143208960475692367828480, 882106482533191605447029340350009049, 147314997388032765439791110273770608260928 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Fers is a leaper [1,1]. LINKS Vaclav Kotesovec, Table of n, a(n) for n = 1..18 V. Kotesovec, Non-attacking chess pieces, 6ed, 2013, p.443 FORMULA Limit n ->infinity (a(n))^(1/n^2) is the hard square entropy constant A085850. CROSSREFS Cf. A067965, A067960, A182407, A212269, A212270. Sequence in context: A279055 A320946 A135868 * A147302 A301861 A112670 Adjacent sequences: A212268 A212269 A212270 * A212272 A212273 A212274 KEYWORD nonn,hard AUTHOR Vaclav Kotesovec, May 12 2012 STATUS approved
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Last modified June 24 13:32 EDT 2021. Contains 345416 sequences. (Running on oeis4.) | 483 | 1,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-25 | latest | en | 0.589121 |
https://www.theochem.ru.nl/~pwormer/Knowino/knowino.org/w/indexa656.html?title=Little_o_notation&oldid=13048 | 1,674,816,111,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494976.72/warc/CC-MAIN-20230127101040-20230127131040-00502.warc.gz | 1,065,354,812 | 7,656 | Little o notation
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The little o notation is a mathematical notation which indicates that the decay (respectively, growth) rate of a certain function or sequence is faster (respectively, slower) than that of another function or sequence. It is often used in particular applications in physics, computer science, engineering and other applied sciences.
More formally, if f and g are real valued functions of the real numbers then the notation f(t) = o(g(t)) (as t tends to plus infinity) indicates that for every real number $\epsilon>0$ there exists a positive real number $T(\epsilon)$ (note the dependence of T on $\epsilon$) such that $|f(t)|\leq \epsilon |g(t)|$ for all $t>T(\epsilon).$
When the function g does not vanish this may be rewritten simply as
$\lim_{t\to\infty} \frac{f(t)}{g(t)} = 0.$
Similarly, if an and bn are two numerical sequences then an = o(bn) means that for any $\varepsilon>0$ and n big enough one has $|a_n|\leq \epsilon |b_n|$ (in case when bn is not zero, this means the limit of the fraction an / bn vanishes in the limit).
The little o notation is also often used to indicate that the absolute value of a real valued function goes to zero around some point at a rate faster than at which the absolute value of another function goes to zero at the same point. For example, suppose that f is a function with f(t0) = 0 for some real number t0. Then the notation f(t) = o(g(tt0)), where g(t) is a function which is continuous at t=0 and with g(0)=0, denotes that for every real number $\epsilon>0$ there exists a neighbourhood $N(\epsilon)$ of t0 such that $|f(t)|\leq \epsilon |g(t-t_0)|$ holds on $N(\epsilon)$. | 444 | 1,723 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-06 | latest | en | 0.895457 |
https://www.convertunits.com/from/ton/minute+%5Bmetric%5D/to/kilogram/hour | 1,604,163,060,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107919459.92/warc/CC-MAIN-20201031151830-20201031181830-00709.warc.gz | 634,004,346 | 11,830 | ## ››Convert ton/minute [metric] to kilogram/hour
ton/minute [metric] kilogram/hour
Did you mean to convert ton/minute [long] ton/minute [metric] ton/minute [short] to kilogram/hour
How many ton/minute [metric] in 1 kilogram/hour? The answer is 1.6666666666667E-5.
We assume you are converting between ton/minute [metric] and kilogram/hour.
You can view more details on each measurement unit:
ton/minute [metric] or kilogram/hour
The SI derived unit for mass flow rate is the kilogram/second.
1 kilogram/second is equal to 0.06 ton/minute [metric], or 3600 kilogram/hour.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between tons/minute and kilograms/hour.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of ton/minute [metric] to kilogram/hour
1 ton/minute [metric] to kilogram/hour = 60000 kilogram/hour
2 ton/minute [metric] to kilogram/hour = 120000 kilogram/hour
3 ton/minute [metric] to kilogram/hour = 180000 kilogram/hour
4 ton/minute [metric] to kilogram/hour = 240000 kilogram/hour
5 ton/minute [metric] to kilogram/hour = 300000 kilogram/hour
6 ton/minute [metric] to kilogram/hour = 360000 kilogram/hour
7 ton/minute [metric] to kilogram/hour = 420000 kilogram/hour
8 ton/minute [metric] to kilogram/hour = 480000 kilogram/hour
9 ton/minute [metric] to kilogram/hour = 540000 kilogram/hour
10 ton/minute [metric] to kilogram/hour = 600000 kilogram/hour
## ››Want other units?
You can do the reverse unit conversion from kilogram/hour to ton/minute [metric], or enter any two units below:
## Enter two units to convert
From: To:
## ››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! | 601 | 2,143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-45 | latest | en | 0.810897 |
https://de.mathworks.com/matlabcentral/profile/authors/9356792-alexandra-kerl | 1,568,978,399,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573988.33/warc/CC-MAIN-20190920092800-20190920114800-00500.warc.gz | 443,477,084 | 17,791 | Community Profile
# Alexandra Kerl
19 total contributions since 2017
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mehr als 2 Jahre ago | 902 | 3,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-39 | latest | en | 0.662588 |
https://www.tutorela.com/math/types-of-triangles | 1,721,484,923,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515164.46/warc/CC-MAIN-20240720113754-20240720143754-00159.warc.gz | 892,787,112 | 51,585 | # Types of Triangles
🏆Practice types of triangles
## Properties of triangles
The triangle is a geometric figure with three sides that form three angles whose sum is always $180^o$ degrees.
Its vertices are called $A,B$ and $C$
The union between these vertices creates the edges $AB,BC$ and $CA$
There are several types of triangles that we will study in this article.
## Test yourself on types of triangles!
In a right triangle, the side opposite the right angle is called....?
In the following section we will present the different types of triangles, along with illustrations and examples.
## Equilateral triangle
An Equilateral triangle is a triangle whose sides have the same length.
### Examples of equilateral triangles
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## Scalene triangle
A scalene triangle is a triangle whose sides are of different lengths (no two edges are the same).
## Isosceles triangle
An isosceles triangle is a triangle in which two of its sides have the same length. One of its properties is that, just as it has two equal edges, also two of its angles are equal.
### Examples of isosceles triangles:
Do you know what the answer is?
## Right triangle
A Right triangle is a triangle in which two sides form an angle of $90^o$ degrees.
## Acute triangle
An acute triangle is a triangle in which all its angles are less than $90^o$ degrees.
## Obtuse triangle
An obtuse triangle is a triangle that has an obtuse angle, that is, greater than $90^o$ degrees, which implies that the remaining two angles are less than $45^o$ degrees.
This is because, as we have already mentioned, the sum of the interior angles of a triangle always equals $180^o$ degrees.
### Examples of obtuse triangles:
Do you want to learn more about triangles? For example, how to calculate their area or perimeter? Watch the complete video with everything you need to know about triangles!
## Exercises on types of triangles and their properties:
### Exercise 1
What is the area of the rectangle?
Solution:
To find the missing side, we will use the Pythagorean Theorem on the triangle above.
Since the triangle is isosceles, we know that the length of the two sides is $7$.
Therefore substituting in the formula of the Pythagorean Theorem we get $A^2+B^2=C^2$:
$7^2+7^2=49+49=98$
Therefore, the measure of side AB is. $\sqrt{98}$
The area of the rectangle is the product of its base and height, therefore:
$\sqrt{98}\times 10=98.99\approx 99u²$
Do you think you will be able to solve it?
### Exercise 2
Given a right triangle:
What is the length of the third side?
Solution:
The picture shows a triangle of which we know the length of two of its sides, and we want to know the value of the third side.
We also know that the triangle shown is a right triangle because the small box indicates which is the right angle.
The Pythagorean theorem says that in a right triangle the following is true:
$17²=8²+X²$
We use the values of our triangle in the Pythagorean Theorem, and get the following equation:
$17²=8²+X²$
$289=64+ x²$
$289-64=x²$
$225=x²$,$\sqrt{}$
Find the square root:
$15=x$
Answer: $15=x$
### Exercise 3
Given the right triangle $\triangle ABC$
The area of the triangle is equal to $38\operatorname{cm}^2$, $AC=8\text{ cm}$
Find the measure of the leg $BC$
Solution:
We will find the length of $BC$ using the formula for finding the area of right triangles:
$\frac{cateto\times cateto}{2}$
$\frac{AC\cdot BC}{2}=$
$\frac{8\operatorname{cm}\cdot BC}{2}=38\operatorname{cm}^2$
We multiply the equation by the common denominator.
$\times2$
We then divide the equation by the coefficient of $BC$
$BC=\frac{76\operatorname{cm}^2}{8\operatorname{cm}}$
$BC=9.5\text{ cm}$
The length of the leg $BC$ is $9.5$ centimeters.
### Exercise 4
The triangle $\triangle ABC$ is a right triangle
The area of the triangle is equal to $6\operatorname{cm}^2$
Calculate $X$ and the length of side $BC$
Solution:
We will use the formula for calculating the area of the right triangle:
$\frac{cateto\times cateto}{2}=\frac{AC\cdot BC}{2}=$
And we will compare the expression with the area of the triangle. $6 \operatorname{cm}^2$
$\frac{4 cm\cdot(X-1)}{2}=6 \operatorname{cm}^2$
We multiply the equation by $2$
$4 cm(X-1)= 12\operatorname{cm}^2$
We will omit the units to perform the operations.
We open the parentheses using the distributive property:
$4X -4+4=12 +4$
$4X=16$
$X=\frac{16}{4}$
$X=4$
We replace $X=4$ in the expression of $BC$ and find that:
$BC=X-1=4-1=3\text{ }$
$BC=3$
Answer: $X=4\operatorname{cm}$ , $BC=3\operatorname{cm}$
### Exercise 5
Calculate which is larger?
Given the right triangle $\triangle ABC$.
Which angle is larger: $∢B$ or $∢A$?
Solution:
It is given to us that the triangle $\triangle ABC$ is a right trignle with $∢A=90°$ and therefore we know that the last $2$ angles are acute angles.
We know this without needing to calculate the exact value of $∢B$
Answer: $∢A>∢B$
Do you know what the answer is?
### Exercise 6
Given the right triangle $\triangle ABC$.
$∢A=20°$
Is it possible to calculate $∢C$?
If possible, calculate it.
Solution:
Given that $\triangle ABC$ is a right triangle.
$∢B=90°$
$∢A=20°$
The sum of the angles $20°+90°+∢C=180°$
$∢C=70°$
Answer: Yes, $∢C=70°$
### Exercise 7
Determine which of the following triangles is obtuse, which is acute, and which is right triangle:
Solution:
1) We will see if the Pythagorean theorem holds for this triangle:
$5²+8²=9²$
$25+64=81$
$89>81$
The sum of the added squares is greater than the third square, therefore it is an acute triangle.
2) Now we will see this triangle:
$7²+7²=13²$
$49+49=169$
$169>98$
The sum of the added squares is a less than the third square, therefore it is an obtuse triangle.
3) $10.6≈\sqrt{113}$
The largest side of the 3 will be treated as the remainder.
$7²+8²=\sqrt{113}²$
$49+64=113$
$113=113$
The Pythagorean theorem works, and therefore triangle 3 is a right triangle.
A-acute triangle B- obtuse triangle C-right triangle.
### Exercise 8
Let's look at $3$ angles:
Angle $A$ is equal to $30°$
Angle $B$ is equal to $60°$
Angle $C$ is equal to $90°$
Do these angles form a triangle?
Solution:
$30°+60°+90°=180°$
The sum of the angles in the triangle is $180°$
therefore these angles form a triangle.
Yes, since the sum of the interior angles of the triangle is $180°$.
### Exercise 9
Angle $A$ is equal to $90^o$
Angle $B$ is equal to $115 ^o$
Angle $C$ equals $35 ^o$
Do these angles form a triangle?
Solution:
$90^o+115^o+35^o=240^o$
The sum of the angles is greater than $180^o$
therefore these angles do not form a triangle.
No, since the sum of the interor angles must be $180^o$ and in this case the angles equal $240^o$
Do you think you will be able to solve it?
## Review questions
### What are the 7 types of triangles?
There are a variety of triangles. According to their sides and angles, we can list the following types:
• Equilateral triangle
• Scalene triangle
• Isosceles triangle
• Rectangular triangle
• Acute triangle
• Obtuse triangle
• Oblique triangle
### How are triangles classified according to their sides?
The different types of angles can be classified according to their sides or angles, let's see the classification according to the sides:
• Equilateral triangle: All of its sides are equal and therefore its angles are equal.
• Isosceles triangle: It has only two equal sides and two equal angles.
• Scalene triangle: All three sides and angles are different.
### What are the sides of a scalene triangle like?
In a scalene triangle, all the sides have different values, that is, no sides are equal.
### What do isosceles triangles look like?
Isosceles triangles have two equal sides and one different side, which gives them two equal angles.
The above triangle is an isosceles triangle, so we can observe that
$AB=AC$
$\sphericalangle B=\sphericalangle C$
Do you know what the answer is?
### What is the sum of the interior angles of a triangle?
One of the properties of triangles is that the sum of its interior angles must be $180^o$
Example:
Calculate the value of the angle $C$, if we have a triangle whose angles have the following values:
$\sphericalangle A=60^o$
$\sphericalangle B=70^o$
Solution:
We know that the sum of the interior angles of a triangle is $180^o$, therefore:
$\sphericalangle A+\sphericalangle B+\sphericalangle C=180^o$
$60^o+70^o+\sphericalangle C=180^o$
$130^o+\sphericalangle C=180^o$
Therefore:
$\sphericalangle C=180^o-130^o$
$\sphericalangle C=50^o$
$\sphericalangle C=50^o$
If you are interested in learning more about triangles, you can visit one of the following articles:
In Tutorela you will find a variety of articles about mathematics.
## examples with solutions for types of triangles
### Exercise #1
What kind of triangle is given in the drawing?
### Step-by-Step Solution
As all the angles of a triangle are less than 90° and the sum of the angles of a triangle equals 180°:
$70+70+40=180$
The triangle is isosceles.
Isosceles triangle
### Exercise #2
Which kind of triangle is given in the drawing?
### Step-by-Step Solution
As we know that sides AB, BC, and CA are all equal to 6,
All are equal to each other and, therefore, the triangle is equilateral.
Equilateral triangle
### Exercise #3
What kid of triangle is the following
### Step-by-Step Solution
Given that in an obtuse triangle it is enough for one of the angles to be greater than 90°, and in the given triangle we have an angle C greater than 90°,
$C=107$
Furthermore, the sum of the angles of the given triangle is 180 degrees so it is indeed a triangle:
$107+34+39=180$
The triangle is obtuse.
Obtuse Triangle
### Exercise #4
What kind of triangle is given in the drawing?
### Step-by-Step Solution
Given that sides AB and AC are both equal to 9, which means that the legs of the triangle are equal and the base BC is equal to 5,
Therefore, the triangle is isosceles.
Isosceles triangle
### Exercise #5
What kind of triangle is given here?
### Step-by-Step Solution
Since none of the sides have the same length, it is a scalene triangle. | 2,775 | 10,274 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 119, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-30 | latest | en | 0.939245 |
http://www.maplesoft.com/support/help/Maple/view.aspx?path=RandomTools/flavor/nonnegint | 1,502,983,553,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103579.21/warc/CC-MAIN-20170817151157-20170817171157-00699.warc.gz | 612,169,429 | 24,323 | RandomTools Flavor: nonnegint - Maple Programming Help
Home : Support : Online Help : Programming : Random Objects : RandomTools package : Flavors : RandomTools/flavor/nonnegint
RandomTools Flavor: nonnegint
describe a flavor of random non-negative integer
Calling Sequence nonnegint nonnegint(opt)
Parameters
opt - equation of the form range = value; specify option for the random non-negative integer
Description
• The flavor nonnegint describes a random non-negative integer.
To describe a flavor of a random non-negative integer, use either nonnegint or nonnegint(opt) (where opt is described following) as the argument to RandomTools[Generate] or as part of a structured flavor.
• By default, the flavor nonnegint describes a random non-negative integer in the range $0..99999999989$, inclusive.
• You can modify the properties a random non-negative integer by using the nonnegint(opt) form of this flavor. The opt argument can contain the following equation.
range = b
This option describes the right endpoint of the range from which the random integer is chosen. The right endpoint must be of type nonnegint and it describes a random integer in the interval $0..b$, inclusive.
Examples
> $\mathrm{with}\left(\mathrm{RandomTools}\right):$
> $\mathrm{Generate}\left(\mathrm{nonnegint}\right)$
${63715876729}$ (1)
> $\mathrm{Generate}\left(\mathrm{nonnegint}\left(\mathrm{range}=7\right)\right)$
${4}$ (2)
> $\mathrm{Generate}\left(\mathrm{list}\left(\mathrm{nonnegint}\left(\mathrm{range}=10\right),10\right)\right)$
$\left[{5}{,}{1}{,}{10}{,}{3}{,}{5}{,}{4}{,}{10}{,}{0}{,}{7}{,}{4}\right]$ (3)
> $\mathrm{seq}\left(\mathrm{Generate}\left(\mathrm{nonnegint}\right),i=1..10\right)$
${7412421905}{,}{85818212663}{,}{43421525586}{,}{73497475450}{,}{62877304839}{,}{42427536557}{,}{49407854984}{,}{54966684858}{,}{65976255360}{,}{22385044556}$ (4)
> $\mathrm{Matrix}\left(3,3,\mathrm{Generate}\left(\mathrm{nonnegint}\left(\mathrm{range}=7\right)\mathrm{identical}\left(x\right)+\mathrm{nonnegint}\left(\mathrm{range}=7\right),\mathrm{makeproc}=\mathrm{true}\right)\right)$
$\left[\begin{array}{ccc}{5}{}{x}{+}{6}& {3}{}{x}{+}{6}& {2}{}{x}{+}{5}\\ {4}{}{x}{+}{6}& {6}{}{x}{+}{2}& {4}{}{x}{+}{3}\\ {2}{}{x}{+}{6}& {7}{}{x}{+}{1}& {2}{}{x}\end{array}\right]$ (5) | 721 | 2,292 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 13, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2017-34 | latest | en | 0.42164 |
https://socratic.org/questions/583ce3bcb72cff28ad575b7f | 1,575,694,296,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540495263.57/warc/CC-MAIN-20191207032404-20191207060404-00386.warc.gz | 539,427,919 | 6,727 | # Question #75b7f
Nov 30, 2016
60 books were donated.
#### Explanation:
What we are looking for is:
48 is 80% of what?
In mathematics "is" means "$=$".
"Percent" or "%" means "out of 100" or "per 100", Therefore 80% can be written as $\frac{80}{100}$.
When dealing with percents the word "of" means "times" or "to multiply".
And we can call "what" $n$.
So putting this altogether gives:
$48 = \frac{80}{100} \cdot \times n$
Solving for $n$ gives:
$\frac{100}{80} \times 48 = \frac{100}{80} \times \frac{80}{100} \times n$
$\frac{4800}{80} = \frac{8000}{8000} x n$
$60 = n$
Nov 30, 2016
The count of donated books was 60
#### Explanation:
The symbol % is a unite of measurement worth $\frac{1}{100}$
So 80% $\to \frac{80}{100}$
Let the total count of books donated be $x$
Given that the count sold is 48
Given that the percentage sold is 80%
$\frac{80}{100} \times x = 48$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
If you multiply $\frac{80}{100} \text{ by } \frac{100}{80}$ you turn it into 1
So this is:
$\frac{80}{100} \times \frac{100}{80} = \frac{80 \times 100}{100 \times 80} = \frac{100 \times 80}{100 \times 80} = \frac{100}{100} \times \frac{80}{80} = 1 \times 1$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Multiply both sides by $\frac{100}{80}$
$1 \times x = \frac{4800}{80} = 60$
$x = 60$ | 445 | 1,352 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 17, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2019-51 | longest | en | 0.692516 |
http://my.opera.com/jehovajah/blog/2012/04/14/compound-interest-and-the-logarithms?cid=87249792 | 1,386,807,799,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164116508/warc/CC-MAIN-20131204133516-00044-ip-10-33-133-15.ec2.internal.warc.gz | 132,173,505 | 11,141 | # Re: Fractal Foundations of mathematics: Axioms notions and the set FS as a model
http://www.fractalforums.com/index.php?action=profile;u=410;sa=showPosts
## Compound interest and the logarithms
http://en.wikipedia.org/wiki/Compound_interest
http://en.wikipedia.org/wiki/Rate_of_return#Logarithmic_or_continuously_compounded_return
http://en.wikipedia.org/wiki/Definitions_of_the_exponential_function
http://en.wikipedia.org/wiki/Natural_logarithm
http://en.wikipedia.org/wiki/Logarithm
http://www-history.mcs.st-andrews.ac.uk/HistTopics/e.html
http://en.wikipedia.org/wiki/E_(mathematical_constant)
http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=3209&pf=1
http://www.math.tamu.edu/~dallen/history/calc1/calc1.html
Compound is not a new notion. It is used when we Aggregate fine material to form a solid mass. The Binomial Theorem is also not new. Indian vedic siddhantas and temple songs were developed using the Binomial pattern to work out intricate and fine rhythms. This knowledge went to The Chinese through cultural exchange then by the Arabs to Europe. Pascal Picked up the triangle version of the binomial theorem and used it in various ideas.
Descartes and De Fermat's algebraic approach to Geometry leads to the development of equations for curves and a new set of puzzles for the "clever" to solve. Particularly tangent to and area under curves. Curves were treated of by the Greeks and areas were found by mechanical means in certain cases. The tangent problem s were gradually being broadened, and methods of solution becoming more general. Gradually solutions stsrt you involve aspects of the Binomial theorem.
Richard Witt's book Arithmeticall Questions, published in 1613, was a landmark in the history of compound interest. It was wholly devoted to the subject (previously called anatocism), whereas previous writers had usually treated compound interest briefly in just one chapter in a mathematical textbook. Witt's book gave tables based on 10% (the then maximum rate of interest allowable on loans) and on other rates for different purposes, such as the valuation of property leases. Witt was a London mathematical practitioner and his book is notable for its clarity of expression, depth of insight and accuracy of calculation, with 124 worked examples.[3][4]
Big interest was generated in binomial theorem by merchant userers. The development of compound interest explained by using the Binomial Theorem , especially by Richard Witt was of great impotance in financing many trade and exploration ventures. The promised returns and the opportunities offered by the Banks, and the Coffee houses, encouraged the speculation needed to venture in the development of trade and the empire. The down side was that those investors were promised fabulous returns which meant that the trade was exploitative, and deliberately so. Compound interest has that effect on those that generate wealth.
The increased interest Led to lots of tables of binomial expansion, and more interest in the Binomial Theorem. Newton was the first to recognise through the connection to the tangent area problem how to generalise it to the binomia;; series.
There were many tables , gradually increasing the number of years or the number of compounding intervals. Newton realised that if n was just allowed to be very large, the formula for each term could be worked out. He found the formula for each term for any n.
Newton's work on the binomial theorem is nothing short of remarkable. He begins, as did Wallis, by making area computations of the curves , and tabulating the results. He noticed the Pascal triangle and reconstructed the formula
for positive integers n.
Now to get to compute , i.e. n=1/2, he simply applied this relation with n=1/2. This of course generated an infinite series because the terms do not terminate.
Next he generalized to function of the form for any n. This gave him the general binomial theorem - but not a proof.
He was able to determine the power series for by integrating the series for , written according as the binomial series. In modern notation, we have
Now integrate to get the series
From this Newton developed his Fluxions, and from the methods of compounding he develops his method of compounding tangents. Along the way he develops logarithmic series for e and sine.
arkieokogy Saturday, April 14, 2012 4:25:57 AM
Jehovajah,
e to the x is the basis of excalibrator logarhythmic floating cybercurrency
It is a negative version of 100% annual interest compounded daily. It is also fractal in nature. Each day each participant in the economy is given an equal share of 1/365 of the total supply of money. This quantity, known as the Daily Bread, grows at the same rate as the total supply of money. The Daily Bread is thus both a quantity that each person receives daily it also retains a constant ratio to total supply of money.
The Babylonians were aware of this number. Money is the basis of the human rationality which makes it so difficult to understand. For us to see money we must look at it logarhythmicly.
Money, the numerical representation of human desire and human interaction, is the foundation of human abstraction. It is in fact another form of matter.
The modeling of human exchange, the Edgeworth Box, is orbital analysis by another name.
excalibrator is fractal money.
Thanks for the forum.
### Write a comment
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December 2013
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https://neighbourhood.agl.com.au/t5/Energy-Accounts-and-Billing/Hot-Water-Gas-Meter-ITRON-UltraMaXX-HW-recording-higher-usage/m-p/10177 | 1,620,988,675,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990449.41/warc/CC-MAIN-20210514091252-20210514121252-00582.warc.gz | 452,795,385 | 43,331 | Energy Accounts and Billing
Help with your AGL Energy account or AGL online tools.
Help with your AGL Energy account or AGL online tools.
Hot Water Gas Meter ITRON UltraMaXX HW recording higher usage
Conductor
Hi All,
In Feb 2018 Jemena has installed new gas hot water meter for my unit. I live in building with 24 blocks and I am the only person having that meter. Since this meter been installed I am getting very higher usage bill, roughly almost \$400 approx for 90 days. We are only 3 person and hardly use hot water just for shower.
I strongly believe its because of wrong billing which AGL doing. Here is my meter pics with read.
They are using all digits from reading, multiplying by 10, then further multiplying by conversion factor .32. So if my meter reading shows 79.375, they are capturing this bill without decimal 7937.50*10*.32 like this and showing total usage 25400 Usage MJ. And this figure get further multiplying by usage and supply charge @0.029 and giving \$736 which I believe is very high.
Question or Doubt what I have -
1) When my meter reading showing only five digit usage 2 integer digit and 3 decimal digit e.g. 79.375 , refer the link above then why AGL showing that figure in their bill like this 7937.50. Changing from 5 digit to 6 digit and moving decimal position.
2) How do I know whether my meter recording water liter in KiloLiter or just Liter? what this 79.375 number represent. I mean measurement Unit? It shows at the end of number cubic meter? What cubic meter represent and how it should calculated for billing?
3) The usage on meter. Is it water usage or gas usage? what exactly it is as the purpose of this meter is to measure Hot water usage running through GAS.
4) If possible share the details how Digital meter reading should capture and calculated. I already got one link where they are showing how you should read Digital Gas Meter reading but I am more interested how calculation should run.
Regards
Manish
23 REPLIES 23
Conductor
Hi RonR,
You correctly said "They think that if you fill a 10 litre bucket your meter ticks over one unit".
In my case I did 10 liter bucket test and my meter reading moves only 1 digit, very last digit move fast and last second digit only move 1 unit. AGL twice checked with Gemena for my case and both ticket Gemena confirmed there is no reading and calculation problem. Its usage only.
Regards.
Manish
Semiconductor
Thanks at RonR.
Agl can’t make up their their mind on how the meter should be read either.
Whether the bill should show the decimal or not. 2 conflicting responses and keeps asking me to resend read.
Woulx love to know outcome. And I might join the que.
Conductor
I'll hopefully get there in the end!
I live in an apartment - so my water usage is drinking, showering, washing, and dishwasher.
Almost all are at least partly hot water.
According to AGL, I use about as much hot water as 9 people (there are 3 of us). According to Sydney water I use just over 2 people's water................can you see just by that that it does not add up.
If you are believing that your meter works fine - which is what essentially the previous person is saying, and I'm not disagreeing - then the calculation from there to the bill is wrong.
As I said before, I'm paying almost 2cents a litre. An average 8 min shower is about 56 litres of hot water (based on a number of hot water tank makers). So my average shower costs around a dollar.
That means a family of 3 can expect to pay for 3 months at say 4 showers a day, 90 days in a billing cycle, 4x90 - \$360 just for showers. Before dishwashers, washing machines and hot water over the sink etc.
So I get bills between \$550 and \$670 per quarter for 3 people - in an apartment - with no leaks!
I use according to them as I said about 9 peoples hot water - it just does not add up - it is not usage. Or if it is, then the cost is absurdly high.
Switched-on
Hi Ron,
Did you get this sorted out. I'm facing the same problem now after the new meter installed in my property. your reply will definitely help me in this matter.
Conductor
Unfortunately yes - but it still does not make me happy or fully explain things.
I went to the ombudsman and they looked into it. I then had a conversation with a nice woman who looked through my past readings and was able to predict (very closely) the reading at the time of the call.
Therefore they find it hard to believe there is a problem with the meter - that I also agree with.
So it appears we use approximately 3 times as much hot water as the average family of 3 (we are a family of 3). I find that hard to believe, but it seems impossible to prove otherwise.
I also calculated that my hot water cost me almost 2c a litre. This is unfortunately meaningless as they don't look at hot water based on cost per litre. (but they measure litres to calculate it!! That's what the meter records). They told me the factor that was being applied to my meter reading to calculate cost was in the typical range (if at the higher end) so my bill made sense to the Ombudsman.
So that leaves nowhere to go - except to look for a better deal on the consumption.
As I live on the 7th floor of a highrise building, if I run a shower at 6am, it actually costs me about 40c just to get the water hot.
So \$1.00 for a shower is a good expectation - 3 people x \$3 x about 90 days in a billing cycle leaves me \$270 before I use hot water anywhere else. When I lived in a much bigger house my gas bill was always lower and it was gas heated (including two gas open fireplaces) - so it still doesn't add up based on experience, but they can do the numbers and the numbers apparently don't lie.......
Tags (1)
Switched-on
Hi RonR and others,
I’ve noted a few things as per my understanding and experience that I would like to share here:
1. The “new Itron ultramaxx hw electronic meter” for me has a unit of m3 (meter cube) with last three digits grouped. That means 1 m3 = 1000 litres. 1litre of hotwater only tapout will move 1 digit in the meter. (Mine looks like one rn13manish has)
2. I noted if I (family of 3) use Supagas 45Kg bottle it lasted 5 months of hotwater shower+dishwash+laundry (has about 2350 Mj energy) - this was in a HOUSE
3. For apartment unit with hotwater meter, there is a common gas hotwater and “efficiency” of hotwater heat exchanger and vacancy in the apartments affects “pressure factor”. Mine is 3. So I’ve to pay thrice my actual use.
4. To increase my efficiency of use, I set the hotwater mixer under bathroom sink to “max” (DISCLAIMER: only do if you know what you’re doing). Close main hotwater tap and slowly open it to get just enough hotwater for one tap use. To begin cold Morning, before starting shower, I start washbasins hot tap, then open cold tap full for couple seconds, and close cold tap. Hotwater now doesn’t get mixed with cold water and starts quick. I really think your 40c/2c=20 litres must come down to say 2-5 if you want to increase your efficiency.
5. If this trick doesn’t work for you, it will be worth to get Tempering valve checked/replaced. This is a safety device to limit “hotness” of water out of “hot” tap, even when cold tap is totally off. They generally have a life of a few years. Please google for more info on tempering valves, e.g. how they work and which one you have.
6. 56 litres of pure hotwater is too much. Even 60 deg hotwater x 56 litres is too much of shower. Try a 20/30 litre bucketful 55deg C hotwater and a mug to take bath one day to actually see your “offline” use. 20 litres warm water should be more than enough, but I’m trying to think with liberty here with 30litres “pure online” hotwater. My (one person) daily shower use is always exactly between 10-11litre. (I turn off shower when applying soap).
I would love to hear your thoughts,
Cheers.
Conductor
Hi - exactly the same problem here - eye watering readings and bills, that remain otherwise unexplained. I am wondering if there has been any resolution? It seems like AGL is the only provider that registers an issue from a web search - I am wondering if other providers have priced the usage for hot water correctly, whereas AGL does not?
Tags (3)
AGL Moderator
Hi Uma!
Welcome to the AGL Community!
Kind regards,
-Nam-
Conductor
Regarding my excessive bills......
The newer meter seemed to produce high numbers - this is just about decimal points it seems.
It can also display different units.....it's all designed to be user unfriendly!
But it seems my usage was always correct. What had happened to me is that I signed up on a deal, and it expired around the same time as I got the new meter - hence the confusion - big bills and a new meter with a different set of values - leading to total confusion.
No-one ever explained this. Including the ombudsman - who only stated that the "factor" or whatever they call the mystery number in the bill calculation was on the high side but within an acceptable range.
Well, finally getting that message, I went to another provider who has almost halved my bill.
The minute I tried to cancel my AGL account someone offered me a large discount - which I refused.
Clearly they could have always done this - but they dug in instead of trying to keep a customer.
The moral is that if your bill is high, and your meter checks out OK, get a competitive quote - as I say, almost a 50% saving.
Tags (1)
Conductor
thanks Ron. My issue is that my gas usage for hot water and a stove top (no heating is a massive 177MJ a day, when a standard 2 bedroom home should use 25MJ a day. The problem is with the stated usage, not the pricing on the plan. Despite the ridiculous level of stated usage, I have been encouraged to take less showers - really?
Can't find what you're looking for? | 2,336 | 9,820 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-21 | latest | en | 0.918622 |
https://community.anaplan.com/discussion/69726/lm2-data-hub-sys13-formula-hint-2 | 1,723,322,214,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640822309.61/warc/CC-MAIN-20240810190707-20240810220707-00738.warc.gz | 144,755,199 | 85,080 | # LM2 - Data Hub SYS13 - Formula Hint 2
Options
Hi Team,
In the Data Hub in module SYS13 Account SKU Filter you setup a Boolean line item to determine whether there is data in DAT03
My formula to be used would of been: TIMESUM(DAT03) > 0 or TIMESUM(DAT03) <> 0
The 2nd Anaplan formula hint is below:
This is confusing as when/why would you use a Number (1/2) to determine the TRUE/FALSE nature rather than use the native Boolean logic of 0 & 1?
Thanks
Mark
• Options
Hi @MarkTurkenburg
I feel there is a misunderstanding here. What are your concerns? I can see 3 concerns here: (correct me if I'm wrong)
1. There is a 1 / 2 in the THEN ELSE parts of the IF statement?
2. A TRUE FALSE should have been used instead
3. Native Boolean Logic of 0 &1
Let me start with the number 3. 0 and 1 and are not native Boolean logic. TRUE/FALSE are.
In the 3 parts of an IF THEN ELSE statement, the only part that should resolve to TRUE or FALSE is the part after the IF. However, the THEN and ELSE sections could resolve to ANYTHING -including Boolean.
The Example provided literally translates to .... IF the condition of x not equal 1000 is met THEN resolve to or enter the value of 1, and if the condition is not met enter the value 2.
an IF statement could be used in a line item of any format, in the example provided that format is NUMBER so whatever comes after the THEN and ELSE should resolve to NUMBER.
• Options
Thanks @einas.ibrahim
I think I wrote the post in haste as I was confused as to my formula was not approved as correct in the LM2 but then looking about at the 2nd post the example provided is not actually for the question & hence they were not using 1/2 as the answer
• Options
Hi, and that formula to be used would be TIMESUM('DAT03 Historic Volumes'.Volumes) <> 0
• Options
This is the formula I thought should work but when I try it I get an error saying "Invalid formula for 'SYS13 Account SKU Filter'.Has Data?: Automatic sum of 'DAT03 Historic Volumes'.Volumes over SKU Flat hierarchy is not possible as SKU Flat does not have a built-in top level"
My DAT03 Historic Volumes module uses the SKU Flat list as instructed in the course. How can I solve this error?
• Options
Hello @aaturnbu
Can you please share the instruction to use P3 SKU Flat in the historical volume module? I don't believe using the flat with this module is correct, and that's the reason for the error you are getting.
• Options
Here's the screenshot from the course:
It clearly says to use the flat list.
I found the problem though, I was not using the flat list in SYS13, I had selected the P3 SKU hierarchical list by mistake. It works now that both modules are using the flat list.
• Options
not enough 😞 if then else is missing for boolean format
IF TIMESUM('DAT03 Historic Volumes'.Volumes) <> 0 THEN TRUE ELSE FALSE
• Options
This is what I had, on the exam I got the question wrong.
• Options
yes but you wrote Historic instead of Historical | 734 | 2,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-33 | latest | en | 0.921351 |
https://rdrr.io/cran/LearnBayes/man/bayesresiduals.html | 1,582,804,027,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146681.47/warc/CC-MAIN-20200227094720-20200227124720-00553.warc.gz | 494,954,461 | 14,068 | # bayesresiduals: Computation of posterior residual outlying probabilities for... In LearnBayes: Functions for Learning Bayesian Inference
## Description
Computes the posterior probabilities that Bayesian residuals exceed a cutoff value for a linear regression model with a noninformative prior
## Usage
`1` ```bayesresiduals(lmfit,post,k) ```
## Arguments
`lmfit` output of the regression function lm `post` list with components beta, matrix of simulated draws of regression parameter, and sigma, vector of simulated draws of sampling standard deviation `k` cut-off value that defines an outlier
## Value
vector of posterior outlying probabilities
Jim Albert
## Examples
```1 2 3 4 5 6 7 8``` ```chirps=c(20,16.0,19.8,18.4,17.1,15.5,14.7,17.1,15.4,16.2,15,17.2,16,17,14.1) temp=c(88.6,71.6,93.3,84.3,80.6,75.2,69.7,82,69.4,83.3,78.6,82.6,80.6,83.5,76.3) X=cbind(1,chirps) lmfit=lm(temp~X) m=1000 post=blinreg(temp,X,m) k=2 bayesresiduals(lmfit,post,k) ```
### Example output
``` 1 2 3 4 5 6
8.650461e-03 1.890880e-01 1.734943e-02 9.350177e-06 3.873963e-11 3.108944e-08
7 8 9 10 11 12
1.527867e-02 2.049203e-12 2.495306e-01 1.725156e-02 8.974623e-03 3.655172e-11
13 14 15
1.560055e-05 1.102365e-06 5.687012e-02
```
LearnBayes documentation built on May 1, 2019, 7:03 p.m. | 521 | 1,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-10 | latest | en | 0.489272 |
https://en.wikipedia.org/wiki/Talk:Diffuse_reflection | 1,493,306,170,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122174.32/warc/CC-MAIN-20170423031202-00026-ip-10-145-167-34.ec2.internal.warc.gz | 781,402,101 | 12,998 | # Talk:Diffuse reflection
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## Lambertian reflectance
In the article it says: "the reflected light will be evenly spread over the hemisphere surrounding the surface (2π steradians)."
Isn't the reflected light scattered according to uniformly distributed surface normal angle and thus the reflected photons are NOT equally distributed over the hemisphere surface? 80.221.29.165 (talk) 22:05, 9 October 2008 (UTC)
Not necessarily. See Lambertian reflectance and Lambert's cosine law. An ideal diffuse scatterer scatters light equally in all directions, without regard to the angle from the surface normal.--Srleffler (talk) 22:36, 9 October 2008 (UTC)
My point is, that if you shoot or reflect photons into random direction, equally in all directions, then the resulting distribution of photons in the hemisphere (photons/m^2) will not be evenly spread as the article says, but top of the hemisphere will be more densely populated than the sides. With ideal diffuse reflector, photons in a solid angle decreases as the cosine of an angle with the surface normal.80.221.29.165 (talk) 11:15, 10 October 2008 (UTC)
OK, you're right, and the article is poorly worded. What they were probably thinking of is that while the intensity (photons per steradian) falls off as cos(θ), the apparent brightness of the surface to an observer remains constant because the apparent area of the surface also falls off as cos(θ). More technically, the radiant intensity for an ideal Lambertian reflector is proportional to cos(θ), while the radiance is independent of θ. This article should probably describe Lambertian surfaces explicitly. They are a pretty important special case. Note that not all surfaces are Lambertian, however. In particular, for some surfaces the intensity falls off much slower than cos(θ).--Srleffler (talk) 15:52, 10 October 2008 (UTC)
## Proposed merge
I oppose the proposed merge with Scattering from rough surfaces. That article is on charged particle scattering. This article is on light. The subject matter is sufficiently distinct for two articles.--Srleffler (talk) 06:01, 13 July 2009 (UTC)
I also oppose it; and the editor who put the tags didn't even bother to put a rationale or start a discussion, so I'd say let's just take them away already. Dicklyon (talk) 02:58, 14 July 2009 (UTC)
## Light scattering
Editors of this article might be interested in the new article (possible content fork) Light scattering. It's possible that the two articles could be merged, although there may be merit in keeping them separate. Discussion at Talk:Light scattering.--Srleffler (talk) 16:53, 27 July 2009 (UTC)
It looks a lot broader than diffuse reflection. I don't like to see articles spring to life fully formed like that, and without URLs for the refs, but that's what we got. Dicklyon (talk) 06:12, 28 July 2009 (UTC)
Discussion is at Talk:Light scattering.--Srleffler (talk) 06:11, 29 July 2009 (UTC)
## Merge with Lambertian reflectance?
This article seems to cover very similar material to Lambertian reflectance to the point that they both use the same intro image. The difference seems to be in emphasis: this article focuses on the physics whereas Lambertian reflectance is primarily about modeling it in computer graphics. The fact that they are different articles seems to imply that they aren't the same thing. —Ben FrantzDale (talk) 13:41, 24 February 2011 (UTC)
Actually they are quite different things: diffuse reflection is the general phenomenon shown by almost any surface we see. Lambert's cosine law is the idealized angular behaviour we expect to find in an ideal, homegeneous, "featureless" surface, for example a white surface with very small, randomly oriented irregularities. Many (or most) diffusing objects are not Lambertian: see, for example, Oren–Nayar reflectance model, or, for celestial bodies, Geometric albedo. The figure here represents just an example (and includes also specular reflection). --GianniG46 (talk) 00:12, 25 February 2011 (UTC)
I agree with Gianni. This is an important point: one cannot assume all diffuse reflectors are Lambertian. It's an idealized model.--Srleffler (talk) 04:44, 25 February 2011 (UTC)
## Who the hell wrote this?
What is the exact mechanism of diffuse reflection? What happens to a photon when it strikes a surface? What causes the photon to reverse direction? Is it the electron, is it a force, is the nucleus?
Welcome to Wikipedia, the encyclopedia that anyone can edit. This article is written by many volunteers working collaboaratively. If you don't like the way an article is written, you are welcome to edit it and try to improve the style or content.
I think the problem is that you are looking for more details on the mechanism of reflection itself, rather than specifically the mechanism for diffuse reflection. Check out Reflection (physics) and see if it answers some of your questions. Particularly the section Mechanism. This article seems to me to provide pretty good information on the mechanism for diffuse reflection, presuming one already knows how reflection works.
Regarding the "aimed at children" comment: keep in mind that this is an encyclopedia, not a physics textbook. For topics that are not very specialized, a good article will have an introduction that is aimed at about the level of a smart 11 year old. A good article will cover topics of general interest first, and then move into more difficult or technical material further down in the body of the article, if necessary. Obviously, some topics demand a more advanced level of exposition than others. --Srleffler (talk) 16:28, 19 August 2012 (UTC)
## Vector Length in Figure 1
In the first paragraph it is said " An illuminated ideal diffuse reflecting surface will have ***equal luminance from all directions*** in the hemisphere surrounding the surface (...)". However, looking at Figure 1, the red arrows (that indicate diffuse reflection) have varying lengths. It can be confusing reading that and then seeing varying length vectors in Figure 1. I suggest to edit the figure in order to explain what the length of the red vectors mean (I personally do not do that because I am also confused .. :). Capagot (talk) 18:46, 10 October 2012 (UTC)
Done. The rays represent luminous intensity rather than luminance. --Srleffler (talk) 04:42, 11 October 2012 (UTC)
## Make it Clear the Light is not "Bouncing"
The wording of this article is technically correct, but does not make it sufficiently clear that photons involved in diffuse reflection are being absorbed and re-emitted. Many pop descriptions of diffuse reflection say incorrectly that light "bounces" off atoms. This is not true. All light involved in diffuse reflection is absorbed and (possibly) re-emitted. The non-absorbed light is scattered by either Thomson Scattering or Resonant Scattered. Either way it is absorbed and re-emitted. It does not "bounce" off the atoms. Thus, for a normal opaque, colored material, like an orange or a leaf, the light we see from the object is composed of newly-created photons, not the photons that originally impinged on the object to illuminate it. This should be made perfectly clear in the article so that there is no confusion about this. John Chamberlain (talk) 10:14, 10 April 2017 (UTC) | 1,752 | 7,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-17 | latest | en | 0.954955 |
https://questions.examside.com/past-years/year-wise/gate/gate-ece/gate-ece-1988?lang=en&redirect=true | 1,721,858,312,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518454.54/warc/CC-MAIN-20240724202030-20240724232030-00138.warc.gz | 416,528,247 | 22,802 | GATE ECE 1988
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## GATE ECE
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NEET | 1,339 | 5,316 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-30 | latest | en | 0.805596 |
https://www.notion.so/77ca1edd9099463594cfae4350082e56 | 1,686,032,887,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652235.2/warc/CC-MAIN-20230606045924-20230606075924-00710.warc.gz | 1,006,727,681 | 3,318 | # Degree
## Definitions
### Regular Graph
If a graph in which every vertex has degree k, for some fixed k, is called a k-regular graph (or a regular graph)
## Theorems
### Handshaking Lemma (Degree-Sum Formula)
$$\underset{v \in V(G)}{\Sigma} deg(v) = 2|E(G)|$$
### The average degree of a vertex in the graph G
$$\frac{2|E(G)|}{|V(G)|}$$
### Corollary
The number of vertices of odd degree in a graph is even.
# Bipartite Graphs
## Definitions
### Bipartite
A graph in which the vertices can be partitioned into two sets A and B, so that all edges join a vertex in A to a vertex in B, is called a bipartite graph, with bipartition (A, B).
## Lemma
• An odd cycle is not bipartite | 205 | 695 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-23 | latest | en | 0.858091 |
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#### Resources tagged with Manipulating algebraic expressions/formulae similar to Crossed Ends:
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Broad Topics > Algebra > Manipulating algebraic expressions/formulae
### Crossed Ends
##### Stage: 3 Challenge Level:
Crosses can be drawn on number grids of various sizes. What do you notice when you add opposite ends?
### Cubes Within Cubes Revisited
##### Stage: 3 Challenge Level:
Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?
### Partitioning Revisited
##### Stage: 3 Challenge Level:
We can show that (x + 1)² = x² + 2x + 1 by considering the area of an (x + 1) by (x + 1) square. Show in a similar way that (x + 2)² = x² + 4x + 4
##### Stage: 3 Challenge Level:
Think of a number and follow my instructions. Tell me your answer, and I'll tell you what you started with! Can you explain how I know?
### Perimeter Expressions
##### Stage: 3 Challenge Level:
Create some shapes by combining two or more rectangles. What can you say about the areas and perimeters of the shapes you can make?
##### Stage: 3 Challenge Level:
List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?
### Multiplication Square
##### Stage: 3 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Perfectly Square
##### Stage: 4 Challenge Level:
The sums of the squares of three related numbers is also a perfect square - can you explain why?
### Pair Products
##### Stage: 4 Challenge Level:
Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice?
### Matchless
##### Stage: 3 and 4 Challenge Level:
There is a particular value of x, and a value of y to go with it, which make all five expressions equal in value, can you find that x, y pair ?
### Sitting Pretty
##### Stage: 4 Challenge Level:
A circle of radius r touches two sides of a right angled triangle, sides x and y, and has its centre on the hypotenuse. Can you prove the formula linking x, y and r?
### System Speak
##### Stage: 4 and 5 Challenge Level:
Solve the system of equations: ab = 1 bc = 2 cd = 3 de = 4 ea = 6
### Nicely Similar
##### Stage: 4 Challenge Level:
If the hypotenuse (base) length is 100cm and if an extra line splits the base into 36cm and 64cm parts, what were the side lengths for the original right-angled triangle?
### Simplifying Doughnut
##### Stage: 4 and 5 Challenge Level:
An algebra task which depends on members of the group noticing the needs of others and responding.
### ' Tis Whole
##### Stage: 4 and 5 Challenge Level:
Take a few whole numbers away from a triangle number. If you know the mean of the remaining numbers can you find the triangle number and which numbers were removed?
### Algebra from Geometry
##### Stage: 3 and 4 Challenge Level:
Account of an investigation which starts from the area of an annulus and leads to the formula for the difference of two squares.
### Salinon
##### Stage: 4 Challenge Level:
This shape comprises four semi-circles. What is the relationship between the area of the shaded region and the area of the circle on AB as diameter?
### Algebra Match
##### Stage: 3 and 4 Challenge Level:
A task which depends on members of the group noticing the needs of others and responding.
### Janine's Conjecture
##### Stage: 4 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Sums of Pairs
##### Stage: 3 and 4 Challenge Level:
Jo has three numbers which she adds together in pairs. When she does this she has three different totals: 11, 17 and 22 What are the three numbers Jo had to start with?”
### Magic Sums and Products
##### Stage: 3 and 4
How to build your own magic squares.
### Orbiting Billiard Balls
##### Stage: 4 Challenge Level:
What angle is needed for a ball to do a circuit of the billiard table and then pass through its original position?
### Always Perfect
##### Stage: 4 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
### Root to Poly
##### Stage: 4 Challenge Level:
Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.
##### Stage: 4 Challenge Level:
If a sum invested gains 10% each year how long before it has doubled its value?
### There and Back
##### Stage: 4 Challenge Level:
Brian swims at twice the speed that a river is flowing, downstream from one moored boat to another and back again, taking 12 minutes altogether. How long would it have taken him in still water?
### Chocolate 2010
##### Stage: 4 Challenge Level:
First of all, pick the number of times a week that you would like to eat chocolate. Multiply this number by 2...
### Back to Basics
##### Stage: 4 Challenge Level:
Find b where 3723(base 10) = 123(base b).
### Really Mr. Bond
##### Stage: 4 Challenge Level:
115^2 = (110 x 120) + 25, that is 13225 895^2 = (890 x 900) + 25, that is 801025 Can you explain what is happening and generalise?
### DOTS Division
##### Stage: 4 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Never Prime
##### Stage: 4 Challenge Level:
If a two digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger, prove the difference can never be prime.
### Cosines Rule
##### Stage: 4 Challenge Level:
Three points A, B and C lie in this order on a line, and P is any point in the plane. Use the Cosine Rule to prove the following statement.
##### Stage: 4 Challenge Level:
Robert noticed some interesting patterns when he highlighted square numbers in a spreadsheet. Can you prove that the patterns will continue?
### The Medieval Octagon
##### Stage: 4 Challenge Level:
Medieval stonemasons used a method to construct octagons using ruler and compasses... Is the octagon regular? Proof please.
### Temperature
##### Stage: 3 Challenge Level:
Water freezes at 0°Celsius (32°Fahrenheit) and boils at 100°C (212°Fahrenheit). Is there a temperature at which Celsius and Fahrenheit readings are the same?
### Triangles Within Squares
##### Stage: 4 Challenge Level:
Can you find a rule which relates triangular numbers to square numbers?
### Always Two
##### Stage: 4 and 5 Challenge Level:
Find all the triples of numbers a, b, c such that each one of them plus the product of the other two is always 2.
### Consecutive Squares
##### Stage: 4 Challenge Level:
The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false?
### Unusual Long Division - Square Roots Before Calculators
##### Stage: 4 Challenge Level:
However did we manage before calculators? Is there an efficient way to do a square root if you have to do the work yourself?
### Lap Times
##### Stage: 4 Challenge Level:
Two cyclists, practising on a track, pass each other at the starting line and go at constant speeds... Can you find lap times that are such that the cyclists will meet exactly half way round the. . . .
### Garfield's Proof
##### Stage: 4 Challenge Level:
Rotate a copy of the trapezium about the centre of the longest side of the blue triangle to make a square. Find the area of the square and then derive a formula for the area of the trapezium. | 1,894 | 7,938 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2015-27 | longest | en | 0.861358 |
https://assignmentgrade.com/question/641098/ | 1,611,618,776,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704792131.69/warc/CC-MAIN-20210125220722-20210126010722-00076.warc.gz | 225,975,766 | 6,904 | All of the following involve waves of electromagnetic energy except A. watching videos on a computer screen B. radios receiving signals through air C. the rumble of thunder during a storm D. sunlight shining on a plant leaf - AssignmentGrade.com
# All of the following involve waves of electromagnetic energy except A. watching videos on a computer screen B. radios receiving signals through air C. the rumble of thunder during a storm D. sunlight shining on a plant leaf
QUESTION POSTED AT 18/04/2020 - 07:16 PM
C involves sound waves not electromagnetic waves (light)
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QUESTION POSTED AT 01/06/2020 - 01:51 PM | 1,792 | 6,987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-04 | latest | en | 0.852284 |
https://avenuewebmedia.com/concave-polygon-cnjvyt/polygons-in-the-coordinate-plane-quizlet-a02aee | 1,618,175,012,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038065492.15/warc/CC-MAIN-20210411204008-20210411234008-00061.warc.gz | 228,821,621 | 8,626 | x9i���Lh�+R#��|МW���Bc vocabulary chapters 1-5 Learn with flashcards, games, and more — for free. This quiz … Topic 9-1 Polygons in the Coordinate Plane Day 2 Class Notes here. 3 eaton drew a parallelogram on a coordinate planeK Two vertices of the parallelogram were located at (1, 1) and (1, 7) The area of the parallelogram is 18 square units Tell whether each statement is True or False. Grade 6 Quick Check Lesson 14.2: Polygons in the Coordinate Plane. by dpalsgrove. \$1.00. perimeter. Practice. In Lesson 9-1 of the textbook on pgs 389-391, do: 1) Find the area of triangle PQR in question 9, 2) Find the areas of triangles ABC and DEF in question 11. 403: 15, 20-25, 28-29, 35. �dB\$!q@ ��3:ƺ�Z��Q�m��⭟'�����cޖ����rN�6z5��'�F����Z\�'��z. nonagon. Finish Editing. B`*�R dg���KN5��Y\�o,95�/� T��,��1�bi� TdI��s\$��b 84% average accuracy. 1) In Lesson 9-1 of the textbook on pgs 389-391, do 18, 20, 21. Live Game Live. In Lesson 9-1 of the textbook on pgs 389-391, do 18, 20, 21. 6th grade . x�\�r�8�����f��͈8� ���ʱ���q�=ݳ���bm+�G�;�Km�����D�E����� ��O��0"S"RZ� �2a�Jr��Qm���*�'�x��R�L[!��*2���L�Y. Topic 9-2 & 9-3 Day 3 - Equation of a Circle Part 2. The x-coordinate of the other two vertices of the parallelogram … decagon. ��rJp�H��9��&���\$� �2`9�N�c�� �S[�9H���Cx9��� ���L`�6���F@=�� ���3TKJ=G�z�iIr�%�-�G��9wn9G�٭�:�@��!�ɠb�5d�&%R�3Dw\3��,+�?�Pr�aҒ��Զ�� 0��OH-pM�9\;"Ϥ�x*�ɿ�ؔ�]�NJn FҔ_��i��8�%_���3B5>d�P��&�����(ҙ��Hu,�Ú:�tό�ڪ�P������@�p t" U�_%�����X�Bi�8�� 3/25 9-3 Circles in the Coordinate Plane Pg. 582. 14. Edit. been assigned), Topic 9-1: Polygons in the Coordinate Plane Class Notes here. mR�u+��XR���Vc��p�h[�]�� 2˄!H�� R��`��0 1�H[�{�-ǾR�NS�s ��� �)>QX\$�Ry!��ј4�)~L-@�{La��K�E" ��x�%2R�E N/A. Topic 9-2 & 9-3 Day 2 - An Analytic Geometry Proof Class Notes here. Textbook Authors: Charles, Randall I., ISBN-10: 0133281159, ISBN-13: 978-0-13328-115-6, Publisher: Prentice Hall Digital Download. �T��� Name three points that are collinear. Consider using notes here and homework here. Features. #p+#d�k A polygon consists of more than two line segments ordered in a clockwise or anti-clockwise fashion. In Lesson 9-1 of the textbook on pgs 385-388, Find the areas of triangles ABC and DEF in question 11. Not Included. an 8 sided polygon. This quiz is … 9th - 10th grade . 3) Find the area of quadrilateral JKLM in question 4. Reasoning GEOMETRY - UNIT 7 CIRCLES Flashcards | Quizlet Homepage - eMathInstruction Match Fishtank - 6th Grade Math - Unit 7: Geometry - Lesson 1 Common Core Geometry.Unit #7.Lesson #3.Dilations and Angles Match Fishtank - Geometry - ... in the coordinate plane. Solo Practice. Find the area of quadrilateral JKLM in question 4. Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side joining points with the same first coordinate or the same second coordinate. Triangle Area Calculator using Coordinates here, Topic 6-1 Polygon Angles Class Notes here, In Lesson 6-1 of the textbook on pgs 250-251, do 17-22, 25, 29, Topic 9-2 & 9-3 Day 1 - Equation of a Circle Class Notes here. 2. Description (a yellow highlight means that this has This quiz is incomplete! Total Pages. Did Example 5 on page 388 in class. Rs�d��F�S59��H a 9 sided polygon. Students also plot distances on the coordinate plane and use reasoning to plot the vertices of parallelograms and triangles on the coordinate plane. 2) In Lesson 9-1 of the textbook on pgs 389-391, do 9, 22-26, 28, Topic 9-1 Polygons in the Coordinate Plane Day 3 Class Notes here. 264 esson 23 Polygons in the Coordinate Plane ©urriculum Associates opying is not permitted Solve. Edit. Quizlet … 3 years ago. 1 page. 1) Spend 10 minutes working on the question asked at the end of class notes here. 2) In Lesson 9-1 of the textbook on pgs 389-391, do 9, 22-26, 28. a. Share practice link. Print; Share; Edit; Delete; Host a game. Add one to cart. %PDF-1.5 0. Then, we find the perimeter and area of polygons in the coordinate plane. Edit. 1) Spend 10 minutes working on the question asked at the end of class notes here. Do the odd questions here. We’ll stop supporting this browser soon. 6.7 Polygons in the Coordinate Plane Flashcards | Quizlet Start studying 6.7 Polygons in the Coordinate Plane. Similarity of polygons and triangles are explored. ,�J��Z`]&�X\$dm�,,�`�Đq nr�D�Ƭ2��S� Finish Editing. Played 132 times. Topic 9-1 Polygons in the Coordinate Plane Day 3 Class Notes here Equation of a Circle – Day 4 Class Notes.pdf, Help for Equation of a Circle Homework.pdf, Topic 9 Coordinate Geometry Overview Class Notes.pdf, Topic 9-1 Polygons in the Coordinate Plane Day 2 Class Notes.pdf, Topic 9-1 Polygons in the Coordinate Plane Day 3 Class Notes.pdf, Topic 9-2 and 9-3 – Day 2 Class Notes - An Analytic Geometry Proof.pdf. 3/21 9-1 Polygons in the Coordinate Plane Pg. To play this quiz, please finish editing it. ��P))�T� *Kahoot - Polygons, Circles, 3D Figures *Review of Benchmark Test - Polygons, Circles, 3D Figures ... *Lesson - Circles on the Coordinate Plane *Quizlet Live! Click here for help. Apply these techniques in the context of solving real-world and mathematical problems. Learn vocabulary, terms, and more with flashcards, games, and … 0. To play this quiz, please finish editing it. Report this resource to let us know if this resource violates TpT’s content guidelines. In Lesson 9-1 of the textbook on pgs 389-391, In Lesson 6-1 of the textbook on pgs 250-251, do 17, 19-22. Triangle similarity postulates and theorems … Report this Resource to TpT. 0. PLAY. STUDY. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Delete Quiz. Geometry: Common Core (15th Edition) answers to Chapter 6 - Polygons and Quadrilaterals - 6-2 Properties of Parallelograms - Practice and Problem-Solving Exercises - Page 364 15 including work step by step written by community members like you. 9 months ago. Play. ... Quizlet. octagon. Concepts & Videos: Hinge Theorem + Converse; Notes & Practice: Teaching Duration. 3/27 9.1-9.4 9.1 and 9.3 Project. been assigned. POLYGONS IN THE COORDINATE PLANE. 0. … the sum of all of the sides of a polygon. Share practice link. This quiz is incomplete! <> Edit. The final module includes an interactive coordinate grapher which is put to work to allow students to explore symmetry on the coordinate plane, including reflecting points across the x-axis, y-axis, and both axes. Solo Practice. Point M is the bisector of ̅̅̅̅. Save. stream Check if a point is on the right or on the left of a line segment . The usual book, fiction, history, novel, scientific Page 1/22. If a point lies left (or right) of all the edges of a polygon whose edges are in anticlockwise (or clockwise) direction then we can say that the point is completely inside the polygon. Answer Key. Reported resources will be reviewed by our team. D� Bɤ�� ����SHInJ��[�ᆀ��Bpc)X�4���B��r�����` �)%IqÖȝ:I��j�@�)=������;��}�-9a��\$ �,)?��KJP"�\�P�f�ܻ��j�Ѐ %���� Live Game Live. vertex (vertices) the point where two sides meet. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Here we find are reminded of polygonal names. Click here for help. Start studying Polygons in the Coordinate Plane. Print; Share; Edit; Delete; Host a game. Mathematics. �h����5�X X`�5�Ca�d��S + Play. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Delete Quiz. Polygons in the coordinate Plane DRAFT. ... *Homework: complete coordinate plane worksheet: 7 *Review coordinate plane worksheet *Lesson - Points, Lines & Planes *begin worksheet in class *Homework: complete worksheet: 8 *review homework … 403: 5-12, 26-27, 33, 34. 2) In Lesson 9-1 of the textbook on pgs 389-391, do 22, 23, 28. 1) Do the even questions here. Drawing polygons with coordinates. Apply these techniques in the context of solving real-world and mathematical problems. 73% average accuracy. 3) In Lesson 9-1 of the textbook on pgs 389-391, find the areas of triangle PQR in question 9 and quadrilateral JKLM in question 4. Buy licenses to … Polygons in the Coordinate Plane DRAFT. Played 86 times. a closed plane figure that has three or more sides and meets at endpoints. For the best experience please update your browser. Our little sister amazon prime. 3/22 9-3 Notes Circles in the Coordinate Plane Pg. Problem 1: Classify the triangle shown in the figure below. Finding Area of a Regular Polygon (given apothem) Developing the formula for the Area of a Circle; ... Quizlet Assignments: 10.4 - perimeter & area in the coordinate plane. Quizlet Live. Save. CCSS.Math.Content.6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side joining points with the same first coordinate or the same second coordinate. a 10 sided polygon. 3/26 9.1, 9.3 9.1 and 9.3 Project 9.1 and 9.3 Worksheet and Key. What is Common Core Math really all about ? Practice. 1 0 obj %HG%�h}�h\�%R��T�����iŰKE����m�)aJ9�����Ue����4��ܸ�հrv2Mi`��rO���(U�t6��V���Fl�ߥ��A9/2�w�4qߑ�9��v4!6��F���&� �qo����{�W�I��qcR�s8�|���W�H�������\$Bq�@B"������M">�~�H������m5�X�ﮗ'����btY}[^��F���� ����h0nd�9��0�#����ֆ������r�g?V�����������o{q[� �:��"p����FV���ed'�v��~V���X~�D�2��@��؉Ui��[DI:���O��ʹ��X\��jV��ƚn�t���J���`�4����:���>hZ-Hw ag���F���_������rz�`tH��kJr Start studying Honors Geometry B Unit 1 Lesson 7: Polygons in the Coordinate Plane. by bbraia. �t�pc>5�����0^�)A� Choose from 500 different sets of test practice 4 3 chapter 1 geometry flashcards on Quizlet. K& polygon. Essential Question: How do we reflect geometric figures in a coordinate plane? Start studying Unit 5 - Polygons and the Coordinate Plane. ... a 7 sided polygon. Obtuse One of the angles is greater than 90° Right One of the angles is equal to 90° Scalene All three sides are of different length Isosceles Two sides are of same length Equilateral All three sides are of same length. 4) In Lesson 6-1 of the textbook on pgs 250-251, do 17, 19-22, 5) On the worksheet here, do 2, 3, 5, 7, 9, 11, 13, Sign in|Recent Site Activity|Report Abuse|Print Page|Powered By Google Sites, a yellow highlight means that this has Polygons in the Coordinate Plane. 1) In Lesson 9-1 of the textbook on pgs 385-388, learn Examples 1-4, 2) In Lesson 9-1 of the textbook on pgs 389-391, do 14-21, Topic 9-1 Polygons in the Coordinate Plane Day 2 Class Notes here. 390: 10, 21-22, 28-30. 2) In Lesson 9-2 of the textbook on pgs 398, do 15, 16, 17, Topic 9-2 & 9-3 Day 3 - Equation of a Circle Part 2 Class Notes here. Standard: MGSE9–12.G.CO.2 Represent transformations in the plane using, e.g., transparencies and geometry software; Compare transformations that preserve distance and angle to those that do not (e.g., translation versus horizontal stretch). Ad. Homework. ʿ����]%ĵ�x2Nl8���w�y��1uwD_f9���hI�FZ��q���9�Ih �Qd��- W�r� �/I)�~�(E��*�;�=���č�ʹS���dO\$�ʸs�R�s8�� �9�e�� �%ףq�9��_�-��PQjC�f�D Learn vocabulary, terms, and more with flashcards, games, and other study … Polygons in the Coordinate Plane Flash Cards Flashcards | Quizlet Start studying Polygons in the Coordinate Plane Flash Cards. Mathematics. 3. Downloads are available in dozens of formats, including EPUB, MOBI, and PDF, and each story … 16. Homework. Area and perimeter on the coordinate plane. 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Super Mario 64 Slide Level, All For The Love Of Dancing Syracuse, Ho Foods Hours, Mr Kat Scp, How To Enable Enter Key In Google Keyboard, Bike Fork Oil Seal Price, Wal Habibi Song Lyrics, Garden Game Pc, Space Engineers Battlecruiser, | 5,953 | 20,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-17 | latest | en | 0.606136 |
http://mathhelpforum.com/advanced-applied-math/180135-ivp-using-leibniz-formula.html | 1,524,508,731,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946120.31/warc/CC-MAIN-20180423164921-20180423184921-00104.warc.gz | 197,030,409 | 10,253 | # Thread: IVP using Leibniz formula
1. ## IVP using Leibniz formula
How does one formulate this integral equation as an initial value problem?
u(t)=1 + integral from 0 to t (s ln (s/t) u(s) ds)
we're supposed to use Leibniz formula but I'm not sure how to apply it. Thanks!
2. So your integral equation is
$\displaystyle u(t)=1+\int_{0}^{t}s\ln\left(\frac{s}{t}\right)u(s )\,ds,$
and you need to convert to an IVP using the Leibniz formula, which is, in this case,
$\displaystyle \frac{d}{dt}\int_{a(t)}^{b(t)}f(s,t)\,ds=\frac{db( t)}{dt}\,f(b(t),t)-\frac{da(t)}{dt}\,f(a(t),t)+\int_{a(t)}^{b(t)} \frac{\partial }{\partial t}\,f(s,t)\,ds.$
So, why not just differentiate the original integral equation with respect to t, and use the Leibniz formula on the integral. What do you get?
3. Originally Posted by Ackbeet
$\displaystyle u(t)=1+\int_{0}^{t}s\ln\left(\frac{s}{t}\right)u(s )\,ds,$
and you need to convert to an IVP using the Leibniz formula, which is, in this case,
$\displaystyle \frac{d}{dt}\int_{a(t)}^{b(t)}f(s,t)\,ds=\frac{db( t)}{dt}\,f(b(t),t)-\frac{da(t)}{dt}\,f(a(t),t)+\int_{a(t)}^{b(t)} \frac{\partial }{\partial t}\,f(s,t)\,ds.$
So, why not just differentiate the original integral equation with respect to t, and use the Leibniz formula on the integral. What do you get?
Why not just note that $\displaystyle \displaystyle \int _0^t s\log\left(\frac{s}{t}\right)u(s)\text{ }ds=\int_0^t s\log(s)u(s)\text{ }ds-\log(t)\int_0^t su(s)\text{ }ds$ from which all you need is the product rule and the FTC?
4. Originally Posted by Drexel28
Why not just note that $\displaystyle \displaystyle \int _0^t s\log\left(\frac{s}{t}\right)u(s)\text{ }ds=\int_0^t s\log(s)u(s)\text{ }ds-\log(t)\int_0^t su(s)\text{ }ds$ from which all you need is the product rule and the FTC?
True enough, but if the OP'er must use the Leibniz formula because of class restraints, then we're back to square one. | 664 | 1,914 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2018-17 | latest | en | 0.773494 |
https://astronomy.stackexchange.com/questions/21623/question-about-3p-c2-in-friedmann-equation/27648 | 1,713,323,773,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817128.7/warc/CC-MAIN-20240417013540-20240417043540-00538.warc.gz | 107,311,948 | 36,940 | # Question about $3p/c^2$ in Friedmann equation
The Friedmann Equation:
$$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3}$$
I want to know, where does the $3p/c^2$ come from. I know it is something like density of energy or something like that but I don't know where it comes from.
• Did you read the Wikipedia article? The $p$ stands for pressure, while $\rho$ is the energy density.
– pela
Jul 13, 2017 at 21:07
• It sounds like you're asking about the origin and significance of that specific term. To really understand that requires tracing through the derivation of the equation. Is that what you're after? Jul 14, 2017 at 14:36
• @zephyr I want to find how $\rho_{rad}$ is equal to $3p/c^2$. Jul 14, 2017 at 18:58
There are two Friedman equations.
• One related to the energy-density component in the Einstein field equations.
• One related to the pressure component in the Einstein field equations.
The Einstein field equations are the following ten equations (16 equations for 4x4 tensor components but only ten of them are independent due to symmetry)
$$G_{\mu\nu} + \Lambda g_{\mu\nu} = \kappa T_{\mu\nu} = \frac{8 \pi G}{c^4} T_{\mu\nu}$$
where $G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} g_{\mu\nu}R$ is the Einstein tensor, $g_{\mu\nu}$ the metric tensor, $\Lambda$ is the cosmological constant, $\kappa = \frac{8 \pi G}{c^4}$ is the Einstein constant, $T_{\mu\nu}$ is the energy-momentum tensor
Friedman was actually assuming that only the matter/energy density component is non-zero $T_{44}= \epsilon = \rho c^2$. He assumed "Die Materie ist inkohärent" (ie non interacting dust with $p=0$) by which the components $T_{ii}$ are non-zero for $i \neq 4$. The other components (energy flux and momentum) are set equal to zero based on the assumption that the motion of energy and matter is considered to be negligibly small.
LeMaître kept the pressure in place arguing that the pressure associated with the kinetic energy of matter might be negligible but this may not be the case for radiation pressure. So he used also $T_{11} = T_{22} = T_{33} = -p$
Using the Robertson Walker metric metric $d s^2 = dt^2 - a(t)^2 \left[ \frac{dr^2}{1-k r^2}+r^2d\Omega^2\right]$ the equations become expressions in terms or the scale factor $a(t)$:
$$\begin{array}{rcrcl} G_{00} &=& c^{-2}\frac{3k}{a^2} + 3c^{-2} \frac{\dot a^2}{a^2} &=& \frac{8 \pi G}{c^4} \rho c^2 - \Lambda \\ -G_{11} = -G_{22} = -G_{33} &=& c^{-2}\frac{k}{a^2} + c^{-2}\frac{\dot a^2}{a^2} + 2 c^{-2} \frac{\ddot a}{a} &=& -\frac{8 \pi G}{c^4} p + \Lambda \end{array}$$
If you subtract the second equations from the first equations
$$G_{00}+G_{11}+G_{22}+G_{33} = -6 c^{-2} \frac{\ddot a}{a} = \frac{8 \pi G}{c^4} (3p+ \rho c^2) + 2\Lambda$$
and rearrange some terms, then you could also write
$$\frac{\ddot a}{a} = -\frac{4 \pi G}{3} \left( \rho + \frac{3p}{c^2} \right) + \frac{ \Lambda c^2}{3}$$
That was a lovely answer. Since in a comment, titansarus asked for the connection between 3p (I'll take c=1, it's just a constant) and the radiation energy density, I will add that it is easy to show that pressure is always E/3 for any isotropic relativistic gas (where E is kinetic energy density and "relativistic" means rest energy is unimportant, so each particle has its kinetic energy equal to the magnitude of its momentum). Pressure is momentum flux per area, so form a ratio of that to the kinetic energy density, for all particles in a given energy bin, integrating over all directions. If that comes out 1/3, you have your answer.
The numerator of that ratio is the integral over direction cosine of the particle density times its momentum times its speed (here 1) times the square of the direction cosine (that's what gives momentum flux, one cosine comes from the normal component of the momentum, and the other from the need to transport it across the boundary via the speed normal to the boundary), and the denominator is exactly the same integral except without the square of the direction cosine (because it's energy density, so has neither of those two reasons to include the cosine). Hence, the ratio is just the integral over the direction cosine of the cosine squared, divided by the integral over the direction cosine. That's 1/3, simple as that. | 1,277 | 4,301 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-18 | latest | en | 0.888726 |
https://www.physicsforums.com/threads/how-far-apart-are-two-stars.290696/ | 1,508,838,235,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828356.82/warc/CC-MAIN-20171024090757-20171024110757-00230.warc.gz | 985,377,321 | 15,029 | # How far apart are two stars
1. Feb 7, 2009
### zyphriss2
1. The problem statement, all variables and given/known data
Two stars 18 light-years away are barely resolved by a 68 -cm (mirror diameter) telescope. How far apart are the stars? Assume \lambda = 540 <units>nm</units> and that the resolution is limited by diffraction.
2. Relevant equations
Theta=(1.22 lambda)/diameter of the lense
9.4605284 × 10^15 meters
3. The attempt at a solution
I have no clue how to do this. I plugged the give info into the equation and got theta to equal 9.6882352941176470588235294117647e-7 then i just plugged this into the Pythagorean equation to get 559491313771834207552834.45286104
2. Feb 7, 2009
### mgb_phys
Roughly from similair triangles:
lamba/D = separation/distance
The angle between the stars is 1.22lambda/D so you can work out this angle (remember is answer in radians) then you have the angle between two stars a distance away so getting the distance between them is easy.
Since the angles are small you can use the apprx theta = sin theta (in radians)
3. Feb 7, 2009
### zyphriss2
I have worked it out both ways and both of the answers i got were wrong
4. Feb 8, 2009
### astrorob
Remember, as mgb_phys stated, Rayleigh's Criterion expresses the angular distance in radians.
If you're still getting the incorrect answer I suggest you explicitly post how you're calculating the distance. | 383 | 1,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2017-43 | longest | en | 0.912008 |
http://www.jiskha.com/members/profile/posts.cgi?name=Lisa&page=14 | 1,498,277,642,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320215.92/warc/CC-MAIN-20170624031945-20170624051945-00147.warc.gz | 557,096,241 | 10,372 | # Posts by Lisa
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Through extensive research, Metlife Insurance claims that the mean footwell intrusions for Toyota Corrola and Toyota Rav4 are equal. Crash test at 40mph were performed at 7 randomly selected Toyota Corrola and 13 randomly Toyota Rav4s selected. The amount that the footwell ...
Can anyone help with this question? Now suppose that, starting with the first row of seats, the floor of the seating area is inclined at an angle of Theta = 25 degrees above the horizontal, and the distance that you sit up the incline is x, as shown in the fi...
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Can anyone help with this question. Now suppose that, starting with the first row of seats, the floor of the seating area is inclined at an angle of Theta = 25 degrees above the horizontal, and the distance that you sit up the incline is x, as shown in the fi...
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Calculus
use law of cosines to show that theta equals inverse cosine of (a^2 + b^2 -484)/(2ab) where a^2=(7+xcos a)^2 + (28-xsin a)^2 and b^2=(7+xcos a)^2 + (xsin a -6)^2
Calculus
Use theta equals inverse tangent of 28/x minus inverse tangent of 6/x to show that theta also equals 22x/(x^2=168).
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do you say women's role or women's roles in society
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W=KE2-KE1 =1/2*mv^2 - 1/2*mv^2 = 30kg(6m/s)^2 - 30kg(2m/s)^2 =1080 - 120 =960 W=(Fe - 30N)*d=960 (Fe - 30)25=960 25Fe - 750=960 25Fe=1710 Fe=68.4N
Revise sentence for better understanding/communication 1.Business has an inordinate influence on governmental operations. 2. Our expectaions are that there will be increasement in commodity value. 3. Can we assertain the types of customers that have a predisposition to utilize...
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If ON=2x-3, LM=7x+2, NM=x+6, and OL=6y-3, find the values of x and y for which LMNO must be parallelogram.
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Through extensive research, Metlife Insurance claims that the mean footwell intrusions for Toyota Corrola and Toyota Rav4 are equal. Crash test at 40mph were performed at 7 randomly selected Toyota Corrola and 13 randomly Toyota Rav4s selected. The amount that the footwell ...
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12?
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College Chemistry
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College Chemistry
Consider the reaction. 4Fe(s) + 3O2(g) ¨ 2Fe2O3(s) If 0.500 mol of Fe reacts with 0.400 mol of O2, what are the limiting reactant and the theoretical yield in moles of Fe2O3 for the reaction, respectively?
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UNIVERSITY OF PHONEX
Wow! Very professional response.
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Four boys sit down to play a game of marbles. They empty the bag of marbles and divide the marbles among themselves, and they find they have one marble left. A fifth boy comes along and wants to play, so they divide the marbles among all five boys and once again find that ...
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washington state history
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# Previous year Questions (2016-20) - Vector Algebra and Three Dimensional Geometry Notes | EduRev
## Commerce : Previous year Questions (2016-20) - Vector Algebra and Three Dimensional Geometry Notes | EduRev
The document Previous year Questions (2016-20) - Vector Algebra and Three Dimensional Geometry Notes | EduRev is a part of the Commerce Course Mathematics (Maths) Class 12.
All you need of Commerce at this link: Commerce
Q.1. A vector lies in the plane of the vectors If bisects the angle between then (2020)
(1)
(2)
(3)
(4)
Ans.
(3)
We have,
On comparing with
We get α = 4 and β = 4
Therefore,
Now, again consider
On comparing with we get
Therefore,
Q.2. Let be three unit vectors such that and then the ordered pair, is equal to (2020)
(1)
(2)
(3)
(4)
Ans.
(4)
Given
...(1)
Taking square of both sides, we get
(since are unit vectors)
Now,
Q.3. Let be two vectors. If is a vector such that then is equal to (2020)
(1) -3/2
(2) 1/2
(3) -1/2
(4) -1
Ans.
(3)
Given
...(1)
Now,
and
Substituting these values in Eq. (1), we get
Hence,
Q.4. The projection of the line segment joining the points (1, −1, 3) and (2, −4, 11) on the line joining the points (−1, 2, 3) and (3, −2, 10) is _____. (2020)
Ans.
(8.00)
We have
Hence, the required projection is
Q.5. If the vectors are coplanar and then the value of λ is _____. (2020)
Ans.
(1.00)
The vectors are coplanar vectors, then
Therefore, ...(1)
...(2)
...(3)
Now, ...(4)
...(5)
Hence,
Q.6. Let be three vectors such that and the angle between If is perpendicular to the vector is equal to ____. (2020)
Ans.
(30.00)
Given,
Now, is perpendicular to the vector then
Hence,
Q.7. If the distance between the plane 23x - 10y - 2z + 48 =0 and the plane containing the lines is equal to then k is equal to _____. (2020)
Ans.
(3.00)
The given lines must be intersecting
Therefore,
(2s - 1, 3s + 3, 8s - 1) = (2t - 3t, t - 2, λt + 1)
Now, 2s - 1 = 2t - 3 ...(1)
3s + 3 = t - 2 ...(2)
8s - 1 = λt + 1 ...(3)
On solving above equations, we get
t = -1, s = -2 and λ = 18
Distance of plane contains given lines from given plane is same as distance between point (–3, –2, 1) from given plane. So,
Q.8. Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then, the image of R in the plane P is (2020)
(1) (6, 5, 2)
(2) (6, 5, −2)
(3) (4, 3, 2)
(4) (3, 4, −2)
Ans.
(2)
The equation of the plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) is
⇒ (x - 2)(1) - (y - 1)(2 - 3) + z(-2) = 0
⇒ x + y - 2z = 3
Let I and F are respectively image and foot of perpendicular of point R in the plane. So, the equation of RI is
(say)
⇒ x = k + 2, y = k + 1 and z = -2k + 6
Hence, the coordinates of point F is
Point F lies in the plane, then
⇒ 3k = 12 ⇒ k = 4
Hence, the image I is (6, 5, -2).
Q.9. If the foot of the perpendicular drawn from the point (1, 0, 3) on a line passing through (α, 7, 1) is then α is equal to _____.
Ans.
(4.00)
We have,
Direction ratio of BP
Direction ratio of AP =
Since AP and BP are perpendicular, then
Q.10. The shortest distance between the lines is (2020)
(1) 2√30
(2) 7/2 (√30)
(3) 3√30
(4) 3
Ans.
(3)
The equations of the lines are
...(1)
...(2)
The distance between the lines is given by
Now,
Hence,
Q.11. Let the volume of a parallelepiped whose coterminous edges are given by and be 1 cu. unit. If θ be the angle between the edges then cos θ can be (2020)
(1)
(2)
(3) 5/7
(4)
Ans.
(2)
The volume of parallelepiped is 1 cu. unit. Therefore,
⇒ λ = 2, 4
Now,
Q.12. The mirror image of the point (1, 2, 3) in a plane is Which of the following points lies on this plane? (2020)
(1) (1, 1, 1)
(2) (1, -1, 1)
(3) (-1, -1, 1)
(4) (-1, -1, -1)
Ans.
(2)
Direction ratios of normal to the plane is
<1 + (7/3), 2 + (4/3),3 + (1/3)> = <1, 1, 1>
The coordinates of mid- point of image and point is
The equation of plane is
⇒ x + y + z = 1
Hence, the point (1, −1, 1) lies on the plane.
Q.13. If for some α and β in the intersection of the following three planes
x + 4y - 2z =1
x + 7y - 5z = β
x + 5y - αz = 5
is a line in then α + β is equal to (2020)
(1) 0
(2) 10
(3) 2
(4) −10
Ans.
(2)
We have
=0
⇒ (7α + 25) - (4α + 10) + (-20 + 14) = 0
⇒ 3α + 9 = 0 ⇒ α = -3
and
Hence, α + β = -3 + 13 = 10
Q.14. Let be a vector such that then is equal to: (2019)
(1) 19/2
(2) 9
(3) 8
(4) 17/2
Ans.
(1)
Q.15. Let and be three vectors such that the projection vector of
If is perpendicular to is equal to: (2019)
(1) √32
(2) 6
(3) √22
(4) 4
Ans.
(2)
Projection of
According to question
⇒ b1 + b2 = 2 ...(1)
Since, is perpendicular to
Hence
⇒ 8 + 5b1 + b2 + 2 = 0 ...(2)
From (1) and (2),
Q.16. Let and be three vectors such that and is perpendicular to Then a possible value of (λ1, λ2, λ3) is: (2019)
(1) (1, 3, 1)
(2) (-1/2, 4, 0)
(3) (1/2, 4, -2)
(4) (1, 5, 1)
Ans.
(2)
∴ 3 - λ2 = 2λ1 ...(1)
...(2)
Since, (-1/2, 4, 0) satisfies equation (1) and (2). Hence, one of possible value of
Q.17. Let be two given vectors where vectors are non-collinear. The value of λ for which vectors are collinear, is: (2019)
(1) -4
(2) -3
(3) 4
(4) 3
Ans.
(1)
Let are collinear for same k
i.e.,
But are non-collinear, then
⇒
λ = -4
Q.18. Let and be coplanar vectors. Then the non-zero vector is: (2019)
(1)
(2)
(3)
(4)
Ans.
(4)
are coplanar
For (Rejected)
Q.19. Let respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is 3/√2, then the sum of all possible values of β is: (2019)
(1) 4
(2) 3
(3) 2
(4) 1
Ans.
(4)
Since, the angle bisector of acute angle between OA and OB would be y = x
Since, the distance of C from bisector = 3/√2
Hence, the sum of all possible value of β = 2 + (-1) = 1
Q.20. The sum of the distinct real values of μ, for which the vectors, are co-planar, is: (2019)
(1) -1
(2) 0
(3) 1
(4) 2
Ans.
(1)
∵ Three vectors and are copalnar.
Therefore, sum of all real values = 1 - 2 = -1
Q.21. Let be three unit vectors, out of which vectors are non-parallel. If α and β are the angles which vector makes with vectors respectively and then is equal to: (2019)
(1) 30°
(2) 90°
(3) 60°
(4) 45°
Ans.
(1)
Since, are three unit vectors
∴
Then,
∴
⇒ β = 60° and α = 90°
Hence,
Q.22. The magnitude of the projection of the vector on the vector perpendicular to the plane containing the vectors and is: (2019)
(1) √3/2
(2) √6
(3) 3√6
(4)
Ans.
(4)
Let
∴ Vector perpendicular to
Now, projection of vector is
Q.23. Let and for some real x. Then is possible if: (2019)
(1)
(2)
(3)
(4)
Ans.
(2)
Given
Now,
Q.24. Let where is parallel to is perpendicular to then is equal to: (2019)
(1)
(2)
(3)
(4)
Ans.
(3)
...(1)
Since, is perpendicular to
Cross product with in equation (1)
Q.25. If a unit vector makes angles π/3 with and θ ∈ (0, π) with then a value of 0 is: (2019)
(1) 5Ï€/6
(2) π/4
(3) 5Ï€/12
(4) 2Ï€/3
Ans.
(4)
Let cos α, cos β, cos γ be direction cosines of a.
Q.26. The distance of the point having position vector from the straight line passing through the point (2, 3, -4) and parallel to the vector, is: (2019)
(1) 7
(2) 4√3
(3) 6
(4) 2√13
Ans.
(1)
Equation of the line is
Let the point M on the line l is (6λ + 2, 3λ + 3, -4λ - 4)
Direction ratio's of PM is (6λ + 3, 3λ + 1, - 4λ- 10)
⇒ (6λ + 3) (6) + (3λ + 1) (3) + (-4λ.- 10) (- 4) = 0
⇒ λ = - 1 ⇒ M = (- 4, 0, 0)
Q.27. Let a = 3i + 2j + 2k and b = i + 2j - 2k be two vectors. If a vector perpendicular to both the vectors has the magnitude 12 then one such vector is: (2019)
(1)
(2)
(3)
(4)
Ans.
(2)
Let vector be
Given that magnitude of the vector is 12.
Q.28. Let α ∈ R and the three vectors and Then the set (2019)
(1) is singleton
(2) is empty
(3) contains exactly two positive numbers
(4) contains exactly two numbers only one of which is positive
Ans.
(2)
Let, three vectors are coplanar, then
∵ no real value of ‘α’ exist,
∴ set S is an empty set.
Q.29. The equation of the line passing through (-4, 3, 1), parallel to the plane x + 2y - z - 5 = 0 and intersecting the line is: (2019)
(1)
(2)
(3)
(4)
Ans.
(3)
Let any point on the intersecting line
is (-3λ - 1, 2λ + 3, -λ + 2)
Since, the above point lies on a line which passes through the point (-4, 3, 1)
Then, direction ratio of the required line
Since, line is parallel to the plane
x + 2y - z - 5 = 0
Then, perpendicular vector to the line is
Now (-3λ + 3)(1) + (2λ)(2) + (-λ + 1)(-1) = 0
⇒ λ = - 1
Now direction ratio of the required line = <6, - 2, 2 > or <3,-1, 1>
Hence required equation of the line is
Q.30. The plane through the intersection of the planes x + y + z = 1 and 2x + 3y-z + 4 = 0 and parallel to j-axis also passes through the point: (2019)
(1) (-3, 0, -1)
(2) (-3, 1, 1)
(3) (3, 3, -1)
(4) (3, 2, 1)
Ans.
(4)
Since, equation of plane through intersection of planes
But, the above plane is parallel to y-axis then
⇒ λ = -3
Hence, the equation of required plane is
-x - 4z + 7 = 0
⇒ x + 4z - 7 = 0
Therefore, (3, 2, 1) the passes through the point.
Q.31. The equation of the plane containing the straight line and perpendicular to the plane containing the straight lines is: (2019)
(1) x - 2y + z = 0
(2) 3x + 2y - 3z = 0
(3) x + 2y - 2z = 0
(4) 5x + 2y - 4z = 0
Ans.
(1)
Let the direction ratios of the plane containing lines
∴ Direction ratio of plane = <-8, 1, 10>.
Let the direction ratio of required plane is <l, m, n>
Then-81 + m + 10n = 0 ...(1)
and 2l + 3m + 4n = 0 ...(2)
From (1) and (2),
∴ D.R.s are <1, -2, 1>
∴ Equation of plane: x - 2y + z = 0
Q.32. The plane passing through the point (4, -1, 2) and parallel to the lines also passes through the point: (2019)
(1) (1, 1, -1)
(2) (1, 1, 1)
(3) (-1, -1,-1)
(4) (-1, -1, 1)
Ans.
(2)
Equation of required plane is
(x - 4)(-3 -4) - (y + 1)(9 - 2) + (z - 2) (6 + 1) = 0
∵ point (1, 1, 1) satisfies this equation
∴ point (1, 1, 1) lies on the plane
Q.33. Let A be a point on the line and B(3, 2, 6) be a point in the space. Then the value of μ for which the vector is parallel to the plane x - 4y + 3z = 1 is: (2019)
(1) 1/4
(2) 1/8
(3) 1/2
(4) -1/4
Ans.
(1)
∵ A be a point on given line.
∴ Position vector of A
Equation of plane is: x - 4y + 3z = 1
∵ is parallel to this plane.
⇒ μ = 1/4
Q.34. The plane which bisects the line segment joining the points (- 3, - 3, 4) and (3, 7, 6) at right angles, passes through which one of the following points? (2019)
(1) (-2, 3, 5)
(2) (4, -1, 7)
(3) (2, 1, 3)
(4) (4, 1, -2)
Ans.
(4)
Q(0, 2, 5)
Since, direction ratios of normal to the plane is
Then, equation of the plane is
(x - 0)6 + (y - 2)10 + (z - 5)2 = 0
3x + 5y - 10 + z - 5 = 0
3x + 5y + z = 15 ...(1)
Since, plane (1) satisfies the point (4, 1, -2)
Hence, required point is (4, 1, -2)
Q.35. On which of the following lines lies the point of inter-section of the line, and the plane, x + y + z = 2? (2019)
(1)
(2)
(3)
(4)
Ans.
(3)
Let any point on the line be A(2λ + 4, 2λ + 5, λ, + 3) which lies on the plane x +y + z = 2
⇒ 2λ + 4 + 2λ + 5 + λ + 3 = 2
⇒ 5λ = -10 ⇒ λ = - 2
Then, the point of intersection is (0, 1, 1)
which lies on the line
Q.36. The plane containing the line and also containing its projection on the plane 2x + 3y- z = 5, contains which one of the following points? (2019)
(1) (2, 2, 0)
(2) (-2, 2, 2)
(3) (0, -2, 2)
(4) (2, 0, -2)
Ans.
(4)
Let normal to the required plane is
⇒ is perpendicular to both vector
⇒ Equation of the required plane is
⇒ (x - 3)(-8) + (y + 2) x 8 + (z - 1) x 8 = 0
⇒ (x - 3) (-1) + (y + 2) x 1 + (z - 1) x 1 = 0
⇒ x - 3 - y - 2 - z + 1 =0
∵ x - y - z = 4 passes through (2, 0, -2)
∴ plane contains (2, 0, -2).
Q.37. The direction ratios of normal to the plane through the points (0, -1, 0) and (0,0, 1) and making an angle π/4 with the plane y - z + 5 = 0 are: (2019)
(1) 2, -1, 1
(2) 2, √2, -√2
(3) √2, 1, -1
(4) 2√3, 1, -1
Ans.
(2, 3)
Let the d.r’s of the normal be (a, b, c)
Equation of the plane is
a(x - 0) + b(y + 1) + c(z - 0) = 0
∵ It passes through (0, 0, 1)
∴ b + c
Also
∴ The d.r’s are √2, 1, -1 or 2, √2, -√2
Q.38. Two lines intersect at the point R. The reflection of R in the xy-plane has coordinates: (2019)
(1) (2, -4, -7)
(2) (2, 4, 7)
(3) (2, -4, 7)
(4) (-2, 4, 7)
Ans.
(1)
Let the coordinate of P with respect to line
and coordinate of P w.r.t.
From above equation : λ = -1, μ = 1 .
∴ Coordinate of point of intersection R = (2, -4, 1).
Image of R w.r.t. xy plane = (2, -4, -7).
Q.39. If the point (2, α, β) lies on the plane which passes through the points (3,4, 2) and (7, 0, 6) and is perpendicular to the plane 2x-5y = 15, then 2α - 3β is equal to: (2019)
(1) 12
(2) 7
(3) 5
(4) 17
Ans.
(2)
Let the normal to the required plane is n, then
∴ Equation of the plane
(x - 3) x 20 + (y - 4) x 8 + (z - 2) x (-12) = 0
5x - 15 + 2y - 8 - 3z + 6 = 0
5x + 2y - 3z - 17 = 0 ...(1)
Since, equation of plane (1) passes through (2, α, β), then
Q.40. The perpendicular distance from the origin to the plane containing the two lines, and is: (2019)
(1) 11√6
(2) 11/√6
(3) 11
(4) 6/√11
Ans.
(2)
∵ plane containing both lines.
∴ D.R. of plane
Now, equation of plane is,
7(x - 1) - 14(y - 4) + 7 (z + 4) = 0
⇒ x - 1 - 2y + 8 + z + 4 = 0
⇒ x - 2y + z + 11 = 0
Hence, distance from (0, 0, 0) to the plane,
Q.41. If an angle between the line, and the plane, x - 2y - kx = 3 is then a value of k is (2019)
(1)
(2)
(3) -3/5
(4) -5/3
Ans.
(1)
Let angle between line and plane is θ, then
Since,
Then,
Q.42. Let S be the set of all real values of λ such that a plane passing through the points {-λ2, 1, 1), (1, -λ2, 1) and (1, 1, -λ2) also passes through the point-(-1,-1, 1). Then S is equal to: (2019)
(1) {√3}
(2) {√3, -√3}
(3) {1, -1}
(4) {3, -3}
Ans.
(2)
Let D(-1, -1, 1) lie on same plane, then
⇒ (λ2 + 1)((1 -λ2)-4) = 0
⇒ (3 - λ2)(λ+ 1) = 0 ⇒ λ2 = 3
Hence, S = {-√3, √3}
Q.43. The equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is: (2019)
(1) x- 3y- 2z = -2
(2) 2x - z = 2
(3) x - y - z = 0
(4) x + 3y + z = 4
Ans.
(3)
Let the equation of required plane be;
(2x - y - 4) + λ(y + 2z - 4) = 0
∴ This plane passes through the point (1, 1, 0) then (2 - 1 - 4) + λ(1 + 0 - 4) = 0
⇒ λ = -1
Then, equation of required plane is,
(2x - y - 4) - (y + 2z - 4) = 0
⇒ 2x - 2y — 2z = 0 ⇒ x - y - z = 0
Q.44. The length of the perpendicular from the point (2, -1, 4) on the straight line, is: (2019)
(1) greater than 3 but less than 4
(2) less than 2
(3) greater than 2 but less than 3
(4) greater than 4
Ans.
(1)
Let P be the foot of perpendicular from point T (2, -1, 4) on the given line. So P can be assumed as P(10λ - 3, -7λ + 2, λ)
DR's of TP is proportional to 10λ - 5, - 7λ + 3, λ - 4
∵ TP and given line are perpendicular, so
10(10λ - 5) -7 (- 7λ + 3)+ 1(λ - 4) = 0
Hence, the length of perpendicular is greater than 3 but less than 4.
Q.45. The vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0 is: (2019)
(1)
(2)
(3)
(4)
Ans.
(4)
Equation of the plane passing through the line of intersection of x + y + z = 1 and 2x + 3y + 4z = 5 is (2x + 3y + 4z - 5) + λ (x + y + z - 1) = 0
∵ plane (i) is perpendicular to the plane x-y + z = 0
∴ (2 + λ)(1) + (3 + λ)(- 1) + (4 + λ)(1) = 0
2 + λ - 3 - λ + 4 + λ = 0
⇒ λ = -3
Hence, equation of required plane is -x + z- 2 = 0 or x - z + 2 = 0
Q.46. If a point R(4, y, z) lies on the line segment joining the points P(2, -3, 4) and Q(8, 0, 10), then distance of R from the origin is: (2019)
(1) 2√14
(2) 2√21
(3) 6
(4) √53
Ans.
(1)
Here, P, Q, R are collinear
Now, or =
Q.47. If the line, meets the plane, x + 2y + 3z = 15 at a point P, then the distance of P from the origin is: (2019)
(1) √5 / 2
(2) 2 / √5
(3) 9/2
(4) 7/2
Ans.
(3)
Let point on line be P (2k + 1, 3k - 1, 4k + 2)
Since, point P lies on the plane x + 2y + 3z = 15
∴ 2k + 1 + 6k - 2 + 12k + 6= 15
⇒ k = 1/2
Then the distance of the point P from the origin is
Q.48. A plane passing through the points (0, -1, 0) and (0, 0, 1) and making an angle π/4 with the plane y - z + 5 = 0, also passes through the point: (2019)
(1) (-√2, 1, -4)
(2) (-√2, -1, 4)
(3) (-√2, -1, -4)
(4) (√2, 1, 4)
Ans.
(4)
Let the required plane passing through the points (0,-1, 0) and (0, 0, 1) be and the given plane is y - z + 5 = 0
Then, the equation of plane is
Then the point (√2, 1, 4) satisfies the equation of plane -√2x - y + z = 1
Q.49. The vertices B and C of a ΔABC lie on the line, such that BC = 5 units. Then the area (in sq. units) of this triangle, given that the point A (1, -1, 2), is: (2019)
(1) 5√17
(2) 2√34
(3) 6
(4) √34
Ans.
(4)
Let a point D on BC = (3λ - 2, 1, 4λ)
Hence,
Area of triangle
Q.50. Let P be the plane, which contains the line of intersection of the planes, x + y + z - 6 = 0 and 2x + 3y + z + 5 = 0 and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to: (2019)
(1) 17/√5
(2) 63√5
(3) 205√5
(4) 11/√5
Ans.
(4)
Let the plane be
P = (2x + 3y + z + 5) + λ(x: +y + z- 6) = 0
∵ above plane is perpendicular to xy plane.
Hence, the equation of the plane is,
P = x + 2y+ 11 = 0
Distance of the plane P from (0, 0, 256)
Q.51. If Q (0, -1, -3) is the image of the point P in the plane 3x - y + 4z = 2 and R is the point (3, -1, -2), then the area (in sq. units) of ΔPQR is: (2019)
(1) 2√13
(2) √91/4
(3) √91/2
(4) √65/2
Ans.
(3)
Image of Q (0, -1, -3) in plane is,
Q.52. Let A(3, 0, -1), B(2, 10, 6) and C (1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2 : 1, then cos (∠GOA)(O being the origin) is equal to: (2019)
(1)
(2) 1/√15
(3)
(4) 1/√30
Ans.
(2)
G is the centroid of ΔABC.
Q.53. If the length of the perpendicular from the point (β, 0, β) (β ≠0) to the line, then β is equal to: (2019)
(1) 1
(2) 2
(3) -1
(4) -2
Ans.
(3)
Given, (let) and point P (β, 0, β)
Any point on line A = (p, 1, -p -1)
Now, DR of AP ≡ <p - β, 1 - 0 - p - 1 - β>
Which is perpendicular to line.
Given that distance AP
∴ β = -1
Q.54. A perpendicular is drawn from a point on the line to the plane x + y + z = 3 such that the foot of the perpendicular Q also lies on the plane x - y + z = 3. Then the co-ordinates of Q are: (2019)
(1) (1, 0, 2)
(2) (2, 0, 1)
(3) (-1, 0, 4)
(4) (4, 0, -1)
Ans.
(2)
Let co-ordinates of Q be (α, β, γ), then
α + β + γ = 3 ...(i)
α - β + γ = 3 ...(ii)
⇒ α + γ = 3 and β = 0
Equating direction ratio's of PQ, we get
Substituting the values of a and y in equation (i), we get
⇒ 5λ + 3 = 3 ⇒ λ = 0
Hence, point is Q (2, 0, 1)
Q.55. If the plane 2x - y + 2z + 3 =0 has the distances 1/2 and 2/3 units from the planes 4x - 2y + 4z + λ = 0 and 2x - y + 2z + μ = 0, respectively, then the maximum value of λ + μ is equal to: (2019)
(1) 9
(2) 15
(3) 5
(4) 13
Ans.
(4)
Let,
Given, distance between P1 and P2 is 1/3
And distance between P1 and P3 is 2/3
Q.56. If the line intersects the plane 2x + 3y - z + 13 = 0 at a point P and the plane 3x + y + 4z = 16 at a point Q, then PQ is equal to: (2019)
(1) 14
(2) √14
(3) 2√7
(4) 2√14
Ans.
(4)
Let points P (3λ + 2, 2λ - 1, - λ + 1) and Q(3μ + 2, 2μ - 1, - μ + 1)
Hence, P (- 1, -3, 2)
Similarly, Q lies on 3x + y + 4z = 16
Hence, Q is (5, 1, 0)
Q.57. A plane which bisects the angle between the two given planes 2x - y + 2z -4 = 0 and x + 2y + 2z - 2 = 0, passes through the point: (2019)
(1) (-1, -4, 1)
(2) (1, 4, -1)
(3) (2, 4, 1)
(4) (2, -4, 1)
Ans.
(4)
The equations of angle bisectors are,
⇒ x - 3y - 2 = 0
or 3x + y + 4z - 6 = 0
(2, -4, 1) lies on the second plane.
Q.58. The length of the perpendicular drawn from the point (2, 1,4) to the plane containing the lines is: (2019)
(1) 3
(2) 1/3
(3) √3
(4) 1/√3
Ans.
(3)
The equation of plane containing two given lines is,
On expanding, we get x - y - z = 0
Now, the length of perpendicular from (2, 1, 4) to this plane
Q.59. Let be a vector coplanar with the vectors and perpendicular to then, is equal to: (2018)
(1) 336
(2) 315
(3) 256
(4) 84
Ans.
(1)
Q.60. If L1 is the line of intersect ion of the planes 2x - 2y + 3z = 0, x - y + z = 0 and L2 is the line of intersection of the planes x + 2y - z - 3 = 0, 3x - y + 2z = 0 then the distance of the origin fro m the plane, containing the lane L1 and L2 , is: (2018)
(1)
(2)
(3)
(4)
Ans.
(2)
Vectors along the given lines L1, L2 are
and
Putting y = 0 in 1st two equation
Point on the plane is (–5, 0, 4) and normal vector of required plane is
Hence, equation of plane is - 7x + 7y - 8z - 3 = 0
Perpendicular distance is
Q.61. The length of the project ion of the line segment joining the points (5, -1, 4) and (4, -1, 3) on the plane, x + y + z = 7 is: (2018)
(1) 2/√3
(2) 2/3
(3) 1/3
(4)
Ans.
(4)
D.R’s of AB = (1, 0, 1)
D.R’s of normal to plane = (1, 1, 1)
AB = √2
Length of projection =
Q.62. Let and be a vector such that and the angle between be 30°. Then is equal to (2017)
(1) 1/8
(2) 25/8
(3) 2
(4) 5
Ans.
(3)
Q.63. If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line, x/1 = y/4 = z/5 is Q, then PQ is equal to (2017)
(1) 6√5
(2) 3√5
(3) 2√42
(4) √42
Ans.
(3)
Equation of PQ,
Let M be (λ+1, 4λ- 2, 5λ+ 3)
As it lies on 2x + 3y – 4z + 22 = 0
λ = 1
For Q, λ = 2
Distance PQ
Q.64. The distance of the point (1, 3, –7) from the plane passing through the point (1, –1, –1), having normal perpendicular to both the lines and is (2017)
(1) 10/√74
(2) 20/√74
(3) 10/√83
(4) 5/√83
Ans.
(3)
Let the plane be
a(x - 1)+ b( y + 1) + c(z +1) = 0
It is perpendicular to the given lines
a – 2b + 3c = 0
Solving, a : b : c = 5 : 7 : 3
∴ The plane is 5x + 7y + 3z + 5 = 0
Distance of (1, 3, –7) from this plane = 10/√83
Q.65. The coodinates of the foot of the perpendicular from the point (1, -2,1) on the plane containing the lines, and is: (2017)
(1) (2, -4, 2)
(2) (1, 1, 1)
(3) (0, 0, 0)
(4) (-1, 2, -`= 1)
Ans.
(3)
Q.66. The line of intersection of the planes and is: (2017)
(1)
(2)
(3)
(4)
Ans.
(3)
13 x = 6
x = 6/13y
y = 5/13
.... is
Q.67. The area (in sq. units) of the parallelogram whosed diagonals are along the vectors and is: (2017)
(1) 20
(2) 65
(3) 52
(4) 26
Ans.
(2)
Q.68. If the line, lies in the plane, 2x - 4y + 3z = 2, then the shortest distance between this line and the line, is: (2017)
(1) 1
(2) 2
(3) 3
(4) 0
Ans.
(4)
pt (3, -2, λ) on pline 2x - 4y +3z - 2 = 0
= 6 + 8 - 3λ - 2 = 0
Now
...(1)
...(2)
p (1, 0, 0)
gives are ditersech - thortest distance = 0
Q.69. If x = a, y = b, z = c is a solution of the system of linear equations
x + 8y + 7z = 0
9x + 2y +3z = 0
y + y + z = 0
such that the point (a, b, c) lies on the plane x + 2y + z = 6, then 2a + b + c equals: (2017)
(1) 2
(2) -1
(3) 1
(4) 0
Ans.
(1)
Q.70. If the vector is written as the sum of a vector parallel to and a vector perpendicular to is equal to: (2017)
(1)
(2)
(3)
(4)
Ans.
(1)
Q.71. If a variable plane, at a distance of 3 units from the origin, intersects the coordinate axes at A, B and C, then the locus of the centroid of ΔABC is. (2017)
(1)
(2)
(3)
(4)
Ans.
(1)
Let Centroid be (h, k, l)
∴ x - intp = 3h Y - intp = 3k, 3 - int = 3l
Q.72. The distance of the point (1, -5, 9) from the plane x - y + z = 5 measured along the line x = y = z is: (2016)
(1) 3√10
(2) 10√3
(3) 10/√3
(4) 20/3
Ans.
(2)
Let Q(1, -5, 9)
Line is (say)
Any pt on line we can take P(r + 1, r - 5, r + 9)
So, Pt satisfy Plane
=>(r + 1) - (r - 5) + (r + 9) = 5 r = -10 So, Point P = (-9, -15, -1)
Distance is PQ =
Q.73. If the line, lies in the plane, lx + my - z = 9, then l2 + m2 is equal to: (2016)
(1) 26
(2) 18
(3) 5
(4) 24
Ans. (4)
Point on line is P = (3, -2, -4)
'P’ lies on lx + my - z = 9
=> 3l - 2m + 4 = 9
3l - 2m = 5 ....(1)
As line lies on plane
=> 2 x l + m x (-1) + 3 x (-1) = 0
2l - m = 3 .....(3)
Solving l = 1, m = -1
So, l+ m= 2
Q.74. Let andbe three unit vectors such that If is not parallel to c , then the angle betweenandis: (2016)
(1) 3Ï€/4
(2) π/2
(3) 2Ï€/3
(4) 5Ï€/6
Ans.
(4)
Equate,
as unit vectors
Q.75. The shortest distance between the lines and lies in the interval: (2016)
(1) (2, 3]
(2) [0, 1)
(3) (3, 4]
(4) [1, 2)
Ans.
(1)
Q.76. The distance of the point (1, -2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x - y + 2z = 3 and 2x - 2y + z + 12 = 0, is (2016)
(1) 1/√2
(2) 2
(3) √2
(4) 2√2
Ans.
(4)
Equation of plane ⊥ to the planes.
x - y + 2z = 3 & 2x - 2y + z + 12 = 0
and passes through (1, 2, 2) is
3(x - 1) + 3(y - 2) = 0
x + y = 3 ..... (1)
distance of plane x + y - 3 = 0 from (1, - 2, 4) is
Q.77. In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively and then the point (p, q) lies on a line (2016)
(1) parallel to y-axis
(2) making an acute angle with the positive direction of x-axis
(3) parallel to x-axis
(4) making an obtuse angle with the position direction of x-axis.
Ans.
(2)
⇒ - 8 + 2(9 - 1) - 3 (p + 1) = 0 ⇒ - 3p + 2q - 13 = 0
⇒ (p, q) lies on line
3x - 2y + 13 = 0
slope = 3/2
Q.78. ABC is a triangle in a plane with vertices A(2, 3, 5), B(-1, 3, 2) and C(λ, 5, µ). If the median through A is equally inclined to the coordinate axes, then the value of (λ3 + µ3 + 5) is (2016)
(1) 676
(2) 1130
(3) 1348
(4) 1077
Ans.
(3)
⇒
Q.79. Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are and respectively, then the position vector of the orthocentre of this triangle, is (2016)
(1)
(2)
(3)
(4)
Ans.
(4)
Position vector of the centroid of
Now we known that centroid divides the line joining orthocentre to circum centre divided by centriod divided by centroid in the ratio in 2 : 1
Q.80. The number of distinct real values of λ for which the lines are coplanar is (2016)
(1) 3
(2) 2
(3) 1
(4) 4
Ans.
(1)
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## Mathematics (Maths) Class 12
209 videos|222 docs|124 tests
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https://www.gamedev.net/forums/topic/661396-not-able-to-move-my-object-in-the-reverse-direction-of-the-x-coordinate/ | 1,542,428,610,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743282.69/warc/CC-MAIN-20181117040838-20181117062838-00226.warc.gz | 889,689,514 | 31,931 | Public Group
# Not able to move my object in the reverse direction of the x coordinate
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I am a beginner in games development and using c++ and opengl to work around with vector class developed by me to move an object.
i have successfully managed to move my object on the positive x direction but i want my object to go in the reverse direction after reaching a defined value. in my project that value is 1.0. here is a piece of my code that will help understand my problem better.
void moveUp()
{
part.velocity.x += 0.01;
part.velocity.y += 0.03;
part.position.x += part.velocity.x;
part.position.y += part.velocity.y;
part.acceleration.x -= 0.0;
part.acceleration.y -= 0.01;
part.velocity.x += part.acceleration.x;
part.velocity.y += part.acceleration.y;
cout << endl <<"this is the position on the x coordinate: " << part.position.x << endl;
if(part.position.y <= 0.15)
{
part.acceleration.x = 0;
part.acceleration.y = 0;
part.velocity.x = 0;
part.velocity.y = 0;
}
cout << endl << "this is the velocity on the x coordinate: " << part.velocity.x << endl;
if(part.position.x > 1)
{
part.velocity.x = 0;
part.velocity.y = 0;
part.velocity.x -= 0.01;
part.velocity.y += 0.03;
part.position.x += part.velocity.x;
part.position.y += part.velocity.y;
part.acceleration.x += 0;
part.acceleration.y -= 0.01;
part.velocity.x += part.acceleration.x;
part.velocity.y += part.acceleration.y;
}
}
after running this code the object keeps bouncing at one place on y coordinate and does not move back on the x coordinate. when i print out the velocity of the object on x coordinate..it gives me the encircled value in the attached picture.
[attachment=23960:screenshot.jpg]
any help is appreciated. Also i want the object to jump once and come to a stationary position instead of using a glutTImerFunc(), that keeps calling the update function again and again after i press a key to make the object jump, the object keeps bouncing.
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part.acceleration.x -= 0.0; IMO.
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part.acceleration.x -=0.0 means no change in acceleration regardless of "+" or "-" because the value increasing or decreasing is 0 which means no change..however i'm changing the value of the velocity..
part.velocity.x -= 0.01;
but doing that just gives me that weird value.
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Your problem likely occurs because you always start out (in the moveUp function) by increasing the x position. The code then decreases the x position when it reaches 1. The result is that you increase to 1, decrease to below 1, increase to 1, etc.
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yes that what the problem is..i figured that out but i can't figure out how to keep decreasing the x position one it reaches the value 1. any idea?
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yes that's what the problem is..i figured that out but i can't figure out how to keep decreasing the x position when it reaches the value 1. any idea?
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First: it's not clear what behavior you want the object to have. You appear to have complicated things a bit with "acceleration," as you increment/decrement it in your code.
Can you describe what behavior you want the object to have, rather than how you want to code it?
Simulating a "jump" is often done with a fixed acceleration (e.g., gravity). At the start of the jump, the object is given an initial upward velocity. Then, each update, the velocity is decreased by acceleration * delta-time. You could simulate that with a simple velocity -= someSmallAmount. That process continues, the velocity eventually changes from positive to negative, and the object falls (still under the same acceleration) until it reaches the "ground."
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float towardsZero(float x, float speed)
{
if (std::abs(x) < speed)
return 0.0f;
else if (part.x > 0.0)
return std::max(0.0, x - speed);
return std::min(0.0, x + speed);
}
Now x will tend towards zero from whichever side of the axis it happens to be on
EDIT: Ok, I understand now
Eg. if speed is (10, 0) you need to figure out what is the highest value you can get away with, say 1.0 (if that is the thinnest terrain you can have)
// number of iterations to run to prevent running through things:
// who cares about rounding, right? just add 1 and be done with it to prevent 0 zero iterations
const int iterations = 1 + speed.length() / 1.0;
// direction player is facing:
const vec2 norm = speed.normalized();
float step_size = 1.0f;
bool moved = false;
while (iterations--)
{
bool collision = test(player, norm * step_size)
if (collision)
// halve step
step_size *= 0.5f;
else
{
player.pos += norm * step_size;
moved = true;
}
}
so what happens above? well it's not a perfect algorithm since you don't guarantee he will move the maximal distance he can go
but it's good enough, for starters
for each iteration, try moving step_size, and if that fails, halve step size
we also take note of if the player ever got moved, since you usually want to use that later on for stuff like animations and gameplay logic
a better algorithm would try halving step size and not decreasing iterations until the player actually moved
but then we have to be careful about not running too many iterations (eg. infinite loop)
Edited by Kaptein
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× | 1,392 | 5,630 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-47 | latest | en | 0.779015 |
https://fr.mathworks.com/help/fusion/ref/assignkbestsd.html | 1,713,200,179,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817002.2/warc/CC-MAIN-20240415142720-20240415172720-00725.warc.gz | 245,473,254 | 21,959 | # assignkbestsd
K-best S-D solution that minimizes total cost of assignment
## Description
[assignments,cost,solutionGap] = assignkbestsd(costmatrix) returns a table of assignments of detections to tracks by finding the best S-D solution that minimizes the total cost of the assignments. The algorithm uses Lagrangian relaxation to convert the S-D assignment problem to a corresponding 2-D assignment problem and then solves the 2-D problem. The cost of each potential assignment is contained in the cost matrix, costmatrix.
costmatrix is an n-dimensional cost matrix where costmatrix(i,j,k ...) defines the cost of the n-tuple (i,j,k, ...) in assignment. The index '1' on all dimensions in costmatrix represents dummy measurement or a false track and is used to complete the assignment problem. The index 1, being a dummy, can be a part of multiple n-tuples. The index can be assigned more than once. A typical cost value for costmatrix(1,1,1,1, ...) is 0.
The function also returns the solution gap, solutionGap, and the cost of assignments, cost.
[assignments,cost,solutionGap] = assignkbestsd(costmatrix,k) also specifies the number, k of K-best S-D solutions. The function finds K optimal solutions that minimize the total cost. First, the function finds the best solution. Then, the function uses the Murty algorithm to generate partitioned cost matrices. Finally, the function obtains the remaining K - 1 minimum cost solutions for each partitioned matrix.
[assignments,cost,solutionGap] = assignkbestsd(costmatrix,k,desiredGap) also specifies the desired maximum gap, desiredGap, between the dual solution and the feasible solution. The gap controls the quality of the solution. Values usually range from 0 to 1. A value of 0 means the dual and feasible solutions are the same.
[assignments,cost,solutionGap] = assignkbestsd(costmatrix,k,desiredGap,maxIterations) also specifies the maximum number of iterations allowed. The desiredGap and maxIterations arguments define the terminating conditions for the S-D algorithm.
[assignments,cost,solutionGap] = assignkbestsd(costmatrix,k,desiredGap,maxIterations,algorithm) also specifies the algorithm for finding the assignments.
## Examples
collapse all
Find the first 5 best assignments of the S-D assignment problem. Set the desired gap to 0.01 and the maximum number of iterations to 100.
Find the 5 best solutions.
[assignments,cost,solutionGap] = assignkbestsd(costMatrix,5,0.01,100)
assignments=5×1 cell array
{2x3 uint32}
{3x3 uint32}
{3x3 uint32}
{3x3 uint32}
{3x3 uint32}
cost = 5×1
-34.7000
-31.7000
-29.1000
-28.6000
-28.0000
solutionGap = 5×1
0
0.0552
0.0884
0.1075
0.1964
## Input Arguments
collapse all
Cost matrix, specified as an n-dimensional array where costmatrix(i,j,k ...) defines the cost of the n-tuple (i,j,k, ...) in an assignment. The index '1' on all dimensions in costmatrix represents a dummy measurement or a false track and is used to complete the assignment problem. The index 1, being a dummy, can be a part of multiple n-tuples. The index can be assigned more than once. A typical cost value for costmatrix(1,1,1,1, ...) is 0.
Data Types: single | double
Number of best solutions, specified as a positive integer.
Data Types: single | double
Desired maximum gap between the dual and feasible solutions, specified as a nonnegative scalar.
Example: 0.05
Data Types: single | double
Maximum number of iterations, specified as a positive integer.
Example: 50
Data Types: single | double
Assignment algorithm for solving the 2-D assignment problem, specified as 'munkres' for the Munkres algorithm, 'jv' for the Jonker-Volgenant algorithm, or 'auction' for the Auction algorithm.
Example: 'jv'
## Output Arguments
collapse all
Assignments of tracks to detections, returned as a K-element cell array. Each cell is an P-by-N list of assignments. Assignments of the type [1 1 Q 1] from a four-dimensional cost matrix can be seen as a Q-1 entity from dimension 3 that was left unassigned. The cost value at (1,1,Q,1) defines the cost of not assigning the (Q-1)th entity from dimension 3.
Total cost of solutions, returned as a K-element vector where K is the number of best solutions. Each element is a scalar value summarizing the total cost of the solution to the assignment problem.
Data Types: single | double
Solution gap, returned as a positive-valued K-element array where K is the number of best solutions. Each element is the duality gap achieved between the feasible and dual solution. A gap value near zero indicates the quality of solution.
Data Types: single | double
## Algorithms
All numeric inputs can be single or double precision, but they all must have the same precision.
## References
[1] Popp, R.L., Pattipati, K., and Bar Shalom, Y. "M-best S=D Assignment Algorithm with Application to Multitarget Tracking". IEEE Transactions on Aerospace and Electronic Systems, 37(1), 22-39. 2001.
[2] Deb, S., Yeddanapudi, M., Pattipati, K., & Bar-Shalom, Y. (1997). "A generalized SD assignment algorithm for multisensor-multitarget state estimation". IEEE Transactions on Aerospace and Electronic Systems, 33(2), 523-538.
## Version History
Introduced in R2018b | 1,273 | 5,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-18 | latest | en | 0.835905 |
https://biz.libretexts.org/Bookshelves/Business/Introductory_Business/Book%3A_Introduction_to_Business_(OpenStax)/15%3A_Understanding_Money_and_Financial_Institutions/15.02%3A_Show_Me_the_Money | 1,721,746,723,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518058.23/warc/CC-MAIN-20240723133408-20240723163408-00185.warc.gz | 115,330,964 | 33,536 | # 15.2: Show Me the Money
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## 1. What is money, what are its characteristics and functions, and what are the three parts of the U.S. money supply?
Money is anything that is acceptable as payment for goods and services. It affects our lives in many ways. We earn it, spend it, save it, invest it—and often wish we had more of it. Businesses and government use money in similar ways. Both require money to finance their operations. By controlling the amount of money in circulation, the federal government can promote economic growth and stability. For this reason, money has been called the lubricant of the machinery that drives our economic system. Our banking system was developed to ease the handling of money.
## Characteristics of Money
For money to be a suitable means of exchange, it should have these key characteristics:
• Scarcity: Money should be scarce enough to have some value but not so scarce as to be unavailable. Pebbles, which meet some of the other criteria, would not work well as money because they are widely available. Too much money in circulation increases prices and inflation. Governments control the scarcity of money by limiting the quantity of money in circulation.
• Durability: Any item used as money must be durable. A perishable item such as a banana becomes useless as money when it spoils. Even early societies used durable forms of money, such as metal coins and paper money, which lasted for a long time.
• Portability: Money must be easily moved around. Large or bulky items, such as boulders or heavy gold bars, cannot be transported easily from place to place.
• Divisibility: Money must be capable of being divided into smaller parts. Divisible forms of money help make transactions of all sizes and amounts possible.
Table 15.1 provides some interesting facts about our money.
## Functions of Money
Using a variety of items as money would be confusing. Thus, societies develop a uniform money system to measure the value of goods and services. For money to be acceptable, it must function as a medium of exchange, as a standard of value, and as a store of value.
As a medium of exchange, money makes transactions easier. Having a common form of payment is much less complicated than having a barter system, wherein goods and services are exchanged for other goods and services. Money allows the exchange of products to be a simple process.
Money also serves as a standard of value. With a form of money whose value is accepted by all, goods and services can be priced in standard units. This makes it easy to measure the value of products and allows transactions to be recorded in consistent terms.
As a store of value, money is used to hold wealth. It retains its value over time, although it may lose some of its purchasing power due to inflation. Individuals may choose to keep their money for future use rather than exchange it today for other types of products or assets.
Did you know . . .
• Currency paper is composed of 25% linen and 75% cotton.
• About 4,000 double folds (first forward and then backwards) are required before a note will tear.
• As of mid-July 2017, there was more than $1.56 trillion in U.S. currency in circulation, with$40 billion in coins.
• 95% of the notes printed each year are used to replace notes already in circulation.
• The largest note ever printed by the Bureau of Engraving and Printing was the $100,000 Gold Certificate, Series 1934. • During fiscal year 2017, it cost approximately 5.4 cents per note to produce nearly 40 billion U.S. paper currency notes. • A stack of currency one mile high would contain over 14 million notes. • If you had 10 billion$1 notes and spent one every second of every day, it would require 317 years for you to go broke.
Table15.1 Source: Bureau of Engraving and Printing, “Resources,” https://www.moneyfactory.gov, accessed September 7, 2017.
## The U.S. Money Supply
The U.S. money supply is composed of currency, demand deposits, and time deposits. Currency is cash held in the form of coins and paper money. Other forms of currency include travelers’ checks, cashier’s checks, and money orders. The amount of currency in circulation depends on public demand. Domestic demand is influenced primarily by prices for goods and services, income levels, and the availability of alternative payment methods such as credit cards. Until the mid-1980s, nearly all U.S. currency circulated only domestically. Today domestic circulation totals only a small fraction of the total amount of U.S. currency in circulation.
Over the past decade, the amount of U.S. currency has doubled to more than $1.56 trillion and is held both inside and outside the country.1 Foreign demand is influenced by the political and economic uncertainties associated with some foreign currencies, and recent estimates suggest that between one-half and two-thirds of the value of currency in circulation is held abroad. Some residents of foreign countries hold dollars as a store of value, whereas others use it as a medium of exchange. Federal Reserve notes make up more than 99 percent of all U.S. currency in circulation. Each year the Federal Reserve Boarddetermines new currency demand and submits a print order to the Treasury’s Bureau of Engraving and Printing (BEP). The order represents the Federal Reserve System’s estimate of the amount of currency the public will need in the upcoming year and reflects estimated changes in currency usage and destruction rates of unfit currency. Table 15.2 shows how long we can expect our money to last on average. How Long Will Your Money Last? Have you ever wondered how quickly money wears out from being handled or damaged? Not surprisingly, smaller denominations have a shorter life span.$1 bill 5.8 years
$5 bill 5.5 years$10 bill 4.5 years
$20 bill 7.9 years$50 bill 8.5 years
$100 bill 15.0 years Table15.2 Source: “How Long Is the Lifespan of U.S. Paper Money?” https://www.federalreserve.gov, accessed September 7, 2017. Demand deposits consist of money kept in checking accounts that can be withdrawn by depositors on demand. Demand deposits include regular checking accounts as well as interest-bearing and other special types of checking accounts. Time deposits are deposits at a bank or other financial institution that pay interest but cannot be withdrawn on demand. Examples are certain savings accounts, money market deposit accounts, and certificates of deposit. Economists use two terms to report on and discuss trends in the U.S. monetary system: M1 and M2. M1 (the M stands for money) is used to describe the total amount of readily available money in the system and includes currency and demand deposits. As of August 2017, the M1 monetary supply was$3.5 trillion. M2includes all M1 monies plus time deposits and other money that is not immediately accessible. In August 2017, the M2 monetary supply was \$13.6 trillion.2 Credit cards, sometimes referred to as “plastic money,” are routinely used as a substitute for cash and checks. Credit cards are not money; they are a form of borrowing. When a bank issues a credit card to a consumer, it gives a short-term loan to the consumer by directly paying the seller for the consumer’s purchases. The consumer pays the credit card company after receiving the monthly statement. Credit cards do not replace money; they simply defer payment.
## CONCEPT CHECK
1. What is money, and what are its characteristics?
2. What are the main functions of money?
3. What are the three main components of the U.S. money supply? How do they relate to M1 and M2?
15.2: Show Me the Money is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by LibreTexts. | 3,409 | 12,035 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.194566 |
https://ccssmathanswers.com/eureka-math-grade-6-module-2-lesson-14/ | 1,716,812,715,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00318.warc.gz | 133,584,226 | 58,452 | ## Engage NY Eureka Math 6th Grade Module 2 Lesson 14 Answer Key
### Eureka Math Grade 6 Module 2 Lesson 14 Example Answer Key
Example 1.
0.5 ÷ 0.1
Rewrite 0. 5 ÷ 0.1 as a fraction.
$$\frac{0.5}{0.1}$$
Express the divisor as a whole number.
$$\frac{0.5}{0.1} \times \frac{10}{10}=\frac{5}{1}=5$$
Example 2.
Evaluate the expression. First, convert the decimal division expression to a fractional division expression in order to create a whole number divisor.
25.2 ÷ 0.72
$$\frac{25.2}{0.72} \times \frac{100}{100}=\frac{2,520}{72}$$
Use the division algorithm to find the quotient.
25.2 ÷ 0.72 = 35
Example 3.
A plane travels 3, 625. 26 miles in 6. 9 hours. What is the plane’s unit rate?
Represent this situation with a fraction.
$$\frac{3,625.26}{6.9}$$
Represent this situation using the same units.
36,252.6 tenths ÷ 69 tenths
Estimate the quotient.
35,000 ÷ 70 = 500
Express the divisor as a whole number.
$$\frac{3,625.26}{6.9} \times \frac{10}{10}=\frac{36,252.6}{69}$$
Use the division algorithm to find the quotient.
Use multiplication to check your work.
525.4 × 69 = 36,252.6
### Eureka Math Grade 6 Module 2 Lesson 14 Exercise Answer Key
Convert the decimal division expressions to fractional division expressions in order to create whole number divisors. You do not need to find the quotients. Explain the movement of the decimal point. The first exercise has been completed for you.
Exercise 1.
18.6 ÷ 2.3
$$\frac{18.6}{2.3} \times \frac{10}{10}=\frac{186}{23}$$
18.6 ÷ 23
I multiplied both the dividend and the divisor by ten, or by one power of ten, so each decimal point moved one place to the right because they grew larger by ten.
Exercise 2.
14.04 ÷ 4.68
$$\frac{14.04}{4.86} \times \frac{100}{100}=\frac{1404}{486}$$
1,404 ÷ 486
I multiplied both the dividend and divisor by one hundred, or by two powers of ten, so each decimal point moved two places to the right because they grew larger by two powers of ten.
Exercise 3.
0.162 ÷ 0036
$$\frac{0.162}{0.036} \times \frac{1,000}{1,000}=\frac{162}{36}$$
162 ÷ 36
I multiplied both the dividend and divisor by one thousand, or three powers often, so each decimal point moved three places to the right because they grew larger by three powers of ten.
Convert the decimal division expressions to fractional division expressions in order to create whole number divisors. Compute the quotients using the division algorithm. Check your work with a calculator.
Exercise 4.
2,000 ÷ 3.2
$$\frac{2,000}{3.2} \times \frac{10}{10}=\frac{20,000}{32}$$
20000 ÷ 32 = 625
625 × 32 = 20000
2,000 ÷ 3.2 = 625
625 × 3.2 = 2,000
Exercise 5.
3,581.9 ÷ 4.9
$$\frac{3,581.9}{4.9} \times \frac{10}{10}=\frac{35,819}{49}$$
35,819 ÷ 49 = 731
731 × 49 = 35,819
3,581.9 ÷ 4.9 = 731
731 × 4.9 = 3,581.9
Exercise 6.
893.76 ÷ 0.21
$$\frac{893.76}{0.21} \times \frac{100}{100}=\frac{89,376}{21}$$
89,376 ÷ 21 = 4,256
4,256 × 21 = 89,376
893.76 ÷ 0.21 = 4,256
4,256 × 0.21 = 893.76
Exercise 7.
6.194 ÷ 0.326
$$\frac{6.194}{0.326} \times \frac{1,000}{1,000}=\frac{6,194}{326}$$
6,194 ÷ 326 = 19
19 × 326 = 6,194
6.194 ÷ 0.326 = 19
19 × 0.326 = 6.194
### Eureka Math Grade 6 Module 2 Lesson 14 Problem Set Answer Key
Convert decimal division expressions to fractional division expressions to create whole number divisors.
Question 1.
35.7 ÷ 0.07
$$\frac{35.7}{0.07} \times \frac{100}{100}=\frac{3570}{7}$$
Question 2.
486.12 ÷ 0.6
$$\frac{486.12}{0.6} \times \frac{10}{10}=\frac{4,861.2}{6}$$
Question 3.
3.43 ÷ 0.035
$$\frac{3.43}{0.035} \times \frac{1,000}{1,000}=\frac{3430}{35}$$
Question 4.
5418.54 ÷ 0.009
$$\frac{5,418.54}{0.009} \times \frac{1,000}{1,000}=\frac{5,418,540}{9}$$
Question 5.
812.5 ÷ 1.25
$$\frac{812.5}{125} \times \frac{100}{100}=\frac{81,250}{125}$$
Question 6.
17.343 ÷ 36.9
$$\frac{17.343}{36.9} \times \frac{10}{10}=\frac{173.43}{369}$$
Estimate quotients. Convert decimal division expressions to fractional division expressions to create whole number divisors. Compute the quotients using the division algorithm. Check your work with a calculator and your estimates.
Question 7.
Norman purchased 3. 5 lb. of his favorite mixture of dried fruits to use in a trail mix. The total cost was $16.87. How much does the fruit cost per pound? Answer: $$\frac{16.87}{3.5} \times \frac{10}{10}=\frac{168.7}{35}$$ Estimate 16 ÷ 4 = 4 The dried fruit costs$4.82 per pound. This is close to my estimate of 4.
Question 8.
Divide: 994.14÷ 18.9
$$\frac{994.14}{18.9} \times \frac{10}{10}=\frac{9,941.4}{189}$$
Estimate: 100 ÷ 2 = 50
The quotient is 52.6. This is close to my estimate of 50.
Question 9.
Daryl spent $4. 68 on each pound of trail mix. He spent a total of$14. 04. How many pounds of trail mix did he purchase?
$$\frac{14.04}{4.68} \times \frac{100}{100}=\frac{1,404}{468}$$
Estimate: 15 ÷ 5 = 3
Daryl purchased 3 pounds of trail mix. This is my estimate.
Question 10.
Mamie saved $161. 25. This is 25% of the amount she needs to save. How much money does Mamie need to save? Answer: $$\frac{161.25}{0.25} \times \frac{100}{100}=\frac{16,125}{25}$$ Estimate: 1,600 ÷ 2 = 800 Mamie needs to save$645. This Is close to my estimate of 800.
Question 11.
Kareem purchased several packs of gum to place in gift baskets for $1.26 each. He spent a total of$8. 82. How many packs of gum did he buy?
$$\frac{8.82}{1.26} \times \frac{100}{100}=\frac{882}{126}$$
Estimate: 9 ÷ 1 = 9
Kareem bought 7 packs of gum. This is close to my estimate of 9.
Question 12.
Jerod is making candles from beeswax. He has 132. 72 ounces of beeswax. If each candle uses 8.4 ounces of beeswax, how many candles can he make? Will there be any wax left over?
$$\frac{132.72}{8.4} \times \frac{10}{10}=\frac{1,327.2}{84}$$
Estimate: 120 ÷ 8 = 15
The quotient is 15.8. This means that Jerod can make 15 candles. This is close to my estimate. There will be wax left over.
Question 13.
There are 20. 5 cups of batter in the bowl. This represents 0.4 of the entire amount of batter needed for a recipe. How many cups of batter are needed?
$$\frac{20.5}{0.4} \times \frac{10}{10}=\frac{205}{4}$$
Estimate: 200 ÷ 4 = 50
The number of cups of batter needed for the recipe is 51.25. This is close to my estimate.
Question 14.
Divide: 159.12 ÷ 6.8
$$\frac{159.12}{6.8} \times \frac{10}{10}=\frac{1,591.2}{68}$$
Estimate: 160 ÷ 8 = 20
The quotient is 23.4. This is close to my estimate.
Question 15.
Divide: 167.67 ÷ 8.1
$$\frac{167.67}{8.1} \times \frac{10}{10}=\frac{1,676.7}{81}$$
Estimate: 160 ÷ 8 = 20
The quotient is 20.7. This is close to my estimate.
### Eureka Math Grade 6 Module 2 Lesson 14 Exit Ticket Answer Key
Estimate quotients. Convert decimal division expressions to fractional division expressions to create whole number divisor. Compute the quotient using the division algorithm. Check your work with a calculator and your estimate.
Question 1.
Lisa purchased almonds for $3. 50 per pound. She spent a total of$24. 50. How many pounds of almonds did she purchase?
$$\frac{24.50}{3.50} \times \frac{100}{100}=\frac{245}{35}$$
Estimate: 270 ÷ 30 = 9
Lisa purchased 7 pounds of almonds. This is close to my estimate of 9.
7 × 35 = 245
7 × 3.5 = 24.5
Question 2.
Divide: 125.01 ÷ 5.4.
$$\frac{125.01}{5.4} \times \frac{10}{10}=\frac{1250.1}{54}$$
Estimate: 125 ÷ 5 = 25
The quotient of 125.01 and 5.4 is 23.15. This is close to my estimate of 25.
23.15 × 54 = 1250.1
23.15 × 5.4 = 125.01
### Eureka Math Grade 6 Module 2 Lesson 14 Opening Exercise Answer Key
Divide $$\frac{1}{2} \div \frac{1}{10}$$ Use a tape diagram to support your reasoning.
This question is asking the following: $$\frac{1}{2} \text { is } \frac{1}{10}$$ of what number?
Relate the model to the invert and multiply rule.
$$\frac{1}{2} \div \frac{1}{10}=\frac{1}{2} \times \frac{10}{1}=\frac{10}{2}=5$$
Another way $$\frac{1}{2} \div \frac{1}{10}$$
→ Let’s look at the Opening Exercise another way. We can represent $$\frac{1}{2} \div \frac{1}{10}$$ using decimals.
→ We can represent the fractions $$\frac{1}{2}$$ and $$\frac{1}{10}$$ with which decimals?
One-half can be represented with 0.5, and one-tenth can be represented by 0.1.
→ Display the following:
$$\frac{1}{2} \div \frac{1}{10}$$
0.5 ÷ 0.1
→ This expression can be represented with the same interpretation as the Opening Exercise: 5 tenths is 1 tenth of what number?
→ Let’s model this question with a tape diagram.
1 unit → 0.5
10 units → 10 × 0.5 = 5
→ What do you notice about the diagram and the quotient?
The diagram is set up exactly the same as in the Opening Exercise with the fraction, and the quotient is the same. Since $$\frac{1}{2}$$ is equivalent to 0.5, and $$\frac{1}{10}$$ is equivalent to 0.1, the quotients must be the same.
→ Look back at the problem: 0.5 ÷ 0.1. Rewrite this division expression as a fraction.
$$\frac{0.5}{0.1}$$
→ How can we express the divisor as a whole number?
Multiply by a fraction equal to one.
→ Choose a fraction to multiply in order to express the divisor as a whole number.
I could multiply by $$\frac{0.5}{0.1}$$ by $$\frac{10}{10}$$ to represent the divisor as the whole number 1.
→ Find the Product of $$\frac{0.5}{0.1} \times \frac{10}{10}$$
$$\frac{0.5}{0.1} \times \frac{10}{10}=\frac{5}{1}=5$$
→ What do you notice about the quotient?
It is the same as when we used a tape diagram to determine the quotient.
→ Why do you think the quotients are the same?
They are the same because I multiplied the divisor and the dividend by the same power of ten.
→ What conjecture can you make?
Because the divisor and the dividend both became ten times greater, when we write the numbers as ten times as much, we move the decimal to the right one place.
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Scroll to Top | 3,431 | 9,754 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2024-22 | latest | en | 0.741736 |
https://aakashdigitalsrv1.meritnation.com/ask-answer/question/abcd-is-a-square-determine-angle-dca/quadrilaterals/1614532 | 1,653,306,664,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00017.warc.gz | 126,384,547 | 8,641 | # ABCD is a square. Determine Angle DCA.
Given: A square ABCD
To Find: ∠DCA i.e., ∠2
Solution: We know all sides of a square are equal.
∴ AB = BC = CD = DA
So, in ∆ ACD, AD = CD
⇒ ∠2 = ∠1 (Angles opposite to equal sides are equal)
Now in ∆ ACD, by angle sum property of ∆,
∠ADC + ∠1 + ∠2 = 180°
⇒ 90° + 2 ∠2 = 180° (Each angle of square = 90°)
⇒ 2 ∠2 = 180° – 90°
⇒ 2 ∠2 = 90°
⇒ ∠2 = 45°
Hence, ∠DCA = 45°.
• 5
obviouslyy 90 degree
• -2
no wrong
• 1
boss i guess its 45 degree!
• 0
oops sorry wrong interpretation!
• 2
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# Problem 2678. Find out sum and carry of Binary adder
Solution 3321863
Submitted on 22 Oct 2020 by chenguang yan
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=1; pc=1; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,1)) assert(isequal(carry,1))
sum = 1 carry = 1
2 Pass
x = 1; y=1; pc=0; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,0)) assert(isequal(carry,1))
sum = 0 carry = 1
3 Pass
x = 1; y=0; pc=0; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,1)) assert(isequal(carry,0))
sum = 1 carry = 0
4 Pass
x = 0; y=0; pc=0; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,0)) assert(isequal(carry,0))
sum = 0 carry = 0
5 Pass
x = 1; y=1; pc=0; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,0)) assert(isequal(carry,1))
sum = 0 carry = 1
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10.40) Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 65.0L and which contains O2 gas at a pressure of 16,500kPa at 23°C. a. What mass of O2 gas does the tank contain? b. What volume would the gas occupy at STP? c. At what temperature would the pressure in the tank equal 150.0atm? d. What would be the pressure of the gas, in kPa, if it were transferred to a container at 24°C whose volume is 55.0L?
10.97) Ammonia, NH3(g), and hydrogen chloride, HCl(g), react to form solid a ammonium chloride, NH4Cl(s):
NH3(g) + HCl(g)  NH4Cl(s)
Two 2.00-L flasks at 25°C are connected by a stopcock. One flask contains 5.00g NH3(g), and the other contains 5.00g HCl(g).
When the stopcock is opened, the gases react until one is completely consumed. a. Which gas will remain in the system after the reaction is complete? b. What will be the final pressure of the system after the reaction is complete? (Neglect the volume of the ammonium chloride formed.)
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11.22) Propyl alcohol (CH3CH2CH2OH) and isopropyl alcohol [(CH3)2CHOH], whose space-filling models have boiling points of 97.2°C and 82.5°C, respectively. Explain why the boiling point of propyl alcohol is higher, even though both have the molecular formula of C3H8O.
11.40) The fluorocarbon composed C2Cl3F3 has a normal boiling point of 47.6°C. The specific heats of C2Cl3F3(l) and C2Cl3F3(g) are 0.91J/g-K and 0.67J/g-K, respectively. The heat of vaporization for the compound is 27.49kJ/mol. Calculate the heat required to convert 25.0g of C2Cl3F3 from a liquid at 5.00°C to a gas at 82.00°C.
11.60) Rutile is a mineral composed of Ti and O. Its unit cell contains Ti atoms at each corner and a Ti atom at the center of the cell. Four O atoms are on the opposite faces of the cell, and two are entirely within the cell. a. What is the chemical formula of this mineral? b. What is the nature of the bonding that holds the solid together?
11.78) For each of the following pairs of substances, predict which will have the higher melting point, and indicate why. a. HF, HCl; b. C (graphite), CH4; c. KBr, Br2; d. LiF, MgF2.
12.28) Write the structural formulas for the two substances that are condensed to form Kevlar ®.
12.38) Silicon carbide, SiC, is a three dimensional structure. Describe how the bonding and structure of SiC lead to its great thermal stability (to 2700°C) and exceptional hardness.
12.84) Although polyethylene can twist and turn in random ways the most stable form is a linear one with the carbon backbone. a. What is the hybridization of orbitals at each carbon atom? What angles do you expect between the bonds? b. Now imagine that the polymer is polypropylene rather than polyethylene. Draw structures for polypropylene in which (i) the CH3 groups all lie on the same side of the plane of the paper; (ii) the CH3 groups lie on alternating sides of the paper; or (iii) the CH3 groups are randomly distributed on either side. Which of these forms would you expect to have the highest crystallinity an melting point, and which the lowest? Explain in terms of intermolecular interactions and molecular shapes. c. Polypropylene fibers have been employed in athletic wear. The product is said to be superior to cotton or polyester clothing in wicking moisture away from the body through fabric to the outside. Explain the differenced between polypropylene and polyester or cotton, in terms of intermolecular interactions with water.
13.44) The density of toluene (C7H8) is 0.867g/mL, and the density of thiophene (C4H4S) is 1.065g/mL. A solution is made by dissolving 10.0g of thiophene in 250.0mL of toluene. a. Calculate the mole fraction of thiophene in the solution. b. Calculate the molality of thiophene in the solution. c. Assuming that the volumes of the solute and solvent are additive, what is the molarity of thiophene in the solution?
13. 72) Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of 5.00g of lauryl alcohol in 0.100kg of benzene freezes at 4.1°C. What is the approximate molar mass of lauryl alcohol?
13.90) Acetonitrile (CH3CN) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1.80 M LiBr solution in acetonitrile is 0.826g/cm3. Calculate the concentration of the solution in a. molality, b. mole fraction of LiBr, c. mass percentage of CH3CN. Solution Summary
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Use this geometry exercise with your third graders to practice recognizing the characteristics of a cylinder and finding real-life examples of the three-dimensional shape. For an added challenge, see just how many different shapes your students can name! | 449 | 2,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2022-33 | latest | en | 0.865301 |
https://bullishbears.com/extrinsic-value/ | 1,716,082,861,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00736.warc.gz | 135,795,135 | 71,678 | # Extrinsic Value in Options
Extrinsic value is one of the main components of options trading. It measures the difference between the market price of an options contract, which is the premium, and its intrinsic value. It is one of the moving parts of options. Extrinsic value is, in essence, value from the outside.
Intrinsic value is the calculated value of a company. It’s found using tangibles and intangibles, also known as fundamental analysis.
You can’t have extrinsic value without intrinsic value. They work hand in hand. EV has other factors that make up the value of an option besides the strike price. At the same time, intrinsic value is the inherent worth of an option contract. Selling options instead of buying them is a very profitable strategy to learn.
Extrinsic value measures the difference in the price of the options, also known as the premium and intrinsic value. The extrinsic value is found by subtracting the intrinsic value from the price of an options contract. However, intrinsic and extrinsic value make up the cost of an options contract.
You can also look at extrinsic value as the risk premium of an option. People like to define their risk. The option writer takes on unlimited risk by creating the option. The buyer of that option has limited risk with unlimited profit potential.
You would have to buy or sell naked calls or puts to have unlimited profit potential. However, strategies like spreads cut down on risk. They also put a cap on profit potential. That isn’t a bad thing, though. You don’t need to hit a home run every single trade.
Breaking down extrinsic value ends up being a lot about intrinsic value. As stated above, they work in tandem. You can’t have one without the other because they make up the price of an option.
For example, let’s say you want to buy a call option with a strike price of \$82. The stock is trading at 82.74 currently. The price of the contract is \$2.46. The option’s intrinsic value is \$0.75, and the extrinsic value is \$1.64. This process becomes easier to learn the more that you paper trade options.
That options contract is in the money. In the money for a call option means the strike price is below the market price. As a result, a call option has value when a stock is trading below the strike price; the premium you’re paying comes from the EV.
There’s also the bearish side to that: put options. If a put option has value when a stock trades above the strike price, the option premium comprises the EV.
### Extrinsic Value Example
This is an example of extrinsic value on the ThinkorSwim platform. The columns are customizable so you can add columns such as volume, open interest, strike price, probability ITM, probability OTM, and the Greeks, delta, gamma, theta, and vega
The options chain is simple to read on most trading platforms. There’s also the ability to look at an options chart.
### Time Value
Did you know that extrinsic value is also known as time value? Time value is very important in an options contract. The time left to expire on an options contract affects its price. Normally, a contract loses value the closer to expiration it gets. That’s why options contracts closest to your purchase date are cheaper. As a result, you pay less money to place the trade.
However, your risk goes up a lot. The profit potential is limited as well. You want to purchase options contracts with a lot of time value. For example, an out-of-the-money contract a couple of months from expiration has more EV than one with five days to expiration.
Implied volatility is another factor that affects EVs. Implied volatility maps out how much a stock can move in a certain amount of time.
The more implied volatility increases, the more the EV increases. As a trader, you want volatility. It means a stock is going to move. When a stock moves, you can profit or lose. It depends on if you’ve made the correct trade.
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## Factors Affecting Extrinsic Value
### 1. Length of the Contract
The length of a contract, or the time to expiration, significantly impacts the EV of an option. Generally, options with longer expiration dates have higher extrinsic values than options with shorter expiration dates. Regarding options, longer timeframes provide a higher chance for the stock price to move and hit the desired price for the option holder.
Here’s how the length of a contract affects extrinsic Value:
### 2. Increased Time Value
Options with longer expiration dates have more time for potential price movements in the underlying stock. This extended time increases the probability of the option becoming profitable, leading to higher time value and extrinsic Value.
### 3. Greater Flexibility
Longer-dated options provide more flexibility to the option holder. They allow adjustments to trading strategies over an extended period, contributing to the option’s EV.
### 4. More Opportunities for Volatility
Options with longer timeframes are more likely to experience significant price fluctuations and changes in implied volatility, which can enhance the option’s EV.
It’s worth noting that options with longer expiration dates have higher extrinsic values but generally have higher premiums than options with shorter expiration dates. This is due to the additional time value and potential for larger price movements that longer-dated options carry.
In summary, the length of a contract, or time to expiration, is a key factor influencing the EV of an option. Longer expiration dates generally result in higher extrinsic values due to increased time value, flexibility, and potential for price fluctuations.
## Extrinsic Values for Calls and Puts
Did you know that the extrinsic values of call (right to buy) and put (right to sell) options with the same strike price – also known as the exercise price – can differ? While call options benefit from higher interest rates through an increase in EV, put options are negatively affected and experience a decrease in extrinsic value.
Despite reducing the extrinsic value of call options, dividends tend to raise the extrinsic value of put options by ensuring that the stock will experience a decline.
The extrinsic values of call and put options for a particular stock generally differ, depending on whether the stock is rising or falling. Due to this, investors tend to purchase more call or put options, thus increasing the EV of these options due to the heightened trading activity.
### Greatest Options Trade of All Time
Legend has it that in 1987, a trader mistakenly believed he had purchased one put option on the S&P 500. However, it turned out that he had bought 1,000 put options instead. This fortunate mistake earned him a profit of well over \$10 million!
### Key Takeaways
• Extrinsic value measures the difference between the market price of an options contract, which is the premium, and its intrinsic value.
• Contracts tend to lose value as they near their end date. This is because there is less time for the prices of securities to move in the holder’s favor.
• The volatility of the underlying security has a direct impact on the EV of an option. If the volatility in the underlying security goes up, the option’s extrinsic value will also go up. And if the volatility decreases, the option’s extrinsic value will also decrease.
• Generally, options with longer expiration dates have higher extrinsic values than options with shorter expiration dates.
• Even when two options have the same strike price and intrinsic value, the EV differs due to their different expiration dates.
### Final Thoughts: Extrinsic Value
Many different factors affect the extrinsic value of options contracts. The Greeks comprise a large component, especially theta, aka time decay. Other factors include economic news, company news, and global economic events.
Extrinsic value makes up an option’s premium. Time matters when opening a trade. Options have more moving parts than shares, so make sure you take the time to study and understand how to trade them.
If you need more help, take our options trading course.
MCD has an expiration date of Jan, 19, 2024. The extrinsic value is \$3.19 on the \$295 strike price. This means that the options contract loses \$3.19 of extrinsic value each day until expiration.
Intrinsic value refers to the in-the-money (ITM) portion of an options contract. Extrinsic value refers to the out-of-the-money (OTM) portion of an options contract.
Yes, extrinsic value can be negative. This occurs when an option's overall premium is lower than its intrinsic value. The option's negative extrinsic value indicates that market factors, such as low volatility or short time to expiration, have diminished its time value component.
In options trading, extrinsic value is calculated by subtracting the market price of an option (a.k.a. its premium) from its intrinsic price.
### Cash Secured Put
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Just choose the course level that you’re most interested in and get started on the right path now. Become a leader, not a follower. When you’re ready you can join our chat rooms and access our Next Level training library. No rush. We’re here to help. | 2,487 | 12,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.933579 |
mkgnu.net | 1,603,822,334,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894426.63/warc/CC-MAIN-20201027170516-20201027200516-00477.warc.gz | 436,046,389 | 3,537 | March 2017, rev. February 2019
I was mistaken about the performance of this approach. Simulating the stack ran ~2x-2.5x slower on my processor.
There is a performance optimization that as far as I know no compiler to date has done. That is to convert recursion to iteration when the recursive function call is not at the end of the function. [1]
The reason to want to do this conversion is that iteration is faster than recursion. Iteration avoids the overhead of creating a new stack frame, while still letting the programmer think in recursion. [2]
Some compilers do a similar conversion called tail-call optimization. When the recursive function is at the end of the function (the tail of the function) the compiler resets the local variables and jumps (using a goto) to the beginning of the function. [3] This has the effect we are after: it avoids creating a new stack frame. But it doesn't work if there's more to do after the call returns. Since the local variables have been wiped out, the program would corrupt its data if it were to reuse them when unwinding from the recursion.
The conversion I'm suggesting is different than tail-call optimization. Although it bypasses the stack, it doesn't throw it out completely. And it applies to the stack of all functions called by a function, rather than just one function called recursively. A good name for it might be: stack unrolling.
The conversion may look familiar. It's a twist to what the processor already does:
For every function call within a function, create a local array to simulate the callee function's stack and set a pointer to the beginning of this array. Before the function call, increment this pointer (like a push) and go to (with a jump or goto) an inline copy of the callee function code. When the callee function code finishes, decrement this pointer (like a pop) and goto the line after the function was called.
This has the same effect as function calls but the stack is simulated. There's a stack per function instead of one stack for all functions. Simulating the stack is faster than making a function call.
All code is inlined in a single function and executed with gotos. With this conversion a whole program can run in a single function. [4]
I can imagine a downside though. If this is applied as a source to source conversion the end result will look like spaghetti code. That's probably why this conversion is rarely done by hand.
All these years we may have been using the stack badly, but we're only one optimization away from not having to use it at all.
How to remove the heap
There is another performance optimization that as far as I know no compiler to date has done. That is to convert automatically a program that builds a control flow graph using pointers into a program that builds a control flow graph using an array. For example, in the recursive descent parser of a compiler.
A good name for this conversion might be: pointer-to-array unrolling.
The reason to want to do this conversion is to avoid thrashing the cache. Programs written with linked lists don't take advantage of the cache, because pointers are often allocated in non-contiguous memory addresses that cause cache misses. Whereas programs written with arrays traversed sequentially benefit not only from cache hits but also from cache prefetching.
When I benchmarked building a parse tree with an array, it was 80% faster than building it with malloc.
The main reason programs are written with linked lists to begin with is to allow for an unknown list size. But I wouldn't be surprised if in most cases this unknown size is small enough to fit in a fixed-size array or an array that's resized dynamically when full. A second reason is to be able to insert elements anywhere in the middle of the list; a fixed-size array can't do this fast. A third reason is it helps to think of the graph as a graph instead of as an array.
Impact
These two conversions touch two different areas of a program. Stack unrolling applies to local variables; these are always on the stack. Pointer-to-array unrolling applies to data structures built on the heap; these could be saved on a single stack.
If as Harold Abelson said "programs must be written for humans to read, and only incidentally for machines to execute," compilers that want to produce faster code for programs written for humans to read must implement these two optimizations.
Notes:
[1] Here is one example of this conversion that trades the stack for the heap in an application instead of a compiler. Here is another that uses an interpreter and requires a scheduler function. [2] I got distracted by this problem while trying to write a fast parser. Parsers are written recursively, where one function may call several others. So this conversion would help when writing the parsers of interpeters and compilers and databases. [3] The name "tail-call" is slightly misleading, because the function call doesn't have to be at the end of the function. It could be in the middle of a switch statement, or in an if clause, or in an expression. It's more accurate to say that a tail call is the last piece of code executed in the function. A better name might be "last-call". [4] A similar example is a feature that never made it into ANSI C: running with jump tables. For gcc to go through the trouble of adding special support for this rare feature there must have been something powerful to gain. The gain is that inlining all code in a single function and running with jump tables runs fast. | 1,139 | 5,532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-45 | latest | en | 0.935725 |
http://mathhelpforum.com/math-topics/201306-gradient-problems-urgent.html | 1,526,939,525,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864544.25/warc/CC-MAIN-20180521200606-20180521220606-00112.warc.gz | 185,418,285 | 10,731 | Find the coordinates of the points on the curve y=x^3 where the gradient is 12
P.S Can people explain this in detail as I am horrible at maths
2. ## Re: Gradient problems URGENT
Gradient means inclination and inclination points to something like 'Slope'
Now, what is slope ?
It is simply $\displaystyle \frac{dy}{dx}$ for any function y=f(x)
so here you have f(x)=x3
$\displaystyle \frac{dy}{dx} = 3x^2$
$\displaystyle x^2 = 4$
$\displaystyle x=+4, -4$
Now you have the value of x. To find the complete set (x,y) just plug the values of x in to f(x)
f(4)=64
f(-4)=-64
3. ## Re: Gradient problems URGENT
Originally Posted by pratique21
Gradient means inclination and inclination points to something like 'Slope'
Now, what is slope ?
It is simply $\displaystyle \frac{dy}{dx}$ for any function y=f(x)
so here you have f(x)=x3
$\displaystyle \frac{dy}{dx} = 3x^2$
$\displaystyle x^2 = 4$
$\displaystyle x=+4, -4$
Now you have the value of x. To find the complete set (x,y) just plug the values of x in to f(x)
f(4)=64
f(-4)=-64
Thanks. This some has been bothering me for some time!!!!
4. ## Re: Gradient problems URGENT
Originally Posted by pratique21
Gradient means inclination and inclination points to something like 'Slope'
Now, what is slope ?
It is simply $\displaystyle \frac{dy}{dx}$ for any function y=f(x)
so here you have f(x)=x3
$\displaystyle \frac{dy}{dx} = 3x^2$
$\displaystyle x^2 = 4$
$\displaystyle x=+4, -4$
Now you have the value of x. To find the complete set (x,y) just plug the values of x in to f(x)
f(4)=64
f(-4)=-64
One more question! How did you get x^2 = 4 in the first place? After getting dy/dx = 3x^2 . You then lost me!
5. ## Re: Gradient problems URGENT
i put the value of dy/dx that is given as 12
so cancelling 3 gives me x^2 = 4
6. ## Re: Gradient problems URGENT
But then after that how did you still find x? If you cancel 3 then its x^2=4
7. ## Re: Gradient problems URGENT
okay, now take the entire thing to one side (i.e subtract 4 from both sides)
so $\displaystyle x^2 = 4$
become: $\displaystyle x^2 - 4 =0$
now use $\displaystyle a^2 - b^2 = (a+b) \cross (a-b)$
so you have:- $\displaystyle (x-2) \cross (x+2)=0$
this is true when either $\displaystyle (x-2)= 0$ or$\displaystyle (x+2)=0$ or both are zero.
So x= $\displaystyle \pm 2$
Thanks dood | 734 | 2,302 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-22 | latest | en | 0.804491 |
https://www.examfriend.in/questions-and-answers/Arithmetic-aptitude/Percentage/General%20Questions/3.html | 1,675,650,905,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500303.56/warc/CC-MAIN-20230206015710-20230206045710-00338.warc.gz | 762,994,660 | 10,659 | <<12345>>
16.
A man loses 12 $\frac{1}{2}$% of his money and after spending 70% of the remainder, has Rs 210 left. At first the man had?
A) Rs.880
B) Rs.720
C) Rs.800
D) Rs.600
E) Rs. 850
17.
Naresh secured 50% marks in Hindi, 60% in English and 70 % in Maths as well in Science. What were his total marks, If the maximum marks obtainable in each of these 4 subjects was 50?
A) 120
B) 125
C) 150
D) 200
E) 250
18.
In an election there were two candidates A and B, 20% of voters did not vote, 10% of polled votes were declared invalid. A received 50% votes of polling and won by 600 votes .What was the total number of voters?
A) 7000
B) 7200
C) 7500
D) 7650
E) 7750
19.
In an examination in which full marks were 500, A got 10% less than B, B 25% more than C and C 20% less than D. If A got 360 marks. Marks obtained by D is?
A) 60%
B) 80%
C) 40%
D) 70%
E) 755
20.
If two numbers are respectively 30% and 40 % more than a third number ,what the percent is the first o the second ?
A) 85%
B) $92\frac{6}{7}$ %
C) $92\frac{7}{6}$
D) $79\frac{6}{7}$
E) $92\frac{7}{6}$
<<12345>> | 402 | 1,111 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-06 | latest | en | 0.938395 |
https://mersenneforum.org/showthread.php?s=2558ef9ae8eaa44c600d6ad0f18fbaa9&p=570833 | 1,619,132,326,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039563095.86/warc/CC-MAIN-20210422221531-20210423011531-00273.warc.gz | 474,318,947 | 10,252 | mersenneforum.org February 2021
Register FAQ Search Today's Posts Mark Forums Read
2021-02-01, 16:03 #1 Xyzzy "Mike" Aug 2002 1F8316 Posts February 2021
2021-02-03, 00:19 #2 Kebbaj "Kebbaj Reda" May 2018 Casablanca, Morocco 83 Posts the first people vaccinated by itself ? Attached Thumbnails
2021-02-03, 09:11 #3
Dieter
Oct 2017
2·5·11 Posts
Quote:
Originally Posted by Kebbaj the first people vaccinated by itself ?
That are 17 people. I have 39.
Example:
(4/1) has only (5/1) as trusted neighbour. (5/1) is vaccinated. So /4/1) will get vaccinated, too - if my understanding of the challenge is ok.
2021-02-03, 10:23 #4 SmartMersenne Sep 2017 32·11 Posts The following people would be vaccinated with no additional convincing other than what is given in the question: 101011000011 100000000001 100100000001 110001000001 100001000001 100000011100 100000011000 000000000000 000000000000 000000101101 100000000001 101110110101
2021-02-03, 11:29 #5
Dieter
Oct 2017
2×5×11 Posts
Quote:
Originally Posted by SmartMersenne The following people would be vaccinated with no additional convincing other than what is given in the question: 101011000011 100000000001 100100000001 110001000001 100001000001 100000011100 100000011000 000000000000 000000000000 000000101101 100000000001 101110110101
That's exactly my result.
2021-02-03, 15:21 #6 SmartMersenne Sep 2017 32×11 Posts Here is a solution set with 60 elements: {(4,7),(6,1),(5,3),(4,4),(7,5),(2,7),(6,2),(6,5),(12,7),(2,11),(4,6),(3,9),(3,2),(9,12),(9,11),(5,2),(6,7),(2,4),(6,3),(11,2),(2,10),(7,10),(10,6),(9,4),(10,8),(9,5),(2,12),(7,9),(4,5),(7,7),(9,1),(4,12),(4,2),(7,4),(2,5),(9,2),(6,11),(11,5),(5,5),(11,3),(9,9),(10,12),(7,8),(4,8),(10,4),(8,10),(7,6),(2,2),(5,9),(11,11),(8,9),(3,6),(8,2),(7,12),(12,4),(3,5),(11,8),(10,10),(4,11),(12,6)} If you start with this set, everyone will get vaccinated after very convincing conversations.
2021-02-03, 23:41 #7
Kebbaj
"Kebbaj Reda"
May 2018
Casablanca, Morocco
1238 Posts
Quote:
Originally Posted by Dieter That are 17 people. I have 39. Example: (4/1) has only (5/1) as trusted neighbour. (5/1) is vaccinated. So /4/1) will get vaccinated, too - if my understanding of the challenge is ok.
I'm talking about the following sentence:
(If the set of trusted neighbors is empty, that person is already convinced and gets vaccinated without further convincing.)
which gives 17 person self vacineted.
For exemple:
0
x0
0
After those who have only one convicton with these 17 will be vaccinated and then become 39 without our inetvention.
I confirm I have the same configuration of 39 people.
Last fiddled with by Kebbaj on 2021-02-03 at 23:44
2021-02-04, 00:33 #8
Kebbaj
"Kebbaj Reda"
May 2018
Casablanca, Morocco
10100112 Posts
Quote:
Originally Posted by SmartMersenne Here is a solution set with 60 elements: {(4,7),(6,1),(5,3),(4,4),(7,5),(2,7),(6,2),(6,5),(12,7),(2,11),(4,6),(3,9),(3,2),(9,12),(9,11),(5,2),(6,7),(2,4),(6,3),(11,2),(2,10),(7,10),(10,6),(9,4),(10,8),(9,5),(2,12),(7,9),(4,5),(7,7),(9,1),(4,12),(4,2),(7,4),(2,5),(9,2),(6,11),(11,5),(5,5),(11,3),(9,9),(10,12),(7,8),(4,8),(10,4),(8,10),(7,6),(2,2),(5,9),(11,11),(8,9),(3,6),(8,2),(7,12),(12,4),(3,5),(11,8),(10,10),(4,11),(12,6)} If you start with this set, everyone will get vaccinated after very convincing conversations.
I checked your 60-item solution,everyone will get vaccinated after very convincing conversations! A lot of negotiation after conviction.
There is much less than 60.
2021-02-04, 02:10 #9
SmartMersenne
Sep 2017
32×11 Posts
Quote:
Originally Posted by Kebbaj I checked your 60-item solution,everyone will get vaccinated after very convincing conversations! A lot of negotiation after conviction. There is much less than 60.
Here is one with 50 items:
{(11,6),(3,2),(10,11),(2,6),(5,11),(10,10),(11,3),(6,1),(4,12),(7,11),(2,3),(9,12),(11,11),(5,2),(3,4),(9,5),(8,8),(2,5),(10,6),(8,12),(8,4),(2,10),(10,9),(6,11),(2,12),(2,7),(4,9),(5,5),(11,9),(4,8),(7,5),(1,4),(7,10),(8,10),(3,5),(8,11),(4,3),(10,12),(3,11),(3,10),(6,4),(4,6),(9,10),(11,7),(12,4),(9,8),(7,12),(5,8),(9,2),(6,3)}
2021-02-04, 08:56 #10
Kebbaj
"Kebbaj Reda"
May 2018
Casablanca, Morocco
5316 Posts
Quote:
Originally Posted by SmartMersenne Here is one with 50 items: {(11,6),(3,2),(10,11),(2,6),(5,11),(10,10),(11,3),(6,1),(4,12),(7,11),(2,3),(9,12),(11,11),(5,2),(3,4),(9,5),(8,8),(2,5),(10,6),(8,12),(8,4),(2,10),(10,9),(6,11),(2,12),(2,7),(4,9),(5,5),(11,9),(4,8),(7,5),(1,4),(7,10),(8,10),(3,5),(8,11),(4,3),(10,12),(3,11),(3,10),(6,4),(4,6),(9,10),(11,7),(12,4),(9,8),(7,12),(5,8),(9,2),(6,3)}
The next answer you will say 40 element?
I give it by optimizing your solution to 40:
{(4,9),(7,5),(11,3),(4,12),(7,11),(11,6),(3,11),(6,3),(10,9),(3,5),(6,4),(10,6),(3,4),(6,11),(10,10),(3,2),(6,1),(10,11),(2,7),(5,8),(9,2),(2,12),(5,5),(9,8),(2,5),(5,2),(9,5),(2,3),(12,4),(4,6),(8,10),(2,6),(11,7),(4,3),(8,4),(1,4),(11,9),(4,8),(8,8),(11,11)}
Who says 30?
Last fiddled with by Kebbaj on 2021-02-04 at 09:07
2021-02-04, 12:58 #11
SmartMersenne
Sep 2017
32·11 Posts
Let's not spoil the problem
Quote:
Originally Posted by Kebbaj The next answer you will say 40 element? I give it by optimizing your solution to 40: {(4,9),(7,5),(11,3),(4,12),(7,11),(11,6),(3,11),(6,3),(10,9),(3,5),(6,4),(10,6),(3,4),(6,11),(10,10),(3,2),(6,1),(10,11),(2,7),(5,8),(9,2),(2,12),(5,5),(9,8),(2,5),(5,2),(9,5),(2,3),(12,4),(4,6),(8,10),(2,6),(11,7),(4,3),(8,4),(1,4),(11,9),(4,8),(8,8),(11,11)} Who says 30?
OK, let's stop here before one of us solves the problem and reveals the answer.
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Thu Apr 22 22:58:46 UTC 2021 up 14 days, 17:39, 0 users, load averages: 2.28, 2.53, 2.48 | 2,469 | 5,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-17 | latest | en | 0.76768 |
https://stats.stackexchange.com/questions/572631/conditional-exponential-family-implies-joint-exponential-family | 1,716,415,543,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00634.warc.gz | 469,462,268 | 39,348 | # Conditional exponential family implies joint exponential family
Suppose $$p(X_j | X_1,\ldots,X_{j-1})$$ comes from some known exponential family for every $$j=1,\ldots,k$$. Does it follow that the joint distribution comes from some (possibly different) exponential family?
An obvious example where this is true is Gaussian RVs. Are there conditions under which this generalizes? (Note that unlike the Gaussian case, I am not requiring that the joint belong to the same exponential family; this is not even really well-defined anyway.)
• Since the conditional probability is a function of the other $X_i,$ you seem to be asserting that a sequence of functions of zero, one, ..., through $k-1$ variables "come from some known exponential family." Could you state more specifically what that might mean?
– whuber
Apr 22, 2022 at 17:53
Let us assume that the conditional density writes as $$(j=1,\ldots,d)$$ $$p_j(x_j|x_{-j})= h_j(x_j)\exp\{A_j(x_j)^\text TB_j(\theta_j,x_{-j})-C_j(\theta,x_{-j})\}$$ where the various functions are arbitrary (but such that the density is integrable with mass one wrt the dominating measure). Then by the Hammersley-Clifford theorem, the joint density can be represented as the (Gibbs) product of conditionals $$p(x)\propto\prod_{j=1}^d \dfrac{p_j(x_j|x^0_{j})}{p_j(x^0_j|x^0_{j})}\tag{1}$$ where $$x⁰$$ is an arbitrary (but possible) value. This identity then implies that \begin{align}p_j(x_j|x_{-j})&\propto \prod_{j=1}^d h_j(x_j) \exp\{A_j(x_j)^\text TB_j(\theta_j,x^0_{j})-C_j(\theta,x^0_{j})\\&\qquad-A_j(x^0_j)^\text TB_j(\theta_j,x^0_{j})+C_j(\theta,x^0_{j})\}\\ &\propto H(\mathbf x)\exp\left\{\sum_{j=1}^d\left[A_j(x_j)^\text TB_j(\theta_j,x^0_{j})-A_j(x^0_j)^\text TB_j(\theta_j,x^0_{j})\right]\right\} \end{align} which does not factorise as an exponential family in general, since the functions $$B_j$$ that appear within the exponential do not necessarily separate as functions of $$\theta$$ times functions of $$\mathbf x$$.
In the somewhat different case mentioned in the question, that is when $$p_j(x_j|x_{ the representation (1) still holds and again does not imply a joint exponential family distribution. If one uses instead the product rule $$p(\mathbf x)\propto\prod_{j=1}^d p_j(x_j|x_{ the joint density writes as $$p(\mathbf x)\propto H(\mathbf x)\exp\left\{\sum_{j=1}^d\left[A_j(x_j)^\text TB_j(\theta_j,x_{ and again there is no generic decomposition as an exponential family density. | 735 | 2,447 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-22 | latest | en | 0.880328 |
https://wildpartyofficial.com/what-is-net-force-in-pe/ | 1,726,507,446,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00355.warc.gz | 576,178,248 | 15,183 | # What is net force in PE?
## What is net force in PE?
The net force is the vector sum of all the forces that act upon an object. That is to say, the net force is the sum of all the forces, taking into account the fact that a force is a vector and two forces of equal magnitude and opposite direction will cancel each other out.
## What is the best definition of net force?
In mechanics, the net force is the vector sum of forces acting on a particle or body. The net force is a single force that replaces the effect of the original forces on the particle’s motion. It is possible for a system of forces to define a torque-free resultant force.
What is an example of a net force?
The net force is the force which is the sum of all the forces acting on an object simultaneously. Net force can accelerate a mass. Some of the other forces act on anybody either at rest or motion. The net force is a term used in a system with multiple forces….The net force formula:
F_Net Net force
a Acceleration
### What is net force Short answer?
The net force is defined as is the sum of all the forces acting on an object. Net force can accelerate a mass. Some other force acts on a body either at rest or motion. The net force is a term used in a system when there is a significant number of forces.
### What is the formula for net force?
Net Force Equation Can Be Stated As Follows Fg = Gravitational force. Net force is when a body is in motion and many forces are active on it like gravitational force Fg, frictional force Ff, and the normal force. FNet = Fa + Fg + Ff + FN.
Is net force and resultant force the same?
Net force, also known as resultant force, is a vector quantity produced when two or more forces act upon one object. The concept of net force is the same as the mathematical concept of vector addition.
#### What unit is net force?
newton
When several forces act on a system it is the net, external force that matters. The SI unit of force is the newton , named after the English physicist and mathematician Isaac Newton (1642–1727).
#### In which direction is the net force?
The force that is “left over” after all of the forces acting on an object are cancelled and/or combined is called the net force.
What’s a balanced force?
Balanced forces are equal in size and opposite in direction. When forces are balanced, there is no change in motion. When the forces on an object are equal and in opposite directions, the forces are balanced, and there is no change in motion.
## When the net force on an object is zero?
When an object is in equilibrium (either at rest or moving with constant velocity), the net force acting on it zero. A vector can only have zero magnitude if all of its components are zero.
## What is the formula of net force?
Net force is when a body is in motion and many forces are active on it like gravitational force Fg, frictional force Ff, and the normal force. Therefore, the net force formula is given by. FNet = Fa + Fg + Ff + FN.
What are the two kinds of forces?
There are 2 types of forces, contact forces and act at a distance force. Every day you are using forces. Force is basically push and pull. When you push and pull you are applying a force to an object.
### Is the net force the sum of all the forces?
It is commonly said that in each situation there is a net force acting upon the object. The net force is the vector sum of all the forces that act upon an object.
### How is the net force of an object determined?
The addition of force vectors can be done in the same manner in order to determine the net force (i.e., the vector sum of all the individual forces). Consider the three situations below in which the net force is determined by summing the individual force vectors that are acting upon the objects.
Is there an unbalanced force or a net force?
In each of the above situations, there is an unbalanced force. It is commonly said that in each situation there is a net force acting upon the object. The net force is the vector sum of all the forces that act upon an object.
#### How to describe force in a picture worksheet?
Forces Worksheet 2 Label the force in each picture as a push or pull. Then describe whether the force is causing a change in speed or direction or both. Forces Worksheet 3 More than one force can act on an object at a time. The forces can push or pull in any direction. | 969 | 4,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-38 | latest | en | 0.938338 |
http://www.gurufocus.com/term/inventory2rev/LINTA/Inventory%252Bto%252BRevenue/Liberty%2BInteractive%2BCorp | 1,410,910,803,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657120057.96/warc/CC-MAIN-20140914011200-00052-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 563,324,289 | 22,296 | Switch to:
Liberty Interactive Corp (NAS:LINTA)
Inventory to Revenue
0.42 (As of Jun. 2014)
Inventory to revenue determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue. Liberty Interactive Corp's inventory for the quarter that ended in Jun. 2014 was \$1,181 Mil. Liberty Interactive Corp's revenue for the three months ended in Jun. 2014 was \$2,818 Mil. Liberty Interactive Corp's inventory to revenue ratio for the quarter that ended in Jun. 2014 was 0.42.
Liberty Interactive Corp's inventory to revenue ratio for the quarter that ended in Jun. 2014 declined from Mar. 2014 (0.45) to Mar. 2014 (0.42)
A decrease in the inventory to revenue ratio from one quarter to next indicates that one of these is occurring:
1. investment in inventory is shrinking in relation to revenue
2. revenue are increasing
No matter which situation is causing the reduction in the inventory to revenue ratio, either one suggests that business's inventory levels and its cash flow are effectively managed.
Days inventory indicates the number of days of goods in sales that a company has in the inventory. Liberty Interactive Corp's days inventory for the three months ended in Jun. 2014 was 68.54.
Inventory can be measured by Days Sales of Inventory (DSI). Liberty Interactive Corp's days sales of inventory (DSI) for the three months ended in Jun. 2014 was 38.14.
Inventory turnover measures how fast the company turns over its inventory within a year. Liberty Interactive Corp's inventory turnover for the quarter that ended in Jun. 2014 was 1.33.
Definition
Inventory to Revenue determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue.
Liberty Interactive Corp's Inventory to Revenue for the fiscal year that ended in Dec. 2013 is calculated as
Inventory to Revenue (A: Dec. 2013 ) = Inventory (A: Dec. 2013 ) / Revenue (A: Dec. 2013 ) = 1135 / 11252 = 0.10
Liberty Interactive Corp's Inventory to Revenue for the quarter that ended in Jun. 2014 is calculated as
Inventory to Revenue (Q: Jun. 2014 ) = Inventory (Q: Jun. 2014 ) / Revenue (Q: Jun. 2014 ) = 1181 / 2818 = 0.42
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Explanation
An increase in inventory to revenue ratio from one quarter to the next indicates that one of the following is happening:
1. investment in inventory is growing more rapidly than revenue
2. revenue are dropping
No matter which situation is causing the problem, an increase in the inventory to revenue ratio may signal an oncoming cash flow problem.
Likewise, a decrease in the inventory to revenue ratio from one quarter to next indicates that one of these is occurring:
1. investment in inventory is shrinking in relation to revenue
2. revenue are increasing
No matter which situation is causing the reduction in the inventory to revenue ratio, either one suggests that business's inventory levels and its cash flow are effectively managed.
More Related Terms:
1. Days Inventory indicates the number of days of goods in sales that a company has in the inventory.
Liberty Interactive Corp's Days Inventory for the three months ended in Jun. 2014 is calculated as:
Days Inventory = Inventory (Q: Jun. 2014 ) / Cost of Goods Sold (Q: Jun. 2014 ) * Days in Period = 1181 / 1568 * 91 = 68.54
2. Inventory can be measured by Days Sales of Inventory (DSI).
Liberty Interactive Corp's Days Sales of Inventory for the three months ended in Jun. 2014 is
Days Sales of Inventory (DSI) = Inventory (Q: Jun. 2014 ) / Revenue (Q: Jun. 2014 ) * Days in Period = 1181 / 2818 * 91 = 38.14
3. Inventory Turnover measures how fast the company turns over its inventory within a year.
Liberty Interactive Corp's Inventory Turnover for the quarter that ended in Jun. 2014 is calculated as
Inventory Turnover = Cost of Goods Sold (Q: Jun. 2014 ) / Average Inventory (Q: Jun. 2014 ) = 1568 / 1181 = 1.33
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Liberty Interactive Corp Annual Data
Dec05 Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 inventory2rev 0.00 0.11 0.11 0.13 0.13 0.12 0.12 0.11 0.11 0.10
Liberty Interactive Corp Quarterly Data
Mar12 Jun12 Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 inventory2rev 0.49 0.46 0.57 0.35 0.42 0.43 0.53 0.33 0.45 0.42
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 1,216 | 4,936 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2014-41 | longest | en | 0.9381 |
https://french.stackexchange.com/questions/29454/could-a-translation-error-lead-to-squares-to-not-be-considered-as-rectangles/29457 | 1,653,064,712,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662533972.17/warc/CC-MAIN-20220520160139-20220520190139-00435.warc.gz | 332,652,459 | 71,378 | # Could a translation error lead to squares to not be considered as rectangles?
I'm reading a certain set of kindergarten/lower primary maths textbooks that is written by American authors for a non-American company.
Whenever students are asked to identify the number of rectangles in a given picture, the answer booklet gives the number of oblongs instead of the number of rectangles.
While the topic may be too advanced for kindergarten students, the maths textbooks indeed explicitly say at the bottom of the first page of a textbook at the very first level to tell students that squares are special types of rectangles, where levels 1-4 are for kindergarten students.
Additionally, the accompany guide for teachers devotes a whole page of discussion as to how to teach that squares are special types of rectangles. There's even a paragraph about teaching to kindergarten students. The authors/some of the co-authors of the teacher guides are also authors/co-authors of the textbooks. They have also said that if students are taught that squares are not rectangles, then they will have misconceptions later.
Perhaps, the ones who wrote the answer booklets were not fluent in English while the ones who wrote the textbooks were.
For example
[picture with 4 circles, 2 triangles, 3 square rectangles, 2 oblong rectangles for a total of 5 rectangles]
Circle ___
Triangle ___
Square ___
Rectangle ___
The answer key would give only the numbers:
4
2
3
2
So, the last line is wrong since it should be 5.
Could this happen in French? Or a French dialect? I mean, is there something specific about the translations of any of the following words 'rectangle, square, oblong, quadrilateral, quadrangle, parallelogram, trapezoid/trapezium, rhombus' that would cause such confusion? I guess the translator/s thought that when English speakers say 'rectangle', it means 'oblong in their language/dialect, but I don't see that as specifically a problem for this particular language.
By the way, are squares considered rectangles in France? Apparently, these things can vary by state, curricula, culture, time, etc. Please provide a document from the education department of your government or something.
P.S. I'm a monolinguist.
Related:
Are kindergartners supposed to be steered from squares being rectangles?
In what curricula are “rectangles” defined so as to exclude squares?
Why do we have circles for ellipses, squares for rectangles but nothing for triangles?
What are/should kids (be) taught about the colour of the sun?
• It's not clear from your question where the translation error could have occurred. Are you saying that the answer booklets were translated while the textbooks were not? Who translated what and why? Mar 23, 2018 at 13:03
• @guillaume31 I'm not sure myself. It could be 1. textbooks english originally then read by non-fluent english 2. textbooks non-english originally then translated then read by non-fluent english 3. non-english textbooks read (either by fluent or non-fluent of that language) though that's not the one i came across 4. in any case, it could be that the ones who wrote the answer booklets were not the ones who wrote the textbooks 5. any valid combination of the previous
– BCLC
Mar 23, 2018 at 14:58
• Now I'm confused. Is it a hypothetical question? It sounded like you'd had these booklets in your hands before. You must surely know which language they are in? Mar 23, 2018 at 15:13
• @guillaume31 I do know the state the company belongs to, but I don't want to say it out of fear of identification. Thus, I'm posting to various language SE sites. Here, we assume American authors wrote this for a French company. I've seen only English versions. I am not aware that these texts have non-English version. So I don't know where exactly the translation error may have taken place. In Spanish SE, a user suggests the people who wrote the answer booklets are not the same as the main (American) authors.
– BCLC
Mar 24, 2018 at 10:22
Yes, the confusion is also possible in French, as is arguably the case with every language that has corresponding terms for square and rectangle as we understand them in English.
The mistake comes from the fact that when a human sees
Square ___
Rectangle ___
they will tend to interpret "Rectangle" as "Rectangles that are not squares, since Squares are accounted for just above".
I'm not sure you'll find Western languages that can make this unambiguous.
Edit
To clarify: Rectangle in French mathematics means "inclusive" rectangle, i.e. squares are a subset of rectangles. That's what you learn in school. See https://fr.wikipedia.org/wiki/Carr%C3%A9.
But the natural tendency to think that Rectangle presented alongside Square means "non-square rectangles", remains. So a French translation of these instructions would remain as ambiguous as the original in English, because on the one hand you've got the formal definition and on the other hand, what your mind will spontaneously interpret.
• It seems you are right about Spanish. I didn't know that. However, I can confirm that rectangle in French does mean "inclusive" rectangle (see my edit). I know it's the same in Dutch (vierkant / rechthoek). Mar 23, 2018 at 9:30
• To answer your question - no, translation errors from English to French are not possible in that regard. The ambiguous English instructions you gave as an example will carry over as equally ambiguous in French. Mar 23, 2018 at 9:37
• 1. "No" to what? My "yes" was a "yes" to : same confusion possible in English and French. 2. and 3. French doesn't have a word for exclusive rectangle, unlike Spanish. French and Spanish are different in that regard. 4. Yes, I disagree with @jiliagre and so does Wikipedia Mar 23, 2018 at 11:26
• Spanish Wikipedia states En algunas traducciones, no aceptadas en general, el cuadrado es un rectángulo de cuatro lados iguales, and there is no mention of a cuadrado being a specialized form of rectangulo, which speaks in favor of an "exclusive" definition of rectangle in that language. Mar 23, 2018 at 11:29
• 1. No, the confusion exists in the original text ; translation into French doesn't add or remove any of it. 5. I don't know enough about Spanish ; guess you'll have to ask on spanish.stackexchange.com ;) Mar 23, 2018 at 11:38
While technically a square is indeed a rectangle too, French kindergarten students (and more generally people outside mathematicians) are very unlikely to consider it as such.
Here is an excerpt of a FAQ for teachers of students from 6 to 9 years old (CP - CE2 / 1st grade to 3rd grade)
Par exemple, il n’est pas d’usage de dire qu’une table carrée est rectangulaire, pourtant, du point de vue mathématique, c’est vrai. En mathématiques on peut dire qu’un carré est un rectangle. Ainsi dès le cycle 1 le nombre de côtés des polygones peut être utilisé pour différencier les formes. On peut également travailler, même si cela peut paraître ambitieux, le fait que les rectangles et les carrés appartiennent à une même famille, celle des quadrilatères (figures à 4 côtés) ayant quatre angles droits.
We do not use rectangle allongé (oblong rectangle) to describe a geometrical form in French. I'd even expect carré allongé (oblong square) to be used in that case.
For example, there is a monument in Nîmes called la maison carrée, which is not squared but rectangular. Until the 16th century, carré was used for all rectangular forms so an oblong rectangle was a carré/quarré long while the modern square was called carré/quarré parfait (perfect square).
A word that encompass oblong rectangles and squares is quadrilatère (quadrilateral), but some quadrilaterals like losanges (rhombus) are not rectangles.
Including carrés in the rectangle class might only be teached at the end of primary (CM2 - grade 5) or possibly later, e.g.:
Nous avons commencé par du calcul mental, ensuite des problèmes de proportionnalités. Puis nous avons travaillé sur la carré, le rectangle et le losange. J’ai appris que le carré était une sorte de rectangle et une sorte de losange. Nous sommes ensuite sortis de la classe.
Here is a also as document stating the fact carrés and rectangles are exclusives concepts in French.
Apport pour les apprentissages de l’explicitation des relations d’inclusion de classes, Laurence Dupuch* et Emmanuel Sander, Université Paris 8
Par exemple, par l’analyse de la phrase «Est-ce un carré ou un rectangle ? » , Politzer (1991) montre que l’usage commun du français place les carrés et les rectangles en relation contrastive : ils forment alors deux classes disjointes si bien qu’un locuteur n’a aucune raison de croire qu’une forme rectangulaire puisse être carrée. Ces deux classes n’ont pas lieu d’être réunies et sont également disjointes de la classe des parallélogrammes et des losanges. L’organisation conceptuelle est alors formée de relations exclusives lexicalisées : quadrilatère – parallélogramme – rectangle – losange – carré. De ce fait, la relation inclusive n’est pas lexicalisée, alors que du point de vue de leur définition mathématique, qui est parfois, mais pas systématiquement, énoncée de façon explicite dans les manuels scolaires, les carrés sont des rectangles : «Un carré est un rectangle qui a des côtés égaux » (Math élèm. manuel CM2, 2000, p. 18).
• jlliagre, thanks! So your answer is that you think that it's possible (not necessarily probable) that this could be a French language issue? Cf: In Mandarin, the term for oblong (exclusive rectangle) is 长方形(的) but the term for rectangle (I guess inclusive rectangle) is 长方形. Thus, for Mandarin, it's possible the answer booklets were mistaken due to language issues rather than geometric misconception issues. Cf: For Spanish, it maybe that it's unlikely, language would be an issue because Spanish doesn't have a word for inclusive rectangle. Now: What about French?
– BCLC
Mar 23, 2018 at 9:17
• I guess many French people and even more young students do not consider a square to be a rectangle at all. Exceptions would likely be due to a recent lecture stating that mathematical fact. Mar 23, 2018 at 10:19
• @jiliagre the last excerpt you added highlights the fact that language definition and mathematical definition are indeed at odds. But maybe we could agree about the answer to BCLC's original question - that a French translation wouldn't change a thing about the comprehension of the exercise by students compared to the English version? Mar 23, 2018 at 13:08
• There are far more chances that booklets don't consider squares to be rectangles because they follow the same intuition that students have than because of language specifics, if you ask me. Mar 23, 2018 at 15:22
• FWIW, I just made a small poll with five kids playing in front of my house, four 10 years old in CM2/last primary grade, and one 8 years old and they were all definitive about the fact squares are not rectangles. One told me a rectangle might be made by two squares joined by one side. None of them accepted rectangle carré as a correct description of anything. Mar 23, 2018 at 17:42 | 2,616 | 11,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-21 | latest | en | 0.951857 |
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# Minimum time required to rot all oranges
Given a matrix of dimension M * N where each cell in the matrix can have values 0, 1 or 2 which has the following meaning:
• 0: Empty cell
• 1: Cells have fresh oranges
• 2: Cells have rotten oranges
Determine what is the minimum time required so that all the oranges become rotten. A rotten orange at index (i,j ) can rot other fresh oranges which are its neighbours (up, down, left and right). If it is impossible to rot every orange then simply return -1.
Examples:
Input: arr[][C] = { {2, 1, 0, 2, 1}, {1, 0, 1, 2, 1}, {1, 0, 0, 2, 1}};
Output: 2
Explanation: At 0th time frame:
{2, 1, 0, 2, 1}
{1, 0, 1, 2, 1}
{1, 0, 0, 2, 1}
At 1st time frame:
{2, 2, 0, 2, 2}
{2, 0, 2, 2, 2}
{1, 0, 0, 2, 2}
At 2nd time frame:
{2, 2, 0, 2, 2}
{2, 0, 2, 2, 2}
{2, 0, 0, 2, 2}
Input: arr[][C] = { {2, 1, 0, 2, 1}, {0, 0, 1, 2, 1}, {1, 0, 0, 2, 1}}
Output: -1
Explanation: At 0th time frame:
{2, 1, 0, 2, 1}
{0, 0, 1, 2, 1}
{1, 0, 0, 2, 1}
At 1st time frame:
{2, 2, 0, 2, 2}
{0, 0, 2, 2, 2}
{1, 0, 0, 2, 2}
At 2nd time frame:
{2, 2, 0, 2, 2}
{0, 0, 2, 2, 2}
{1, 0, 0, 2, 2}
The 1 at the bottom left corner of the matrix is never rotten.
Recommended Practice
Naive Approach:
The idea is very basic. Traverse through all oranges in multiple rounds. In every round, rot the oranges to the adjacent position of oranges that were rotten in the last round.
Follow the steps below to solve the problem:
• Create a variable no = 2 and changed = false.
• Run a loop until there is no cell of the matrix which is changed in an iteration.
• Run a nested loop and traverse the matrix:
• If the element of the matrix is equal to no then assign the adjacent elements to no + 1 if the adjacent element’s value is equal to 1, i.e. not rotten, and update changed to true.
• Traverse the matrix and check if there is any cell that is 1
• If 1 is present return -1
• Else return no – 2.
Below is the implementation of the above approach.
## C++14
`// C++ program to rot all oranges when you can move` `// in all the four direction from a rotten orange` `#include ` `using` `namespace` `std;` `const` `int` `R = 3;` `const` `int` `C = 5;` `// Check if i, j is under the array limits of row and column` `bool` `issafe(``int` `i, ``int` `j)` `{` ` ``if` `(i >= 0 && i < R && j >= 0 && j < C)` ` ``return` `true``;` ` ``return` `false``;` `}` `int` `rotOranges(``int` `v[R][C])` `{` ` ``bool` `changed = ``false``;` ` ``int` `no = 2;` ` ``while` `(``true``) {` ` ``for` `(``int` `i = 0; i < R; i++) {` ` ``for` `(``int` `j = 0; j < C; j++) {` ` ``// Rot all other oranges present at` ` ``// (i+1, j), (i, j-1), (i, j+1), (i-1, j)` ` ``if` `(v[i][j] == no) {` ` ``if` `(issafe(i + 1, j)` ` ``&& v[i + 1][j] == 1) {` ` ``v[i + 1][j] = v[i][j] + 1;` ` ``changed = ``true``;` ` ``}` ` ``if` `(issafe(i, j + 1)` ` ``&& v[i][j + 1] == 1) {` ` ``v[i][j + 1] = v[i][j] + 1;` ` ``changed = ``true``;` ` ``}` ` ``if` `(issafe(i - 1, j)` ` ``&& v[i - 1][j] == 1) {` ` ``v[i - 1][j] = v[i][j] + 1;` ` ``changed = ``true``;` ` ``}` ` ``if` `(issafe(i, j - 1)` ` ``&& v[i][j - 1] == 1) {` ` ``v[i][j - 1] = v[i][j] + 1;` ` ``changed = ``true``;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// if no rotten orange found it means all` ` ``// oranges rottened now` ` ``if` `(!changed)` ` ``break``;` ` ``changed = ``false``;` ` ``no++;` ` ``}` ` ``for` `(``int` `i = 0; i < R; i++) {` ` ``for` `(``int` `j = 0; j < C; j++) {` ` ``// if any orange is found to be` ` ``// not rotten then ans is not possible` ` ``if` `(v[i][j] == 1)` ` ``return` `-1;` ` ``}` ` ``}` ` ``// Because initial value for a rotten` ` ``// orange was 2` ` ``return` `no - 2;` `}` `// Driver function` `int` `main()` `{` ` ``int` `v[R][C] = { { 2, 1, 0, 2, 1 },` ` ``{ 1, 0, 1, 2, 1 },` ` ``{ 1, 0, 0, 2, 1 } };` ` ``cout << ``"Max time incurred: "` `<< rotOranges(v);` ` ``return` `0;` `}`
## Java
`// Java program to rot all oranges when you can move` `// in all the four direction from a rotten orange` `class` `GFG {` ` ``static` `int` `R = ``3``;` ` ``static` `int` `C = ``5``;` ` ``// Check if i, j is under the array` ` ``// limits of row and column` ` ``static` `boolean` `issafe(``int` `i, ``int` `j)` ` ``{` ` ``if` `(i >= ``0` `&& i < R && j >= ``0` `&& j < C)` ` ``return` `true``;` ` ``return` `false``;` ` ``}` ` ``static` `int` `rotOranges(``int` `v[][])` ` ``{` ` ``boolean` `changed = ``false``;` ` ``int` `no = ``2``;` ` ``while` `(``true``) {` ` ``for` `(``int` `i = ``0``; i < R; i++) {` ` ``for` `(``int` `j = ``0``; j < C; j++) {` ` ``// Rot all other oranges present at` ` ``// (i+1, j), (i, j-1), (i, j+1), (i-1,` ` ``// j)` ` ``if` `(v[i][j] == no) {` ` ``if` `(issafe(i + ``1``, j)` ` ``&& v[i + ``1``][j] == ``1``) {` ` ``v[i + ``1``][j] = v[i][j] + ``1``;` ` ``changed = ``true``;` ` ``}` ` ``if` `(issafe(i, j + ``1``)` ` ``&& v[i][j + ``1``] == ``1``) {` ` ``v[i][j + ``1``] = v[i][j] + ``1``;` ` ``changed = ``true``;` ` ``}` ` ``if` `(issafe(i - ``1``, j)` ` ``&& v[i - ``1``][j] == ``1``) {` ` ``v[i - ``1``][j] = v[i][j] + ``1``;` ` ``changed = ``true``;` ` ``}` ` ``if` `(issafe(i, j - ``1``)` ` ``&& v[i][j - ``1``] == ``1``) {` ` ``v[i][j - ``1``] = v[i][j] + ``1``;` ` ``changed = ``true``;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// If no rotten orange found it means all` ` ``// oranges rottened now` ` ``if` `(!changed)` ` ``break``;` ` ``changed = ``false``;` ` ``no++;` ` ``}` ` ``for` `(``int` `i = ``0``; i < R; i++) {` ` ``for` `(``int` `j = ``0``; j < C; j++) {` ` ``// If any orange is found to be` ` ``// not rotten then ans is not possible` ` ``if` `(v[i][j] == ``1``)` ` ``return` `-``1``;` ` ``}` ` ``}` ` ``// Because initial value for a rotten` ` ``// orange was 2` ` ``return` `no - ``2``;` ` ``}` ` ``// Driver Code` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``int` `v[][] = { { ``2``, ``1``, ``0``, ``2``, ``1` `},` ` ``{ ``1``, ``0``, ``1``, ``2``, ``1` `},` ` ``{ ``1``, ``0``, ``0``, ``2``, ``1` `} };` ` ``System.out.println(``"Max time incurred: "` ` ``+ rotOranges(v));` ` ``}` `}` `// This code is contributed by divyesh072019`
## Python3
`# Python3 program to rot all` `# oranges when you can move` `# in all the four direction` `# from a rotten orange` `R ``=` `3` `C ``=` `5` `# Check if i, j is under the` `# array limits of row and` `# column` `def` `issafe(i, j):` ` ``if` `(i >``=` `0` `and` `i < R ``and` ` ``j >``=` `0` `and` `j < C):` ` ``return` `True` ` ``return` `False` `def` `rotOranges(v):` ` ``changed ``=` `False` ` ``no ``=` `2` ` ``while` `(``True``):` ` ``for` `i ``in` `range``(R):` ` ``for` `j ``in` `range``(C):` ` ``# Rot all other oranges` ` ``# present at (i+1, j),` ` ``# (i, j-1), (i, j+1),` ` ``# (i-1, j)` ` ``if` `(v[i][j] ``=``=` `no):` ` ``if` `(issafe(i ``+` `1``, j) ``and` ` ``v[i ``+` `1``][j] ``=``=` `1``):` ` ``v[i ``+` `1``][j] ``=` `v[i][j] ``+` `1` ` ``changed ``=` `True` ` ``if` `(issafe(i, j ``+` `1``) ``and` ` ``v[i][j ``+` `1``] ``=``=` `1``):` ` ``v[i][j ``+` `1``] ``=` `v[i][j] ``+` `1` ` ``changed ``=` `True` ` ``if` `(issafe(i ``-` `1``, j) ``and` ` ``v[i ``-` `1``][j] ``=``=` `1``):` ` ``v[i ``-` `1``][j] ``=` `v[i][j] ``+` `1` ` ``changed ``=` `True` ` ``if` `(issafe(i, j ``-` `1``) ``and` ` ``v[i][j ``-` `1``] ``=``=` `1``):` ` ``v[i][j ``-` `1``] ``=` `v[i][j] ``+` `1` ` ``changed ``=` `True` ` ``# if no rotten orange found` ` ``# it means all oranges rottened` ` ``# now` ` ``if` `(``not` `changed):` ` ``break` ` ``changed ``=` `False` ` ``no ``+``=` `1` ` ``for` `i ``in` `range``(R):` ` ``for` `j ``in` `range``(C):` ` ``# if any orange is found` ` ``# to be not rotten then` ` ``# ans is not possible` ` ``if` `(v[i][j] ``=``=` `1``):` ` ``return` `-``1` ` ``# Because initial value` ` ``# for a rotten orange was 2` ` ``return` `no ``-` `2` `# Driver function` `if` `__name__ ``=``=` `"__main__"``:` ` ``v ``=` `[[``2``, ``1``, ``0``, ``2``, ``1``],` ` ``[``1``, ``0``, ``1``, ``2``, ``1``],` ` ``[``1``, ``0``, ``0``, ``2``, ``1``]]` ` ``print``(``"Max time incurred: "``,` ` ``rotOranges(v))` `# This code is contributed by Chitranayal`
## C#
`// C# program to rot all oranges when you can move` `// in all the four direction from a rotten orange` `using` `System;` `class` `GFG {` ` ``static` `int` `R = 3;` ` ``static` `int` `C = 5;` ` ``// Check if i, j is under the array` ` ``// limits of row and column` ` ``static` `bool` `issafe(``int` `i, ``int` `j)` ` ``{` ` ``if` `(i >= 0 && i < R && j >= 0 && j < C)` ` ``return` `true``;` ` ``return` `false``;` ` ``}` ` ``static` `int` `rotOranges(``int``[, ] v)` ` ``{` ` ``bool` `changed = ``false``;` ` ``int` `no = 2;` ` ``while` `(``true``) {` ` ``for` `(``int` `i = 0; i < R; i++) {` ` ``for` `(``int` `j = 0; j < C; j++) {` ` ``// Rot all other oranges present at` ` ``// (i+1, j), (i, j-1), (i, j+1), (i-1,` ` ``// j)` ` ``if` `(v[i, j] == no) {` ` ``if` `(issafe(i + 1, j)` ` ``&& v[i + 1, j] == 1) {` ` ``v[i + 1, j] = v[i, j] + 1;` ` ``changed = ``true``;` ` ``}` ` ``if` `(issafe(i, j + 1)` ` ``&& v[i, j + 1] == 1) {` ` ``v[i, j + 1] = v[i, j] + 1;` ` ``changed = ``true``;` ` ``}` ` ``if` `(issafe(i - 1, j)` ` ``&& v[i - 1, j] == 1) {` ` ``v[i - 1, j] = v[i, j] + 1;` ` ``changed = ``true``;` ` ``}` ` ``if` `(issafe(i, j - 1)` ` ``&& v[i, j - 1] == 1) {` ` ``v[i, j - 1] = v[i, j] + 1;` ` ``changed = ``true``;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// if no rotten orange found it means all` ` ``// oranges rottened now` ` ``if` `(!changed)` ` ``break``;` ` ``changed = ``false``;` ` ``no++;` ` ``}` ` ``for` `(``int` `i = 0; i < R; i++) {` ` ``for` `(``int` `j = 0; j < C; j++) {` ` ``// if any orange is found to be` ` ``// not rotten then ans is not possible` ` ``if` `(v[i, j] == 1)` ` ``return` `-1;` ` ``}` ` ``}` ` ``// Because initial value for a rotten` ` ``// orange was 2` ` ``return` `no - 2;` ` ``}` ` ``static` `void` `Main()` ` ``{` ` ``int``[, ] v = { { 2, 1, 0, 2, 1 },` ` ``{ 1, 0, 1, 2, 1 },` ` ``{ 1, 0, 0, 2, 1 } };` ` ``Console.Write(``"Max time incurred: "` ` ``+ rotOranges(v));` ` ``}` `}` `// This code is contributed by divyeshrabadiya07`
## Javascript
``
Output
`Max time incurred: 2`
Time Complexity: O((R*C) * (R *C)),
• The matrix needs to be traversed again and again until there is no change in the matrix, that can happen max(R *C)/2 times.
• So time complexity is O((R * C) * (R *C)).
Auxiliary Space: O(1), No extra space is required.
## Minimum time required to rot all oranges using Breadth First Search:
The idea is to use Breadth First Search. The condition of oranges getting rotten is when they come in contact with other rotten oranges. This is similar to a breadth-first search where the graph is divided into layers or circles and the search is done from lower or closer layers to deeper or higher layers.
In the previous approach, the idea was based on BFS but the implementation was poor and inefficient. To find the elements whose values are no the whole matrix had to be traversed. So time can be reduced by using this efficient approach of BFS.
Follow the steps below to solve the problem:
• Create an empty queue Q
• Find all rotten oranges and enqueue them to Q. Also, enqueue a delimiter to indicate the beginning of the next time frame.
• Run a loop While Q is not empty and do the following while the delimiter in Q is not reached
• Dequeue an orange from the queue, and rot all adjacent oranges.
• While rotting the adjacent, make sure that the time frame is incremented only once. And the time frame is not incremented if there are no adjacent oranges.
• Dequeue the old delimiter and enqueue a new delimiter. The oranges rotten in the previous time frame lie between the two delimiters.
• Return the last time frame.
Illustration:
Below is the implementation of the above approach.
## C++
`// C++ program to find minimum time required to make all` `// oranges rotten` `#include ` `#define R 3` `#define C 5` `using` `namespace` `std;` `// This function finds if it is possible to rot all oranges` `// or not. If possible, then it returns minimum time` `// required to rot all, otherwise returns -1` `int` `rotOranges(vector >& grid)` `{` ` ``int` `n = grid.size(); ``// row size` ` ``int` `m = grid[0].size(); ``// column size` ` ``// delrow and delcol are used to traverse in` ` ``// up,right,bottom and left respectively.` ` ``int` `delrow[] = { -1, 0, 1, 0 };` ` ``int` `delcol[] = { 0, 1, 0, -1 };` ` ``// visited matrix to keep track of the visited cell.` ` ``int` `vis[n][m];` ` ``// queue stores rowIndex,colIndex and time taken to rot` ` ``// respectively.` ` ``queue, ``int``> > q;` ` ``// counter to keep track of fresh cells.` ` ``int` `cntfresh = 0;` ` ``for` `(``int` `i = 0; i < n; i++) {` ` ``for` `(``int` `j = 0; j < m; j++) {` ` ``if` `(grid[i][j] == 2) {` ` ``q.push({ { i, j },` ` ``0 }); ``// already rotten hence 0` ` ``// time to rot.` ` ``vis[i][j]` ` ``= 2; ``// visited cell marked as rotten.` ` ``}` ` ``else` `{` ` ``vis[i][j] = 0; ``// unvisited` ` ``}` ` ``if` `(grid[i][j] == 1)` ` ``cntfresh++; ``// maintaining count for fresh` ` ``// oranges.` ` ``}` ` ``}` ` ``int` `cnt = 0, ``tm` `= 0;` ` ``while` `(!q.empty()) {` ` ``int` `row = q.front().first.first; ``// row index` ` ``int` `col = q.front().first.second; ``// col index` ` ``int` `t = q.front().second; ``// time an orange at a` ` ``// cell takes to rot.` ` ``q.pop();` ` ``tm` `= max(``tm``, t);` ` ``// checking for adjacent nodes in 4 directions.` ` ``for` `(``int` `i = 0; i < 4; i++) {` ` ``int` `nrow = row + delrow[i];` ` ``int` `ncol = col + delcol[i];` ` ``// checking the validity of a node and also` ` ``// vis[nrow][ncol] !=2` ` ``if` `(nrow >= 0 && nrow < n && ncol >= 0` ` ``&& ncol < m && grid[nrow][ncol] == 1` ` ``&& vis[nrow][ncol] != 2) {` ` ``vis[nrow][ncol] = 2; ``// adj orange is rotten` ` ``q.push({ { nrow, ncol },` ` ``t + 1 }); ``// incrementing time for` ` ``// that orange by 1` ` ``cnt++;` ` ``}` ` ``}` ` ``}` ` ``return` `(cnt == cntfresh) ? ``tm` `: -1;` `}` `// Driver program` `int` `main()` `{` ` ``vector > arr` ` ``= { { 0, 1, 2 }, { 0, 1, 2 }, { 2, 1, 1 } };` ` ``int` `ans = rotOranges(arr);` ` ``if` `(ans == -1)` ` ``cout << ``"All oranges cannot rotn"``;` ` ``else` ` ``cout << ``"Time required for all oranges to rot => "` ` ``<< ans << endl;` ` ``return` `0;` `}`
## Java
`// Java program to find minimum time required to make all` `// oranges rotten` `import` `java.util.LinkedList;` `import` `java.util.Queue;` `public` `class` `RotOrange {` ` ``public` `final` `static` `int` `R = ``3``;` ` ``public` `final` `static` `int` `C = ``5``;` ` ``// structure for storing coordinates of the cell` ` ``static` `class` `Ele {` ` ``int` `x = ``0``;` ` ``int` `y = ``0``;` ` ``Ele(``int` `x, ``int` `y)` ` ``{` ` ``this``.x = x;` ` ``this``.y = y;` ` ``}` ` ``}` ` ``// function to check whether a cell is valid / invalid` ` ``static` `boolean` `isValid(``int` `i, ``int` `j)` ` ``{` ` ``return` `(i >= ``0` `&& j >= ``0` `&& i < R && j < C);` ` ``}` ` ``// Function to check whether the cell is delimiter` ` ``// which is (-1, -1)` ` ``static` `boolean` `isDelim(Ele temp)` ` ``{` ` ``return` `(temp.x == -``1` `&& temp.y == -``1``);` ` ``}` ` ``// Function to check whether there is still a fresh` ` ``// orange remaining` ` ``static` `boolean` `checkAll(``int` `arr[][])` ` ``{` ` ``for` `(``int` `i = ``0``; i < R; i++)` ` ``for` `(``int` `j = ``0``; j < C; j++)` ` ``if` `(arr[i][j] == ``1``)` ` ``return` `true``;` ` ``return` `false``;` ` ``}` ` ``// This function finds if it is possible to rot all` ` ``// oranges or not. If possible, then it returns minimum` ` ``// time required to rot all, otherwise returns -1` ` ``static` `int` `rotOranges(``int` `arr[][])` ` ``{` ` ``// Create a queue of cells` ` ``Queue Q = ``new` `LinkedList<>();` ` ``Ele temp;` ` ``int` `ans = ``0``;` ` ``// Store all the cells having rotten orange in first` ` ``// time frame` ` ``for` `(``int` `i = ``0``; i < R; i++)` ` ``for` `(``int` `j = ``0``; j < C; j++)` ` ``if` `(arr[i][j] == ``2``)` ` ``Q.add(``new` `Ele(i, j));` ` ``// Separate these rotten oranges from the oranges` ` ``// which will rotten due the oranges in first time` ` ``// frame using delimiter which is (-1, -1)` ` ``Q.add(``new` `Ele(-``1``, -``1``));` ` ``// Process the grid while there are rotten oranges` ` ``// in the Queue` ` ``while` `(!Q.isEmpty()) {` ` ``// This flag is used to determine whether even a` ` ``// single fresh orange gets rotten due to rotten` ` ``// oranges in the current time frame so we can` ` ``// increase the count of the required time.` ` ``boolean` `flag = ``false``;` ` ``// Process all the rotten oranges in current` ` ``// time frame.` ` ``while` `(!isDelim(Q.peek())) {` ` ``temp = Q.peek();` ` ``// Check right adjacent cell that if it can` ` ``// be rotten` ` ``if` `(isValid(temp.x + ``1``, temp.y)` ` ``&& arr[temp.x + ``1``][temp.y] == ``1``) {` ` ``if` `(!flag) {` ` ``// if this is the first orange to` ` ``// get rotten, increase count and` ` ``// set the flag.` ` ``ans++;` ` ``flag = ``true``;` ` ``}` ` ``// Make the orange rotten` ` ``arr[temp.x + ``1``][temp.y] = ``2``;` ` ``// push the adjacent orange to Queue` ` ``temp.x++;` ` ``Q.add(``new` `Ele(temp.x, temp.y));` ` ``// Move back to current cell` ` ``temp.x--;` ` ``}` ` ``// Check left adjacent cell that if it can` ` ``// be rotten` ` ``if` `(isValid(temp.x - ``1``, temp.y)` ` ``&& arr[temp.x - ``1``][temp.y] == ``1``) {` ` ``if` `(!flag) {` ` ``ans++;` ` ``flag = ``true``;` ` ``}` ` ``arr[temp.x - ``1``][temp.y] = ``2``;` ` ``temp.x--;` ` ``Q.add(``new` `Ele(` ` ``temp.x,` ` ``temp.y)); ``// push this cell to Queue` ` ``temp.x++;` ` ``}` ` ``// Check top adjacent cell that if it can be` ` ``// rotten` ` ``if` `(isValid(temp.x, temp.y + ``1``)` ` ``&& arr[temp.x][temp.y + ``1``] == ``1``) {` ` ``if` `(!flag) {` ` ``ans++;` ` ``flag = ``true``;` ` ``}` ` ``arr[temp.x][temp.y + ``1``] = ``2``;` ` ``temp.y++;` ` ``Q.add(``new` `Ele(` ` ``temp.x,` ` ``temp.y)); ``// Push this cell to Queue` ` ``temp.y--;` ` ``}` ` ``// Check bottom adjacent cell if it can be` ` ``// rotten` ` ``if` `(isValid(temp.x, temp.y - ``1``)` ` ``&& arr[temp.x][temp.y - ``1``] == ``1``) {` ` ``if` `(!flag) {` ` ``ans++;` ` ``flag = ``true``;` ` ``}` ` ``arr[temp.x][temp.y - ``1``] = ``2``;` ` ``temp.y--;` ` ``Q.add(``new` `Ele(` ` ``temp.x,` ` ``temp.y)); ``// push this cell to Queue` ` ``}` ` ``Q.remove();` ` ``}` ` ``// Pop the delimiter` ` ``Q.remove();` ` ``// If oranges were rotten in current frame than` ` ``// separate the rotten oranges using delimiter` ` ``// for the next frame for processing.` ` ``if` `(!Q.isEmpty()) {` ` ``Q.add(``new` `Ele(-``1``, -``1``));` ` ``}` ` ``// If Queue was empty than no rotten oranges` ` ``// left to process so exit` ` ``}` ` ``// Return -1 if all arranges could not rot,` ` ``// otherwise ans` ` ``return` `(checkAll(arr)) ? -``1` `: ans;` ` ``}` ` ``// Driver program` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``int` `arr[][] = { { ``2``, ``1``, ``0``, ``2``, ``1` `},` ` ``{ ``1``, ``0``, ``1``, ``2``, ``1` `},` ` ``{ ``1``, ``0``, ``0``, ``2``, ``1` `} };` ` ``int` `ans = rotOranges(arr);` ` ``if` `(ans == -``1``)` ` ``System.out.println(``"All oranges cannot rot"``);` ` ``else` ` ``System.out.println(` ` ``"Time required for all oranges to rot => "` ` ``+ ans);` ` ``}` `}` `// This code is contributed by Sumit Ghosh`
## Python3
`# Python3 program to find minimum time required to make all` `# oranges rotten` `from` `collections ``import` `deque` `# function to check whether a cell is valid / invalid` `def` `isvalid(i, j):` ` ``return` `(i >``=` `0` `and` `j >``=` `0` `and` `i < ``3` `and` `j < ``5``)` `# Function to check whether the cell is delimiter` `# which is (-1, -1)` `def` `isdelim(temp):` ` ``return` `(temp[``0``] ``=``=` `-``1` `and` `temp[``1``] ``=``=` `-``1``)` `# Function to check whether there is still a fresh` `# orange remaining` `def` `checkall(arr):` ` ``for` `i ``in` `range``(``3``):` ` ``for` `j ``in` `range``(``5``):` ` ``if` `(arr[i][j] ``=``=` `1``):` ` ``return` `True` ` ``return` `False` `# This function finds if it is` `# possible to rot all oranges or not.` `# If possible, then it returns` `# minimum time required to rot all,` `# otherwise returns -1` `def` `rotOranges(arr):` ` ``# Create a queue of cells` ` ``Q ``=` `deque()` ` ``temp ``=` `[``0``, ``0``]` ` ``ans ``=` `1` ` ``# Store all the cells having` ` ``# rotten orange in first time frame` ` ``for` `i ``in` `range``(``3``):` ` ``for` `j ``in` `range``(``5``):` ` ``if` `(arr[i][j] ``=``=` `2``):` ` ``temp[``0``] ``=` `i` ` ``temp[``1``] ``=` `j` ` ``Q.append([i, j])` ` ``# Separate these rotten oranges` ` ``# from the oranges which will rotten` ` ``# due the oranges in first time` ` ``# frame using delimiter which is (-1, -1)` ` ``temp[``0``] ``=` `-``1` ` ``temp[``1``] ``=` `-``1` ` ``Q.append([``-``1``, ``-``1``])` ` ``# print(Q)` ` ``# Process the grid while there are` ` ``# rotten oranges in the Queue` ` ``while` `False``:` ` ``# This flag is used to determine` ` ``# whether even a single fresh` ` ``# orange gets rotten due to rotten` ` ``# oranges in current time` ` ``# frame so we can increase` ` ``# the count of the required time.` ` ``flag ``=` `False` ` ``print``(``len``(Q))` ` ``# Process all the rotten` ` ``# oranges in current time frame.` ` ``while` `not` `isdelim(Q[``0``]):` ` ``temp ``=` `Q[``0``]` ` ``print``(``len``(Q))` ` ``# Check right adjacent cell that if it can be rotten` ` ``if` `(isvalid(temp[``0``] ``+` `1``, temp[``1``]) ``and` `arr[temp[``0``] ``+` `1``][temp[``1``]] ``=``=` `1``):` ` ``# if this is the first orange to get rotten, increase` ` ``# count and set the flag.` ` ``if` `(``not` `flag):` ` ``ans, flag ``=` `ans ``+` `1``, ``True` ` ``# Make the orange rotten` ` ``arr[temp[``0``] ``+` `1``][temp[``1``]] ``=` `2` ` ``# append the adjacent orange to Queue` ` ``temp[``0``] ``+``=` `1` ` ``Q.append(temp)` ` ``temp[``0``] ``-``=` `1` `# Move back to current cell` ` ``# Check left adjacent cell that if it can be rotten` ` ``if` `(isvalid(temp[``0``] ``-` `1``, temp[``1``]) ``and` `arr[temp[``0``] ``-` `1``][temp[``1``]] ``=``=` `1``):` ` ``if` `(``not` `flag):` ` ``ans, flag ``=` `ans ``+` `1``, ``True` ` ``arr[temp[``0``] ``-` `1``][temp[``1``]] ``=` `2` ` ``temp[``0``] ``-``=` `1` ` ``Q.append(temp) ``# append this cell to Queue` ` ``temp[``0``] ``+``=` `1` ` ``# Check top adjacent cell that if it can be rotten` ` ``if` `(isvalid(temp[``0``], temp[``1``] ``+` `1``) ``and` `arr[temp[``0``]][temp[``1``] ``+` `1``] ``=``=` `1``):` ` ``if` `(``not` `flag):` ` ``ans, flag ``=` `ans ``+` `1``, ``True` ` ``arr[temp[``0``]][temp[``1``] ``+` `1``] ``=` `2` ` ``temp[``1``] ``+``=` `1` ` ``Q.append(temp) ``# Push this cell to Queue` ` ``temp[``1``] ``-``=` `1` ` ``# Check bottom adjacent cell if it can be rotten` ` ``if` `(isvalid(temp[``0``], temp[``1``] ``-` `1``) ``and` `arr[temp[``0``]][temp[``1``] ``-` `1``] ``=``=` `1``):` ` ``if` `(``not` `flag):` ` ``ans, flag ``=` `ans ``+` `1``, ``True` ` ``arr[temp[``0``]][temp[``1``] ``-` `1``] ``=` `2` ` ``temp[``1``] ``-``=` `1` ` ``Q.append(temp) ``# append this cell to Queue` ` ``Q.popleft()` ` ``# Pop the delimiter` ` ``Q.popleft()` ` ``# If oranges were rotten in` ` ``# current frame than separate the` ` ``# rotten oranges using delimiter` ` ``# for the next frame for processing.` ` ``if` `(``len``(Q) ``=``=` `0``):` ` ``temp[``0``] ``=` `-``1` ` ``temp[``1``] ``=` `-``1` ` ``Q.append(temp)` ` ``# If Queue was empty than no rotten oranges left to process so exit` ` ``# Return -1 if all arranges could not rot, otherwise return ans.` ` ``return` `ans ``+` `1` `if``(checkall(arr)) ``else` `-``1` `# Driver program` `if` `__name__ ``=``=` `'__main__'``:` ` ``arr ``=` `[[``2``, ``1``, ``0``, ``2``, ``1``],` ` ``[``1``, ``0``, ``1``, ``2``, ``1``],` ` ``[``1``, ``0``, ``0``, ``2``, ``1``]]` ` ``ans ``=` `rotOranges(arr)` ` ``if` `(ans ``=``=` `-``1``):` ` ``print``(``"All oranges cannot rotn"``)` ` ``else``:` ` ``print``(``"Time required for all oranges to rot => "``, ans)` ` ``# This code is contributed by mohit kumar 29`
## C#
`// C# program to find minimum time` `// required to make all oranges rotten` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG {` ` ``public` `const` `int` `R = 3;` ` ``public` `const` `int` `C = 5;` ` ``// structure for storing` ` ``// coordinates of the cell` ` ``public` `class` `Ele {` ` ``public` `int` `x = 0;` ` ``public` `int` `y = 0;` ` ``public` `Ele(``int` `x, ``int` `y)` ` ``{` ` ``this``.x = x;` ` ``this``.y = y;` ` ``}` ` ``}` ` ``// function to check whether a cell` ` ``// is valid / invalid` ` ``public` `static` `bool` `isValid(``int` `i, ``int` `j)` ` ``{` ` ``return` `(i >= 0 && j >= 0 && i < R && j < C);` ` ``}` ` ``// Function to check whether the cell` ` ``// is delimiter which is (-1, -1)` ` ``public` `static` `bool` `isDelim(Ele temp)` ` ``{` ` ``return` `(temp.x == -1 && temp.y == -1);` ` ``}` ` ``// Function to check whether there` ` ``// is still a fresh orange remaining` ` ``public` `static` `bool` `checkAll(``int``[][] arr)` ` ``{` ` ``for` `(``int` `i = 0; i < R; i++) {` ` ``for` `(``int` `j = 0; j < C; j++) {` ` ``if` `(arr[i][j] == 1) {` ` ``return` `true``;` ` ``}` ` ``}` ` ``}` ` ``return` `false``;` ` ``}` ` ``// This function finds if it is possible` ` ``// to rot all oranges or not. If possible,` ` ``// then it returns minimum time required` ` ``// to rot all, otherwise returns -1` ` ``public` `static` `int` `rotOranges(``int``[][] arr)` ` ``{` ` ``// Create a queue of cells` ` ``LinkedList Q = ``new` `LinkedList();` ` ``Ele temp;` ` ``int` `ans = 0;` ` ``// Store all the cells having rotten` ` ``// orange in first time frame` ` ``for` `(``int` `i = 0; i < R; i++) {` ` ``for` `(``int` `j = 0; j < C; j++) {` ` ``if` `(arr[i][j] == 2) {` ` ``Q.AddLast(``new` `Ele(i, j));` ` ``}` ` ``}` ` ``}` ` ``// Separate these rotten oranges from` ` ``// the oranges which will rotten` ` ``// due the oranges in first time frame` ` ``// using delimiter which is (-1, -1)` ` ``Q.AddLast(``new` `Ele(-1, -1));` ` ``// Process the grid while there are` ` ``// rotten oranges in the Queue` ` ``while` `(Q.Count > 0) {` ` ``// This flag is used to determine` ` ``// whether even a single fresh` ` ``// orange gets rotten due to rotten` ` ``// oranges in current time frame so` ` ``// we can increase the count of the` ` ``// required time.` ` ``bool` `flag = ``false``;` ` ``// Process all the rotten oranges` ` ``// in current time frame.` ` ``while` `(!isDelim(Q.First.Value)) {` ` ``temp = Q.First.Value;` ` ``// Check right adjacent cell that` ` ``// if it can be rotten` ` ``if` `(isValid(temp.x + 1, temp.y)` ` ``&& arr[temp.x + 1][temp.y] == 1) {` ` ``if` `(!flag) {` ` ``// if this is the first orange` ` ``// to get rotten, increase` ` ``// count and set the flag.` ` ``ans++;` ` ``flag = ``true``;` ` ``}` ` ``// Make the orange rotten` ` ``arr[temp.x + 1][temp.y] = 2;` ` ``// push the adjacent orange to Queue` ` ``temp.x++;` ` ``Q.AddLast(``new` `Ele(temp.x, temp.y));` ` ``// Move back to current cell` ` ``temp.x--;` ` ``}` ` ``// Check left adjacent cell that` ` ``// if it can be rotten` ` ``if` `(isValid(temp.x - 1, temp.y)` ` ``&& arr[temp.x - 1][temp.y] == 1) {` ` ``if` `(!flag) {` ` ``ans++;` ` ``flag = ``true``;` ` ``}` ` ``arr[temp.x - 1][temp.y] = 2;` ` ``temp.x--;` ` ``// push this cell to Queue` ` ``Q.AddLast(``new` `Ele(temp.x, temp.y));` ` ``temp.x++;` ` ``}` ` ``// Check top adjacent cell that` ` ``// if it can be rotten` ` ``if` `(isValid(temp.x, temp.y + 1)` ` ``&& arr[temp.x][temp.y + 1] == 1) {` ` ``if` `(!flag) {` ` ``ans++;` ` ``flag = ``true``;` ` ``}` ` ``arr[temp.x][temp.y + 1] = 2;` ` ``temp.y++;` ` ``// Push this cell to Queue` ` ``Q.AddLast(``new` `Ele(temp.x, temp.y));` ` ``temp.y--;` ` ``}` ` ``// Check bottom adjacent cell` ` ``// if it can be rotten` ` ``if` `(isValid(temp.x, temp.y - 1)` ` ``&& arr[temp.x][temp.y - 1] == 1) {` ` ``if` `(!flag) {` ` ``ans++;` ` ``flag = ``true``;` ` ``}` ` ``arr[temp.x][temp.y - 1] = 2;` ` ``temp.y--;` ` ``// push this cell to Queue` ` ``Q.AddLast(``new` `Ele(temp.x, temp.y));` ` ``}` ` ``Q.RemoveFirst();` ` ``}` ` ``// Pop the delimiter` ` ``Q.RemoveFirst();` ` ``// If oranges were rotten in current` ` ``// frame than separate the rotten` ` ``// oranges using delimiter for the` ` ``// next frame for processing.` ` ``if` `(Q.Count > 0) {` ` ``Q.AddLast(``new` `Ele(-1, -1));` ` ``}` ` ``// If Queue was empty than no rotten` ` ``// oranges left to process so exit` ` ``}` ` ``// Return -1 if all arranges could` ` ``// not rot, otherwise ans` ` ``return` `(checkAll(arr)) ? -1 : ans;` ` ``}` ` ``// Driver Code` ` ``public` `static` `void` `Main(``string``[] args)` ` ``{` ` ``int``[][] arr` ` ``= ``new` `int``[][] { ``new` `int``[] { 2, 1, 0, 2, 1 },` ` ``new` `int``[] { 1, 0, 1, 2, 1 },` ` ``new` `int``[] { 1, 0, 0, 2, 1 } };` ` ``int` `ans = rotOranges(arr);` ` ``if` `(ans == -1) {` ` ``Console.WriteLine(``"All oranges cannot rot"``);` ` ``}` ` ``else` `{` ` ``Console.WriteLine(``"Time required for all "` ` ``+ ``"oranges to rot => "` `+ ans);` ` ``}` ` ``}` `}` `// This code is contributed by Shrikant13`
## Javascript
`// JS program to find minimum time required to make all` `// oranges rotten` `// function to check whether a cell is valid / invalid` `function` `isvalid(i, j)` `{` ` ``return` `(i >= 0 && j >= 0 && i < 3 && j < 5)` `}` `// Function to check whether the cell is delimiter` `// which is (-1, -1)` `function` `isdelim(temp)` `{` ` ``return` `(temp[0] == -1 && temp[1] == -1)` `}` `// Function to check whether there is still a fresh` `// orange remaining` `function` `checkall(arr)` `{` ` ``for` `(``var` `i = 0; i < 3; i++)` ` ``for` `(``var` `j = 0; j < 5; j++)` ` ``if` `(arr[i][j] == 1)` ` ``return` `true` ` ``return` `false` `}` `// This function finds if it is` `// possible to rot all oranges or not.` `// If possible, then it returns` `// minimum time required to rot all,` `// otherwise returns -1` `function` `rotOranges(arr)` `{` ` ``// Create a queue of cells` ` ``let Q = []` ` ``let temp = [0, 0]` ` ``let ans = 1` ` ``// Store all the cells having` ` ``// rotten orange in first time frame` ` ``for` `(``var` `i = 0; i < 3; i++)` ` ``{` ` ``for` `(``var` `j = 0; j < 5; j++)` ` ``{` ` ``if` `(arr[i][j] == 2)` ` ``{` ` ``temp[0] = i` ` ``temp[1] = j` ` ``Q.push([i, j])` ` ``}` ` ``}` ` ``}` ` ``// Separate these rotten oranges` ` ``// from the oranges which will rotten` ` ``// due the oranges in first time` ` ``// frame using delimiter which is (-1, -1)` ` ``temp[0] = -1` ` ``temp[1] = -1` ` ``Q.push([-1, -1])` ` ``// print(Q)` ` ``// Process the grid while there are` ` ``// rotten oranges in the Queue` ` ``while` `(``false``)` ` ``{` ` ``// This flag is used to determine` ` ``// whether even a single fresh` ` ``// orange gets rotten due to rotten` ` ``// oranges in current time` ` ``// frame so we can increase` ` ``// the count of the required time.` ` ``flag = ``false` ` ``console.log(Q.length)` ` ``// Process all the rotten` ` ``// oranges in current time frame.` ` ``while` `(!isdelim(Q[0]))` ` ``{` ` ``temp = Q[0]` ` ``console.log(Q.length)` ` ``// Check right adjacent cell that if it can be rotten` ` ``if` `(isvalid(temp[0] + 1, temp[1]) && arr[temp[0] + 1][temp[1]] == 1)` ` ``{` ` ``// if this is the first orange to get rotten, increase` ` ``// count && set the flag.` ` ``if` `(!flag)` ` ``{` ` ``ans = ans + 1` ` ``flag = ``true` ` ``}` ` ``// Make the orange rotten` ` ``arr[temp[0] + 1][temp[1]] = 2` ` ``// append the adjacent orange to Queue` ` ``temp[0] += 1` ` ``Q.push(temp)` ` ``temp[0] -= 1 ``// Move back to current cell` ` ``}` ` ``// Check left adjacent cell that if it can be rotten` ` ``if` `(isvalid(temp[0] - 1, temp[1]) && arr[temp[0] - 1][temp[1]] == 1)` ` ``{` ` ``if` `(!flag)` ` ``{` ` ``ans = ans + 1` ` ``flag = ``true` ` ``}` ` ``arr[temp[0] - 1][temp[1]] = 2` ` ``temp[0] -= 1` ` ``Q.push(temp) ``// append this cell to Queue` ` ``temp[0] += 1` ` ``}` ` ``// Check top adjacent cell that if it can be rotten` ` ``if` `(isvalid(temp[0], temp[1] + 1) && arr[temp[0]][temp[1] + 1] == 1)` ` ``{` ` ``if` `(!flag)` ` ``{` ` ``ans++;` ` ``flag = ``true``;` ` ``} ` ` ``arr[temp[0]][temp[1] + 1] = 2` ` ``temp[1] += 1` ` ``Q.push(temp) ``// Push this cell to Queue` ` ``temp[1] -= 1` ` ``}` ` ``// Check bottom adjacent cell if it can be rotten` ` ``if` `(isvalid(temp[0], temp[1] - 1) && arr[temp[0]][temp[1] - 1] == 1)` ` ``{` ` ``if` `(!flag)` ` ``{` ` ``ans ++;` ` ``flag = ``true``;` ` ``}` ` ``arr[temp[0]][temp[1] - 1] = 2` ` ``temp[1] -= 1` ` ``Q.push(temp) ``// append this cell to Queue` ` ``}` ` ``Q.shift()` ` ``}` ` ``// Pop the delimiter` ` ``Q.shift()` ` ``// If oranges were rotten in` ` ``// current frame than separate the` ` ``// rotten oranges using delimiter` ` ``// for the next frame for processing.` ` ``if` `(Q.length == 0)` ` ``{` ` ``temp[0] = -1` ` ``temp[1] = -1` ` ``Q.push(temp)` ` ``}` ` ``// If Queue was empty than no rotten oranges left to process so exit` ` ``}` ` ` ` ``// Return -1 if all arranges could not rot, otherwise return ans.` ` ``if` `(checkall(arr))` ` ``return` `ans + 1 ` ` ``return` `-1` `}` `// Driver program` `let arr = [[2, 1, 0, 2, 1],` ` ``[1, 0, 1, 2, 1],` ` ``[1, 0, 0, 2, 1]]` `let ans = rotOranges(arr)` `if` `(ans == -1)` ` ``console.log(``"All oranges cannot rotn"``)` `else` ` ``console.log(``"Time required for all oranges to rot => "``, ans)` `// This code is contributed by phasing17`
Output
`Time required for all oranges to rot => 2`
Time Complexity: O( R *C), Each element of the matrix can be inserted into the queue only once so the upper bound of iteration is O(R*C)
Auxiliary Space: O(R*C), To store the elements in a queue.
Thanks to Gaurav Ahirwar for suggesting the above solution. | 15,028 | 44,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-23 | latest | en | 0.842573 |
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### Lesson 2: The Law of Momentum Conservation
Momentum Conservation in Collisions
Eq'ns as a "Recipe" for Problem Solving
Eq'ns as a Guide to Thinking
## Lesson 2: The Law of Momentum Conservation
### Using Equations as a Guide to Thinking
The three problems on the previous page illustrate how the law of momentum conservation can be used to solve problems in which the after-collision velocity of an object is predicted based on mass and velocity information. There are similar practice problems (with accompanying solutions) lower on this page which are worth the practice. However, let's first take a more qualitative approach to some collision problems. The questions which follow provide a real test of your conceptual understanding of momentum conservation in collisions.
Suppose that you have joined NASA and are enjoying your first space walk. You are outside the space shuttle when your fellow astronaut of approximately equal mass is moving towards you at 2 m/s (with respect to the shuttle). If she collides with you and holds onto you, then how fast (with respect to the shuttle do you both move after the collision?
This problem could be solved in the usual manner with a momentum table; the variable m could be used for the mass of the astronauts or any random number could be used for the mass of the astronauts (provided each astronaut had the same mass). In the process of solving the problem, the mass would cancel out of the momentum conservation equation and the post-collision velocities could be determined. However, there is a more conceptual means of solving this problem. In order for the momentum before the collision to be equal to the momentum after the collision, the after collision velocity must be smaller than the before collision velocity. How many times smaller must it be? By what factor must the velocity be decreased? Before the collision, the amount of mass in motion is m; after the collision, the amount of mass in motion is 2•m. The amount of mass in motion has doubled as the result of the collision. If the mass is increased by a factor of two, then the velocity must be decreased by a factor of 2. The before-collision velocity was 2 m/s so the after-collision velocity must be one-half this value: 1 m/s. Each astronaut is moving with a velocity of 1 m/s after the collision.
The process of solving this problem involved using a conceptual understanding of the equation for momentum (p=m*v). This equation becomes a guide to thinking about how a change in one variable effects a change in another variable. The constant quantity in a collision is the momentum (momentum is conserved). For a constant momentum value, mass and velocity are inversely proportional. Thus, an increase in mass results in a decrease in velocity.
A twofold increase in mass, results in a twofold decrease in velocity (the velocity is one-half its original value); a threefold increase in mass results in a threefold decrease in velocity (the velocity is one-third its original value); etc. Of course, it is instructive to point out that this form of problem-solving is limited to situations in which one of the two objects is at rest before the collision and both objects move at the same speed after the collision. To further test your understanding of this type of quantitative reasoning, try the following two questions.
A large fish is in motion at 2 m/s when it encounters a smaller fish which is at rest. The large fish swallows the smaller fish and continues in motion at a reduced speed. If the large fish has three times the mass of the smaller fish, then what is the speed of the large fish (and the smaller fish) after the collision? Click the button to view answer.
A railroad diesel engine has five times the mass of a boxcar. A diesel coasts backwards along the track at 4 m/s and couples together with the boxcar (initially at rest). How fast do the two trains cars coast after they have coupled together? Click the button to view answer.
Express your understanding of the concept and mathematics of momentum conservation by answering the following questions. Assume isolated systems and momentum conservation for each problem. Click the See Answer button to view answer. (If necessary, return to the instructional page on solving collision analysis problems.)
1. In a physics lab, 0.500-kg cart (Cart A) moving rightward with a speed of 92.8 cm/s collides with a 1.50-kg cart (Cart B) moving leftward with a speed of 21.6 cm/s. The two carts stick together and move as a single object after the collision. Determine the post-collision speed of the two carts.
See Solution
2. A 25.0-gram bullet enters a 2.35-kg watermelon and embeds itself in the melon. The melon is immediately set into motion with a speed of 3.82 m/s. The bullet remains lodged inside the melon. What was the entry speed of the bullet? (CAUTION: Be careful of the units on mass.)
See Solution
3. A 25.0-gram bullet enters a 2.35-kg watermelon with a speed of 217 m/s and exits the opposite side with a speed of 109 m/s. If the melon was originally at rest, then what speed will it have as the bullet leaves its opposite side? (CAUTION: Be careful of the units on mass.)
See Solution
4. In a physics lab, a 0.500-kg cart (Cart A) moving with a speed of 129 cm/s encounters a magnetic collision with a 1.50-kg cart (Cart B) which is initially at rest. The 0.500-kg cart rebounds with a speed of 45 cm/s in the opposite direction. Determine the post-collision speed of the 1.50-kg cart.
See Solution
5. A 3000-kg truck moving rightward with a speed of 5 km/hr collides head-on with a 1000-kg car moving leftward with a speed of 10 km/hr. The two vehicles stick together and move with the same velocity after the collision. Determine the post-collision velocity of the car and truck. (CAREFUL: Be cautious of the +/- sign on the velocity of the two vehicles.)
See Solution
6. During a goal-line stand, a 75-kg fullback moving eastward with a speed of 8 m/s collides head-on with a 100-kg lineman moving westward with a speed of 4 m/s. The two players collide and stick together, moving at the same velocity after the collision. Determine the the post-collision velocity of the two players. (CAREFUL: Be cautious of the +/- sign on the velocity of the two players.)
See Solution
1. The problem can be solved using a momentum table:
2. The problem can be solved using a momentum table:
3. The problem can be solved using a momentum table:
4. The problem can be solved using a momentum table:
5. The problem can be solved using a momentum table:
6. The problem can be solved using a momentum table:
### Lesson 2: The Law of Momentum Conservation
Momentum and Collisions: Chapter Outline || About the Tutorial || Tutorial Topics || Usage Policy || Feedback | 1,527 | 6,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2018-05 | latest | en | 0.937587 |
https://numbermatics.com/n/634/ | 1,721,690,386,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517927.60/warc/CC-MAIN-20240722220957-20240723010957-00291.warc.gz | 361,850,025 | 6,470 | # 634
## 634 is an even composite number composed of two prime numbers multiplied together.
What does the number 634 look like?
This visualization shows the relationship between its 2 prime factors (large circles) and 4 divisors.
634 is an even composite number. It is composed of two distinct prime numbers multiplied together. It has a total of four divisors.
## Prime factorization of 634:
### 2 × 317
See below for interesting mathematical facts about the number 634 from the Numbermatics database.
### Names of 634
• Cardinal: 634 can be written as Six hundred thirty-four.
### Scientific notation
• Scientific notation: 6.34 × 102
### Factors of 634
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 2
• Sum of prime factors: 319
### Divisors of 634
• Number of divisors d(n): 4
• Complete list of divisors:
• Sum of all divisors σ(n): 954
• Sum of proper divisors (its aliquot sum) s(n): 320
• 634 is a deficient number, because the sum of its proper divisors (320) is less than itself. Its deficiency is 314
### Bases of 634
• Binary: 10011110102
• Base-36: HM
### Squares and roots of 634
• 634 squared (6342) is 401956
• 634 cubed (6343) is 254840104
• The square root of 634 is 25.1793566241
• The cube root of 634 is 8.5907237279
### Scales and comparisons
How big is 634?
• 634 seconds is equal to 10 minutes, 34 seconds.
• To count from 1 to 634 would take you about five minutes.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 634 cubic inches would be around 0.7 feet tall.
### Recreational maths with 634
• 634 backwards is 436
• The number of decimal digits it has is: 3
• The sum of 634's digits is 13
• More coming soon!
#### Copy this link to share with anyone:
MLA style:
"Number 634 - Facts about the integer". Numbermatics.com. 2024. Web. 22 July 2024.
APA style:
Numbermatics. (2024). Number 634 - Facts about the integer. Retrieved 22 July 2024, from https://numbermatics.com/n/634/
Chicago style:
Numbermatics. 2024. "Number 634 - Facts about the integer". https://numbermatics.com/n/634/
The information we have on file for 634 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 634, math, Factors of 634, curriculum, school, college, exams, university, Prime factorization of 634, STEM, science, technology, engineering, physics, economics, calculator, six hundred thirty-four.
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Some bits of this website may not work unless you switch it on. | 805 | 3,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-30 | latest | en | 0.884289 |
https://www.dubapk.com/download-apk/free-graphing-calculator-2/com.jockusch.freegraphingcalculator/ | 1,638,059,496,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358323.91/warc/CC-MAIN-20211127223710-20211128013710-00030.warc.gz | 789,857,848 | 25,940 | #### Screenshot
A powerful, flexible graphing calculator . . . and it's free!
Does far more than most of the paid calculators out there . . . let alone the free ones.
Features:
1) Scientific Calculator. Simple to grasp and easy to use, but powerful features are available when you need them. Available functions include the following:
• the usual arithmetic functions and exponentiation.
• square root, cube root, nth root, natural log, log base 10, log of arbitrary base, absolute value, factorial, permutations (nPr), combinations (nCr), modulus, random integer, bell curve, cumulative normal distribution, decimal to fraction.
2) Graphing. Capabilities:
• Graph up to four equations at once.
• Graphs are labeled.
• You can drag the graph or pinch to zoom in or out.
• Calculator can find roots and intersections.
• Graph in polar coordinates.
• Graph parametric equations
• Can graph implicit functions, such as x^2+y^2-4=0. Most calculator apps can't do this!
3) A unit converter. With a tap, you can enter the result of your conversion into the calculator. Currently converts different units of the following: acceleration, angle, area, density, distance, energy, force, mass, power, pressure, speed, temperature, time, and volume. Great for doing physics homework!
4) Constants for scientific calculations -- speed of light, strength of gravity at Earth's surface, etc. etc. etc. Tapping on a constant will insert it into your calculation -- i.e, you don't have to key in the value. Again, great for doing physics homework!
5) It can make a table of the values of a function you enter. You can choose the starting x value of the table, as well as how much x increases for each successive row.
6) Help screens linked directly to many of the available functions and constants. Tap the disclosure arrow to see the definition.
7) Forgot the quadratic formula? Or the double-angle formulas for sine and cosine? The math/science reference hits the high points of various subjects. Currently includes algebra, differential and integral calculus, geometry, trigonometry, vectors, vector calculus, and classical mechanics.
8) Statistics -- enter data and make a histogram, box and whisker plot, or scatter plot with optional regression line. Also statistical distributions (under functions > statistics) -- beta, Cauchy, chi, chi squared, continuous uniform, Erlang, exponential, Fisher-Snedecor, gamma, inverse gamma, Laplace, log-normal, normal, Pareto, Rayleigh, Student's T, and Weibull.
I'd love to hear your comments or suggestions. Because of spammers, my email address is not in this description, but it is inside the app, and it's great to hear from real users.
From version Free Graphing Calculator 2 10.4:
Attempted to fix launch issue where one has to switch tabs to get it to work | 619 | 2,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-49 | longest | en | 0.863162 |
https://betterlesson.com/lesson/572468/my-unique-perimeter-performance-task?from=consumer_breadcrumb_dropdown_lesson | 1,542,398,861,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743184.39/warc/CC-MAIN-20181116194306-20181116220306-00204.warc.gz | 570,841,750 | 23,660 | # My Unique Perimeter Performance Task
9 teachers like this lesson
Print Lesson
## Objective
SWBAT create unique words and shapes and determine the perimeter of each one.
#### Big Idea
I can create things that have their own unique perimeter.
## Warm-Up
5 minutes
Today I have a quick activity for you to do to wake-up your brain and get ready for more fun with perimeter. Who can remind me what perimeter is? When could I use it in my life? How can I find the perimeter of things? Great! Off you go!
I expect that students will discuss how we can measure items, count unit squares along the outside of a figure or add the labeled sides of geometric shapes
## Activity
30 minutes
We have been on a journey with perimeter these past few days and I have enjoyed watching you discover that perimeter is all around you! Today we will get to use our creativity to create some things with unique perimeter!
Today your imaginations get to have some fun, because we are going to be creating with perimeter. I drew my name in big letters on this piece of grid paper (show) and I was able to find the perimeter of each letter.
I also had a little fun making a monster, and then finding the perimeter of my new friend. Now, it's your turn to use what you know about perimeter to create on grid paper, and find each unique thing's perimeter.
Students use the graph paper to create unique figures (MP5) and solve to find their perimeter. I also expect students to label, show their work and be able to discuss how they determined the perimeter of each figure (MP6).
## Share Out
5 minutes
It was exciting to see the new monsters you created with perimeter, and to see how your names came to life with their unique perimeter! Who can share something they noticed today as they were working? | 382 | 1,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-47 | latest | en | 0.954576 |
https://math.answers.com/Q/What_is_height_of_the_rectangle_if_the_area_is_108m_base_18m_side_20m | 1,685,607,914,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647639.37/warc/CC-MAIN-20230601074606-20230601104606-00147.warc.gz | 426,467,679 | 50,077 | 0
# What is height of the rectangle if the area is 108m base 18m side 20m?
Wiki User
2012-07-28 06:13:14
The question cannot be answered because there is an error in the information provided.
If the base of a rectangle is 18m, and the side is 20 m, then the area cannot be 108m. (It cannot be any number of metres since area must be measured in square metres, but that is another matter)
Wiki User
2012-07-28 06:13:14
Study guides
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2523 Reviews | 145 | 482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-23 | latest | en | 0.896986 |
https://progassignments.com/who-can-help-me-with-understanding-and-implementing-algorithms-for-computational-history-in-c | 1,713,848,094,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818464.67/warc/CC-MAIN-20240423033153-20240423063153-00891.warc.gz | 419,584,837 | 18,614 | # Who can help me with understanding and implementing algorithms for computational history in C++?
Who can help me with understanding and implementing algorithms for computational history in C++? I just came across one of my C++ blog posts. It’s about the history of time and memory usage and I can’t believe that someone didn’t mentioned how much time an algorithm was. And this article just popped up around the 3rd week of October. A lot of discussion here: Why does the algorithm exist? I can’t think of anything else. I found this article on Wikipedia, but my current interest in it seems somewhat off-topic. What does that suggest? I have seen articles discussing this; one by Thomas Gault, a professor in Houston, et. al. from 2002, where Charles Lin-Wunsch, a mathematician, was called one of the most infamous culprits who kept track of the past. Lin-Wunsch put forward a 20-year-cure theory showing what happened to computer science on a daily basis. The major figures in the research were Albert Einstein, Charles Cling, Alexander Celestis. Someone I would not be surprised to know was out for the same reason. What is the mathematical basis of a C++ program? Your algorithm/library is specifically designed to do programming, and every line you make makes the algorithm their website in memory real, long enough to run as it should. It’s good for understanding the historical machinery of programming/using the algorithms to implement/develop new algorithms. It’s all pretty cool, though. In a very few seconds, you’ll get that logic working: Programming time: When you calculate hours, use the formula 1 minute = 100/1000 and solve for hours; you can subtract seconds from hours to compute hours: A more efficient method using the work set of the algorithm’s computation would be: where M = 20 seconds, n = 300 for 5 seconds, and n is the number of minutes to run, which is 3 minutes. Who can help me with understanding and implementing algorithms for computational history in C++? or you can at least elaborate the terminology correctly, and so let’s get off the work. A basic model for using an ariasche of dates can easily be translated into a hierarchical, but time efficient multivariate time machine (TM) framework. We’ll here explore one such configuration. Let’s use the traditional four-year special info window to sample from a real world scenario. Over the past 16 months of 2016, I downloaded 20. | 518 | 2,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-18 | latest | en | 0.947219 |
https://studentshare.net/logic-programming/11406-zemax-ee-software-and-programming | 1,513,311,243,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948563629.48/warc/CC-MAIN-20171215040629-20171215060629-00722.warc.gz | 664,399,964 | 16,832 | StudentShare solutions
# Zemax EE Software and Programming - Dissertation Example
## Extract of sampleZemax EE Software and Programming
Aside from investigating properties such as reflection coefficient, absorption coefficient, temperature variation, and thermal conductivity, additional theoretical estimations are made. An example is the determination of focal length, exposed effective areas, and light collection and concentration for all components. A number of designs are evaluated to examine the effect of different geometrical shapes of light pipes for refraction studies and light cups for reflection studies. Zemax software is used for a large part of this study. Therefore, this chapter details the basics of Zemax software for optical design and ray tracing. Software programs for lens design and optical studies have been of immense utility in the field of optics. Their advent has substantially simplified and enhanced our understanding of optical design and analysis. As stated by Winston, Minano, and Benitez (2005), The design of imaging optical systems is a classic ?eld of research that has achieved a high level of development. There are on the market optical design programs that permit the numerical optimization of the design parameters, allowing us to obtain results that were unattainable with the analytical tools used before the development of computers (p. 219). Software programs are based on a number of optical principles, such as geometrical optics and ray tracing that function according to the basic laws of optics. The following sections discuss these basic principles along with an introduction to Zemax software that is used in this study. 4.2. Ray Tracing Programs and Geometrical Optics The basic tool required in designing any imaging or non-imaging optical system is geometrical optics (Winston, Minano, & Benitez, 2005). Geometrical optics is based on the universal laws of reflection, refraction, and transmission of light. The incident light on a reflective surface and the reflected light from that surface always make equal angles to the normal, and they lie on the same plane. In case of transmitted light, the direction of the refracted and transmitted ray changes according to Snell’s law of refraction, according to which the sine of the angle between the incident ray and the normal is in constant ratio with the sine of the angle between the refracted ray and the normal, with all three being coplanar (Winston, Minano, & Benitez). Based on these universal laws of light, the behavior of a light ray in any optical system can be predicted. Simple optical systems can be predicted manually, but complex ones require sophisticated computer programs that can easily predict the behavior of light through computerized ray tracing. Even the analysis and design of solar concentrators requires such a program. Ray tracing is the process of constructing or following the paths of light rays through an optical system consisting of refracting and reflecting surfaces (Winston, Minano, & Benitez, 2005). For instance, the transmission of a specific concentrator can be determined through ray tracing as follows: N rays enter a concentrator’s aperture at a ? angle of incidence at the entry aperture of the concentrator, as shown in Figure 4.1. Fig. 4.1: Ray tracing of the rays transmitted through a concentrator to determine transmission (Winston, Minano, & Benitez, 2005). After the rays are traced through the optical system, N’ rays appear from the concentrator’s exit aperture. The dimensions of the exit aperture are determined based on the required concentration ratio of the system. The remaining N and N’ rays are lost through various processes such as ray scattering. The transmission power of the system for a different angle of incidence ...Show more
## Summary
This dissertation aims to use Zemax software to understand the light paths through ray tracing for various light shapes and samples with different geometries made from glass, polymeric materials, and Al metal…
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Broad Topics > Using, Applying and Reasoning about Mathematics > Making and testing hypotheses
### Fit These Shapes
##### Age 5 to 11 Challenge Level:
What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?
### Marvellous Matrix
##### Age 7 to 11 Challenge Level:
Follow the directions for circling numbers in the matrix. Add all the circled numbers together. Note your answer. Try again with a different starting number. What do you notice?
### Count the Digits
##### Age 5 to 11 Challenge Level:
In this investigation we are going to count the number of 1s, 2s, 3s etc in numbers. Can you predict what will happen?
### Logic Block Collections
##### Age 5 to 7 Challenge Level:
What do you think is the same about these two Logic Blocks? What others do you think go with them in the set?
### Seven Pots of Plants
##### Age 7 to 11 Challenge Level:
There are seven pots of plants in a greenhouse. They have lost their labels. Perhaps you can help re-label them.
### Sorting Logic Blocks
##### Age 5 to 11 Challenge Level:
This interactivity allows you to sort logic blocks by dragging their images.
### Different Sizes
##### Age 5 to 11 Challenge Level:
A simple visual exploration into halving and doubling.
### Folding, Cutting and Punching
##### Age 7 to 11 Challenge Level:
Exploring and predicting folding, cutting and punching holes and making spirals.
### The Mathemagician's Seven Spells
##### Age 7 to 11 Challenge Level:
"Tell me the next two numbers in each of these seven minor spells", chanted the Mathemagician, "And the great spell will crumble away!" Can you help Anna and David break the spell?
### National Flags
##### Age 7 to 11 Challenge Level:
This problem explores the shapes and symmetries in some national flags.
### Spot Thirteen
##### Age 7 to 11 Challenge Level:
Choose 13 spots on the grid. Can you work out the scoring system? What is the maximum possible score?
### Troublesome Triangles
##### Age 7 to 14 Challenge Level:
Many natural systems appear to be in equilibrium until suddenly a critical point is reached, setting up a mudslide or an avalanche or an earthquake. In this project, students will use a simple. . . .
### Poly Plug Rectangles
##### Age 5 to 14 Challenge Level:
The computer has made a rectangle and will tell you the number of spots it uses in total. Can you find out where the rectangle is?
### Shapes in the Alphabet
##### Age 5 to 7 Challenge Level:
In this problem, we're going to find sets of letter shapes that go together.
### I Like ...
##### Age 5 to 7 Challenge Level:
Mr Gilderdale is playing a game with his class. What rule might he have chosen? How would you test your idea?
### Light the Lights Again
##### Age 7 to 11 Challenge Level:
Each light in this interactivity turns on according to a rule. What happens when you enter different numbers? Can you find the smallest number that lights up all four lights?
### Light the Lights
##### Age 5 to 7 Challenge Level:
Investigate which numbers make these lights come on. What is the smallest number you can find that lights up all the lights?
### Our Numbers
##### Age 5 to 7 Challenge Level:
These spinners will give you the tens and unit digits of a number. Can you choose sets of numbers to collect so that you spin six numbers belonging to your sets in as few spins as possible?
### White Box
##### Age 7 to 18 Challenge Level:
This game challenges you to locate hidden triangles in The White Box by firing rays and observing where the rays exit the Box.
### Which Numbers? (2)
##### Age 7 to 11 Challenge Level:
I am thinking of three sets of numbers less than 101. Can you find all the numbers in each set from these clues?
### Which Numbers? (1)
##### Age 7 to 11 Challenge Level:
I am thinking of three sets of numbers less than 101. They are the red set, the green set and the blue set. Can you find all the numbers in the sets from these clues?
### Observing the Sun and the Moon
##### Age 7 to 14 Challenge Level:
How does the time of dawn and dusk vary? What about the Moon, how does that change from night to night? Is the Sun always the same? Gather data to help you explore these questions.
### Table Patterns Go Wild!
##### Age 7 to 11 Challenge Level:
Nearly all of us have made table patterns on hundred squares, that is 10 by 10 grids. This problem looks at the patterns on differently sized square grids.
### Ishango Bone
##### Age 7 to 18 Challenge Level:
Can you decode the mysterious markings on this ancient bone tool? | 1,092 | 4,813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-39 | latest | en | 0.848333 |
https://www.cleancss.com/convert-units/units-of-length/pc/dm | 1,656,338,107,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103334753.21/warc/CC-MAIN-20220627134424-20220627164424-00447.warc.gz | 739,555,943 | 18,365 | # Convert PC to DM
1
Parsecs
=
3.0856775814914E+17
Decimeters
How To Calculate Parsecs To Decimeters
To convert parsecs to decimeters you simply multiply your parsecs by 3.0856775814914E+17. The formula would look like this:
Ydm = Xpc * 3.0856775814914E+17
#### 1 Parsecs equals
Meters 3.08568e+16 Kilometers 3.08568e+13 Decimeters 3.08568e+17 Centimeters 3.08568e+18 Milimeters 3.08568e+19 Micrometers 3.08568e+22 Nanometers 3.08568e+25 Picometers 3.08568e+28 Inchs 1.21483e+18 Feet 1.01236e+17 Yards 3.37454e+16 Miles 1.91735e+13 Hands 3.03708e+17 Lightyears 3.262 Astronomical Units 206265
#### Parsecs To Decimeters Conversion Table
From To
1 pc3.0856775814914E+17 dm
2 pc6.1713551629827E+17 dm
3 pc9.2570327444741E+17 dm
4 pc1.2342710325965E+18 dm
5 pc1.5428387907457E+18 dm
6 pc1.8514065488948E+18 dm
7 pc2.159974307044E+18 dm
8 pc2.4685420651931E+18 dm
9 pc2.7771098233422E+18 dm
10 pc3.0856775814914E+18 dm
11 pc3.3942453396405E+18 dm
12 pc3.7028130977896E+18 dm
13 pc4.0113808559388E+18 dm
14 pc4.3199486140879E+18 dm
15 pc4.6285163722371E+18 dm
16 pc4.9370841303862E+18 dm
17 pc5.2456518885353E+18 dm
18 pc5.5542196466845E+18 dm
19 pc5.8627874048336E+18 dm
20 pc6.1713551629827E+18 dm
30 pc9.2570327444741E+18 dm
40 pc1.2342710325965E+19 dm
50 pc1.5428387907457E+19 dm
60 pc1.8514065488948E+19 dm
70 pc2.159974307044E+19 dm
80 pc2.4685420651931E+19 dm
90 pc2.7771098233422E+19 dm
100 pc3.0856775814914E+19 dm | 637 | 1,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-27 | longest | en | 0.424773 |
https://www.kenyaplex.com/videolessons/400-karani-bought-4-pencils-and-6-biro-pens-for-sh-66-and-tachora-bought-2-pencils-and-5-biro-pens-for-sh-51.aspx | 1,555,991,516,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578586680.51/warc/CC-MAIN-20190423035013-20190423061013-00328.warc.gz | 722,123,461 | 12,632 | # Karani bought 4 pencils and 6 biro pens for sh. 66 and Tachora bought 2 pencils and 5 biro pens for sh. 51
Class: Form 1
Subject: Mathematics
Topic: Linear Equations
## Lesson Summary
Karani bought 4 pencils and 6 biro pens for sh. 66 and Tachora bought 2 pencils and 5 biro pens for sh. 51
a) Find the price of each item.
b) Musoma spent sh. 228 to buy the same type of pencils and biro pens. If the number of biro pens he bought were 4 more than the number of pencils, find the number of pencils he bought.
a) Let the cost of biro pen be b
Let the price of a pencil be p
4š+6š=66ā¦ā¦..(1)
2š+5š=51ā¦ā¦ā¦(2)
By elimination method
Multiply (2) by 2 to eliminate p
4š+6š=66
2(2š+5š=51)
4š+6š=66
-4š+10š=102
-4p-4=-36-4
š=š ā.9
Substitute b = 9 in (2)
2(š)+5(9)=51
2š+45=51
2š=51ā45
2p2=62
p=3
B= sh.9 p = sh.3
b)
B=x + 4
P = x
9(x + 4) + 3x = 228
9x + 36 + 3x = 228
12x = 228 ā 36
12x = 192
12x12=19212
X = 16
## More Video Lessons
View all video lessons | 465 | 1,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2019-18 | latest | en | 0.753552 |
https://www.funwithpuzzles.com/2023/09/mistake-finding-visual-picture-riddle.html | 1,695,996,268,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510516.56/warc/CC-MAIN-20230929122500-20230929152500-00423.warc.gz | 846,084,811 | 34,924 | If you're a fan of Gin Rummy and love a good puzzle challenge, you're in for a treat! This mistake-finding picture riddle features an image taken from the Gin Rummy Card Game, but there's a hidden mistake lurking within.
Your task is to put your sharp observation skills to the test and find the mistake in this puzzle picture as quickly as possible. It might not be obvious at first glance, so take your time and examine every detail closely.
Whether you're a Gin Rummy enthusiast or simply enjoy solving picture puzzles, this one will keep you engaged and entertained. Challenge yourself to find the error in the quickest time, and if you're feeling competitive, invite your friends to join in the fun too!
Once you've successfully spotted the mistake, feel free to share your triumph in the comments. So, are you ready to embark on this Gin Rummy mistake-finding adventure? Let's see how fast you can find the mistake in this picture riddle!
Can you find the mistake?
The answer to this "Mistake Finding Picture Riddle", can be viewed by clicking on the button. Please do give your best try before looking at the answer.
Finding the Mistake in the given picture will test your observational skills. Please check out the below-listed fun puzzles and brain teasers to test your visual skills and logical analytical ability.
## List of Fun Brain Teasers and Puzzles
Tough Picture Puzzles for Adults: It contains very tough picture puzzles in which your challenge is to find the hidden animals in the given picture puzzle images.
What is Next Number In The Sequence?: These are Maths and Logic puzzles in which your challenge is to find the next number in the series of number which follows a particular pattern.
Quick Solving Maths Brain Teasers with Answers for Kids: These are very easy Maths puzzles that kids will love to enjoy solving.
Logical Pictures Brain Teasers with Answers: At last set of some logical picture puzzles which will twist your brain. | 403 | 1,970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-40 | longest | en | 0.940691 |
https://www.slideserve.com/presley/welcome-to-econ-420-applied-regression-analysis | 1,529,462,563,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863407.58/warc/CC-MAIN-20180620011502-20180620031502-00126.warc.gz | 911,367,442 | 14,534 | Welcome to Econ 420 Applied Regression Analysis
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# Welcome to Econ 420 Applied Regression Analysis - PowerPoint PPT Presentation
Welcome to Econ 420 Applied Regression Analysis. Study Guide Week Nine. Some of you did not do as well as I expected on Assignment 2. So here is what we will do. I explain the assignment. You will redo Question 3 & 4 plus a couple of other questions.
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### Welcome to Econ 420 Applied Regression Analysis
Study Guide
Week Nine
Some of you did not do as well as I expected on Assignment 2
• So here is what we will do.
• I explain the assignment.
• You will redo Question 3 & 4 plus a couple of other questions.
• You will utilize the discussion board of the WebCT to ask questions.
Assignment 7 Key (40 points)
• #4, page 112
This regression has fallen into the dummy variable trap. CFP and NOCFP have a perfect linear relationship because CFP + NOCFP = 1 no matter what. This violates the assumption that two independent variables cannot have a perfectly linear relationship. The regression cannot be estimated. One of the two dummy variables, probably NOCFP, should be left out of the regression.
2. #5, page 112
a. Model ii is the best model. Model i has fallen into the dummy variable trap; one dummy variable must be omitted from the regression. Model iii does not have an intercept.
b. Looking at model ii, the hypothesis that the seasons of the year don’t affect EMPLOYMENT would be H0: B2=B3=B4=0. Use an F-test to test this null hypothesis versus the alternative hypothesis that at least one of the seasons matter. For the test, Model ii is the unrestricted model. The restricted model is EMPLOYMENT = B0 + B1GDP + e.
3. Use the data set dvd4 and EViews to test the hypothesis that at high levels of income people are less sensitive to the price of dvd than at low levels of income. Use 5 percent level of significance
• Need to add an interaction variable to our equation:
• DVDEXP = B^0 + B^1INCOME + B^2PRICE + B^3RAINFALL + B^4(inctimesprice)
• The interaction variable is inctimesprice which is INCOME * PRICE
• Partial derivative (slope) of DVDEXP with respect to price is
• d(DVDEXP)/d (Price) = B^2 + B^4 Income
• You expect B^2to be negative
• If the hypothesis is true then you also expect B^4 to be positive (and smaller than B^2)
• If B^4 turns out to be significantly bigger than zero, then you have found evidence that at as income goes up the slope of DVDEXP with respect to PRICE goes down (becomes less of a negative)
To create the new variable “incomeXprice” in EViews:
• Go to proc
• Click on generate series
• Enter the equation: inctimesprice= Income * Price
• Note: you can name the new variable whatever you would like. I have called it inctimesprice
Step 1:
• Null hypothesis: H0: B4 ≤ 0
• Alt. hypothesis: HA: B4 > 0
• Step 2:
• We will use 5% as our level of significance. Degrees of freedom = n – k – 1 = 30 – 4 – 1 = 25; tc = 1.708 (from pg. 312). Decision rule: reject null hypothesis if t > tc, that is if t > 1.708
• Step 3:
• Run the regression:
• The estimated coefficient on INCOMEXPRICE = 0.001973 and the t-stat =1.19
• Step 4:
• Because our t-statistic, 1.19 is not greater than 1.708, the null hypothesis B4 ≤ 0 cannot be rejected at a 5% significance level; we cannot say that at high levels of income people are less sensitive to the price of a dvd than those at low levels of income.
4, #13, page 113
a. H0: B1=B2, HA: B1 is not equal to B2
b. You will need to design your own F-test. The unrestricted model is:
• CONSUMPTION = B0 + B1 INCOME + B2 WEALTH+ e
• The restricted model is found by forcing the null hypothesis to be true. If B1=B2 then
• CONSUMPTION = B0 + B1 INCOME + B1 WEALTH+ e
• This is the same as:
• CONSUMPTION = B0 + B1 (INCOME + WEALTH) + e
• where (INCOME + WEALTH) is the only independent variable in the regression.
• Under EViews you will need to generate a new variable call it “incwealth”
• Incwealth = INCOME + WEALTH
c. After estimating the restricted and unrestricted models, you should get a residual sum of squares for the restricted model of 5.28 x 109 or 5,280,000,000. The residual sum of squares for the unrestricted model is 4.33 x 109 or 4,330,000,000. q, the number of restrictions, is 1. B1 can take any value, and then B2 must take the same value as B1 for the null hypothesis to be true, so B2 has a restricted value if the null hypothesis is true. The null hypothesis used here only imposes one restriction.
5.92
The critical value for F1,27 with a 1% error level is 7.68. Since the calculated value of the F-statistic is lower than the critical value, we cannot reject the null hypothesis that B1=B2. (Note that the critical value for a 5% error level is 4.21, so the null hypothesis could be rejected with a 5% error level).
• Redo Question 3 of Assignment 7
• Redo Question 4 of Assignment 7
• Question 10, Page 113
• Question 12, Page 113 | 1,490 | 5,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-26 | latest | en | 0.88321 |
https://www.teacherspayteachers.com/Product/Resources-For-Teaching-3-Digit-Addition-Subtraction-With-Regrouping-117278 | 1,505,849,105,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685993.12/warc/CC-MAIN-20170919183419-20170919203419-00339.warc.gz | 840,797,640 | 23,311 | Total:
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# Resources For Teaching 3-Digit Addition & Subtraction With Regrouping
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"Three Digit Subtraction With Regrouping Sample Problem-5 Steps"
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\$10.00 | 561 | 2,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-39 | longest | en | 0.792315 |
https://bookvea.com/can-the-decimal-representation-of-a-number-be-non-terminating-non-repeating/ | 1,675,514,698,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500126.0/warc/CC-MAIN-20230204110651-20230204140651-00487.warc.gz | 170,909,417 | 21,741 | Can the decimal representation of a number be non terminating non repeating?
Can the decimal representation of a number be non terminating non repeating?
A non-terminating, non-repeating decimal is a decimal number that continues endlessly, with no group of digits repeating endlessly. Decimals of this type cannot be represented as fractions, and as a result are irrational numbers. Pi is a non-terminating, non-repeating decimal.
Is the decimal representation of an irrational number?
Decimal representation of an irrational number is always non terminating non repeating. When an irrational number is changed into a decimal, the resulting number is a non terminating, nonrecurring decimal.
What Cannot be the decimal representation of a rational number?
The decimal representation of a rational number cannot be non-terminating non-repeating because the decimal expansion of rational numbers is either terminating or non-terminating recurring. The non-Terminating and non-repeating decimals are said to be Irrational numbers.
What’s a non repeating non terminating decimal called?
irrational numbers
How do you write a non terminating non repeating decimal?
The decimal representation of a rational number cannot be non-terminating non-repeating because the decimal expansion of rational numbers is either terminating or non-terminating recurring. | 251 | 1,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2023-06 | latest | en | 0.839376 |
http://math.stackexchange.com/questions/169994/question-about-proof-of-arzel%c3%a0-ascoli | 1,469,647,108,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827077.13/warc/CC-MAIN-20160723071027-00201-ip-10-185-27-174.ec2.internal.warc.gz | 158,449,640 | 18,493 | # Question about proof of Arzelà-Ascoli
(Arzelà-Ascoli, $\Longleftarrow$) Let $K$ be a compact metric space. Let $S \subset (C(K), \|\cdot\|_\infty)$ be closed, bounded and equicontinuous. Then $S$ is compact, that is, for a sequence $f_n$ in $S$ we can find a convergent subsequence (conv. in $\|\cdot\|_\infty)$.
Proof: Note that a compact metric space is separable. Let $D$ be a dense, countable subset of $K$. By assumption, $S$ is bounded, hence there exists $M$, such that $\|f\|_\infty \leq M$ for all $f$ in $S$, in particular, for $f_n$.
Now consider the space of all functions from $D$ to $[-M,M]$. By Tychonoff, $[-M,M]^D$ is compact. Define $g_n = f_n\mid_D$. Then $g_n$ is a sequence in $[-M,M]^D$, hence has a convergent subsequence $g_{n_k}$.
This is what it says in my notes. Now I've been thinking about what exactly "convergent" means in the last sentence. I'm quite sure it means pointwise. But theoretically, I can endow $[-M,M]^D$ with a norm or metric (that induces the product topology). So "convergent" could mean convergent with respect to that metric, no? Or does that not make sense?
-
You even need a metric to conclude a convergent subsequence. And the metric $e$ for the product topology is obvious as $D$ is countable, i.e. $e(f,g) = \sum_{d\in D} |f(d)-g(d)|$. – Vobo Jul 12 '12 at 16:31
Ooh, of course. Thank you! Why don't you post it as an answer, then I can upvote and accept. – Rudy the Reindeer Jul 12 '12 at 16:35
@Vobo How do we know $\sum_{d\in D} |f(d)-g(d)| < \infty$? – Rudy the Reindeer Jul 12 '12 at 16:36
I forgot the converging factor, see the answer below. – Vobo Jul 12 '12 at 16:42
I am quite sure that if you read the rest of the proof in your notes, the answer to your question will be become clear. – wildildildlife Jul 12 '12 at 18:44
You even need a metric to conclude a convergent sub*sequence*. And the metric d for the product topology is obvious for e.g. $D=\{ x_n| n\in N\}$, i.e. \begin{align} d(f,g)=\sum_{n=1}^\infty 2^{-n}|f(x_n)−g(x_n)|. \end{align}
-
For this to be finite, each term in the sum would have to be at least smaller than $1$, that is, $|f(d) - g(d)| < 2^n$. Why do we know that this is the case? – Rudy the Reindeer Jul 12 '12 at 16:48
You just need a uniform bound. Since $S$ is bounded, this is true. – copper.hat Jul 12 '12 at 16:51
You don’t actually need a metric to conclude that there’s a convergent subsequence: first countability is sufficient. Of course you have a metric, so you might as well use it! @Matt: $|f(x_n)-g(x_n)|\le 2M$ for each $n$, so $d(f,g)\le 2M$. – Brian M. Scott Jul 12 '12 at 20:40
Nice, thank you everyone, Vobo, @BrianM.Scott and @copper.hat! – Rudy the Reindeer Jul 13 '12 at 6:02 | 895 | 2,701 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2016-30 | latest | en | 0.884742 |
https://in.mathworks.com/matlabcentral/answers/580146-euler-s-method | 1,680,426,860,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950422.77/warc/CC-MAIN-20230402074255-20230402104255-00682.warc.gz | 359,028,283 | 27,312 | # Euler's Method
4 views (last 30 days)
Mikela Petersen on 16 Aug 2020
I'm getting an error that reads "Error in myeuler (line 22)
y(i+1) = y(i) + (dx * f(x(i),y(i)));"
I don't exactly know what I'm doing wrong or even where to look in this line to see what's wrong.
% must call from edrive.m or other
% required input arguments: RHS function of two variables f,
% vector x of length n+1,
% init val c
% computed output argument: y\in\R^{n+1} approximate solution
function y = myeuler(f,x,c)
% approximates sol to y' = f(x,y) over [a,b] with y(a)=c
% via n steps of Euler's method
for n = [10, 20, 40, 80, 160]
a = 0; b = .5; c = .25;
x = [a:n+1:b];
f = @(x,y) (x.^3);
dx = (x(end)-x(1))/n;
y = [c ; zeros(n,1)];
for i = 1: n
y(i+1) = y(i) + (dx * f(x(i),y(i)));
end
end
##### 2 CommentsShowHide 1 older comment
Mikela Petersen on 16 Aug 2020
the error is the first thing that was listed:
>> myeuler
Index exceeds the number of array elements (1).
Error in myeuler (line 22)
y(i+1) = y(i) + (dx * f(x(i),y(i)));
Walter Roberson on 16 Aug 2020
for n = [10, 20, 40, 80, 160]
a = 0; b = .5; c = .25;
x = [a:n+1:b];
Consider the first case, n = 10 . Then
a = 0; b = .5; c = .25;
x = [0:10+1:.5]
but 0:11:.5 is going to give you just 0 .
The number in the middle in a a:b:c operation is the increment, not the number of elements to generate. To generate a particular number of elements,
x = linspace(a, b, n+1);
Rafael Hernandez-Walls on 16 Aug 2020
for n = [10, 20, 40, 80, 160]
a = 0; b = .5; c = .25;
x = linspace(a,b,n);%[a:n+1:b];
f = x.^3;
dx = x(2)-x(1);%(x(end)-x(1))/n;
y = [c ; zeros(n-1,1)];
for i = 1: n-1
y(i+1) = y(i) + dx * f(i);
end
plot(x,y),hold on
end | 656 | 1,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2023-14 | longest | en | 0.777876 |
http://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-02-825275-6&chapter=4&lesson=6&title=scq | 1,369,401,334,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704658856/warc/CC-MAIN-20130516114418-00048-ip-10-60-113-184.ec2.internal.warc.gz | 495,726,986 | 3,727 | 1. What is the measure of each angle of an equilateral triangle? A. 45° B. 90° C. 30° D. 60° Hint 2. In isosceles with base and Find the measure of A. 42 B. 96 C. 52 D. 34 Hint 3. In is 10 more than 3 times a number, is 8 less than 5 times the same number. Find A. 37 B. 48 C. 106 D. 116 Hint 4. Find the value of x. A. 10 B. 20 C. 13 D. 26 Hint 5. In isosceles PQR with base , PQ = 2x + 3, and PR = 9x - 11. What is the value of x? A. B. 2 C. -7 D. 7 Hint | 194 | 456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2013-20 | longest | en | 0.822004 |
https://altoalsimce.org/what-is-the-gcf-of-15-and-25/ | 1,660,456,274,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571996.63/warc/CC-MAIN-20220814052950-20220814082950-00585.warc.gz | 124,489,765 | 5,433 | GCF that 15 and also 25 is the largest feasible number that divides 15 and also 25 specifically without any type of remainder. The components of 15 and 25 are 1, 3, 5, 15 and also 1, 5, 25 respectively. There space 3 frequently used approaches to uncover the GCF of 15 and 25 - element factorization, lengthy division, and also Euclidean algorithm.
You are watching: What is the gcf of 15 and 25
1 GCF the 15 and also 25 2 List the Methods 3 Solved Examples 4 FAQs
Answer: GCF of 15 and 25 is 5.
Explanation:
The GCF of two non-zero integers, x(15) and also y(25), is the biggest positive essence m(5) the divides both x(15) and also y(25) without any type of remainder.
The techniques to find the GCF of 15 and also 25 are explained below.
Using Euclid's AlgorithmListing common FactorsPrime factorization Method
### GCF the 15 and also 25 by Euclidean Algorithm
As per the Euclidean Algorithm, GCF(X, Y) = GCF(Y, X mod Y)where X > Y and also mod is the modulo operator.
Here X = 25 and also Y = 15
GCF(25, 15) = GCF(15, 25 mode 15) = GCF(15, 10)GCF(15, 10) = GCF(10, 15 mod 10) = GCF(10, 5)GCF(10, 5) = GCF(5, 10 mode 5) = GCF(5, 0)GCF(5, 0) = 5 (∵ GCF(X, 0) = |X|, where X ≠ 0)
Therefore, the worth of GCF of 15 and 25 is 5.
### GCF of 15 and 25 by Listing typical Factors
Factors the 15: 1, 3, 5, 15Factors that 25: 1, 5, 25
There space 2 common factors of 15 and 25, that are 1 and also 5. Therefore, the greatest common factor of 15 and also 25 is 5.
### GCF of 15 and also 25 by prime Factorization
Prime factorization of 15 and also 25 is (3 × 5) and (5 × 5) respectively. Together visible, 15 and 25 have only one common prime element i.e. 5. Hence, the GCF of 15 and 25 is 5.
## GCF the 15 and also 25 Examples
Example 1: The product of 2 numbers is 375. If their GCF is 5, what is their LCM?
Solution:
Given: GCF = 5 and also product of number = 375∵ LCM × GCF = product the numbers⇒ LCM = Product/GCF = 375/5Therefore, the LCM is 75.
Example 2: For two numbers, GCF = 5 and LCM = 75. If one number is 15, find the various other number.
Solution:
Given: GCF (y, 15) = 5 and LCM (y, 15) = 75∵ GCF × LCM = 15 × (y)⇒ y = (GCF × LCM)/15⇒ y = (5 × 75)/15⇒ y = 25Therefore, the other number is 25.
Example 3: find the GCF of 15 and also 25, if your LCM is 75.
Solution:
∵ LCM × GCF = 15 × 25⇒ GCF(15, 25) = (15 × 25)/75 = 5Therefore, the greatest common factor that 15 and also 25 is 5.
Show solution >
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## FAQs ~ above GCF of 15 and 25
### What is the GCF of 15 and also 25?
The GCF the 15 and also 25 is 5. To calculate the greatest usual factor the 15 and 25, we need to element each number (factors of 15 = 1, 3, 5, 15; determinants of 25 = 1, 5, 25) and choose the greatest element that exactly divides both 15 and also 25, i.e., 5.
### What is the Relation between LCM and also GCF of 15, 25?
The complying with equation deserve to be used to express the relation in between Least common Multiple and also GCF that 15 and 25, i.e. GCF × LCM = 15 × 25.
### What room the techniques to uncover GCF that 15 and also 25?
There room three commonly used techniques to uncover the GCF the 15 and 25.
By Euclidean AlgorithmBy element FactorizationBy long Division
### How to discover the GCF the 15 and 25 by Long department Method?
To uncover the GCF the 15, 25 using long department method, 25 is separated by 15. The equivalent divisor (5) when remainder equals 0 is taken together GCF.
See more: 2004 Cadillac Deville Door Panel Removal, How To Take Apart A Cadillac Door Panel
### How to find the GCF the 15 and 25 by element Factorization?
To discover the GCF the 15 and 25, we will discover the prime factorization that the given numbers, i.e. 15 = 3 × 5; 25 = 5 × 5.⇒ because 5 is the only common prime factor of 15 and also 25. Hence, GCF (15, 25) = 5.☛ element Numbers
### If the GCF of 25 and 15 is 5, uncover its LCM.
GCF(25, 15) × LCM(25, 15) = 25 × 15Since the GCF the 25 and 15 = 5⇒ 5 × LCM(25, 15) = 375Therefore, LCM = 75☛ Greatest usual Factor Calculator | 1,414 | 4,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2022-33 | latest | en | 0.926914 |
https://www.memrise.com/course/700001/learn-mathematics/209/ | 1,590,844,505,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347409171.27/warc/CC-MAIN-20200530102741-20200530132741-00124.warc.gz | 820,711,515 | 6,348 | Level 208 Level 210
Level 209
## Ignore words
Check the boxes below to ignore/unignore words, then click save at the bottom. Ignored words will never appear in any learning session.
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a+b=b+a
ab=ba
Commutative Property of Multiplication
(a+b)+c=a+(b+c)
(AB)C = A(BC)
Associative Property of Multiplication
a(b+c)=ab+ac
Distributive Property
For all real numbers x, y, and z, if x = y, then x + z = y + z.
Subtraction Property of Equality
For all real numbers x, y, and z, if x = y, then x - z = y - z.
Multiplication Property of Equality
For all real numbers x, y, and z, if x = y, then xz = yz.
Division Property of Equality
For all real numbers x, y, and z, if x = y, and z ≠ 0, then x/z = y/z. You can divide each side of an equation by the same non-zero number and not change its truth value.
Transitive Property of Equality
For all real numbers x, y, and z , if x = y and y = z, then x = z.
a=a
Reflexive Property of Equality
Symmetric Property
when two segments or two angles are congruent, you can flip them over and they will still be congruent
Substitution property
a(b) = (ab)
Equation to Inequality Property
If a > 0 and b > 0 and a+b=c then c>a and c>b
Trichotomy Property
For every x and y, one and only one of the following conditions holds: x < y, x=y, x > y
Transitive Property of Inequalities
If x > y and y < z, the x < z | 401 | 1,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2020-24 | latest | en | 0.691601 |
https://www.physicsforums.com/threads/free-fall.52697/ | 1,544,619,895,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823872.13/warc/CC-MAIN-20181212112626-20181212134126-00186.warc.gz | 1,019,455,525 | 12,816 | # Homework Help: Free fall
1. Nov 14, 2004
### clh7871
A stone is dropped off a cliff of height h. at the same instant a ball is thrown straight up from the base of the cliff with initial velocity (vi). assuming the ball is thrown hard enough, at what time t will stone and ball meet? (neglect air resistance)
I just don't even know how to approach this problem??? could someone help me??
2. Nov 14, 2004
### jamesrc
Write the position of each object as a function of time, then set the two expressions equal to each other to solve for time. You should find that they are equal at time t = 0 (the trivial solution to the problem) and some other time that will be your answer. You can find the position of each using the same formula for motion of an object with constant acceleration provided you understand what each term in the formula means:
$$y(t) = y_o + v_{oy}t + \frac 1 2 at^2$$
3. Nov 14, 2004
### Diane_
Small amendment - the stones don't start from the same place, so the trivial solution t = 0 won't be there. The technique suggested will, of course, still work.
4. Nov 15, 2004
### jamesrc
That's true. Thank you; I must have read the problem too fast.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 319 | 1,262 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-51 | latest | en | 0.929213 |
https://www.gradesaver.com/textbooks/math/trigonometry/CLONE-68cac39a-c5ec-4c26-8565-a44738e90952/chapter-1-trigonometric-functions-section-1-1-angles-1-1-exercises-page-9/90 | 1,721,063,953,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00452.warc.gz | 688,229,028 | 12,792 | ## Trigonometry (11th Edition) Clone
1. Since this number is larger than 360°, to find its smallest positive coterminal positive angle, we will subtract 360 to get positive angle less than 360 $699^{\circ}-360^{\circ}= 339^{\circ}$ | 65 | 232 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-30 | latest | en | 0.781578 |
https://www.wishfin.com/business-loan-calculator/rbl-bank-business-loan-emi-calculator-1851/ | 1,719,151,010,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862474.84/warc/CC-MAIN-20240623131446-20240623161446-00447.warc.gz | 921,112,218 | 13,234 | # RBL Bank Business Loan EMI Calculator
Last Updated : May 28, 2021, 9:13 p.m.
RBL Bank Business Loan EMI Calculator is the best tool for finding out the EMI that you will pay on a certain amount of business loans. The needed information for the calculation of EMI is the rate of interest, the principal amount and the tenure for which the loan has been taken. Pre-calculation of EMIs is helpful in deciding the loan amount to be taken and in deciding the repayment period.
Results
Loan Amount Interest Rate Tenure(Months) Monthly Instalment (EMI) Total Interest Amount Total Amount (Principal + Interest)
Year Principal Interest Balance Amount
## Factors that Affects the Amount of EMI
Principal Amount:- Principal amount is basically the loan amount that is borrowed by the applicant. So it is obvious that the higher the loan amount, the higher will be the EMI and the lower the loan amount, the lower will be the EMI. Therefore the loan or the principal amount is responsible for the fluctuation in the EMI.
Rate of Interest:- Business Loan Interest Rates also helps to calculate the EMI of the loan. No financial institution grants a business loan without charging a rate of interest. So higher interest rates result in the computation of high EMIs.
Time Period:- The time period for which the loan has been taken is also an important factor. You can reduce your EMIs by taking a loan for a longer period. But it will increase the overall interest amount charged on the loan.
### Manual Formula for RBL Bank Business Loan EMI Calculation
EMI = P x R x (1+R)^n/((1+R)^n-1)
P represents the Principal Amount.
R is the Rate of Interest.
Small “n” is the tenure given by the bank
## Drawbacks of Using the Manual Method for Calculating EMI
Time Taking:- The manual calculation of EMI is really difficult. Therefore the only way left is to use the Business Loan EMI Calculator . It will save a lot of time. On the other hand, the manual method is time consuming and needs a lot of attention.
Chances of Errors:- In the manual method of calculating EMI there are chances of errors. So it is better to use the online RBL Bank Business Loan EMI Calculator for computing the EMI. A small negligence can result in the calculation of wrong EMIs.
Needs Mathematical Expertise:- As you can see the formula used for calculating EMI is complex and it needs a lot of mathematical expertise. It will be difficult for you to derive the exact EMI using the formula or the manual method.
Lengthy Formula:- The formula used in the EMI calculation is too lengthy. It will be difficult for you to memorise the formula. Even a wrong placement of a single character in the formula will calculate the wrong EMI.
RBL stands for Ratnakar Bank Limited that is a private banking institution. It grants loans to the business having less capital at a competitive rate of interest. One can easily get an RBL Bank Business Loan up to 10 Lacs without collateral. The processing fee is between the range of 2% to 3% on the loan amount. So if you are willing to take a business loan then you can consider the RBL Bank as an option.
Now, you know why RBL Bank Business Loan EMI Calculator is better than manual calculation. This online calculator displays all the information in a graphical form as well. You can easily differentiate between the principal amount and the interest amount in your calculated EMI. The main factors that fluctuate the amount of EMI are the loan amount, rate of interest and the time period. | 764 | 3,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-26 | latest | en | 0.936188 |
https://www.homeownershub.com/maintenance/glass-fuse-565636-.htm | 1,493,566,632,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125654.80/warc/CC-MAIN-20170423031205-00185-ip-10-145-167-34.ec2.internal.warc.gz | 909,892,459 | 13,829 | glass fuse
The house is circa 1960. The electrical panel is glass fuses. The whole house runs off the box of 8 glass fuses. They are 30 amp. I wonder if the electrical fuses should be 20's and someone has replaced them with 30's. How do I determine what is the correct size? I have just recently become familiar with the house...it belonged to my father. The tenant blew a fuse, running a leaf blower and fryer all at the same time. I told her to try a 20 amp fuse first, but she said it sparked and wouldn't physically screw in. (did she have it crossthreaded, I don't know) But the 30 was fine. After a couple of hours after replacing the fuse, I asked her to feel the panel box and notice if anything was warm or hot...she said no. Are the 20's and 30's the same size? If the box was built for 30's and I replace them with 20's, what's the damage? And vice versa. How tough is it to replace a glass fuse breaker box with a modern breaker box? Perry
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I bet most of them should really be 15s. 20 would usually be the max for 120v circuits. 15a is 14ga wire, 20a is 12ga wire and 30s need 10ga wire (copper)
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Perry Templeton wrote:
That's a very real possibility, as the wire going through the house is almost guaranteed to be 20 Amp capacity at best, likely 15 Amp in some places.
More on that later...
Way too many unknowns in that description, sorry.
She kept the leaf blower and fryer running all of that time? Something tells me probably not.
Standard glass fuses, yes. There is a type where there are adapters that *do* vary in size by amperage rating. If that's what you have then they should/could be different.
It's not "the box", it's the wiring running through the walls, but to answer -- none at all. You would be being "extra safe"
Exactly the opposite. Warning, danger, alert the fire department, they ma want to keep a truck stationed at the nearest hydrant. No kidding.
(the "more later" that I promised) Not at all for a competent professional. Please don't take it personally, but by the very scope of your questions, this is obviously not an area of expertise for you. There may be more involved than just replacing the panel (probably everything out through and including the meter base and the wiring up to the utility co's connection point.) \$ Ballpark? 800-1500 bucks, that's for the 'going rates" in this part of the country, your mileage may vary. At the *very* least, pay an electrician for a couple of hours work to open up a few outlets/fixtures and the box, check some wire sizes and determine what fuses are *required* to make it all safe. Then lock your tenants out of the panel, if that's legal in your municipality.
Good luck.
--
The real Tom Pendergast [ So if you meet me, have some courtesy,
aka I-zheet M'drurz [ have some sympathy, and some taste.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Thank you for all of your replies. Nothing was taken personally, I 100% admit that I am an electrical numbskull. Just some things that sounded and looked out of whack with the electrical. I have been around enough old houses and older people that I have seen the fuses replaced with too big fuses and knew that probably didn't make good sense. It will certainly become a priority on the "to-do" list. Thanks again. Perry | 855 | 3,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-17 | longest | en | 0.968029 |
http://www.ck12.org/book/Basic-Speller-Student-Materials/r1/section/15.12/ | 1,490,277,154,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186895.51/warc/CC-MAIN-20170322212946-00414-ip-10-233-31-227.ec2.internal.warc.gz | 455,621,065 | 31,198 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 15.12: More Practice with the Third Vowel Rule
Difficulty Level: At Grade Created by: CK-12
## More Practice with the Third Vowel Rule
1. The Third Vowel Rule. The third vowel sound from the end of a word will often be _______ if it is ___________, even if it is the first vowel in a _______ string.
2. In sixteen of the words below the vowel in bold type is covered by the Third Vowel Rule. In the other eight words the vowel in bold type is not covered by the Third Vowel Rule — sometimes because it is not stressed, sometimes because it is not the third vowel sound from the end of the word. In each word put an accent mark over the vowel that has stress on it, and put a ‘3’ under the vowel letter that spells the third vowel sound from the end of the word. If a word does not have three vowels sounds, do not put a number under it. We have given you a start with xerography and committees:
\begin{align*}&\text{xer}\acute{\mathbf{o}}\text{graphy} && \text{r}\mathbf{e}\text{medy} && \text{acc}\mathbf{o}\text{mplish} &&\text{c}\mathbf{a}\text{lculate}\\ & \quad \ 3\\ &\text{c}\mathbf{o}\text{mmíttees} && \mathbf{e}\text{nergy} && \text{pr}\mathbf{e}\text{sident} && \text{sc}\mathbf{i}\text{ssors}\\ &\text{3}\\ &\text{s}\mathbf{o}\text{lvable} && \text{h}\mathbf{e}\text{sitate} && \text{t}\mathbf{e}\text{lephone} &&\text{v}\mathbf{e}\text{nerate}\\ &\text{person}\mathbf{a}\text{lity} && \text{s}\mathbf{y}\text{mphony} && \mathbf{e}\text{xcessive} && \text{s}\mathbf{a}\text{tisfy}\\ &\text{alt}\mathbf{e}\text{rnate} && \mathbf{o}\text{bjective} && \text{d}\mathbf{e}\text{finite} &&\text{t}\mathbf{o}\text{lerate}\\ &\text{amb}\mathbf{a}\text{ssador} && \mathbf{e}\text{lephant} && \text{aff}\mathbf{e}\text{ction} && \text{m}\mathbf{i}\text{grant}\end{align*}
3. Sort the words into the two groups described below. Remember that for one of these vowels to be covered by the Third Vowel Rule, it must have an accent mark over it and a ‘3’ under it. In the Reason column show why the vowels in bold type in the eight words are not covered by the rule: Put “No stress” if they are not stressed or “Not #3” if they are not spelling the third vowel sound from the end of the word:
4. In the sixteen words in which the vowel in bold type is covered by the Third Vowel Rule, eleven of the bold vowels are the first vowel in a VCV string; five are in a VCC string. Sort the sixteen words into these two groups:
5. Are the vowels in the VCV strings in the eleven words long or short? ___________
Why? _________________________________________________
______________________________________________________
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Subjects: | 851 | 2,887 | {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-13 | longest | en | 0.568824 |
https://www.brighthubeducation.com/preschool-lesson-plans/39933-kite-lesson-plan-to-teach-patterns-and-shapes/ | 1,601,541,829,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402124756.81/warc/CC-MAIN-20201001062039-20201001092039-00272.warc.gz | 723,386,221 | 14,684 | # Preschool Kite Lesson Plan: Teach Patterns and Shapes with a Preschool Kite Craft
Page content
Materials
• Book, Kite Flying by Grace Lin or other book about kites
• Pictures of different kinds of kites
• Unifix Cubes
• Large poster of a diamond-shaped kite with an long string for the tail drawn on it.
• Die cut medium triangle shapes in different colors to put on the kite tail.
• Large sheet (12” x 18”) of light blue construction paper, one for each child
• White diamond for the kite, one for each child
• Die cut small triangle shapes in different colors, twelve per child
• Glue
• Crayons
Prior Knowledge
Before beginning this preschool kite lesson plan, read the book Kite Flying by Grace Lin. Talk about kites and show the class some pictures of different kinds of kites, including a diamond-shaped kite. This can be done a day or two or earlier or during story time the day of the lesson. Students should also be familiar with patterns and shapes, specifically diamond and triangle.
Teach
Review the shapes diamond and triangle. Talk about the number of sides and corners. Show the students how you can put two triangles together to make a diamond. Review patterns if you have taught them before.
Procedure
Begin by asking students if they know what a pattern is. Using Unifix cubes make a simple pattern such as red, blue, red, blue. Explain that a pattern is something that repeats. Show them a few more simple patterns. You can try a harder one such as red, blue, blue, red, blue, blue if you think they are ready. Tell them that they are going to play a quick pattern game. Give each table a basket of Unifix cubes in different colors. Use the Unifix cubes to start a pattern, hold it up for the class to see and have the students find the color cube that goes next and hold it up. Do this a few times to review or practice patterns.
Now show the class your kite poster. Point out the diamond shape of the kite and the empty tail. Tell them that you have some triangle that you want to use to make the bows for tail. Give each student two triangles and give them a few minutes to see if they can make a bow out of the two triangle. You might remind them how to make a diamond shape with the two triangles first. After they have had some time to try, show the whole class how to make a bow and let them make one with their triangles.
Tell the class that you want to make a pattern out of the bows for the tail for your kite. Show the the colors you have and let a few students give ideas for the pattern. Choose one and show them how to glue the triangles into bows on the poster and make a pattern.
Now it is the students turn to make a kite picture. Give each student a light blue sheet of construction paper, a white diamond, crayons and glue. Have them decorate their kite and glue it on the paper near the top. Show them how to draw a line for the kite tail. Give each table an assortment of different colored triangles and tell them to use them to make the bows on their kite tail. Remind them that the bows should be arranged into a pattern.
Assess
A quick and easy method for assessing this kite lesson plan is to simply look at the finished kites. Could they arrange the triangles into the bow shape? Did they make a pattern?
Extend
These kites would make a cute spring bulletin board. To continue working on patterns have the students play the pattern game with the cubes in pairs of two. Play the same game with pattern blocks or other math manipulatives.
You preschool students will have lots of fun learning about patterns and shapes with this kite lesson plan. Try this kite lesson plan too! | 791 | 3,638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-40 | latest | en | 0.915928 |
https://uk.mathworks.com/matlabcentral/cody/problems/8-add-two-numbers/solutions/501939 | 1,597,260,940,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738913.60/warc/CC-MAIN-20200812171125-20200812201125-00187.warc.gz | 522,768,670 | 15,544 | Cody
# Problem 8. Add two numbers
Solution 501939
Submitted on 16 Sep 2014 by PARTH PATEL
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_correct = 3; assert(isequal(add_two_numbers(a,b),c_correct))
2 Pass
%% a = 17; b = 2; c_correct = 19; assert(isequal(add_two_numbers(a,b),c_correct))
3 Pass
%% a = -5; b = 2; c_correct = -3; assert(isequal(add_two_numbers(a,b),c_correct)) | 169 | 524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-34 | latest | en | 0.63643 |
https://publications.parliament.uk/pa/cm200910/cmhansrd/cm100118/debtext/100118-0012.htm | 1,726,511,620,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00244.warc.gz | 439,512,947 | 11,831 | 18 Jan 2010 : Column 71
This is a dangerous matter, because juries take DNA evidence as the golden bullet. Too many of them have watched "CSI", "Bones" or "Cold Case Files" and believe that DNA evidence is perfect. I shall ambush the Home Secretary on his way home on the train on Thursday and give him a copy of this week's New Scientist. I recommend that the other Ministers read it, too. It contains a long overdue piece about the doubts that many scientists have about the veracity of the presumption that there is a one in 7 million chance of DNA evidence being wrong. There are all sorts of reasons why that should not be the case, and I shall not bore the House with the technicalities today-my first degree is too long ago for that. It is likely that the probability of a mismatch is much greater than we think.
We come up against what is known as the "birthday syndrome". The chance of me and the Home Secretary having the same birthday is one in 365-that is pretty straightforward to work out. The chance of me and one of the entire Home Office ministerial and Parliamentary Private Secretary team having the same birthday is about 25 per cent. If we add in the people in all their private offices, it is about 50 per cent. Actually, the chance of my having the same birthday as the Home Secretary is zero, because his is in May, but the point is that the probability of a mismatch rises steeply the greater the database gets.
The presumption among those who are tied to databases is that the bigger the database, the better it will be. No, the bigger the database, the geometrically bigger the chance of a mismatch. That is the mathematics. Members should not believe me, they should go and get themselves a mathematician to explain it to them. It is straightforward, and it is called the birthday syndrome. A bigger database carries a greater chance of a mismatch. If a quarter or a fifth of those on the database are innocent, there is a serious risk of a miscarriage of justice.
DNA is a powerful and effective tool, but we make an awful lot of presumptions if we jump to the conclusion that it is perfect. The Government are in a position to release the information from the database to the scientific community so that it can make the judgments that I have mentioned. The American Government have been approached to do that and turned down the request. I suggest that if the British Government really want to do something in their last few months in office, they should accede to it. They are right in that DNA is a powerful tool that can protect British citizens, but it can also create serious miscarriages of justice. It is in their interest to ensure that that does not happen.
6.25 pm
Mr. Neil Gerrard (Walthamstow) (Lab): I apologise for having been absent for part of the debate, but I had to attend a statutory instrument Committee upstairs.
I listened with a great deal of interest to what the right hon. Member for Haltemprice and Howden (David Davis) said about DNA, about which I shall make a few comments later. I wish to talk first about another part of the Bill, which in some ways relates to the problems that he mentioned of the effect on communities. That is the part that deals with stop and search.
18 Jan 2010 : Column 72
The Bill proposes changes to how stop and search is recorded. I absolutely understand the reasoning-we do not want excessive police bureaucracy, and we want the police to spend as much time as they possibly can on the street dealing with crime, rather than filling in cumbersome forms or records. However, the use of stop and search can be controversial and it is important that we get it absolutely right.
Stop and search can take several different forms, and there are three main powers to exercise it. There is section 1 of the Police and Criminal Evidence Act 1984, which is the commonest form, but there is also section 44 of the Terrorism Act 2000 and section 60 of the Criminal Justice and Public Order Act 1994. These last two do not require a police officer to have reasonable suspicion about an individual before they conduct a stop and search. Section 60 of the 1994 Act, in particular, is often ignored when there is discussion of stop and search. There was a recently a prominent court case about section 44 of the 2000 Act, and there has been a great deal of publicity about how it has been used in relation to recent demonstrations in central London. However, when we consider the regulations governing stop and search during the passage of this Bill, we need to examine what happens under section 60.
The proposal in the Bill is to reduce the recording requirements under PACE. There will not be any need to record whether anything was found during the stop and search, whether any injury or damage were caused to the person stopped, or their name. Monitoring will continue in relation to ethnicity, but not age. We need to consider what the consequences of that might be. If someone wished to make a complaint that a stop and search was unlawful, or wished to show in their defence in a case that they had been stopped and searched two or three times in the same day-that sometimes happens-they would find it very difficult without names and other details being recorded.
We need to think much more carefully about how we monitor the use of stop and search. There is obviously an issue with section 44 of the 2000 Act following the court case. It must be reviewed and there is likely to be an appeal-there is of course no guarantee that that will succeed-but it is absolutely certain that all the reservations about the use of section 44 given in the judgment apply equally well to section 60 of the Criminal Justice and Public Order Act 1994.
I was in the House when that Bill was debated back in 1994, as were one or two others in the Chamber today. In the debate on section 60, we were told that the power would be used only in exceptional circumstances, when a superintendent considered that there was the possibility of serious violence in their area. However, the power is being used absolutely routinely in certain police forces, as a method of doing stop and search without having to have suspicion regarding the individuals concerned. In the Met, in 2000-01, there were 2,800 recorded uses of section 60; in 2002-03, the figure was up to 8,600; and in 2003-04, it was 4,400; but by 2007-08, it had risen to 17,000. The latest figures show that in just one year, in one London borough-Newham-there were 25,500 searches using section 60. That happens to be the borough that uses section 60 most heavily, but in my borough, Waltham Forest, there were 6,000 in one year, which is more than there were in the whole of the Met just a few years ago.
18 Jan 2010 : Column 73
The excuse for those figures is Operation Blunt 2. That was when the big increase occurred. However, the usage of section 60 is simply not being monitored properly. In fact, there is very little relationship between knife crime and the number of searches under section 60. Like the DNA provisions, the use of section 60 is a source of resentment among young men-it is generally young men-who are stopped again and again. It is about time we looked again at how section 60 is used and monitored, and we should be looking to amend how it operates. The Bill gives us an opportunity to do that.
I welcome the fact that the law on DNA retention is being looked at and reformed, but I am afraid I cannot welcome how it is being done. There are some very basic issues. The fact that three quarters of young black men between the ages of 16 and 34 are on that database has been mentioned by the right hon. Member for Haltemprice and Howden and others. There is evidence that there is a higher rate of arrest among young black men, but often a lower rate of conviction than among the corresponding white cohort.
David T.C. Davies: I have also heard the criticism that there are proportionately more young black men in prison and other institutions than white males, but there are all sorts of possible reasons for that, one being that the former group may be committing more crime. I accept that there are many other possible reasons.
Mr. Gerrard: I would not pretend that statistics relating to ethnicity and crime are not complex. We need to look at the causes of crime and the age groups involved and all sorts of issues. However, it should concern us that there are considerable disproportionalities between one population and another. That certainly happens with the stop and search, as I mentioned. A young black male is much more likely to be stopped and searched, and in London in particular, in recent years, the stop and search of young Asian men has increased. In turn, that is reflected in the DNA database.
The retention of the DNA data of someone who has not been charged with, let alone convicted of, an offence is a basic issue. The right hon. Member for Haltemprice and Howden gave some specific examples, so let me give a couple concerning people I have dealt with in the past few months. One 15-year-old girl was accused, exactly as the right hon. Gentleman suggested, by another girl in the same school. There was an argument at school and there was perhaps a little bit of pushing one way or the other, and that girl ended up being arrested and is now on the DNA database. When her mother and I asked for her data to be removed, the answer was no. I looked at the procedure that is currently used in the Metropolitan police. Its guidance on removing people from the database states that
"exceptional cases will be extremely rare"
and gives examples of when that might happen, which include
"where the original arrest was found to be unlawful"
or when "no offence" whatever "existed". The reality is that virtually no one is removed. That 15-year-old will remain on the database.
I have had two examples in the past few months of Asian men-one in his 30s, one in his 50s-who had been on holiday to visit relatives in the US. On their way back through Gatwick and Heathrow, they were stopped
18 Jan 2010 : Column 74
under anti-terrorism powers and their DNA samples were taken. In neither case has there been any charge, nor is there going to be. That is quite clear. However, from what the Home Secretary said in his opening speech, those two men, who have not been charged or convicted, and who have no previous record whatever, may remain on the database for years, because they were stopped and questioned under anti-terrorism powers.
I cannot see what the hard evidence is to support the proposals in the Bill. I heard the statistics the Home Secretary mentioned, but the basis for his argument from those statistics is extreme flimsy-it was based on thin research that has not decided anything definitively. That the data are thin has been admitted, and yet they are the basis on which we are going to bring in primary legislation that specifies the number of years for which data can be held. At the very least, we should be looking at a measure that allows the number of years to be changed quickly and easily if we get the definitive research that I believe we need.
It is almost a no-smoke-without-fire argument: the fact that somebody has been arrested at one point leads to the assumption that they are more likely to commit a crime than somebody who has not been arrested. The evidence to back that up that I have seen-the hard evidence, based on solid data going back over a significant period-seems remarkably thin. I do not think we should be going down that road on the basis of data of that standard.
Finally, I welcome another part of the Bill that the Home Secretary mentioned-the measures to regulate wheel-clamping. That has been a real nightmare to deal with in many areas, and certainly in my constituency. I have seen a number of cases of the absolute abuse of power by private companies carrying out wheel-clamping on private land. I have seen cases of people who work in shops-not shop owners-being conned into signing a contract with a company, which then comes along and clamps people in the back alleyway behind the shop. I have seen cases of people being clamped on a garage forecourt after they have simply gone into the shop to buy something. They have come out to find that they are being charged £400 or some other ridiculous amount of money to release their car. I welcome measures to deal with that. The Bill contains the power to regulate, and what we need as soon as possible is the statutory code of practice that will put the detail into effect. That is what is needed, not just the requirement for the companies to register.
Alongside that, I hope that we can also use this Bill to consider some of the other activities of private security companies. I am really concerned about how many areas they operate in. I dealt recently with the case of a constituent who had had a problem in a department store. The private security company involved handled her somewhat roughly, she was thrown out of the store and banned from entering it again. She was also accused of being racist, but that was totally untrue, as the store eventually admitted. However, she had no comeback against the store, because the private security company was responsible. Such companies operate in far too many places, and they seem to operate without any significant regulation. They may have to be registered and obtain licences from the Security Industry Authority, but if they behave badly it is difficult to get anything done about it.
18 Jan 2010 : Column 75
Tony Baldry: Is not the real problem that when the police indicate that they are no longer willing to investigate problems such as shoplifting, the private sector intervenes, but it is unregulated and some companies can act like cowboys? People receive letters threatening prosecution or telling them that they have been put on lists, and it is very intimidating. There is no redress in the magistrates courts, and no supervision or overview.
Mr. Gerrard: That is right, although the police have not withdrawn. In the street market and shopping centre in my constituency, there is a local police team which takes an active interest in what goes on. There is however a growing trend to employ private security, especially in stores, and then if something goes wrong, there is no redress. If people have a problem with the police, there is a complaints procedure and-if necessary and in major cases-it can end up with the Independent Police Complaints Commission. As with wheel-clamping, for which we are considering codes of practice and increased regulation, we should consider other areas in which private security companies operate. This Bill may be an opportunity to do that.
The Bill deals with some big issues-the reforms to stop and search, the DNA database and private security. I am disappointed by some aspects, especially those to stop and search, on which we need more monitoring, not less, and the DNA database. As the Bill makes progress, I hope that we will be able to improve those aspects.
6.44 pm
Mr. Humfrey Malins (Woking) (Con): I begin, as always, by declaring an interest as a lawyer, a Crown court recorder and a part-time district judge. I wish to address an aspect of the Bill that has not yet been raised in any great detail-the clauses on domestic violence and domestic violence protection notices and orders. I wish to speak from a practical point of view and tell the House and the Minister-I have no expectation that the outside world, or even any other hon. Members, will have the slightest interest in what I say-about some problems that will need to be considered carefully in Committee.
We all agree that domestic violence is a very serious issue. The official figure for domestic violence is 14 per cent. of violent offences, but anecdotally it seems higher in the cases that have come before me. Such incidents usually involve a man being violent to a woman, although not always. They are, sadly, under-reported, but the Government have taken the issue seriously over the years. As well as introducing several important initiatives for the police, they have also set up specialist domestic violence courts. The Government recognise the importance of the issue and have tried to act on it as best they can.
I once heard it said that domestic violence is even worse than stranger-to-stranger violence, because it involves a breach of trust and therefore has a longer lasting impact. Be that as it may, we all agree that domestic violence is terrible, and everything that we can do as legislators to stop it should be done. We should punish those of either sex who are guilty of it.
However, I am not sure that the provisions in the Bill are necessary. The Government are nearly always well intentioned, but they have had a tendency to legislate a bit too much and forget about how the legislation will
18 Jan 2010 : Column 76
work in practice-what it will mean in extra bureaucracy for those who have to enforce it and what the results will be on the ground. I refer to clause 21 onwards and the provisions on domestic violence protection notices-the ability of the police to issue a notice to someone who has been violent or threatened violence to another in a domestic situation. Such a notice could have some nasty results for the person who received it. However, if we are dealing with violence-the Bill specifically mentions someone who "has been violent towards" another, as well as threatening violence-I contend that we have enough provision in the criminal law to deal with the perpetrator, without adding another layer of statute.
Let us say that I am cohabiting with a woman and I am violent towards her, and she calls the police. The Bill would provide that I could be issued with a domestic violence protection notice, but the police can already arrest me, take me to the police station and charge me with a criminal offence. They can charge me with common assault, which is only triable summarily in a magistrates court and carries six months in prison. They could charge me with actual bodily harm, which is triable on indictment, or-if the violence is very nasty-with grievous bodily harm or malicious wounding with intent. If I threaten to kill my partner, they can charge me with threatening to kill. They cannot charge me under the Public Order Acts because those offences apply only outside dwellings, but they have several options for dealing with me. If the police are not happy with those options, they can move on to the Protection from Harassment Act 1997, which I think was introduced by this Government. It would apply if I had embarked on a course of threatening or abusive conduct-the Act requires such conduct to be carried out only twice before an offence is committed. That Act also gives the courts the fullest powers to impose non-molestation or restraining orders on me.
Mr. Llwyd: I am following the hon. Gentleman's argument carefully and I fully agree with him. I would make the further point that, with the various assaults to which he has referred, there could be a remedy whereby the person who had committed the assault could be freed on bail, albeit on conditions that state that he should not return to the property or go within so many yards of the person concerned.
Mr. Malins: The hon. Gentleman is absolutely right. He has extensive legal experience and makes the point, which I was about to make, that the ability of the police to act in relation to charges of common assault, ABH and so on is not restricted to just charging a person. Rather, they can deal with the issue in exactly the same way that is proposed in the Bill, namely through the granting of bail or not. So if I go to the police station and I am charged, the police can make it a condition of bail that I do not attend the premises or go within 100 metres of it, and so on. Therefore, my direct question is: what exactly is not covered in the law as it stands today that the Bill seeks to cover? | 4,182 | 20,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-38 | latest | en | 0.969388 |
http://rna.colostate.edu/dokuwiki/doku.php?id=exam2&image=mirnas.fa.gz&ns=&tab_details=view&do=media | 1,566,045,832,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313259.30/warc/CC-MAIN-20190817123129-20190817145129-00270.warc.gz | 161,759,111 | 5,113 | # Computational biology at CSU
exam2
## Exam (100 pts)
INSTRUCTIONS
You cannot use any modules that have to be imported into your code.
Upload a single file named exam_yourname.py containing your functions. Any content outside of the function must be commented (#).
1. [20 pts] Write a function, `gc_content(sequence)`, that calculates the proportion of nucleotides in a sequence that are either G or C. For example, the sequence ATGC would return 0.5.
2. [20 pts] Write a function, `word_count(input_file)`, that returns the number of lines and characters (all characters including white space, except new lines) in a file. The return value should be a tuple. The function should work with any size file (i.e. don't read the entire file into memory). The function should exit gracefully if the file can't be opened and should not leave the file open.
For example:
```>seq1
ATG
>seq2
CTA```
would return (4, 16) (i.e. 4 lines, 16 characters)
3. [20] Write a function, `matrix_mean(matrix)`, that returns the mean value for each row in a matrix as a list. For example, the matrix [[1,2,3], [5,6,7]] would return [2, 6].
4. [20] Write a function, `element_counter(some_list)` that returns a dictionary of each unique element and the corresponding number of occurrences of the element in the list. For example, the list ['ATG', 'TAG', 'TTT', 'TAG', 'TTT'] would return {'ATG': 1, 'TAG': 2, 'TTT': 2}
5. [20] Write a function, `motif_coordinates(sequence, motif)`, that returns the start and end position of the first occurrence of a motif within a sequence as a tuple. Report the position relative to a starting position of 1 not the first python index of 0. For example, given the sequence ATGCTGTTAGCAG and the motif CAG, the function would return (11, 13).
exam2.txt · Last modified: 2018/11/05 09:24 by dokuroot | 487 | 1,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-35 | latest | en | 0.818582 |
http://www.thescienceforum.com/mathematics/15590-statistics-plz-help-kings-paradox.html | 1,674,771,065,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494826.88/warc/CC-MAIN-20230126210844-20230127000844-00292.warc.gz | 83,676,750 | 21,254 | # Thread: statistics plz help with the kings paradox
1. im a king
yes the king of my house
as so is my father and his father and his father and so on for 8000 generations, the time we left caves
i suppose you have not trouble in believing theres an interrupted lines of kings of the house in my family
after all is just natural to have a father and a grandfather and a grandgrand father and so on till very far
in fact the fact i descend from an uniterrupted line of males is 100%
but lets play with statistics lets look it the other way around:
what are the chances that my always paternal ancestor from 8000 generations away had an uninterrupted linage in which every generation happen to have a son
lets suppose there are 50% chances to have a boy for every generation
so that every generation has a boy the statistics say it will be 0.5 elevated at 8000
if you introudece that number in a calculator it says chances are 0
so in one hand the chances i can trace myself an uninterrapted male lineage is 100%
but the chances my paternal antecesor form 8000 generations away had a line in which every generation had males is 0
plz can somebody help me with my confusion?
oh maybe im not confused and the statistics im here is zero, and that together with being the winning spermatozoid of millions makes me quite unique, and always descendant of winning spermatozoids
and the statistic we are all here is 0/6 billion
2.
3. The first problem is assuming that the chance would be 50/50. If it's just "has a male child" instead of "first child is male," the chance is complicated, but better than 50/50. The second problem is in trusting your calculator so completely. is a very small number, but not 0.
Edit: Cut some stuff out I wasn't quite sure of, being too sleepy.
4. my point is to prove statistics fail for being deterministic in analizing an undeterministic fact as reproduction, if you allow me ill think aloud:
what are the chances that being everything fine humanity survives 8000 generations, 120000 years?
lets suppose humanity survives for 8000 generations, all males from the future can trace a line dad grandad grand grandad and so on with males for 12000 years
this means with no doubt that theres at least one person now with at least one line that goes 8000 genearations into the future all males, from grand grand father to grand father to father to son
so for the survival of humanity during 8000 geneartions is indispensable that now theres at least one person with at least one line that continues to the future with every generation having at least one boy
so lets calculate the chances this requisite is acomplished in the present so humanity can last 120000 years at least:
that one man from nowadays have children theres a 50%, and that that man has at least one boy will be the 50%of 50%, a 25% of posibilities for a man to have a boy
now what are the posibilities that a line for 8000 generations all generations have at least one boy always
it would be a posibility of 0.25 elevated to 8000, that multiplied by 3000 millions of men there are now
gives a posibility of 3000 millions*(0.25 elevated to 8000)
so for the requisite that humanity exists in 12000 years is proved that it must be acomplished, that there is now at least one person who will have a 8000 generations of all males genealogy, if this is not acomplished there just wont be humanity
and the chances of this are almost 0 with which statistics contradicts facts
with which the deterministic aproach of statistics fails on analizing nature since nature is indeterministic
5. It's like shuffling a deck of cards and then asking what are the odds of getting that particular order of cards. The answer is "close to zero," and yet there it is.
6. yes the chances of getting that determined order is close to 0 but the chances of getting any order is 100%
i think i proved using statistics that the posibility of future existance of humanity with reproduction is close to 0
i didnt calculate the posibilities of getting a certain person instead of other, more similar to your analogy
but the posibilities of getting any person more similar of my analogy of getting any combination
i used statistics to calculate chances of existance of people in 8000 generations
if statistics says the posibilities are close to zero statistics are wrong
7. what are the odds that, out of children, EVERY one will be female? You're fallacy comes from assuming that an unbroken line of males is seemingly impossible, when, given an average of, say, 2 children, there's a 75% chance that there will be a son; an average of 3 children gives an 87.5% and so on and so forth.
allow it to be 2 children, we have so the odds are massively better, comparatively speaking. Allow it to be 3 children and we have which is, again, massively better. Let's say that for 4000 generations the average is much high than the other 4000 generations, where there is an 8 child per couple average. We have great odds that atleast one child is a male now: alright. That's kinda low, but not nearly as low as you're original figure, and you multiply that with what ever the average would be of the second half.
8. Originally Posted by luxtpm
yes the chances of getting that determined order is close to 0 but the chances of getting any order is 100%
Yes, just like the odds of getting SOME male line out of all the people in the world is close to 100%, even though the odds of getting any SPECIFIC line is very low.
9. according statistics the chances i have some male line uninterrupted during the next 8000 generations:
supposing me and every male has a 99%posibility of having at least one son
the chances every generation during 8000 generations all happen to have at least one male ( a must for the survival of the specy during 120000 years) is:
0.99elevated to 8000
and the chances of all the males from today to have some male line that lasts 8000 generations is
3000 millions*(0.99elevated to 8000)=
3.6*10 elevated to -21
those are the chances per 1 of survival of this humanity via reproduction using statistics and being very generous, though i know that with no disaster posibilities are almost 100% of survival
i wished i had a 99% chance of having children
i see the issue better an better thanks:
human reproduction implies there will be a male chain(as well female) ...grandfather father son,if the chain breaks we can not exist
the chances of a chain like this all males to exist is equal to the chances to have borned a boy elevated to the number of generations that extend in the future
10. No, the chances of you having an unbroken male line for the next 8000 generations is not the same as the chances of there being some unbroken male line for the previous 8000 generations. (You wouldn't be here if there wasn't some line.)
Even so, no specific line has to remain unbroken for the next 8000 generations for there to be males in each generation. Again, an unbroken line of ancestors isn't the same as an unbroken line of descendants, at least not from a probability point of view.
11. Originally Posted by luxtpm
according statistics the chances i have some male line uninterrupted during the next 8000 generations:
supposing me and every male has a 99%posibility of having at least one son
the chances every generation during 8000 generations all happen to have at least one male ( a must for the survival of the specy during 120000 years) is:
0.99elevated to 8000
and the chances of all the males from today to have some male line that lasts 8000 generations is
3000 millions*(0.99elevated to 8000)=
3.6*10 elevated to -21
You are calculating it as though each male in a generation could only produce either a single male child or zero male children, in which case the male population would indeed slowly shrink. Fortunately people can have more than one child.
More realistic numbers would be along the lines of each of the 3*10^9 males today producing an average of 1.1 male children, so we have 3.3*10^9 males in the next generation.
12. im not talking the probability we survided the last 120000 years, this is 100% probability
im talking of the probabilty we survive another 120000 years via reproduction being everything fine
if we are going to survive that will mean there will go 8000 generations
all those people from the future will be able to tell they all had at least one relative from our present exclusivily related by paternity ... the grandfather of the father of the son
so that theres humanity in the future implies that now theres at least one person whos gonna have a linage of 8000 generation in which in every generation it doesnt fail to have at least one boy,the grandfather of the father of the son
to realize of the improbability of a long line all male think of kings who have only daughters
or family names inherited only from father to son how soon they get lost
and as explained that humanity survives implies theres now at least one person whos gonna have an unbroken line of males, father son, not father daughter who marries another man
this probaility is extreamly low yet we know intutivily that the posibilities humanity srvives there being everything fine is almost 100%
13. "You are calculating it as though each male in a generation could only produce either a single male child or zero male children, in which case the male population would indeed slowly shrink. Fortunately people can have more than one child. "
well its true im ignoring that
but what im searching is the posibility that from now i develope a timeline all male, which is a must for reproduction since we evolved from one and only one timeline all males besides many others
and once found my posibilities of developing a time line all males that lasts multiply it by all males of humanity
i think we should agree that the posibilities of my family name only inherited from father to last 120,000 years is quite remote not to mention 2 million years
14. The odds of one specific family line being all male is somewhere between 0 and 100. The odds of humanity ending for lack of males is nearly 0.
Let's do a few examples. Let's say each couple has 2 children, and generations are somehow synchronized, for simplification sake.
In the first question, we start with a specific couple. There's a 25% chance that the line ends there, when they have two daughters. There's a 50% chance of there being exactly one son, and a 25% chance of there being two.
If there's one son, the next generation is an exact repeat of this scenario. If there's two though, the chances of the line continuing improves. Both sub-lines would have to end for the whole line to end.
Let P be the probability of the line continuing. Then P = 0.25 * 0 + 0.5 * P + 0.25 * (1 - (1 - P)^2). Rewriting gives -P^2/4 = 0, or P=0. In other words, under these simple assumptions, the chance of a particular individual's male line continuing forever is 0. That doesn't say anything about how long it'll last though.
On the other hand, the only way humanity would end would be if every child in one generation was female (or male). (Actually, all females could probably survive on artificial methods much better than an all male generation, but that's beside the point.) Let's say there are 4 generations alive at any one time and 6,000,000,000 people on the planet. Then each generation would be 1,500,000,000 people. The chance of the next generation being all female would be (1/2)^(1500000000). The chance of this happening in the next 10000 generations would be, 1 - (1 - (1/2)^(1500000000))^(10000). Again, this is a simplification, since the size of each generation would be dependent on the last, but you can start to see why this isn't likely.
Edit: To approximate that number, (1 + 1/x)^(nx) is about e^n. Using this, (1 - (1/2)^(1500000000))^(10000), n would be -10000/1500000000 or -1/150000, so that probability would be about 1 - e^(-1/150000), if I did my math right at least.
Edit again: Oh yeah. I should also point out that the fewer males, the more likely their lines are to continue, assuming humanity continues. If there was only one male, the entire next generation would be of his line, as would any future generations, so his line would be guaranteed to continue as long as humanity did.
15. " The odds of humanity ending for lack of males is nearly 0. "
this is not what im saying
what are the chances my last name inherited only from the father lasts 2 million years?
the same chance i have to have a line that goes from father to son unbroken, always males, not a father that has a daughter who marries another man,but father to son during 2 million years
quite a remote chance
then how comes everybody in the future has one and only one exclusivily paternal line, the oddity
hadnt we say that the chances of a line exclusivily paternal for 2 million years is almost 0?
16. "More realistic numbers would be along the lines of each of the 3*10^9 males today producing an average of 1.1 male children, so we have 3.3*10^9 males in the next generation."
this is the most challenging answer:
that 0.1 of extra male children are gonna die and their line extint if the popultation is to remain constant, most probable due to limited resources
if they die before the 8000th generation they have a 0% of acomplish exclusive paternality through 8000 generations so they just dont count
17. luxt, that doesn't even make sense
18. Uh, males have to have sex with females in order to reproduce. Also, each new generation is so much larger because they have new families changing the probability of having a male.
So unless your entire line of ancestry only reproduced via incest, what you are saying doesn't seem to have any meaning.
EDIT: What I am trying to say is, on family can easily make more than one person which then goes on to make another family, which can also produce more than one person. Each time there is a new generation, it is usually larger. This exponential growth is much faster than the exponential shrinking of the probability of having a male.
19. :-D
i find extreamly odd you have a father of your father of your father of your father... that goes back 2 million years
you must find this stupid but i find it odd
a line of father son father son never unbroken with father daughter seems very rare to me
what are the chances i create a line that extends 2 million years in the future of unbroken father son?
hell i even have remote chances of just making a line that lasts that long, then what are the chances this line is all father son?
20. i think ive found the answer to my question:
my father has an original last name that he took from my name as he signed his maths books with this name
what are the chances this last name lasts undefinly? or what is the same what are the chances my father breeds unbroken lines of males from father to son and not from father to daughter?
first if humanity population is to remain constant everybody will have a media of two children plus a 0.1 who will die childless so i just wont count them
so my father tosses a coin two times heads a girl tails a boy
for every time he gets a tails he has two extra tosses, like for every male he has this male has two chances of having a boy, two more coin tosses
try this game to find your chances of having an ubroken lines of males so your last name lasts
you start with only two tosses and for every tails you get you have two extra tosses, when you have no more tosses you lose
try it and youll see the chances we all have of having an unbroken lines of males is something like 75%
then the thing about the mithocondrial eve is bs
since for this she should be the only female of her time to have an unbroken lines till our days of females, mother to daughter while every female of her time had a chance of doing this as well of 75%
21. The mitocondrial eve is pretty well established. Sometime long, long ago (I forget how long) there existed a woman that all living humans could trace their ancestry back to. This doesn't mean she was the only woman alive at the time, just that her family line (mitocondria are passed down from the mother) came to dominate the others.
For your game, I've already worked out chances of the game lasting forever. It worked out to 0. Check my math though, I might have done it wrong. Like I said though, that doesn't say anything about how long it will last.
22. still wonder if probability can be aplied to human reproduction:
on the one hand to know the probaility of every female to have a matriarcal line till our days she could play that tossing the coin game, it seems to me a quite high probability
on the other hand as you go up on the line every time two sisters have a same mother it will reduce the number of females candidates to have a matriarchal line
so either theres a contradiction here or my tossing the coin game is not the accurate representation of my posibilities of stablishing a patribility line and that my last name lasts
any suggestion on how to find my posibilities that my last name lasts?
23. Yes, probability can be applied to these questions, but the answers are much more complicated than what you're thinking they'll be.
As for the question, will your last name last, your coin tossing game is basically the right idea, but you need to do your math more carefully. Actually, for the question, will it last forever, I already did the math (your game and my scenario are equivalent) and came up with a 0 probability. I haven't answered the related (but much harder to work) problem of will it last n generations, or how many generations would you expect it to last.
Edit: I should point out that something that goes to 0 at infinity might still be positive everywhere (consider 1/x). So even if there's a 0 probability of your line lasting forever, there might still be a chance of it lasting any finite amount of time.
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# 9.3: Einstein A Coefficient
Although either oscillator strength or Einstein $$A$$ coefficient could be used to describe either an emission line or an absorption line, oscillator strength is more appropriate when talking about absorption lines, and Einstein $$A$$ coefficient is more appropriate when talking about emission lines.
We think of an atom as an entity that can exist in any of a number of discrete energy levels. Only the lowest of these is stable; the higher levels are unstable with lifetimes of the order of nanoseconds. When an atom falls from an excited level to a lower level, it emits a quantum of electromagnetic radiation of frequency $$ν$$ given by
$\label{9.3.1}h\nu=\Delta E,$
where $$∆E = E_2 - E_1 ,\, E_2\text{ and }E_1$$ respectively being the energies of the upper (initial ) and lower (final) levels. The number of downward transitions per unit time is supposed to be merely proportional to the number of atoms, $$N_2$$, at a given time in the upper level. The number of downward transitions per unit time is $$-\dot N_2,$$ since $$\dot N_2$$ in calculus means the rate at which $$N_2$$ is increasing. Thus
$\label{9.3.2}-\dot N_2=A_{21}N_2.$
The proportionality constant $$A_{21}$$ is the Einstein coefficient for spontaneous emission for the transition from $$E_2\text{ to }E_1$$. It is equivalent to what, in the study of radioactivity, would be called the decay constant, usually given the symbol $$λ$$. It has dimensions $$\text{T}^{-1}\text{ and SI units s}^{-1}$$. Typically for electric dipole transitions, it is of order $$10^8 \text{s}^{-1}$$. As in radioactivity, integration of the above Equation shows that if, at time zero, the number of atoms in the upper level is $$N_2(0)$$, the number remaining after time $$t$$ will be
$\label{9.3.3}N_2(t)=N_2 (0)\mathsf{e}^{-A_{21}t}.$
Likewise, as will be familiar from the study of radioactivity (or of first-order chemical reactions, if you are a chemist), the mean lifetime in the upper level is $$1/A_{21}$$ and the halflife in the upper level is $$(\ln 2)/A_{21}$$. This does presume, however, that there is only one lower level below $$E_2$$. We return to this point in a moment, when we consider the situation when there is a choice of more than one lower level to which to decay from $$E_2$$.
Since there are $$A_{21}N_2$$ downward transitions per units time from $$E_2\text{ to }E_1$$, and each transition is followed by emission of an energy quantum $$hν$$, the rate of emission of energy from these $$N_2$$ atoms, i.e. the radiant power or radiant flux (see chapter 1) is
$\label{9.3.4}\Phi = N_2 A_{21}h\nu \qquad \text{watts}.$
(For absolute clarity, we could append the subscript 21 to the frequency $$ν$$ in order to make clear that the frequency is the frequency appropriate to the transition between the two energy levels; but a surfeit of subscripts might be too distracting to the point of actually making it less clear.) Provided the radiation is emitted isotropically, the intensity is
$\label{9.3.5}I=\frac{N_2 A_{21}h\nu}{4\pi}\quad \text{W sr}^{-1}.$
The emission coefficient (intensity per unit volume) is
$\label{9.3.6}j=\frac{n_2A_{21}h\nu}{4\pi}\qquad \text{W m}^{-3}\text{sr}^{-1}.$
If we are looking at a layer, or slice, or slab, of gas, the radiance is
$\label{9.3.7}L=\frac{\mathcal{N}_2A_{21}h\nu}{4\pi}.\qquad \text{W m}^{-2}\text{sr}^{-1}.$
Here, I have been obliged to use $$I\text{ and }L$$ correctly for intensity and radiance, rather than follow the unorthodox astronomical custom of using $$I$$ for radiance and calling it "intensity". I hope that, by giving the SI units, I have made it clear, though the reader may want to refer again to the definitions of the various quantities described in chapter 1. I am using the symbols described in section 9.1 of the present chapter for $$N,\, n\text{ and }\mathcal{N}$$. I should also point out that Equations 9.3.4-7 require the gas to be optically thin.
Equation \ref{9.3.2} and \ref{9.3.3} assume that the atom, starting from level 2, can decay to only one lower level. This may sometimes be the case, or, even if it is not, transitions to one particular lower level are far more likely than decay to any or all of the others. But in general, there will be a choice (with different branching ratios) of several lower levels. The correct form for the decay constant under those circumstances is $$λ = ∑A_{21}$$, the sum to be taken over all the levels below $$E_2$$ to which the atom can decay, and the mean lifetime in level 2 is $$1/∑A_{21}$$. Nowadays it is possible to excite a particular energy level selectively and follow electronically on a nanosecond timescale the rate at which the light intensity falls off with time. This tells us the lifetime (and hence the sum of the relevant Einstein coefficients) in a given level, with great precision without having to measure absolute intensities or the number of emitting atoms. This is a great advantage, because the measurement of absolute intensities and determination of the number of emitting atoms are both matters of great experimental difficulty, and are among the greatest sources of error in laboratory determinations of oscillator strengths. The method does not by itself, however, give the Einstein coefficients of individual lines, but only the sum of the Einstein coefficients of several possible downward transitions. Measurements of (or theoretical calculations of) relative oscillator strengths or branching ratios (which do not require absolute intensity measurements or determinations of the number of emitting or absorbing atoms), combined with lifetime measurements, however, can result in relatively reliable absolute oscillator strengths or Einstein coefficients.
We shall deal in section 9.4 with the relation between oscillator strength and Einstein coefficient.
If the optically thin layer of gas described by Equation \ref{9.3.7} is in thermodynamic equilibrium, then $$\mathcal{N}_2$$ is given by Boltzmann's Equation, so that Equation \ref{9.3.7} becomes
$\label{9.3.8}L=\frac{\mathcal{N}hcϖ_2A_{21}\mathsf{e}^{-E_2/(kT)}}{4\piλ u}.$
The common logarithm of this is
$\label{9.3.9}\log \left ( \frac{Lλ}{ϖ_2A_{21}}\right ) =\log \frac{hc}{4\pi}+\log \mathcal{N}-\log u -\frac{eV_2}{kT}\log \mathsf{e} .$
If everything is in SI units, this becomes
$\label{9.3.10}\log \left (\frac{Lλ}{ϖ_2A_{21}}\right )=-25.801 +\log \mathcal{N}-\log u-\theta V_2.$
Thus a graph of $$\log \left ( \frac{Lλ}{ϖ_2A_{21}}\right )$$ versus the upper excitation potential $$V_2$$ will yield (for optically thin lines) the temperature and the column density of atoms from the slope and intercept. I leave it to the reader to work out the procedure for determining the electron density in a manner similar to how we did this for absorption lines in developing Equation 9.2.11.
The radiance of a line is, of course, the sum of the radiances of its Zeeman components, and, since the radiance is proportional to $$ϖ_2A_{21}$$, one can say, following a similar argument to that given in the penultimate paragraph of section 9.2, that the Einstein coefficient of a line is equal to the average of the Einstein coefficients of its components.
At this stage, you may be asking yourself if there is a relation between oscillator strength and Einstein coefficient. There is indeed, but I crave your patience a little longer, and I promise to address this in Section 9.4.
“Transition Probability” ($$\mathfrak{die Übergangswahrscheinlichkeit}$$.) The expression “transition probability” is often used for the Einstein $$A$$ coefficient, and it is even sometimes defined as “the probability per second that an atom will make a spontaneous downward transition from level 2 to level 1”. Both are clearly wrong.
In probability theory (especially in the theory of Markov chains) one sometimes has to consider a system that can exist in any of several states (as indeed an atom can) and the system, starting from one state, can make a transition to any of a number of other possible states. The probability of making a particular transition is called, not unnaturally, the transition probability. The transition probability so defined is a dimensionless number in the range zero to one inclusive. The sum of the transition probabilities to all possible final states is, of course unity. “Branching ratio” is another term often used to describe this concept, although perhaps “branching fraction” might be better. In any case, the reader must be aware that in many papers on spectroscopy, the phrase “transition probability” is used when what is intended is the Einstein $$A$$ coefficient.
The reader will have no difficulty in showing (from Equation \ref{9.3.3}) that the probability that an atom, initially in level 2, will make a spontaneous downward transition to level 1 in time $$t,\text{ is }1-e^{A_{21}t}$$, and that the probability that it will have made this transition in a second is $$1-e^{-A_{21}}$$. With $$A_{21}$$ being typically of order $$10^8\text{ s}^{-1}$$, this probability is, unsurprisingly, rather close to one! | 2,355 | 9,103 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-22 | latest | en | 0.862431 |
https://www.futilitycloset.com/2016/05/31/the-fifth-card/ | 1,718,779,479,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861806.64/warc/CC-MAIN-20240619060341-20240619090341-00401.warc.gz | 714,689,169 | 12,433 | # The Fifth Card
I hand you an ordinary deck of 52 cards. You inspect and shuffle it, then choose five cards from the deck and hand them to my assistant. She looks at them and passes four of them to me. I name the fifth card.
At first this appears impossible. The hidden card is one of 48 possibilities, and by passing me four cards in some order my assistant can have sent me only 1 of 4! = 24 messages. How am I able to name the card?
Part of the secret is that my assistant gets to choose which card to withhold. The group of five cards that you’ve chosen must contain two cards of the same suit. My assistant chooses one of these to be the hidden card and passes me the other one. Now I know the suit of the hidden card, and there are 12 possibilities as to its rank. But my assistant can pass me only three more cards, with 3! = 6 possible messages, so the task still appears impossible.
The rest of the secret lies in my assistant’s choice as to which of the two same-suit cards to give me. Think of the 13 card ranks arranged in a circle (with A=1, J=11, Q=12, and K=13). Given two ranks, it’s always possible to get from one to the other in at most 6 steps by traveling “the short way” around the circle. So we agree on a convention beforehand: We’ll imagine that the ranks increase in value A-K, and the suits as in bridge (or alphabetical) order, clubs-diamonds-hearts-spades. This puts the whole deck into a specified order, and my assistant can pass me the three remaining cards in one of six ways:
{low, middle, high} = 1
{low, high, middle} = 2
{middle, low, high} = 3
{middle, high, low} = 4
{high, low, middle} = 5
{high, middle, low} = 6
So if my assistant knows that I’ll always travel clockwise around the imaginary circle, she can choose the first card to establish the suit of the hidden card and to specify one point on the circle, and then order the remaining three cards to tell me how many clockwise steps to take from that point to reach the hidden rank.
“If you haven’t seen this trick before, the effect really is remarkable; reading it in print does not do it justice,” writes mathematician Michael Kleber. “I am forever indebted to a graduate student in one audience who blurted out ‘No way!’ just before I named the hidden card.”
It first appeared in print in Wallace Lee’s 1950 book Math Miracles. Lee attributes it to William Fitch Cheney, a San Francisco magician and the holder of the first math Ph.D. ever awarded by MIT.
(Michael Kleber, “The Best Card Trick,” Mathematical Intelligencer 24:1 [December 2002], 9-11.) | 628 | 2,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-26 | latest | en | 0.954048 |
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Pick a box! (Posted on 2002-03-28)
You are shown three boxes, and told that one of them contains a prize. You are then asked to pick one box, and if that box is the one with the prize, you will win it. After picking a box, you are shown that one of the other two boxes is empty, and offered a chance to change your selection.
Should you do this? Would changing your choice to the other remaining box affect your odds of winning? Why or why not?
See The Solution Submitted by levik Rating: 4.2857 (14 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
re: Flat out wrong logic in solution. | Comment 32 of 42 |
(In reply to Flat out wrong logic in solution. by jduval)
John, you might be interested in this problem, which is quite similar to this one.
Consider the following analogy:
An opaque box has two white marbles and a black marble. I want to pick the black marble. After I pick the marble, but before I look at its color, one of the marbles accidentally falls out of the box. It just so happens to be a white marble. I realize that the marble in my hand now has a 1/2 probability of being black.
After recounting the incident to my friend, he concocts an ingenious plan to increase the odds of getting a black marble when I play the game again. After I pick the marble, but before looking at it, my friend would take a random marble out of the box and show it to me. We realized before long that this didn't work, since there was a chance of picking a black marble out of the box, in which case my probability of winning dropped to 0 rather than going up to 1/2. The probabilities averaged out to a disappointing 1/3.
So we tried another idea, where my friend would look inside the box and pick a white marble out of the box, not telling me whether he saw any black marbles. Of course, it was 100% likely that he would find a white marble to show me. Therefore my chance of winning would be 1/2, right? It didn't happen. Try it yourself.
Edit: while I'm at it, I might as well link Levik's simulation for this problem, even though it was linked in a below comment.
Edited on March 4, 2005, 3:14 am
Posted by Tristan on 2005-03-03 22:20:52
Search: Search body:
Forums (0) | 582 | 2,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-30 | latest | en | 0.948475 |
https://coffeetearoom.com/how-many-oz-in-a-12-cup-coffee-pot/ | 1,702,019,638,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100724.48/warc/CC-MAIN-20231208045320-20231208075320-00146.warc.gz | 200,771,620 | 14,717 | # how many oz in a 12 cup coffee pot?
## How to make the perfect pot of coffee?
• 2 or 3 x 3 cm fresh ginger
• 1 cinnamon stick, broken in half
• 2 cloves
• 1 scant tsp soft brown sugar or honey
• 500 ml kombucha, chilled
• 5 or 6 star anise, to serve (optional)
More…
## How many tablespoons of coffee for 12 cups?
Considering that you are trying to determine how many tablespoons of coffee for 12 cups, then you just need to do a simple math operation: 12 cups of coffee = 24 tablespoons of coffee. Check out how to easily know how many tablespoons in a cup.
More…
## How much Folgers coffee to make 12 cups?
More so, you should also think of whether they like their coffee light or strong. Generally speaking, if you are making coffee for 12 cups, you will need 4.5 ounces of coffee, which will be equal to 127.6 grams, 72 teaspoons , 24 table spoons, or 1.5 cups.
More…
## How much coffee grounds do need for 12 cups?
The Quantity Of Grounded Coffee For 12 Cups. According to the online recommendations for some top coffee makers, we found the following recommendations: 5 fl. oz multiplied by 12, we get 60 fl. oz; 60 fl. oz, also we get 1800 ml; Based on the 1:18 golden ratio, 12 cups of coffee require 100 grams.
More…
## Good Wednesday Morning – Carol Underhill
· They gifted me with a coffee pot that has a 12 cup carafe on one side and a one-cup serving on the other side. Instead of a Keurig and K-cups, it uses your own coffee grounds. My son’s girlfriend gifted me with flavored coffees. I hate to think how many cups I drank over the weekend as I tested out that one-cup pot and all the different flavors.
## This Denver Coffee Company Helps Women in Recovery ‘Stay …
· The future of a cup of coffee Wagon Coffee currently carries over 12 varieties of coffee beans in its online store. They range from dark roasts from coffee farms in Indonesia to the lightest roast …
## Hamilton beach drip coffee maker instructions – Canadian …
· Make 12 cups of delicious coffee in less than 12 minutes with the Hamilton Beach 12 cup percolator, elegant stainless steel will compliment every setting. Single Serve Coffee Makers. 2-Way FlexBrew® Coffeemaker (49985C) You’re all signed up! Thank you for your interest in Hamilton Beach Brands, Inc.
## Natural and organic Nicaragua Un Paraiso – The Sapphire …
· Yet, the town proper features no men and women of obvious African heritage almost. My partner and i thought a 12 pot caffeine pan would be the norm and with two caffeine users in the household I just look much like we get two glasses each (Granted my drink is relatively significant) and We demand to come up with another marijuana.
## Not Everyone’s Cup of Tea – The Silver Teapot
· Also, the teacups from China in the early days of Tea had no handles, so that one would drink Tea with fingers spread around the cup with the pinkie finger raised for balance. Many more customs are considered good manners when we think about attending or hosting an Afternoon Tea. “Afternoon tea needn’t stand on ceremony.
## Sam Adams Octoberfest nutrition – uw favoriete producten …
1 package (19 oz.) Johnsonville® Beer ‘n Bratwurst Brats. 4 bratwurst buns. 1⁄2 cup Kroger® Shredded Sauerkraut. 1⁄4 cup Private Selection™ Whole Grain Garlic Mustard. 2 cups Private Selection™ Smoked Gouda & Garlic Ripple-Cut Kettle Chips. 4 cans (12 oz.) Samuel Adams® Octoberfest beer.
## cantaloupe nutrition facts 100g? – All About Food
· A cup of this delicious melon contains less than 13 grams of sugar. This may be a bit higher than other fruits, but keep in mind that a 12 ounce can of soda has nearly 40 grams of sugar, and very little nutritional value. Why does cantaloupe make me poop? Regulates blood pressure: Cantaloupes are loaded with potassium.
## Does Costco have Chamba chai? – Pursuantmedia.com
· A typical cup of chai tea prepared as directed contains approximately 40mg of caffeine (4 oz of black tea) compared to roughly 120mg in an average cup of coffee. Why do I poop immediately after tea? 04/6 Intestine muscle contractions Impact the movement of food from one place to another inside the intestinal tract and caffeine helps in speeding …
## Dragon fruit slurpee 711? – All About Food
· What Slurpee flavors Does 711 have? Slurpee® Blue Raspberry. Blueberry Lemonade Bliss. Cherry. Coca-Cola. MTN DEW® Citrus. Peach Perfect. Pina Colada. Pineapple Whip. What is a Slurpuccino Slurpee? The Slurpuccino is back! Slurpee® meets coffee in a dairy-free match made in heaven. Made with real coffee beans, every sip is bursting with coffee …
## John Wayne Breakfast Casserole / Instant Pot John Wayne …
· Place chilies and cheese in a 2 qt. Our simple keto breakfast casserole comes together with just a few basic ingredients. Pour in 9 x 12 inch pan. 16 oz container of biscuits 6 eggs ½ cup milk 1½ cups of shredded cheddar/jack mix 2 cups cut up breakfast sausage (about 8 frozen patties) salt and pepper to taste Click here for the full recipe: | 1,228 | 4,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-50 | latest | en | 0.903643 |
https://apple.stackexchange.com/questions/329794/how-can-i-describe-three-different-situations-on-a-cell-in-numbers | 1,571,096,802,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655554.2/warc/CC-MAIN-20191014223147-20191015010647-00100.warc.gz | 373,333,453 | 27,378 | # How can I describe three different situations on a cell in Numbers
We have business rules as follows:
• If it is bigger than £35 we charge 75% of it
• If it is smaller than £20 we charge 70% of it
• If is is between 20 to 35 we charge 72% of it
I don’t know how to set up a formula to automate this calculation in numbers. Is there a way to do this?
• What is it? A cell? – JBis Jul 10 '18 at 12:33
``````=IF(B3>35,B3×0.75,IF(B3<25,B3×0.7,B3×0.72)) | 154 | 455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-43 | latest | en | 0.945928 |
https://powerquery.how/list-repeat/ | 1,716,958,407,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059206.29/warc/CC-MAIN-20240529041915-20240529071915-00175.warc.gz | 392,046,965 | 74,307 | # List.Repeat
Updated on
List.Repeat is a Power Query M function that creates a list containing a specified number of repetitions of the original list. The function returns the newly created list with the repeated elements.
Compatible with: Power BI Service Power BI Desktop Excel Microsoft 365
## Syntax
``````List.Repeat(
list as list,
count as number,
) as list``````
## Description
The `List.Repeat` function creates a new list by repeating the elements of a given list for a specified number of times. It takes two arguments: the `list` of values to be repeated and the `count`, which represents the number of repetitions. After performing the operation this function returns a single list with the values repeated consecutively.
## Examples
Let’s look at a few examples of the List.Repeat function.
You can create a list that has {1, 2, 3 } repeated 3 times.
``````// Output: { 1, 2, 3, 1, 2, 3, 1, 2, 3 }
List.Repeat( { 1, 2, 3 }, 3 )``````
If you’re working on a mathematical or algorithmic problem, you might need to generate a sequence pattern. For instance, if you want a list that repeats the sequence 1 to 5 for 4 times:
``````// Output: { 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5 }
List.Repeat( List.Numbers( 1, 5 ), 4 )``````
The List.Repeat function also works with other values, like text. Suppose you’re categorizing items into groups, and you want to label each group with a repeated identifier. For example, if you have 15 items and you want to label them in groups of 5 with “Group A”, “Group B”, and “Group C”:
``````/* "Group A", "Group A", "Group A", "Group A", "Group A", "Group B", "Group B",
"Group B", "Group B", "Group B", "Group C", "Group C", "Group C", "Group C", "Group C" */
List.Combine(
{
List.Repeat( { "Group A" }, 5 ),
List.Repeat( { "Group B" }, 5 ),
List.Repeat( { "Group C" }, 5 )
}
)``````
These are just a few examples, but the potential applications are numerous. The function can be particularly useful when combined with other Power Query functions. | 593 | 2,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-22 | latest | en | 0.840209 |
https://core.ac.uk/search?q=author:(Zannier,%20Umberto) | 1,717,013,880,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059408.76/warc/CC-MAIN-20240529200239-20240529230239-00725.warc.gz | 151,588,064 | 23,600 | 144 research outputs found
### Hilbert Irreducibility above algberaic groups
The paper offers versions of Hilbert's Irreducibility Theorem for the lifting of points in a cyclic subgroup of an algebraic group to a ramified cover. A version of Bertini Theorem in this context is also obtained.Comment: 22 page
### On the Hilbert Property and the Fundamental Group of Algebraic Varieties
We review, under a perspective which appears different from previous ones, the so-called Hilbert Property (HP) for an algebraic variety (over a number field); this is linked to Hilbert's Irreducibility Theorem and has important implications, for instance towards the Inverse Galois Problem. We shall observe that the HP is in a sense `opposite' to the Chevalley-Weil Theorem, which concerns unramified covers; this link shall immediately entail the result that the HP can possibly hold only for simply connected varieties (in the appropriate sense). In turn, this leads to new counterexamples to the HP, involving Enriques surfaces. We also prove the HP for a K3 surface related to the above Enriques surface, providing what appears to be the first example of a non-rational variety for which the HP can be proved. We also formulate some general conjectures relating the HP with the topology of algebraic varieties.Comment: 24 page
### A lower bound for the height of a rational function at $S$-unit points
Let $\Gamma$ be a finitely generated subgroup of the multiplicative group \G_m^2(\bar{Q}). Let p(X,Y),q(X,Y)\in\bat{Q} be two coprime polynomials not both vanishing at $(0,0)$; let $\epsilon>0$. We prove that, for all $(u,v)\in\Gamma$ outside a proper Zariski closed subset of $G_m^2$, the height of $p(u,v)/q(u,v)$ verifies $h(p(u,v)/q(u,v))>h(1:p(u,v):q(u,v))-\epsilon \max(h(uu),h(v))$. As a consequence, we deduce upper bounds for (a generalized notion of) the g.c.d. of $u-1,v-1$ for $u,v$ running over $\Gamma$.Comment: Plain TeX 18 pages. Version 2; minor changes. To appear on Monatshefte fuer Mathemati
### Integral points, divisibility between values of polynomials and entire curves on surfaces
We prove some new degeneracy results for integral points and entire curves on surfaces; in particular, we provide the first example, to our knowledge, of a simply connected smooth variety whose sets of integral points are never Zariski-dense (and no entire curve has Zariski-dense image). Some of our results are connected with divisibility problems, i.e. the problem of describing the integral points in the plane where the values of some given polynomials in two variables divide the values of other given polynomials.Comment: minor changes, two references adde
### Rational points in periodic analytic sets and the Manin-Mumford conjecture
We present a new proof of the Manin-Mumford conjecture about torsion points on algebraic subvarieties of abelian varieties. Our principle, which admits other applications, is to view torsion points as rational points on a complex torus and then compare (i) upper bounds for the number of rational points on a transcendental analytic variety (Bombieri-Pila-Wilkie) and (ii) lower bounds for the degree of a torsion point (Masser), after taking conjugates. In order to be able to deal with (i), we discuss (Thm. 2.1) the semi-algebraic curves contained in an analytic variety supposed invariant for translations by a full lattice, which is a topic with some independent motivation.Comment: 12 page
• ⊠| 827 | 3,452 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 11, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-22 | latest | en | 0.839916 |
http://agenda.fest.ansible.com/vezovyfu67679.html | 1,642,590,763,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301309.22/warc/CC-MAIN-20220119094810-20220119124810-00281.warc.gz | 1,410,925 | 5,820 | # Question: How many centimeters are there in a 40 inches length of fabric?
Contents
## How many centimeters are there in 40 inches length of fabric?
Inch to Centimeter Conversion TableInchesCentimeters3793.98 cm3896.52 cm3999.06 cm40101.6 cm36 more rows
## How many centimeters are there in a inches length of fabric?
Explanation: One inch is equal to roughly 2.54 centimeters, so converting inches to centimeters means multiplying a value in inches by 2.54.
## How many centimeters are there in a 60 inches length of fabrics?
In 60 in there are 152.4 cm .
## How many centimeters are there in a 30 inches length of fabric?
Inches to Centimeters Calculation We know that one inch is equal to 2.54 centimeters. Thus, to convert 30 inches in cm, we have to multiply 30 inches by 2.54. (i.e.) 30 x 2.54 = 76.2 centimeters.
## What is a 1/2 yard of fabric?
The fabric is sold by the yard, so if you got one yard of fabric it would 36 inches long by 44 inches wide. If you got a half yard it would be 18 inches long by 44 inches wide. If you wanted a quarter of a yard it would be 9 inches long by 44 inches wide. | 305 | 1,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-05 | latest | en | 0.928173 |
https://gmatclub.com/forum/v03-94995.html | 1,495,503,615,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607245.69/warc/CC-MAIN-20170523005639-20170523025639-00459.warc.gz | 753,945,520 | 45,804 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
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# V03 #13
Author Message
Intern
Joined: 03 Jul 2009
Posts: 9
Followers: 0
Kudos [?]: 0 [0], given: 1
### Show Tags
29 May 2010, 13:02
The cost of gas-burning grills has dropped nearly 40% in the last year, prompting an increase in sales. Hank’s Hardware responded to the popularity of gas grills with large displays and deeply-discounted sale prices. Sales of gas grills at Hank’s Hardware went up by a third, but at the end of the quarter managers were shocked to find they had lost money on gas grills.
Which of the following, if true, is the best explanation for why Hank’s Hardware’s profits on gas grills are suffering?
(a) Sales volume has been miscalculated, surprising managers with a loss in profits.
(b) The further discounts offered by sale prices account for a larger dollar amount than the increase in sales, leading to a loss of profits.
(c) The reduction in cost is caused by shoddy construction, which is why many people who purchased gas grills are returning them, leading to a loss in profits.
(d) Wood-burning grills are more popular than gas-grills; since Hank’s carries mostly gas grills, their profits are suffering
(e) Customers only bought gas grills when they were on sale; once the sale was over, they stopped buying the grills, leading to a loss in profits.
OA (b)
maybe i'm missing something here, but (b) can't explain why the store is losing money (there's a difference between losing money, as in the situation discribed in the pasage, and decreased profits, which (b) suggests)
the only answer that can explain loosing money is (c) but it's a bit lame
maybe "they had lost money on gas grills" should be rephrased to "their profits declined"?
CEO
Joined: 15 Aug 2003
Posts: 3454
Followers: 67
Kudos [?]: 874 [0], given: 781
### Show Tags
06 Jun 2010, 12:43
hi,
I think your observation is mostly a nuance. Lost money on gas grills means loss in profits. What else would losing money mean?
Intern
Joined: 03 Jul 2009
Posts: 9
Followers: 0
Kudos [?]: 0 [0], given: 1
### Show Tags
06 Jun 2010, 13:13
right, right, right
too tired, sorry
Re: V03 #13 [#permalink] 06 Jun 2010, 13:13
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# V03 #13
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 914 | 3,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-22 | latest | en | 0.954339 |
https://math.answers.com/questions/What_is_the_LCM_of_15_45_and_90 | 1,719,235,729,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00642.warc.gz | 324,718,317 | 48,087 | 0
# What is the LCM of 15 45 and 90?
Updated: 9/20/2023
Wiki User
6y ago
The LCM is: 90
It is 90 because 6*15 = 90, 2*45 = 90 and 1*90 = 90
Wiki User
6y ago
Anonymous
Lvl 1
3y ago
you told us the hcf
Anonymous
Lvl 1
3y ago
The LCM of 15, 45 and 90 is 135. | 124 | 267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-26 | latest | en | 0.923967 |
https://betsat28.com/womboota/one-real-life-application-of-each-class-of-lever.php | 1,611,557,347,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565376.63/warc/CC-MAIN-20210125061144-20210125091144-00179.warc.gz | 236,021,878 | 13,573 | ## What are some examples of levers in everyday life?
Mechanical Advantage and Simple Machines Encyclopedia.com. 13/12/2011В В· For my project I am supposed to draw one example for all of The lever can be any class. I need some real life examples of the simple machines, please?, The Six Types of Simple Machines. An ax blade is one example of a wedge. Lever - Any tool that pries something loose is a lever. VEXpro Application Examples..
### Spur Gears Gear Lever
What are some examples of levers in everyday life?. • examples of the three classes of levers I need one example of a class one lever and one of resistance and force is for each one Answers:First-class, Source for information on Mechanical Advantage and Simple Machines: REAL-LIFE APPLICATIONS The Lever. the handle that one operates is a Class II,.
Class 1 levers have . you can choose one of the three classes of levers and simulate its operation. A real-life realization of each type is also illustrated. The Six Types of Simple Machines. An ax blade is one example of a wedge. Lever - Any tool that pries something loose is a lever. VEXpro Application Examples.
Examples of second class levers include doors, staplers, wheelbarrows and can openers. In a second class lever, the load is found between the effort and the fulcrum Kinematic Concepts + Levers study guide by kaschoen includes 58 questions in real life/competition, using our levers more efficiently is 1st Class Levers:
A lever is a mechanism that can be used to exert a large force over a small distance at one end of the lever by exerting a small force over a greater distance at the Levers work to create movement in the human body First-class levers in the human body are rare. One example is the joint between the In a third-class lever,
19/01/2007В В· What are some examples of a lever? I need to come up with one for school but my mind See http://en.wikipedia.org/wiki/Lever for info on each class. Open the other two paper clips and tape them opposite each other on one end of the can class lever? Real World Application: real life applications
• examples of the three classes of levers I need one example of a class one lever and one of resistance and force is for each one Answers:First-class Real World Science: Simple Machines with the students’ life experiences, real-ities, Newsroom is to actively involve each class member in a whole learning
They are found everywhere and it is one of the most useful simple machines. There are three classes of levers. sometimes called a Third Class Lever. A lever is one of the most common tools we use to help us lift Levers: Definition, Classes & Examples. A class 3 lever is a lever that has the input force in
Applications of circles In real life and a circle picture. application of a circle 1] the outer circle is not made from one piece of steel but rather several real life examples of a third class lever. distance from either the application of the force to the axis or the Third class levers increase _____ and
Planning Sheet for Single Lessons Explain one other example of a third class lever. Use concrete examples from previous lessons of each class of lever This is also termed mechanical advantage, and is one example of the and application point of the force applied at each end of the lever. class levers
A class 3 lever does not have the mechanical advantage of class-one levers and class-two levers, effort and fulcrum for each class of lever. real life examples of a third class lever. distance from either the application of the force to the axis or the Third class levers increase _____ and
### Life Application Study Bible В® Devotion Learn to
What are some examples of levers in everyday life?. Second class levers are similar to first class levers. Second class levers always multiply the input force, however they do not change its direction., 1st Class Lever. The first class lever is one of three classes of levers and is one possible arrangement of muscles, bones, and joints found in the human body..
### Kinematic Concepts + Levers Flashcards Quizlet
Life Application Study Bible В® Devotion Learn to. Second class levers are similar to first class levers. Second class levers always multiply the input force, however they do not change its direction. What are some examples of levers in everyday life? Update Cancel. Answer Wiki. What is a first-class lever? What are the real life examples of classes of levers?.
• examples of the three classes of levers I need one example of a class one lever and one of resistance and force is for each one Answers:First-class Second class levers are similar to first class levers. Second class levers always multiply the input force, however they do not change its direction.
We experience many Boyle's law applications in real life Ideal gas molecules do not attract or repel each and reminds me of chemistry class in the good 16/05/2012В В· Top Ten Real Life Examples of weights you need to increase the lever arm the block over is because there is not only one force acting on
Levers work to create movement in the human body First-class levers in the human body are rare. One example is the joint between the In a third-class lever, Levers work to create movement in the human body First-class levers in the human body are rare. One example is the joint between the In a third-class lever,
We experience many Boyle's law applications in real life Ideal gas molecules do not attract or repel each and reminds me of chemistry class in the good Ions and Ionization - Real-life applications Photo by: Each ends up possessing eight valence electrons, a class of minerals that contain aluminum,
Levers work to create movement in the human body First-class levers in the human body are rare. One example is the joint between the In a third-class lever, Open the other two paper clips and tape them opposite each other on one end of the can class lever? Real World Application: real life applications
10/02/2009В В· Can anybody give me 3 examples of the 3 types of levers? (First class, Need Examples of levers? resistance and force is for each one. Follow . 5 Open the other two paper clips and tape them opposite each other on one end of the can class lever? Real World Application: real life applications
The difference between first, second and third for the third class lever, we have the fulcrum at one What Is the Difference Between First, Second & Third answer to What is the real life example of mean median and mode? Suppose in your class there are 11 students and one of So Mean pocket Money for each one of
answer to What is the real life example of mean median and mode? Suppose in your class there are 11 students and one of So Mean pocket Money for each one of Second class levers are similar to first class levers. Second class levers always multiply the input force, however they do not change its direction.
real life examples of a third class lever. distance from either the application of the force to the axis or the Third class levers increase _____ and Wheel barrow is one good example of second class lever. All That Matters В» Different classes of levers in everyday life. Structure and application;
Place one weight rod on each side of the GROUP ACTIVITY & CLASS DISCUSSION: Each of the students discuss some real life applications of moments & couples. Did you know your body is full of simple machines? This project looks at one of the third-class levers in the human body: your arm!
## What are some examples of levers in everyday life?
Kinematic Concepts + Levers Flashcards Quizlet. Resources for connecting Math class to the real Connecting to Math in Real Life An activity chart lists the math skills tapped by each activity. Real World, • examples of the three classes of levers I need one example of a class one lever and one of resistance and force is for each one Answers:First-class.
### Spur Gears Gear Lever
I need some real life examples of the simple machines. First class levers in the body are rare, In each of these exercises, Change your life with MyPlate by LIVE STRONG.COM., A class 3 lever does not have the mechanical advantage of class-one levers and class-two levers, effort and fulcrum for each class of lever..
What levers does your body use example of a Class 1 lever where the rotation directions to each other. If a load tries to turn the lever They are found everywhere and it is one of the most useful simple machines. There are three classes of levers. sometimes called a Third Class Lever.
9/06/2012В В· An understanding of the action and principle of levers is of considerable use when considering the application class lever has the fulcrum at one Place one weight rod on each side of the GROUP ACTIVITY & CLASS DISCUSSION: Each of the students discuss some real life applications of moments & couples.
Second class levers are similar to first class levers. Second class levers always multiply the input force, however they do not change its direction. 13/12/2011В В· For my project I am supposed to draw one example for all of The lever can be any class. I need some real life examples of the simple machines, please?
Kinematic Concepts + Levers study guide by kaschoen includes 58 questions in real life/competition, using our levers more efficiently is 1st Class Levers: 10/02/2009В В· Can anybody give me 3 examples of the 3 types of levers? (First class, Need Examples of levers? resistance and force is for each one. Follow . 5
Kinematic Concepts + Levers study guide by kaschoen includes 58 questions in real life/competition, using our levers more efficiently is 1st Class Levers: There are relatively few occurrences of second-class levers in the body. In the second class lever, application of lifting force to a each. one example of
10/02/2009 · Can anybody give me 3 examples of the 3 types of levers? (First class, Need Examples of levers? resistance and force is for each one. Follow . 5 • examples of the three classes of levers I need one example of a class one lever and one of resistance and force is for each one Answers:First-class
The speed of a sprocket and chain system is determined by comparing the number of teeth on each 1.CLASS 1 LEVER Sketch at least one real life application This is also termed mechanical advantage, and is one example of the and application point of the force applied at each end of the lever. class levers
Resources for connecting Math class to the real Connecting to Math in Real Life An activity chart lists the math skills tapped by each activity. Real World ... they will be shown how the musculoskeletal system is a system of levers and one example of each type of lever. of each lever: 1st class real muscles but
In the use of a small force to overcome a large one the lever The human forearm is an application of the third-class lever, Because each style has its 13/12/2011В В· For my project I am supposed to draw one example for all of The lever can be any class. I need some real life examples of the simple machines, please?
Exercises With First-, Second- and Third-Class Levers. In each of these exercises, Biceps curls employ a third-class lever, Source for information on Mechanical Advantage and Simple Machines: REAL-LIFE APPLICATIONS The Lever. the handle that one operates is a Class II,
In class 2 levers, the fulcrum lies at one end, the effort is applied at the other end, and the load is placed at the middle. The closer the load is to the fulcrum How a Lever Works Share Flipboard we see that if we double the distance on one side of the lever, the mechanical advantage of a class 3 lever is always less
ADVANTAGES OF LEVERS ACTIVITY GUIDE applications in: LIFE SCIENCES – Structure and Function/Adaptations class of lever, one … A lever is one of the most common tools we use to help us lift Levers: Definition, Classes & Examples. A class 3 lever is a lever that has the input force in
9/06/2012В В· An understanding of the action and principle of levers is of considerable use when considering the application class lever has the fulcrum at one 1st Class Lever. The first class lever is one of three classes of levers and is one possible arrangement of muscles, bones, and joints found in the human body.
First class levers in the body are rare, In each of these exercises, Change your life with MyPlate by LIVE STRONG.COM. What levers does your body use example of a Class 1 lever where the rotation directions to each other. If a load tries to turn the lever
The Six Types of Simple Machines. An ax blade is one example of a wedge. Lever - Any tool that pries something loose is a lever. VEXpro Application Examples. Mechanical advantage is a The use of velocity in the static analysis of a lever is an application of which means the mechanical advantage of the real system
answer to What is the real life example of mean median and mode? Suppose in your class there are 11 students and one of So Mean pocket Money for each one of Did you know your body is full of simple machines? This project looks at one of the third-class levers in the human body: your arm!
1st Class Lever. The first class lever is one of three classes of levers and is one possible arrangement of muscles, bones, and joints found in the human body. They are found everywhere and it is one of the most useful simple machines. There are three classes of levers. sometimes called a Third Class Lever.
Levers are the simple of simple machines and are used often in everyday life. Levers all pivot around a second class, and third class. Each lever has an input 19/08/2018В В· and third class levers. Third class levers are these real-life with the application of very little force. Class three levers in the
### Spur Gears Gear Lever
I need some real life examples of the simple machines. Kinematic Concepts + Levers study guide by kaschoen includes 58 questions in real life/competition, using our levers more efficiently is 1st Class Levers:, The speed of a sprocket and chain system is determined by comparing the number of teeth on each 1.CLASS 1 LEVER Sketch at least one real life application.
What are some examples of a lever? Yahoo Answers. ... they will be shown how the musculoskeletal system is a system of levers and one example of each type of lever. of each lever: 1st class real muscles but, Source for information on Mechanical Advantage and Simple Machines: REAL-LIFE APPLICATIONS The Lever. the handle that one operates is a Class II,.
### Three Classes of Levers Wolfram Demonstrations Project
Real World Science Simple Machines NHPBS. Second class levers are similar to first class levers. Second class levers always multiply the input force, however they do not change its direction. This is also termed mechanical advantage, and is one example of the and application point of the force applied at each end of the lever. class levers.
Heavy Lifting with a Lever. Class 1 levers usually have a beam that • Cut two strips of tape that are each about 2.5 centimeters longer than the width of A class 3 lever does not have the mechanical advantage of class-one levers and class-two levers, effort and fulcrum for each class of lever.
Open the other two paper clips and tape them opposite each other on one end of the can class lever? Real World Application: real life applications Levers work to create movement in the human body First-class levers in the human body are rare. One example is the joint between the In a third-class lever,
First class levers in the body are rare, In each of these exercises, Change your life with MyPlate by LIVE STRONG.COM. ... of levers and one example of each type of lever. Levers: How the Human Body Uses them to and identifying the parts of each lever: 1st class
Did you know your body is full of simple machines? This project looks at one of the third-class levers in the human body: your arm! A lever is a mechanism that can be used to exert a large force over a small distance at one end of the lever by exerting a small force over a greater distance at the
Explanation about the Three Lever Classes. List of Topics. Each has its own uses and Class 1 lever. The effort in a class 1 lever is in one direction, Simple Machines Terri Wakild South Haven Public Schools April, second and third class levers. First class lever one example of each type of simple machine
Did you know your body is full of simple machines? This project looks at one of the third-class levers in the human body: your arm! Mechanical advantage is a The use of velocity in the static analysis of a lever is an application of which means the mechanical advantage of the real system
A class 3 lever does not have the mechanical advantage of class-one levers and class-two levers, effort and fulcrum for each class of lever. Wheel barrow is one good example of second class lever. All That Matters В» Different classes of levers in everyday life. Structure and application;
The Six Types of Simple Machines. An ax blade is one example of a wedge. Lever - Any tool that pries something loose is a lever. VEXpro Application Examples. 10/02/2009В В· Can anybody give me 3 examples of the 3 types of levers? (First class, Need Examples of levers? resistance and force is for each one. Follow . 5
... they will be shown how the musculoskeletal system is a system of levers and one example of each type of lever. of each lever: 1st class real muscles but Explanation about the Three Lever Classes. List of Topics. Each has its own uses and Class 1 lever. The effort in a class 1 lever is in one direction,
The Six Types of Simple Machines. An ax blade is one example of a wedge. Lever - Any tool that pries something loose is a lever. VEXpro Application Examples. • examples of the three classes of levers I need one example of a class one lever and one of resistance and force is for each one Answers:First-class
Levers are the simple of simple machines and are used often in everyday life. Levers all pivot around a second class, and third class. Each lever has an input 13/12/2011В В· For my project I am supposed to draw one example for all of The lever can be any class. I need some real life examples of the simple machines, please?
A second-class lever has a fulcrum at one end and a load in the The types of the levers and the importance of each of them; "Levers Used in Everyday Life." real life examples of a third class lever. distance from either the application of the force to the axis or the Third class levers increase _____ and
Wheel barrow is one good example of second class lever. All That Matters В» Different classes of levers in everyday life. Structure and application; In class 2 levers, the fulcrum lies at one end, the effort is applied at the other end, and the load is placed at the middle. The closer the load is to the fulcrum
What levers does your body use example of a Class 1 lever where the rotation directions to each other. If a load tries to turn the lever Real World Science: Simple Machines with the students’ life experiences, real-ities, Newsroom is to actively involve each class member in a whole learning
They are found everywhere and it is one of the most useful simple machines. There are three classes of levers. sometimes called a Third Class Lever. answer to What is the real life example of mean median and mode? Suppose in your class there are 11 students and one of So Mean pocket Money for each one of
• examples of the three classes of levers I need one example of a class one lever and one of resistance and force is for each one Answers:First-class What are some examples of levers in everyday life? Update Cancel. Answer Wiki. What is a first-class lever? What are the real life examples of classes of levers?
A terrific example of a lever is a seesaw. When one child sits down on one end of the seesaw, Here are some examples of third-class levers. Explanation about the Three Lever Classes. List of Topics. Each has its own uses and Class 1 lever. The effort in a class 1 lever is in one direction,
Examples of second class levers include doors, staplers, wheelbarrows and can openers. In a second class lever, the load is found between the effort and the fulcrum Levers are the simple of simple machines and are used often in everyday life. Levers all pivot around a second class, and third class. Each lever has an input
4722106 | 4,523 | 20,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-04 | latest | en | 0.904853 |
https://www.jiskha.com/questions/1599865/a-bouncy-ball-rebounds-to-90-of-the-height-of-the-preceding-bounce-jason-drops-a-bouncy | 1,601,249,948,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401582033.88/warc/CC-MAIN-20200927215009-20200928005009-00797.warc.gz | 876,754,176 | 5,052 | # Math
a bouncy ball rebounds to 90% of the height of the preceding bounce. jason drops a bouncy ball from initial height of 25 feet
(a) write out the sequence of the height of the first 4 bounces.
(b) derive an explicit formula for the rebound height of a bouncy ball dropped from an intial height of 25 feet.
1. 👍 0
2. 👎 0
3. 👁 764
1. initial height = 25 ft
height after 1 bounce = 25(.9)
height after 2 bounces = 25(9)^2
height after 3 bounces = 25(.9)^3
height after 4 bounces = 25(.9)^4
..
height of ball after nth bounce = 25(.9)^n
1. 👍 0
2. 👎 0
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http://brainden.com/forum/topic/15566-two-fuses/ | 1,603,915,712,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107900860.51/warc/CC-MAIN-20201028191655-20201028221655-00716.warc.gz | 17,607,199 | 16,723 | BrainDen.com - Brain Teasers
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# Two Fuses
Go to solution Solved by bushindo,
## Question
Here is a fun puzzle a college room mate once gave to me:
You have two lengths of fuse, and a book of matches. You know that each fuse length has (somehow) been cut so that if the fuse is allowed to burn from end to end, it will take precisely 1 minute to burn down. However, the fuses will not necessarily burn at an even pace; all you know is the total combustion time of each fuse. Using nothing but the two lengths of fuse and the matches to ignite them with, how can you measure an interval of 45 seconds?
## Recommended Posts
• 0
• Solution
Approach
Call the two fuses A and B. The strategy is as follows
1) Light A from both ends and simultaneously light 1 end of B
2) When A burns out after 30 sec, immediately light the other end of B.
3) B will burn out at the 45 sec mark.
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