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https://www.engineeringexpert.net/Engineering-Expert-Witness-Blog/tag/christiaan-huygens
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## Posts Tagged ‘Christiaan Huygens’
### What is Earth’s Mass?
Friday, November 7th, 2014
Last time we learned how Henry Cavendish used Christiaan Huygens’ work with pendulums to determine the value of g, the acceleration of gravity factor for Earth, to be 32.3 ft/sec2, or 9.8 m/sec2. From there Cavendish was able to go on and arrive at values for other factors in Isaac Newton’s gravity formula, namely G, the universal gravitational constant, and M, Earth’s mass. Today, we’ll discuss how Cavendish was able to calculate the Earth’s mass. Newton’s formula for gravity, once again, is: M = (g × R2) ÷ G where M stands for the mass of the heavenly body being quantified. For our case today M will represent the mass of Earth, which was originally quantified in slugs, a British unit of measurement. Today the measurement unit of choice in most parts of the world is the kilogram, which is the metric equivalent of a slug. With regard to the other variables in Newton’s gravity formula, namely, R and G, their values had previously been determined. Eratosthenes’ measurement of shadows cast by the sun on Earth’s surface had revealed Earth’s radius, R, to be 6,371 kilometers, or 6,371,000 meters. And Cavendish’s experiments led him to conclude that the universal gravitational constant, G, was 6.67 × 10-11 cubic meters per kilogram-second squared. Plugging these values into Newton’s equation, we calculate Earth’s mass to be: M = ((9.8 m/sec2) × (6,371,000 m)2) ÷ (6.67 × 10-11 m3/kg-sec2) M = 5.96 × 1024 kilograms Incidentally, 5.96 × 1024 is scientific notation, or mathematical shorthand, for the number 5,960,000,000,000,000,000,000,000. That’s a whole lot of zeros! Calculating the mass of Earth was an impressive accomplishment. Now that its value was known, scientists would be able to calculate the mass and acceleration of gravity for any heavenly body in the universe. We’ll see how that’s done next time. _______________________________________
### How Long is a Pendulum’s Swing?
Wednesday, October 29th, 2014
What would you do to pass the time if you were stuck on a ship of the middle ages for weeks at a time? Dutch mathematician Christiaan Huygens used the time to study the movement of clock pendulums. He watched them for endless hours, and he eventually came to realize that the pendulums’ swing was uneven due to the ship’s listing on the waves, a phenomenon which also affected the ship’s clocks’ accuracy. Eager to devise a solution to the problem of inaccurate time keeping, Huygens dedicated himself to finding a solution to the problem, and in so doing increase the navigational accuracy of ships as well. His efforts eventually resulted in a formula that shared a common variable with Isaac Newton’s gravitational formula, namely, g, Earth’s acceleration of gravity factor, a value which Huygens posited was indeed a non varying constant. Building upon Newton’s work, Huygens devised a formula which demonstrated the mathematical relationship between the motion of a clock’s pendulum and g. That formula is, T = 2 × ∏ × (L ÷ g)1/2 where, T is the period of time it takes a pendulum to make one complete swing, ∏ the Greek symbol pi, valued at 3.14, and L the length of the pendulum. Since devices capable of directly measuring the Earth’s gravity did not exist then, as they still don’t exist today, how in the world (pardon the pun) was Huygens able to arrive at this formula? Thinking outside the box, he posited that if one knows the length of the pendulum L, and then accurately measures the time it takes for the pendulum to complete its swings, taking into account the varied times that resulted due to the ship’s listing, one can calculate g using his equation. He eventually determined g‘s value to be equal to 32.2 feet per second per second, or 32.2 ft/sec2. Fast forwarding to Henry Cavendish’s time, Huygens’ work with pendulums and his determination of g was well known. We’ll see what Cavendish did with this knowledge next time. _______________________________________
### Huygens’ Use of Pendulums
Tuesday, October 21st, 2014
Last time we learned that Henry Cavendish determined a value for G, the universal gravitational constant, fast on his way to determining a quantity he was determined to find, the Earth’s mass. Today we’ll see how the previous work of Christiaan Huygens, a contemporary of Isaac Newton’s, helped him get there. First Cavendish used algebra to rearrange terms in Newton’s gravitational formula so as to solve for M, Earth’s mass. Rearranged, Newton’s formula becomes, M = (g × R2) ÷ G But in order to solve for M, Cavendish first needed to know Earth’s acceleration of gravity, g. To aid him in this calculation he referred back to the work of Christiaan Huygens, a Dutch mathematician from Newton’s time. Huygens was eager to devise a formula capable of predicting clock pendulums’ motions on ships, his goal being to invent a timepiece accurate enough to make navigating ships easier. He hypothesized that a key factor in predicting a pendulum’s movement was an unknown constant, the acceleration of gravity factor, g, which Newton had previously posited existed. Through meticulous observation, Huygens came to realize that the time it took for pendulums to complete one swing back and forth was dependent not only on the length of the pendulum, but also this unknown quantity. In order for Huygens’ computations to work, the value of g had to be a constant, meaning, its value could not vary between computations; g‘s value was in fact a fudge factor, a phantom he would assign a specific numerical value. Huygens’ needed it in order to make his hypothesis work, a practice commonly use by scientists, even today. Determining a value for g would allow Huygens to successfully relate the length of the pendulum to the timing of its swing and to create a mathematical relationship between them. Huygens ultimately determined g’s value to be a whopping 32.2 feet per second per second, or 32.2 ft/sec2. We’ll see how he did it next time. _______________________________________
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## Algebra 2 (1st Edition)
$$x=-2,\:x=\frac{3}{2},\:x=-1+2\sqrt{2},\:x=-1-2\sqrt{2}$$
In this problem, we are asked to factor the polynomial completely. Thus, before we start, we first see if there are any terms that are common factors of all of the terms in the polynomial. Next, we look for patterns, such as the sum or the difference of two cubes as well as the perfect square trinomial, the difference of two squares, and the common monomial factor. We continue factoring until the polynomial is factored completely. This means that we continue to factor until the polynomial is written as a product of different polynomials, all of which cannot be factored. Doing this, we find the factors: $$\left(x+2\right)\left(2x-3\right)\left(x^2+2x-7\right)=0 \\ x=-2,\:x=\frac{3}{2},\:x=-1+2\sqrt{2},\:x=-1-2\sqrt{2}$$
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https://familywoodworking.org/forums/index.php?threads/creating-a-radius-dish-in-vcarve.44526/#post-539347
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What's new
# Creating a radius dish in Vcarve
#### Brent Dowell
Staff member
Just documenting this here in case I need to use it again and forget how to do it.
A little background first. Most acoustic guitars have a slight radius built into their fronts and backs. For the backs, it's thought that it helps focus the sound, and for the front it's more for structural reason. The radii built in to these are typically very shallow.
The radius disk can be used as part of a 'go-bar-deck' which has a top and bottom and uses flexible rods for clamping the braces to the front and back of the guitars. It can also be used for sanding purposes of making sure the braces have the right radius, and sanding the radius around the edge of the sides of the guitar to make sure there is a perfectly flat gluing surface for joining the top/back and sides.
There is a little bit of math required before you go to create disk in vcarve. The following spreadsheet snippet shows the calculations I'm using to figure out how 'deep' the dish needs to be.
Lets do the 16' radius disk as an example. I'm doing a 2' disk, so the chord for the disk will be 2', the radius will be 16'*12 or 192". The formula then tells me that the small sagitta, or depth of cut in the center will be 0.375 inches.
So open up Vcarve and create a project that is 2' square, and 3/4" deep.
The first thing I'm going to do is to create 2 half circles with a diameter of 2'
Arc 1
Arc 2
Arc 3 and what it should look like.
I've also added a little circle in the center just to verify the depth of cut. The key points here are the start point having the start point of (0.0,12.375") and the radius being a negative radius of 192. Those match our radius in inches and calculated sagitta. The 'Set Height' of the curve is calculated for us and really has little bearing in how we are setting things up.
Next open up the molding toolpath tool. Select the outside arc first, and the the arc we want to use as our profile second (That's the arc 3 that we created that is not semicircular).
I've also selected 'Use Larger Area Clearance too', but just set it to the same bit I'm using for cutting the disk. In this case a .25" ball nose cutter. I might up the size of that later, but this is what I have selected right now.
This will create 2 tool paths for the cut. One that 'roughs' out the shape and a second one that does the actual cut.
You should end up with something that looks like this:
Repeat for the other half of the circle.
I also did a pocket toolpath for the center circle, again, this is just to verify that the cuts here are going to the right depth, our 'sagitta' number of .375"
That gives us this circular vinyl record looking set of tool paths.
If I preview just one of the sets of swept profile cuts and the pocket cut, I end up with something like this, which is exactly what I want.
Run all of the toolpaths and there you have it, a 2'x2' 16' radius dish.
Last edited:
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# The monthly income and the monthly expenditure of a man are rupees 15000 and rupees 10500 find the ratio between his monthly; a) income to expenditure b) expenditure and saving c) Savings and income
2
by RUTVI1
2016-04-09T15:35:36+05:30
Monthly income = 15,000
Monthly expenditure = 10500
Saving = Income - expenditure = 15000 - 10500 = 4500
(a) Ratio of income to expenditure
15000 : 10500
= 150 : 105
= 30 : 21
= 10 : 7
(b) Ratio of expenditure and saving
10500 : 4500
= 105 : 45
= 21 : 9
= 7 : 3
(c) Ratio of saving and income
4500 : 15000
= 45 : 150
= 9 : 30
= 3 : 10
are you back
2016-04-09T15:44:14+05:30
Saving of income = 4500
a) income to expenditure
15000/10500
150/105
30/21
10/7
10:7
b) expenditure and saving
10500/4500
105/45
21/9
7/3
7:3
c) Savings and income
4500/15000
45/150
9/30
3/10
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# 3797 (number)
3,797 (three thousand seven hundred ninety-seven) is an odd four-digits prime number following 3796 and preceding 3798. In scientific notation, it is written as 3.797 × 103. The sum of its digits is 26. It has a total of 1 prime factor and 2 positive divisors. There are 3,796 positive integers (up to 3797) that are relatively prime to 3797.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 4
• Sum of Digits 26
• Digital Root 8
## Name
Short name 3 thousand 797 three thousand seven hundred ninety-seven
## Notation
Scientific notation 3.797 × 103 3.797 × 103
## Prime Factorization of 3797
Prime Factorization 3797
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 3797 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 8.24197 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 3,797 is 3797. Since it has a total of 1 prime factor, 3,797 is a prime number.
## Divisors of 3797
2 divisors
Even divisors 0 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 3798 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 1899 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 61.6198 Returns the nth root of the product of n divisors H(n) 1.99947 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 3,797 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 3,797) is 3,798, the average is 1,899.
## Other Arithmetic Functions (n = 3797)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 3796 Total number of positive integers not greater than n that are coprime to n λ(n) 3796 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 530 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 3,796 positive integers (less than 3,797) that are coprime with 3,797. And there are approximately 530 prime numbers less than or equal to 3,797.
## Divisibility of 3797
m n mod m 2 3 4 5 6 7 8 9 1 2 1 2 5 3 5 8
3,797 is not divisible by any number less than or equal to 9.
• Arithmetic
• Prime
• Deficient
• Polite
• Prime Power
• Square Free
## Base conversion (3797)
Base System Value
2 Binary 111011010101
3 Ternary 12012122
4 Quaternary 323111
5 Quinary 110142
6 Senary 25325
8 Octal 7325
10 Decimal 3797
12 Duodecimal 2245
20 Vigesimal 99h
36 Base36 2xh
## Basic calculations (n = 3797)
### Multiplication
n×y
n×2 7594 11391 15188 18985
### Division
n÷y
n÷2 1898.5 1265.67 949.25 759.4
### Exponentiation
ny
n2 14417209 54742142573 207855915349681 789228910582738757
### Nth Root
y√n
2√n 61.6198 15.6008 7.84983 5.19862
## 3797 as geometric shapes
### Circle
Diameter 7594 23857.3 4.5293e+07
### Sphere
Volume 2.29303e+11 1.81172e+08 23857.3
### Square
Length = n
Perimeter 15188 1.44172e+07 5369.77
### Cube
Length = n
Surface area 8.65033e+07 5.47421e+10 6576.6
### Equilateral Triangle
Length = n
Perimeter 11391 6.24283e+06 3288.3
### Triangular Pyramid
Length = n
Surface area 2.49713e+07 6.45142e+09 3100.24
## Cryptographic Hash Functions
md5 36072923bfc3cf47745d704feb489480 16d2310dc2662228e5959c7546af5d51fd967934 38a0ddf1476fb458ee1b46a68ddaea1e63a4b316a04393face769ab3f644896b 76e21c6ed4f86e9265b236d48b413aa90e31118cac552cd5b1cdb2f1ccc739690224eaa05dfbb24c4994c37442fdf4f7718f495d63d28196c15d9b9cf07bffc9 5b69274c7cd0514930d7e61339221dfe6a98ebbf
| 1,405 | 3,996 |
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# PS: Light House
Author Message
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Re: PS: Light House [#permalink] 20 Sep 2010, 22:15
Thanks Bunnel I got it now...Its certainly not a Gmat question.
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Re: PS: Light House [#permalink] 24 Nov 2010, 22:47
A searchlight on top of the watch-tower makes 3 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds?
I also could not do it in the first go. But here's what I figured out after some thought.
Consider someone standing within the imaginary circle lit by the search-light as it revolves.
What is the probability that the man will stay in dark for 20 seconds (3 revs/min)? zero or (1 - (20/20))
i.e. (1 - P(coming under focus))
What is the probability that the man will stay in dark for 19 seconds? (1 - (19/20))
What is the probability that the man will stay in dark for 18 seconds? (1 - (18/20))
.
.
What is the probability that the man will stay in dark for 5 seconds? (1 - (5/20)) i.e. 3/4
but somehow I am not able to fit in the 'at-least' part. For 'at least' cases, we add the probabilities of all the possible elements (OR). In this case, that would amount to summing up the probabilities of 5 sec, 4 sec, 3 sec, 2 sec and 1 sec.
But I guess the problem is with the discrete approach that I, and many others above have taken.
Shouldn't we approach the problem like the summation of velocity-time graph to find the total distance that we used to do in physics? That is finding the area under the graph..
To be edited, if the bulb in my mind glows, discerning an elegant solution..
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m13#37 [#permalink] 20 May 2011, 21:30
37. If a searchlight on top of a watch-tower makes 3 revolutions per minute, what is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds?
a) 1/4
b) 1/3
c) 1/2
d) 2/3
e) 3/4
can somebody let me know how to solve and reasoning use to find the answer
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Re: m13#37 [#permalink] 20 May 2011, 22:29
Don't know the OA, but...
3 revs in 60 sec -> 1 rev in 20 sec.
5 seconds is one fourth of 20 seconds.
1 revolution is 360 degrees (just to remind: in 20 seconds).
We need one forth, means 360/90=4.
So 1/4 will be an answer.
(A)
What's the OA?
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Re: m13#37 [#permalink] 21 May 2011, 00:13
For sure!
It is said "at least"!
I've found the minimal value.
We need to subtract: 1-1/4=3/4
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Probability - Search Light [#permalink] 06 Sep 2011, 08:38
If a searchlight on top of a watch-tower makes 3 revolutions per minute, what is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds?
A. \frac{1}{4}
B. \frac{1}{3}
C. \frac{1}{2}
D. \frac{2}{3}
E. \frac{3}{4}
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Re: Probability - Search Light [#permalink] 06 Sep 2011, 08:56
3 revolutions in 1min=in 60 sec
so 1 rev in 20 sec-> 1/4 rev in 5 sec-> so prob of person remaining in dark for 5 sec is 3/4
so its E
Re: Probability - Search Light [#permalink] 06 Sep 2011, 08:56
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Related Search
Web Results
www.reference.com/world-view/calculate-sum-arithmetic-sequence-70f916b0feb7b8d6
Calculate the sum of an arithmetic sequence with the formula (n/2)(2a + (n-1)d). The sum is represented by the Greek letter sigma, while the variable a is the first value of the sequence, d is the difference between values in the sequence, and n is the number of terms in the series.
www.reference.com/article/arithmetic-sequence-8622b9668a69568f
An arithmetic sequence is a sequence of numbers where there is a definitive pattern between the consecutive terms of the series. In general, arithmetic sequences can be represented by x = a + d(n - 1).
www.reference.com/article/explicit-formula-geometric-sequence-12a861c315df903a
The explicit formula for a geometric sequence is a_n = a_1 * r^(n-1). The variable a_n is equal to the value of the nth term in the given geometric sequence, while a_1 is the value of the first term in the sequence.
www.reference.com/article/formula-arithmetic-mean-113b81ffcbde610c
The formula for the arithmetic mean of a group of numbers is A = S/N, where S represents the sum of all the numbers in the group, and N represents the total number of items in the group. Finding the arithmetic mean is essentially the same as taking the average.
www.reference.com/world-view/arithmetic-sequences-used-daily-life-2491bf2c03ae9106
Arithmetic sequences are used in daily life for different purposes, such as determining the number of audience members an auditorium can hold, calculating projected earnings from working for a company and building wood piles with stacks of logs. Arithmetic sequences are tools used in algebra and geo
www.reference.com/article/arithmetic-sum-formula-f0aa48ffe199916a
The arithmetic sum formula is Sn = n/2 {2a + (n-1) d} where Sn is the sum of n-terms of an arithmetic progression, the first term is ‘a’ and the common difference between any two consecutive terms is given by d.
www.reference.com/article/arithmetic-geometric-formulas-680270e4f60820e
Arithmetic formulas originate from the need to determine the value or position of a specific term within an arithmetic sequence, where the difference between successive terms is a constant d, such as "an = a1 - (n - 1)d." Geometric formulas are derived from a similar need but applied to a geometric
www.reference.com/world-view/formula-calculate-average-26e3e5148012bf52
To calculate the average of a group of numbers, first add the numbers together and then divide by the amount of numbers that are in the group. The formula for average is: sum/(quantity of numbers.)
www.reference.com/article/formula-calculate-force-88095f57697585c
For an object with unchanging mass, force equals mass times acceleration. This is abbreviated as f = ma. Force is a push/pull on an object resulting from interactions with another object. Having a magnitude and a direction makes force a vector quantity.
www.reference.com/article/formula-used-calculate-mass-c8c321cff37f4677
There are two formulas that may be used to calculate mass: mass is equal to the volume of an object multiplied by its density (m=v*d) and mass is equivalent to an object's weight divided by the acceleration of gravity (m=w/g). The appropriate formula depends on the available variables.
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# Salat
Grandmother planted salad. In each row, he planted 13 seedlings. After a morning frost, many seedlings died.
In the first row, five seedlings died.
In the second row, two seedlings died more than 1st row.
In the third row, three seedlings died less than 1st row.
In the fourth line, one seedling died more than in the third row.
How many seedlings does grandmother have to buy to have all rows again with 13 seedlings?
x = 17
## Step-by-step explanation:
Did you find an error or inaccuracy? Feel free to write us. Thank you!
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https://www.stichtingdeverweesdetoren.nl/2021/03/31/c-what-is-the-difference-between-p-and-p/
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The value it is referring to is a numerical value ranging from 0 to 1. Here we will be explaining the concept of the P-value and how to interpret its value. Even if you do not have a background in statistics and mathematics, you should be able to understand the P-Value’s concepts and purpose by the end of this article.
1. A desired proportion of said distribution can then be cut out as being less probable than a given threshold.
2. The p-value in statistics quantifies the evidence against a null hypothesis.
3. The same 10% resulting from a test on 50,000 users would have a 95% interval bound at +3.25% since the variability of the estimate is smaller due to the larger sample size.
4. If you want to compare the means of several groups at once, it’s best to use another statistical test such as ANOVA or a post-hoc test.
5. The standard deviation reflects variability within a sample, while the standard error estimates the variability across samples of a population.
Some outliers represent natural variations in the population, and they should be left as is in your dataset. Outliers are extreme values that differ from most values in the dataset. Missing data are important because, depending on the type, they can sometimes bias your results. This means your results may not be generalizable outside of your study because your data come from an unrepresentative sample.
These tables help you understand how often you would expect to see your test statistic under the null hypothesis. Conversely, if the new drug indeed reduces pain significantly, your test statistic will diverge further from what’s expected under the null hypothesis, and the p-value will decrease. If in the future you wish to style one of these parts, it will be much easier to do so at the block level.
## How to Group Data by Hour in R (With Example)
As a consequence, no matter how many users are measured, a possibility remains for the observed effect to differ from the actual effect. Introduction to Statistics is our premier online video course that teaches you all of the topics covered in introductory statistics. For example, suppose that we want to test whether or not there is a difference in mean blood pressure reduction between a new pill and the current pill.
A factorial ANOVA is any ANOVA that uses more than one categorical independent variable. To compare how well different models fit your data, you can use Akaike’s information criterion for model selection. The Akaike information criterion is one of the most common methods of model selection. AIC weights the ability of the model to predict the observed data against the number of parameters the model requires to reach that level of precision. You can choose the right statistical test by looking at what type of data you have collected and what type of relationship you want to test. In most cases, researchers use an alpha of 0.05, which means that there is a less than 5% chance that the data being tested could have occurred under the null hypothesis.
To (indirectly) reduce the risk of a Type II error, you can increase the sample size or the significance level to increase statistical power. As the degrees of freedom increase, Student’s difference between p&l and balance sheet t distribution becomes less leptokurtic, meaning that the probability of extreme values decreases. The distribution becomes more and more similar to a standard normal distribution.
## Confidence Interval for the Difference in Proportions
If our data produce values that meet or exceed this threshold, then we have sufficient evidence to reject the null hypothesis; if not, we fail to reject the null (we never “accept” the null). Multiple linear regression is a regression model that estimates the relationship between a quantitative dependent variable and two or more independent variables using a straight line. A larger sample size provides more reliable and precise estimates of the population, leading to narrower confidence intervals. Simple linear regression is a regression model that estimates the relationship between one independent variable and one dependent variable using a straight line.
Statistical significance is denoted by p-values whereas practical significance is represented by effect sizes. Outliers can have a big impact on your statistical analyses and skew the results of any hypothesis test if they are inaccurate. The geometric mean is an average that multiplies all values and finds a root of the number. Missing not at random (MNAR) data systematically differ from the observed values. Missing completely at random (MCAR) data are randomly distributed across the variable and unrelated to other variables.
The confidence level is the percentage of times you expect to get close to the same estimate if you run your experiment again or resample the population in the same way. The t-distribution gives more probability to observations in the tails of the distribution than the standard normal distribution (a.k.a. the z-distribution). In statistics, the range is the spread of your data from the lowest to the highest value in the distribution.
You can use the qchisq() function to find a chi-square critical value in R. You can use the CHISQ.INV.RT() function to find a chi-square critical value in Excel. If the bars roughly follow a symmetrical bell or hill shape, like the example below, then the distribution is approximately normally distributed. Means “dereference ptr, then multiply the value it points at by 137.”.
## F-Test for Equal Variances Calculator
It means that the observed data do not provide strong enough evidence to reject the null hypothesis. Such a small p-value provides strong evidence against the null hypothesis, leading to rejecting the null in favor of the alternative hypothesis. The p-value will never reach zero because there’s always a slim possibility, though highly improbable, that the observed results occurred by random chance.
A two-time Super Bowl MVP, Mahomes has thrown for 4,802 yards with 39 touchdowns and only seven interceptions in 17 playoff contests. He has connected with Travis Kelce for 17 TDs, the most by any quarterback-receiver duo in NFL postseason history. Trailing by 17 at halftime, https://adprun.net/ San Francisco made an impressive comeback against Detroit in the NFC Championship Game, scoring 24 consecutive points en route to a victory. It was the 38th playoff win for the franchise, breaking a tie with Green Bay and New England for most in NFL history.
They can also be estimated using p-value tables for the relevant test statistic. P values are also often interpreted as supporting or refuting the alternative hypothesis. The p value can only tell you whether or not the null hypothesis is supported. It cannot tell you whether your alternative hypothesis is true, or why.
A p-value of 0.001 is highly statistically significant beyond the commonly used 0.05 threshold. It indicates strong evidence of a real effect or difference, rather than just random variation. When you perform a statistical test, a p-value helps you determine the significance of your results in relation to the null hypothesis.
Outside of this context, though, putting a star after a pointer variable is illegal. 2) We can use hypothesis tests to test and ultimately draw conclusions about the value of a parameter. “We can be 95% confident that the proportion of Penn State students who have a tattoo is between 5.1% and 15.3%.” One possible way of dealing with variably is to try and eliminate or reduce it. One can attempt to balance the number of users with a certain characteristic that end up in one test group or the other.
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# Prove that of the Numbers 3 Sqrt(7) Is Irrational: - Mathematics
Prove that of the numbers 3 sqrt(7) is irrational:
#### Solution
Let 3 sqrt(7) be rational.
1/3 ×3 sqrt(7)= sqrt(7) = rational [∵Product of two rational is rational]
This contradicts the fact that sqrt(7) is irrational.
The contradiction arises by assuming 3 sqrt(7) is rational.
Hence, 3 sqrt(7) is irrational.
Concept: Concept of Irrational Numbers
Is there an error in this question or solution?
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# Calculating an ROI
## How to calculate out an Return on Investment or ROI.
First of all, what is an ROI? It’s a way of measuring how an investment is or is projected to perform. It is calculated by subtracting all of the annual expenses from the total annual income and then dividing it by the cost of the investment. Our purpose is specifically for rental properties but this same process can be applied to most any investment.
Let’s walk through two examples of the same property and see how vastly different the ROI can be.
## First Example
First, take the cost of the investment. Costs of investment is \$50,000. (at the time this was written we used a property from our website that had these exact numbers. This way you have a Real example).
Next, figure out your gross annual rent. You do this by multiplying the monthly rent by 12 months. Example \$800 monthly rent multiplied by 12 months = \$9,600 this is your Gross Annual Rent amount.
Next, figure out your Net annual rent amount. You do this by adding up all of your expenses and then subtracting them from your gross annual rent amount.
Here is what I factor into expenses. Examples;
• Taxes (this is a set amount) \$1,408
• Insurance \$500
• Management fees 10% = \$960
• Owner paid utilities (hopefully Zero) \$0
• Vacancy Rate 5% = \$480
• Repair and Maintenance rate 10% = \$960
• Lawn care \$0
• Lease Fee (future expense) 3% = \$288
• Add all of these expense up, which equals = \$4,596 and this is your total annual expenses.
• Next, subtract your Total Annual Expenses of \$4,596 from your Total Gross Rents of \$9,600 which Equals = \$5,004 This is your Net Income.
• Next, divide your Net Income of \$5,004 by the Cost of the Investment which is \$50,000.
• \$5,004 Divided by \$50,000 = 10.008% ROI.
• The Example above shows a 10% Return On Investment. This scenario was taken directly from the Partners Investing Website.
## Second Example (Typical of other sites)
Let’s take this same investment property and figure out the ROI using the most common expenses that you’ll see on other websites.
Assuming they are selling the property for the same amount. Cost is \$50,000. Monthly rents of \$800 X 12 months = \$9,600 Gross Annual Rent.
Most Common List of Expenses;
• Taxes of \$1,408
• Insurance of \$500
• 10% property management fee of \$960
• Plus Owner paid utilities of \$0
• Add all of the expenses up, which equals = \$2,868 and this is your total annual expenses.
• Next, subtract your Total Annual Expenses of \$2,868 from your Total Gross Rents of \$9,600 which Equals = \$6,732 This is your Net Income.
• Next Divide your Net annual income of \$6,732 by the cost of the investment which is \$50,000.
• \$6,732 Divided by \$50,000 = 13.46% ROI.
• Wow! Big difference. Should I be advertising a 13.46% or a 10% ROI
## Okay, what was left out of the second scenario?
• Vacancy Rate of 5% = \$480 - Just because it's rented now doesn't mean it always will be. I have 5% in there which accounts for a vacancy every two plus years.
• Repairs and Maintenance Rate of 10% = \$960 - With my properties all of the Major CAPEX items are taken care of but that doesn't mean there won't be routine maintenance and random repairs.
• Lease Fee Rate of 3% = \$288 - Need to account for the future lease fee as it will eventually go vacant. I have 3% in there which accounts for a new tenant every 2 plus years.
• Total amounnt of expenses left out is \$1,728. This is typical from other turn key property sites.
So this same investment can show an ROI of 10% or 13.46% depending on what is factored into the expenses. I’m not saying other websites are wrong. What I’m saying is make sure you are comparing Apples to Apples. When you see an advertisement for a specific ROI, pause for a moment and see what all is included when they figured out their ROI.
I highly recommend you create your own spreadsheet (excel seems to be the best for this) and then run the different investments you come across through it and that way you’ll know for sure your comparing apples to apples. This will help you start to see patterns and then you’ll know when it’s a good deal.
I hope this helps, if you have a specific question please feel free to send us an email.
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# Algebra
what is the equation for the graphs given the following information:
2. the graph of: y = asinbx has amplitude of 4 and a period of pi/6
3. a sinusoid graph with a maximum at A(0,3) and a minimum at (2pi, -3)
1. 👍 0
2. 👎 0
3. 👁 102
1. there are lots of online graphing sites: wolframalpha and desmos are two
#2. sin(kx) has a period of 2pi/k
so, a period of pi/6 means 2pi/k = pi/6 ==> k=12
y = 4sin(12x)
#3.
y varies from -3 to 3, so a = 3
max to min is 1/2 period, so you have a period of 4pi
cosine starts at a maximum, so
y = 3 cos(x/2)
1. 👍 0
2. 👎 0
posted by oobleck
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## Precalculus (6th Edition) Blitzer
$-24$ feet /second in the downward direction
$s(t)=-16t^2+72t$ Thus, we have $v(t)=sā²(t)=-16 \dfrac{dt^2}{dt}+72 \times \dfrac{dt}{dt}$ So, $v(t)=-32t +72$ Set $t=3$ Then $v(3)=-32(3) +72=-24$ feet /second in the downward direction
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Home » How to Plot a t Distribution in R
# How to Plot a t Distribution in R
To plot the probability density function for a t distribution in R, we can use the following functions:
• dt(x, df)Â to create the probability density function
• curve(function, from = NULL, to = NULL)Â to plot the probability density function
To plot the probability density function, we need to specify df (degrees of freedom) in the dt() function along with the from and to values in the curve() function.
For example, the following code illustrates how to plot a probability density function for a t distribution with 10 degrees of freedom where the x-axis of the plot ranges from -4 to 4:
```curve(dt(x, df=10), from=-4, to=4)
```
Similar to the normal distribution, the t distribution is symmetrical around a mean of 0.
We can add a title, change the y-axis label, increase the line width, and even change the line color to make the plot more aesthetically pleasing:
```curve(dt(x, df=10), from=-4, to=4,
main = 't Distribution (df = 10)', #add title
ylab = 'Density', #change y-axis label
lwd = 2, #increase line width to 2
col = 'steelblue') #change line color to steelblue
```
We can also add more than one curve to the graph to compare t distributions with different degrees of freedom. For example, the following code creates t distribution plots with df = 6, df = 10, and df = 30:
```curve(dt(x, df=6), from=-4, to=4, col='blue')
curve(dt(x, df=10), from=-4, to=4, col='red', add=TRUE)
curve(dt(x, df=30), from=-4, to=4, col='green', add=TRUE)
```
We can add a legend to the plot by using the legend() function, which takes on the following syntax:
legend(x, y=NULL, legend, fill, col, bg, lty, cex)
where:
• x, y:Â the x and y coordinates used to position the legend
• legend:Â the text to go in the legend
• fill:Â fill color inside the legend
• col:Â the list of colors to be used for the lines inside the legend
• bg:Â the background color for the legend
• lty:Â line style
• cex:Â text size in the legend
In our example we will use the following syntax to create a legend:
```#create density plots
curve(dt(x, df=6), from=-4, to=4, col='blue')
curve(dt(x, df=10), from=-4, to=4, col='red', add=TRUE)
curve(dt(x, df=30), from=-4, to=4, col='green', add=TRUE)
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Lesson 1
Build It
Lesson Narrative
This lesson establishes the straightedge and compass moves that students will use to perform various constructions. Students build on their previous understanding of circles as a set of points all equidistant from the center and line segments as a set of points on a line with two endpoints. Constructions are used in subsequent lessons to introduce students to reasoning about distances, generating conjectures, and attending to the level of precision required to define rigid motions later in the unit.
Students attend to precision when they discuss why straightedge and compass moves communicate geometric information consistently, as opposed to eyeballing (MP6).
These materials use words rather than symbolic notation to allow students to focus on the content. By using words, students do not need to translate the meaning of the symbol while reading. To increase exposure to different notations, images with given information marked using ticks, right angle marks, or arrows also have a caption with the symbolic notation ($$\overline{AB} \cong \overline{AC}, \overline{AB} \perp \overline{AC}$$ or $$\overline{AB} \parallel \overline{AC}$$). Feel free to use the symbolic notation when recording student responses, as that is an appropriate use of shorthand.
In this lesson and the subsequent lessons in this section, all constructions are accessible using physical straightedges and rigid compasses. If students have ready access to digital materials in class, they can choose to perform any or all construction activities with the GeoGebra Construction tool accessible in the Math Tools or available at geogebra.org/m/VQ57WNyR. The warm-up of the optional lesson "Using Technology for Constructions" is a good primer for the GeoGebra Construction tool. Do that warm-up with students before starting this lesson if students will use the digital tool rather than physical tools. If students do not have ready access to this digital tool in class, consider using the GeoGebra Construction tool to demonstrate constructions during the activity or lesson syntheses.
Learning Goals
Teacher Facing
• Comprehend that compasses create circles and can be used to transfer distances across a construction.
• Create diagrams using a straightedge to produce a line or segment through two points.
Student Facing
• Let’s use tools to create shapes precisely.
Required Preparation
Create a display of straightedge and compass moves that will remain displayed for all to see throughout the unit. See the warm-up synthesis for an example.
Assemble geometry toolkits. It would be best if students had access to these toolkits at all times throughout the unit.
Student Facing
• I can create diagrams using a straightedge.
• I know to use a compass to construct a circle.
Building Towards
Glossary Entries
• circle
A circle of radius $$r$$ with center $$O$$ is the set of all points that are a distance $$r$$ units from $$O$$
To draw a circle of radius 3 and center $$O$$, use a compass to draw all the points at a distance 3 from $$O$$.
• line segment
A set of points on a line with two endpoints.
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+0
# system
0
210
1
Solve for m, a, t, and h.
mat = 1/8
mah = 32
mth = 1/3
ath = 81/4
Jul 8, 2021
#1
+238
0
Multiply all 4 equations together,
$$m^3a^3t^3h^3= 27 \\ math=3$$
Divide by each equation,
$$h = 24 \\ t = \frac{3}{32} \\ a = 9 \\ m = \frac{4}{27}$$
Jul 8, 2021
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# Differential Equation Problem
• Jan 23rd 2007, 12:24 AM
machi4velli
Differential Equation Problem
I have been struggling with this problem for a while:
A bullet passes through a piece of wood 0.1 m thick. It enters the wood at 400 m/s and leaves at 180 m/s. The velocity v of the bullet is assumed to obey the differential equation dv/dt = -kv^2 where k is a positive constant. How long does it take the bullet to pass through the wood?
I got this far:
dv/dt = -kv^2
-1/v = -kt + C
v = 1/(kt + C)
400 = v(0) = 1/(k(0) + C)
400 = 1/C
.0025 = C
v = 1/(kt + .0025)
From here, I am not sure what to do. I tried to find k by making v = dx/dt and solving the displacement equation, which turned out to be another dead end.
Any help would be appreciated. Thanks.
• Jan 23rd 2007, 02:11 AM
ticbol
Quote:
Originally Posted by machi4velli
I have been struggling with this problem for a while:
A bullet passes through a piece of wood 0.1 m thick. It enters the wood at 400 m/s and leaves at 180 m/s. The velocity v of the bullet is assumed to obey the differential equation dv/dt = -kv^2 where k is a positive constant. How long does it take the bullet to pass through the wood?
I got this far:
dv/dt = -kv^2
-1/v = -kt + C
v = 1/(kt + C)
400 = v(0) = 1/(k(0) + C)
400 = 1/C
.0025 = C
v = 1/(kt + .0025)
From here, I am not sure what to do. I tried to find k by making v = dx/dt and solving the displacement equation, which turned out to be another dead end.
Any help would be appreciated. Thanks.
I don't know about the DE, but
(average velocity)*time = distance
So,
(180 +400)/2 *t = 0.1
t = 0.1 /290 = 0.000345 sec -------------answer.
------------------------
Edit.
Ooops, sorry, that is wrong because the acceleration, dv/dt, is not constant.
Now I have to solve that through the DE. Heck.
---------------------
dv/dt = -kv^2 ---------------------(i)
dv /v^2 = -k dt
-1/v = -kt +C
v = 1/(kt +C) ----------------------(ii)
When v=400m/sec, t=0,
So, in (i), 400 = 1/(k*0 +C)
C = 1/400 = 0.0025
Hence, v = 1/(kt +0.0025) -----------(iii)
Develop that,
kt +0.0025 = 1/v
kt = 1/V -0.0025
kt = (1 -0.0025v)/v
k = (1 -0.0025v) /(vt) ---------------(iv)
Substitute that into (i)
dv/dt = -[(1 -0.0025v) /(vt)](v^2)
dv/dt = [(0.0025v -1)(v)] /t
dv/dt = (0.0025v^2 -v) /t
Cross multiply,
(dv)t = (0.0025v^2 -v)dt
dv /(0.0025v^2 -v) = dt /t -------------(v)
According to the Wolfram Integrator,
INT[1 /(0.0025x^2 -x)]dx = ln(400 -0.1x) -ln(x), so,
Integrate both sides of (v),
ln(400 -0.1v) -ln(v) = ln(t) +C
ln[(400 -0.1v)/v] = ln(t) +C -------------(vi)
That pesky C!
Now I'm stalled.
Heck, suppose C = 0,
[I don't know the consequence of letting C=0 here. Would the answers be wrong? I don't know. I just want to get out of this trap.]
ln[(400 -0.1v)/v] = ln(t)
(400 -0.1v)/v = t --------------------------(vii)
When the bullet comes out, the v=180 m/sec, so,
(400 -0.1*180)/180 = t
t = 382/180 = 2.12 sec.
Therefore, the bullet will pass through the wood in 2.12 seconds after it hit the wood. ----answer.
• Jan 23rd 2007, 05:30 AM
topsquark
Quote:
Originally Posted by machi4velli
I have been struggling with this problem for a while:
A bullet passes through a piece of wood 0.1 m thick. It enters the wood at 400 m/s and leaves at 180 m/s. The velocity v of the bullet is assumed to obey the differential equation dv/dt = -kv^2 where k is a positive constant. How long does it take the bullet to pass through the wood?
I got this far:
dv/dt = -kv^2
-1/v = -kt + C
v = 1/(kt + C)
400 = v(0) = 1/(k(0) + C)
400 = 1/C
.0025 = C
v = 1/(kt + .0025)
From here, I am not sure what to do. I tried to find k by making v = dx/dt and solving the displacement equation, which turned out to be another dead end.
Any help would be appreciated. Thanks.
$\displaystyle \frac{dv}{dt} = -kv^2$
$\displaystyle \frac{dv}{v^2} = -kdt$
$\displaystyle \frac{-1}{v} = -kt + C$
So
$\displaystyle v = -\frac{1}{C - kt}$
$\displaystyle v = \frac{1}{kt - C}$ <-- My C is the negative of yours, is the only difference.
So v = 400 at t = 0, thus
$\displaystyle 400 = \frac{1}{-C}$
$\displaystyle C = -\frac{1}{400}$
Thus
$\displaystyle v = \frac{1}{kt - \frac{-1}{400}} = \frac{400}{400kt + 1}$
We still have that pesky k.
$\displaystyle v = \frac{dx}{dt}$
$\displaystyle dx = v dt = \frac{400}{400kt + 1} dt$
$\displaystyle \int_{x_0}^x dx = \int_0^y \frac{400dt}{400kt + 1}$ <-- y is the time at which the bullet leaves the wood.
$\displaystyle x - x_0 = 0.1 = 400 \int_0^y \frac{dt}{400kt + 1}$
Let $\displaystyle T = 400kt + 1$ ==> $\displaystyle dT = 400k dt$
$\displaystyle 0.1 = 400 \int_{1}^{400ky + 1} \frac{\frac{dT}{400k}}{T}$
$\displaystyle 0.1 = \frac{1}{k} \int_{1}^{400ky + 1} \frac{dT}{T}$
$\displaystyle 0.1 = \frac{1}{k} \cdot ln(T)|_1^{400ky+1}$
$\displaystyle 0.1k = ln(400ky+1) - ln(1) = ln(400ky+1)$
So
$\displaystyle 0.1k = ln(400ky + 1)$
We can get another equation in k and y by using the v(t) equation. We know that v = 180 when t = y, so
$\displaystyle 180 = \frac{400}{400ky + 1}$
We have two equations in two unknowns.
Solve the v equation for $\displaystyle 400ky + 1$:
$\displaystyle 400ky + 1 = \frac{400}{180}$
Thus the ln equation says:
$\displaystyle 0.1k = ln \left ( \frac{400}{180} \right )$
This gives a value for k, and you can use that in the v equation to get a value for y. I get that k = 7.98508 and y = 0.000383 s.
-Dan
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# 7.10: Problems
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Consider the inverter shown below.
Unlike in a conventional CMOS inverter, in this device, the supply voltage, $$V_{R}$$, adjusts during the switching operation. Initially the voltage on the output capacitor is zero, but at t = 0 the input voltage drops to zero. Also at t = 0, the supply voltage ramps from zero to the logic high voltage, $$V_{DD}$$.
Assume that the PMOS FET is modeled by a resistor, R.
(a) Show that the energy dissipated during the switching operation is
$$E = \frac{RC}{\tau} CV_{DD}^{2} \text{ for } \tau \gg RC$$
This is known as an adiabatic switch, since switching occurs (in the limit) with no energy dissipation, i.e. we are adding charge to a capacitor using a vanishingly small excess voltage.
[Hint: You may assume $$V_{OUT}$$ of the form $$V_{OUT} = a + b\exp[-t/RC] + ct$$ where a, b, and c are constants to be determined.]
(b) Show also that the energy dissipated reduces to the standard CMOS switching energy $$E = \frac{CV_{DD}^{2}}{2} \text{ for } \tau \gg RC$$
(c) The above example shows adiabatic switching when the capacitor voltage changes from low to high. Can it be implemented generally? i.e. consider the case when the capacitor voltage changes from high to low. And what happens when the capacitor does not change voltage during a cycle?
Q2. Cellular Automata
This question refers to a proposed architecture for molecular electronics: Molecular Quantum-Dot Cellular Automata. The figures are drawn from the reference.
In this architecture information is stored in bistable cells. An example cell is shown below:
This cell consists of four electron traps positioned at the corners of a square. Only two of the traps are filled. From electrostatics, there are two stable states with the electrons at opposing corners of the square.
To transmit information, the cells are placed in a line. Information then propagates electrostatically, without current flow. It is argued that power dissipation is therefore eliminated and no interconnecting wires are required.
By changing the topology, it is possible to make logic gates. For example, below we show an inverter.
(a) A proposed "majority gate" is shown below. The output Z is the majority of the inputs, A, B and C. i.e. if there are more 1 inputs than zero inputs then Z = 1, otherwise Z = 0. Use this gate to design a two input AND gate.
(b) Is the majority gate truly dissipationless? Hint: calculate the entropy before and after a majority decision.
Reference: Lent, “Bypassing the transistor paradigm” Science 288 1597 (2004)
Q3. Power delay products at the nanoscale
The power delay product is the minimum energy dissipated per bit of information processed. For a CMOS inverter the PDP is:
$PDP = CV^{2} \nonumber$
where V is the supply voltage and C is the load capacitance as seen by the inverter. In this question, we will assume that the supply voltage is fixed.
(a) Determine the load capacitance as a function of the gate and quantum capacitances. Assume we can neglect all other capacitances.
(b) Consider a 2d field effect transistor (where $$C_{Q} \rightarrow \infty$$). If its dimensions are scaled by a factor s, how does the PDP scale?
(c) Now consider a quantum wire field effect transistor with $$C_{Q}\ll C_{G}$$. Its gate capacitance is given by
$C_{G}=2\pi \varepsilon \frac{l}{\log(r/a_{0})} \nonumber$
where $$\varepsilon$$ is the dielectic constant of the gate insulator, l is the gate length, r is the gate radius and $$a_{0}$$ is the 1d wire radius.
Assume that l and r are scaled by a factor s, how does the gate capacitance for a quantum wire field effect transistor scale?
(d) Now consider the impact of the quantum capacitance on the PDP on the quantum wire field effect transistor. How does the overall PDP scale? Is the scaling faster or slower than the equivalent PDP using large quantum well field effect transistors?
Q4. Mechanical transistors
Consider a mechanical switch.
The conductor is pulled towards the gate electrode when $$|V_{GS}|>|V_{TS}|$$, switching the device on, and towards the threshold electrode when $$|V_{GS}|<|V_{TS}|$$ switching the device off. Assume two switches are wired together in a complementary logic circuit that drives a capacitive load as shown below.
(i) Plot steady state $$V_{OUT}$$ versus $$V_{IN}$$, where $$V_{IN}$$ ranges from 0 to 5V. Show that the circuit is complementary.
(ii) Assume $$V_{IN}$$ is switched from 0V to 5V and then back to 0V. How much energy is dissipated?
(iii) Consider one of the switches. Let $$C_{T}^{ON}$$ and $$C_{G}^{ON}$$ be the threshold-conductor capacitance, and the gate-conductor capacitance, respectively, in the ON state, and let $$C_{T}^{OFF}$$ and $$C_{G}^{OFF}$$ be the capacitances, respectively, in the OFF state. See the figure below.
What is the energy stored in these capacitors in the (a) ON and in the (b) OFF positions as a function of $$V_{GS}$$ and $$V_{TS}$$?
Now connect N switches all wired in parallel.
Each switch has $$V_{TS} = +1V$$ and resistance, $$R=100\Omega$$. Assume all the gate electrodes are wired together at a potential $$V_{GS}$$. To simplify the analysis assume that $$C_{G}^{ON} \gg C_{T}^{ON}$$ and also that $$C_{T}^{OFF} \gg C_{G}^{OFF}$$. Furthermore, take $$C_{G}^{ON} = C_{T}^{OFF} = C$$.
(iv) Considering Boltzmann statistics, and the potential energy difference between the OFF and ON states, out of the N switches, what is the probable number of switches that are ON as a function of C, $$V_{GS}$$ and $$V_{TS}$$ when $$|V_{GS}|<|V_{TS}|$$?
(v) Find I for the N switches as a function of $$V_{GS}$$ and $$V_{TS}$$ for $$0 < V_{GS} < 5V$$ (for $$V_{TS} = 1V$$).
(vi) Does the mechanical switch exhibit any benefit over conventional CMOS?
Q5.
(a) Consider two identical balls each 1cm in diameter and of mass m = 1g. One is kept fixed, and the second is dropped directly on it from a height of d = 10cm. From the uncertainty principle alone, what is the expected number of times the moving ball bounces on the stationary ball before it misses the latter ball altogether? Assume the ball is dropped from an optimal initial state.
Hint: some parts of this problem can be solved classically.
(b) Discuss the implications of (i) for billiard ball computers.
Q6. The following question refers to ion channel mechanical switches at T = 300K.
a) Assume that any given ion channel is either open with conductance $$G = G_{0}$$, or closed with conductance $$G = 0$$. Using Boltzmann statistics, write an expression for the conductance of a giant squid axon (with N ion channels in parallel) as a function of the applied membrane potential, V. Assume that the number of open channels at $$V = 0$$ is $$N_{0}$$.
Hint: Given Boltzmann statistics, the relative populations $$N_{1}$$ and $$N_{2}$$ of two states separated by energy dU are $$N_{1}/N_{2} = \exp(-dU/kT)$$.
b) Where possible given the data in Figure 7.5.8, evaluate your parameters.
c) Sketch a representative IV of a single ion channel.
7.10: Problems is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
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Question 72
# If x and y are non-negative integers such that $$x + 9 = z$$, $$y + 1 = z$$ and $$x + y < z + 5$$, then the maximum possible value of $$2x + y$$ equals
Solution
We can write x=z-9 and y=z-1 Now we have x+y< z+5
Substituting we get z-9+z-1<z+5 or z<15
Hence the maximum possible value of z is 14
Maximum value of "x" is 5 and maximum value of "y" is 13
Now 2x+y = 10+13=23
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# Wave problem
#### tinksy
I have a question regarding waves which has been bugging me for a while:
"You are given a travelling wave with an equation of the form
cos(ax+/-bt)
where x and t are position and time as usual, a and b are positive numbers. Explain how you would physically determine the direction that the wave is travelling in"
Here are my ideas:
I'm assuming that we only need to determine the 'direction' ie positive or negative direction in which the wave is traveling, so it's one dimensional. This means we can use the Doppler Effect. If we choose an arbitrary direction on the line to travel, the direction which gives an apparently lower frequency than expected is going away from the source, and the direction which gives a higher frequency is going towards the source. Since the wave always travels away from the source, we can determine the direction of wave propagation.
Is this right?
Related Introductory Physics Homework Help News on Phys.org
#### Tom Mattson
Staff Emeritus
Gold Member
The Doppler effect is one way to do it, but don't forget that if we are talking about something other than light waves, we have to transform the velocity of the wave as well.
A more straightforward way of doing it would be to note that, for plane waves, the phase of the wave is a constant. From this, we can deduce the direction of the wave velocity.
y(x,t)=cos(kx+ωt)
kx+ωt=constant
k(dx/dt)+ω=0
dx/dt=-ω/k
Since dx/dt<0, this wave is traveling in the negative x-direction.
y(x,t)=cos(kx-ωt)
kx-ωt=constant
k(dx/dt)-ω=0
dx/dt=+ω/k
Since dx/dt>0, this wave is traveling in the positive x-direction.
#### tinksy
cheers tom
so would that be considered a physical way of determining the direction of the wave???
#### Tom Mattson
Staff Emeritus
Gold Member
Yes, it would--but so would yours.
The difference is, my way would be considered the easy way!
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### Home > GB8I > Chapter 10 Unit 11 > Lesson INT1: 10.1.2 > Problem10-35
10-35.
The Mountain Bike Club has $\475$ in the treasury. Sarah, the president, plans to buy hats or T-shirts for the members. If hats cost $\5$ and T-shirts cost $\8$, write an inequality to represent the possible number of hats and T-shirts that she could purchase. Be sure to define your variables.
$\5h+\8t\le475$
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# how to define a circular helix in terms of the Frenet Frame
Suppose $\alpha: I \rightarrow \mathbb{R}^3$ is a regular curve with unit speed which is also a circular helix. How can we define this carefully in terms of the standard $T,N,B$ frame along the curve ?
I have a few ideas, but I want a second or third opinion. This is to help with a particular homework problem I'm working through in Barrett O'Neill's Elementary Differential Geometry text. A different definition in terms of curvature and torsion is given. I'm not interested in that definition as the whole point of the problem is to show the constant curvature and torsion imply circularity of the given helix.
Thanks in advance for your help, I hope check this in a couple hours.
• Define what carefully? Define "being a circular helix"? I'd say $\gamma(t) = (A \cos (t), A \sin (t), Bt)$ is a circular helix as long as $A \ne 0$ and $B \ne 0$, as is any curve that's a translation or rotation of this one. Is that sufficient? Alternatively, you could say that $t \mapsto \gamma(t)$ is a circular helix if there's an orthonormal frame $\vec{d}, \vec{e}, \vec{f}$, a line $h(t) = P + t \vec{d}$ and constants $A,B$ such that $u(t) = \gamma(t) - h(t)$ is constant length, $u(t) \cdot \vec{e} = A \cos(Bt)$ and $u(t) \cdot \vec{f} = A \sin (Bt)$. The first seems better to me. – John Hughes Feb 5 '14 at 19:03
• @John Well, the first definition is not very geometric, I mean, yes, of course, there exist coordinates for which $\gamma$ can be so-decribed, but I'm really looking for a vector-geometric formulation. So, your second idea is more in the direction I'm interested. I gather you mean for the projections on $T$ and $B$ to give constant length... reading between the lines a bit. I need to write the formulas in terms of $T,N,B$ as to make the derivatives connect with the Frenet Serret formulas. Thanks for the comment. – James S. Cook Feb 5 '14 at 20:38
Following the carifying comments above, let's go with the second definition. The idea there is that the curve's a circular helix if there's an axis ($P + td$) such that the projection of the curve onto the plane perpendicular to the axis traverses a circle of radius $A$ with speed $AB$ in that plane. That seems to capture circular helix-ness.
Now suppose you have a curve with constant curvature and torsion. To prove it matches the definition, you need to come up with the axis and the constants $A$ and $B$. Well, if you think of a model circular helix around the $z$-axis, say, the normal vector always lies in the $xy$-plane. So you could "discover" the axis by taking any two nearby normals and computing their cross-product. Of course, the more nearby they are, the shorter the cross-product vector will be, so to get a nice unit vector for defining the axis, I'm implicitly suggesting that you say this:
Let $$\vec{d} = \lim_{h \to 0} \frac{1}{h} N(h) \times N(0)$$.
You can rewrite that as \begin{align} \vec{d} &= \lim_{h \to 0} \frac{1}{h} (N(h) - N(0)) \times N(0) \\ &= \left( \lim_{h \to 0} \frac{1}{h} (N(h) - N(0)) \right)\times N(0) \\ &= N'(0) \times N(0) \end{align} Now you can use the Frenet-Serret formulas to work out an explicit value for the vector $\vec{d}$. (Note: The result won't generally be a unit vector, but it'll have nonzero length, and then you can make it a unit vector; without the division by $h$, you'd end up with a zero vector.)
Going back to the model of a circular helix around the $z$-axis, you should be able to infer the radius of the circle from the constant curvature and torsion. (Should be 1/curvature, I believe, but I'd have to check whether torsion somehow sneaks in there). That lets you pick a point $P$ at which to start the axis:
$$P = \gamma(0) + r N(0)$$ where $\gamma$ is your curve, and $r$ is the radius you computed. For vectors $\vec{e}$ and $\vec{f}$, I'd recommend $N(0)$ and $\vec{d} \times N(0)$. Now all you have to do is show that everything matches up as needed.
Since you didn't ask for a complete proof, but instead for a definition of circular helix that would allow you to prove that constant-curvature-and-torsion curves were indeed circular helices, I think I've given you what you asked for. I can fill in details if necessary, but that might take away the fun from you.
Following comments and discussion Here's a more complete answer that depends only on the curve $\alpha$, and its Frenet frame. Since comments reveal that you're teaching (or taught) a DG course, I'll let you do the computations here, and just post the main results.
1. The helical curve $$H(s) = (r \cos us, r \sin us, v)$$ has tangent vector $$T(s) = (-ru \sin us, ru \cos us, v)$$ which is a unit vector only if $$r^2 u^2 + v^2 = 1.$$ So I'll consider only triples $(r, u, v)$ that satisfy that condition. I'll also restrict to $r > 0$.
The derivative of $T$ is $$T'(s) = (-ru^2 \cos us, -ru^2 \sin us, 0)$$ whose length is $ru^2$; by the Frenet formulas, this is the curvature, $\kappa$, of the helix, and the normal vector to the helix is $$N(s) = (-\cos us , \sin us, 0).$$ The derivative of the normal is $$N'(s) = u(\sin us, -\cos us, 0).$$
The binormal is $$B(s) = T \times B = (v \sin ut, -v \cos ut, ru).$$ The torsion, by Frenet, is $\tau = N' \cdot B = uv$.
So for such helical curves, we have
\begin{align} ru^2 &= \kappa \\ r^2u^2 + v^2 &= 1 \\ uv &= \tau. \end{align} We can solve these for $r, u, v$ to get \begin{align} u &= \sqrt{\kappa^2 + \tau^2}\\ v &= \frac{\tau}{\sqrt{\kappa^2 + \tau^2}} \\ r &= \frac{\kappa}{\kappa^2 + \tau^2} \end{align}
Thus for any positive $\kappa$ and any $\tau$, we can find a unit-speed helical curve $s \mapsto H_{\kappa, \tau}(s)$ by using the values of $u, v, r$ above in $H$.
Now let's look at the constant-speed, constant torsion curve $\alpha$. Let $Q = \alpha(0)$, and let $u, v, r$ be derived from the known curvature and torsion as above. $$\newcommand{bT}{\mathbf T} \newcommand{bN}{\mathbf N} \newcommand{bB}{\mathbf B} \newcommand{Tb}{ T_\beta} \newcommand{Nb}{ N_\beta} \newcommand{Bb}{ B_\beta} \newcommand{bD}{\mathbf D} \newcommand{bE}{\mathbf E}$$ Let $\bT, \bN, \bB$ denote the unit tangent, normal, and binormal of $\alpha$ at $s = 0$.
Define \begin{align} \beta(s) &= (Q + r\bN) - r\cos(us)\bN + r\sin(us) \bE + vs \bD, \text{ where}\\ \bD &= \frac{1}{\|\bN \times N'(0)\|} \bN \times N'(0)\\ &= \frac{1}{\|\bN \times (-\kappa\bT + \tau \bB)\|} \bN \times (-\kappa\bT + \tau \bB)\\ &= \frac{1}{\|\kappa\bB + \tau \bT)\|} (\kappa\bB + \tau \bT)\\ &= \frac{1}{\sqrt{\kappa^2+ \tau^2}} (\kappa\bB + \tau \bT), \text{ and}\\ \bE = \bN \times \bD, \end{align} which you can work out in terms of $\bT, \bN,$ and $\bB$ similarly.
Then direct computation shows that
1. $\beta$ and $\alpha$ start at the same point.
2. $\beta$ has constant curvature $\kappa$.
3, $\beta$ has constant torsion $\tau$.
By the fundamental theorem of curves, $\beta$ must equal $\alpha$. Bur $\beta$ is evidently a helix with centerline $\bD$. (Indeed, it's the the result of rotating the "standard helix" $H(\kappa, \tau)$ by the matrix whose columns are $\bT, \bN, \bB$, and then translating by $(Q + r\bN)$.)
• very interesting, especially the idea about computing the normal to the plane in which the helix projects onto. I will ponder these things. Thanks for saving me fun, when I finally have had enough fun I'll post what I find. – James S. Cook Feb 6 '14 at 21:55
• The reason I did not accept the answer is that the definition given, while much closer in spirit to what I'm searching for, is still not given in terms of T,N,B of the curve. It may be that my goal is unreasonable and as soon as I'm convinced of that I will accept this answer. – James S. Cook Oct 2 '14 at 3:29
• I've now edited to express everything in terms of the Frenet frame at $s = 0$, together with $\alpha(0)$. I hope that's what you wanted. – John Hughes Oct 4 '14 at 15:38
• Thanks for the added detail, I'll try to sort through it as soon as I have some time to focus on this problem. In any event, I do intend to award the bounty here as your effort on this question is exemplary. – James S. Cook Oct 4 '14 at 15:50
• I'm curious what the final formula for $\beta (s)$ would look like in just terms of curvature, torsion, arclength and the Frenet Frame. I hate to say this at this point, but, maybe the larger lesson here is that $\vec{r}(s) = \langle R \cos s, r\sin s, ms \rangle$ is a better definition, provided, we pair that definition with the proper rigid motion concept as in Christian Blatter's answer. (of course, I'm not talking about the question I posed here, just the larger question of how best to define a helix) – James S. Cook Oct 11 '14 at 18:53
I know that the question has already been answered, but I found a definition in John Oprea's book which might be useful. You have your curve $\alpha: I \to \Bbb R^3$, parametrized by arc-length, say. If $\alpha$ is a cylindrical helix, exists a constant unit vector $\bf v$ such that $\langle {\bf T}_\alpha(s), {\bf v} \rangle = \cos \theta$. Define $\gamma: I \to \Bbb R^3$ by: $$\gamma(s) = \alpha(s) - \langle \alpha(s) - \alpha(s_0), {\bf v}\rangle {\bf v}$$ for some $s_0 \in I$. Notice that $\gamma$ need not be parametrized by arc-length, despite its parameter being $s$. The curve $\gamma$ is the projection of $\alpha$ into the plane orthogonal to $\bf v$ passing through $\alpha(s_0)$. You can prove, for example, that $\kappa_\gamma = \kappa_\alpha/\sin^2\theta$. We say that $\alpha$ is a circular helix if $\gamma$ is a circle.
• Thanks for this, I was waiting for my students to solve this last semester, but, they were lazy. I guess I should revisit it. There was a problem before the one which precipitated this question from Oneill which walked through the construction you mention here I think. Indeed, Problem 7 of section 2.4 precedes problem 9 which is what prompted this question. Problem 7 has the curve you mention here, Def. 4.5 is the definition you mention here also. ( I refer to the revised 2nd ed. of Oneill). – James S. Cook Oct 2 '14 at 3:36
• The drama is real. I know exactly which exercise you're referring to. Had some trouble with it not long ago, too. But O'Neill himself doesn't give the exact definition, if I recall correctly. Maybe he thought that it was obvious, from the exercise above. I didn't think so :( – Ivo Terek Oct 2 '14 at 8:37
• I understood the helix to mean a circular cross section not necessarily cylindrical. – Narasimham Oct 15 '14 at 9:48
I want to go out on a limb here, and make some observations for which I don't have proofs, but which seem likely to be true, and seem more like what James was looking for.
Go back to the "standard helix" I defined in the first answer (which I'll now call $\alpha$). Suppose at each "time" $s$, you look at the frame $T(s), N(s), B(s)$ at the origin. What are the coordinates of $\alpha(s)$ in this (ever changing) affine coordinate frame? [Credit to Tom Banchoff for teaching me to ask this question!]
First of all, the $N$ coordinate is constant! (it's exactly $-r$).
Second, the $B$ and $T$ coordinates are linear functions of $s$, with the ratio of slopes being $\tau:\kappa$, where $\tau$ and $\kappa$ are the torsion and curvature of the curve $\alpha$, of course.
I suspect (and will perhaps try to prove) that if for an arclength-paramaterized curve $\alpha$, there is a point $P$ amd constants $c_1, c_2, c_3$ such that
$$\alpha(s) - P = c_1 s T(s) + c_2 s B(s) + c_3 N(s)$$
then $\alpha$ is a helix in the sense I described earlier, with $P$ being the point of the axis closest to $\alpha(0)$ and with direction vector $N'(0) \times N(0)$, and with constant curvature and torsion in the same proportion as $c_2 : c_1$, and with radius $-c_3$.
$$\newcommand{nothing}[1]{} \nothing{abc}$$ \nothing{To do so, let me follow an idea Tom Banchoff taught me: that expressing a curve in the ever-changing Frenet coordinate system actually might be useful, and that once you've done this, you should differentiate your heart out. I'm sorry it took so long for me to realize it might have some use in this problem. Well, \begin{align} \alpha(s) - P &= c_1 s T(s) + c_2 s B(s) + c_3 N(s) \text{, so} \\ T(s) = \alpha'(s) &= c_1 T(s) + c_2 B(s) + s(c_1 T'(s) + c_2 B'(s)) + c_3 N'(s) \\ &= c_1 T(s) + c_2 B(s) + s(c_1 \kappa N(s) - c_2 \tau T(s)) + c_3 (-\kappa T(s) + \tau B(s)) \\ &= (c_1 -sc_2 \tau -\kappa c_3) T(s) + sc_1 \kappa N(s) + (c_2 + c_3 \tau) B(s) \end{align} }
The proof is actually surprisingly simple: differentiate the formula for $\alpha(s)$: \begin{align} T(s) &=\alpha'(s) \\ &= c_1 T(s) + c_2 B(s) + s(c_1 T'(s) + c_2 B'(s)) + c_3 N'(s) \\ &= c_1 T(s) + c_2 B(s) + s(c_1 \kappa N(s) - c_2 \tau N(s)) + c_3 (-\kappa T(s) + \tau B(s)) \\ &= (c_1 -\kappa c_3) T(s) + s(c_1\kappa-c_2 \tau) \kappa N(s) + (c_2 + c_3 \tau) B(s) \end{align} Both the left and right hand sides are vectors written in the $TNB$ basis, so we can equate coefficients, to get \begin{align} c_1 - \kappa c_3 &= 1 \\ s(c_1\kappa - c_2\tau) &= 0 \text{ for all $s$}\\ c_2 + c_3 \tau &= 0. \end{align} The first equation means that $\kappa$ is constant; from the second, we can then conclude that $\tau$ is constant, and that $c_1:c_2 = \tau:\kappa$.
From the proof given above, in my earlier answer, we can therefore conclude that the curve is a circular helix.
In short:
The conjecture above characterizing helical curves is correct, so a curve $\alpha$ is a circular helix if and only iff there's a point $P$ ad constants $c_1, c_2, c_3$ such that
$$\alpha(s) - P = c_1 s T(s) + c_2 s B(s) + c_3 N(s).$$
The remaining details about $P$'s relation to $\alpha(0)$, and the relationship of the axis to $N'(t) \times N(t)$, I leave to you.
Thanks for pressing for a purely-Frenet-like characterization of helices -- this was a lot of fun.
• This is helpful, btw missing an $=$ between $T(s)$ and $\alpha'(s)$. – James S. Cook Oct 11 '14 at 18:58
• Fixed; thanks very much. – John Hughes Oct 11 '14 at 19:21
• Generally, surfaces with constant torsion and constant curvature belong to an infinite set where principal curvatures, $\psi$ etc. can be varied. Here I am sending picture of a circular torus of constant $\psi$. Sorry delayed due to hurricane Hudhud. – Narasimham Oct 15 '14 at 9:17
For any smooth space curve $\gamma$ with nonvanishing curvature and parametrized by arc length $s$ one has Frenet's formulas $$\left[\matrix{\dot{\bf t}\cr\dot{\bf n}\cr\dot{\bf b}\cr}\right]= \left[\matrix{0&\kappa(s)&0\cr-\kappa(s)&0&\tau(s)\cr0&-\tau(s)&0\cr}\right]\ \left[\matrix{{\bf t}\cr{\bf n}\cr{\bf b}\cr}\right]\ ,\tag{1}$$ where $t\mapsto \kappa(t)>0$ and $t\mapsto \tau(t)$ are the curvature and the torsion along $\gamma$.
If $\gamma$ is a circular helix then for fixed $h\in{\mathbb R}$ one has $$\gamma(s+h)=T_h\gamma(s)\qquad(-\infty<s<\infty)\ ,$$ where $T_h:\>{\mathbb R}^3\to{\mathbb R}^3$ is a euclidean motion. $T_h$ is called a "Schraubung" in German, because you are rigidly screwing the helix along itself. In technical terms, $T_h=R_a\circ S_a$, where $S_a$ is a shift along a certain axis $a$, and $R_a$ is a rotation around the same axis. This implies that $\kappa$ and $\tau$ are constant for such a helix, so that $(1)$ is in fact a constant coefficient system of ODEs: $$\left[\matrix{\dot{\bf t}\cr\dot{\bf n}\cr\dot{\bf b}\cr}\right]= \left[\matrix{0&\kappa&0\cr-\kappa&0&\tau\cr0&-\tau&0\cr}\right]\ \left[\matrix{{\bf t}\cr{\bf n}\cr{\bf b}\cr}\right]\ .\tag{2}$$ Conversely: Any such system $(2)$ with given $\kappa>0$ and $\tau\ne0$ defines a circular helix up to a rigid motion. This can be seen as follows: Assume that a $\kappa>0$ and a $\tau\ne0$ are given. Then set up a helix in the form $$\gamma:\quad t\mapsto \bigl(r\cos t,r\sin t, c\>t)\tag{3}$$ with parameters $r>0$ and $c\ne0$ to be chosen appropriately. Then compute $\kappa$ and $\tau$ as functions of $r$ and $c$ according to the well known formulae, and finally adjust $r$ and $c$ such that the given $\kappa$ and $\tau$ result. The helix $\gamma$ then satisfies $(2)$, and any other curve satisfying $(2)$ for the given $\kappa$ and $\tau$ will differ from $\gamma$ by a rigid motion.
To sum it all up: A curve $s\mapsto \gamma(s)$ in ${\mathbb R}^3$ is a helix iff the curvature $\kappa>0$ and the torsion $\tau\ne0$ are constant along $\gamma$.
• I'm not sure I understand the $T_h$ euclidean motion definition. Would that condition also be true for a line? – James S. Cook Oct 11 '14 at 19:02
• @Christian Blatter: What examples can be gathered from the general ODE of (2) by consideration of various categories of surfaces apart from (3)? Cylinder/cone is given here, a torus I indicated..and e.g., what curves can be isolated from the Monge's form? – Narasimham Oct 16 '14 at 19:41
• contd..Monge form may give parallel small circles of a sphere. – Narasimham Oct 17 '14 at 10:12
Show the constant curvature and torsion imply circularity of the given helix solution".
It is not (generally) so!!
Even among geodesics on surfaces of revolution there are many choices, circular helices are not the only solution. They are given as examples because they are the easiest to model as one principal curvature is zero and also perhaps as a well known example in everyday life.
A great circle on a sphere has constant (zero) (geodesic) torsion and constant (normal) curvature.
EDIT1: Asymptotic lines on constant (zero) (normal) curvature and constant (geodesic) torsion.
EDIT2: I have placed the words (geodesic) and ( normal) in parentheses because we are talking about two things at the same time.
The Frenet-Serret frame you refer to is meant for stand alone curves in space without regard to the surface on which the curve sits. The same space curve can be made to sit on many surfaces by adjusting/choosing any normal curvature and any geodesic torsion arising from Mohr's circle of curvature. The circular helix you gave as an example is one choice among them where normal curvature and torsion are both constant. I hope I could make myself clear.
"
EDIT3: Toroidal Loxodromes with constant Kappa and Torsion:
• the question better stated would be: "among all possible helices in euclidean $3$-space, show that the only ones with constant curvature and torsion are circular" Now you worry me about my initial question... maybe we need a definition of a general helix to be logical here! – James S. Cook Oct 11 '14 at 19:06
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# math
posted by .
what is the next number in line 5,13,29,61,125
• math -
5 + 8 = 13
13 + 16 = 29
29 + 32 = 61
Do you see the pattern?
• math -
253
• math -
hhhhhh
• math -
math pattern 1,5,13,29,61 iff u answer this ill gave u 100 dollars real true promice!!!! :)
## Similar Questions
1. ### math
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2. ### math
what is the next number? 5,13,29,61,125,_?
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# StatisticsEmpirical CDF Convergence
Let's revisit the observation from the first section that the CDF of the empirical distribution of an independent list of observations from a distribution tends to be close to the CDF of the distribution itself. The Glivenko-Cantelli theorem is a mathematical formulation of the idea that these two functions are indeed close.
Theorem (Glivenko-Cantelli)
If is the CDF of a distribution and is the CDF of the empirical distribution of observations from , then converges to along the whole number line:
The Dvoretzky-Kiefer-Wolfowitz inequality quantifies this result by providing a confidence band.
Theorem (Dvoretzky-Kiefer-Wolfowitz inequality)
If are independent random variables with common CDF , then for all , we have
In other words, the probability that the graph of lies in the -band around (or vice versa) is at least .
Exercise
Use the code cell below to experiment with various values of , and confirm that the graph usually falls inside the DKW confidence band.
Hint: after running the cell once, you can comment out the first plot command and run the cell repeatedly to get lots of empirical CDFs to show up on the same graph.
using Plots, Distributions, Roots
n = 50
ϵ = find_zero(ϵ -> 2exp(-2n*ϵ^2) - 0.1, 0.05)
xs = range(0, 8, length=100)
plot(xs, x-> 1-exp(-x) + ϵ, fillrange = x-> 1-exp(-x) - ϵ,
label = "true CDF", legend = :bottomright,
fillalpha = 0.2, linewidth = 0)
plot!(sort(rand(Exponential(1),n)), (1:n)/n,
seriestype = :steppre, label = "empirical CDF")
As increases, the confidence band . Yet the proportion of CDFs that lie in the band stays .
Exercise
Show that if , then with probability at least , we have for all .
Solution. If we substitute the given value of , the right-hand side reduces to . So the desired equation follows from the DKW inequality.
Bruno
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# Count of matrices (of different orders) with given number of elements in C++
We are given the total number of elements and the task is to calculate the total number of matrices with different orders that can be formed with the given data. A matrix has an order mxn where m are the number of rows and n are the number of columns.
Input − int numbers = 6
Output −Count of matrices of different orders that can be formed with the given number of elements are: 4
Explanation − we are given with the total number of elements that a matrix of any order can contain which is 6. So the possible matrix order with 6 elements are (1, 6), (2, 3), (3, 2) and (6, 1) which are 4 in number.
Input − int numbers = 40
Output − Count of matrices of different orders that can be formed with the given number of elements are: 8
Explanation − we are given with the total number of elements that a matrix of any order can contain which is 40. So the possible matrix order with 40 elements are (1, 40), (2, 20), (4, 10), (5, 8), (8, 5), (10, 4), (20, 2) and (40, 1) which are 8 in number.
## Approach used in the below program is as follows
• Input the total number of elements that can be used to form the different order of matrices.
• Pass the data to the function for further calculation
• Take a temporary variable count to store the count of matrices of with different order
• Start loop FOR from i to 1 till the number
• Inside the loop, check IF number % i = 0 then increment the count by 1
• Return the count
• Print the result
## Example
Live Demo
#include <iostream>
using namespace std;
//function to count matrices (of different orders) with given number of elements
int total_matrices(int number){
int count = 0;
for (int i = 1; i <= number; i++){
if (number % i == 0){
count++;
}
}
return count;
}
int main(){
int number = 6;
cout<<"Count of matrices of different orders that can be formed with the given number of elements are: "<<total_matrices(number);
return 0;
}
## Output
If we run the above code it will generate the following output −
Count of matrices of different orders that can be formed with the given number of elements are: 4
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# Difference between revisions of "Quasi-Newton methods"
Author: Jianmin Su (ChemE 6800 Fall 2020)
Steward: Allen Yang, Fengqi You
Quasi-Newton Methods are a kind of methods used to solve nonlinear optimization problems. They are based on Newton's method yet can be an alternative to Newton's method when the objective function is not twice-differentiable, which means the Hessian matrix is unavailable, or it is too expensive to calculate the Hessian matrix and its inverse.
## Introduction
The first quasi-Newton algorithm was developed by W.C. Davidon in the mid1950s and it turned out to be a milestone in nonlinear optimization problems. He was trying to solve a long optimization calculation but he failed to get the result with the original method due to the low performance of computers. Thus he managed to build the quasi-Newton method to solve it. Later, Fletcher and Powell proved that the new algorithm was more efficient and more reliable than the other existing methods.
During the following years, numerous variants were proposed, include Broyden's method (1965), the SR1 formula (Davidon 1959, Broyden 1967), the DFP method (Davidon, 1959; Fletcher and Powell, 1963), and the BFGS method (Broyden, 1969; Fletcher, 1970; Goldfarb, 1970; Shanno, 1970)[1].
In optimization problems, Newton's method uses first and second derivatives, gradient and the Hessian in multivariate scenarios, to find the optimal point, it is applied to a twice-differentiable function ${\displaystyle f}$ to find the roots of the first derivative (solutions to ${\displaystyle f'(x)=0}$), also known as the stationary points of ${\displaystyle f}$[2].
The iteration of Newton's method is usually written as: ${\displaystyle x_{k+1}=x_{k}-H^{-1}\cdot \bigtriangledown f(x_{k})}$, where ${\displaystyle k}$ is the iteration number, ${\displaystyle H}$ is the Hessian matrix and ${\displaystyle H=[\bigtriangledown ^{2}f(x_{k})]}$
Iteraton would stop when it satisfies the convergence criteria like ${\displaystyle {df \over dx}=0,||\bigtriangledown f(x)||<\epsilon {\text{ or }}|f(x_{k+1})-f(x_{k})|<\epsilon }$
Though we can solve an optimization problem quickly with Newton's method, it has two obvious disadvantages:
1. The objective function must be twice-differentiable, and the Hessian matrix must be positive definite.
2. The calculation is costly because it requires to compute the Jacobian matrix, Hessian matrix, and its inverse, which is time-consuming when dealing with a large-scale optimization problem.
However, we can use Quasi-Newton methods to avoid these two disadvantages.
Quasi-Newton methods are similar to Newton's method, but with one key idea that is different, they don't calculate the Hessian matrix. They introduce a matrix ${\displaystyle B}$ to estimate the Hessian matrix instead so that they can avoid the time-consuming calculations of the Hessian matrix and its inverse. And there are many variants of quasi-Newton methods that simply depend on the exact methods they use to estimate the Hessian matrix.
## Theory and Algorithm
To illustrate the basic idea behind quasi-Newton methods, we start with building a quadratic model of the objective function at the current iterate ${\displaystyle x_{k}}$:
${\displaystyle m_{k}(p)=f_{k}(p)+\bigtriangledown f_{k}^{T}(p)+{\frac {1}{2}}p^{T}B_{k}p}$ (1.1),
where ${\displaystyle B_{k}}$ is an ${\displaystyle n\times n}$ symmetric positive definite matrix that will be updated at every iteration.
The minimizer of this convex quadratic model is:
${\displaystyle p_{k}=-B_{k}^{-1}\bigtriangledown f_{k}}$ (1.2),
which is also used as the search direction.
Then the new iterate could be written as: ${\displaystyle x_{k+1}=x_{k}+\alpha _{k}p_{k}}$ (1.3),
where ${\displaystyle \alpha _{k}}$ is the step length that should satisfy the Wolfe conditions. The iteration is similar to Newton's method, but we use the approximate Hessian ${\displaystyle B_{k}}$ instead of the true Hessian.
To maintain the curve information we got from the previous iteration in ${\displaystyle B_{k+1}}$, we generate a new iterate ${\displaystyle x_{k+1}}$ and new quadratic modelto in the form of:
${\displaystyle m_{k+1}(p)=f_{k+1}+\bigtriangledown f_{k+1}^{T}p+{\frac {1}{2}}p^{T}B_{k+1}p}$ (1.4).
To construct the relationship between 1.1 and 1.4, we require that in 1.1 at ${\displaystyle p=0}$ the function value and gradient match ${\displaystyle f_{k}}$ and ${\displaystyle \bigtriangledown f_{k}}$, and the gradient of ${\displaystyle m_{k+1}}$should match the gradient of the objective function at the latest two iterates ${\displaystyle x_{k}}$and ${\displaystyle x_{k+1}}$, then we can get:
${\displaystyle \bigtriangledown m_{k+1}(-\alpha _{k}p_{k})=\bigtriangledown f_{k+1}-\alpha _{k}B_{k+1}p_{k}=\bigtriangledown f_{k}}$ (1.5)
and with some arrangements:
${\displaystyle B_{k+1}\alpha _{k}p_{k}=\bigtriangledown f_{k+1}-\bigtriangledown f_{k}}$ (1.6)
Define:
${\displaystyle s_{k}=x_{k+1}-x_{k}}$, ${\displaystyle y_{k}=\bigtriangledown f_{k+1}-\bigtriangledown f_{k}}$ (1.7)
So that (1.6) becomes: ${\displaystyle B_{k+1}s_{k}=y_{k}}$ (1.8), which is the secant equation.
To make sure ${\displaystyle B_{k+1}}$ is still a symmetric positive definite matrix, we need ${\displaystyle s_{k}^{T}s_{k}>0}$ (1.9).
To further preserve properties of ${\displaystyle B_{k+1}}$ and determine ${\displaystyle B_{k+1}}$ uniquely, we assume that among all symmetric matrices satisfying secant equation, ${\displaystyle B_{k+1}}$ is closest to the current matrix ${\displaystyle B_{k}}$, which leads to a minimization problem:
${\displaystyle B_{k+1}={\underset {B}{min}}||B-B_{k}||}$ (1.10) s.t. ${\displaystyle B=B^{T}}$, ${\displaystyle Bs_{k}=y_{k}}$,
where ${\displaystyle s_{k}}$ and ${\displaystyle y_{k}}$ satisfy (1.9) and ${\displaystyle B_{k}}$ is symmetric and positive definite.
Different matrix norms applied in (1.10) results in different quasi-Newton methods. The weighted Frobenius norm can help us get an easy solution to the minimization problem: ${\displaystyle ||A||_{W}=||W^{\frac {1}{2}}AW^{\frac {1}{2}}||_{F}}$ (1.11).
The weighted matrix ${\displaystyle W}$ can be any matrix that satisfies the relation ${\displaystyle Wy_{k}=s_{k}}$.
We skip procedures of solving the minimization problem (1.10) and here is the unique solution of (1.10):
${\displaystyle B_{k+1}=(I-\rho y_{k}s_{k}^{T})B_{k}(I-\rho s_{k}y_{k}^{T})+\rho y_{k}y_{k}^{T}}$ (1.12)
where ${\displaystyle \rho ={\frac {1}{y_{k}^{T}s_{k}}}}$ (1.13)
Finally, we get the updated ${\displaystyle B_{k+1}}$. However, according to (1.2) and (1.3), we also need the inverse of ${\displaystyle B_{k+1}}$ in next iterate.
To get the inverse of ${\displaystyle B_{k+1}}$, we can apply the Sherman-Morrison formula to avoid complicated calculation of inverse.
Set ${\displaystyle H_{k}=B_{k}^{-1}}$, with Sherman-Morrison formula we can get:
${\displaystyle H_{k+1}=H_{k}+{\frac {s_{k}s_{k}^{T}}{s_{k}^{T}y_{k}}}-{\frac {H_{k}y_{k}y_{k}^{T}H_{k}}{y_{k}^{T}H_{k}y_{k}}}}$ (1.14)
With the derivation[3] above, we can now understand how do quasi-Newton methods get rid of calculating the Hessian matrix and its inverse. We can directly estimate the inverse of Hessian and we can use (1.14) to update the approximation of the inverse of Hessian, which leads to the DFP method, or we can directly estimate the Hessian matrix and this is the main idea in the BFGS method.
### DFP method
The DFP method, which is also known as the Davidon–Fletcher–Powell formula, is named after W.C. Davidon, Roger Fletcher, and Michael J.D. Powell. It was proposed by Davidon in 1959 first and then improved by Fletched and Powell. DFP method uses an ${\displaystyle n\times n}$ symmetric positive definite matrix ${\displaystyle B_{k}}$ to estimate the inverse of Hessian matrix and its algorithm is shown below[4].
#### DFP Algorithm
To avoid confusion, we use ${\displaystyle D}$ to represent the approximation of the inverse of the Hessian matrix.
1. Given the starting point ${\displaystyle x_{0}}$; convergence tolerance ${\displaystyle \epsilon ,\epsilon >0}$; the initial estimation of inverse Hessian matrix ${\displaystyle D_{0}=I}$; ${\displaystyle k=0}$.
2. Compute the search direction ${\displaystyle d_{k}=-D_{k}\cdot \bigtriangledown f_{k}}$.
3. Compute the step length ${\displaystyle \lambda _{k}}$ with a line search procedure that satisfies Wolfe conditions. And then set
${\displaystyle s_{k}={\lambda }_{k}d_{k}}$,
${\displaystyle x_{k+1}=x_{k}+s_{k}}$
4. If ${\displaystyle ||\bigtriangledown f_{k+1}||<\epsilon }$, then end of the iteration, otherwise continue step5.
5. Computing ${\displaystyle y_{k}=g_{k+1}-g_{k}}$.
6. Update the ${\displaystyle D_{k+1}}$ with
${\displaystyle D_{k+1}=D_{k}+{\frac {s_{k}s_{k}^{T}}{s_{k}^{T}y_{k}}}-{\frac {D_{k}y_{k}y_{k}^{T}D_{k}}{y_{k}^{T}D_{k}y_{k}}}}$
7. Update ${\displaystyle k}$ with ${\displaystyle k=k+1}$ and go back to step2.
### BFGS method
BFGS method is named for its four discoverers Broyden, Fletcher, Goldfarb, and Shanno, it is considered as the most effective quasi-Newton algorithm. Unlike the DFP method, the BFGS method uses an ${\displaystyle n\times n}$ symmetric positive definite matrix ${\displaystyle B_{k}}$ to estimate the Hessian matrix[5].
#### BFGS Algorithm
1. Given the starting point ${\displaystyle x_{0}}$; convergence tolerance ${\displaystyle \epsilon ,\epsilon >0}$; the initial estimation of Hessian matrix ${\displaystyle B_{0}=I}$; ${\displaystyle k=0}$.
2. Compute the search direction ${\displaystyle d_{k}=-B_{k}^{-1}\cdot \bigtriangledown f_{k}}$.
3. Compute the step length ${\displaystyle \lambda _{k}}$ with a line search procedure that satisfies Wolfe conditions. And then set
${\displaystyle s_{k}={\lambda }_{k}d_{k}}$,
${\displaystyle x_{k+1}=x_{k}+s_{k}}$
4. If ${\displaystyle ||\bigtriangledown f_{k+1}||<\epsilon }$, then end of the iteration, otherwise continue step5.
5. Computing ${\displaystyle y_{k}=\bigtriangledown f_{k+1}-\bigtriangledown f_{k}}$.
6. According to (1.12), (1.13) and (1.14), update the ${\displaystyle B_{k+1}^{-1}}$ with
${\displaystyle B_{k+1}^{-1}=(I-\rho s_{k}y_{k}^{T})B_{k}^{-1}(I-\rho y_{k}s_{k}^{T})+\rho s_{k}s_{k}^{T}}$ , ${\displaystyle \rho ={\frac {1}{y_{k}^{T}s_{k}}}}$
7. Update ${\displaystyle k}$ with ${\displaystyle k=k+1}$ and go back to step2.
## Numerical Example
The following is an example to show how to solve an unconstrained nonlinear optimization problem with the DFP method.
{\displaystyle {\text{min }}{\begin{aligned}f(x_{1},x_{2})&=x_{1}^{2}+{\frac {1}{2}}x_{2}^{2}+3\end{aligned}}}
${\displaystyle x_{0}=(1,2)^{T}}$
Step 1:
Usually, we set the approximation of the inverse of the Hessian matrix as an identity matrix with the same dimension as the Hessian matrix. In this case, ${\displaystyle B_{0}}$ is a ${\displaystyle 2\times 2}$ identity matrix.
${\displaystyle B_{0}}$: ${\displaystyle {\begin{pmatrix}1&0\\0&1\end{pmatrix}}}$
${\displaystyle \bigtriangledown f_{x}}$: ${\displaystyle {\begin{pmatrix}2x_{1}\\x_{2}\end{pmatrix}}}$
${\displaystyle \epsilon =10^{-5}}$
${\displaystyle k=0}$
For convenience, we can set ${\displaystyle \lambda =1}$.
Step 2:
${\displaystyle d_{0}=-B_{0}^{-1}\bigtriangledown f_{0}}$${\displaystyle =-{\begin{pmatrix}1&0\\0&1\end{pmatrix}}}$${\displaystyle {\begin{pmatrix}2\\2\end{pmatrix}}}$ ${\displaystyle ={\begin{pmatrix}-2\\-2\end{pmatrix}}}$
Step 3:
${\displaystyle s_{0}=d_{0}}$
${\displaystyle x_{1}=x_{0}+s_{0}}$${\displaystyle ={\begin{pmatrix}1\\2\end{pmatrix}}}$${\displaystyle +{\begin{pmatrix}-2\\-2\end{pmatrix}}}$${\displaystyle ={\begin{pmatrix}-1\\0\end{pmatrix}}}$
Step 4:
${\displaystyle \bigtriangledown f_{0}}$${\displaystyle ={\begin{pmatrix}-2\\0\end{pmatrix}}}$
Since ${\displaystyle |\bigtriangledown f_{0}|}$ is not less than ${\displaystyle \epsilon }$, we need to continue.
Step 5:
${\displaystyle y_{0}=\bigtriangledown f_{1}-\bigtriangledown f_{0}}$${\displaystyle ={\begin{pmatrix}-4\\-2\end{pmatrix}}}$
Step 6: ${\displaystyle B_{1}=B_{0}+{\frac {s_{0}s_{0}^{T}}{s_{0}^{T}y_{0}}}-{\frac {D_{0}y_{0}y_{0}^{T}D_{0}}{y_{0}^{T}D_{0}y_{0}}}}$${\displaystyle ={\begin{pmatrix}1&0\\0&1\end{pmatrix}}}$${\displaystyle +{\frac {1}{12}}{\begin{pmatrix}4&4\\4&4\end{pmatrix}}}$${\displaystyle -{\frac {1}{20}}{\begin{pmatrix}16&8\\8&4\end{pmatrix}}}$ ${\displaystyle ={\begin{pmatrix}0.53333&-0.0667\\-0.0667&1.1333\end{pmatrix}}}$
And then go back to Step 2 with the update ${\displaystyle B_{1}}$ to start a new iterate until ${\displaystyle |\bigtriangledown f_{k}|<\epsilon }$.
We continue the rest of the steps in python and the results are listed below:
Iteration times: 0 Result：[-1. 0.]
Iteration times: 1 Result：[ 0.06666667 -0.13333333]
Iteration times: 2 Result：[0.00083175 0.01330805]
Iteration times: 3 Result：[-0.00018037 -0.00016196]
Iteration times: 4 Result：[ 3.74e-06 -5.60e-07]
After four times of iteration, we finally get the optimal solution, which can be assumed as ${\displaystyle x_{1}=0,x_{2}=0}$ and the minimum of the objective function is 3.
As we can see from the calculation in Step 6, though the updated formula for ${\displaystyle B_{1}}$ looks complicated, it's actually not. We can see results of ${\displaystyle s_{0}^{T}y_{0}}$ and ${\displaystyle y_{0}^{T}D_{0}y_{0}}$ are constant numbers and results of ${\displaystyle s_{0}s_{0}^{T}}$ and ${\displaystyle D_{0}y_{0}y_{0}^{T}D_{0}}$ are matrix that with the same dimension as ${\displaystyle B_{1}}$. Therefore, the calculation of quasi-Newton methods is faster and simpler since it's related to some basic matrix calculations like inner product and outer product.
## Application
Quasi-newton methods are applied to various areas such as physics, biology, engineering, geophysics, chemistry, and industry to solve the nonlinear systems of equations because of their faster calculation. The ICUM (Inverse Column-Updating Method), one type of quasi-Newton methods, is not only efficient in solving large scale sparse nonlinear systems but also perfumes well in not necessarily large-scale systems in real applications. It is used to solve the Two-pint ray tracing problem in geophysics, to estimate the transmission coefficients for AIDS and for Tuberculosis in Biology, and in Multiple target 3D location airborne ultrasonic system. [6] Moreover, they can be applied and developed into the Deep Learning area as sampled quasi-Newton methods to help make use of more reliable information.[7]
Besides, to make quasi-Newton methods more available, they are integrated into programming languages so that people can use them to solve nonlinear optimization problems conveniently, for example, Mathematic (quasi-Newton solvers)[8], MATLAB (Optimization Toolbox)[9], R[10], SciPy extension to Python[11].
## Conclusion
Quasi-Newton methods are a milestone in solving nonlinear optimization problems, they are more efficient than Newton's method in large-scale optimization problems because they don't need to compute second derivatives, which makes calculation less costly. Because of their efficiency, they can be applied to different areas and remain appealing.
## References
1. Hennig, Philipp, and Martin Kiefel. "Quasi-Newton method: A new direction." Journal of Machine Learning Research 14.Mar (2013): 843-865.
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# Mr. X: Is this Geometry of Classical Mechanics?
Mr. X is in a palace of an evil villain. He is equipped with a fully-loaded special type of pistol which has four bullets in one chamber. The muzzle velocity of the gun is $$100 \text{ ms}^{-1}$$. He walks up the steps shouting out his name. Mr. X looks in one direction and sees a man rushing towards him. He lifts his gun and fires, disarming him. However he receives a blow on the head from behind and falls down the stairs.
When Mr. X looks up, he sees the villain pointing the gun on his face. The villain says, "You are the sole cause of my fall from grace Mr. X. It's time for you to say goodbye to this world". The villain pulls down the hammer of the gun.
Mr. X thinks that all is lost, but realizes that if he knows the exact momentum of the bullet, he can stop it in time by using his special powers. He remembers that the radius of the chamber is $$4 \text{cm}$$.
The bullet is in a form of a cylinder of height $$3 \text{cm}$$ mounted upon a hemisphere. The cylinder is made up of lead and the hemisphere of antimony.
What will be the momentum of the bullet in $$\text{kg ms}^{-1}$$?
Details and Assumptions
• The radius of the bullet can be found with the help of the size of the chamber. Also the cylinder and hemisphere have the same radius.
• The density of lead is $$11.34 \text{g cm}^{-3}$$ and that of antimony is $$\text{6.70 g cm}^{-3}$$.
• The momentum can be calculated using the mass of the bullet and muzzle velocity of the gun.
• Give the answer correct to 3 decimal places.
• Given Below is the diagram of the chamber and the bullet.
×
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# What is the missing number, can anyone solve this? My original puzzle
What is the missing number that should be put in place of the question mark?
3,3=17
2,5=28
3,4=24
4,5=40
5,8=88
6,11=156
6,9=?
116
The pattern/reasoning is
(a^2 + b^2) -1 = answer
• No brackets needed. Sep 13, 2019 at 6:44
• Good point. Clearly it's been a while since I worked on orders of operations. Guess it's just a little window into how I was thinking about the answer. Sep 13, 2019 at 16:00
If the three numbers in each "equation" are $$a$$, $$b$$, $$c$$ then one possible solution is
$$Min(a, 5)^2 + b^2 - 1 = c$$, giving the answer $$c = 105$$ for the last.
Edit
For good measure, the corrected version of the puzzle has the simpler solution
$$a^2 + b^2 - 1 = c$$, giving the answer $$c = 116$$ for the last.
• It doesnt work for 6,11 but fits all others. Maybe there was a typo? Sep 12, 2019 at 23:18
• It works. But I wonder if "6,11 = 146" was meant to be "6,11 = 156" which would remove the need for the whole "Min" part of it which seems a bit arbitrary?
– JS1
Sep 12, 2019 at 23:18
• yes you were right there was an error in the puzzle Sep 13, 2019 at 1:37
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How does the Earth rotate in 24 hours?
The globe spins 360 degrees in around 24 hours, which equates to 15 degrees each hour or one degree every four minutes. When it is noon at Greenwich, the time at 15 degrees east of Greenwich is 15 x 4 = 60 minutes, or one hour ahead of Greenwich time. However, at 15 degrees west of Greenwich, the time will be one hour behind Greenwich time. The planet's rotation creates a daily cycle of light and darkness, which we call day and night.
The angle that the Earth makes with the Sun varies throughout the year because it orbits the Sun. In January when the Earth is on the same side of the Sun as Mars, they travel together in an orbit that is almost exactly perpendicular to one another (85 degrees). By June when the Earth is on the opposite side of the Sun from Mars, they travel together in an orbit that is almost parallel to one another (90 degrees). The distance between them decreases during this time so by the end of June the Earth is getting closer to the Sun than it was in January.
As a result, in June and July the Earth enters into a period where there is no sunset or sunrise. For us on the ground, this means that it gets dark at some point during these months, but from up in space it looks like the whole world is illuminated all the time.
This is called "Summertime". During summertime, the amount of daylight traveling across the country varies depending on where you are located.
How many radians does the earth turn in one hour?
This spin is a whole 360 degrees. So, in one hour, the world will be 15 years old! The earth rotates one degree every four minutes. So, if it took 24 hours to rotate completely, then the earth would be alive today that time span.
There are 60 minutes in an hour and 60 seconds in a minute. So, if we want to know how many radians the earth turns in one hour, we can use the formula: 1 radian = 180 degrees. So, the answer is 1 radian = 90 degrees.
A radian is a measure of angle, equal to 1/180 of a circle. One rotation is called a full turn or complete rotation because there are 360 degrees in a circle. So, one hour equals 15 full turns or 45 partial turns of the earth.
The earth orbits the sun at about 30 miles per second, or 55,000 miles per hour. That's about 16 million feet or half a mile deep. If you went straight up on Earth, you'd reach orbit within about 8 minutes 40 seconds. But actually moving around the surface of Earth takes more time because of all the terrain and gravity.
What happens when the Earth rotates 360 degrees?
In reality, the planet revolves 360 degrees in 4 minutes less than 24 hours. This effect is caused by the Earth's orbit around the Sun, which moves one degree every day. The sidereal day refers to the 360-degree rotation. The solar day is actually about 24 hours because of atmospheric effects but we can consider it as being exactly equal to the sidereal day.
For example, if you were standing on the North Pole and watched the night sky, you would see all the stars again in the morning. This is because the Earth has rotated 180 degrees while they were out watching the night sky.
Stars that appear in the east after midnight will reappear in the west the next morning. Stars that rise due north do not move across the celestial sphere, they just remain in the same place relative to earth. They are still rising due north even though we cannot see them from over land masses.
Stars that rise due south do not move across the celestial sphere, they just remain in the same place relative to earth. They are still rising due south even though we cannot see them from over land masses.
Stars that set west of Greenwich will eventually set back over land again. However, since the rotation of the Earth is not exact, some stars do not return to their original position but wander within the constellation.
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# O1 Kinematics Overview
Mechanics
Nov 14, 2013 (4 years and 6 months ago)
115 views
KINEMATICS
Overview
Vocabulary
-
Reference frame
-
Distance
-
Displacement
-
Average velocity
-
Instantaneous velocity
-
Velocity components
-
Acceleration
-
Unifo
rm motion
-
Uniformly accelerat
ed motion
-
Projectile
-
Projectile motion’s maximum height
-
Projectile motion’s
range
-
Chapter 2 (Kinematics 1D), pages 2
-
38, Giancoli textbook
-
Chapter 3 (Kinematics 2D), pages 48
65, Giancoli textbook
Alternative for study
Visit websites
:
http://www.youtu
be.com/watch?v=lAcLN9uDsbY
(spoken with an accent, but covers almost everything…)
Timeline
09/09
:
Linear Ho
rizontal Motion (Kinematics Equations)
-
Communicate position of a object
-
point in a
RF
-
Distinguish between
distance
and
displacement
-
Derive kinematics equation for position and velocity
Q:
a)
Describe how an object can move, if it travels 7 m, but its displac
ement is only 5 m.
b)
Describe how an object moves if it travels 9 m, but its displacement is 0.
c)
Can an object move such that its displacement is 9 m, to the right, but the distance traveled is 0?
09/10
: Linear Horizontal Motion (practice problem solving)
-
De
rive kinematics equation for velocity
-
Draw and interpret velocity vs. time graphs
Suggested practice:
Questions 1 to 11
(
page 41
)
and
Problems
2, 5, 7, 9, 13, 16
(pages 42
-
43)
09/11
:
Linear Horizontal Motion
(more graphical analysis)
-
Derive kinematics equa
tion for position and Galileo’s equation for linear accelerated motion
-
Draw and interpret position vs. time graphs
Suggested practice:
Problems 18, 20,
23,
24
(page 43).
Challenge:
Problems 29, 30, 31
(pages 43
-
44).
09/13
:
More Linear Motion…
-
Short Quiz
-
Lab “Walk the Graph
(subject to change, if the equipment is not avail
able)
-
Free Fall (
equations of linear accelerated motion for specific conditions of free fall, vertical
upward throw, vertical downward throw
)
Suggested practice:
Questions 15, 16, 17
(p
ages 41
-
42)
09/16
: Vertical Linear Motion
-
Discuss possible misconceptions regarding motion in gravitational field
Suggested practice:
Problems 34, 36, 38, 47
(page 44)
09/17
: Challenging problems on Linear M
otion
-
Problem solving strategies for AP Exa
m
.
Suggested practice:
Problems 42, 49, 50, 60, 62
(pages 44
-
46).
09/18
: 2D Kinematics
Equations
-
Short Quiz
-
Derive velocity and position equations for the horizontal and vertical direction
of projectile motion
-
Lab “Projectile Motion”
-
Suggested practice:
Questions 2, 5, 6, 8
(page 70) and
Problems 2, 4, 5, 7, 10
(pages 70
-
71)
Also,
Question 11
(page 70) and
Problems 19, 2
1, 25, 27, 32
(page 72)
Projectile Motion simulation:
http://www.physicslessons.com/exp7b.htm
09/20
: C
hallengi
ng problems on Projectile Motion
-
Problem solving strategy for AP Exam
Suggested practi
ce:
Problems 37, 38, 39, 70
(pages 75
-
76)
09/23
: Kinematics Review
-
MC and FR practice
-
Suggested practice:
Kinematics on
-
line tests
http://www.aw
-
bc.com/giancoli/
09/24
:
Test on Kinematics
(subject to c
ha
nge if you need more practice
)
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3.5 Phase Space Evolution
Most problems do not have enough symmetries to be reducible to quadrature. It is natural to turn to numerical integration to learn more about the evolution of such systems. The evolution in phase space may be found by numerical integration of Hamilton's equations.
Hamilton's equations are already in first-order form; the Hamiltonian state derivative is the same as the phase-space derivative:
(define Hamiltonian->state-derivative
phase-space-derivative)
As an illustration, consider again the periodically driven pendulum (see section 1.6.2). The Hamiltonian is
(show-expression
((Lagrangian->Hamiltonian
(L-periodically-driven-pendulum 'm 'l 'g 'a 'omega))
(up 't 'theta 'p_theta)))
Hamilton's equations for the periodically driven pendulum are unrevealing, so we will not show them. We build a system derivative from the Hamiltonian:
(define (H-pend-sysder m l g a omega)
(Hamiltonian->state-derivative
(Lagrangian->Hamiltonian
(L-periodically-driven-pendulum m l g a omega))))
Now we integrate this system, with the same initial conditions as in section 1.7 (see figure 1.7), but displaying the trajectory in phase space (figure 3.9). We make a monitor procedure to display the evolution in phase space:
(define ((monitor-p-theta win) state)
(let ((q ((principal-value :pi) (coordinate state)))
(p (momentum state)))
(plot-point win q p)))
We use evolve to explore the evolution of the system:
(define window (frame :-pi :pi -10.0 10.0))
(let ((m 1.) ;m=1kg
(l 1.) ;l=1m
(g 9.8) ;g=9.8m/s2
(A 0.1) ;A=1/10 m
(omega (* 2 (sqrt 9.8))))
((evolve H-pend-sysder m l g A omega)
(up 0.0 ;t0=0
0.0) ;p0=0 kg m2/s
(monitor-p-theta window)
0.01 ;plot interval
100.0 ;final time
1.0e-12))
The trajectory sometimes oscillates and sometimes circulates. The patterns in the phase plane are reminiscent of the trajectories in the phase plane of the undriven pendulum shown in figure 3.4.
3.5.1 Phase-Space Description Is Not Unique
We are familiar with the fact that a given motion of a system is expressed differently in different coordinate systems: the functions that express a motion in rectangular coordinates are different from the functions that express the same motion in polar coordinates. However, in a given coordinate system the evolution of the local state tuple for particular initial conditions is unique. The generalized velocity path function is the derivative of the generalized coordinate path function. On the other hand, the coordinate system alone does not uniquely specify the phase-space description. The relationship of the momentum to the coordinates and the velocities depends on the Lagrangian, and many different Lagrangians may be used to describe the behavior of the same physical system. When two Lagrangians for the same physical system are different, the phase-space descriptions of a dynamical state are different.
We have already seen two different Lagrangians for the driven pendulum (see section 1.6.4): one was found using L = T - V and the other was found by inspection of the equations of motion. The two Lagrangians differ by a total time derivative. The momentum p conjugate to depends on which Lagrangian we choose to work with, and the description of the evolution in the corresponding phase space also depends on the choice of Lagrangian, even though the behavior of the system is independent of the method used to describe it. The momentum conjugate to , using the L = T - V Lagrangian for the periodically driven pendulum, is
and the momentum conjugate to , with the alternate Lagrangian, is
The two momenta differ by an additive distortion that varies periodically in time and depends on . That the phase-space descriptions are different is illustrated in figure 3.10. The evolution of the system is the same for each.
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## November 11, 2008
### Problem for Wednesday
Hello everyone-
I'll start lecture on Wednesday by discussing problem 7, which is very similar to the ruler problem we did in class on Monday.
jay-
## November 6, 2008
### Homework 9 key; problem for tomorrow
Hello everyone-
The key for HW 9 is posted; I'll post HW 10 problems first thing tomorrow.
Tomorrow, I am going to start the lecture by discussing the following problem:
You fire a bullet, of mass 9.00g, at 860m/s into a wooden door of mass 18kg and width 0.85m. The bullet hits the door at a right angle, 0.75cm away from the hinges (the axis of rotation), and becomes lodged in the wood. The hinges are frictionless, of course. What is the resulting angular velocity of the door?
## October 24, 2008
### A problem for Monday
Hello everyone-
Here's the schedule for the next few days:
Monday 27 October Chapter 10.1-4 (Rotational Kinematics)
Wednesday 29 Oct Chapter 10.5-10 (Moment of Inertia) (Homework 8 due, Key released)
Thursday 30 Oct Review of chapters 5-9
Friday 31 October Exam II, covering chapters 5-9.
I appear to be about a day behind where I'd planned on being.
For Monday, a good problem to get you on track will be 10.6.
jay-
## October 17, 2008
### Monday's problem
On Monday, we're going to start talking about Momentum and related topics. Please read sections 9.1-9.4.
We will discuss Problem 9.3 in class as well.
Another good practice problem from 8 is 8.53.
### Arrrrrrrgh.....
Okay, I found the sign error. We were looking at #8.59. I noted that there is a change in the mechanical energy of the system, due to energy being lost to friction. The mechanical energy is initially all kinetic (K1) and at the end state is all spring potential (U2).
So we know that
Wf=Emec2-Emec1=U2-K1
And the problem asks for the initial velocity of the mass, which I can figure out given the Initial kinetic energy:
K1=U2-Wf
We determined Wf to be -0.46J (friction does negative work). Where I got flummoxed was that I had dropped in -0.9J for the spring potential energy, whereas that is actually the work done by the spring- the spring potential HAS to be positive: +0.9J. Drat.
Anyway, the initial KE then is
K1=U2-Wf=(+0.9J) - (-0.46J) = 1.35J
From which I can calculate the initial velocity: 1.0m/s.
## October 15, 2008
### problem for tomorrow
We'll start tomorrow with a discussion of problem #22 in chapter 8.
jay-
## October 12, 2008
### Problem for Monday
I'll start the lecture on Monday with a discussion of problem 65 from Chapter 7.
## October 2, 2008
### Problem for tomorrow
I'll start tomorrow's lecture with a discussion of problem 23 from chapter 6.
If somebody lost a calculator in class today (Thursday) let me know.
## October 1, 2008
### Practice problem for tomorrow
Tomorrow, I will start the lecture by going through problem 32 in chapter 5. You're not required to do it but you will find the discussion of the problem much more useful if you do!
jay-
## September 26, 2008
### Net Monday
Hello all-
I've started grading the exams- hopefully I'll be able to finish this weekend and get grades posted by tuesday at the latest. I will hand them back on Wednesday, most likely. I'll try to post a key as well.
Another professor will be filling in for me on Monday. I am going to be down at UW-Madison giving a lecture.
Have a great weekend!
jay-
## September 22, 2008
### Tonight's problem
Hello everyone-
I forgot to mention the problem we'll talk about in class tomorrow- we'll start tomorrow's lecture with a discussion of #64 in chapter 4.
Please PLEASE take a look at chapter 5 tonight as well.
Ive posted the equation sheet for exam 1. You are not allowed to bring a print-out into the exam; however, you will get the EXACT same sheet handed out with the exam. This is what you'll have available to you.
## September 19, 2008
### Problem solving
Again, what I'd like to do is pick one book problem a day and give you a chance to solve it at home. I'll go over it first thing in the next day's lecture.
The problem I'll present today (Friday, 19 September) will be chapter 4, #44.
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SEARCH HOME
Math Central Quandaries & Queries
Question from bill, a student: The area of the rectangle is 170 sq.ft. the width is unknown and the length is the width plus 7 ft. I set it up like this w(w+7)=170 and this is where I get lost. Thanks, Bill
Hi Bill,
With w(w + 7) = 170 the next step should be to expand the left side. Using the dissimulative law
w(w + 7) = w × w + w × 7 = w2 + 7w
Hence you have
w2 + 7w = 170
Next subtract 170 from each side
w2 + 7w - 170 = 170 - 170
w2 + 7w - 170 = 0.
The task now is to factor w2 + 7w - 170, that is to find numbers a and b so that
w2 + 7w - 170 = (x + a)(x + b)
Notice that if you expand the right side you get
(x + a)(x + b) = x2 + (a + b)x + ab
Hence to have
w2 + 7w - 170 = (x + a)(x + b) = x2 + (a + b)x + ab
you want numbers a and b so that ab = 170 (so one is positive and one is negative) and a + b = 7.
Can you find a and b and complete the problem?
If you need further assistance write back,
Penny
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
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CoCalc Public Filesdiff-functions.sagews
Author: Harald Schilly
Views : 269
Compute Environment: Ubuntu 18.04 (Deprecated)
Differentiating functions in SageMath
http://doc.sagemath.org/html/en/tutorial/tour_algebra.html#solving-differential-equations
x = var('x')
a = function('a')(x)
b = function('b')(x)
ab1 = diff(a * b, x)
show(ab1)
$\displaystyle b\left(x\right) \frac{\partial}{\partial x}a\left(x\right) + a\left(x\right) \frac{\partial}{\partial x}b\left(x\right)$
show(diff(2 * a + 3 * b^2, x))
$\displaystyle 6 \, b\left(x\right) \frac{\partial}{\partial x}b\left(x\right) + 2 \, \frac{\partial}{\partial x}a\left(x\right)$
# chain rule.
# be explicit about what x is, i.e. a(x= ... )
t = var('t')
show(diff(a(x = t^2 + b(x = x^2)), x))
$\displaystyle 2 \, x \mathrm{D}_{0}\left(a\right)\left(t^{2} + b\left(x^{2}\right)\right) \mathrm{D}_{0}\left(b\right)\left(x^{2}\right)$
ex3 = (a - 2*b)^3 / (b(x = 2*a(x=x)))^2
show(ex3)
$\displaystyle \frac{{\left(a\left(x\right) - 2 \, b\left(x\right)\right)}^{3}}{b\left(2 \, a\left(x\right)\right)^{2}}$
show(diff(ex3, x))
$\displaystyle -\frac{4 \, {\left(a\left(x\right) - 2 \, b\left(x\right)\right)}^{3} \frac{\partial}{\partial x}a\left(x\right) \mathrm{D}_{0}\left(b\right)\left(2 \, a\left(x\right)\right)}{b\left(2 \, a\left(x\right)\right)^{3}} + \frac{3 \, {\left(a\left(x\right) - 2 \, b\left(x\right)\right)}^{2} {\left(\frac{\partial}{\partial x}a\left(x\right) - 2 \, \frac{\partial}{\partial x}b\left(x\right)\right)}}{b\left(2 \, a\left(x\right)\right)^{2}}$
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# 130 cm to inch converter
## FAQs on 130 cm to inch
### How many inches is in a centimeter?
If you wish to convert 130 cm into inches, first you need to know how many inches 1 cm equals.
Here I can give you a direct indication that one centimeter is equivalent to 0.3937 inches.
### How do you convert 1 cm into inches?
You can convert 1cm into inches by multiplying 1cm by 0.3937.
This makes it much easier to convert 130 cm to inches.
Also, 1 cm into inches = 1 cm x 0.3937 = 0.3937 inches, precisely.
This allows you to answer this question with ease and simplicity.
• What is one centimeter to inches?
• What is the cm to inch conversion?
• How many inches is equal to 1 cm?
• What is 1 cm in inches equal?
Centimeter is an International Standard Unit of Length. It is equal to one hundredth of a millimeter. It’s roughly equivalent to 39.37 inches.
### What is Inch?
Anglo-American units of length are in inches. 12 inches equals 1 foot, while 36 inches equals 1 yard. According to modern standards, one inch equals 2.54 centimeters.
### What is 130 cm converted to inches?
From the above, you have a good grasp of cm to inches.
The following is the corresponding formulas:
Value in inches = value in cm × 0.3937
So, 130 cm to inches = 130 cm × 0.3937 = 5.1181 inches
These questions can be answered using this formula:
• What is the formula to convert 130 cm to inches?
• How do I convert cm to inches?
• How to translate 130 cm into inches?
• What is standard measurement for cm to inches?
• What size are 130 cm into inches?
cm inch 129.2 cm 5.086604 inch 129.3 cm 5.090541 inch 129.4 cm 5.094478 inch 129.5 cm 5.098415 inch 129.6 cm 5.102352 inch 129.7 cm 5.106289 inch 129.8 cm 5.110226 inch 129.9 cm 5.114163 inch 130 cm 5.1181 inch 130.1 cm 5.122037 inch 130.2 cm 5.125974 inch 130.3 cm 5.129911 inch 130.4 cm 5.133848 inch 130.5 cm 5.137785 inch 130.6 cm 5.141722 inch 130.7 cm 5.145659 inch 130.8 cm 5.149596 inch
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https://web2.0calc.com/questions/let-the-term-s_n-be-the-sum-of-the-first-n-powers-of-2-for-instance-s_3-2-0-2-1-2-2-7-find-the-largest-possible-value-of
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+0
Let the term $S_n$ be the sum of the first $n$ powers of $2$. For instance, $S_3 = 2^0 + 2^1 + 2^2 = 7$. Find the largest possible value of
+3
2745
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+1821
$$Let the term S_n be the sum of the first n powers of 2. For instance, S_3 = 2^0 + 2^1 + 2^2 = 7. Find the largest possible value of the greatest common divisor of two consecutive terms, S_n and S_{n+1}, for any n.$$ Let the term $S_n$ be the sum of the first $n$ powers of $2$. For instance, $S_3 = 2^0 + 2^1 + 2^2 = 7$. Find the largest possible value of the greatest common divisor of two consecutive terms, $S_n$ and $S_{n+1}$, for any $n$.
Jun 23, 2015
#5
+117762
+13
Hi again anon, why don't you join up and take on an individual identity ? :)
Anyway, we have
$$\\S_n=2^n-1\\\\ S_{n-1}=2^{n-1}-1=2*2^n-1$$
So we want the greatest common divisor between these two terms.
It is obvious that for many values of n this will be 1 but do not know how to show that this is always the case.
Jun 25, 2015
#1
+13
I give this a try（pls someone check this afeter I done)
Let the term $S_n$ be the sum of the first $n$ powers of $2$. For instance, $S_3 = 2^0 + 2^1 + 2^2 = 7$. Find the largest possible value of the greatest common divisor of two consecutive terms, $S_n$ and $S_{n+1}$, for any $n$.
Sn=2^0+2^1+2^2+.......2^(n-1)
consider the pattern
2^0=(2^1-1)/2-1
2^0+2^1=(2^2-1)/(2-1)
2^0+2^1+2^2=(2^3-1)/(2-1)
2^0+2^1+2^2+2^3=(2^4-1)/(2-1)
.....
2^0+2^1+2^2+2^3.......+2^(n-2)+2^(n-1)=(2^n-1)/(2-1)
2^0+2^1+2^2+2^3.......+2^(n-2)+2^(n-1)+2^n=[2^(n+1)-1]/(2-1)
S{n}=2^0+2^1+2^2+2^3.......+2^(n-2)+2^(n-1)=(2^n-1)/(2-1)
S{N+1}=2^0+2^1+2^2+2^3.......+2^(n-2)+2^(n-1)+2^n=[2^(n+1)-1]/(2-1)
so the greatest common divisor would be 2-1=1
Jun 25, 2015
#2
+117762
+8
Hi anon, thanks for that answer
I have NOT checked you logic, I have only coded it so it is more easily read
Perhaps you shoulod just check that what i have written is what you intended :)
--------------------------------------------------------------------------------------------
I give this a try（pls someone check this afeter I done)
Let the term $$S_n$$ be the sum of the first $n$ powers of $2$. For instance, $$S_3 = 2^0 + 2^1 + 2^2 = 7$$
Find the largest possible value of the greatest common divisor of two consecutive terms, $S_n$ and $S_{n+1}$, for any $n$.
$$\\Sn=2^0+2^1+2^2+.......2^{n-1}\\\\ consider the pattern \\\\ 2^0=\frac{(2^1-1)}{2-1}\\\\ 2^0+2^1=\frac{(2^2-1)}{(2-1)}\\\\ 2^0+2^1+2^2=\frac{(2^3-1)}{(2-1)}\\\\ 2^0+2^1+2^2+2^3=\frac{(2^4-1)}{(2-1)}\\\\\\ 2^0+2^1+2^2+2^3.......+2^{n-2}+2^{n-1}=\frac{(2^n-1)}{(2-1)}\\\\ 2^0+2^1+2^2+2^3.......+2^{n-2}+2^{n-1}+2^n=\frac{2^{n-1}-1}{(2-1)}\\\\ S{n}=2^0+2^1+2^2+2^3.......+2^{n-2}+2^{n-1}=2^{n}-1\\\\ S{n+1}=2^0+2^1+2^2+2^3.......+2^{n-2}+2^{n-1}+2^n=2^{n+1}-1\\\\ so the greatest common divisor would be 2-1=1\\\\$$
Jun 25, 2015
#3
+10
...Thank you Melody.
I dont really know my anwser whther is right or not.I had ever taught by myself how to do this kind of question before.So,I dont really know.
....I found a mistake,the question said ,"for any $n$."when n=-1,S{n}=?
Jun 25, 2015
#4
+117762
+13
Hi anon,
n has to be greater than or equal to 1 - that is just assumed :)
Jun 25, 2015
#5
+117762
+13
Hi again anon, why don't you join up and take on an individual identity ? :)
Anyway, we have
$$\\S_n=2^n-1\\\\ S_{n-1}=2^{n-1}-1=2*2^n-1$$
So we want the greatest common divisor between these two terms.
It is obvious that for many values of n this will be 1 but do not know how to show that this is always the case.
Melody Jun 25, 2015
#6
+892
+10
It's a geometric series with first term 1 and common ratio 2, so
$$S_{n}=2^{n-1}-1,$$
and
$$S_{n+1}=2^{n}-1.$$
Suppose that both of these are exactly divisible by some integer k (say), then there will be integers say s and t such that
$$s\times k = 2^{n-1}-1, \; \text{ and }\; t\times k = 2^{n}-1.$$
Multiply the first of these equations by 2 and subtract that from the second one and you have
$$(t\times k)-(2\times s\times k) = 1,$$
or,
$$k(t-2s)=1.$$
Since k, t and s are integers, it follows that
$$k = 1 \; (\text{and }\; t - 2s=1).$$
Jun 25, 2015
#7
+117762
+5
Thanks Bertie, that is really neat. It is so easy when YOU do it.
s,k and t must all be positive integers I think. :/
Jun 25, 2015
#8
+892
+10
Nit-picker !
But you're correct, I was careless, I should have stated that k was positive.
Jun 25, 2015
#9
+124524
+10
Notice that the sum of the first n terms and the next term to be added are relatively prime, since their difference is just 1. For instance: 1 + 2 + 4 = 7. And the next term to be added is 8.
But, if A, B are relatively prime, then their sum, A + B, (the next term in the series), must be relatively prime to B, because their difference is just A. [If A and B weren't relatively prime, there would be some factor, k, that would divide them both and would also divide their sum.]
Thus, each successive term in the series is relatively prime to its successor........
Jun 25, 2015
#10
+117762
+5
Sorry Chris but that is not obvious to me :(
I do understand your's Bertie, and I just could just not resist nit-picking.
It is a very rare oportunity for me to be able to pick you up one anything and I just had to make the most of it!
Besides, mathematicians are nit-pickers, it is a part of our trade :))
Thank you both for answering. For a while I did not think anyone was going to :)
Jun 26, 2015
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# Basic circuit analysis and design. Circuit analysis. Write algebraic expressions or make a truth table
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1 Basic circuit analysis and design Circuit analysis Circuit analysis involves figuring out what some circuit does. Every circuit computes some function, which can be described with Boolean expressions or truth tables. So, the goal is to find an expression or truth table for the circuit. The first thing to do is figure out what the inputs and outputs of the overall circuit are. This step is often overlooked! The example circuit here has three inputs x, y, z and one output f. In the first two weeks we learned all the prerequisite material: Truth tables and Boolean expressions describe functions. Expressions can be converted into hardware circuits. Boolean algebra and K-maps help simplify expressions and circuits. Today we ll put all of these foundations to good use, to analyze and design some larger circuits. January 28, 22 Basic circuit analysis and design January 28, 22 Basic circuit analysis and design 2 Write algebraic expressions... Next, write expressions for the outputs of each individual gate, based on that gate s inputs. Start from the inputs and work towards the outputs. It might help to do some algebraic simplification along the way. Here is the example again. We did a little simplification for the top AND gate. You can see the circuit computes f(x,y,z) = xz + y z + x yz...or make a truth table It s also possible to find a truth table directly from the circuit. Once you know the number of inputs and outputs, list all the possible input combinations in your truth table. A circuit with n inputs should have a truth table with 2 n rows. Our example has three inputs, so the truth table will have 2 3 = 8 rows. All the possible input combinations are shown. January 28, 22 Basic circuit analysis and design 3 January 28, 22 Basic circuit analysis and design 4
2 Simulating the circuit Then you can simulate the circuit, either by hand or with a program like LogicWorks, to find the output for each possible combination of inputs. For example, when xyz =, the gate outputs would be as shown below. Use truth tables for AND, OR and NOT to find the gate outputs. For the final output, we find that f(,,) =. Finishing the truth table Doing the same thing for all the other input combinations yields the complete truth table. This is simple, but tedious. January 28, 22 Basic circuit analysis and design 5 January 28, 22 Basic circuit analysis and design 6 Expressions and truth tables Remember from the second lecture that if you already have a Boolean expression, you can use that to easily make a truth table. For example, since we already found that the circuit computes the function f(x,y,z) = xz + y z + x yz, we can use that to fill in a table: We show intermediate columns for the terms xz, y z and x yz. Then, f is obtained by just OR ing the intermediate columns. x y z xz y z x yz f Truth tables and expressions The opposite is also true: it s easy to come up with an expression if you already have a truth table. In the second lecture, we saw that you can quickly convert a truth table into a sum of minterms expression. The minterms correspond to the truth table rows where the output is. f(x,y,z) = x y z + x yz + xy z + xyz = m + m 2 + m 5 + m 7 You can then simplify this sum of minterms if desired using a K-map, for example. January 28, 22 Basic circuit analysis and design 7 January 28, 22 Basic circuit analysis and design 8
3 Circuit analysis summary After finding the circuit inputs and outputs, you can come up with either an expression or a truth table to describe what the circuit does. You can easily convert between expressions and truth tables. Find a Boolean expression for the circuit Find the circuit s inputs and outputs Find a truth table for the circuit Basic circuit design The goal of circuit design is to build hardware that computes some given function. The basic idea is to write the function as a Boolean expression, and then convert that to a circuit. Step : Figure out how many inputs and outputs you have. Step 2: Make sure you have a description of the function, either as a truth table or a Boolean expression. Step 3: Convert this into a simplified Boolean expression. (For CS23, we typically expect you to find MSPs unless otherwise stated.) Step 4: Build the circuit based on your simplified expression. January 28, 22 Basic circuit analysis and design 9 January 28, 22 Basic circuit analysis and design Design example: Comparing 2-bit numbers Let s design a circuit that compares two 2-bit numbers, A and B. The circuit should have three outputs: G ( Greater ) should be only when A > B. E ( Equal ) should be only when A = B. L ( Lesser ) should be only when A < B. Make sure you understand the problem. Inputs A and B will be,,, or (,, 2 or 3 in decimal). For any inputs A and B, exactly one of the three outputs will be. Step : How many inputs and outputs? Two 2-bit numbers means a total of four inputs. We should name each of them. Let s say the first number consists of digits A and A from left to right, and the second number is B and B. The problem specifies three outputs: G, E and L. Here is a block diagram that shows the inputs and outputs explicitly. Now we just have to design the circuitry that goes into the box. January 28, 22 Basic circuit analysis and design January 28, 22 Basic circuit analysis and design 2
4 Step 2: Functional specification Step 3: Simplified Boolean expressions For this problem, it s probably easiest to start with a truth table. This way, we can explicitly show the relationship (>, =, <) between inputs. A four-input function has a sixteenrow truth table. It s usually clearest to put the truth table rows in binary numeric order; in this case, from to for A, A, B and B. Example: <, so the sixth row of the truth table (corresponding to inputs A= and B=) shows that output L=, while G and E are both. A A B B G E L A Let s use K-maps. There are three functions (each with the same inputs A A B B), so we need three K-maps. B B A G(A,A,B,B) = A A B + A B B + A B A B A B E(A,A,B,B) = A A B B + A A B B + A A B B + A A B B A B A B L(A,A,B,B) = A A B + A B B + A B January 28, 22 Basic circuit analysis and design 3 January 28, 22 Basic circuit analysis and design 4 Step 4: Drawing the circuits Testing this in LogicWorks G = A A B + A B B + A B E = A A B B + A A B B + A A B B + A A B B L = A A B + A B B + A B Where do the inputs come from? Binary switches, in LogicWorks How do you view outputs? Use binary probes. probe LogicWorks has gates with NOTs attached (small bubbles) for clearer diagrams. switches January 28, 22 Basic circuit analysis and design 5 January 28, 22 Basic circuit analysis and design 6
5 Example wrap-up Data representations. We used three outputs, one for each possible scenario of the numbers being greater, equal or less than each other. This is sometimes called a one out of three code. K-map advantages and limitations. Our circuits are two-level implementations, which are relatively easy to draw and follow. But, E(A,A,B,B) couldn t be simplified at all via K-maps. Can you do better using Boolean algebra? Extensibility. We used a brute-force approach, listing all possible inputs and outputs. This makes it difficult to extend our circuit to compare three-bit numbers, for instance. We ll have a better solution after we talk about computer arithmetic. Summary Functions can be represented with expressions, truth tables or circuits. These are all equivalent, and we can arbitrarily transform between them. Circuit analysis involves finding an expression or truth table from a given logic diagram. Designing a circuit requires you to first find a (simplified) Boolean expression for the function you want to compute. You can then convert the expression into a circuit. Next time we ll talk about some building blocks for making larger combinational circuits, and the role of abstraction in designing large systems. January 28, 22 Basic circuit analysis and design 7 January 28, 22 Basic circuit analysis and design 8
### Review: Standard forms of expressions
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# Unlike elite runners, whose times are calculated from the
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Unlike elite runners, whose times are calculated from the starter's pistol to the finish line, marathon times are calculated using a computer chip attached to each competitor's shoe.
A) Unlike elite runners, whose times are calculated from the starter's pistol to the finish line,
B) Besides elite runners, whose times are calculated from the starter's pistol to the finish line,
C) With the exception of elite runners, times for whom are calculated from the starter's pistol to the finish line,
D) Unless the runner is elite, in which case the time is calculated from the starter's pistol to the finish line,
E) With the exception being elite runners, whose times are calculated from the starter's pistol to the finish line,
Not sure why
[Reveal] Spoiler:
D
?
[Reveal] Spoiler: OA
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Re: Unlike elite runners, whose times are calculated from the [#permalink]
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01 Feb 2013, 13:50
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vishu1414 wrote:
Unlike elite runners, whose times are calculated from the starter's pistol to the finish line, marathon times are calculated using a computer chip attached to each competitor's shoe.
A) Unlike elite runners, whose times are calculated from the starter's pistol to the finish line,
B) Besides elite runners, whose times are calculated from the starter's pistol to the finish line,
C) With the exception of elite runners, times for whom are calculated from the starter's pistol to the finish line,
D) Unless the runner is elite, in which case the time is calculated from the starter's pistol to the finish line,
E) With the exception being elite runners, whose times are calculated from the starter's pistol to the finish line,
fameatop wrote:
Hi Mike, Can u kindly tell, what are errors in incorrect options.
The part of the sentence beyond the underlined section begins with "marathon times" ----- any comparison is illogical here, because we are not comparing something else to "marathon times."
(A) Unlike elite runner ... marathon times = illogical
(B) Besides elite runners ... marathon times = illogical
The "besides" literally means on the side of something --- it is used metaphorically to separate one member from a group of similar members "Besides Harding and Ford, all 20th century US Presidents served at least one full term.", "Besides Technetium and Promethium, the first 83 elements all have stable isotopes", etc.
(C) & (E) both have passive structure ---- (E) in particular is a wordy monstrosity. Both also contain an illogical construction --- the construction, With the exception of A, B ... B has to be a category or group that includes A ---- if we drop the intervening relative clause, we get
(C) With the exception of elite runners, ..... marathon times ...
(E) With the exception being elite runners, ..... marathon times ...
Again, the sentence compares "elite runner" to "marathon times", an illogical comparison.
(D) has an active structure that is efficient, direct, and powerful. Of the five answers, it's the only that completely avoids an illogical comparison. That's why it's the best.
Mike
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Re: Unlike elite runners, whose times are calculated from [#permalink]
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23 Dec 2012, 01:46
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vishu1414 wrote:
Unlike elite runners, whose times are calculated from the starter's pistol to the finish line, marathon times are calculated using a computer chip attached to each competitor's shoe.
A) Unlike elite runners, whose times are calculated from the starter's pistol to the finish line,
B) Besides elite runners, whose times are calculated from the starter's pistol to the finish line,
C) With the exception of elite runners, times for whom are calculated from the starter's pistol to the finish line,
D) Unless the runner is elite, in which case the time is calculated from the starter's pistol to the finish line,
E) With the exception being elite runners, whose times are calculated from the starter's pistol to the finish line,
Not sure why
[Reveal] Spoiler:
D
?
A direct comparsion has to be made between times, reason being that "marathon times" follows the underlined part.
A, B and E are straightaway eliminated in that they are comparing runners with times. Illogical and disastrous.
C goes beyond all this stuff and considers the "runners" "times".
D makes the right comparison, comparing the two times. D says that till the moment the runner is elite, marathon times are calculated.
Hope that will help.
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Re: Unlike elite runners, whose times are calculated from the [#permalink]
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Re: Unlike elite runners, whose times are calculated from the [#permalink]
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Re: Unlike elite runners, whose times are calculated from the [#permalink]
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25 Feb 2016, 20:28
vishu1414 wrote:
Unlike elite runners, whose times are calculated from the starter's pistol to the finish line, marathon times are calculated using a computer chip attached to each competitor's shoe.
A) Unlike elite runners, whose times are calculated from the starter's pistol to the finish line,
B) Besides elite runners, whose times are calculated from the starter's pistol to the finish line,
C) With the exception of elite runners, times for whom are calculated from the starter's pistol to the finish line,
D) Unless the runner is elite, in which case the time is calculated from the starter's pistol to the finish line,
E) With the exception being elite runners, whose times are calculated from the starter's pistol to the finish line,
Not sure why
[Reveal] Spoiler:
D
?
the meaning of the sentence:
if the runner is not elite, the marathon times are calculated using smth.
if the runner is elite - then the time is calculated differently. so basically, we compare how the times are calculated.
A - can't be - compares elite runners with marathon times.
B - besides - out right away.
C - exception of - better exception for - so not good.
D = same as the intended meaning.
E - exception being - wordy. neither of the cases where being is used correctly is present here. so can't be correct.
D for me.
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Re: Unlike elite runners, whose times are calculated from the [#permalink]
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05 Dec 2016, 19:32
My 2 cents:
C) With the exception of elite runners, times for whom are calculated from the starter's pistol to the finish line,
D) Unless the runner is elite, in which case the time is calculated from the starter's pistol to the finish line,
...marathon times are calculated using a computer chip attached to each competitor's shoe.
The i choose D instead of C was that the structure of D was somewhat sensible - sub-ordinate clause, modifier, main clause
In C we have - modifier, non-essential part, main clause
& also wrong comparison. Thus C is wrong!!!
Re: Unlike elite runners, whose times are calculated from the [#permalink] 05 Dec 2016, 19:32
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# Unlike elite runners, whose times are calculated from the
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Alternating Current and Voltage MCQ Test - ObjectiveBooks
Practice Test: Question Set - 03
1. A 20 kHz pulse waveform consists of pulses that are 15 μs wide. The duty cycle
(A) Is 1%
(B) Is 30%
(C) Is 100%
(D) Cannot be determined
2. The designation rms means repetitions measured per second.
(A) True
(B) False
3. The average value of a 12 V peak sine wave over one complete cycle is
(A) 0 V
(B) 1.27 V
(C) 7.64 V
(D) 6.37 V
4. A square wave has a period of 60 μs. The first odd harmonic is
(A) 5 kHz
(B) 50 kHz
(C) 500 kHz
(D) 33.33 kHz
5. A sine wave with a period of 4 ms is changing at a faster rate than a sine wave with a period of
(A) 0.0045 s
(B) 2 ms
(C) 1.5 ms
(D) 3,000 μs
6. Two series resistors are connected to an ac source. If there are 7.5 V rms across one resistor and 4.2 V rms across the other, the peak source voltage is
(A) 16.54 V
(B) 1.65 V
(C) 10.60 V
(D) 5.93 V
7. A waveform has a baseline of 3 V, a duty cycle of 20%, and an amplitude of 8 V. The average voltage value is
(A) 4 V
(B) 4.6 V
(C) 1.6 V
(D) 11 V
8. A signal with a 400 μs period has a frequency of
(A) 250 Hz
(B) 2,500 Hz
(C) 25,000 Hz
(D) 400 Hz
9. Average value of a sine wave is 0.707 times the peak value.
(A) True
(B) False
10. A saw-tooth wave has a period of 10 ms. Its frequency is
(A) 10 Hz
(B) 50 Hz
(C) 100 Hz
(D) 1,000 Hz
11. The length of a phasor represents the amplitude.
(A) True
(B) False
12. A phasor represents
(A) The magnitude and a quantity direction
(B) The width of a quantity
(C) The phase angle
(D) The magnitude of a quantity
13. A 1 kHz signal has a period of 1 ms.
(A) True
(B) False
14. The average half-cycle value of a sine wave with a 40 V peak is
(A) 25.48 V
(B) 6.37 V
(C) 14.14 V
(D) 50.96 V
15. Duty cycle is the characteristic of a pulse waveform that indicates the high time versus the low time.
(A) True
(B) False
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# How many calories is 200 grams of chicken pasta?
200 grams of chicken pasta contains approximately 360 calories. The exact calorie count can vary depending on the specific ingredients and preparation method. Chicken pasta is a nutritious meal that provides protein, carbohydrates, and other important nutrients. When consumed in moderation as part of a balanced diet, 200 grams of chicken pasta can be part of a healthy lifestyle.
## Calculating Calories in Chicken Pasta
To determine the calorie count of chicken pasta, we need to look at the calories contained in each of the main ingredients:
### Pasta
– Dry pasta contains approximately 350 calories per 100 grams. This varies slightly between different types of pasta – for example 100g of dry spaghetti contains 360 calories while 100g of penne contains 368 calories.
– When pasta is cooked, it absorbs water which increases the weight. 100g of cooked pasta contains approximately 160 calories.
– For 200g of cooked pasta, this would equal 320 calories.
### Chicken
– 100g of cooked chicken breast (with no skin) contains approximately 165 calories.
– The chicken in pasta dishes is often diced or shredded. This can make it hard to determine the exact weight, but we can estimate that 200g of cooked pasta contains around 80-100g of chicken.
– So 80-100g of chicken would add 130-165 calories.
### Sauce
– Tomato or cream based sauces contain around 100 calories per 100ml.
– A 200g pasta portion may have around 50-100ml of sauce.
– This would add approximately 50-100 calories.
### Other Ingredients
– Onions, garlic, herbs and spices have a relatively low calorie content. For a 200g portion they may contribute up to 50 calories.
– Ingredients like cheese or olive oil can significantly increase the calorie content. 100g of parmesan cheese contains approximately 400 calories for example.
## Total Calories in 200g Chicken Pasta
Adding up the estimated calories from each ingredient:
– Pasta: 320 calories
– Chicken: 130-165 calories
– Sauce: 50-100 calories
– Other ingredients: up to 50 calories
This gives us an estimated total of 550-635 calories for a 200g portion of chicken pasta.
For a simpler estimate, we can assume the following averages:
– 200g pasta: 320 calories
– 100g chicken: 165 calories
– 75ml sauce: 75 calories
– Total: Approximately 560 calories
So for a round number approximation, 200g of chicken pasta contains approximately 360 calories.
This number can vary based on the specific recipe. Factors like the following can increase or decrease the calorie count:
– Type of pasta used – wholewheat or gluten-free pasta may be slightly higher in calories
– Lean chicken breast vs chicken thigh which has higher fat content
– Type of sauce – a creamy alfredo sauce may be higher in calories than a tomato sauce
– Added vegetables like onions, peppers, mushrooms etc which increase volume without adding many calories
– Use of additional ingredients like cheese, oil, butter, cream etc
– Cooking method – frying in oil increases calorie content compared to boiling
So while we can estimate an average of 360 calories for a chicken pasta portion, the specifics of the recipe can result in this varying from around 300 to over 500 calories.
## Nutritional Breakdown of Chicken Pasta
While calories provide an indication of the energy content of chicken pasta, looking at the full nutritional breakdown can give a better understanding of its nutritional value:
### Protein
– Chicken and pasta both provide protein.
– 100g of chicken breast contains around 31g of protein.
– 100g of pasta contains around 12g of protein.
– For 200g pasta with 100g chicken, this equals around 43g of protein.
– Protein is important for building and repairing muscles and tissues in the body.
### Carbohydrates
– Pasta is mostly made up of carbs from wheat flour.
– 200g of pasta contains around 50-60g of carbs.
– Carbs provide the body with glucose which is used for energy.
### Fat
– Chicken breast is a lean source of protein and contains little fat.
– The sauce contributes most of the fat content. A tomato sauce has around 5g of fat per 100ml.
– So a 200g chicken pasta portion likely contains 10-15g of fat.
– Fat helps with absorption of fat-soluble vitamins and provides fatty acids for health.
### Fiber
– Wholewheat or high fiber pasta varieties contain around 4g of fiber per 100g.
– So 200g of pasta would provide around 8g of fiber.
– Fiber promotes good digestion and bowel function.
### Vitamins and Minerals
Chicken pasta can contribute B vitamins like niacin, vitamins A, E, K, iron, selenium, phosphorus and other micronutrients. Tomatoes are a good source of vitamin C and lycopene.
So in summary, 200g of chicken pasta provides a nutritious blend of protein, carbs, vitamins and minerals. It is a balanced meal when eaten as part of a healthy diet.
## Benefits of Chicken Pasta
Here are some of the main benefits that chicken pasta offers as part of a balanced diet:
### High in Protein
The chicken provides a lean source of protein which helps to build and maintain muscle mass. Protein also helps you to feel fuller for longer after eating.
### Complex Carbohydrates
The pasta provides a good source of energizing complex carbohydrates. Wholewheat or high fiber pasta varietes have a lower glycemic index, meaning they do not spike blood sugar levels as dramatically.
### Nutrient Dense
Chicken pasta can contain a range of vitamins and minerals from ingredients like tomatoes, chicken, onions and garlic. Choose recipes with lots of vegetables to maximize nutrition.
### Versatile Ingredients
Both chicken and pasta are versatile ingredients that can be adapted in many ways, allowing for lots of variations in flavors, textures and nutrition.
### Child Friendly
Many children love pasta and chicken, so this can be an easy way to provide balanced nutrition in kid friendly meals.
### Cost Effective
Chicken and pasta are relatively affordable ingredients, making this a budget friendly meal option. Dried pasta and frozen chicken allow for even more cost savings.
So in moderation as part of a healthy lifestyle, chicken pasta can be a nutritious and tasty choice.
## Healthiest Ways to Prepare Chicken Pasta
Here are some tips for making chicken pasta as healthy as possible:
### Choose Wholegrain/High Fiber Pasta
Opt for wholewheat or multigrain varieties over plain white pasta. These have a lower glycemic index and provide more nutrients and fiber.
### Use Lean Protein
Skinless chicken breast is the leanest option. Thighs or legs are higher in fat and calories.
Onions, garlic, mushrooms, tomatoes, spinach, broccoli etc all add nutrients without many extra calories.
### Make your own Tomato Based Sauce
Homemade sauce allows you to control nutrients and avoid added sugar. Canned tomatoes and fresh herbs make quick healthy sauce.
### Limit Oil
Saute vegetables in broth instead of oil when making the sauce to limit excess calories.
### Go Easy on Cheese
Parmesan adds flavor but is high in calories – sprinkle lightly rather than smothering.
### Choose Healthy Cooking Methods
Steaming, boiling or baking instead of frying to avoid excess oil.
### Control Portions
Stick to around 200g dry pasta and 100g chicken per adult portion to keep calories in check.
### Balance with Vegetables
Serve pasta with a green salad or steamed broccoli to increase nutrition.
So in summary, be choosy about each ingredient and preparation method to maximize the nutritional value of your chicken pasta.
## Common Questions
Here are answers to some common questions about the calories and nutrition in chicken pasta:
### Is chicken pasta healthy?
Yes, when prepared with lean protein, whole grains, lots of vegetables and healthy cooking methods, chicken pasta can be a very healthy meal choice as part of a balanced diet.
### Is chicken pasta good for weight loss?
Chicken pasta can be an excellent choice for weight loss diets when portion sizes are controlled. Protein and fiber keep you feeling fuller longer while calories are kept reasonable.
### How many calories in chicken pesto pasta?
A 200g portion of pesto chicken pasta contains around 400-550 calories depending on the specific recipe. The olive oil and parmesan cheese in pesto sauce increases the calorie content compared to a tomato based sauce.
### What is the healthiest pasta?
For the healthiest choice, pick a pasta made from wholewheat, multigrain, or legume flour. These have more fiber and nutrients than plain white pasta. Popular options include wholewheat spaghetti, penne or fusilli.
### Is pasta high carb?
Yes, pasta is primarily a high carb food since it is made from wheat flour. However, it has a lower glycemic index than many other refined carbs. When combined with protein from chicken and healthy fats it can be part of a balanced diet.
### Is chicken pasta gluten free?
Traditional pasta contains gluten from wheat flour so is not gluten free. Gluten free alternatives include pasta made from rice, quinoa, chickpeas or lentils. This can be used to make gluten free chicken pasta.
So in summary, yes chicken pasta can be a nutritious choice when prepared mindfully. Limit portions, choose healthy ingredients, and balance with vegetables and plant foods.
## Chicken Pasta Recipes
Here are some healthy and delicious chicken pasta recipe ideas:
### Chicken Tomato Pasta
Saute chicken with onions and garlic. Add diced tomatoes, tomato paste, italian herbs and seasonings. Toss with wholewheat pasta and garnish with basil.
### Chicken Pesto Zoodles
Spiralized zucchini noodles provide a lighter carb alternative. Toss with chicken, homemade basil pesto, cherry tomatoes and parmesan.
### Chicken Veggie Pasta Bake
Mix pasta, chicken and steamed vegetables like broccoli, carrots, peas. Top with tomato sauce and mozzarella and bake until browned and crispy.
### Creamy Chicken and Mushroom Pasta
Saute chicken with mushrooms and garlic. Add to pasta with a light mornay sauce made from milk and parmesan.
### Mediterranean Chicken Pasta
Use greek yogurt instead of cream to make a creamy sauce. Add chicken, sun dried tomatoes, spinach, artichokes and feta cheese.
### Thai Peanut Chicken Pasta
Stir fry chicken and vegetables like carrots, peppers and snap peas. Toss through pasta with a Thai peanut sauce.
So get creative with the variations to enjoy lots of flavorful and nutritious chicken pasta meals.
## Chicken Pasta for Weight Loss
Here are some tips for turning chicken pasta into a weight loss friendly meal:
### Choose Portion Controlled Pasta
Look for wholewheat pasta that comes in portion controlled bags or measure out 200g dried pasta per serving.
### Bulk up on Vegetables
Double the amount of vegetables to pasta – choose low calorie options like zucchini, broccoli, tomatoes, mushrooms etc.
### Use Lean Protein
Skinless chicken breast or thigh fillets instead of fattier cuts with skin. Tofu is a plant based alternative.
### Make a Light Sauce
Tomato based sauces are lower in calories than creamy sauces. Enjoy pesto or cream sauces as an occasional treat.
### Avoid Extra Cheese
Grate parmesan lightly rather than adding big chunks of full fat cheese. Leave it off altogether.
### Use Healthy Fats
Saute chicken and vegetables in a little olive oil instead of slathering in butter.
### Fill Up On Protein and Fiber
The protein and fiber will keep you feeling satisfied while limiting overall calorie intake.
So with smart ingredient choices and controlled portions, chicken pasta can be an effective component of a weight loss diet.
## Conclusion
To summarize key points:
– 200 grams of chicken pasta contains approximately 360 calories, though this can vary based on specific ingredients used.
– Chicken breast provides lean protein. Wholegrain pasta offers complex carbohydrates. And vegetables add key micronutrients.
– When prepared with healthy ingredients and cooking methods, chicken pasta can be a nutritious choice as part of a balanced diet.
– Controlling portion sizes and pairing with vegetables makes chicken pasta ideal for weight loss diets.
– There are endless healthy and tasty ways to adapt chicken pasta recipes by varying vegetables, seasonings and sauces.
– Chicken pasta makes for a versatile, budget friendly, child friendly and satisfying meal that can be customized to suit your nutritional needs.
So enjoy experimenting with healthy chicken pasta recipes that work for your lifestyle!
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NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4.
Board CBSE Textbook NCERT Class Class 10 Subject Maths Chapter Chapter 13 Chapter Name Surface Areas and Volumes Exercise Ex 13.4 Number of Questions Solved 5 Category NCERT Solutions
## NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4
Unless stated otherwise, take π = $$\frac { 22 }{ 7 }$$
Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular .ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:
Given: upper diameter = 4 cm ⇒ upper radius = $$\frac { 1 }{ 2 }$$ = 2 cm = R
lower diameter = 2 cm ⇒ lower radius = $$\frac { 2 }{ 2 }$$ = 1 cm = r
height of glass = 14 cm
Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
Given: upper circumference of the frustum = 18 cm
Slant height (l) = 4 cm
We have C.S.A of the frustum = π (r1 + r2)l
Putting values from equation (i) and (ii), we get
Curved surface area = (πr1 + πr2)l = (9 + 3) x 4 = 12 x 4 = 48 cm²
Question 3.
A fez, the cap used by the Turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.
Solution:
Radius of open side (r1) = 10 cm
Radius of upper base (r2) = 4 cm
Slant height (l) = 15 cm
Total surface area of the cap = C.S.A. of the frustum + Area of upper base
= 660 cm² + 50.28 cm² = 710.28 cm²
Question 4.
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 8 per 100 cm2. (Take π = 3.14)
Solution:
Radius of the lower end (r1) = 8 cm
Radius of the upper end (r2) = 20 cm
Height of the frustum (h) = 16 cm
Question 5.
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter $$\frac { 1 }{ 16 }$$ cm, find the length of the wire.
Solution:
Let ADC is a cone with vertical angle 600.
Now, cone is cut into two parts, parallel to its base at height 10 cm.
Radius of larger end of the frustum = R1
A wire be formed having diameter $$\frac { 1 }{ 16 }$$ cm and length be H cm
Volume of wire so obtained = πr²H
We hope the NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4, drop a comment below and we will get back to you at the earliest.
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In mathematics, an addition chain for computing a positive integer n can be given by a sequence of natural numbers starting with 1 and ending with n, such that each number in the sequence is the sum of two previous numbers. The length of an addition chain is the number of sums needed to express all its numbers, which is one less than the cardinality of the sequence of numbers.[1]
## Examples
As an example: (1,2,3,6,12,24,30,31) is an addition chain for 31 of length 7, since
2 = 1 + 1
3 = 2 + 1
6 = 3 + 3
12 = 6 + 6
24 = 12 + 12
30 = 24 + 6
31 = 30 + 1
Addition chains can be used for addition-chain exponentiation. This method allows exponentiation with integer exponents to be performed using a number of multiplications equal to the length of an addition chain for the exponent. For instance, the addition chain for 31 leads to a method for computing the 31st power of any number n using only seven multiplications, instead of the 30 multiplications that one would get from repeated multiplication, and eight multiplications with exponentiation by squaring:
n2 = n × n
n3 = n2 × n
n6 = n3 × n3
n12 = n6 × n6
n24 = n12 × n12
n30 = n24 × n6
n31 = n30 × n
## Methods for computing addition chains
Calculating an addition chain of minimal length is not easy; a generalized version of the problem, in which one must find a chain that simultaneously forms each of a sequence of values, is NP-complete.[2] There is no known algorithm which can calculate a minimal addition chain for a given number with any guarantees of reasonable timing or small memory usage. However, several techniques are known to calculate relatively short chains that are not always optimal.[3]
One very well known technique to calculate relatively short addition chains is the binary method, similar to exponentiation by squaring. In this method, an addition chain for the number ${\displaystyle n}$ is obtained recursively, from an addition chain for ${\displaystyle n'=\lfloor n/2\rfloor }$. If ${\displaystyle n}$ is even, it can be obtained in a single additional sum, as ${\displaystyle n=n'+n'}$. If ${\displaystyle n}$ is odd, this method uses two sums to obtain it, by computing ${\displaystyle n-1=n'+n'}$ and then adding one.[3]
The factor method for finding addition chains is based on the prime factorization of the number ${\displaystyle n}$ to be represented. If ${\displaystyle n}$ has a number ${\displaystyle p}$ as one of its prime factors, then an addition chain for ${\displaystyle n}$ can be obtained by starting with a chain for ${\displaystyle n/p}$, and then concatenating onto it a chain for ${\displaystyle p}$, modified by multiplying each of its numbers by ${\displaystyle n/p}$. The ideas of the factor method and binary method can be combined into Brauer's m-ary method by choosing any number ${\displaystyle m}$ (regardless of whether it divides ${\displaystyle n}$), recursively constructing a chain for ${\displaystyle \lfloor n/m\rfloor }$, concatenating a chain for ${\displaystyle m}$ (modified in the same way as above) to obtain ${\displaystyle m\lfloor n/m\rfloor }$, and then adding the remainder. Additional refinements of these ideas lead to a family of methods called sliding window methods.[3]
## Chain length
Let ${\displaystyle l(n)}$ denote the smallest ${\displaystyle s}$ so that there exists an addition chain of length ${\displaystyle s}$ which computes ${\displaystyle n}$. It is known that
${\displaystyle \log _{2}(n)+\log _{2}(\nu (n))-2.13\leq l(n)\leq \log _{2}(n)+\log _{2}(n)(1+o(1))/\log _{2}(\log _{2}(n))}$,
where ${\displaystyle \nu (n)}$ is the Hamming weight (the number of ones) of the binary expansion of ${\displaystyle n}$.[4]
One can obtain an addition chain for ${\displaystyle 2n}$ from an addition chain for ${\displaystyle n}$ by including one additional sum ${\displaystyle 2n=n+n}$, from which follows the inequality ${\displaystyle l(2n)\leq l(n)+1}$ on the lengths of the chains for ${\displaystyle n}$ and ${\displaystyle 2n}$. However, this is not always an equality, as in some cases ${\displaystyle 2n}$ may have a shorter chain than the one obtained in this way. For instance, ${\displaystyle l(382)=l(191)=11}$, observed by Knuth.[5] It is even possible for ${\displaystyle 2n}$ to have a shorter chain than ${\displaystyle n}$, so that ${\displaystyle l(2n); the smallest ${\displaystyle n}$ for which this happens is ${\displaystyle n=375494703}$,[6] which is followed by ${\displaystyle 602641031}$, ${\displaystyle 619418303}$, and so on (sequence A230528 in the OEIS).
## Brauer chain
A Brauer chain or star addition chain is an addition chain in which each of the sums used to calculate its numbers uses the immediately previous number. A Brauer number is a number for which a Brauer chain is optimal.[5]
Brauer proved that
l*(2n−1) ≤ n − 1 + l*(n)
where ${\displaystyle l^{*}}$ is the length of the shortest star chain. For many values of n, and in particular for n < 12509, they are equal:[7] l(n) = l*(n). But Hansen showed that there are some values of n for which l(n) ≠ l*(n), such as n = 26106 + 23048 + 22032 + 22016 + 1 which has l*(n) = 6110, l(n) ≤ 6109. The smallest such n is 12509.
## Scholz conjecture
The Scholz conjecture (sometimes called the Scholz–Brauer or Brauer–Scholz conjecture), named after Arnold Scholz and Alfred T. Brauer), is a conjecture from 1937 stating that
${\displaystyle l(2^{n}-1)\leq n-1+l(n).}$
This inequality is known to hold for all Hansen numbers, a generalization of Brauer numbers; Neill Clift checked by computer that all ${\displaystyle n\leq 5784688}$ are Hansen (while 5784689 is not).[6] Clift further verified that in fact ${\displaystyle l(2^{n}-1)=n-1+l(n)}$ for all ${\displaystyle n\leq 64}$.[5]
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# IP3 05 Density, Pressure and Upthrust
## Manometer Simulation using GeoGebra
After trying to create a javascript simulation for the manometer using ChatGPT, I decided to return to GeoGebra, a platform that I am more used to, to create this interactive with more customisation. It comes with a ruler and a self-assessment feature for students to key in the value of pressure.
https://www.geogebra.org/m/qyuncz3h
A U-tube manometer is a fundamental device used to measure pressure differences in fluid systems with simplicity and precision. It consists of a U-shaped tube typically made from transparent glass or plastic, allowing for clear observation of fluid levels within the tube. The tube is partially filled with a manometric fluid such as mercury or water, chosen based on the expected pressure range and application. The manometer has a graduated scale for measuring the height difference between the fluid levels in the two arms of the U-tube. The device is connected at two points where the pressure difference needs to be assessed, such as in gas pipelines, liquid tanks, or other systems requiring pressure monitoring.
When the manometer is connected to the pressure sources, the fluid within the U-tube will adjust its levels until equilibrium is achieved, with one side rising and the other side falling. The difference in height between the two columns of fluid (h) indicates the pressure difference. This difference is read using the graduated scale and is used to calculate the pressure difference (ΔP) with the formula ΔP = ρgh, where ρ is the density of the manometric fluid, and g is the acceleration due to gravity.
## U-tube Manometer – a ChatGPT-Generated Simulation
I wanted a simple manometer simulation for the Sec 3 topic of Pressure and decided to try generating one with ChatGPT3.5. The first attempt, using just words, resulted in many errors such as single lines instead of 2D objects being used to represent the tubes and coloured bars that move in the wrong direction.
However, after I switched to ChatGPT4 and uploaded an image for reference, it was then able to produce a proper design consisting of glass tubes and coloured columns that move up and down with pressure changes in a flask. With a bit of UI changes using further prompts, this was the final product.
The following is the screenshot showing the image that was uploaded as well as the initial prompt.
## Pythagorean Cup
This is a 3D printed Pythagorean cup, otherwise known as a greedy cup, where if one pours far too much water or wine or whatever your greedy heart desires, all the contents in the cup will leak out through the bottom.
This is based on the design by “jsteuben” on Thingiverse (https://www.thingiverse.com/thing:123252). The siphoning effect kicks in when the water level is above the internal “tube” printed and hidden into the walls of the cup.
I printed another cup based on a more conventional design as well, but due to the wrong settings given when I prepared the gcode file, the cup was rather leaky when the water level was low. This design by “MonzaMakers” has a protruding siphon tube. (https://www.thingiverse.com/thing:562790)
Explaining how the siphon works is easier with the second cup. When the water level is lower than the highest point in the siphoning tube, it remains in the cup. When it exceeds the highest point of the tube, water begins to flow down the part of the tube leading to the opening at the bottom of the cup. The falling water column creates a suction effect and continuously draws the rest of the water in, until the cup is dry.
## Magdeburg Hemispheres
As promised, I am sharing another purchase made during this mid-term break for my kids.
Magdeburg hemispheres are used to demonstrate the power of atmospheric pressure. This simple demonstration kit consists of two plastic hemispheres, a rubber ring, a one-way valve, a syringe and some rubber tubing.
First, the one-way valve and the syringe are attached to the hemisphere that has a nozzle.
The two hemispheres are then placed together with the rubber ring between them. The rubber ring will serve as a seal as the hemispheres press against it when the air is pumped out.
As the syringe is pulled, the pressure inside the sphere decreases. This results in the atmospheric pressure being significantly larger than the internal pressure and thus, the hemispheres can not be pulled apart by hand.
To separate the hemispheres, remove the tubing and the hemispheres will simply fall apart as the internal pressure rises and reaches an equilibrium with atmospheric pressure.
The kit can be bought for less than S$3 here. There are other sellers that seem to offer lower prices, which I realised while doing a search for the keywords “Magdeburg Hemispheres” only after making the purchase because I was thinking it could not get any lower. ## Hydraulic Elevator This is a hydraulic lift kit for kids that was purchased online for only S$2.10 from Shopee, with free shipping! I am not, in anyway, affiliated to this, but simply sharing about one of several fun and cheap educational sets that I bought to occupy my kids during this mid-term break.
Other than the syringe, joints and tube, the parts are mainly laser-cut from a piece of wood with a thickness of two millimetres. The instructions come with pictures for each step so even though the words are in Chinese, there is no need to read them.
This kit demonstrates Pascal’s principle which states that a pressure change in one part of a closed container is transmitted without loss to every part. Hence, the pressure change is transmitted from one syringe to another, allowing work to be done. Do not expect it to lift up very heavy weights, though as the syringes are not perfectly sealed.
I shall share about other kits that I bought for this break soon, including a \$6.62 Tesla coil that I am looking forward to testing.
## Hydraulic Press Simulation
This simulation can be used for O-level Physics, for the topic of Pressure. I created it as it was relevant to our school’s IP3 physics as well.
It demonstrates the working principle of a hydraulic press. By adjusting the cross-section areas (A) of the two cylinders, you only need a small amount of force at the narrow piston to exert a large amount of force at the wider piston. This is how, when driving, the force applied by one’s foot is enough to supply a large force to apply the brake pads on a car’s wheels.
The advantage of using GeoGebra is that one can create such simple simulations within a couple of hours and it can be readily embedded into SLS – a wonderful tool to have during this period of full home-based learning.
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Related Articles
Maximize the expression (A AND X) * (B AND X) | Bit Manipulation
• Difficulty Level : Easy
• Last Updated : 04 Nov, 2019
Given two positive integers A and B such that A != B, the task is to find a positive integer X which maximizes the expression (A AND X) * (B AND X).
Example:
Input: A = 9 B = 8
Output: 8
(9 AND 8) * (8 AND 8) = 8 * 8 = 64 (maximum possible)
Input: A = 11 and B = 13
Output: 9
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Naive approach: One can run a loop from 1 to max(A, B) and can easily find X which maximizes the given expression.
Efficient approach: It is known that,
(a – b)2 ≥ 0
which implies (a + b)2 – 4*a*b ≥ 0
which implies a * b ≤ (a + b)2 / 4
Hence, it concludes that a * b will be maximum when a * b = (a + b)2 / 4
which implies a = b
From the above result, (A AND X) * (B AND X) will be maximum when (A AND X) = (B AND X)
Now X can be found as:
A = 11 = 1011
B = 13 = 1101
X = ? = abcd
At 0th place: (1 AND d) = (1 AND d) implies d = 0, 1 but to maximize (A AND X) * (B AND X) d = 1
At 1st place: (1 AND d) = (0 AND d) implies c = 0
At 2nd place: (0 AND d) = (1 AND d) implies b = 0
At 3rd place: (1 AND d) = (1 AND d) implies a = 0, 1 but to maximize (A AND X) * (B AND X) a = 1
Hence, X = 1001 = 9
Below is the implementation of the above approach:
## C++
`// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` ` `#define MAX 32 ` ` ` `// Function to find X according ` `// to the given conditions ` `int` `findX(``int` `A, ``int` `B) ` `{ ` ` ``int` `X = 0; ` ` ` ` ``// int can have 32 bits ` ` ``for` `(``int` `bit = 0; bit < MAX; bit++) { ` ` ` ` ``// Temporary ith bit ` ` ``int` `tempBit = 1 << bit; ` ` ` ` ``// Compute ith bit of X according to ` ` ``// given conditions ` ` ``// Expression below is the direct ` ` ``// conclusion from the illustration ` ` ``// we had taken earlier ` ` ``int` `bitOfX = A & B & tempBit; ` ` ` ` ``// Add the ith bit of X to X ` ` ``X += bitOfX; ` ` ``} ` ` ` ` ``return` `X; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `A = 11, B = 13; ` ` ` ` ``cout << findX(A, B); ` ` ` ` ``return` `0; ` `} `
## Java
`// Java implementation of the approach ` `class` `GFG ` `{ ` `static` `int` `MAX = ``32``; ` ` ` `// Function to find X according ` `// to the given conditions ` `static` `int` `findX(``int` `A, ``int` `B) ` `{ ` ` ``int` `X = ``0``; ` ` ` ` ``// int can have 32 bits ` ` ``for` `(``int` `bit = ``0``; bit < MAX; bit++) ` ` ``{ ` ` ` ` ``// Temporary ith bit ` ` ``int` `tempBit = ``1` `<< bit; ` ` ` ` ``// Compute ith bit of X according to ` ` ``// given conditions ` ` ``// Expression below is the direct ` ` ``// conclusion from the illustration ` ` ``// we had taken earlier ` ` ``int` `bitOfX = A & B & tempBit; ` ` ` ` ``// Add the ith bit of X to X ` ` ``X += bitOfX; ` ` ``} ` ` ``return` `X; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String []args) ` `{ ` ` ``int` `A = ``11``, B = ``13``; ` ` ` ` ``System.out.println(findX(A, B)); ` `} ` `} ` ` ` `// This code is contributed by 29AjayKumar `
## Python3
`# Python3 implementation of the approach ` `MAX` `=` `32` ` ` `# Function to find X according ` `# to the given conditions ` `def` `findX(A, B) : ` ` ` ` ``X ``=` `0``; ` ` ` ` ``# int can have 32 bits ` ` ``for` `bit ``in` `range``(``MAX``) : ` ` ` ` ``# Temporary ith bit ` ` ``tempBit ``=` `1` `<< bit; ` ` ` ` ``# Compute ith bit of X according to ` ` ``# given conditions ` ` ``# Expression below is the direct ` ` ``# conclusion from the illustration ` ` ``# we had taken earlier ` ` ``bitOfX ``=` `A & B & tempBit; ` ` ` ` ``# Add the ith bit of X to X ` ` ``X ``+``=` `bitOfX; ` ` ` ` ``return` `X; ` ` ` `# Driver code ` `if` `__name__ ``=``=` `"__main__"` `: ` ` ` ` ``A ``=` `11``; B ``=` `13``; ` ` ``print``(findX(A, B)); ` ` ` `# This code is contributed by AnkitRai01 `
## C#
`// C# implementation of the approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` `static` `int` `MAX = 32; ` ` ` `// Function to find X according ` `// to the given conditions ` `static` `int` `findX(``int` `A, ``int` `B) ` `{ ` ` ``int` `X = 0; ` ` ` ` ``// int can have 32 bits ` ` ``for` `(``int` `bit = 0; bit < MAX; bit++) ` ` ``{ ` ` ` ` ``// Temporary ith bit ` ` ``int` `tempBit = 1 << bit; ` ` ` ` ``// Compute ith bit of X according to ` ` ``// given conditions ` ` ``// Expression below is the direct ` ` ``// conclusion from the illustration ` ` ``// we had taken earlier ` ` ``int` `bitOfX = A & B & tempBit; ` ` ` ` ``// Add the ith bit of X to X ` ` ``X += bitOfX; ` ` ``} ` ` ``return` `X; ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main(String []args) ` `{ ` ` ``int` `A = 11, B = 13; ` ` ` ` ``Console.WriteLine(findX(A, B)); ` `} ` `} ` ` ` `// This code is contributed by 29AjayKumar `
Output:
```9
```
My Personal Notes arrow_drop_up
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## Inverse to a Curious Partition Problem Correct Solution
In ΔABC, ∠B = 80°. The bisector of angle B meets AC in D such that AD = BC + BD. Find angle C.
The first attempt at the problem has been found to contain a mistake - making an assumption that is equivalent to the statement one has to prove. Here's a correct
Solution
In ΔABC, ∠B = 80°. The bisector of angle B meets AC in D such that AD = BC + BD. Find angle C.
Let F on AB be such that DF||BC. Then
∠BDF = ∠CBD = ∠DBF = 40°,
making, in particular, ΔBDF isosceles:
BF = DF.
Now there are three possibilities: CD = BF, CD < BF, and CD > BF. In the first case, there is nothing to prove. The trapezoid BCDF is isosceles so that ∠BCD = ∠CBF = 80°.
Let's denote ∠BCD = γ. If CD < BF, then γ > 80°; otherwise, if CD > BF, then γ < 80°.
For both cases we shall define a point E on where ∠DEF = 40°. This choice makes triangles BCD and EDF similar, because also ∠EDF = ∠BCD.
Assume, CD < BF. Then also CD < DF. All sides of Δ EDF are greater than their counterparts in ΔBCD. In particular, EF > BD and also DE > BC. Since AD = BC + BD, AE < BD < EF. Thus in ΔAEF, ∠AFE < ∠EAF. Since the sum of the two angles is 40°, ∠EAF > 20°.
However, the assumption CD < BF implies that γ > 80° and we get a contradiction since angles in ΔABC add to more than 180°.
The case CD > BF similarly leads to a contradiction.
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Test: Combinatory- 3 - Computer Science Engineering (CSE) MCQ
# Test: Combinatory- 3 - Computer Science Engineering (CSE) MCQ
Test Description
## 15 Questions MCQ Test GATE Computer Science Engineering(CSE) 2024 Mock Test Series - Test: Combinatory- 3
Test: Combinatory- 3 for Computer Science Engineering (CSE) 2024 is part of GATE Computer Science Engineering(CSE) 2024 Mock Test Series preparation. The Test: Combinatory- 3 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Test: Combinatory- 3 MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Combinatory- 3 below.
Solutions of Test: Combinatory- 3 questions in English are available as part of our GATE Computer Science Engineering(CSE) 2024 Mock Test Series for Computer Science Engineering (CSE) & Test: Combinatory- 3 solutions in Hindi for GATE Computer Science Engineering(CSE) 2024 Mock Test Series course. Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free. Attempt Test: Combinatory- 3 | 15 questions in 45 minutes | Mock test for Computer Science Engineering (CSE) preparation | Free important questions MCQ to study GATE Computer Science Engineering(CSE) 2024 Mock Test Series for Computer Science Engineering (CSE) Exam | Download free PDF with solutions
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Test: Combinatory- 3 - Question 1
### How many onto (or surjective) functions are there from an n-element (n >= 2) set to a 2-element set?
Detailed Solution for Test: Combinatory- 3 - Question 1
Total possible number of functions is 2n. In mathematics, a function f from a set X to a set Y is surjective (or onto), or a surjection, if every element y in Y has a corresponding element x in X such that f(x) = y (Source: http://en.wikipedia.org/wiki/Surjective_function) There are total 2 functions out of 2n that are NOT onto: one that maps to all 1s and other that maps to all 2s. Therefore total number of onto functions is 2n - 2.
Test: Combinatory- 3 - Question 2
### What is the possible number of reflexive relations on a set of 5 elements?
Detailed Solution for Test: Combinatory- 3 - Question 2
Number of reflexive relations is 2n2-n which is 220 for n = 5
Test: Combinatory- 3 - Question 3
### What is the maximum number of different Boolean functions involving n Boolean variables?
Detailed Solution for Test: Combinatory- 3 - Question 3
No of inputs sequences possible for a n variable Boolean function = 2n Each input sequence can give either T or F as output ( 2 possible values ) So, Total no of Boolean functions are -
2X2X2X2X2X2X.............X2X2X2X2X2X2 <-------------------- 2n Times --------------> 22n
Test: Combinatory- 3 - Question 4
Suppose that a robot is placed on the Cartesian plane. At each step it is allowed to move either one unit up or one unit right, i.e., if it is at (i,j) then it can move to either (i+1,j) or (i,j+1). How many distinct paths are there for the robot to reach the point (10,10) starting from the initial position (0, 0)
Detailed Solution for Test: Combinatory- 3 - Question 4
At each move, robot can move either 1 unit right, or 1 unit up, and there will be 20 such moves required to reach (10,10) from (0,0). So we have to divide these 20 moves, numbered from 1 to 20, into 2 groups: right group and up group. Right group contains those moves in which we move right, and up group contains those moves in which we move up. Each group contains 10 elements each. So basically, we have to divide 20 things into 2 groups of 10 10 things each, i.e., we need to find all possible arrangements of {r, r, r, r, r, r, r, r, r, r, u, u, u, u, u, u, u, u, u, u} where r represents right move and u represents up move. The arrangements can can be done in 20! / (10!∗10!) = 20C10 ways. So option (A) is correct.
Test: Combinatory- 3 - Question 5
Consider the data given in above question. Suppose that the robot is not allowed to traverse the line segment from (4,4) to (5,4). With this constraint, how many distinct paths are there for the robot to reach (10,10) starting from (0,0)?
Detailed Solution for Test: Combinatory- 3 - Question 5
Since we are not allowed to traverse from (4,4) to (5,4), we subtract all those paths which were passing through (4,4) to (5,4). To count number of paths passing through (4,4) to (5,4), we find number of paths from (0,0) to (4,4), and then from (5,4) to (10,10).
So option (D) is correct.
Test: Combinatory- 3 - Question 6
Let ∑ be a finite non-empty alphabet and let be the power set of ∑*. Which one of the following is TRUE?
Test: Combinatory- 3 - Question 7
The number of different n × n symmetric matrices with each element being either 0 or 1 is: (Note: power(2, x) is same as 2x)
Detailed Solution for Test: Combinatory- 3 - Question 7
We are considering a symmetric matrix (given in question). So, we need to look at half of elements ie. either upper or lower traingle i.e. A[i][j] = A[j][i] Hence, No. of elements = (n^2 + n)/2 Since, we have only two elements : 0 & 1 No. of choices = 2 Therefore, No. of possibilities = 2 ^ (No. of elements) = 2 ^ ((n^2 + n)/2) = power (2 , (n^2 + n)/2)
Test: Combinatory- 3 - Question 8
Mala has a colouring book in which each English letter is drawn two times. She wants to paint each of these 52 prints with one of k colours, such that the colour-pairs used to colour any two letters are different. Both prints of a letter can also be coloured with the same colour. What is the minimum value of k that satisfies this requirement ?
Detailed Solution for Test: Combinatory- 3 - Question 8
This question is slightly ambiguous. So first let us understand what question is asking. So in a book, we have letters A-Z and each letter is printed twice, so there are 52 letters. Now we have to color each letter, so we need a pair of colors for that, because each letter is printed twice. Also in a pair, both colors can be some. Now condition is that a pair of colors can't be used more than once. So suppose Mala has 3 colors : Red, Blue, Green. She can color as follows : (A,A) : (Red,Red), (B,B) : (Blue,Blue), (C,C) : (Green,Green), (D,D) : (Red,Blue), (E,E) : (Red,Green), (F,F) : (Blue,Green). Now we don't have more pairs of colors left, we have used all pairs, but could color only 6 letters out of 26. So question is to find minimum no. of colors, so that we could color all 26 letters. So if Mala has k colors, she can have k pairs of same colors, thus coloring k letters, then kC2 other pairs of colors, thus coloring kC2 more letters. So total no. of letters colored = k+kC2=k+k(k−1)2=k(k+1)2. So we want k(k+1)2≥26 i.e. k(k+1)≥52, so k≥7, so option (C) is correct.
Test: Combinatory- 3 - Question 9
Let A be a sequence of 8 distinct integers sorted in ascending order. How many distinct pairs of sequences, B and C are there such that
(i) each is sorted in ascending order,
(ii) B has 5 and C has 3 elements, and
(iii) the result of merging B and C gives A?
Detailed Solution for Test: Combinatory- 3 - Question 9
Suppose you have selected 3 elements from 8 in 8C3 ways, the remaining elements are treated as another array and merging both the arrays gives the sorted array. Here, you can select either 3 or 5. => 8C3 = 8C5 = 8!/(3!5!) = 7*8 = 56 Ways.
Test: Combinatory- 3 - Question 10
n couples are invited to a party with the condition that every husband should be accompanied by his wife. However, a wife need not be accompanied by her husband. The number of different gatherings possible at the party is
Detailed Solution for Test: Combinatory- 3 - Question 10
There are three options for every couple.
1) Nobody goes to gathering
2) Wife alone goes
3) Both go
Since there are n couples, total possible ways of gathering are 3n
Test: Combinatory- 3 - Question 11
m identical balls are to be placed in n distinct bags. You are given that m ≥ kn, where, k is a natural number ≥ 1. In how many ways can the balls be placed in the bags if each bag must contain at least k balls?
Detailed Solution for Test: Combinatory- 3 - Question 11
This is very simple application of stars and bars. Since we want atleast k balls in each bag, so first we put kn balls into bags, k balls in each bag. Now we are left with m - kn balls, and we have to put them into n bags such that each bag may receive 0 or more balls. So applying theorem 2 of stars and bars with m - nk stars and n bars, we get number of ways to be m−kn+n-1 Cn−1. So option (B) is correct.
Test: Combinatory- 3 - Question 12
How many 4-digit even numbers have all 4 digits distinct?
Detailed Solution for Test: Combinatory- 3 - Question 12
This is a basic permutation combination question. Considering two cases : numbers ending with 0 and not ending with 0: Numbers ending with 0 1{first place: 0} ∗9{fourth place: 9 possibilities, 1-9} ∗8{third place: 8 possibilities left} ∗7{second place: 7 possibilities left} = 504 Numbers ending with non-0 4{first place: 2,4,6,8} ∗8{fourth place: 8 possibilities left, 1-9}∗8{third place: 8 possibilities left b/w 0-9} ∗7{second place: 7 possibilities left b/w 0-9} = 2592 Total = 2296
Test: Combinatory- 3 - Question 13
For a set A, the power set of A is denoted by 2A. If A = {5, {6}, {7}}, which of the following options are True.
Detailed Solution for Test: Combinatory- 3 - Question 13
I is true, as φ belongs to Powerset.
II is true, as an empty set is a subset of every set.
III is true as {5, {6}} belongs to Powerset.
IV is false, {5, {6}} is not a subset, but {{5, {6}}} is.
Test: Combinatory- 3 - Question 14
The number of 4 digit numbers having their digits in non-decreasing order (from left to right) constructed by using the digits belonging to the set {1, 2, 3} is ____.
Detailed Solution for Test: Combinatory- 3 - Question 14
{1, 1, 1, 1} {1, 1, 1, 2} {1, 1, 1, 3} {1, 1, 2, 2} {1, 1, 2, 3} {1, 1, 3, 3} {1, 2, 2, 2} {1, 2, 2, 3} {1, 2, 3, 3} {1, 3, 3, 3} {2, 2, 2, 2} {2, 2, 2, 3} {2, 2, 3, 3} {2, 3, 3, 3} {3, 3, 3, 3}
Test: Combinatory- 3 - Question 15
How many substrings of different length (non-zero) can be formed from a character string of length n ?
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Info
Lifehacks
# How many square feet are in a square of 3 tab shingles?
## How many square feet are in a square of 3 tab shingles?
If the shingles you are using come three bundles to a square, calculating the number of bundles you’ll need is simple. Each bundle covers 33.3 sq. ft. of roof area—close enough to the 32 sq.
## How many squares of shingles do I need for 1000 square feet?
For the number of ridge caps, multiply the length of the hip or ridge by twelve and divide it by five. This house is 1000 square feet. It requires 18 squares of shingles.
## How many square feet is 20 squares of shingles?
Please share if you found this tool useful:
Conversions Table
10 Roofing Squares to Square Feet = 1000 800 Roofing Squares to Square Feet = 80000
20 Roofing Squares to Square Feet = 2000 900 Roofing Squares to Square Feet = 90000
30 Roofing Squares to Square Feet = 3000 1,000 Roofing Squares to Square Feet = 100000
## How big is a square of roofing shingles?
A square is the number of shingles it would take to cover 100 square feet of roof area with each row of shingles covering the tabs of the layer below it. Shingles are packaged in bundles that include a number of strips of roofing.
## How many bundles of shingles are there in a square?
It typically takes three bundles of standard three-tab asphalt shingles to cover one square on a roof. Therefore, a 40-square roof would require approximately 120 bundles of shingles. A “roof square” is equal to 100 square feet, so each bundle covers roughly 33 1/3 square feet.
## How many square feet do you need to Shingle a garage roof?
Three bundles are equal to one square. Each square equates to 100 square feet. In the previous example, three squares, or 300 square feet, of shingles will be needed to cover the garage roof. After calculating the square footage needed, always add in an extra 10 percent to the total for shingle breakage.
## How many nails per square of shingles do you need?
High wind areas and other types of shingles may need more. 320 nails will be needed to install a square of standard 3-tab shingles assuming four nails per shingle and 80 shingles per square. Keep in mind that a drip-edge will be needed to protect the edge of the roof from rot, vents, and other materials such as flashing or caulk.
## How much does square of roof shingles cost?
Across the US, the average price for roof shingles is \$3.20 – 3.40 per square foot , including materials and installation. This is equivalent to \$320 – 340 per roof square (1 roof square = 100 sq.ft.) You can use Roof Shingles Calculator to get a more accurate estimate for your home.
## How many shingles are in a square?
The most used shingles type is called strip shingle (three-tab shingle) which is typically packed with three bundles for a square. So, that should answer your question on how many bundles of shingles in a square: there are three bundles per square.
## How do you calculate shingles on roof?
To estimate how many shingles you’ll need, first estimate the total square footage of your roof’s surface. To do this, measure the length and width of each plane on the roof, including dormers. Then, multiply length x width to get the square footage of each plane.
## How much does square of shingles weigh?
At the bottom, one square is equivalent to three bundles of shingles. It indicates that each bundle is comprised of 33 square feet. The thickets bundle of shingles usually contains 4 bundles. The general term confirms that one square of shingles actually weighs between 150 and 240 lbs.
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### Author Topic: TT2 problem 3 (Read 3430 times)
#### Victor Ivrii
• Elder Member
• Posts: 2607
• Karma: 0
##### TT2 problem 3
« on: March 12, 2015, 07:15:37 PM »
Consider the diffusion equation
\begin{equation*}
\end{equation*}
with the boundary conditions
\begin{equation*}
\end{equation*}
and the initial condition
\begin{equation*}
u(x,0)=|\cos (x)|.
\end{equation*}
a. Write the associated eigenvalue problem.
b. Find all eigenvalues and corresponding eigenfunctions.
c. Show that the eigenfunctions associated to 2 different eigenvalues are orthogonal.
d. Write the solution in the form of a series expansion.
e. Write a formula for the coefficients of the series expansion.
« Last Edit: March 12, 2015, 07:18:31 PM by Victor Ivrii »
#### Victor Ivrii
• Elder Member
• Posts: 2607
• Karma: 0
##### Re: TT2 problem 3
« Reply #1 on: March 14, 2015, 08:53:25 AM »
a-b. Separation of variables brings $X''+\lambda X=0$, $X(0)=X(3\pi)=0$ and then $\lambda_n= -n^2/9$, $X_n=\sin (nx/3)$.
c. We know that $\sin (\pi nx/l)$ orthogonal on $[0,l]$.
d. From separation of variables we also have $T'/T=-k\lambda$ and then $T_n=A_n e^{-k n^2t/9}$. Then
u(x,t)= \sum_{n=1}^\infty A_n e^{-k n^2t/9} \sin (nx/3),
\label{K}
e. Then $u(x,0)=\sum_{n=1}^\infty A_n \sin (nx/3)=|\cos (x)|$. We need to decompose $|\cos (x)|$ into series. Then
A_n = \frac{2}{3\pi} \int_0^{3\pi} |\cos (x)|\sin \bigl(\frac{nx}{3}\bigr)\,dx
\label{L}
which does not look very easy since $\cos (x)$ changes sign several times on $(0,3\pi)$. Because those who did not make calculations got no reduction.
But there is still an easy way to calculate $A_n$. Since $|\cos (x)$ is $\pi$-periodic
\begin{gather*}
A_{n} = \frac{2}{3\pi} \int_{\pi}^{2\pi} |\cos (x)|\bigl[ \sin \bigl(\frac{n(x-\pi)}{3}\bigr)+
\sin \bigl(\frac{nx}{3}\bigr) + \sin \bigl(\frac{n(x+\pi)}{3}\bigr)\bigr]\,dx=\\
\frac{2}{3\pi} \int_{\pi}^{2\pi} |\cos (x)| \sin \bigl(\frac{nx}{3}\bigr)
\bigl[ 2\cos \bigl(\frac{n\pi}{3}\bigr)+1\bigr) \bigr]\,dx
\end{gather*}
where we used formula for $\sin(\alpha)+\sin(\beta)$. Observe that $|\cos (x)|$ is even with respect to $\frac{3\pi}{2}$ and $\sin (n x/3)$ is even as $n$ is odd and odd as $n$ is even. Then $A_{2m}=0$ and
\begin{equation*}
A_{2m+1} =
\frac{2}{3\pi} \bigl[ 2\cos \bigl(\frac{(2m+1)\pi}{3}\bigr)+1\bigr) \bigr]
\int_{\frac{3\pi}{2}}^{2\pi} \bigl[ \sin\bigl( \frac{(m+2)x}{3} \bigr) + \sin\bigl( \frac{(m-1)x}{3} \bigr) \bigr]\,dx
\end{equation*}
where we used formula for $2\sin(\alpha)\cos (\beta)$. This integral is easy.
« Last Edit: March 16, 2015, 10:39:11 AM by Victor Ivrii »
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# 0.2 Exponentials
Page 1 / 2
## Introduction
In this chapter, you will learn about the short cuts to writing $2×2×2×2$ . This is known as writing a number in exponential notation .
## Definition
Exponential notation is a short way of writing the same number multiplied by itself many times. For example, instead of $5×5×5$ , we write ${5}^{3}$ to show that the number 5 is multiplied by itself 3 times and we say “5 to the power of 3”. Likewise ${5}^{2}$ is $5×5$ and ${3}^{5}$ is $3×3×3×3×3$ . We will now have a closer look at writing numbers using exponential notation.
Exponential Notation
Exponential notation means a number written like
${a}^{n}$
where $n$ is an integer and $a$ can be any real number. $a$ is called the base and $n$ is called the exponent or index .
The ${n}^{\mathrm{th}}$ power of $a$ is defined as:
${a}^{n}=a×a×\cdots ×a\phantom{\rule{2.em}{0ex}}\left(\mathrm{n times}\right)$
with $a$ multiplied by itself $n$ times.
We can also define what it means if we have a negative exponent $-n$ . Then,
${a}^{-n}=\frac{1}{a×a×\cdots ×a\phantom{\rule{2.em}{0ex}}\left(\mathrm{n times}\right)}$
Exponentials
If $n$ is an even integer, then ${a}^{n}$ will always be positive for any non-zero real number $a$ . For example, although $-2$ is negative, ${\left(-2\right)}^{2}=-2×-2=4$ is positive and so is ${\left(-2\right)}^{-2}=\frac{1}{-2×-2}=\frac{1}{4}$ .
## Laws of exponents
There are several laws we can use to make working with exponential numbers easier. Some of these laws might have been seen in earlier grades, but we will list all the laws here for easy reference and explain each law in detail, so that you can understand them and not only remember them.
$\begin{array}{ccc}\hfill {a}^{0}& =& 1\hfill \\ \hfill {a}^{m}×{a}^{n}& =& {a}^{m+n}\hfill \\ \hfill {a}^{-n}& =& \frac{1}{{a}^{n}}\hfill \\ \hfill {a}^{m}÷{a}^{n}& =& {a}^{m-n}\hfill \\ \hfill {\left(ab\right)}^{n}& =& {a}^{n}{b}^{n}\hfill \\ \hfill {\left({a}^{m}\right)}^{n}& =& {a}^{mn}\hfill \end{array}$
## Exponential law 1: ${a}^{0}=1$
Our definition of exponential notation shows that
$\begin{array}{ccc}\hfill {a}^{0}& =& 1\phantom{\rule{1.em}{0ex}},\phantom{\rule{1.em}{0ex}}\left(a\ne 0\right)\hfill \end{array}$
To convince yourself of why this is true, use the fourth exponential law above (division of exponents) and consider what happens when $m=n$ .
For example, ${x}^{0}=1$ and ${\left(1\phantom{\rule{0.277778em}{0ex}}000\phantom{\rule{0.277778em}{0ex}}000\right)}^{0}=1$ .
## Application using exponential law 1: ${a}^{0}=1,\left(a\ne 0\right)$
1. ${16}^{0}$
2. $16{a}^{0}$
3. ${\left(16+a\right)}^{0}$
4. ${\left(-16\right)}^{0}$
5. $-{16}^{0}$
## Exponential law 2: ${a}^{m}×{a}^{n}={a}^{m+n}$
Our definition of exponential notation shows that
$\begin{array}{cccc}\hfill {a}^{m}×{a}^{n}& =& 1×a×...×a\hfill & \left(\mathrm{m times}\right)\hfill \\ \hfill & & \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}×1×a×...×a\hfill & \left(\mathrm{n times}\right)\hfill \\ \hfill & =& 1×a×...×a\hfill & \left(\mathrm{m}+\mathrm{n times}\right)\hfill \\ \hfill & =& {a}^{m+n}\hfill & \end{array}$
For example,
$\begin{array}{ccc}\hfill {2}^{7}×{2}^{3}& =& \left(2×2×2×2×2×2×2\right)×\left(2×2×2\right)\hfill \\ & =& {2}^{7+3}\hfill \\ & =& {2}^{10}\hfill \end{array}$
## Interesting fact
This simple law is the reason why exponentials were originally invented. In the days before calculators, all multiplication had to be done by hand with a pencil and a pad of paper. Multiplication takes a very long time to do and is very tedious. Adding numbers however, is very easy and quick to do. If you look at what this law is saying you will realise that it means that adding the exponents of two exponential numbers (of the same base) is the same as multiplying the two numbers together. This meant that for certain numbers, there was no need to actually multiply the numbers together in order to find out what their multiple was. This saved mathematicians a lot of time, which they could use to do something more productive.
## Application using exponential law 2: ${a}^{m}×{a}^{n}={a}^{m+n}$
1. ${x}^{2}·{x}^{5}$
2. ${2}^{3}·{2}^{4}$ [Take note that the base (2) stays the same.]
3. $3×{3}^{2a}×{3}^{2}$
## Exponential law 3: ${a}^{-n}=\frac{1}{{a}^{n}},\phantom{\rule{1.em}{0ex}}a\ne 0$
Our definition of exponential notation for a negative exponent shows that
#### Questions & Answers
general equation for photosynthesis
6CO2 + 6H2O + solar energy= C6H1206+ 6O2
Anastasiya
meaning of amino Acids
a diagram of an adult mosquito
what are white blood cells
white blood cell is part of the immune system. that help fight the infection.
MG
what about tissue celss
Mlungisi
Cells with a similar function, form a tissue. For example the nervous tissue is composed by cells:neurons and glia cells. Muscle tissue, is composed by different cells.
Anastasiya
I need further explanation coz celewi anything guys,,,
hey guys
Isala
on what?
Anastasiya
hie
Lish
is air homogenous or hetrogenous
homogenous
Kevin
why saying homogenous?
Isala
explain if oxygen is necessary for photosynthesis
explain if oxygen is necessary for photosynthesis
Yes, the plant does need oxygen. The plant uses oxygen, water, light, and produced food. The plant use process called photosynthesis.
MG
By using the energy of sunlight, plants convert carbon dioxide and water into carbohydrates and oxygen by photosynthesis. This happens during the day and sunlight is needed.
NOBLE
no. it s a product of the process
Anastasiya
yet still is it needed?
NOBLE
no. The reaction is: 6CO2+6H20+ solar energy =C6H12O6(glucose)+602. The plant requires Carbon dioxyde, light, and water Only, and produces glucose and oxygen( which is a waste).
Anastasiya
what was the question
joining
Godfrey
the specific one
NOBLE
the study of non and living organism is called.
Godfrey
Is call biology
Alohan
yeah
NOBLE
yes
Usher
what Is ecology
what is a cell
A cell is a basic structure and functional unit of life
Ndongya
what is biolgy
is the study of living and non living organisms
Ahmed
may u draw the female organ
i dont understand
Asal
:/
Asal
me too
DAVID
anabolism and catabolism
Anabolism refers to the process in methabolism in which complex molecules are formed "built" and requires energy to happen. Catabolism is the opposite process: complex molecules are deconstructed releasing energy, such as during glicolysis.
Anastasiya
Explain briefly independent assortment gene .
hi
Amargo
hi I'm Anatalia
Joy
what do you mean by pituitary gland
Digambar
Got questions? Join the online conversation and get instant answers!
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# Should you pay points to get a lower mortgage rate?
Peter Miller
The Mortgage Reports Contributor
February 25, 2018 - 4 min read
## Is it worth it to pay points?
Whenever mortgage rates go up, borrowers always wonder if it makes sense pay points and thus reduces the rate. The answer is sometimes yes, sometimes no. Here’s how to tell the difference.
## What is a point?
A “point” or “loan discount fee” is equal to 1 percent of the mortgage amount. If you buy a house for \$200,000 with 10 percent down (\$20,000) the mortgage amount will be \$180,000. As a result, one percent of \$180,000 – one point – is \$1,800.
A point is paid at closing. The general idea is that if you pay a point at closing, the interest rate will be reduced. How much is a matter of negotiation with the lender?
If you have a 30-year, \$180,000 mortgage at 4.25 percent, the monthly cost for principal and interest is \$885.49.
But suppose that your lender offers you an alternative: In exchange for paying one point upfront, the lender will reduce the interest rate to 3.875 percent.
5 ways to shave .25 percent from your mortgage rate
You now have a \$180,000 mortgage at 3.875 percent. The monthly cost for principal and interest is \$846.43. You save \$39.06 per month (\$885.49 less \$846.43). Divide \$1,800 – the cost of a point in this example – by \$39.06 and in basic terms, you will have to own the property for 46 months to recover your money.
Paying one point decreases your stated interest rate, the rate used to calculate your payment. But you also want to consider your annual percentage rate or APR. APR incorporates the cost of the points as well as your interest paid.
APR considers the fact that you borrowed \$180,000, but since you paid \$1,800 in discount points, you actually got \$178,200. When you apply the \$846.43 payment to an actual loan amount of \$178,200, your APR is 3.9579 percent — considerably lower than 4.25 percent.
Is this a good deal? Here are some questions to consider.
## Pay points and cash
In our example, the borrower paid \$1,800 upfront. For marginal borrowers with little in savings \$1,800 on top of the other closing costs may be an impossible burden. As a result, points may be out of the picture. Alternatively, for strong savers, paying points can be worth a look.
How to buy a home: 3 concessions that are better than a price cut
If you’re buying a home, you may be better off negotiating seller-paid points instead of a lower purchase price. So for a home listed at \$200,000, instead of offering \$196,000 (98 percent of the purchase price), it might be better to pay \$200,000 and get seller-paid discount points costing 2 percent of the loan amount. Here’s what it looks like with a price reduction:
• Price: \$196,000
• Loan amount: \$176,400
• Points: 0
• Rate: 4.50 percent
• Payment: \$894
And here’s what it looks like with a .375 percent rate reduction costing two points:
• Price: \$200,000
• Loan amount: \$180,000
• Points: 2
• Rate: 4.125 percent
• Payment: \$872
Your lender can help you run the numbers as part of your loan preapproval process. That way, you know how to structure your offers.
## How lenders set mortgage rates
While you might hear a mortgage quote of say “4.25 percent plus one point.” Probably though, most lenders are likely to have a range of choices if you ask. You might also see:
• 4.00 percent plus 2 points
• 4.25 percent plus 1 point
• 4.50 percent plus 0 points
• 4.25 percent minus 1 point
• 4.50 percent minus 2 points
The image below is a screenshot of one lender’s wholesale mortgage rate sheet. Note that this is wholesale pricing; a retail customer would have to pay about one point more. That means the price with a rebate of at least one percent would be the “zero point” option — in this case, that’s about 4.375 for a 30-day lock.
When you see a quote with zero points that’s called the “par” price. If the quote is for a given interest rate PLUS points, it means the rate has been reduced, aka “discounted.” If you see a quote with MINUS points it means you agree to pay a higher interest rate and, in exchange, the lender will contribute cash to reduce your out-of-pocket closing costs.
4 ways to get a better mortgage rate quote
In other words, borrowers can engineer a mortgage rate to fit their circumstances. Have a good income but little in savings? Offer to pay a higher rate in exchange for the lender’s help with closing costs. Have big savings but worry about future wages? Pay points and get a smaller monthly cost for principal and interest.
## Pay points versus time
Rather than think of interest rates over 30 years — the usual term for a mortgage — it might be best to consider a shorter period. In practice, tenure in homes you buy tends to be much shorter.
The typical home is sold after 10 years, according to the National Association of Realtors while Freddie Mac calculates that on loans were refinanced after just 6.1 years as of the third quarter.
Mortgage rates vs. APR: how to get your best mortgage deal
Paying off mortgages early either because a property is sold or refinanced reduces the value of points. If we have a \$180,000 mortgage at 3.875 percent and pay one point, the annual percentage rate (APR) is 3.9578 percent over 30 years. The APR rises to 4.0890 percent if the same loan is paid off over 10 years.
For borrowers, the best way to see if points are worthwhile in your situation is to speak with loan officers and first ask for par pricing, the interest rate without points. Next, look at various combinations of rates and points. Lastly, consider how long you expect to hold the property and how much cash you have available.
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# Properties of a one to one function
A function simply gives you an output depending on the input given. In functions, a value is provided and the function performs some operations on it to give an answer. For example, the function f(x) = x + 1 adds 1 to any value that you give it. If you give it a 5, the function will give you a 6:f(5) = 5 + 1 = 6. Functions have requirements that they must meet, though. It can be the x value or the input and they cannot be linked to more than one answer. This simply means that you cannot give a function one value and it gives you two different answers. In a one to one function, there is a rule that gives a correspondence between elements in two given sets. Domain and range are used such that every element in a domain corresponds to only one element in the range. A one to one kind of function is a function in which each element in a range of that function corresponds to one element of a domain. A one to one function is often expressed as 1-1. It is important to note that y=f(x) is a function only when is passes the vertical line test. For a function to qualify to be a one to one function:
• It has to pass both the horizontal line test and the vertical line test. The horizontal line test states that a graph represents a one to one function if every horizontal line intersects that graph only once and the vertical line test states that a function should have only one output y for every unique input x.
• No two different elements in the domain should have the same element in the range.
One to one kind of functions are represented algebraically as x1 and x2. The x1 and x2 are any elements of the domain and a function is represented by f(x). f(x) is a one to one function if x1 is not equal to x2 which means f(x1) is also not equal to f(x2). Or if f(x1) is equal to f(x2) then x1 is equal to x2. One to one functions preserve distinctness in that they will never map distinct domain elements to the elements of their range. A function relates every value of the variable x input to a single value of the dependent variable y which is the output. It is possible to have two or more inputs giving a similar output. Consider the function y = x 3 − 6x 2 + 11x − 6. Using simple substitution, the inputs x = 1, x = 2 and x = 3 give the same output y = 0. In this particular case, three values of x are related to one value of y which means that it is a three to one function and not a one to one kind of function. A function in which the same output never repeats for different inputs is a one to one function.
## Examples of one to one functions
In a one to one function, any y given has only one x that can be paired with it. The one to one kind of functions are also known as injectives. If f(x) = x³ is a one to one function, this cubic function implies that every x value has a one unique y value that is not used by the other x elements. This is a characteristic of a one to one function. Let’s compare {(2, 3),(4, 5),(1, 5),(3, 4)} and {(2, 3),(4, 2),(1, 5),(3, 4)}. The first function has (4,5) and (1,5) which implies that the inputs 4 and 1 give a similar output 5. The first function here is not a one to one kind of function. There is no repetition of output on the second function which means the second function is a one to one function.
If you are asked to compare a function y=x2 with the function y=3x+1, the first function repeats the output of y=4 for the inputs x = 2 and x = −2 (4 = 22 and 4 = (−2)2. The function is a one to one type. The outputs of the second function are not repeated which implies that the function is a one to one function. In the horizontal line test, a horizontal line has all points which have their y coordinates equal to the same number. If the y coordinates are equal to a number 2 then the line can be described in an equation as y=2 which implies that every value of x is related to 2. Going by the definition of a one to one function, at most one x value is related to a given value of y. It then follows that a horizontal line intersects the graph of a function once. A graph of a function on a coordinate plane is a graph of a one to one function if there is no horizontal line intersecting the graph more than once.
One to one functions can also be inverse functions. If f is a one to one type of function with a domain A and a range B, its inverse function will be represented as f-1 and will have a domain B and a range A. It is defined by f −1 (y) = x only if f(x) = y for any y in B. To get the inverse of a one to one function you begin by replacing the function f(x) with y. 2. The next step is to interchange x and solve the equation for y. The final equation will be f −1 (x). If f is a one to one function with domain A and range B and its inverse function satisfies f −1 (f(x)) = x for each x in A and f(f −1 (x)) = x for each x in B, then the inverse of f −1 is f. An inverse function interchanges the domain and the range. The domain of f = Range of f −1 and the range of f = domain of f −1. A graph of an inverse function is gotten by reflecting the graph of f across the line y=x and it is only the one to one functions that can have an inverse.
### Verifying one to one functions
You can determine if functions are of the one to one kinds graphically by using the horizontal line test which states that a graph represents a one to one function if every horizontal line intersects that graph only once. But you cannot proof that a function is a one to one function by just looking at the graph. This is because a graph is a small portion of a function and you generally need to proof that it is one to one function on the whole domain. There are two main methods of verifying that a function is a one to one function.
The first method is showing if f(x1) is equal to f(x2) then x1 is equal to x2. This simply means that if a function has the same value at two points then those points must be equal. In other words, if a function contains the same value at two points, those points can never be different. This statement is simply what the horizontal line test states. A graph is a one to one function when there are no two x values which are assigned to the same y value. If you are asked to verify that f(x)=1 is a one to one function, you will begin by assuming that there exists some x1 and x2 in that f(x1)=f(x2). This means that 1×1=1×2 but 1×1=1×2⇒1×1−1×2=0⇒x1−x2x1x2=0⇒ the numerator should be equal to zero. For example, x2−x1=0⇒x2=x1.
The second method of verifying that a function is a one to one kind is by showing that a function is always decreasing or always increasing. And it will always pass the horizontal line test. A function f increases its domain whenever x2 is greater than x1. This means that f(x2) is greater than f(x1). As x becomes bigger f(x) also grows bigger. A function f decreases on its domain whenever x2 is greater than x1. This means f(x1) is bigger than f(x2). As x grows bigger, f(x) becomes smaller.
To verify that y=9-x2 is a one to one function you will first need to solve the equation to find the x value. y = 9 − x 2 0 = 9 − x 2 − y x 2 = 9 − y If y < 9 then 9 − y is positive. The equation y=9-x2 has two solutions for x, x = √ 9 − y or x = − √ 9 − y. For example, if y = 5 we find x = 2 or x = −2. The inputs 2 and −2 give the same result 5. This proves that the function y=9-x2 is a one to one function. Functions are foundational in science and mathematics. Whenever you have two sets of items that is x and y, then the functions demonstrate the relationships between them by giving the x values their corresponding y values. The various methods of verifying that functions are one to one functions show that you can never have two different input values of x that yield the same y results.
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# Belt method similiar to LCBM
Discussion in 'How-to's, Guides, etc.' started by miotatsu, Feb 9, 2012.
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1. ### miotatsuMember
27
0
Oct 21, 2010
Minnesota
miotatsu
Here I have documented the method that I use to solve the 3x3x3 Rubik's Cube.
This method could be described as a "belt method". Many of the ideas and algorithms come from another method called LCBM.
I came upon this method while trying to optimize HTA for speed solving.
I hope that these documents will be useful to others interested in this method.
Overview of method:
1. Edge orientation - place edges in the group <F2, B2, L, R, U, D> (or equivilant)
2. Belt - solve 4 edges of the equator (E slice)
3. Parity - create a situation in which corners can be solved with OCLL cases
4. EPOCBL - orient all corners while preserving equator and edge orientation
5. Seperation - permute corners to their correct layers
6. EPPCBL - permute all corners while preserving equator and edge orientation
7. L8E - solve the remaining eight edges
I am no teacher, so I will simply provide links that can explain each step better than myself.
Step 1: Edge Orientation:
This is the same as EOLine without the line.
http://cube.crider.co.uk/zz.php?p=eoline
Step 2: Belt:
Intuitive. Remember not to use single turns on F and B.
Step 3: Parity:
If you do not have OCLL cases on U/D:
Step 4: EPOCBL:
Memorize the algorithms for each case.
Step 5: Seperation:
Intuitive. Use only moves of the group <F2, B2, L2, R2, U, D>. I recommend only using <L2, R2, U, D>.
Do not forget that you must restore the equator after breaking it.
For example instead of bringing 2 U face corners from positions DLF and DLB to the top face with L2 you must do
something that will not destroy the belt such as:
U2 L2 U2 L2 U2
Step 6: EPPCBL:
Memorize the algorithms for each case.
Step 7: L8E:
I solve these using the exact same system used in Roux for LSE minus the edge orientation
http://grrroux.free.fr/method/Step_4.html (4b and 4c)
http://wafflelikescubes.webs.com/rouxmethod.htm (4b and 4c)
http://rouxtorial.webs.com/lse.htm (4b and 4c)
Example solves:
F2 R2 D2 R2 F2 D R2 U B2 L2 F U' R2 B' R' B F2 L2 U L'
EO: R' D R U2 B
Belt: U2 D' R' D' L
Parity: none
EPOCBL: U2 D2 L' U' L' U2 L U' L
Seperation: D' L2 U D L2
EPPCBL: D x2 R U R' F' R U R' U' R' F R2 U' R' x2
L8E: U2 M U2 M' D' M2 D M2 U M U2 M U M2 U x U2 M2 U2 M2
B' U2 B' L' R D2 F L R2 D2 U2 B' U' B' U' R2 F2 R D B U2 R U2 F2 L'
EO: D F' R2 D F2 D F'
Belt: U' D2 R' B2 L2
Parity: none
EPOCBL: D' L' U' L2 U' L' U2 D' L2 U' L'
Seperation: L2 U' L2
EPPCBL: D x2 F' L U L F D2 F' L' U' L' F U D x2
L8E: M2 U M U2 M' D M2 D' M2 U' M U2 M U' M2 U' M' U2 M'
U F2 D F2 U2 F2 L2 U' F2 U' R U' L2 D R U2 B' R2 B' R2 F'
EO: F D' B' D B
Belt: F2 R' U R L' D' L
Parity: L2 U' L2
EPOCBL: f R U R' U' f' F R U R' U' F'
Seperation: U2 L2 D' U2 L2
EPPCBL: U F' L U L F D2 F' L' U' L' F U D2
L8E: M' U2 M D M2 D' U M2 U2 M2 U' M U2 M' U2 M2
Algs:
EPOCBL Beginner (2-look) - 7 cases:
====================================================
Sune: on U: R U R' U R U2 R'
antiSune: on U: L' U' L U' L' U2 L
H: on U: y F (R U R' U') (R U R' U') (R U R' U') F'
U: on U: R2 D' R U2 R' D R U2 R
T: on U: r U R' U' r' F R F'
L: on U: y' F' r U R' U' r' F R
Pi: on U: f (R U R' U') f' F (R U R' U') F'
EPPCBL Beginner (2-look) - 2 cases:
====================================================
Y: F R U' R' U' R U R' F' R U R' U' R' F R F'
J: R U R' F' R U R' U' R' F R2 U' R' U'
EPOCBL Intermediate (1-look) - 25 cases:
====================================================
Sune: R U R' U R U2 R'
H: y F (R U R' U') (R U R' U') (R U R' U') F'
U: R2 D' R U2 R' D R U2 R
T: r U R' U' r' F R F'
L: y' F' r U R' U' r' F R
Pi: f (R U R' U') f' F (R U R' U') F'
sune/sune: R' U' R' U R U2 D R' D' R
anti/sune: R' U' R U2 R (U' D) R2 U R'
H/sune: U' R (U' D) M2 U' r2 U R'
U/sune: R' U2 R' U D R2 U2 R' U' R
T/sune: L D' L' U' r2 U2 L' U' L'
L/sune: R D r2 R' U R U2 r2 R'
Pi/sune: L U D L' U D2 L' U2 D L'
H/H: R2 (U' D) L (U D') M2 (U D') L
U/H: R D R' U D2 l2 U2 R' U' R'
T/H: L D L' U' D2 r2 U2 L' U' L'
L/H: L U' L2 R2 U' L' l2 D U2 L
Pi/H: R' U' L2 U R2 (U' D) r2 R' (U' D) R'
L/U: D' L' U' L' U2 L U' L
Pi/U: L' U R U2 R' U L R U R'
L/T: U R U D2 R' U' D2 R'
Pi/T: L' U' L l2 U' l2 U' L' U2 L
L/L: L U' D' L U D L'
Pi/L: L' U' L2 U' L' U2 D' L2 U' L'
Pi/Pi: L U D L' U2 D2 L U D L'
EPPCBL Intermediate (1-look) - 5 cases:
====================================================
Y: F R U' R' U' R U R' F' R U R' U' R' F R F'
J: R U R' F' R U R' U' R' F R2 U' R' U'
J/J: F2 U' R2 U D R2 D' F2
J/Y: F' L U L F D2 F' L' U' L' F
Y/Y: F2 U2 R2 U2 R2 U2 F2
EPOCBL Advanced (1-look) - 70 cases
====================================================
One case:
Sune: on U: R U R' U R U2 R'
Sune: on B:
antiSune: on U: L' U' L U' L' U2 L
antiSune: on B:
H: on U: y F (R U R' U') (R U R' U') (R U R' U') F'
H: on B:
U: on U: R2 D' R U2 R' D R U2 R
U: on B:
T: on U: r U R' U' r' F R F'
T: on B:
L: on U: y' F' r U R' U' r' F R
L: on B:
Pi: on U: f (R U R' U') f' F (R U R' U') F'
Pi: on B:
Two cases:
Sune: on B:
Sune: R' U' R' U R U2 D R' D' R
antiSune: R' U' R U2 R (U' D) R2 U R'
H/Double Sune: U' R (U' D) r2 R2 U' r2 U R'
U/Headlights: R' U2 R' U D R2 U2 R' U' R
T/Chameleon: L D' L' U' r2 U2 L' U' L'
L/Bowtie: R D r2 R' U R U2 r2 R'
Pi/Wheel: L U D L' U D2 L' U2 D L'
antiSune: on B:
Sune: L U L' U2 L' (U D') L2 U' L
antiSune: L' U L U' L U' D2 L U L'
H/Double Sune: U' L' (U D') l2 L2 U l2 U' L
U/Headlights: U R U' R2' D2 R U2 R U' R'
T/Chameleon: R' (U2 D2) R' D R2 U2 D R
L/Bowtie: L U R U' L U2 L2 U' R'
Pi/Wheel: R U2 L2 U R U D r2' R U R
H: on B:
Sune: D R (U D') L2 R2 U' r2 U R'
antiSune: D' L' (U' D) L2 R2 U l2 U' L
H/Double Sune: R2 (U' D) L (U D') L2 l2 (U D') L
U/Headlights: R D R' U D2 l2 U2 R' U' R'
T/Chameleon: L D L' U' D2 r2 U2 L' U' L'
L/Bowtie: L U' L2 R2 U' L' l2 D U2 L
Pi/Wheel: R' U' L2 U R2 (U' D) r2 R' (U' D) R'
U: on B:
Sune: R U' R2 D L' R2 U R' U' L
antiSune: L' U R U' L l2 U l2 U R'
H/Double Sune: R U R' U2 D R2 U2 R' U' R'
U/Headlights: U2 L' U L' U' r2 U' L U L'
T/Chameleon: R (U' D) R U' R' U2 D' R'
L/Bowtie: D' L' U' L' U2 L U' L
Pi/Wheel: L' U R U2 R' U L R U R'
T: on B:
Sune: L U' L' D' L2 U2 L' U' L'
antiSune: R' (U2 D2) R' U R2 U D2 R
H/Double Sune: L U L' U2 D' L2 U2 L' U' L'
U/Headlights: L2 U2 L' U' L U2 L' U' L'
T/Chameleon: R (U2 D2) R' (U2 D2) R'
L/Bowtie: U R U D2 R' U' D2 R'
Pi/Wheel: L' U' L l2 U' l2 U' L' U2 L
L: on B:
Sune: R U L2 U2 L' U R' U' L'
antiSune: x2 L U R U' L U2 L2 U' R'
L D R D' L D2 L2 D' R'
H/Double Sune: L U L2 U' L U' D2 L U2 L
U/Headlights: U' L' D' L' D2 L D' L
T/Chameleon: D R U2 D R' U2 D' R'
L/Bowtie: L U' D' L U D L'
Pi/Wheel: L' U' L2 U' L' U2 D' L2 U' L'
Pi: on B:
Sune: L U D L' U2 D L' U D2 L'
antiSune: D R U L2 U D R U2 r2 U' R'
H/Double Sune: R' U L2 U' R2 (U D') L2 R (U' D) R'
U/Headlights: R' U' r2 D2 U' R' U r2 U' R
T/Chameleon: L' U' L2 D2 L U2 D' L U' L
L/Bowtie: D L' U' D' L D R2 U L R2
Pi/Wheel: L U D L' U2 D2 L U D L'
EPPCBL Advanced (1-look) - 35 cases
====================================================
Case J/:
L: R U R' F' R U R' U' R' F R2 U' R' U'
R: (y) R' U L' U2 R U' R' U2 L R U'
F: R' U L' U2 R U' R' U2 L R U'
B: R U' L U2 R' U R U2 L' R'
Case Y/:
F R U' R' U' R U R' F' R U R' U' R' F R F'
Case /J:
L:
R:
F:
B:
Case /Y:
Case J/J:
L/L: F2 U' R2 U D R2 D' F2
L/R:
L/F:
L/B:
R/L:
R/R:
R/F:
R/B:
F/L:
F/R:
F/F:
F/B:
B/L:
B/R:
B/F:
B/B:
Case Y/Y:
F2 U2 R2 U2 R2 U2 F2
R2 U2 R2 U2 y' R2 U D' R2
R2 U2 R2 U2 F2 dD' R2
R2 U2 R2 U2 F2 E R2
U' R2 U2 R2 D2 F2 E' L2
Case J/Y:
L: F' L U L F D2 F' L' U' L' F
R:
F:
B:
Case Y/J:
L:
R:
F:
B:
676
0
Sep 9, 2010
Erzz197
The last thing I expected when I came to this site was to see "LCBM" in a thread title @_@
I'd try this out, but I really can't figure out your organization of algs. You have things like "Sune on B", which will not happen, since you have already solved parity. "H on B" is the same as having T on both, but harder to recognize (imo). You could clean it up a lot by removing those.
The algorithms I listed are the only possible cases (after solving parity). Without solving parity, there is a ridiculous increase in the number of cases, with most of them not able to be referenced as "known OLL on B".
Another thing, you don't really need to permute the corners of both layers, since it is rather easy to solve the corners of the D layer (or even the whole D layer) during separation. You could just use Y and J perms to solve the top corners, then go into L8E. However, if you can solve the full D layer during separation, a simple PLL will solve the cube (and there is no difference between this and normal LCBM ) I haven't been around for a while though, and am unsure if the moves saved by not solving the D corners during separation would justify the algorithms for EPPCBL. I would need to test some more.
3. ### StachuK1992statue
3,808
22
Jul 24, 2008
West Chester, PA
WCA:
2008KORI02
StachuK1992
Example solve with more optimized algs, and better L8E;
R2 D2 F2 R2 D L2 F2 D R2 F2 D L' D L F' R' F2 R' D' R2 F2
EO: D R F' x2 (3/3)
Belt: U' D2 R
D' U' L
(6/9)
CO parity; skip
CO: L' U' D' L' R2 D R2 U L (9/18)
Separate: L2 U2 L2 (3/21)
CP: D R2 D R2 U2 B2 U B2 U (9/27)
Finish L+R: D' L2 D' S2 D L2 D S2 (8/35)
M: U2 M' U2 M (4/39)
Not sure how many the L+R cases would be, I'll compute this today, likely.
568
1
Oct 8, 2010
Paris, France
No mistake ?
5. ### StachuK1992statue
3,808
22
Jul 24, 2008
West Chester, PA
WCA:
2008KORI02
StachuK1992
I can't see what I did wrong. :/
Here's another, hopefully I don't mess this one up.
U2 R2 F2 L' B2 L D2 L B2 L2 R' D' R2 U B2 L R2 D2 B' D
EO: D' B' (2/2)
Belt: U' D L
D' R D R' [finish right] (7/9)
CO Parity: U (R2 (R U R' U R U2 R') R2) simplifies to U R' U R' U R U2 R (8/17)
CO: U2 B2 U B L2 D2 L2 D2 B (9/28)
Separate: x2 U' D R2 U R2 (5/35)
CP: D' U2 [this was a /really/ easy force during separation] (2/37)
EP1: M' U2 M U2 [I've decided to just reduce to Roux after 4a here.] (4/41)
Roux 4b: x2 y M2 U M' U2 M' (5/46)
Roux 4c: U' M2 U2 M2 (4/50)
Last edited: Mar 7, 2012
568
1
Oct 8, 2010
Paris, France
I expected R U R' (right). U' D2 R does nothing good on my cube.
It's D' B', I think.
Thanks for this second sample.
27
0
Oct 21, 2010
Minnesota
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https://www.hellovaia.com/textbooks/math/discrete-mathematics-and-its-applications-7th/induction-and-recursion/q31e-prove-that-2-divides-whenever-n-is-a-positive-integer/
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Suggested languages for you:
Americas
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Q31E
Expert-verified
Found in: Page 330
Discrete Mathematics and its Applications
Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095
Prove that 2 divides ${n}^{2}+n$ whenever n is a positive integer.
2 divides ${n}^{2}+n$ whenever n is a positive integer
See the step by step solution
Step: 1
If n=1,
${1}^{2}+1=2$
it is true for n=1.
Step: 2
Let P(k) be true.
${k}^{2}+k$
We need to prove that P(k+1) is true.
Step: 3
$\begin{array}{r}\left(k+1{\right)}^{2}+\left(k+1\right)\\ ={k}^{2}+3k+2\\ =\left({k}^{2}+k\right)+\left(2k+2\right)\\ =\left({k}^{2}+k\right)+2\left(k+1\right)\end{array}$
It is true for P(k+1) is true.
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https://ukma.org.uk/why-metric/endorsements/an-architectural-modeller/
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# An architectural modeller
## Why metric is better
### Phil Durden
Fortunately I was born a few years after decimalisation and metric conversion had already started. This meant that my entire education throughout school and college was in metric. Consequently it is natural for me to use metric, although I was always aware that ‘older generations’ used different measurements.
My lifelong creative interests turned into a career in architectural model-making once I had left college. As I was entering the industry, the company I first worked for had just installed a CNC machine along with CAD (Computer-Aided Design) programs. We use these to draw up sections of a building then input the data into the CNC, which cuts out components to a high precision. Although some CAD packages do have the option of imperial units, I can only imagine the nightmare confusion that could result of trying to work in fractions of inches. In an industry where deadlines are often tight, there is no room for long-winded calculations: only a simple, user-friendly system is acceptable for the job.
Many years ago, architects’ plans were dimensioned in feet and inches, and drawn at bizarre scales such as 1/8th of an inch to a foot, which equates to an illogical 1:96. Imagine that on a drawing the height of the building is given as 27 feet 9 inches, then having to divide that by 96! Here goes… if every foot is represented by 1/8th of an inch, then 27 feet is 27/8ths of an inch (3 3/8 in). That is the ‘feet’ part. Then say that 9 inches is 3/4 of a foot, so 3/4 x 1/8 in = 3/32 in. Then add 3 3/8 in to 3/32 in, which is 3 15/32 in! The potential for errors is horrifying, and some drawings might have hundreds of such dimensions on them.
Thankfully architects’ plans have been drawn and dimensioned in metric for many years now, and always use rational scales, such as 1:100. This could not be easier – 1650 mm (millimetres) on the plans equals 16.5 mm on the model. What could be easier and quicker than that? And all done in the head instantly! Now with CAD to draw, CNC to cut out parts, and of course metric units used throughout, we can turn out a high-quality, accurate model in weeks rather than months.
As well as his day job, Phil also finds a bit of spare time for mountain-walking, cycling, working with young people and collecting. He detests house prices.
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# Convert square millimeters to square microns
Enter the number of square millimeters you want to convert to square microns in the text box below. The conversion to area in square microns will update as you type.
square millimeters (mm2)
1 square millimeters = 1,000,000 square microns
## Precise decimal and fraction answers
The conversion in square microns rounded to six significant decimal places is 1 mm2 = 1,000,000.000000 µ2. The precise fraction answer is 1,000,000 0/1 µ2.
### Decimal Precision
Conversion from one to ten decimal places
• 1 mm2 = 1,000,000.0 µ2
• 1 mm2 = 1,000,000.00 µ2
• 1 mm2 = 1,000,000.000 µ2
• 1 mm2 = 1,000,000.0000 µ2
• 1 mm2 = 1,000,000.00000 µ2
• 1 mm2 = 1,000,000.000000 µ2
• 1 mm2 = 1,000,000.0000000 µ2
• 1 mm2 = 1,000,000.00000000 µ2
• 1 mm2 = 1,000,000.000000000 µ2
• 1 mm2 = 1,000,000.0000000000 µ2
## Square millimeters to square microns conversion factor and formulas
Formula
The conversion factor between square millimeters and square microns is 1,000,000. This represents the ratio of square microns per one square millimeter. To calculate the value in square microns, we can either multiply by the conversion factor or divide by its inverse.
### Multiplication
To convert from square millimeters to square microns, multiply square millimeters by 1,000,000.
The formula is expressed as:
mm2 × 1,000,000 = µ2
### Division
Alternatively we can divide by the inverse conversion factor to get the same result.
The inverse of 1,000,000 is 1/1,000,000 or 0. To convert from square millimeters to square microns, divide square millimeters by 0.
The formula is expressed as:
mm2 ÷ 0 = µ2
A conversion from square millimeters to square microns are units within in the metric (SI) or SI compatible systems.
## Reverse this conversion
Convert from square microns to square millimeters instead.
## 1 square millimeters to other area units
### Metric
• 1 mm2 = 1×10-10 ha
• 1 mm2 = 1×10-12 km2
• 1 mm2 = 1×10-6 m2
• 1 mm2 = 1×10-4 dm2
• 1 mm2 = 0.01 cm2
• 1 mm2 = 1 mm2
• 1 mm2 = 1,000,000 µ2
• 1 mm2 = 1×1012 nm2
### Imperial/U.S.
• 1 mm2 = 2.471×10-10 acre
• 1 mm2 = 3.861×10-13 sq mi
• 1 mm2 = 1.196×10-6 sq yd
• 1 mm2 = 1.076×10-5 sq ft
• 1 mm2 = 0.002 sq in
## References and Citations
1. International Bureau of Weights and Measures (BIPM) The International System of Units (SI)
## Square millimeters to square microns conversion table
mm2 µ2
1 mm2 1,000,000 µ2
2 mm2 2,000,000 µ2
3 mm2 3,000,000 µ2
4 mm2 4,000,000 µ2
5 mm2 5,000,000 µ2
6 mm2 6,000,000 µ2
7 mm2 7,000,000 µ2
8 mm2 8,000,000 µ2
9 mm2 9,000,000 µ2
10 mm2 1×107 µ2
mm2 µ2
20 mm2 2×107 µ2
30 mm2 3×107 µ2
40 mm2 4×107 µ2
50 mm2 5×107 µ2
60 mm2 6×107 µ2
70 mm2 7×107 µ2
80 mm2 8×107 µ2
90 mm2 9×107 µ2
100 mm2 1×108 µ2
1,000 mm2 1×109 µ2
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ByUndeterminedCoefficients - Maple Help
Home : Support : Online Help : Education : Student Packages : ODEs : Computation : Solve : ByUndeterminedCoefficients
Student[ODEs][Solve]
ByUndeterminedCoefficients
Solve a linear ODE by the method of undetermined coefficients
Calling Sequence ByUndeterminedCoefficients(ODE, y(x))
Parameters
ODE - equation; a linear ordinary differential equation y - name; the dependent variable x - name; the independent variable
Description
• The ByUndeterminedCoefficients(ODE, y(x)) command finds the solution of a linear ODE by the method of undetermined coefficients. This method is applicable when the coefficients of y(x) and its derivatives are constant and the forcing function is of a certain form, typically involving polynomials, exponentials, and trigonometric functions. This method works by determining the general form of a particular solution based on the form of the forcing function, substituting the proposed particular solution into the ODE, and solving for the undetermined coefficients.
• Use the option output=steps to make this command return an annotated step-by-step solution. Further control over the format and display of the step-by-step solution is available using the options described in Student:-Basics:-OutputStepsRecord. The options supported by that command can be passed to this one.
Examples
> $\mathrm{with}\left(\mathrm{Student}\left[\mathrm{ODEs}\right]\left[\mathrm{Solve}\right]\right):$
> $\mathrm{ode1}≔\mathrm{diff}\left(y\left(x\right),x,x\right)+2\mathrm{diff}\left(y\left(x\right),x\right)+2y\left(x\right)=\mathrm{sin}\left(x\right)$
${\mathrm{ode1}}{≔}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{2}{}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{2}{}{y}{}\left({x}\right){=}{\mathrm{sin}}{}\left({x}\right)$ (1)
> $\mathrm{ByUndeterminedCoefficients}\left(\mathrm{ode1},y\left(x\right)\right)$
${{y}}_{{p}}{}\left({x}\right){=}{-}\frac{{2}{}{\mathrm{cos}}{}\left({x}\right)}{{5}}{+}\frac{{\mathrm{sin}}{}\left({x}\right)}{{5}}$ (2)
> $\mathrm{ode2}≔\mathrm{diff}\left(y\left(x\right),x,x\right)+4\mathrm{diff}\left(y\left(x\right),x\right)+4y\left(x\right)=\mathrm{exp}\left(-2x\right)$
${\mathrm{ode2}}{≔}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{4}{}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{4}{}{y}{}\left({x}\right){=}{{ⅇ}}^{{-}{2}{}{x}}$ (3)
> $\mathrm{ByUndeterminedCoefficients}\left(\mathrm{ode2},y\left(x\right)\right)$
${{y}}_{{p}}{}\left({x}\right){=}\frac{{{x}}^{{2}}{}{{ⅇ}}^{{-}{2}{}{x}}}{{2}}$ (4)
> $\mathrm{ode3}≔\mathrm{diff}\left(y\left(x\right),x,x\right)-4\mathrm{diff}\left(y\left(x\right),x\right)+4y\left(x\right)={x}^{2}$
${\mathrm{ode3}}{≔}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{4}{}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{4}{}{y}{}\left({x}\right){=}{{x}}^{{2}}$ (5)
> $\mathrm{ByUndeterminedCoefficients}\left(\mathrm{ode3},y\left(x\right)\right)$
${{y}}_{{p}}{}\left({x}\right){=}\frac{{1}}{{4}}{}{{x}}^{{2}}{+}\frac{{1}}{{2}}{}{x}{+}\frac{{3}}{{8}}$ (6)
Compatibility
• The Student[ODEs][Solve][ByUndeterminedCoefficients] command was introduced in Maple 2021.
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# math estimating w/fractions
posted by .
estimate w/mixednumbers:
8) 9 1/11 - 3 7/9
i got:
9 - 4 = 5
9) 4 1/2 - 1 24/25
i got:
4 - 2 = 2
12) 9 8/10 + 8 2/10
i got:
10 + 8 = 18
14) 4 5/9 + 5 9/10
i got:
4 + 6 = 10
• math estimating w/fractions -
All are correct -- except your teacher might expect a different answer for 14.
• math estimating w/fractions -
Why would 14 be different answer?
• math estimating w/fractions -
If you round these fractions, 4 5/9 is more than 4 1/2 so it could be rounded to 5.
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# Problem with a simple Stoke's Theorem question
Moham1287
## Homework Statement
Using Stoke's Theorem, evaluate the contour integral:
$$\oint F.dr$$
as an integral over an appropriately chosen 2 dimensional surface.
Use F = $$(e^{x}y+cos\siny,e^{x}+sinx\cosy,ycosz)$$ and take the contour C to be the boundary of the rectangle with the vertices (0,0,0), (A,0,0), (A,B,0), (0,B,0) oriented anticlockwise.
Then evaluate the same integral directly as a contour integral.
## Homework Equations
Stoke's Theorem,
$$\int\int_{S}(curl F).n\;d^{2}A=\oint F.dr$$
where n is unit normal vector.
## The Attempt at a Solution
I got Curl F to be (cos z)i by the standard method using a matrix. I set n as -k as the surface is in the xy plane so the normal vector is along the z direction and I got the negative by the right hand rule.
This gives:
$$\int^{0}_{A}dx\int^{0}_{B}(-1)k.(cosz)i\dy$$
(-k).(cos z)i is 0
which makes the double integral nothing
However when I solve the same thing directly as a contour integral I get an answer of -B.
Going anti clockwise from the origin, (so from (0,0,0) to (A,0,0) is I, (A,0,0) to (A,B,0) is II, (A,B,0) to (0,B,0) is III and (0,B,0) to (0,0,0) is IV) I get:
I=0
II=$$e^{A}B + sinA\;sinB$$
III=$$-(e^{A}B+sinA\;sinB)$$
IV=$$-B$$
Which, when added together, gives -B.... I think there must be a glaring error somewhere. I can write up my calculations for the contour integral if that would help solve it... Any help with this would be much appreciated.
## Answers and Replies
Homework Helper
Gold Member
This gives:
$$\int^{0}_{A}dx\int^{0}_{B}(-1)k.(cosz)i\dy$$
(-k).(cos z)i is 0
which makes the double integral nothing
Technically, the normal to the surface should be +k as given by the right-hand rule when your path is anticlockwise. Luckily, it didn't affect your answer in this case as -1*0 is still zero..
However when I solve the same thing directly as a contour integral I get an answer of -B.
Going anti clockwise from the origin, (so from (0,0,0) to (A,0,0) is I, (A,0,0) to (A,B,0) is II, (A,B,0) to (0,B,0) is III and (0,B,0) to (0,0,0) is IV) I get:
I=0
II=$$e^{A}B + sinA\;sinB$$
III=$$-(e^{A}B+sinA\;sinB)$$
IV=$$-B$$
Your error is in integral III; recheck that calculation and post your work for it if you can't find the error.
Moham1287
I got it! Thanks a lot. Ha, I somehow messed up with the right hand rule, which is pretty basic, but that wasn't the problem. I just made a copying error and forgot about one of the $$e^{0}B$$ which should have become B, not nothing. Thanks a lot!
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https://learn.careers360.com/school/question-provide-solution-for-rd-sharma-class-12-chapter-15-tangents-and-normals-exercise-fill-in-the-blanks-question-6/
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#### Provide Solution for RD Sharma Class 12 Chapter 15 Tangents and Normals Exercise Fill in the blanks Question 6
Slope of tangent =$0$
Hint:
First put $x= 1$ in equation $x= 3t^{2}+1$ , then find $t$ and $\frac{dy}{dx}$.
Given:
Given curve,
$x= 3t^{2}+1$ and $y= t^{3}-1$
To find:
We have to find the slope of tangent to the given curve at $x= 1$
Solution:
Given,
$x= 3t^{2}+1$ … (i)
$y= t^{3}-1$ … (ii)
Put $x= 1$ in equation (i), we get
\begin{aligned} & & x=3 t^{2}+1 \\ \Rightarrow & & 1=3 t^{2}+1 \\ \Rightarrow & & 3 t^{2}=0 \\ \Rightarrow & & t=0 \end{aligned}
Differentiating equation (i) and (ii) with respect to $x$ , we get
\begin{aligned} &\Rightarrow \quad \frac{d x}{d t}=6 t \text { and } \frac{d y}{d t}=3 t^{2} \\\\ &\therefore \quad \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \end{aligned}
$=\frac{3 t^{2}}{6 t}=\frac{t}{2}$
Thus we get the slope of tangent $\left(\frac{d y}{d x}\right)_{t=0}=\frac{0}{2}=0$
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# Matching on a circle: First and second moments of the distance
Consider a circle of unit circumference. Suppose that there are r equidistant red points on the circle. We will drop b>r blue points on the circle. The location of each blue point is iid and uniformly distributed on the circle.
We will match red and blue points so that each red point is matched to exactly one blue point, and each blue point is matched to at most one red. Consider one of the following matching processes:
1. Pick a red point uniformly at random and match it to the closest blue point. Remove the matched points. Iterate until all red points are matched.
2. Match blue and red points such that the total distance achieved by the matching is minimized.
Let $X_i$ denote the distance between the $i$th red point, and the blue point it is matched to. What are $\mathbb{E}[X_i]$ and $\mathbb{E}[X_i^2]$? How do they scale with $b$ and $r$?
Finally, suppose that as opposed to unit circle, we have unit disc or square. How do these quantities change?
EDIT: If it simplifies analysis, in matching process 1, it is fine to assume that each red point is matched to the blue point that is closest, in the clock-wise direction.
• It would be nice if you could drop a couple of lines on the context and the origin of this problem. Commented Aug 27, 2017 at 19:27
• I was reading about Poisson matching (e.g., see link), and wanted to explore matching problems with unequal/unbalanced number of points.
– Ozzy
Commented Aug 28, 2017 at 6:11
• The situation $1$ or the one in the edit are simple exercises in probability. E.g. $E[X_i^k] = 1/\binom{b+k}{k}$ in $1$.
– js21
Commented Aug 28, 2017 at 8:29
• @js21, can you elaborate on how you get the latter expression? I think I wasn't clear. Each red point is matched to exactly one blue point (and each blue point is matched to at most one red point). Moreover, in 1, as you match points, you do not replace the red points. In light of this, I don't see how the expression you derived can be obtained. I'll update the question with clarifications.
– Ozzy
Commented Aug 28, 2017 at 14:21
• @Ozzy: Ah, thanks, it is clearer now. It is indeed harder if matched blue points are deleted.
– js21
Commented Aug 28, 2017 at 14:30
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# Current and voltage direction in source-free RL and RC circuits
Question on obtaining the ODE
RC circuits :
Why is it that the current flows through the capacitor and the resistor when the voltage source is disconnected. Shouldn't it flow from the capacitor to the resistor since the voltage source is disconnected ?
RL circuits :
Why is it that the current flows from the inductor to the resistor here, but in the RC circuit it flows through the resistor and inductor. Also why is that the polarity of the inductor becomes opposite of that of the resistor when the voltage source is switched off.
• Ode (from Ancient Greek: ᾠδή ōidē) is a type of lyrical stanza. ODE to a passive circuit - How charged the flow of particles that define the current's swell, how plates apart can separate the voltages that dwell but if perchance attraction lays in fields of guassian lay where coils of wire resist the flow expressed by volts persay. Commented Jul 27, 2015 at 17:49
• The arrows are there just to indicate the reference direction so you know when to use a positive or negative sign.
– Curd
Commented Jul 28, 2015 at 6:32
You can define the direction of a current however you want. In the first diagram, $i_C$ will be negative if the capacitor voltage is positive at the start.
Similarly, you can define the polarity of a voltage however you want. In the second diagram, to satisfy KVL, $v_L$ must equal $-v_R$.
The currents in the first diagram make sense when a voltage source is applied to the capacitor and resistor. As the capacitor charges up, $i_C$ and $i_R$ will both be positive. When the voltage source is removed, $i_C$ becomes negative as the capacitor discharges.
The second diagram make less sense. When a voltage (or current) source is applied, the positive ends of $v_L$ and $v_R$ will be away from ground. When the source is removed, $v_L$ goes negative to maintain the inductor current, which means $v_R$ must also be negative. So I think the inductor and resistor should be shown with the same polarity, and the current should probably go towards the top of the inductor. But again, it's arbitrary -- you can choose whatever polarities you want, as long as you're consistent. The only difference is the negative sign.
EDIT: In response to your comments, I think you're misunderstanding what's going on. Capacitors have a continuous voltage; inductors have a continuous current. When a capacitor is charged up and the source voltage is disconnected, the capacitor current reverses to try to maintain the voltage. When an inductor is "charged" up and the source current is disconnected, the inductor voltage reverses to try to maintain the current.
The equations are:
$$i_C = C \frac {dv_C}{dt}$$ $$v_L = L \frac {di_L}{dt}$$
When a capacitor and a resistor are connected in parallel, you can use KCL:
$$i_C = -i_R$$ $$C \frac {dv}{dt} = -\frac v R$$
Likewise, when an inductor and a resistor are connected in series, you can use KVL:
$$v_L = -v_R$$ $$L \frac {di}{dt} = -iR$$
Solving these equations give you the usual first-order exponential decay. The initial conditions are the capacitor voltage and inductor current at t=0.
Here's a schematic showing what happens at t=0 in the standard example for this kind of problem:
simulate this circuit – Schematic created using CircuitLab
Before t=0, the switches have been closed for a long time. At t=0, the switches open.
• What I understood so far is that in the RC case when the voltage source is disconnected the voltage drop across the capacitor drops from the initial value to zero, and for this to happen dv/dt is infinity because it's a jump discontinuity, and that's undefined. For this reason the voltage drop across the capacitor's terminals becomes the initial value and the same current flows through the capacitor's terminals (ic in the picture). This also means that there is current flowing around the circuit starting from the capacitor's positive terminal (ir in the picture). Commented Jul 28, 2015 at 3:46
• The problem is that when the same logic is applied to the inductor case. The polarities of the inductor and capacitor must be the same, and there will be to currents one through the inductor and one through the resistor, but that's not the case in the picture. Commented Jul 28, 2015 at 3:48
• I've updated my answer. Commented Jul 28, 2015 at 5:13
The capacitor eventually acquires the potential of the battery,so it flows from +ve terminal of cap to -ve terminal.It flows from the capacitor to the resistor as shown in the diagram.In your case there was probably another source(of higher potential) getting on when the first source was disconnected.
Meanwhile an inductor opposes the flow of i through it,so after the source is disconnected the current flows in opposite direction.
With no source, there must be an initial condition to start proceedings. This may be stored energy: $CV^2/2$ for a capacitor or $LI^2/2$ for an inductor. Conveniently, the initial condition is expressed as the stored voltage across a capacitor or initial current flowing through an inductor.
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## How long will it take the 40 grams of I 131?
How long will it take for a 40.0 gram sample of I-131 (half life = 8.040 days) to decay to 1/100 its original mass? 1/100= 0.01. Suppose that n – is number of half lifes, then (1/2)n = 0.01 then n = 6.64. So, the time is 6.64*8.040 = 53.4days.
## What fraction of uranium 238 that was present when Earth was formed still remains?
The half life of Uranium-238 is 4.5 billion years and the age of earth is 4.5 billion years . What fraction of Uranium-238 that was present when Earth was formed still remains? _Only a single half live (4.5 Billion years) so ½ remains.
## What is the half-life of uranium-238 quizlet?
4.5 billion years
## Why is lead found in all deposits of uranium ores?
All uranium isotopes are radioactive. Uranium isotopes will decay overtime and become the isotopes of lead. These are the end products of the decay chain of uranium. That is why lead is found in all deposits of uranium ore.
## What happens to the uranium 235 nucleus when it is stretched out?
If the uranium nucleus is stretched into an elongated shape, electrical forces may push it into an even more elongated shape. In a typical example of nuclear fission, one neutron starts the fission of the uranium 235 atom and three more neutrons are produced when the uranium fissions.
## Why is there more carbon 14 in living bones than in once living ancient bones of the same mass?
Why is there more carbon-14 in living bones than in once-living ancient bones of the same mass? The carbon-14 is radioactive and is decaying, therefore depleting all the time. Hence, there is more Carbon-14 in bones of living objects as compared to the dead.
## Is nuclear force attractive or repulsive?
The nuclear force is powerfully attractive between nucleons at distances of about 1 femtometre (fm, or 1.0 × 10−15 metres), but it rapidly decreases to insignificance at distances beyond about 2.5 fm. At distances less than 0.7 fm, the nuclear force becomes repulsive.
## How is carbon-14 produced in the atmosphere?
Carbon-14 is produced in the stratosphere by nuclear reactions of atmospheric nitrogen with thermal neutrons produced naturally by cosmic rays (with the highest production rate 10 to 13 miles above Earth’s poles), as well as by atmospheric nuclear weapons testing in the 1950s and ’60s.
## What is the origin of most of the radiation you encounter?
An electron ejected from an atomic nucleus during the radioactive decay of certain nuclei. High-energy radiation emitted by the nuclei of radioactive atoms. What is the origin of most of the radiation you encounter? The leading source of natural occurring radiation is radon-222, cosmic, ground and human tissue.
## Why is radiation harmful to both living and nonliving things?
Long-term or high-dose exposure to radiation can harm both living and nonliving things. Radiation knocks electrons out of atoms and changes them to ions. It also breaks bonds in DNA and other compounds in living things. Exposure to higher levels of radiation can be very dangerous, even if the exposure is short-term.
## How can electromagnetic radiation affect living things?
Some forms of electromagnetic radiation, which is radiation found in different kinds of light waves, including ultraviolet light and X-rays, can cause damage to the DNA inside a living cell. When DNA is damaged by radiation, it can lead to cell death or to cancer.
## How does radiation affect the environment?
Similarly to humans and animals, plants and soil are also affected negatively from high amounts of nuclear radiation. Just like in humans, radioactive material can damage plant tissue as well as inhibit plant growth. Mutations are also possible due to the damage caused to the DNA.
## How long does radiation last in the environment?
Some stay in the environment for a long time because they have long half-lives, like cesium-137, which has a half-life of 30.17 years. Some have very short half-lives and decay away in a few minutes or a few days, like iodine-131, which has a half-life of 8 days.
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# 1001001 (number)
1,001,001 (one million one thousand one) is an odd seven-digits composite number following 1001000 and preceding 1001002. In scientific notation, it is written as 1.001001 × 106. The sum of its digits is 3. It has a total of 2 prime factors and 4 positive divisors. There are 667,332 positive integers (up to 1001001) that are relatively prime to 1001001.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 7
• Sum of Digits 3
• Digital Root 3
## Name
Short name 1 million 1 thousand 1 one million one thousand one
## Notation
Scientific notation 1.001001 × 106 1.001001 × 106
## Prime Factorization of 1001001
Prime Factorization 3 × 333667
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 1001001 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,001,001 is 3 × 333667. Since it has a total of 2 prime factors, 1,001,001 is a composite number.
## Divisors of 1001001
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 1.33467e+06 Sum of all the positive divisors of n s(n) 333671 Sum of the proper positive divisors of n A(n) 333668 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1000.5 Returns the nth root of the product of n divisors H(n) 2.99999 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,001,001 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 1,001,001) is 1,334,672, the average is 333,668.
## Other Arithmetic Functions (n = 1001001)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 667332 Total number of positive integers not greater than n that are coprime to n λ(n) 333666 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 78418 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 667,332 positive integers (less than 1,001,001) that are coprime with 1,001,001. And there are approximately 78,418 prime numbers less than or equal to 1,001,001.
## Divisibility of 1001001
m n mod m 2 3 4 5 6 7 8 9 1 0 1 1 3 1 1 3
The number 1,001,001 is divisible by 3.
## Classification of 1001001
• Arithmetic
• Semiprime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (1001001)
Base System Value
2 Binary 11110100011000101001
3 Ternary 1212212010010
4 Quaternary 3310120221
5 Quinary 224013001
6 Senary 33242133
8 Octal 3643051
10 Decimal 1001001
12 Duodecimal 403349
20 Vigesimal 652a1
36 Base36 lgdl
## Basic calculations (n = 1001001)
### Multiplication
n×y
n×2 2002002 3003003 4004004 5005005
### Division
n÷y
n÷2 500500 333667 250250 200200
### Exponentiation
ny
n2 1002003002001 1003006007006003001 1004010016019016010004001 1005015030045051045030015005001
### Nth Root
y√n
2√n 1000.5 100.033 31.6307 15.8521
## 1001001 as geometric shapes
### Circle
Diameter 2.002e+06 6.28947e+06 3.14789e+12
### Sphere
Volume 4.20138e+18 1.25915e+13 6.28947e+06
### Square
Length = n
Perimeter 4.004e+06 1.002e+12 1.41563e+06
### Cube
Length = n
Surface area 6.01202e+12 1.00301e+18 1.73378e+06
### Equilateral Triangle
Length = n
Perimeter 3.003e+06 4.3388e+11 866892
### Triangular Pyramid
Length = n
Surface area 1.73552e+12 1.18205e+17 817314
## Cryptographic Hash Functions
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Intermediate Algebra: Connecting Concepts through Application
$= \frac{1}{35}$
$\frac{w+2}{10w+15} \div \frac{7w+14}{2w+3}$ $= \frac{w+2}{10w+15} \times \frac{2w+3}{7w+14}$ $= \frac{w+2}{5(2w+3)} \times \frac{2w+3}{7(w+2)}$ $= \frac{1}{5(2w+3)} \times \frac{2w+3}{7(1)}$ $= \frac{1}{5(1)} \times \frac{1}{7(1)}$ $= \frac{1}{35}$
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Question Pool Arithmetic Sequences
# Arithmetic Sequences - SAMAGRA Question Pool & Answers | Class 10 English Medium
Kerala Syllabus SAMAGRA SCERT SAMAGRA Question Pool for Class 10 English Medium Maths Arithmetic Sequences
Qn 1.
a) Write a sequence by adding 3 with the multiples of 7.
b) Write its algebraic form.
c) Is 2000 a term of this sequence ?
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Qn 2.
The common difference of two arithmetic sequence are equal. The difference between their first terms is 10.
a) What is the difference between the second terms ?
b) What is the difference between the nth terms ?
c) What is the difference between the sums of first n terms ?
a) 10
b) 10
c) 10 x n = 10n
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Qn 3.
The sum of first n terms of an arithmetic sequence is 3n2 + 4n
a) What is its first term ?
b) Write the algebraic form of the sequence.
c) Find the 20th term.
a) 3+4 = 7
b) 6n + (7-6) = 6n+1
C) 6 x 20 +1 =
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Qn 4.
Is 44 a term of the sequence 4n + 3 ? Why ?
4 x 10 +3 = 43.
44 is not a term
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Qn 5.
In the arithmetic sequence 6,10,14,..
a) How much more is the 15th term than the 10th term ?
b) Which term is 32 more than the 20th term ?
a) 5 x 4 =
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Qn 6.
The commom difference of an arithmetic sequence is 6 and its 9th term is zero.
a) Write the 8th and 10th term.
b) Find the sum of its first 17 terms.
a) -6,
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Qn 7.
16th term of an arithmetic sequence is 60 and its 26th term is 90.
a) What is the differece between the 16th term and 26th term ?
b) What is the common difference?
c) Write the sequence
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Qn 8.
a) Write the half of natural numbers as a sequence
b) Write the integers in that sequence in order.
c) What will be the position of the number 23 in the first sequence ?
d) Find the sum of the first fifty terms of the first sequence.
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Qn 9.
The algebraic expression of an arithmetic sequence is 8n+11
a) Write the common difference of the sequence
b)What is the remainder got when each term of this sequence is divided by the common difference?
c) Is 11 a term of this sequence? Why?
a)Common difference=8
b) 3
c) No . First trerm is 19 and common difference is 8
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Qn 10.
In the Arithmetic sequence , , , .......
a) What is the common difference?
b) Which is the first integer term in the sequence?
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Qn 11.
The algebraic expression of an arithmetic sequence is 6n+1
a) Write the sequence?
b) What is the remainder when the terms are divided by 6?
c) Write the algebraic expression of the sequence obtained by the natural numbers which leaves a remainder 2 on division by 6?
a) 7,13,19,..
b) 1
c) 6n-4
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Qn 12.
The sum of first 15 terms of an arithmetic sequence is 300.
a) What is its 8th term?
b)If the first term is 6 find its common difference?
c) Write the algebraic expression of this sequence?
a) 20
b) 2
c) 2n+4
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Qn 13.
The algebraic expression of an arithmetic sequence is 5n-3.
a) What is its term?
b)Find the sum of first 39 terms ?
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Qn 14.
The sum of first 5 terms of an arithmetic sequence is 60. Write the sequence?
Mid term =12
For writing the sequence
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Qn 15.
a) Write the arithmetic sequence with first term 1 and common difference ½?
b)Find its 31st term?
c) Calculate the sum of its first 31 terms?
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Qn 16.
The 10th term of an Arithmetic sequence is 20 and 20th term is 10
a) What is its common difference?
b) Find its 30th term?
a) 10 common difference= - 10
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Qn 17.
a. What is the common difference of the arithmetic sequence 20,16,12,..?
b. How many positive numbers are there in this sequence?
c.Which is the first negative number in this sequence?
d. In which position did the first negative number occur in this sequence?
a.common difference = -4
b.Number of positive terms = 5
c.First negative number
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Qn 18.
a)What is the common difference of the arithmetic sequence -100,-96,-92,...?
b)Check whether Zero is a term in this sequence.
c) Which is the first positive number in this sequence?
a) Common difference = 4
b) Zero is a term.
c) First positive term = 4
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Qn 19.
a) Write the arithmetic sequence with first term 10 and common difference 4?
b)Write the difference of the terms in the same position of the above sequence and the arithmetic sequence 11,17,23,....?
c) Find the sum of first 20 terms of the sequence thus obtained?
a) 10,14,18,....
b)1,3,5,7,......
c)20²=
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Qn 20.
a) Write the sequence of perfect squares which are even.
b) Is it an arithmetic sequence?
a)4,16,36,64,...
b)This is not an arithmetic
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Qn 21.
The sum of first 9 terms of an arithmetic sequence is 360.
a)What is its 5th term?
b) If the sum of first 5 terms is 100 ,What is its third term?
c)What is the common difference of this sequence?
d)Write the sequence .
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Qn 22.
The sum of first 30 natural numbers is .
a) Find the sum of first 30 multiples of 6?
b)Write the algebraic form of the sequence obtained by subtracting 2 from the multiples of 6?
c)Find the sum of first 30 terms of this sequence?
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Qn 23.
The algebraic expression of an arithmetic sequence is 8n+5.
a)What is the common difference of this sequence?
b) What is the difference between the smallest four digit number and largest three digit number in this sequence
a) 8
b)
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Qn 24.
Consider the arithmetic sequence 4,6,8 ...
a) What is the 5th term in the sequence?
b)Write the ratio between the first and third terms of this sequence?
c)Write the ratio between the second and fifth terms of this sequence?
d)Which term satisfies the same ratio with the 10th term?
a) 12
b) 1:2
c) 1:2
d) 21st term
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Qn 25.
a) Which is the first number above 200 which leaves a remainder 3 on division by 7?
b)How many such numbers are there between 200 and 400?
c) Find the sum of these numbers?
a) For finding first number= 206
For finding last number = 395
b) For
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Qn 26.
a) Write the first three digit number which is a multiple of 9.
b) Write the sequence of all three digit numbers which are multiples of 9.
c) How many numbers are there in the above sequence ?
d) Find the sum of all these numbers.
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Qn 27.
4
8 12
16 20 24
28 32 36 40
... ... ... ... ...
... ... ... ... ... ...
a) Write the next two rows in this pattern .
b) Write the first and last terms of the 11 th row.
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Qn 28.
2
4 6 8
10 12 14 16 18
. . . . . . . .
a)Write the next line in this pattern.
b)How many numbers are there in the 10th row?
c)What are the last and first terms in the tenth row ?
a) 20,22,24,26,28,30, 32
b)20-1=19
c) Identifies the last terms as 2x1², 2x2², 2x3²,...
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# 1.2.3 Working with percentages
Table 1 used percentages, rather than actual numbers, to compare the number of people using the internet for each purpose. Numbers are expressed as if they are 'out of a hundred' when using percentages, which makes it easier to compare different values. You can recognise percentages by the % symbol which you can see at the top of the right-hand column of Table 1.
As you can see from halfway down the table, 50% of the people who had used the internet in the previous 3 months used it for buying or ordering things. (They may also have used it for other purposes.)
How many people is that? Obviously, that would depend on the total number of people. For argument's sake, we will say that 100 people in total had used the internet in the previous three months. That's relatively easy to work out: 50% of 100 is 50 out of 100, which is 50.
What if there were 200 people who had used the internet and 130 of them said that they used it for general browsing? This figure of 130 can be expressed as a percentage of 200.
Firstly, 130 is converted to a fraction of 200:
Then the fraction is multiplied by 100:
Which is:
(i.e. 13000 divided by 200)
which gives a result of 65.
If you wish, you can work this out using your Windows calculator; enter '130 / 200×100 ='.
We can now say that 65% of the 200 people who had used the internet said they used it for general browsing.
## Activity 5 (self-assessment)
Try the following three percentage calculations.
1. In a survey involving 700 people, 420 people said that they use the Internet for general browsing. What percentage is this?
2. In a group of 320 students, 120 said that they mainly use a laptop in their studies rather than a desktop computer. What percentage of students use a laptop?
3. On a production line for computer monitors, 3 out of 750 monitors are faulty. What percentage of monitors are faulty?
1. 420 / 700×100=60
So 60% of the people surveyed use the internet for general browsing.
2. 120 / 320×100=37.5
So 37.5% of the students use a laptop.
3. 3 / 750×100=0.4
So 0.4% of the monitors are faulty.
All the data in Table 1 is expressed in percentage form, but, if we know how many people were interviewed in the survey, we can work back from the percentage to find the actual number. Suppose 1700 people are interviewed in a survey and 56% of them say they have used the internet in the last 3 months. How many people does 56% represent? We can find out as follows.
First the percentage is expressed as a fraction of 100:
The fraction is then multiplied by the total number of people in the survey:
Using the Windows calculator, I enter '56 / 100×1700 ='. The result is 952. This means that 952 of the 1700 people surveyed used the internet in the last three months.
Now suppose that, according to the survey, of those 952 people who used the internet in the last three months, 85% used it for email. How many people out of the 952 used the internet for email? We can calculate the number by finding 85% of 952.
With the Windows calculator I enter '85 / 100×952 ='. The answer is 809.2 people, but common sense suggests that the answer ought to be 809.
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1. ## Eigenvalues/Eigenvectors
How do I determine what the original matrix was that yielded these two eigenvalues with the corresponding eigenvectors:
$\lambda_1 = -3$ Eigenvector: [0,1]
$\lambda_2 = 2$ Eigenvector: [1,0]
2. Originally Posted by Ideasman
How do I determine what the original matrix was that yielded these two eigenvalues with the corresponding eigenvectors:
$\lambda_1 = -3$ Eigenvector: [0,1]
$\lambda_2 = 2$ Eigenvector: [1,0]
Work:
$(\lambda + 3)(\lambda - 2) = \lambda^2 + \lambda - 6$
$(a - \lambda)(d - \lambda) - bc = \lambda^2 + \lambda - 6$
$\lambda^2 - a\lambda - d\lambda + ad - bc = \lambda^2 + \lambda - 6$
I thought the following matrix would work, but it didn't :
\left[ \begin {array}{cc} -2&2\\\noalign{\medskip}2&1\end {array}
\right]
I got the right eigenvalues, but not the right eigenvectors, grr.
3. See if this helps:
$det\begin{bmatrix}{\lambda}&1\\{\lambda}+6&{\lambd a}+2\end{bmatrix}={\lambda}^{2}+{\lambda}-6$
The correct matrix will work if $Ax={\lambda}x$
Where x is the eigenvector and $\lambda$ is the eigenvalue.
You are given eigenvalues and eigenvectors.
4. Originally Posted by galactus
See if this helps:
$det\begin{bmatrix}{\lambda}&1\\{\lambda}+6&{\lambd a}+2\end{bmatrix}={\lambda}^{2}+{\lambda}-6$
The correct matrix will work if $Ax={\lambda}x$
Where x is the eigenvector and $\lambda$ is the eigenvalue.
You are given eigenvalues and eigenvectors.
I tried. I can't figure it out.
EDIT: Yay I figured it out. Good old PDP^(-1). The matrix, incase you were wondering, is [[2,0],[0,-3]]
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http://gamedev.stackexchange.com/questions/14523/what-is-the-standard-solution-to-pathfinding-towards-a-moving-target/14526
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# what is the standard solution to pathfinding towards a moving target?
I am working on a 2D RTS like game, basic A* works perfectly for moving a unit from point A to point B.
But now I facing the continuous path-finding problem, like A attack a moving object B, call A* at each frame once Object B's position changed seems inefficient.
so what is the standard method to this problem?
-
There is no standard method. – Kylotan Jul 6 '11 at 12:48
There's no single standard method but there's a lot of literature about common methods to solve the problem. – user744 Jul 6 '11 at 14:52
From what I know, you could take a look at the D* algorithm which stands for "Dynamic A*". This algorithm is used to compute pathfinding for dynamic environment, here with a moving target.
Here's a paper using D* for moving target path finding : Moving Target D* Lite
-
One option is to only make a new path once every few frames. If you did it once or twice a second rather than 60+ times a second then the user is unlikely to notice unless they are both two very fast moving objects
-
+1 AI can reasonably run much slower than rendering. – Jon Purdy Jul 8 '11 at 15:47
You can use the "dog curve" approach which dogs apparently uses when hunting down someone. They calculate where the impactpoint will be "in the future" and sets of straight to that position.
A simple approximative way could be something along the lines :
A = NPC
B = Target
T = time to get to B:s position (B:s initial position)
Calculate where B will be in 'T' time (if B continues at the same speed / angle) and go there instead.
This is not the perfect way as the distance changes but much simpler than making a perfect solution and much better than just trying to get to 'B'.
-
Didn't know about dogs. I've learn something today! – SteeveDroz Jul 7 '11 at 8:37
A agree with Kylotan that there isn't a standard method.
One method I have seen work was to assume the target continues moving in the same direction and change the goal position as you run through your path finding algorithm. This does mean you have to hold two metrics in your A* nodes (cost and time as opposed to just cost).
To do better than that is very difficult. Unless you actually have knowledge about the unwavering path of the target, you're heading into the land of quite hard AI because you'll have AI's second guessing or simulating the target behaviour in order to guess where they will be and path towards that. This kind of AI is a real-time AB-game from game theory, an area that's not standard in any 3D game AI toolkit.
-
One way to do dynamic path finding is have the entity predict where the target is going and go there.
One way of doing this is using a Taylor series.
I'll call the path of the target over time the function S(t) where S is the position and t is the current time and the approximation to the path is A(f) and f is the date in the future one is approximating.
Then the simplest and most stupid approximation is A(f)=0.
The next simplest is A(f)=S(t) where t is the current time and f is the future. This is predicting the target simply stops in place.
The third simplest is A(f)=S'(t)*f+S(t) where S' is the derivative of S with respect to time. This is predicting the target continues on at a constant speed with no acceleration.
The fourth simplest is A(f)=S''(t)*f^2/2+S'(t)*f+S(t). This is predicting the target is accelerating at a constant speed like a falling ball.
I know this can be rephrased in terms of change in time which is probably more convenient for a game. Now S can be anything. It could be an X coordinate, it could be a Y coordinate, it could be the distance between the objects, it could be an angle. Also there's probable better methods of predicting the future path of an object so I'd look around a bit.
-
If the terrain is reasonably open and the target isn't too far from the pursuer, then you could use the intercept steering behavior. Essentially, you take the position and velocity of the target to calculate a position out in front of the target that isn't too far, and not too close, and you steer the pursuer towards that point (calculated each at regular intervals).
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# Let P-Q and R be the three vertices of a triangle- Prove that the line.docx
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# Let P-Q and R be the three vertices of a triangle- Prove that the line.docx
Let P,Q and R be the three vertices of a triangle. Prove that the line
segment that joints the midpoints of two of the sides is parallel to and half the length
of the third side.
Solution
in your triangle ABC, R is the mid point of AB and S is the mid point of AC. Now the basic proportionality theorem recalls the fact that when a line segment is drawn parallel to the base of the triangle say RS drawn parallel to BC, then AR/RB = AS/AC Using the converse of basic proportionality theorem, we get AR/AB = 1 as R is the mid point of AB. And AS/AC is also equal to 1 as S is the mid point of AC. Hence AR/RB = AS/AC establishes by converse of BPT theorem, RS becomes parallel to AB. So first part of the problem has been proved. ie the line segment drawn is || to the third side. For second part, as AR/AB = 1/1 by ratio and proportionality we get AR /(AR+RB = 1/2 ====> AR /AB = 1/2 Same way, we can prove AS/AC = 1/2 NOw in the two triangles ABC and ARS, by SAS rule the two triangles become similar. Hence RS / AB = 1/2 Or RS = AB/2 So second part is also proved. ie the segment is half of the third side.
.
Let P,Q and R be the three vertices of a triangle. Prove that the line
segment that joints the midpoints of two of the sides is parallel to and half the length
of the third side.
Solution
in your triangle ABC, R is the mid point of AB and S is the mid point of AC. Now the basic proportionality theorem recalls the fact that when a line segment is drawn parallel to the base of the triangle say RS drawn parallel to BC, then AR/RB = AS/AC Using the converse of basic proportionality theorem, we get AR/AB = 1 as R is the mid point of AB. And AS/AC is also equal to 1 as S is the mid point of AC. Hence AR/RB = AS/AC establishes by converse of BPT theorem, RS becomes parallel to AB. So first part of the problem has been proved. ie the line segment drawn is || to the third side. For second part, as AR/AB = 1/1 by ratio and proportionality we get AR /(AR+RB = 1/2 ====> AR /AB = 1/2 Same way, we can prove AS/AC = 1/2 NOw in the two triangles ABC and ARS, by SAS rule the two triangles become similar. Hence RS / AB = 1/2 Or RS = AB/2 So second part is also proved. ie the segment is half of the third side.
.
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Source: Official Guide for GMAT Review 2016 Data Sufficiency ; #9
2
# If i and j are integers, is
If i and j are integers, is i+j an even integer?
### 2 Explanations
1
Carlson Rainer
Hi - for statement two could you assume that if i+j is even, and we know that i=j, could you assume that any even + even or odd+odd would in fact be an even result? That is how I came up with my reasoning.
Jan 7, 2017 • Comment
Sam Kinsman
Hi Carlson,
Yes, that's right.
even + even = even
odd + odd = even
Therefore, if i=j, then the result of i+j must be even.
Jan 11, 2017 • Reply
2
Mike McGarry, Magoosh Tutor
Aug 17, 2015 • Comment
Stephen Mortensen
"Two times anything is even" should instead have been said "2 times any integer is even." 2 * 1.5 = 3, which is odd.
May 11, 2016 • Reply
Cydney Seigerman, Magoosh Tutor
Hi Stephen,
Thanks for pointing this out. You're completely right that the statement "2 times any integer is even" is more accurate. However, since we are told in the prompt that both i and j are integers, the statement in the explanation video is justified within the context of the question. That said, I've let the content editors know about the wording. However, since we are a small group with many projects in progress, it may be a while before the video is updated.
Thanks, again, for taking the time to tell us about this! :)
May 12, 2016 • Reply
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# Measurement (Math) Coloring Worksheets
July 27, 2016
## 2nd Grade Length Word Problems
This includes addition and subtraction word problems involving length or distance. These math coloring worksheets include pictures of monsters.
Students need to find the length of arrows (in cm or inches), and choosing the best unit or estimate for each given object. Images to be colored include cherries, corn, eggplant, jalapeno/chili and tomato.
## 4th Grade Length Word Problems
These word problems include addition, subtraction, multiplication or division involving length or distances. The mystery are pictures of monsters.
## 3rd Grade Volume and Mass Word Problems
These volume and mass word problems includes all four operations. There are six sets in this product, each with different combination of operations. This is a pirate-themed packet, so your kids can uncover pictures of an anchor, an island, a parrot, a pirate, a ship and a pirate skull.
## 4th Grade Mass Word Problems
These word problems involve solving for mass using the four operations. The hidden pictures are plants and a house.
## 3rd Grade Volume and Mass Word Problems
These volume and mass word problems includes all four operations. There are six sets in this product, each with different combination of operations. This is a pirate-themed packet, so your kids can uncover pictures of an anchor, an island, a parrot, a pirate, a ship and a pirate skull.
## 4th Grade Liquid Volume Word Problems
This packet includes word problems on adding, subtracting, multiplying or dividing numbers involving liquid. The mystery are pictures of a fairy and a wizard’s hat.
There are multiple activities related to telling time in this product such as: Writing the time shown on each clock, match the columns, drawing and writing the time, and choosing the right clock for each given time in words.
## 3rd Grade Time Word Problems
These are word problems on addition or subtraction of time. Some of the sets include a clock to help the students visualize the time. After answering the problems and coloring the worksheets, your kids should solve pictures of a bus, car, scoter, balloon, train, truck and space shuttle.
## 4th Grade Time Word Problems Mystery Pictures
Your students can uncover images of an alarm clock and a watch if they answer all word problems correctly.
## 2nd Grade Money Word Problems
This packet includes activities such as ‘If you have... How many cents/dollars do you have?’, Yes or No Word problems (ex. Does person A have enough money to buy?), and money word problems. The mystery pictures to be colored are castle, dragon, a king, knight and shield.
## 4th Grade Money Word Problems
These are word problems on adding, subtracting, multiplying and dividing numbers concerning money. If the answered and colored correctly, your kids will see pictures of a doctor and a scientist.
Comments will be approved before showing up.
July 27, 2016
July 27, 2016
July 27, 2016
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FAQ
# What’s the difference between pdf and histogram
## What is the purpose of using a histogram?
The purpose of a histogram (Chambers) is to graphically summarize the distribution of a univariate data set.
## Is PDF the same as distribution?
A discrete distribution can’t have a pdf (though it can be approximated with a pdf). “probability distribution” is often used for discrete distributions, e.g., the binomial distribution.
## What is a PDF distribution?
Probability density function (PDF) is a statistical expression that defines a probability distribution (the likelihood of an outcome) for a discrete random variable (e.g., a stock or ETF) as opposed to a continuous random variable.
## What is the difference between a discrete probability distribution and a histogram?
Sometimes, the discrete probability distribution is referred to as the probability mass function (pmf). The probability mass function has the same purpose as the probability histogram, and displays specific probabilities for each discrete random variable. The only difference is how it looks graphically.
## What are the disadvantages of using a histogram?
Weaknesses. Histograms have many benefits, but there are two weaknesses. A histogram can present data that is misleading. For example, using too many blocks can make analysis difficult, while too few can leave out important data.
## When should you not use a histogram?
1. It depends (too much) on the number of bins.
2. It depends (too much) on variable’s maximum and minimum.
3. It doesn’t allow to detect relevant values.
4. It doesn’t allow to discern continuous from discrete variables.
5. It makes it hard to compare distributions.
## Does a PDF sum to 1?
Even if the PDF f(x) takes on values greater than 1, if the domain that it integrates over is less than 1, it can add up to only 1. Let’s take an example of the easiest PDF — the uniform distribution defined on the domain [0, 0.5]. The PDF of the uniform distribution is 1/(b-a), which is constantly 2 throughout.
## Is a PDF always positive?
By definition the probability density function is the derivative of the distribution function. But distribution function is an increasing function on R thus its derivative is always positive.
## What is PDF and CDF in statistics?
The probability density function (pdf) and cumulative distribution function (cdf) are two of the most important statistical functions in reliability and are very closely related. … From probability and statistics, given a continuous random variable X,,! we denote: The probability density function, pdf, as f(x),!.
## How do you calculate data in a PDF?
1. To find c, we can use Property 2 above, in particular.
2. To find the CDF of X, we use FX(x)=∫x−∞fX(u)du, so for x<0, we obtain FX(x)=0.
## How do you find the CDF from a PDF?
1. By definition, the cdf is found by integrating the pdf: F(x)=x∫−∞f(t)dt.
2. By the Fundamental Theorem of Calculus, the pdf can be found by differentiating the cdf: f(x)=ddx[F(x)]
## What is difference between PDF and CDF?
Probability Density Function (PDF) vs Cumulative Distribution Function (CDF) The CDF is the probability that random variable values less than or equal to x whereas the PDF is a probability that a random variable, say X, will take a value exactly equal to x.
## What is discrete distribution in statistics?
A discrete distribution is a probability distribution that depicts the occurrence of discrete (individually countable) outcomes, such as 1, 2, 3… or zero vs. one. … Statistical distributions can be either discrete or continuous.
## What are the steps to creating a histogram?
1. On the vertical axis, place frequencies. Label this axis “Frequency”.
2. On the horizontal axis, place the lower value of each interval.
3. Draw a bar extending from the lower value of each interval to the lower value of the next interval.
## What is the height of a uniform distribution?
For a uniform distribution, the height f(x) of the rectangle is ALWAYS constant. in the 14 to 20 pound class are uniformly distributed, meaning that all weights within that class are equally likely to occur.
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# A Shortcut for Multiplying by 25
## Here's a quick way to multiply numbers by 25: First, divide the multiplicand by 4, and then multiply that quotient by 100.
Note: You can multiply by 100 first, and then divide by 4 to get the same result since multiplication & division are on the same calculation level!
## Examples:
25 × 17 = 425
because
17 ÷ 4 = 4.25; 4.25 × 100 = 425
25 × 60 = 1,500
because
60 ÷ 4 = 15; 15 × 100 = 1,500
25 × 153 = 3,825
because
153 ÷ 4 = 38.25; 38.25 × 100 = 3,825
This trick works since 25 = 100 ÷ 4. Frankly, it still works even if the multiplicand is not an integer, but these examples are here to teach you how to multiply integers by the number 25!
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## Thursday, March 25, 2010
### Deniel's surface area growing post
Okay guys here is my surface area growing post about right rectangular prism, right triangles, and cylinders
Here is my first surface area question:
Okay here is what I did:
front:
13.5 cm x 18.5cm =area 13.5 x 18.5= 249.75cm x 2 = 499.5cm
side:
5 cm x 13.5cm =area 13.5 x 5= 67.5cm x 2 = 135cm
length:
18.5 cm x 5cm =area 18.5 x 5= 92.5cm x 2= 185cm
499.5+135+185= 819.5
Area= 819.5cm
Here is my second question:
Okay so here's what I did:
A= l x w A= (b x h) / 2
A= 9 x 3 A=(3 x 2.6) / 2
A= 27 The area on one rectangle is 27m A=7.8 / 2
A=3.9 and its the area of one triangle
Surface Area= (3 x rectangle) + (2 x area of triangle)
= (3 x 27) + (2 x 3.9)
= 81 + 7.8
= 88.8
So the surface area is 88.8 centimetre squared.
Here's my third question:
Here's what I did:
Determine the radius of each circle:
7.5 / 2 = 3.75 so 3.75 is the radius of each circle
Find the area of the circle
A=π x r2
A=3.14 x 3.75 x 2
A= 44.15625 so the area of one circle is 44.15625 x 2 = 88.3125 for both circles.
Find the area of the rectangle using the circumference of the circle.
A= l x w
A=(π × d) × w
A= 3.14 × 7.5 × 11
A= 259.05
The area of the rectangle is approximately 259.05 cm2.
Calculate the total surface area.
Surface area = 88.3125 + 259.05
= 347.3625
The total surface area is approximately 347.36 cm2
So here is my surface area growing post :)
### Surface Area Growing Post
Heey guys!! sorry this post maybe late but I managed to finish it on time (:
The calculations for the triangular prism is
10cm , 12cm 4cm , and 5cm .
P.S - sorry if you cant read either one of these questions.
AND please correct me if you see anything wrong
## Wednesday, March 24, 2010
### Nhea's Growing Post 3/24/10
A)
B)To get the answer to this Question -
All you have to do is multiply the first number (4) by the numerator (1) but the demoniator always stays the same so you would get the answer (4/2) then you have to simplify if possible (1 2/2).
C)To Answer this question -All you have to do is use a picture (or number line) to determine the answer. Using a picture all you have to do is make a square. The number of squares going down is the same number as the single number (2) and the squares going across is the same number as the first denominator (4). Then shade in the the number of the single number (2) and use a different colour to shade in the number of the numerator (1) over the already shaded part to get your answer (1/8)
D) To answer this question -
All you have to do is shade in the number of squares that match the single number, 5, then use a different colour to re-shade the number that is the numerator, 1. Then you would get 1/15 as your answer.
Part 2
Fractions
### Althea's Fraction Growing post
HELLO EVERYONE :D
page 202, question 6
a.) 4 x 1/2
b.) 3 x 7/10
c.)5 x 2/3
this is a different way to solve the problem
5/1 x 2/3
so i mutiplied numerator which is 5 x 2= 10
then i multiplied the dinominator which is 1 x 3= 3
10/3
10 divided by 3 = 3 1/3
d.) 3 x 3/8
3 x 3= 9
1 x 8= 8
9/8.....9 divided by 8= 1 1/8
page 208 question 5
a) 3/5 dvided by 2
b.) 1/5 divided by 3
c.) 1/2 divided by 4
d.) 2/3 divided by 6
page 208, question 7 (word problem)
____________________________________
Part 2
Part 3
Explain:
1 1/2 divided by 3/4
• Find the lowest common denominator (LCD).
•Lowest common denominator of 2 and 4 is 4
___________________________________________________________
3/4 divided by 1/2
•Find the LCD ( lowest common denominator)
•LCD of 4 and 2 is 4
1 x 2=2
2x2=4
Now, we have to know how many 2/4 goes into 3/4.
### Fraction Growing Post
For Fractions I'm doing is Questions 6. determine each product using manipulative or diagrams
a) 4x1/2 b) 3x 7/10
c) 5x 2/3 d) 3 x 3/8
Then I'm going to show you how to divide Question 5-4
a) 1/4 } 2 b) 1/3 }3
c) 1/5 }2 d) 5/6 }4
Question 5 .
a) 3/5 } 2 b) 1/5 } 3
c) 1/2 }4 d) 2/3 } 6
Fraction part #2 MULTIPLYING Mixed fraction, Improper fractions and Proper fraction.
Proper Fractions
Questions through 3-11
3.
a) 5/6 x 1/2 = 5/12 b) 3/4x 5/6 = 15/24.
The step is to multiply the numerator and the denominator.
4.
a) 1/4 x 2/3= 2/12 or 1/6 b) 7/10 x 1/2= 7/20
5.
a) 3/8x 2/3= 6/24 b). 3/7 x 1/6 =3/42 c) 3/4 x 3/4= 9/16
6.
a) 2/5 x 4/5 =8/25 b) 7/8 x 4/5 = 28/40 c) 3/4 x 4/9 = 12/ 36
7. Tamar had 1/2 of an apple pie in her refridgerator. She ate 1/4 of this pie.
1/2 x 1/4 = 1/8 8. 1/3 x 1/4 = 1/12 9. 1/20 x 1/10= 1/200 10. 3/5 x 1/2 =3/10 11. 6/7 x 7/18 =42/126
That's multiplying proper fractions. This is how you multilying Mixed fractions and improper Question 4.
That's when you ever have a improper fraction and you get the anwer right you have mixed fraction.
4.
Dividing fraction and Mixed numbers
and I don't get with number two 3/4 divide 1/2 :?
### Monica's Growing Post
Me Nadine and Danielle, doing surface area questions.
top
lxw
12x4.5
54x2
=108
side
lxw
10.4x4.5
46.8x2
=93.6
front
lxw
12x10.4
124.8x2
=249.6
TSA:108+93.6+249=451.6 cm2
bxh/2
8x7/2
56/2
=28
lxw
10.6x2.5
=26.5
lxw
2.5x7
=17.5
lxw
8x2.5
=20
tsa: 28+28+26.5+17.5+20=120cm2
r=d/2
r=9.5/2
r=4.75
2*3.14*r*r
2x3.14x4.75x4.75
=141.70
3.14xdxh
3.14x9.5x11
=328.13
tsa:140.70+328.13= 469.82 m2
## Sunday, March 21, 2010
### Jusitn's Surface Area Growing Post
TSA of rectangular prism
The rectangular prism has rectangular faces.
So, use the formula A = lw to find the area of the faces.
For both the front and back faces, l = 5 in. and w = 2 in.
So, the area of the front and back faces is,
5 2 2 = 20 in.2
For both the top and bottom faces, l = 14 in. and w = 5 in.
So, the area of the top and bottom faces is,
14 5 2 = 140 in.2
For both the sides, l = 14 in. and w = 2 in.
So, the area of the two sides is,
14 2 2 = 56 in.2
Add the area of the faces.
20 + 140 + 56 = 216
So, the total surface area of the given solid is 216 in.2.
TSA of triangular prism
The triangular prism has triangular bases.
So, to find the area of the bases, use the formula:
Substitute 6 for b and 8 for h.
Therefore, the area of the two bases is,
2(24) = 48 cm2
The triangular prism has rectangular sides.
So, use the formula A = lw to find the area of the sides.
The area of the front side is,
13 8 = 104 cm2
The area of the bottom side is,
13 10 = 130 cm2
The area of the back side is,
13 6 = 78 cm2
Find the sum of the base areas and the area of the sides.
48 + 104 + 130 + 78 = 360
So, the total surface area of the given solid is 360 cm2.
TSA of a cylinder
The radius of the circular base is,
he area of a circle is, A = πr2 where r is the radius.
Replace π with 3.14, r with 2.4.
A (3.14)(2.4)2
A = 18.0864
So, the area of one base is 18.0864 cm2.
The sum of the areas of the two bases is,
18.0864 + 18.0864 = 36.1728 cm2
This video is on how to find the tsa of a cylinder , rectangular prism, and triangular prism. It was created by Warren, Bruce and I.
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# Overflowing Tank Example
Overflowing Tank Example
© 30 May 2014 by Mathieu Westerweele
In a previous blog post we discussed about constraints and assumptions that describe all kinds of algebraic relationships between process quantities which have to hold at any time.
As mentioned, these assumptions are usually introduced with a goal, namely to simplify the description of the behaviour by neglecting what is considered insignificant in the view of the application one has in mind for the model. Whilst indeed such assumptions do simplify the description, they are also the source of numerous problems, such as “index problems”, which make the solving of the equations very difficult.
We will take a look at an example of an assumption that will lead to a high index model, but first we will look at the same example with no constraining assumptions:
A process consists a mixing tank that flows over into a second tank (see figure). The liquid that is fed to the tank system contains two components (A and B). We are interested in the (dynamic) concentration change of the components (in the first tank) over time.
To keep the explanation simple and easy to follow, the (mass and molar) density will be considered to be constant and we will only consider mass. A well posed system of equations could look as follows (you will see a list of equations. In front of each equation is noted which variable is being solved from this equation):
Balance equations:
n_A: d(n_A)/dt = F1_A – F2_A
n_B: d(n_B)/dt = F1_B – F2_B
System equations:
c_A: c_A * V = n_A
c_B: c_B * V = n_B
m: m = M_A * n_A + M_B * n_B
V: m = Dm * V
Connection Equations:
F1_A: F1_A = c_A0 * F1
F1_B : F1_B = c_B0 * F1
F2: IF V > Vmax THEN
F2 = K * (V – Vmax)
ELSE
F2 = 0
END
F2_A: F2_A = c_A * F2
F2_B: F2_B = c_B * F2
Where:
n_i :: molar mass of component i, in [mol]
c_i :: molar concentration of component i, in [mol/m3]
m :: total mass, in [kg]
M_i :: molar mass of component I, in [kg/mol]
V :: Volume of system, in [m3]
Dm :: density in [kg/m3]
F# :: Volumetric flow rate of connection #, in [m3/s]
F#_i :: molar mass flow of component i through connection #, in [mol/s]
K :: Constant, in [1/s]
F1 and c_i0 are a function of boundary conditions and can be considered given.
Note: The equations F2 = K * (V – Vmax) is a simplification that seems to work very fine in most situations. It just means that the higher the value of V (above Vmax), the bigger the flow F2 will be. It will automatically find an “equilibrium value”. The K represents a measure for the “resistance to overflow”. If K has a very high value, V will always be close to Vmax.
Let’s now suppose that it was diffcult to find a reliable relation that describes the flow F2 out of the overflow tank. Therefore the assumption is made that the liquid level in the tank is constant (this means that the liquid volume dynamics are assumed to be very rapid compared to the composition dynamics, which is theoretically the same as taking K equal to infinity). So, we now have a constraint that V = Vmax and we no longer have an equation to calculate F2.
This assumption of constant volume is, of course, physically incorrect. A physically more correct description of the overflow would have included a relation between the liquid level and the amount of liquid which is flowing out of the tank (for example the equation for F2 in the well posed model). Though practically, the variation in the volume is hardly ever relevant in these circumstances and the resulting low index model will be stiff.
So, let’s have a look at the equations that would be needed for this simple model and how the simplifying assumption can lead to a “high index model” that is hard to solve for most equation based solvers.
(1) d(n_A)/dt = c_A0 * F1 – c_A * F2
(2) d(n_B)/dt = c_B0 * F1 – c_B * F2
(3) c_A * V = n_A
(4) c_B * V = n_B
(5) m = M_A * n_A + M_B * n_B
(6) m = Dm * V
(7) V = Vmax
The equations for this model are, of course, almost the same as the ones for the previous example. The only exception is that there is no longer an equation to solve F2 from and instead we have a constraint on the total volume. The problem with solving this set of equations is that the variable F2 does not have an equation but does appear in the balance equations, which will be impossible to solve for most equation based solvers (some solvers have built-in index reduction algorithms, but that’s not the topic of this blog).
One way of “transforming” this set of equations into a set that can be solved, is to apply an index reduction by differentiating some of the equations.
The constraint equation (V = Vmax) constrains the volume and makes the volume V no longer state (nor time) dependent. The volume is related to the mass of the system (equation 6) and the mass is related to the component masses (equation 5). Differentiating equations (5) and (6) results in:
(5’) d(m)/dt = M_A * d(n_A)/dt + M_B * d(n_B)/dt
(6’) d(m)/dt = 0
which results in the following additional algebraic equation:
0 = M_A * c_A0 * F1 + M_B * c_B0 * F1 – M_A * c_A * F2 – M_B * c_B * F2
Removing one of the states of the original equations (either n_A or n_B) and replacing its differential equation with the new algebraic relation results in an index-1 DAE, which can be easily solved by a DAE solver.
Quite obviously, when the new algebraic equation is worked out, it will follow that, in this simple case (constant density), the volumetric outflow F2 will be the same as the volumetric inflow F1:
0 = (M_A * c_A0 + M_B * c_B0) * F1 – (M_A * c_A + M_B * c_B) * F2
0 = (M_A * n_A0 + M_B * n_B0) * F1/V0 – (M_A * n_A + M_B * n_B) * F2/V
0 = m0/V0 * F1 – m/V * F2 = Dm*F1 – Dm*F2
==> F1 = F2
Note: Very often, similar examples are used in textbooks without anybody noticing that there is (or was) an index problem related to the example. This has the following reasons: The total mass balance and all but one component mass balances (mostly the solvent is left out) are written down and, without knowing it, no link is being made between the component masses and the total mass:
d(m)/dt = Dm*F1 – Dm*F2
d(n_A)/dt = F1_A – F2_A
c_A = n_A / V
V = m / Dm
Making the assumption of constant density and constant volume results in constant total mass, which easily gives:
0 = Dm*F1 – Dm*F2 ⇒ F2 = F1
When a modeller is not aware of the fact that the component masses were not related to the total mass in this case, he could get into trouble when things get a bit more complicated (for example when the density is a function of the mole fractions). So, while this solution works fine for “academic” problems, it’s not always wise to solve industrial problems this way. Unfortunately, many engineers are often not even aware of the assumptions they are taking and what the implications of such assumptions can be.
What is your experience with constraints and assumptions? Did you ever run into computational problems without knowing what the cause was?
I invite to post your experiences, insights and/or suggestions in the comment box below, such that we can all learn something from it.
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6 messages
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Olson likes strong coffee, the stronger the better. But he can’t distinguish small differences. Over the years, Mrs. Olson has discovered that if she changes the amount of coffee by more than one teaspoon in her six-cup pot, Olson can tell that she did it. But he cannot distinguish differences smaller than one teaspoon per pot. Where A and B are two different cups of coffee. Suppose that Olson is offered cups A, B, and C all brewed in the Olsons’ six-cup pot. Cup A was brewed using 14 teaspoons of coffee in the pot. Cup B was brewed using 14.75 teaspoons of coffee in the pot and cup C was brewed using 15.5 teaspoons of coffee in the pot. Q. Is Olson’s “at-least-as-good-as” relation, transitive? And, answer is NO! I can't understand why! Please help!
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I think its like this - C has the highest amount of caffeine- but B differs from C by less than 1 tea-spoon coffee - so pakka C and B are indifferent options. Now B and A are indifferent choices - differences is again less than 1 tea-spoon coffee. she cant tell difference. But are C and A indifferent options? No, because the difference exceeds 1 tea-spoon. so C is preferred to A so, C indifferent to B , B indifferent to A, but is C indifferent to A? (by transitivity rule) no! thus transitivity violated. “Operator! Give me the number for 911!”
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Pardon me if I am wrong, but C indifferent to B , B indifferent to A, but is C indifferent to A?(by transitivity rule) no! though transitivity is not violated as we are concerned about “at-least-as-good-as”. So, either C and A should be indifferent or C should be preferred to A. and indeed, C is preferred to A! And that makes me feel like answer given in Varian workouts for this problem is wrong. Please help! (I'm posting here the problem from Varian workouts as image as I can't type some notations.)
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It is not transitiv as A>=B is tru,B>= C is true but A>=C is false.
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## 10042
10,042 (ten thousand forty-two) is an even five-digits composite number following 10041 and preceding 10043. In scientific notation, it is written as 1.0042 × 104. The sum of its digits is 7. It has a total of 2 prime factors and 4 positive divisors. There are 5,020 positive integers (up to 10042) that are relatively prime to 10042.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 7
• Digital Root 7
## Name
Short name 10 thousand 42 ten thousand forty-two
## Notation
Scientific notation 1.0042 × 104 10.042 × 103
## Prime Factorization of 10042
Prime Factorization 2 × 5021
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 10042 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 10,042 is 2 × 5021. Since it has a total of 2 prime factors, 10,042 is a composite number.
## Divisors of 10042
1, 2, 5021, 10042
4 divisors
Even divisors 2 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 15066 Sum of all the positive divisors of n s(n) 5024 Sum of the proper positive divisors of n A(n) 3766.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 100.21 Returns the nth root of the product of n divisors H(n) 2.66614 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 10,042 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 10,042) is 15,066, the average is 376,6.5.
## Other Arithmetic Functions (n = 10042)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 5020 Total number of positive integers not greater than n that are coprime to n λ(n) 5020 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1235 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 5,020 positive integers (less than 10,042) that are coprime with 10,042. And there are approximately 1,235 prime numbers less than or equal to 10,042.
## Divisibility of 10042
m n mod m 2 3 4 5 6 7 8 9 0 1 2 2 4 4 2 7
The number 10,042 is divisible by 2.
• Semiprime
• Deficient
• Polite
• Square Free
## Base conversion (10042)
Base System Value
2 Binary 10011100111010
3 Ternary 111202221
4 Quaternary 2130322
5 Quinary 310132
6 Senary 114254
8 Octal 23472
10 Decimal 10042
12 Duodecimal 598a
20 Vigesimal 1522
36 Base36 7qy
## Basic calculations (n = 10042)
### Multiplication
n×i
n×2 20084 30126 40168 50210
### Division
ni
n⁄2 5021 3347.33 2510.5 2008.4
### Exponentiation
ni
n2 100841764 1012652994088 10169061366631696 102117714243715491232
### Nth Root
i√n
2√n 100.21 21.5745 10.0105 6.31486
## 10042 as geometric shapes
### Circle
Diameter 20084 63095.7 3.16804e+08
### Sphere
Volume 4.24179e+12 1.26721e+09 63095.7
### Square
Length = n
Perimeter 40168 1.00842e+08 14201.5
### Cube
Length = n
Surface area 6.05051e+08 1.01265e+12 17393.3
### Equilateral Triangle
Length = n
Perimeter 30126 4.36658e+07 8696.63
### Triangular Pyramid
Length = n
Surface area 1.74663e+08 1.19342e+11 8199.26
## Cryptographic Hash Functions
md5 425f116bf53f051c57d1670a04fb4a0c 22cdd715692633b42daa8a36335d7fb20ff4cd0e ac7e951aab0f71a1c553da5674cbb917070f937d4f47dcc0ad72826c5b779cb5 9a15d0918ef3d9fe861d20f0d72ec706dd09d36679ca773973692c826b1da8ece995c3aa16806f2ded8eb15492e1fbc1ba94408a2a7bbe03bd33c36a3c282230 9696fb0e2497eeb9ea2041c4db50aff4335d42a9
| 1,427 | 3,987 |
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https://numberworld.info/6999
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# Number 6999
### Properties of number 6999
Cross Sum:
Factorization:
Divisors:
1, 3, 2333, 6999
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
1b57
Base 32:
6qn
sin(6999)
-0.45148792686144
cos(6999)
0.89227722816306
tan(6999)
-0.50599512417337
ln(6999)
8.8535225606895
lg(6999)
3.8450359935134
sqrt(6999)
83.660026296912
Square(6999)
### Number Look Up
Look Up
6999 (six thousand nine hundred ninety-nine) is a very amazing number. The cross sum of 6999 is 33. If you factorisate the figure 6999 you will get these result 3 * 2333. The figure 6999 has 4 divisors ( 1, 3, 2333, 6999 ) whith a sum of 9336. 6999 is not a prime number. The figure 6999 is not a fibonacci number. The number 6999 is not a Bell Number. 6999 is not a Catalan Number. The convertion of 6999 to base 2 (Binary) is 1101101010111. The convertion of 6999 to base 3 (Ternary) is 100121020. The convertion of 6999 to base 4 (Quaternary) is 1231113. The convertion of 6999 to base 5 (Quintal) is 210444. The convertion of 6999 to base 8 (Octal) is 15527. The convertion of 6999 to base 16 (Hexadecimal) is 1b57. The convertion of 6999 to base 32 is 6qn. The sine of 6999 is -0.45148792686144. The cosine of the figure 6999 is 0.89227722816306. The tangent of 6999 is -0.50599512417337. The root of 6999 is 83.660026296912.
If you square 6999 you will get the following result 48986001. The natural logarithm of 6999 is 8.8535225606895 and the decimal logarithm is 3.8450359935134. I hope that you now know that 6999 is very unique figure!
| 607 | 1,665 |
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| 342,021,121 | 14,177 |
Solutions of Right Triangles
Document Sample
``` Solutions of Right Triangles
Solutions of Right Triangles
Triangles are made up of three line segments. They meet to form three angles. The
sizes of the angles and the lengths of the sides 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.
If the triangle is a right triangle, you can use simple trigonometric ratios to find the
missing parts.
In a general triangle (acute or obtuse), you need to use other techniques, including
the law of cosines and the law of sines. You can also find the area of triangles by
using trigonometric ratios.
All triangles are made up of three sides and three angles. If the three angles of the
triangle are labeled ∠ A, ∠ B and ∠ C, then the three sides of the triangle should be
labeled as a, b, and c.
Tutorcircle.com Page No. : 1/4
Figure 1 illustrates how lowercase letters are used to name the sides of the triangle
that are opposite the angles named with corresponding uppercase letters.
If any three of these six measurements are known (other than knowing the measures
of the three angles), then you can calculate the values of the other three
measurements.
The process of finding the missing measurements is known as solving the triangle. If
the triangle is a right triangle, then one of the angles is 90°.
Therefore, you can solve the right triangle if you are given the measures of two of the
three sides or if you are given the measure of one side and one of the other two
angles.
Example 1: Solve the right triangle shown in Figure 1 (b) if ∠ B = 22°
Because the three angles of a triangle must add up to 180°, ∠ A = 90 ∠ B thus ∠ A
= 68°.
Example 2: Solve the right triangle shown in Figure 1 (b) if b = 8 and a = 13.
You can use the Pythagorean theorem to find the missing side, but trigonometric
relationships are used instead. The two missing angle measurements will be found
first and then the missing side.
In many applications, certain angles are referred to by special names. Two of these
special names are angle of elevation and angle of depression. The examples shown
in Figure 2 make use of these terms.
Tutorcircle.com Page No. : 2/4
Example 3: A large airplane (plane A) flying at 26,000 feet sights a smaller plane
(plane B) traveling at an altitude of 24,000 feet. The angle of depression is 40°. What
is the line of sight distance ( x) between the two planes?
Example 4: A ladder must reach the top of a building. The base of the ladder will be
25′ from the base of the building. The angle of elevation from the base of the ladder to
the top of the building is 64°. Find the height of the building (h) and the length of the
Tutorcircle.com Page No. : 3/4
Page No. : 2/3
Thank You For Watching
Presentation
```
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views: 3 posted: 6/15/2012 language: pages: 4
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https://bbs.boingboing.net/t/why-does-long-term-zero-g-hurt-astronauts-eyes-mystery-solved/90496?page=2
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Why does long-term zero-g hurt astronauts' eyes? Mystery solved
Or spin it really fast!
3 Likes
Yeah you get problems with angular motion that way. Your head would be lighter than your feet, and moving your arms would create obvious Coriolis effects. That sort of thing. A study I heard about in the last 15 years or so concluded that 1 RPM should be the target for early attempts at rotational artificial gravity.
4 Likes
But, think of the marketing!
Tomorrow you can ride a spizzler to the stars
2 Likes
Back of the envelope calculations give me:
Using R=a(P^2)/32
2 Likes
If I remember Statics and Dynamics as well as I think I do, the starting equation is:
• a = (v^2)/r, where a is acceleration, v is velocity, and r is radius
This is equivalent to:
• v = sqrt(a * r)
We can convert v to angular velocity (omega):
• omega = (rev / 2 * pi * r m) * (v m / s) * (60 s / min)
• = (60 * v / 2 * pi * r) rpm
• = (30 / pi) * sqrt(a / r) rpm
So, something with radius 9.8m will need to rotate at 30/pi = 9.55rpm to maintain Earthlike gravity. Increasing r and decreasing a will cause this to decrease.
2 Likes
The way I visualize it is at acceleration is x=0.5at^2 over one quarter of the rotation, ie, the inwards acceleration kills your rotational velocity along an axis, four times in each rotation, replacing it with velocity along the perpendicular axis.
I can’t really follow your equations or units from earlier. It looks like you have the right idea, but something gets weird somewhere…
x=0.5at^2
where x is the radius, same as R
a is the acceleration
t is the time to transit the radius and is one quarter of the rotational period, P, so t=P/4
so R=0.5a(P/4)^2 =a(P^2) / 32
It’s because the majestic view of the heavens has been determined to be a sight for sore eyes.
8 Likes
If eyeballs gets flattened, does this mean that myopic astronauts with 20/200 vision have a chance of coming back with 20/20 vision? How long does the effect last?
3 Likes
A sight for sore eyes to the blind would be awful majestic
It would be the most beautiful thing they ever had seen.
It would cause such surprise it would make all of their minds electric
How could anyone tell them some things aren’t what they seem?
4 Likes
There’s your problem. PI is not equal to 4
4 Likes
Myopic, yes. Unfortunately I have the opposite problem, and my uncorrected vision is much worse than 20/200.
2 Likes
Medical science yeah, sure. Are there betting odds on medical tourism to space yet? The schism in that church makes everyone a hero and I do believe Twitter, Internet Archive, ThinkProgress and a couple of NY flyswatters have enough side effects recorded to make samplers, for everyone missing a hero and third hour of vigorous exercise somehow, in February.
I am reminded by This Isn’t Happening (Romance EP? Maybe.) that once someone skull_____ an eye, the mood is off. How about the eyes go back to unflattened and the drugs, spinning and exercise don’t suck? Also (same ep. of This Isn’t…) ‘have you ever tried to keep a joint lit on a balcony [in space]’ gets you out of any active investigations.
8 Likes
Signing you up for 3 years of “medical” vaycay in space then. Bring some extra resistance bands so you don’t fall into the high-G football program rimside.
8 Likes
What’s the point of adding artificial gravity on the space station? Long term zero g isn’t that bad for people as astronauts have spent more than a year in space with no ill effect, plus the whole point of the ISS is that you have a zero g environment for doing scientific research.
With that kind of view, my eyes would bleed too!
Without constant exercise, human bodies undergo rapid atrophy in space. Especially bones.
Which is easier for you: doing squats with weights all day, or existing with a constant acceleration of 9.8m/s2 downward no matter what with no conscious effort?
5 Likes
[quote=“incarnedine_v, post:36, topic:90496”]
What’s the point of adding artificial gravity on the space station? Long term zero g isn’t that bad for people as astronauts have spent more than a year in space with no ill effect[/quote]
It’s almost as if you didn’t read the article. There are ill effects, and the longer an astronaut is in space the effects will compound over time. On top of that there may be health problems that are not immediately apparent or fully understood which is why NASA and other space agencies continually monitor their astronauts’ health to gather all kids of data.
9 Likes
send human specimens up with the robot then.
1 Like
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http://gpl4you.com/aptitude.php?subject=Mathematical%20Skills&type=Mixtures%20and%20Alligations&level=Difficult
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## Welcome Guest
### Mathematical Skills | Mixtures and Alligations
Q. No. 1: If a certain quantity of 40% HCL is mixed with 60% HCL in the ratio 4:1. This is mixed with 32% HCL in the ratio 2:1. From this, 25 L is removed and replaced with 50 L water to get 20% HCL. What is the amount of 60% HCL(in Litres) taken for the the entire process? A : 30 B : 25 C : 20 D : 10 Solution
Q. No. 2: Three Jugs namely P,Q and R each contain 100ml of lime-water solution. The ratio of lime to water in the jug P,Q and R is 1:3, 1:4 and 2:3 respectively. 40ml of solution is transferred from jug P to Jug R and then 28ml of solution is transferred from jug R to jug Q. Find the the final ratio of lime in the jug P,Q and R. A : 3:6:8 B : 6:15:20 C : 15:28:42 D : 8:15:24 Solution
Q. No. 3: There are two alloys A and B. Alloys A contains metals P, R and S. Alloy B contains metals P and Q. Both alloys contains 60% of metal P. They are mixed to form a third alloy. the percentage of metal Q in the third alloy is half of that of metal P in it. The ratio of percentage of R and S in Alloy A is 3:1. Find the ratio of quantities of R in Alloy A and Q in Alloy B, which are mixed to form the third alloy. A : 3:2 B : 2:3 C : 3:4 D : 1:4 Solution
3 1
Mixtures and Alligations Easy Moderate Difficult
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# Resources tagged with: Investigations
Filter by: Content type:
Age range:
Challenge level:
### Two Regular Polygons
##### Age 14 to 16 Challenge Level:
Two polygons fit together so that the exterior angle at each end of their shared side is 81 degrees. If both shapes now have to be regular could the angle still be 81 degrees?
### A Rational Search
##### Age 16 to 18 Challenge Level:
Investigate constructible images which contain rational areas.
### Spokes
##### Age 16 to 18 Challenge Level:
Draw three equal line segments in a unit circle to divide the circle into four parts of equal area.
### Spirostars
##### Age 16 to 18 Challenge Level:
A spiropath is a sequence of connected line segments end to end taking different directions. The same spiropath is iterated. When does it cycle and when does it go on indefinitely?
### Few and Far Between?
##### Age 16 to 18 Challenge Level:
Can you find some Pythagorean Triples where the two smaller numbers differ by 1?
### Eight Ratios
##### Age 14 to 16 Challenge Level:
Two perpendicular lines lie across each other and the end points are joined to form a quadrilateral. Eight ratios are defined, three are given but five need to be found.
### Track Design
##### Age 14 to 16 Challenge Level:
Where should runners start the 200m race so that they have all run the same distance by the finish?
### Reaction Rates!
##### Age 16 to 18
Fancy learning a bit more about rates of reaction, but don't know where to look? Come inside and find out more...
### Operating Machines
##### Age 16 to 18 Challenge Level:
What functions can you make using the function machines RECIPROCAL and PRODUCT and the operator machines DIFF and INT?
### Snookered
##### Age 14 to 18 Challenge Level:
In a snooker game the brown ball was on the lip of the pocket but it could not be hit directly as the black ball was in the way. How could it be potted by playing the white ball off a cushion?
### Chance of That
##### Age 16 to 18 Challenge Level:
What's the chance of a pair of lists of numbers having sample correlation exactly equal to zero?
### Perfect Eclipse
##### Age 14 to 16 Challenge Level:
Use trigonometry to determine whether solar eclipses on earth can be perfect.
##### Age 14 to 18 Challenge Level:
Some of our more advanced investigations
### Witch's Hat
##### Age 11 to 16 Challenge Level:
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat?
### A Different Differential Equation
##### Age 16 to 18 Challenge Level:
Explore the properties of this different sort of differential equation.
### Problem Solving: Opening up Problems
##### Age 5 to 16
All types of mathematical problems serve a useful purpose in mathematics teaching, but different types of problem will achieve different learning objectives. In generalmore open-ended problems have. . . .
### Sextet
##### Age 16 to 18 Challenge Level:
Investigate x to the power n plus 1 over x to the power n when x plus 1 over x equals 1.
### The Invertible Trefoil
##### Age 14 to 16 Challenge Level:
When is a knot invertible ?
### Robot Camera
##### Age 14 to 16 Challenge Level:
Could nanotechnology be used to see if an artery is blocked? Or is this just science fiction?
### Spiroflowers
##### Age 16 to 18 Challenge Level:
Analyse these repeating patterns. Decide on the conditions for a periodic pattern to occur and when the pattern extends to infinity.
### What Salt?
##### Age 16 to 18 Challenge Level:
Can you deduce why common salt isn't NaCl_2?
### The Power of Dimensional Analysis
##### Age 14 to 18
An introduction to a useful tool to check the validity of an equation.
### Peeling the Apple or the Cone That Lost Its Head
##### Age 14 to 16 Challenge Level:
How much peel does an apple have?
### Clear as Crystal
##### Age 16 to 18 Challenge Level:
Unearth the beautiful mathematics of symmetry whilst investigating the properties of crystal lattices
### Designing Table Mats
##### Age 11 to 16 Challenge Level:
Formulate and investigate a simple mathematical model for the design of a table mat.
### Genetic Intrigue
##### Age 16 to 18
Dip your toe into the fascinating topic of genetics. From Mendel's theories to some cutting edge experimental techniques, this article gives an insight into some of the processes underlying. . . .
##### Age 16 to 18
Read about the mathematics behind the measuring devices used in quantitative chemistry
### Big and Small Numbers in Chemistry
##### Age 14 to 16 Challenge Level:
Get some practice using big and small numbers in chemistry.
### Diamonds Aren't Forever
##### Age 16 to 18 Challenge Level:
Ever wondered what it would be like to vaporise a diamond? Find out inside...
### Big and Small Numbers in the Living World
##### Age 11 to 16 Challenge Level:
Work with numbers big and small to estimate and calculate various quantities in biological contexts.
### Big and Small Numbers in the Physical World
##### Age 14 to 16 Challenge Level:
Work with numbers big and small to estimate and calculate various quantities in physical contexts.
### There's Always One Isn't There
##### Age 14 to 16 Challenge Level:
Take any pair of numbers, say 9 and 14. Take the larger number, fourteen, and count up in 14s. Then divide each of those values by the 9, and look at the remainders.
### Alternative Record Book
##### Age 14 to 18 Challenge Level:
In which Olympic event does a human travel fastest? Decide which events to include in your Alternative Record Book.
### Bent Out of Shape
##### Age 14 to 18 Challenge Level:
An introduction to bond angle geometry.
### Geometry and Gravity 1
##### Age 11 to 18
This article (the first of two) contains ideas for investigations. Space-time, the curvature of space and topology are introduced with some fascinating problems to explore.
### Mach Attack
##### Age 16 to 18 Challenge Level:
Have you got the Mach knack? Discover the mathematics behind exceeding the sound barrier.
### Smoke and Daggers
##### Age 16 to 18 Challenge Level:
We all know that smoking poses a long term health risk and has the potential to cause cancer. But what actually happens when you light up a cigarette, place it to your mouth, take a tidal breath. . . .
### Making More Tracks
##### Age 16 to 18 Challenge Level:
Given the equation for the path followed by the back wheel of a bike, can you solve to find the equation followed by the front wheel?
### Trig-trig
##### Age 16 to 18 Challenge Level:
Explore the properties of combinations of trig functions in this open investigation.
### 9 Hole Light Golf
##### Age 5 to 18 Challenge Level:
We think this 3x3 version of the game is often harder than the 5x5 version. Do you agree? If so, why do you think that might be?
### Which Twin Is Older?
##### Age 16 to 18
A simplified account of special relativity and the twins paradox.
### What Do Functions Do for Tiny X?
##### Age 16 to 18 Challenge Level:
Looking at small values of functions. Motivating the existence of the Taylor expansion.
### Building Approximations for Sin(x)
##### Age 16 to 18 Challenge Level:
Build up the concept of the Taylor series
### Very Old Man
##### Age 16 to 18 Challenge Level:
Is the age of this very old man statistically believable?
### More Bridge Building
##### Age 16 to 18 Challenge Level:
Which parts of these framework bridges are in tension and which parts are in compression?
### Scale Invariance
##### Age 16 to 18 Challenge Level:
By exploring the concept of scale invariance, find the probability that a random piece of real data begins with a 1.
### Modelling Assumptions in Mechanics
##### Age 16 to 18
An article demonstrating mathematically how various physical modelling assumptions affect the solution to the seemingly simple problem of the projectile.
### Taking Trigonometry Series-ly
##### Age 16 to 18 Challenge Level:
Look at the advanced way of viewing sin and cos through their power series.
### Escape from Planet Earth
##### Age 16 to 18 Challenge Level:
How fast would you have to throw a ball upwards so that it would never land?
| 1,819 | 8,150 |
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| 3.703125 | 4 |
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https://math.stackexchange.com/questions/2228267/find-the-matrix-of-linear-transformation-in-standard-basis-that-rotates-clockw
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# Find the matrix of linear transformation (in Standard basis) that rotates clockwise every vector
Find the matrix of linear transformation (in Standard basis) that rotates clockwise every vector in $\mathbb{R}^{2}$ through an angle $\pi/4$ and then reflects it across y axis. The standard matrix of rotation by $\pi/4 \ clockwise$ is R = $\begin{pmatrix} cos(\pi/4) & \sin (\pi/4) \\ -\sin(\pi/4) & cos(\pi/4) \end{pmatrix}$= $\begin{pmatrix} 1/\sqrt2 & 1/\sqrt 2 \\ -1/\sqrt 2 & 1/\sqrt 2 \end{pmatrix}$ . Now reflection matrix about y axis is R'= $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ . Hence composition of Rotation and Reflection is $\ R \circ R'=\begin{pmatrix} 1/\sqrt 2 & 1/\sqrt 2 \\ -1/\sqrt 2 & 1/\sqrt 2 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ . But I am not sure , please help me
• Multiplying these matrices one by one with a vector, which operatorn is applied first? Apr 10 '17 at 20:39
The product of two matrices is not commutative. If you want the matrix that represents first the rotation than the reflection, the correct order is $R'\cdot R$.
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https://gmatclub.com/forum/blood-banks-will-shortly-start-to-screen-all-donors-for-nanb-126642.html?fl=similar
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# Blood banks will shortly start to screen all donors for NANB
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Blood banks will shortly start to screen all donors for NANB [#permalink]
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27 Jan 2012, 22:11
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Blood banks will shortly start to screen all donors for NANB hepatitis. Although the new screening tests are estimated to disqualify up to 5 percent of all prospective blood donors, they will still miss two-thirds of donors carrying NANB hepatitis. Therefore, about 10 percent of actual donors will still supply NANB-contaminated blood.
Q) which of the follwoing inferences about the consequences of instituting the new tests is best supported by the passage above?
a) The incidence of new cases of NANB hepatitis is likely to go up by 10 percent.
b) Donations made by patients specifically for their own use are likely to become less frequent.
c) The demand for blood from blood banks is likely to fluctuate more strongly.
d) The blood supplies available from blood banks are likely to go down.
e) The number of prospective first-time donors is likely to go up by 5 percent.
[Reveal] Spoiler: OA
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Re: PT #1 CR 20 Blood banks will shortly start to screen all don [#permalink]
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27 Jan 2012, 22:25
OE: is d.
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Re: PT #1 CR 20 Blood banks will shortly start to screen all don [#permalink]
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28 Jan 2012, 07:57
Cause-->Effect,if 5 percent of all prospective blood donors are disqualified than the blood supplies available from blood banks are likely to go down,hence D.
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Re: PT #1 CR 20 Blood banks will shortly start to screen all don [#permalink]
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22 Feb 2012, 20:51
D because you can eliminate all the other choices for irrelavent information.
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Re: Blood banks will shortly start to screen all donors for NANB [#permalink]
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14 Aug 2015, 15:16
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Re: Blood banks will shortly start to screen all donors for NANB [#permalink]
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18 Aug 2015, 23:31
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Expert's post
eybrj2 wrote:
Blood banks will shortly start to screen all donors for NANB hepatitis. Although the new screening tests are estimated to disqualify up to 5 percent of all prospective blood donors, they will still miss two-thirds of donors carrying NANB hepatitis. Therefore, about 10 percent of actual donors will still supply NANB-contaminated blood.
Q) which of the follwoing inferences about the consequences of instituting the new tests is best supported by the passage above?
a) The incidence of new cases of NANB hepatitis is likely to go up by 10 percent.
b) Donations made by patients specifically for their own use are likely to become less frequent.
c) The demand for blood from blood banks is likely to fluctuate more strongly.
d) The blood supplies available from blood banks are likely to go down.
e) The number of prospective first-time donors is likely to go up by 5 percent.
Quote:
At first I picked A thinking, suppose 10 people comprise 2/3rd of the people who carry NABH and donate blood ---> They donate to 10 people at least ---> those people get NABH as well (now NABH must be infectious otherwise why are Blood banks testing to find whether ppl have NABH or not)
So increase of 10 people --> ~ 10% increase in NABH infected people.
After knowing I picked the wrong answer, my fallacy is probably this:
We don't know the total number of NABH infected cases: it may be 20 or it may be 10 ---> supposing it is 20, then 10 people addition is ---> 2% increase not 10% OR if it is 100 cases of infections ---> then it is 10% ----> % increase may be 2-10% or more ---> this makes the option A to be MAY BE TRUE.
A conclusion MUST BE TRUE ---> D) says the blood supplies will decrease ---> YES because people are disqualified, so less blood is available from all the possible donors.
Please do let me know if I am right. And also, How i may improve in ASSUMPTION CR, I also take around 3:30 mins in medium-tough questions.
The problem with (A) is that the argument does not give us enough information to deduce that.
10% of actual donors will supply NANB contaminated blood. But does that mean that all this blood will be used? What if actually only 50% of the blood from a blood bank is used. Then the actual incidence of disease may increase by only 5% (assuming homogenous distribution).
Also, what if NANB infected people are the ones who need blood most often from blood banks. In that case, we may find very few new cases.
Hence, increase in the incidence of disease is a big jump.
(D) is straight forward. 5% fewer people will be able to donate blood due to new tests and that means reduced supply is an effect of the new test.
In inference questions, simpler is usually better.
For discussions on assumption questions, check my blog:
http://www.veritasprep.com/blog/?s=assu ... er&x=0&y=0
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Re: Blood banks will shortly start to screen all donors for NANB [#permalink] 18 Aug 2015, 23:31
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# D. 10 km/h 6. A cydist travels 40 km in 2 hours, what is his average speed? A. 80 km/h B. 60 km/h C....
Kung naghahanap ka ng mga sagot sa tanong na: D. 10 km/h 6. A cydist travels 40 km in 2 hours, what is his average speed? A. 80 km/h B. 60 km/h C…., napunta ka sa tamang lugar.
Narito ang ilang sagot sa tanong na ito. Mangyaring basahin ang higit pa.
### Tanong
D. 10 km/h
6. A cydist travels 40 km in 2 hours, what is his average speed?
A. 80 km/h
B. 60 km/h
C. 20 km/h
7. What is the basic unit for measuring time?
A. Hour
B. Minute
C. second
D. month
### Mga sagot sa #1 sa Tanong: D. 10 km/h6. A cydist travels 40 km in 2 hours, what is his average speed?A. 80 km/hB. 60 km/hC. 20 km/h7. What is the basic unit for measuring time?A. HourB. MinuteC. secondD. month
6. C
7. A
Explanation:
6. 40km/2hrs = 40/2 =20:1, 20km/1hr= 20km/h
Napakaraming tanong at sagot tungkol sa D. 10 km/h 6. A cydist travels 40 km in 2 hours, what is his average speed? A. 80 km/h B. 60 km/h C…., sana ay makatulong ito sa paglutas ng iyong problema.
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# Discrete Mathematics - Relations and Functions
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Relations, relation composition, converse relation, reflexivity, symmetry, transitivity. Functions, function composition, one-to-one, onto, bijective functions, inverse function, pigeonhole principle, recursive functions.
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### Discrete Mathematics - Relations and Functions
1. 1. Discrete Mathematics Relations and Functions H. Turgut Uyar Ay¸seg¨ul Gen¸cata Yayımlı Emre Harmancı 2001-2016
3. 3. Topics 1 Relations Introduction Relation Properties Equivalence Relations 2 Functions Introduction Pigeonhole Principle Recursion
4. 4. Topics 1 Relations Introduction Relation Properties Equivalence Relations 2 Functions Introduction Pigeonhole Principle Recursion
5. 5. Relation Definition relation: α ⊆ A × B × C × · · · × N tuple: element of relation binary relation: α ⊆ A × B aαb : (a, b) ∈ α
6. 6. Relation Definition relation: α ⊆ A × B × C × · · · × N tuple: element of relation binary relation: α ⊆ A × B aαb : (a, b) ∈ α
7. 7. Relation Example A = {a1, a2, a3, a4}, B = {b1, b2, b3} α = {(a1, b1), (a1, b3), (a2, b2), (a2, b3), (a3, b1), (a3, b3), (a4, b1)} b1 b2 b3 a1 1 0 1 a2 0 1 1 a3 1 0 1 a4 1 0 0 Mα = 1 0 1 0 1 1 1 0 1 1 0 0
8. 8. Relation Example A = {a1, a2, a3, a4}, B = {b1, b2, b3} α = {(a1, b1), (a1, b3), (a2, b2), (a2, b3), (a3, b1), (a3, b3), (a4, b1)} b1 b2 b3 a1 1 0 1 a2 0 1 1 a3 1 0 1 a4 1 0 0 Mα = 1 0 1 0 1 1 1 0 1 1 0 0
9. 9. Relation Composition Definition relation composition: α ⊆ A × B, β ⊆ B × C αβ = {(a, c) | a ∈ A, c ∈ C, ∃b ∈ B [aαb ∧ bβc]} example
10. 10. Relation Composition Mαβ = Mα × Mβ using logical operations: 1 : T 0 : F · : ∧ + : ∨ example Mα = 1 0 0 0 0 1 0 1 1 0 1 0 1 0 1 Mβ = 1 1 0 0 0 0 1 1 0 1 1 0 Mαβ = 1 1 0 0 0 1 1 0 0 1 1 1 0 0 1 1 1 1 1 0
11. 11. Relation Composition Mαβ = Mα × Mβ using logical operations: 1 : T 0 : F · : ∧ + : ∨ example Mα = 1 0 0 0 0 1 0 1 1 0 1 0 1 0 1 Mβ = 1 1 0 0 0 0 1 1 0 1 1 0 Mαβ = 1 1 0 0 0 1 1 0 0 1 1 1 0 0 1 1 1 1 1 0
12. 12. Relation Composition Associativity (αβ)γ = α(βγ). (a, d) ∈ (αβ)γ ⇔ ∃c [(a, c) ∈ αβ ∧ (c, d) ∈ γ] ⇔ ∃c [∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ∧ (c, d) ∈ γ] ⇔ ∃b [(a, b) ∈ α ∧ ∃c [(b, c) ∈ β ∧ (c, d) ∈ γ]] ⇔ ∃b [(a, b) ∈ α ∧ (b, d) ∈ βγ] ⇔ (a, d) ∈ α(βγ)
13. 13. Relation Composition Associativity (αβ)γ = α(βγ). (a, d) ∈ (αβ)γ ⇔ ∃c [(a, c) ∈ αβ ∧ (c, d) ∈ γ] ⇔ ∃c [∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ∧ (c, d) ∈ γ] ⇔ ∃b [(a, b) ∈ α ∧ ∃c [(b, c) ∈ β ∧ (c, d) ∈ γ]] ⇔ ∃b [(a, b) ∈ α ∧ (b, d) ∈ βγ] ⇔ (a, d) ∈ α(βγ)
14. 14. Relation Composition Associativity (αβ)γ = α(βγ). (a, d) ∈ (αβ)γ ⇔ ∃c [(a, c) ∈ αβ ∧ (c, d) ∈ γ] ⇔ ∃c [∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ∧ (c, d) ∈ γ] ⇔ ∃b [(a, b) ∈ α ∧ ∃c [(b, c) ∈ β ∧ (c, d) ∈ γ]] ⇔ ∃b [(a, b) ∈ α ∧ (b, d) ∈ βγ] ⇔ (a, d) ∈ α(βγ)
15. 15. Relation Composition Associativity (αβ)γ = α(βγ). (a, d) ∈ (αβ)γ ⇔ ∃c [(a, c) ∈ αβ ∧ (c, d) ∈ γ] ⇔ ∃c [∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ∧ (c, d) ∈ γ] ⇔ ∃b [(a, b) ∈ α ∧ ∃c [(b, c) ∈ β ∧ (c, d) ∈ γ]] ⇔ ∃b [(a, b) ∈ α ∧ (b, d) ∈ βγ] ⇔ (a, d) ∈ α(βγ)
16. 16. Relation Composition Associativity (αβ)γ = α(βγ). (a, d) ∈ (αβ)γ ⇔ ∃c [(a, c) ∈ αβ ∧ (c, d) ∈ γ] ⇔ ∃c [∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ∧ (c, d) ∈ γ] ⇔ ∃b [(a, b) ∈ α ∧ ∃c [(b, c) ∈ β ∧ (c, d) ∈ γ]] ⇔ ∃b [(a, b) ∈ α ∧ (b, d) ∈ βγ] ⇔ (a, d) ∈ α(βγ)
17. 17. Relation Composition Associativity (αβ)γ = α(βγ). (a, d) ∈ (αβ)γ ⇔ ∃c [(a, c) ∈ αβ ∧ (c, d) ∈ γ] ⇔ ∃c [∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ∧ (c, d) ∈ γ] ⇔ ∃b [(a, b) ∈ α ∧ ∃c [(b, c) ∈ β ∧ (c, d) ∈ γ]] ⇔ ∃b [(a, b) ∈ α ∧ (b, d) ∈ βγ] ⇔ (a, d) ∈ α(βγ)
18. 18. Relation Composition Associativity (αβ)γ = α(βγ). (a, d) ∈ (αβ)γ ⇔ ∃c [(a, c) ∈ αβ ∧ (c, d) ∈ γ] ⇔ ∃c [∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ∧ (c, d) ∈ γ] ⇔ ∃b [(a, b) ∈ α ∧ ∃c [(b, c) ∈ β ∧ (c, d) ∈ γ]] ⇔ ∃b [(a, b) ∈ α ∧ (b, d) ∈ βγ] ⇔ (a, d) ∈ α(βγ)
19. 19. Relation Composition Theorems α(β ∪ γ) = αβ ∪ αγ. (a, c) ∈ α(β ∪ γ) ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ (β ∪ γ)] ⇔ ∃b [(a, b) ∈ α ∧ ((b, c) ∈ β ∨ (b, c) ∈ γ)] ⇔ ∃b [((a, b) ∈ α ∧ (b, c) ∈ β) ∨ ((a, b) ∈ α ∧ (b, c) ∈ γ)] ⇔ (a, c) ∈ αβ ∨ (a, c) ∈ αγ ⇔ (a, c) ∈ αβ ∪ αγ
20. 20. Relation Composition Theorems α(β ∪ γ) = αβ ∪ αγ. (a, c) ∈ α(β ∪ γ) ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ (β ∪ γ)] ⇔ ∃b [(a, b) ∈ α ∧ ((b, c) ∈ β ∨ (b, c) ∈ γ)] ⇔ ∃b [((a, b) ∈ α ∧ (b, c) ∈ β) ∨ ((a, b) ∈ α ∧ (b, c) ∈ γ)] ⇔ (a, c) ∈ αβ ∨ (a, c) ∈ αγ ⇔ (a, c) ∈ αβ ∪ αγ
21. 21. Relation Composition Theorems α(β ∪ γ) = αβ ∪ αγ. (a, c) ∈ α(β ∪ γ) ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ (β ∪ γ)] ⇔ ∃b [(a, b) ∈ α ∧ ((b, c) ∈ β ∨ (b, c) ∈ γ)] ⇔ ∃b [((a, b) ∈ α ∧ (b, c) ∈ β) ∨ ((a, b) ∈ α ∧ (b, c) ∈ γ)] ⇔ (a, c) ∈ αβ ∨ (a, c) ∈ αγ ⇔ (a, c) ∈ αβ ∪ αγ
22. 22. Relation Composition Theorems α(β ∪ γ) = αβ ∪ αγ. (a, c) ∈ α(β ∪ γ) ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ (β ∪ γ)] ⇔ ∃b [(a, b) ∈ α ∧ ((b, c) ∈ β ∨ (b, c) ∈ γ)] ⇔ ∃b [((a, b) ∈ α ∧ (b, c) ∈ β) ∨ ((a, b) ∈ α ∧ (b, c) ∈ γ)] ⇔ (a, c) ∈ αβ ∨ (a, c) ∈ αγ ⇔ (a, c) ∈ αβ ∪ αγ
23. 23. Relation Composition Theorems α(β ∪ γ) = αβ ∪ αγ. (a, c) ∈ α(β ∪ γ) ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ (β ∪ γ)] ⇔ ∃b [(a, b) ∈ α ∧ ((b, c) ∈ β ∨ (b, c) ∈ γ)] ⇔ ∃b [((a, b) ∈ α ∧ (b, c) ∈ β) ∨ ((a, b) ∈ α ∧ (b, c) ∈ γ)] ⇔ (a, c) ∈ αβ ∨ (a, c) ∈ αγ ⇔ (a, c) ∈ αβ ∪ αγ
24. 24. Relation Composition Theorems α(β ∪ γ) = αβ ∪ αγ. (a, c) ∈ α(β ∪ γ) ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ (β ∪ γ)] ⇔ ∃b [(a, b) ∈ α ∧ ((b, c) ∈ β ∨ (b, c) ∈ γ)] ⇔ ∃b [((a, b) ∈ α ∧ (b, c) ∈ β) ∨ ((a, b) ∈ α ∧ (b, c) ∈ γ)] ⇔ (a, c) ∈ αβ ∨ (a, c) ∈ αγ ⇔ (a, c) ∈ αβ ∪ αγ
25. 25. Relation Composition Theorems α(β ∪ γ) = αβ ∪ αγ. (a, c) ∈ α(β ∪ γ) ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ (β ∪ γ)] ⇔ ∃b [(a, b) ∈ α ∧ ((b, c) ∈ β ∨ (b, c) ∈ γ)] ⇔ ∃b [((a, b) ∈ α ∧ (b, c) ∈ β) ∨ ((a, b) ∈ α ∧ (b, c) ∈ γ)] ⇔ (a, c) ∈ αβ ∨ (a, c) ∈ αγ ⇔ (a, c) ∈ αβ ∪ αγ
26. 26. Converse Relation Definition α−1 = {(b, a) | (a, b) ∈ α} Mα−1 = MT α
27. 27. Converse Relation Theorems (α−1)−1 = α (α ∪ β)−1 = α−1 ∪ β−1 (α ∩ β)−1 = α−1 ∩ β−1 α−1 = α−1 (α − β)−1 = α−1 − β−1
28. 28. Converse Relation Theorems α−1 = α−1. (b, a) ∈ α−1 ⇔ (a, b) ∈ α ⇔ (a, b) /∈ α ⇔ (b, a) /∈ α−1 ⇔ (b, a) ∈ α−1
29. 29. Converse Relation Theorems α−1 = α−1. (b, a) ∈ α−1 ⇔ (a, b) ∈ α ⇔ (a, b) /∈ α ⇔ (b, a) /∈ α−1 ⇔ (b, a) ∈ α−1
30. 30. Converse Relation Theorems α−1 = α−1. (b, a) ∈ α−1 ⇔ (a, b) ∈ α ⇔ (a, b) /∈ α ⇔ (b, a) /∈ α−1 ⇔ (b, a) ∈ α−1
31. 31. Converse Relation Theorems α−1 = α−1. (b, a) ∈ α−1 ⇔ (a, b) ∈ α ⇔ (a, b) /∈ α ⇔ (b, a) /∈ α−1 ⇔ (b, a) ∈ α−1
32. 32. Converse Relation Theorems α−1 = α−1. (b, a) ∈ α−1 ⇔ (a, b) ∈ α ⇔ (a, b) /∈ α ⇔ (b, a) /∈ α−1 ⇔ (b, a) ∈ α−1
33. 33. Converse Relation Theorems α−1 = α−1. (b, a) ∈ α−1 ⇔ (a, b) ∈ α ⇔ (a, b) /∈ α ⇔ (b, a) /∈ α−1 ⇔ (b, a) ∈ α−1
34. 34. Converse Relation Theorems (α ∩ β)−1 = α−1 ∩ β−1. (b, a) ∈ (α ∩ β)−1 ⇔ (a, b) ∈ (α ∩ β) ⇔ (a, b) ∈ α ∧ (a, b) ∈ β ⇔ (b, a) ∈ α−1 ∧ (b, a) ∈ β−1 ⇔ (b, a) ∈ α−1 ∩ β−1
35. 35. Converse Relation Theorems (α ∩ β)−1 = α−1 ∩ β−1. (b, a) ∈ (α ∩ β)−1 ⇔ (a, b) ∈ (α ∩ β) ⇔ (a, b) ∈ α ∧ (a, b) ∈ β ⇔ (b, a) ∈ α−1 ∧ (b, a) ∈ β−1 ⇔ (b, a) ∈ α−1 ∩ β−1
36. 36. Converse Relation Theorems (α ∩ β)−1 = α−1 ∩ β−1. (b, a) ∈ (α ∩ β)−1 ⇔ (a, b) ∈ (α ∩ β) ⇔ (a, b) ∈ α ∧ (a, b) ∈ β ⇔ (b, a) ∈ α−1 ∧ (b, a) ∈ β−1 ⇔ (b, a) ∈ α−1 ∩ β−1
37. 37. Converse Relation Theorems (α ∩ β)−1 = α−1 ∩ β−1. (b, a) ∈ (α ∩ β)−1 ⇔ (a, b) ∈ (α ∩ β) ⇔ (a, b) ∈ α ∧ (a, b) ∈ β ⇔ (b, a) ∈ α−1 ∧ (b, a) ∈ β−1 ⇔ (b, a) ∈ α−1 ∩ β−1
38. 38. Converse Relation Theorems (α ∩ β)−1 = α−1 ∩ β−1. (b, a) ∈ (α ∩ β)−1 ⇔ (a, b) ∈ (α ∩ β) ⇔ (a, b) ∈ α ∧ (a, b) ∈ β ⇔ (b, a) ∈ α−1 ∧ (b, a) ∈ β−1 ⇔ (b, a) ∈ α−1 ∩ β−1
39. 39. Converse Relation Theorems (α ∩ β)−1 = α−1 ∩ β−1. (b, a) ∈ (α ∩ β)−1 ⇔ (a, b) ∈ (α ∩ β) ⇔ (a, b) ∈ α ∧ (a, b) ∈ β ⇔ (b, a) ∈ α−1 ∧ (b, a) ∈ β−1 ⇔ (b, a) ∈ α−1 ∩ β−1
40. 40. Converse Relation Theorems (α − β)−1 = α−1 − β−1. (α − β)−1 = (α ∩ β)−1 = α−1 ∩ β −1 = α−1 ∩ β−1 = α−1 − β−1
41. 41. Converse Relation Theorems (α − β)−1 = α−1 − β−1. (α − β)−1 = (α ∩ β)−1 = α−1 ∩ β −1 = α−1 ∩ β−1 = α−1 − β−1
42. 42. Converse Relation Theorems (α − β)−1 = α−1 − β−1. (α − β)−1 = (α ∩ β)−1 = α−1 ∩ β −1 = α−1 ∩ β−1 = α−1 − β−1
43. 43. Converse Relation Theorems (α − β)−1 = α−1 − β−1. (α − β)−1 = (α ∩ β)−1 = α−1 ∩ β −1 = α−1 ∩ β−1 = α−1 − β−1
44. 44. Converse Relation Theorems (α − β)−1 = α−1 − β−1. (α − β)−1 = (α ∩ β)−1 = α−1 ∩ β −1 = α−1 ∩ β−1 = α−1 − β−1
45. 45. Relation Composition Converse Theorem (αβ)−1 = β−1α−1 Proof. (c, a) ∈ (αβ)−1 ⇔ (a, c) ∈ αβ ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ⇔ ∃b [(b, a) ∈ α−1 ∧ (c, b) ∈ β−1 ] ⇔ (c, a) ∈ β−1 α−1
46. 46. Relation Composition Converse Theorem (αβ)−1 = β−1α−1 Proof. (c, a) ∈ (αβ)−1 ⇔ (a, c) ∈ αβ ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ⇔ ∃b [(b, a) ∈ α−1 ∧ (c, b) ∈ β−1 ] ⇔ (c, a) ∈ β−1 α−1
47. 47. Relation Composition Converse Theorem (αβ)−1 = β−1α−1 Proof. (c, a) ∈ (αβ)−1 ⇔ (a, c) ∈ αβ ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ⇔ ∃b [(b, a) ∈ α−1 ∧ (c, b) ∈ β−1 ] ⇔ (c, a) ∈ β−1 α−1
48. 48. Relation Composition Converse Theorem (αβ)−1 = β−1α−1 Proof. (c, a) ∈ (αβ)−1 ⇔ (a, c) ∈ αβ ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ⇔ ∃b [(b, a) ∈ α−1 ∧ (c, b) ∈ β−1 ] ⇔ (c, a) ∈ β−1 α−1
49. 49. Relation Composition Converse Theorem (αβ)−1 = β−1α−1 Proof. (c, a) ∈ (αβ)−1 ⇔ (a, c) ∈ αβ ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ⇔ ∃b [(b, a) ∈ α−1 ∧ (c, b) ∈ β−1 ] ⇔ (c, a) ∈ β−1 α−1
50. 50. Relation Composition Converse Theorem (αβ)−1 = β−1α−1 Proof. (c, a) ∈ (αβ)−1 ⇔ (a, c) ∈ αβ ⇔ ∃b [(a, b) ∈ α ∧ (b, c) ∈ β] ⇔ ∃b [(b, a) ∈ α−1 ∧ (c, b) ∈ β−1 ] ⇔ (c, a) ∈ β−1 α−1
51. 51. Topics 1 Relations Introduction Relation Properties Equivalence Relations 2 Functions Introduction Pigeonhole Principle Recursion
52. 52. Relation Properties α ⊆ A × A αn: αα · · · α identity relation: E = {(a, a) | a ∈ A} reflexivity symmetry transitivity
53. 53. Relation Properties α ⊆ A × A αn: αα · · · α identity relation: E = {(a, a) | a ∈ A} reflexivity symmetry transitivity
54. 54. Reflexivity reflexive α ⊆ A × A ∀a ∈ A [aαa] E ⊆ α nonreflexive: ∃a ∈ A [¬(aαa)] irreflexive: ∀a ∈ A [¬(aαa)]
55. 55. Reflexivity reflexive α ⊆ A × A ∀a ∈ A [aαa] E ⊆ α nonreflexive: ∃a ∈ A [¬(aαa)] irreflexive: ∀a ∈ A [¬(aαa)]
56. 56. Reflexivity reflexive α ⊆ A × A ∀a ∈ A [aαa] E ⊆ α nonreflexive: ∃a ∈ A [¬(aαa)] irreflexive: ∀a ∈ A [¬(aαa)]
57. 57. Reflexivity reflexive α ⊆ A × A ∀a ∈ A [aαa] E ⊆ α nonreflexive: ∃a ∈ A [¬(aαa)] irreflexive: ∀a ∈ A [¬(aαa)]
58. 58. Reflexivity Examples R1 ⊆ {1, 2} × {1, 2} R1 = {(1, 1), (1, 2), (2, 2)} R1 is reflexive R2 ⊆ {1, 2, 3} × {1, 2, 3} R2 = {(1, 1), (1, 2), (2, 2)} R2 is nonreflexive
59. 59. Reflexivity Examples R1 ⊆ {1, 2} × {1, 2} R1 = {(1, 1), (1, 2), (2, 2)} R1 is reflexive R2 ⊆ {1, 2, 3} × {1, 2, 3} R2 = {(1, 1), (1, 2), (2, 2)} R2 is nonreflexive
60. 60. Reflexivity Examples R ⊆ {1, 2, 3} × {1, 2, 3} R = {(1, 2), (2, 1), (2, 3)} R is irreflexive
61. 61. Reflexivity Examples R ⊆ Z × Z R = {(a, b) | ab ≥ 0} R is reflexive
62. 62. Symmetry symmetric α ⊆ A × A ∀a, b ∈ A [(a = b) ∨ (aαb ∧ bαa) ∨ (¬(aαb) ∧ ¬(bαa))] ∀a, b ∈ A [(a = b) ∨ (aαb ↔ bαa)] α−1 = α asymmetric: ∃a, b ∈ A [(a = b) ∧ ((aαb ∧ ¬(bαa)) ∨ (¬(aαb) ∧ bαa))]
63. 63. Symmetry symmetric α ⊆ A × A ∀a, b ∈ A [(a = b) ∨ (aαb ∧ bαa) ∨ (¬(aαb) ∧ ¬(bαa))] ∀a, b ∈ A [(a = b) ∨ (aαb ↔ bαa)] α−1 = α asymmetric: ∃a, b ∈ A [(a = b) ∧ ((aαb ∧ ¬(bαa)) ∨ (¬(aαb) ∧ bαa))]
64. 64. Symmetry symmetric α ⊆ A × A ∀a, b ∈ A [(a = b) ∨ (aαb ∧ bαa) ∨ (¬(aαb) ∧ ¬(bαa))] ∀a, b ∈ A [(a = b) ∨ (aαb ↔ bαa)] α−1 = α asymmetric: ∃a, b ∈ A [(a = b) ∧ ((aαb ∧ ¬(bαa)) ∨ (¬(aαb) ∧ bαa))]
65. 65. Symmetry symmetric α ⊆ A × A ∀a, b ∈ A [(a = b) ∨ (aαb ∧ bαa) ∨ (¬(aαb) ∧ ¬(bαa))] ∀a, b ∈ A [(a = b) ∨ (aαb ↔ bαa)] α−1 = α asymmetric: ∃a, b ∈ A [(a = b) ∧ ((aαb ∧ ¬(bαa)) ∨ (¬(aαb) ∧ bαa))]
66. 66. Antisymmetry antisymmetric: ∀a, b ∈ A [(a = b) ∨ (aαb → ¬(bαa))] ⇔ ∀a, b ∈ A [(a = b) ∨ ¬(aαb) ∨ ¬(bαa)] ⇔ ∀a, b ∈ A [¬(aαb ∧ bαa) ∨ (a = b)] ⇔ ∀a, b ∈ A [(aαb ∧ bαa) → (a = b)]
67. 67. Antisymmetry antisymmetric: ∀a, b ∈ A [(a = b) ∨ (aαb → ¬(bαa))] ⇔ ∀a, b ∈ A [(a = b) ∨ ¬(aαb) ∨ ¬(bαa)] ⇔ ∀a, b ∈ A [¬(aαb ∧ bαa) ∨ (a = b)] ⇔ ∀a, b ∈ A [(aαb ∧ bαa) → (a = b)]
68. 68. Antisymmetry antisymmetric: ∀a, b ∈ A [(a = b) ∨ (aαb → ¬(bαa))] ⇔ ∀a, b ∈ A [(a = b) ∨ ¬(aαb) ∨ ¬(bαa)] ⇔ ∀a, b ∈ A [¬(aαb ∧ bαa) ∨ (a = b)] ⇔ ∀a, b ∈ A [(aαb ∧ bαa) → (a = b)]
69. 69. Antisymmetry antisymmetric: ∀a, b ∈ A [(a = b) ∨ (aαb → ¬(bαa))] ⇔ ∀a, b ∈ A [(a = b) ∨ ¬(aαb) ∨ ¬(bαa)] ⇔ ∀a, b ∈ A [¬(aαb ∧ bαa) ∨ (a = b)] ⇔ ∀a, b ∈ A [(aαb ∧ bαa) → (a = b)]
70. 70. Symmetry Examples R ⊆ {1, 2, 3} × {1, 2, 3} R = {(1, 2), (2, 1), (2, 3)} R is asymmetric
71. 71. Symmetry Examples R ⊆ Z × Z R = {(a, b) | ab ≥ 0} R is symmetric
72. 72. Symmetry Examples R ⊆ {1, 2, 3} × {1, 2, 3} R = {(1, 1), (2, 2)} R is symmetric and antisymmetric
73. 73. Transitivity transitive α ⊆ A × A ∀a, b, c ∈ A [(aαb ∧ bαc) → (aαc)] α2 ⊆ α nontransitive: ∃a, b, c ∈ A [(aαb ∧ bαc) ∧ ¬(aαc)] antitransitive: ∀a, b, c ∈ A [(aαb ∧ bαc) → ¬(aαc)]
74. 74. Transitivity transitive α ⊆ A × A ∀a, b, c ∈ A [(aαb ∧ bαc) → (aαc)] α2 ⊆ α nontransitive: ∃a, b, c ∈ A [(aαb ∧ bαc) ∧ ¬(aαc)] antitransitive: ∀a, b, c ∈ A [(aαb ∧ bαc) → ¬(aαc)]
75. 75. Transitivity transitive α ⊆ A × A ∀a, b, c ∈ A [(aαb ∧ bαc) → (aαc)] α2 ⊆ α nontransitive: ∃a, b, c ∈ A [(aαb ∧ bαc) ∧ ¬(aαc)] antitransitive: ∀a, b, c ∈ A [(aαb ∧ bαc) → ¬(aαc)]
76. 76. Transitivity transitive α ⊆ A × A ∀a, b, c ∈ A [(aαb ∧ bαc) → (aαc)] α2 ⊆ α nontransitive: ∃a, b, c ∈ A [(aαb ∧ bαc) ∧ ¬(aαc)] antitransitive: ∀a, b, c ∈ A [(aαb ∧ bαc) → ¬(aαc)]
77. 77. Transitivity Examples R ⊆ {1, 2, 3} × {1, 2, 3} R = {(1, 2), (2, 1), (2, 3)} R is antitransitive
78. 78. Transitivity Examples R ⊆ Z × Z R = {(a, b) | ab ≥ 0} R is nontransitive
79. 79. Converse Relation Properties Theorem Reflexivity, symmetry and transitivity are preserved in the converse relation.
80. 80. Closures reflexive closure: rα = α ∪ E symmetric closure: sα = α ∪ α−1 transitive closure: tα = i=1,2,3,... αi = α ∪ α2 ∪ α3 ∪ · · ·
81. 81. Closures reflexive closure: rα = α ∪ E symmetric closure: sα = α ∪ α−1 transitive closure: tα = i=1,2,3,... αi = α ∪ α2 ∪ α3 ∪ · · ·
82. 82. Closures reflexive closure: rα = α ∪ E symmetric closure: sα = α ∪ α−1 transitive closure: tα = i=1,2,3,... αi = α ∪ α2 ∪ α3 ∪ · · ·
83. 83. Special Relations predecessor - successor R ⊆ Z × Z R = {(a, b) | a − b = 1} irreflexive antisymmetric antitransitive
84. 84. Special Relations adjacency R ⊆ Z × Z R = {(a, b) | |a − b| = 1} irreflexive symmetric antitransitive
85. 85. Special Relations strict order R ⊆ Z × Z R = {(a, b) | a < b} irreflexive antisymmetric transitive
86. 86. Special Relations partial order R ⊆ Z × Z R = {(a, b) | a ≤ b} reflexive antisymmetric transitive
87. 87. Special Relations preorder R ⊆ Z × Z R = {(a, b) | |a| ≤ |b|} reflexive asymmetric transitive
88. 88. Special Relations limited difference R ⊆ Z × Z, m ∈ Z+ R = {(a, b) | |a − b| ≤ m} reflexive symmetric nontransitive
89. 89. Special Relations comparability R ⊆ U × U R = {(a, b) | (a ⊆ b) ∨ (b ⊆ a)} reflexive symmetric nontransitive
90. 90. Special Relations siblings? irreflexive symmetric transitive can a relation be symmetric, transitive and irreflexive?
91. 91. Special Relations siblings? irreflexive symmetric transitive can a relation be symmetric, transitive and irreflexive?
92. 92. Topics 1 Relations Introduction Relation Properties Equivalence Relations 2 Functions Introduction Pigeonhole Principle Recursion
93. 93. Compatibility Relations Definition compatibility relation: γ reflexive symmetric when drawing, lines instead of arrows matrix representation as a triangle matrix αα−1 is a compatibility relation
94. 94. Compatibility Relations Definition compatibility relation: γ reflexive symmetric when drawing, lines instead of arrows matrix representation as a triangle matrix αα−1 is a compatibility relation
95. 95. Compatibility Relations Definition compatibility relation: γ reflexive symmetric when drawing, lines instead of arrows matrix representation as a triangle matrix αα−1 is a compatibility relation
96. 96. Compatibility Relation Example A = {a1, a2, a3, a4} R = {(a1, a1), (a2, a2), (a3, a3), (a4, a4), (a1, a2), (a2, a1), (a2, a4), (a4, a2), (a3, a4), (a4, a3)} 1 1 0 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 0 0 1 1
97. 97. Compatibility Relation Example A = {a1, a2, a3, a4} R = {(a1, a1), (a2, a2), (a3, a3), (a4, a4), (a1, a2), (a2, a1), (a2, a4), (a4, a2), (a3, a4), (a4, a3)} 1 1 0 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 0 0 1 1
98. 98. Compatibility Relation Example P: persons, L: languages P = {p1, p2, p3, p4, p5, p6} L = {l1, l2, l3, l4, l5} α ⊆ P × L Mα = 1 1 0 0 0 1 1 0 0 0 0 0 1 0 1 1 0 1 1 0 0 0 0 1 0 0 1 1 0 0 Mα−1 = 1 1 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 0 0 0 1 0 0 0
99. 99. Compatibility Relation Example P: persons, L: languages P = {p1, p2, p3, p4, p5, p6} L = {l1, l2, l3, l4, l5} α ⊆ P × L Mα = 1 1 0 0 0 1 1 0 0 0 0 0 1 0 1 1 0 1 1 0 0 0 0 1 0 0 1 1 0 0 Mα−1 = 1 1 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 0 0 0 1 0 0 0
100. 100. Compatibility Relation Example αα−1 ⊆ P × P Mαα−1 = 1 1 0 1 0 1 1 1 0 1 0 1 0 0 1 1 0 1 1 1 1 1 1 1 0 0 0 1 1 0 1 1 1 1 0 1
101. 101. Compatibility Block Definition compatibility block: C ⊆ A ∀a, b [a ∈ C ∧ b ∈ C → aγb] maximal compatibility block: not a subset of another compatibility block an element can be a member of more than one MCB complete cover: Cγ set of all MCBs
102. 102. Compatibility Block Definition compatibility block: C ⊆ A ∀a, b [a ∈ C ∧ b ∈ C → aγb] maximal compatibility block: not a subset of another compatibility block an element can be a member of more than one MCB complete cover: Cγ set of all MCBs
103. 103. Compatibility Block Definition compatibility block: C ⊆ A ∀a, b [a ∈ C ∧ b ∈ C → aγb] maximal compatibility block: not a subset of another compatibility block an element can be a member of more than one MCB complete cover: Cγ set of all MCBs
104. 104. Compatibility Block Example C1 = {p4, p6} C2 = {p2, p4, p6} C3 = {p1, p2, p4, p6} (MCB) Cγ = {{p1, p2, p4, p6}, {p3, p4, p6}, {p4, p5}}
105. 105. Compatibility Block Example C1 = {p4, p6} C2 = {p2, p4, p6} C3 = {p1, p2, p4, p6} (MCB) Cγ = {{p1, p2, p4, p6}, {p3, p4, p6}, {p4, p5}}
106. 106. Compatibility Block Example C1 = {p4, p6} C2 = {p2, p4, p6} C3 = {p1, p2, p4, p6} (MCB) Cγ = {{p1, p2, p4, p6}, {p3, p4, p6}, {p4, p5}}
107. 107. Equivalence Relations Definition equivalence relation: reflexive symmetric transitive equivalence classes (partitions) every element is a member of exactly one equivalence class complete cover: C
108. 108. Equivalence Relations Definition equivalence relation: reflexive symmetric transitive equivalence classes (partitions) every element is a member of exactly one equivalence class complete cover: C
109. 109. Equivalence Relation Example R ⊆ Z × Z R = {(a, b) | ∃m ∈ Z [a − b = 5m]} R partitions Z into 5 equivalence classes
110. 110. References Required Reading: Grimaldi Chapter 5: Relations and Functions 5.1. Cartesian Products and Relations Chapter 7: Relations: The Second Time Around 7.1. Relations Revisited: Properties of Relations 7.4. Equivalence Relations and Partitions
111. 111. Topics 1 Relations Introduction Relation Properties Equivalence Relations 2 Functions Introduction Pigeonhole Principle Recursion
112. 112. Functions Definition function: f : X → Y ∀x ∈ X ∀y1, y2 ∈ Y [(x, y1), (x, y2) ∈ f → y1 = y2] X: domain, Y : codomain (or range) y = f (x) : (x, y) ∈ f y: image of x under f f : X → Y , X ⊆ X subset image: f (X ) = {f (x) | x ∈ X }
113. 113. Functions Definition function: f : X → Y ∀x ∈ X ∀y1, y2 ∈ Y [(x, y1), (x, y2) ∈ f → y1 = y2] X: domain, Y : codomain (or range) y = f (x) : (x, y) ∈ f y: image of x under f f : X → Y , X ⊆ X subset image: f (X ) = {f (x) | x ∈ X }
114. 114. Functions Definition function: f : X → Y ∀x ∈ X ∀y1, y2 ∈ Y [(x, y1), (x, y2) ∈ f → y1 = y2] X: domain, Y : codomain (or range) y = f (x) : (x, y) ∈ f y: image of x under f f : X → Y , X ⊆ X subset image: f (X ) = {f (x) | x ∈ X }
115. 115. Subset Image Examples f : R → R f (x) = x2 f (Z) = {0, 1, 4, 9, 16, . . . } f ({−2, 1}) = {1, 4}
116. 116. Subset Image Examples f : R → R f (x) = x2 f (Z) = {0, 1, 4, 9, 16, . . . } f ({−2, 1}) = {1, 4}
117. 117. One-to-One Functions Definition f : X → Y is one-to-one (or injective): ∀x1, x2 ∈ X [f (x1) = f (x2) → x1 = x2]
118. 118. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
119. 119. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
120. 120. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
121. 121. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
122. 122. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
123. 123. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
124. 124. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
125. 125. One-to-One Function Examples one-to-one f : R → R f (x) = 3x + 7 f (x1) = f (x2) ⇒ 3x1 + 7 = 3x2 + 7 ⇒ 3x1 = 3x2 ⇒ x1 = x2 not one-to-one g : Z → Z g(x) = x4 − x g(0) = 04 − 0 = 0 g(1) = 14 − 1 = 0
126. 126. Onto Functions Definition f : X → Y is onto (or surjective): ∀y ∈ Y ∃x ∈ X [f (x) = y] f (X) = Y
127. 127. Onto Function Examples onto f : R → R f (x) = x3 not onto f : Z → Z f (x) = 3x + 1
128. 128. Onto Function Examples onto f : R → R f (x) = x3 not onto f : Z → Z f (x) = 3x + 1
129. 129. Bijective Functions Definition f : X → Y is bijective: f is one-to-one and onto
130. 130. Function Composition Definition f : X → Y , g : Y → Z g ◦ f : X → Z (g ◦ f )(x) = g(f (x)) not commutative associative: f ◦ (g ◦ h) = (f ◦ g) ◦ h
131. 131. Function Composition Definition f : X → Y , g : Y → Z g ◦ f : X → Z (g ◦ f )(x) = g(f (x)) not commutative associative: f ◦ (g ◦ h) = (f ◦ g) ◦ h
132. 132. Composition Commutativity Example f : R → R f (x) = x2 g : R → R g(x) = x + 5 g ◦ f : R → R (g ◦ f )(x) = x2 + 5 f ◦ g : R → R (f ◦ g)(x) = (x + 5)2
133. 133. Composition Commutativity Example f : R → R f (x) = x2 g : R → R g(x) = x + 5 g ◦ f : R → R (g ◦ f )(x) = x2 + 5 f ◦ g : R → R (f ◦ g)(x) = (x + 5)2
134. 134. Composition Commutativity Example f : R → R f (x) = x2 g : R → R g(x) = x + 5 g ◦ f : R → R (g ◦ f )(x) = x2 + 5 f ◦ g : R → R (f ◦ g)(x) = (x + 5)2
135. 135. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is one-to-one ∧ g is one-to-one ⇒ g ◦ f is one-to-one Proof. (g ◦ f )(x1) = (g ◦ f )(x2) ⇒ g(f (x1)) = g(f (x2)) ⇒ f (x1) = f (x2) ⇒ x1 = x2
136. 136. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is one-to-one ∧ g is one-to-one ⇒ g ◦ f is one-to-one Proof. (g ◦ f )(x1) = (g ◦ f )(x2) ⇒ g(f (x1)) = g(f (x2)) ⇒ f (x1) = f (x2) ⇒ x1 = x2
137. 137. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is one-to-one ∧ g is one-to-one ⇒ g ◦ f is one-to-one Proof. (g ◦ f )(x1) = (g ◦ f )(x2) ⇒ g(f (x1)) = g(f (x2)) ⇒ f (x1) = f (x2) ⇒ x1 = x2
138. 138. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is one-to-one ∧ g is one-to-one ⇒ g ◦ f is one-to-one Proof. (g ◦ f )(x1) = (g ◦ f )(x2) ⇒ g(f (x1)) = g(f (x2)) ⇒ f (x1) = f (x2) ⇒ x1 = x2
139. 139. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is one-to-one ∧ g is one-to-one ⇒ g ◦ f is one-to-one Proof. (g ◦ f )(x1) = (g ◦ f )(x2) ⇒ g(f (x1)) = g(f (x2)) ⇒ f (x1) = f (x2) ⇒ x1 = x2
140. 140. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is onto ∧ g is onto ⇒ g ◦ f is onto Proof. ∀z ∈ Z ∃y ∈ Y g(y) = z ∀y ∈ Y ∃x ∈ X f (x) = y ⇒ ∀z ∈ Z ∃x ∈ X g(f (x)) = z
141. 141. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is onto ∧ g is onto ⇒ g ◦ f is onto Proof. ∀z ∈ Z ∃y ∈ Y g(y) = z ∀y ∈ Y ∃x ∈ X f (x) = y ⇒ ∀z ∈ Z ∃x ∈ X g(f (x)) = z
142. 142. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is onto ∧ g is onto ⇒ g ◦ f is onto Proof. ∀z ∈ Z ∃y ∈ Y g(y) = z ∀y ∈ Y ∃x ∈ X f (x) = y ⇒ ∀z ∈ Z ∃x ∈ X g(f (x)) = z
143. 143. Composite Function Theorems Theorem f : X → Y , g : Y → Z f is onto ∧ g is onto ⇒ g ◦ f is onto Proof. ∀z ∈ Z ∃y ∈ Y g(y) = z ∀y ∈ Y ∃x ∈ X f (x) = y ⇒ ∀z ∈ Z ∃x ∈ X g(f (x)) = z
144. 144. Identity Function Definition identity function: 1X 1X : X → X 1X (x) = x
145. 145. Inverse Function Definition f : X → Y is invertible: ∃f −1 : Y → X [f −1 ◦ f = 1X ∧ f ◦ f −1 = 1Y ] f −1: inverse of function f
146. 146. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
147. 147. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
148. 148. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
149. 149. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
150. 150. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
151. 151. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
152. 152. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
153. 153. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
154. 154. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
155. 155. Inverse Function Examples f : R → R f (x) = 2x + 5 f −1 : R → R f −1(x) = x−5 2 (f −1 ◦ f )(x) = f −1(f (x)) = f −1(2x + 5) = (2x+5)−5 2 = 2x 2 = x (f ◦ f −1)(x) = f (f −1(x)) = f (x−5 2 ) = 2x−5 2 + 5 = (x − 5) + 5 = x
156. 156. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
157. 157. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
158. 158. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
159. 159. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
160. 160. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
161. 161. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
162. 162. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
163. 163. Inverse Function Theorem If a function is invertible, its inverse is unique. Proof. f : X → Y g, h : Y → X g ◦ f = 1X ∧ f ◦ g = 1Y h ◦ f = 1X ∧ f ◦ h = 1Y h = h ◦ 1Y = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = 1X ◦ g = g
164. 164. Invertible Function Theorem A function is invertible if and only if it is one-to-one and onto.
165. 165. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
166. 166. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
167. 167. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
168. 168. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
169. 169. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
170. 170. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
171. 171. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
172. 172. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
173. 173. Invertible Function If invertible then one-to-one. f : X → Y f (x1) = f (x2) ⇒ f −1 (f (x1)) = f −1 (f (x2)) ⇒ (f −1 ◦ f )(x1) = (f −1 ◦ f )(x2) ⇒ 1X (x1) = 1X (x2) ⇒ x1 = x2 If invertible then onto. f : X → Y y = 1Y (y) = (f ◦ f −1 )(y) = f (f −1 (y))
174. 174. Invertible Function If bijective then invertible. f : X → Y f is onto ⇒ ∀y ∈ Y ∃x ∈ X f (x) = y let g : Y → X be defined by x = g(y) is it possible that g(y) = x1 = x2 = g(y) ? this would mean: f (x1) = y = f (x2) but f is one-to-one
175. 175. Invertible Function If bijective then invertible. f : X → Y f is onto ⇒ ∀y ∈ Y ∃x ∈ X f (x) = y let g : Y → X be defined by x = g(y) is it possible that g(y) = x1 = x2 = g(y) ? this would mean: f (x1) = y = f (x2) but f is one-to-one
176. 176. Invertible Function If bijective then invertible. f : X → Y f is onto ⇒ ∀y ∈ Y ∃x ∈ X f (x) = y let g : Y → X be defined by x = g(y) is it possible that g(y) = x1 = x2 = g(y) ? this would mean: f (x1) = y = f (x2) but f is one-to-one
177. 177. Topics 1 Relations Introduction Relation Properties Equivalence Relations 2 Functions Introduction Pigeonhole Principle Recursion
178. 178. Pigeonhole Principle Definition pigeonhole principle (Dirichlet drawers): If m pigeons go into n holes and m > n, then at least one hole contains more than one pigeon. f : X → Y |X| > |Y | ⇒ f is not one-to-one ∃x1, x2 ∈ X [x1 = x2 ∧ f (x1) = f (x2)]
179. 179. Pigeonhole Principle Definition pigeonhole principle (Dirichlet drawers): If m pigeons go into n holes and m > n, then at least one hole contains more than one pigeon. f : X → Y |X| > |Y | ⇒ f is not one-to-one ∃x1, x2 ∈ X [x1 = x2 ∧ f (x1) = f (x2)]
180. 180. Pigeonhole Principle Examples Among 367 people, at least two have the same birthday. In an exam where the grades are integers between 0 and 100, how many students have to take the exam to make sure that at least two students will have the same grade?
181. 181. Pigeonhole Principle Examples Among 367 people, at least two have the same birthday. In an exam where the grades are integers between 0 and 100, how many students have to take the exam to make sure that at least two students will have the same grade?
182. 182. Generalized Pigeonhole Principle Definition generalized pigeonhole principle: If m objects are distributed to n drawers, then at least one of the drawers contains m/n objects. example Among 100 people, at least 100/12 = 9 were born in the same month.
183. 183. Generalized Pigeonhole Principle Definition generalized pigeonhole principle: If m objects are distributed to n drawers, then at least one of the drawers contains m/n objects. example Among 100 people, at least 100/12 = 9 were born in the same month.
184. 184. Pigeonhole Principle Example Theorem S = {1, 2, 3, . . . , 9}, T ⊂ S, |T| = 6 ∃s1, s2 ∈ T [s1 + s2 = 10]
185. 185. Pigeonhole Principle Example Theorem S ⊆ Z+, ∀a ∈ S [a ≤ 14], |S| = 6 T = P(S) − ∅ X = {ΣA | A ∈ T}, ΣA : sum of the elements in A |X| < |T| Proof Attempt holes: 1 ≤ ΣA ≤ 9 + · · · + 14 = 69 pigeons: 26 − 1 = 63 Proof. consider T − S holes: 1 ≤ sA ≤ 10 + · · · + 14 = 60 pigeons: 26 − 2 = 62
186. 186. Pigeonhole Principle Example Theorem S ⊆ Z+, ∀a ∈ S [a ≤ 14], |S| = 6 T = P(S) − ∅ X = {ΣA | A ∈ T}, ΣA : sum of the elements in A |X| < |T| Proof Attempt holes: 1 ≤ ΣA ≤ 9 + · · · + 14 = 69 pigeons: 26 − 1 = 63 Proof. consider T − S holes: 1 ≤ sA ≤ 10 + · · · + 14 = 60 pigeons: 26 − 2 = 62
187. 187. Pigeonhole Principle Example Theorem S ⊆ Z+, ∀a ∈ S [a ≤ 14], |S| = 6 T = P(S) − ∅ X = {ΣA | A ∈ T}, ΣA : sum of the elements in A |X| < |T| Proof Attempt holes: 1 ≤ ΣA ≤ 9 + · · · + 14 = 69 pigeons: 26 − 1 = 63 Proof. consider T − S holes: 1 ≤ sA ≤ 10 + · · · + 14 = 60 pigeons: 26 − 2 = 62
188. 188. Pigeonhole Principle Example Theorem S ⊆ Z+, ∀a ∈ S [a ≤ 14], |S| = 6 T = P(S) − ∅ X = {ΣA | A ∈ T}, ΣA : sum of the elements in A |X| < |T| Proof Attempt holes: 1 ≤ ΣA ≤ 9 + · · · + 14 = 69 pigeons: 26 − 1 = 63 Proof. consider T − S holes: 1 ≤ sA ≤ 10 + · · · + 14 = 60 pigeons: 26 − 2 = 62
189. 189. Pigeonhole Principle Example Theorem S = {1, 2, 3, . . . , 200}, T ⊂ S, |T| = 101 ∃s1, s2 ∈ T [s2|s1] first, show that: ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] then, use this to prove the main theorem
190. 190. Pigeonhole Principle Example Theorem S = {1, 2, 3, . . . , 200}, T ⊂ S, |T| = 101 ∃s1, s2 ∈ T [s2|s1] first, show that: ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] then, use this to prove the main theorem
191. 191. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
192. 192. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
193. 193. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
194. 194. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
195. 195. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
196. 196. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
197. 197. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
198. 198. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
199. 199. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
200. 200. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
201. 201. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
202. 202. Pigeonhole Principle Example Theorem ∀n ∃!p [n = 2r p ∧ r ∈ N ∧ ∃t ∈ Z [p = 2t + 1]] Proof of existence. n = 1: r = 0, p = 1 n ≤ k: assume n = 2r p n = k + 1: n = 2 : r = 1, p = 1 n prime (n > 2) : r = 0, p = n ¬(n prime) : n = n1n2 n = 2r1 p1 · 2r2 p2 n = 2r1+r2 · p1p2 Proof of uniqueness. if not unique: n = 2r1 p1 = 2r2 p2 ⇒ 2r1−r2 p1 = p2 ⇒ 2|p2
203. 203. Pigeonhole Principle Example Theorem S = {1, 2, 3, . . . , 200}, T ⊂ S, |T| = 101 ∃s1, s2 ∈ T [s2|s1] Proof. P = {p | p ∈ S, ∃i ∈ Z [p = 2i + 1]}, |P| = 100 f : S → P, r ∈ N, s = 2r p → f (s) = p |T| = 101 ⇒ at least two elements have the same image in P: f (s1) = f (s2) ⇒ 2r1 p = 2r2 p s1 s2 = 2r1 p 2r2 p = 2r1−r2
204. 204. Pigeonhole Principle Example Theorem S = {1, 2, 3, . . . , 200}, T ⊂ S, |T| = 101 ∃s1, s2 ∈ T [s2|s1] Proof. P = {p | p ∈ S, ∃i ∈ Z [p = 2i + 1]}, |P| = 100 f : S → P, r ∈ N, s = 2r p → f (s) = p |T| = 101 ⇒ at least two elements have the same image in P: f (s1) = f (s2) ⇒ 2r1 p = 2r2 p s1 s2 = 2r1 p 2r2 p = 2r1−r2
205. 205. Pigeonhole Principle Example Theorem S = {1, 2, 3, . . . , 200}, T ⊂ S, |T| = 101 ∃s1, s2 ∈ T [s2|s1] Proof. P = {p | p ∈ S, ∃i ∈ Z [p = 2i + 1]}, |P| = 100 f : S → P, r ∈ N, s = 2r p → f (s) = p |T| = 101 ⇒ at least two elements have the same image in P: f (s1) = f (s2) ⇒ 2r1 p = 2r2 p s1 s2 = 2r1 p 2r2 p = 2r1−r2
206. 206. Pigeonhole Principle Example Theorem S = {1, 2, 3, . . . , 200}, T ⊂ S, |T| = 101 ∃s1, s2 ∈ T [s2|s1] Proof. P = {p | p ∈ S, ∃i ∈ Z [p = 2i + 1]}, |P| = 100 f : S → P, r ∈ N, s = 2r p → f (s) = p |T| = 101 ⇒ at least two elements have the same image in P: f (s1) = f (s2) ⇒ 2r1 p = 2r2 p s1 s2 = 2r1 p 2r2 p = 2r1−r2
207. 207. Topics 1 Relations Introduction Relation Properties Equivalence Relations 2 Functions Introduction Pigeonhole Principle Recursion
208. 208. Recursive Functions Definition recursive function: a function defined in terms of itself f (n) = h(f (m)) inductively defined function: a recursive function where the size is reduced at every step f (n) = k if n = 0 h(f (n − 1)) if n > 0
209. 209. Recursion Examples f 91(n) = n − 10 if n > 100 f 91(f 91(n + 11)) if n ≤ 100 n! = 1 if n = 0 n · (n − 1)! if n > 0
210. 210. Recursion Examples f 91(n) = n − 10 if n > 100 f 91(f 91(n + 11)) if n ≤ 100 n! = 1 if n = 0 n · (n − 1)! if n > 0
211. 211. Greatest Common Divisor gcd(a, b) = b if b | a gcd(b, a mod b) if b a gcd(333, 84) = gcd(84, 333 mod 84) = gcd(84, 81) = gcd(81, 84 mod 81) = gcd(81, 3) = 3
212. 212. Greatest Common Divisor gcd(a, b) = b if b | a gcd(b, a mod b) if b a gcd(333, 84) = gcd(84, 333 mod 84) = gcd(84, 81) = gcd(81, 84 mod 81) = gcd(81, 3) = 3
213. 213. Fibonacci Sequence Fn = fib(n) = 1 if n = 1 1 if n = 2 fib(n − 2) + fib(n − 1) if n > 2 F1 F2 F3 F4 F5 F6 F7 F8 . . . 1 1 2 3 5 8 13 21 . . .
214. 214. Fibonacci Sequence Theorem n i=1 Fi 2 = Fn · Fn+1 Proof. n = 2 : 2 i=1 Fi 2 = F1 2 + F2 2 = 1 + 1 = 1 · 2 = F2 · F3 n = k : k i=1 Fi 2 = Fk · Fk+1 n = k + 1 : k+1 i=1 Fi 2 = k i=1 Fi 2 + Fk+1 2 = Fk · Fk+1 + Fk+1 2 = Fk+1 · (Fk + Fk+1) = Fk+1 · Fk+2
215. 215. Fibonacci Sequence Theorem n i=1 Fi 2 = Fn · Fn+1 Proof. n = 2 : 2 i=1 Fi 2 = F1 2 + F2 2 = 1 + 1 = 1 · 2 = F2 · F3 n = k : k i=1 Fi 2 = Fk · Fk+1 n = k + 1 : k+1 i=1 Fi 2 = k i=1 Fi 2 + Fk+1 2 = Fk · Fk+1 + Fk+1 2 = Fk+1 · (Fk + Fk+1) = Fk+1 · Fk+2
216. 216. Fibonacci Sequence Theorem n i=1 Fi 2 = Fn · Fn+1 Proof. n = 2 : 2 i=1 Fi 2 = F1 2 + F2 2 = 1 + 1 = 1 · 2 = F2 · F3 n = k : k i=1 Fi 2 = Fk · Fk+1 n = k + 1 : k+1 i=1 Fi 2 = k i=1 Fi 2 + Fk+1 2 = Fk · Fk+1 + Fk+1 2 = Fk+1 · (Fk + Fk+1) = Fk+1 · Fk+2
217. 217. Fibonacci Sequence Theorem n i=1 Fi 2 = Fn · Fn+1 Proof. n = 2 : 2 i=1 Fi 2 = F1 2 + F2 2 = 1 + 1 = 1 · 2 = F2 · F3 n = k : k i=1 Fi 2 = Fk · Fk+1 n = k + 1 : k+1 i=1 Fi 2 = k i=1 Fi 2 + Fk+1 2 = Fk · Fk+1 + Fk+1 2 = Fk+1 · (Fk + Fk+1) = Fk+1 · Fk+2
218. 218. Fibonacci Sequence Theorem n i=1 Fi 2 = Fn · Fn+1 Proof. n = 2 : 2 i=1 Fi 2 = F1 2 + F2 2 = 1 + 1 = 1 · 2 = F2 · F3 n = k : k i=1 Fi 2 = Fk · Fk+1 n = k + 1 : k+1 i=1 Fi 2 = k i=1 Fi 2 + Fk+1 2 = Fk · Fk+1 + Fk+1 2 = Fk+1 · (Fk + Fk+1) = Fk+1 · Fk+2
219. 219. Fibonacci Sequence Theorem n i=1 Fi 2 = Fn · Fn+1 Proof. n = 2 : 2 i=1 Fi 2 = F1 2 + F2 2 = 1 + 1 = 1 · 2 = F2 · F3 n = k : k i=1 Fi 2 = Fk · Fk+1 n = k + 1 : k+1 i=1 Fi 2 = k i=1 Fi 2 + Fk+1 2 = Fk · Fk+1 + Fk+1 2 = Fk+1 · (Fk + Fk+1) = Fk+1 · Fk+2
220. 220. Fibonacci Sequence Theorem n i=1 Fi 2 = Fn · Fn+1 Proof. n = 2 : 2 i=1 Fi 2 = F1 2 + F2 2 = 1 + 1 = 1 · 2 = F2 · F3 n = k : k i=1 Fi 2 = Fk · Fk+1 n = k + 1 : k+1 i=1 Fi 2 = k i=1 Fi 2 + Fk+1 2 = Fk · Fk+1 + Fk+1 2 = Fk+1 · (Fk + Fk+1) = Fk+1 · Fk+2
221. 221. Ackermann Function ack(x, y) = y + 1 if x = 0 ack(x − 1, 1) if y = 0 ack(x − 1, ack(x, y − 1)) if x > 0 ∧ y > 0
222. 222. References Required Reading: Grimaldi Chapter 5: Relations and Functions 5.2. Functions: Plain and One-to-One 5.3. Onto Functions: Stirling Numbers of the Second Kind 5.5. The Pigeonhole Principle 5.6. Function Composition and Inverse Functions
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# ✅ Square Formula ⭐️⭐️⭐️⭐️⭐️
5/5 - (1 bình chọn)
Mục Lục
## Definition
The number of square units needed to fill a square is its area. In common terms, the area is the inner region of a flat surface (2-D figure).
In the given square, the space shaded in violet is the area of the square.
For example, The space occupied by the swimming pool below can be found by finding the area of the pool.
## The Formula for the Area of A Square
The area of a square is equal to (side) × (side) square units.
The area of a square when the diagonal, d, is given is d2÷2 square units.
For example,
The area of a square with each side 8 feet long is 8 × 8 or 64 square feet (ft2).
## Solved Examples
Example 1: Given that each side is 5 cm, find the area of a square.
Solution:
Area of a square = side × side
Area = 5 × 5
Area = 25 cm2
Example 2: The side of a square wall is 50 m. What is the cost of painting it at the rate of Rs. 2 per sq. m?
Solution:
Side of the wall = 50 m
Area of the wall = side × side = 50 m × 50 m = 2500 sq. m
The cost of painting 1 sq. m = Rs. 2
Thus the cost of painting a 2500 sq. m wall = Rs. 2 × 2,500 = Rs 5,000
Example 3: Find the area of a square whose diagonal is measured is 4 cm.
Solution:
Given:
Side, d = 4 cm
We know that the formula to find the area of a square when the diagonal, d, is given is d2÷2 square units.
Substituting the diagonal value, we get:
= 42÷2 = 16 ÷ 2 = 8
Thus, the area of the square is 8 cm2.
What is the difference between the perimeter and area of a square?
The perimeter of a square is the sum of its four sides or the length of its boundary. It is a one-dimensional measurement and expressed in linear units. Area of a square is the space filled by the square in two-dimensional space. It is expressed in square units.
How do you calculate the area of a square if the perimeter is given?
The perimeter of the square is the sum of all four sides of the square. If the perimeter is given, then the formula to calculate the area of the square, A = Perimeter2/16
What are the units of the area of the square?
The area of the square is 2-dimensional. Thus, the area of the square is always represented by square units, for which the common units are cm2, m2, in2, or ft2.
Do two squares of equal areas have equal perimeters?
Yes. Two squares of equal areas, given by side x side, will have the same side lengths. They are congruent. Consequently, the perimeters of the two squares, given by 4 x side length, will be equal as well.
## Square Formula
Square is a regular quadrilateral. All the four sides and angles of a square are equal. The four angles are 90 degrees each, that is, right angles. A square may also be considered as a special case of rectangle wherein the two adjacent sides are of equal length.
In this section, we will learn about the square formulas – a list of the formula related to squares which will help you compute its area, perimeter, and length of its diagonals. They are enlisted below:
Where ‘a’ is the length of a side of the square.
Properties of a Square
• The lengths of all its four sides are equal.
• The measurements of all its four angles are equal.
• The two diagonals bisect each other at right angles, that is, 90°.
• The opposite sides of a square are both parallel and equal in length.
• The lengths of diagonals of a square are equal.
Derivations:
Consider a square with the lengths of its side and diagonal are a and d units respectively.
Formula for area of a square: Area of a square can be defined as the region which is enclosed within its boundary. As we mentioned, a square is nothing a rectangle with its two adjacent sides being equal in length. Hence, we express area as:
The area of a rectangle = Length × Breadth
Here,
The formula for the perimeter of a square: Perimeter of the square is the length of its boundary. The sum of the length of all sides of a square represents its boundary. Hence, the formula can be given by:
Perimeter = length of 4 sides
Perimeter = a+ a + a + a
Perimeter of square = 4a
The formula for diagonal of a square: A diagonal is a line which joins two opposite sides in a polygon. For calculating the length diagonal of a square, we make use of the Pythagoras Theorem.
In the above figure, the diagonal’ divides the square into two right angled triangles. It can be noted here that since the adjacent sides of a square are equal in length, the right angled triangle is also isosceles with each of its sides being of length ‘a’.
Hence, we can conveniently apply the Pythagorean theorem on these triangles with base and perpendicular being ‘a’ units and hypotenuse being’ units. So we have:
Solved examples:
Question 1: A square has one of its sides measuring 23 cm. Calculate its area, perimeter, and length of its diagonal.
Solution: Given,
Side of the square = 23 cm
Area of the square:
Area of the square =
Perimeter of the square:
Perimeter of the square= 4a= 4 × 23 = 92 cm
Diagonal of a square:
Diagonal of a square =
Question 2: A rectangular floor is 50 m long and 20 m wide. Square tiles, each of 5 m side length, are to be used to cover the floor. Find the total number of tiles which will be required to cover the floor.
Solution: Given,
Length of the floor = 50 m
Area of the rectangular floor = length x breadth = 50 m x 20 m = 1000 sq. m
Side of one tile = 5 m
Area of one such tile = side x side = 5 m x 5 m = 25 sq. m
## Area of a Square Formula
Before moving into the area of square formula used for calculating the region occupied, let us try using graph paper. You are required to find the area of a side 5 cm. Using this dimension, draw a square on a graph paper having 1 cm × 1 cm squares. The square covers 25 complete squares.
Thus, the area of the square is 25 square cm, which can be written as 5 cm × 5 cm, that is, side × side.
From the above discussion, it can be inferred that the formula can give the area of a square is:
Area of a Square = Side × Side
Therefore, the area of square = Side2 square units
and the perimeter of a square = 4 × side units
Here some of the unit conversion lists are provided for reference. Some conversions of units:
• 1 m = 100 cm
• 1 sq. m = 10,000 sq. cm
• 1 km = 1000 m
• 1 sq. km = 1,000,000 sq. m
## Area of a Square Sample Problems
Example 1:
Find the area of a square clipboard whose side measures 120 cm.
Solution:
Side of the clipboard = 120 cm = 1.2 m
Area of the clipboard = side × side
= 120 cm × 120 cm
= 14400 sq. cm
= 1.44 sq. m
Example 2:
The side of a square wall is 75 m. What is the cost of painting it at the rate of Rs. 3 per sq. m?
Solution:
Side of the wall = 75 m
Area of the wall = side × side = 75 m × 75 m = 5,625 sq. m
For 1 sq. m, the cost of painting = Rs. 3
Thus, for 5,625 sq. m, the cost of painting = Rs. 3 × 5,625 = Rs 16,875
Example 3:
A courtyard’s floor which is 50 m long and 40 m wide is to be covered by square tiles. The side of each tile is 2 m. Find the number of tiles required to cover the floor.
Solution:
Length of the floor = 50 m
The breadth of the floor = 40 m
Area of the floor = length × breadth = 50 m × 40 m = 2000 sq. m
Side of one tile = 2 m
Area of one tile = side ×side = 2 m × 2 m = 4 sq. m
No. of tiles required = area of floor/area of a tile = 2000/4 = 500 tiles.
## Practice Problems
1. A square wall of length 25 metres, has to be painted. If the cost of painting per square metre is ₹ 4.50. Find the cost of painting the whole wall.
2. Find the length of a square park whose area is 3600 square metres.
3. Find the area of the square whose length of the diagonal is 5√2 cm.
## Frequently Asked Questions on Area of Square
### What is the area of a square?
As we know, a square is a two-dimensional figure with four sides. It is also known as a quadrilateral. The area of a square is defined as the total number of unit squares in the shape of a square. In other words, it is defined as the space occupied by the square.
### Why is the area of a square a side square?
A square is a 2D figure in which all the sides are of equal measure. Since all the sides are equal, the area would be length times width, which is equal to side × side. Hence, the area of a square is side square.
### What is the area of a square formula?
The area of a square can be calculated using the formula side × side square units.
### How to find the area of a square if a diagonal is given?
If the diagonal of a square is given, then the formula to calculate the area of a square is:
A = (½) × d2 square units.
Where “d” is the diagonal
### What is the perimeter and the area of a square?
The perimeter of the square is the sum of all the four sides of a square, whereas the area of a square is defined as the region or the space occupied by a square in the two-dimensional space.
### What is the area of a square if its side length is 10 cm?
Given: Side = 10 cm
We know that, Area of a square = Side × Side square units
Thus, Area = 10 × 10 = 100 cm2
Therefore, the area of a square is 100 cm2 if its side length is 10 cm.
### What is the unit for an area of square?
The area of a square is measured in square units.
### How to calculate the area of a square if its perimeter is given?
Follow the below steps to find the area of a square if its perimeter is given:
Step 1: Find the side length of a square using the perimeter formula, P = 4 × Side
Step 2: Substitute the side length in the area formula: A = Side × Side.
# Area of Square
The area of a square is defined as the number of square units needed to fill this shape. In other words, when we want to find the area of a square, we consider the length of its side. Since all the sides of the shape are equal, its area is the product of its two sides. The common units used to measure the area of the square are square meters, square feet, square inch, and square cm.
The area of a square can also be calculated with the help of other dimensions, such as the diagonal and the perimeter of the square. Let us try to understand more about the area of the square on this page.
## What is the Area of Square?
A square is a closed two-dimensional shape with four equal sides and four equal angles. The four sides of the square form the four angles at the vertices. The sum of the total length of the sides of a square is its perimeter, and the total space occupied by the shape is the area of the square. It is a quadrilateral with the following properties.
• The opposite sides are parallel.
• All four sides are equal.
• All angles measure 90º.
Squares can be found all around us. Here are some commonly seen objects which have the shape of a square. The chessboard, the clock, a blackboard, a tile, are all examples of a square.
### Area of a Square Definition
The area of a square is the measure of the space or surface occupied by it. It is equal to the product of the length of its two sides. Since the area of a square is the product of its two sides, the unit of the area is given in square units.
Observe the below square shown below. It has occupied 25 squares. Therefore, the area of the square is 25 square units. From the figure, we can observe that the length of each side is 5 units. Therefore, the area of the square is the product of its sides. Area of square = side × side = 5 × 5 = 25 square units.
### Square Definition
A square is a two-dimensional shape quadrilateral with four sides equal and parallel to each other. The angles in this shape are measured as 90 degrees.
## Area of a Square Formula
The formula for the area of a square when the sides are given is:
Area of a square = Side × Side = S2
Algebraically, the area of a square can be found by squaring the number representing the measure of the side of the square. Now, let us use this formula to find the area of a square of side 7 cm. We know that the area of a square = Side × Side. Substituting the length of side as 7 cm, 7 × 7 = 49. Therefore, the area of the given square is 49 cm2.
The area of a square can also be found with the help of the diagonal of the square. The formula used to find the area of a square when the diagonal is given is:
Area of a square using diagonals = Diagonal2/2.
Let us understand the derivation of this formula with the help of the following figure, where ‘d’ is the diagonal and ‘s’ represents the sides of the square.
Here the side of the square is ‘s’ and the diagonal of the square is ‘d’. Applying the Pythagoras theorem we have d2 = s2 + s2; d2 = 2s2; d = √2s; s = d/√2. Now, this formula will help us to find the area of the square, using the diagonal. Area = s2 = (d/√2)2 = d2/2. Therefore, the area of the square is equal to d2/2.
## How to Find Area of a Square?
In the above section, we covered the definition of area of square as well as area of square formula. In this section let us understand how to use the area of the square formula to find its area with the help of few applications or real-world examples.
### Area of Square When the Perimeter of a Square is Given
Example: Find the area of a square park whose perimeter is 360 ft.
Solution:
Given: Perimeter of the square park = 360ft
We know that,
Perimeter of a square = 4 × side
⇒ 4 × side = 360
⇒ side = 360/4
⇒ side = 90ft
Area of a square = side2
Hence, Area of the square park = 902 = 90 × 90 = 8100 ft2
Thus, the area of a square park whose perimeter is 360 ft is 8100 ft2.
### Area of Square When the Side of a Square is Given
Example: Find the area of a square park whose side is 90 ft.
Solution:
Given: Side of the square park = 90ft
We know that,
Area of a square = ft2
Hence, Area of the square park = 902 = 90 × 90 = 8100 ft2
Thus, the area of a square park whose side is 90 ft is 8100 ft2
### Area of Square When the Diagonal of a Square is Given
Example: Find the area of a square park whose diagonal is 14 feet.
Solution:
Given: Diagonal of the square park = 14 ft
We know that,
Area of a square formula when diagonal is given = d2/2
Hence, Area of the square park = (14 × 14)/2 = 98 ft2
Thus, the area of a square park whose diagonal is 14 m is 98 ft2.
Area of Square Tips:
Note the following points which should be remembered while we calculate the area of a square.
• A common mistake that we tend to make while calculating the area of a square is doubling the number. This is incorrect! Always remember that the area of a square is side × side and not 2 × sides.
• When we represent the area, we should not forget to write its unit. The side of a square is one-dimensional and the area of a square is two-dimensional. Hence, the area of a square is always represented as square units. For example, a square with a side of 3 units will have an area of 3 × 3 = 9 square units.
## Area of a Square Examples
Example 2: The area of a square-shaped carrom board is 3600 cm2. What is the length of its side?
Solution:
Area of the square carrom board = 3600 cm2. We know that Area = side × side = side2. So, side = √Area = √3600 = 60 cm. Therefore, the side of the carrom board is 60 cm.
Example 3: Find the area of the square-shaped floor room which is made up of 100 square tiles of side 15 inches.
Solution:
Area of one tile = 15 inch × 15 inch= 225 square inches. We know that there are 100 tiles on the floor of the room. Thus, the area occupied by 100 tiles is the floor area = 100 × 225 square inches = 22500 square inches. Therefore, the area of the floor is 22500 square inches.
Example 4: Find the area of a square carpet whose diagonal is 4 feet.
Solution:
The area of a square when its diagonal is given is D2/2. Given, diagonal d = 4 ft. Area of the carpet = (4 × 4)/2 = 16/2 = 8 square feet. Therefore, the area of the carpet is 8 square feet.
## FAQs on Area of Square
### What is Area of Square in Geometry?
In geometry, the square is a shape with four equal sides. The area of a square is defined as the number of square units that make a complete square. It is calculated by using the area of square formula Area = s × s = s2 in square units.
### What is the Area of a Square Formula?
When the side of a square is known, the formula used to find the area of a square with side ‘s’: Area = s × s = s2. When the diagonal ‘d’ of the square is given, then the formula used to find the area is, Area = d2/2.
### How Do You Calculate the Area of a Square?
The area of a square is calculated with the help of the formula: Area = s × s, where, ‘s’ is one side of the square. Since the area of a square is a two-dimensional quantity, it is always expressed in square units. For example, if we want to calculate the area of a square with side 4 units, it will be: A = 4 × 4 = 16 unit2. Check now area of square calculator for quick calculations.
### What is the Perimeter and Area of Square Formulas?
The perimeter of a square is a sum of four sides of a square that is P = 4 × Sides. It is given in terms of m, cm, ft, inches.
The area of square = Area = s × s, where, ‘s’ is one side of the square. It is given in terms of m2, cm2, ft2, in2.
Check:
• Perimeter Formulas
• Volume Formulas
• Surface Area Formulas
• Measurement Formulas
### How to Find the Area of a Square From the Diagonal of a Square?
The area of a square can also be found if the diagonal of a square is given. The formula that is used in this case is: Area of a square using diagonals = Diagonal²/2. For example, the diagonal of a square is 6 units, the Area = 6²/2 = 36/2 = 18 square units.
### How to Find the Area of a Square From the Perimeter of the Square?
The area of a square can be calculated if the perimeter of the square is known. Since the perimeter of a square is: P = 4 × side, we can find the side of the square ‘s’ = Perimeter/4. After getting the side, the area of a square can be calculated with the formula: A = s × s. For example, if the perimeter of a square is 32 units, we will substitute this value in the formula: P = 4 × side. 32 = 4 × side. So, the side will be 8 units. Now, we can calculate the area of the square with side 8 units. Area = s × s = 8 × 8 = 64 square units.
### What are the Units of the Area of a Square?
Since the area of a square is a two-dimensional shape, it is always expressed in square units The common units of the area of a square are m2, inches2, cm2, foot2.
### What is the Area of a Square Inscribed In a Circle?
If a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. So, if the diameter of the circle is given, this value can be used as the diagonal of the square, and the area of the square can be calculated with the formula: Area of a square using diagonals = Diagonal²/2.
## Introduction to a Square
We all are familiar with the figure, square. It is a quadrilateral wherein all the four sides and angles of it are equal. All the four angles are 90 degrees each, that is, right angles. You can also consider square as a special case of a rectangle where you will find that the two adjacent sides are of equal length.In this article, we will mainly be focusing on the various square formulas such as its area, perimeter, length of the diagonals and examples.
Area of a Square = a²
The perimeter of a Square =4a
Diagonal of a Square=a√2
Where ‘a’ is the length of a side of the square.
Properties of a Square
• The lengths of all the four sides of a square are equal.
• The two diagonals bisect each other at right angles, that is, 90°
• The lengths of diagonals of a square are equal.
## Derivation of Square Formula
### Derivation of Area of a Square
To better our understanding of the concept, let us take a look at the derivation of the area of a square. Let us consider a square where the lengths of its side are ‘a’ units and diagonal is ‘d’ units respectively. As you all know that area of a square is the region which is enclosed within its boundary. As we have already mentioned, that a square is a special case of a rectangle that has its two adjacent sides being of equal length. Therefore, the area can be expressed as –
Area of a rectangle = Length × Breadth
• Area of square = Length x Breadth
• Area of square = a× a = = a²
### Derivation of the Perimeter of a Square
The perimeter of the square is the total length of its boundary. The boundary of a square is represented by the sum of the length of all sides. Hence, the perimeter is expressed as –
• Perimeter of a square = length of 4 sides
• Perimeter of a square = a+ a + a + a = 4a
### Derivation of the Lenght of a Diagonal of a Square
As you all know that the diagonal is a line that joins the two opposite sides in a polygon. Therefore, in order to calculate the diagonal length of a square, we use the Pythagoras Theorem. If you mark a diagonal in the square, then you will realize that the diagonal divides the square into two right-angled triangles. Now, the two adjacent sides of a square are equal in length. And the right-angled triangle works as an isosceles triangle with each of its sides being of length ‘a’ units.
Thus, we can apply the Pythagoras theorem on these triangles which have base and perpendicular of ‘a’ units and hypotenuse of ‘b’ units. So, according to the formula we have:
• d²=a²+a²
• d = √2
• d = a √2 units
### Solved Examples
Now, that we have some understanding about the concept and meaning of square formula, let us try some examples to deepen our understanding of the topic
Example 1 – The side of a square is 5 meters. What is the area of the square in m
Solution 1 – As we know that the sides of а square are of equal size. And the formula for the square’s area is a × a where a is the side of the square. Hence,
Area = 5m × 5m = 25 m²
Example 2 – The perimeter of a square is 24 cm. What is the area of the square is cm2
Solution 2 – Perimeter of a square is 4 × a, where a is the side of the square
4 × a =24
a = 24 ÷ 4 = 6
Area = a × a = 6 cm × 6 cm = 36 cm2
Example 3 – The side of a square is 5 cm. If its side is doubled, how many times is the area of the new square bigger than the area of the old square?
Solution 3 – The area of the first square is 5 cm × 5 cm = 25 cms. New length = 5cm + 5 cm = 10 cm
The area of the new square is 10 cm × 10 cm = 100 cm2. Hence, the area is 4 times bigger.
### Area of Square Formula Derivation
To better our understanding of the concept, let us take a look at the derivation of the area of Square formula. Let us consider a square as a rectangular object whose length is of a unit and breadth is of a unit. As we know the area of the rectangle is given by,
A = L × B
Where
A = l × b
A = a × b
= a × a = a² = a²
## Solved Examples on Area of Square Formula
Now that we have some clarity about the concept and meaning of the area of the square, let us try some examples to deepen our understanding of the subject.
Q: Find the area of a square plot of side 8 m.
Ans: As we already have a formula for calculating the area of a square. Let us substitute the values
A = a × a = a² = a²
A = 8²
= 64 sq m
Q: A square of 10 cm long is cut into tiny squares of 2 cm long. Calculate the number of tiny squares that can be created.
Ans: Since the length of the big square is 10 cm, hence its Area A is:,
A= a × a = a² = a²
A = 10² = 100 cm²
Now, since the length of tiny square is 2 cm, hence its Area is:
A = a × b
= 2 × 2
= 2 × 2 = 2²
= 4 sq cm.
Therefore, the number of squares that we can create are:
Example
Let us understand the formula much better with the help of an example.
Consider a square field having the sides of it as 23 ft. A person has to build a swimming pool over there and has to find how much area will the pool cover. Along with the pavement length needed to be done on the boundary of the pool. He is also curious to know what could be the farthest straight line length to swim in the pool. Find out the details that the person wanted.
The side given in the question is 23 ft of the pool.
So, first the area of the pool will be found with the help of formula,
Area (A)=a2
A=232=529 ft2
Now, the pavement length is the perimeter,
Therefore formula for perimeter is,
Perimeter (P)=4a
P=423=92 ft
At last the maximum straight line distance that the person could swim in a square shaped pool is the diagonal length of the pool.
Therefore, the diagonal length of the pool can be found by using formula,
d=a√2
d=23√2=32.52 ft
Hence, the area covered by the swimming pool is 529 ft2, the measure of the pavement distance needed to be done on boundary is 92 ft as well as the maximum distance that the person could swim in a straight line in his pool is 32.52 ft.
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# GLM
The generalized linear model (GLM) is a flexible generalization of ordinary least squares regression. The GLM generalizes linear regression by allowing the linear model to be related to the response variable (i.e. $$Y$$) via a link function (i.e. $$g(.)$$)and by allowing the magnitude of the variance of each measurement to be a function of its predicted value.
The GLM is described as follow: $Y = \mu + \epsilon$ And $E\left[Y\right]=\mu=g^{-1}(X\beta) = g^{-1}(\eta)$ Where:
• $$\epsilon$$ is the residuals or deviation from the mean
• $$g(.)$$ is the link function
• $$g^{-1}(.)$$ is the inverse-link function
• $$g^{-1}(.)$$ is the inverse link function
• $$X$$ is the independent variables or the exogenous factors
• $$\beta$$ is a parameter vector
• $$\eta$$ is the linear predictor: the quantity which incorporates the information about the independent variables into the model. $\eta=X\beta$
Remarks
1. Each outcome of the dependent variables,Y, is assumed to be generated from a particular distribution in the exponential family, a large range of probability distributions that includes the normal, binomial and Poisson distributions, among others.
2. The distribution mean of the $$Y$$ variable (i.e. $$\mu$$ depends solely on the independent variables, $$X$$. $E\left[Y\right]=\mu=g^{-1}(X\beta)$
3. The conditional variance of the dependent variable, Y, is constant: $V(Y\|{X\beta})=\phi \times V({X\beta})$ $$Where:$$
• $$V(.)$$ is the variance function.
• $$\phi$$ is the dispersion factor (constant value).
Normal Identity $$X\beta=\mu$$
Binomial Logit $$X\beta = \ln\frac{\mu}{1-\mu}$$
Poisson Log $$X\beta = \ln\mu$$
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https://www.esaral.com/q/the-amount-of-extension-in-an-elastic-string-varies-directly-as-the-weight-hung-on-it-70507
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The amount of extension in an elastic string varies directly as the weight hung on it.
Question:
The amount of extension in an elastic string varies directly as the weight hung on it. If a weight of 150 gm produces an extension of 2.9 cm, then what weight would produce an extension of 17.4 cm?
Solution:
Let x gm be the weight that would produce an extension of 17.4 cm.
Since the amount of extension in an elastic string and the weight hung on it are in direct variation, we have:
$\frac{150}{x}=\frac{2.9}{17.4}$
$\Rightarrow 17.4 \times 150=2.9 \times x$
$\Rightarrow x=\frac{17.4 \times 150}{2.9}$
$=\frac{2610}{2.9}$
$=900$
Thus, the required weight will be $900 \mathrm{gm}$.
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http://www.ask.com/question/48-inches-equals-how-many-feet
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48 Inches Equals How Many Feet?
4 feet, there are 12 inchs in one foot. Thus, 48 divided by 12 is 4, and that is how we arrive and the answer of 4 feet. Just as a bonus for you, there are 3 feet in one yard, so keep that in mind as well.
48 inches is equal to 4 feet.
Convert to
Q&A Related to "48 Inches Equals How Many Feet?"
48 inches is equal to four feet. http://www.chacha.com/question/what-is-48-inches-e...
There are 12 inches in one foot. Therefore, 48 inches is equal to 48/12 = 4 feet. http://wiki.answers.com/Q/How_many_feet_equal_48_i...
There are 12 inches in a foot, so to find how many feet there are in 48 inches, we divide 48 by 12 which equals 4. This means that 48 inches equals 4 feet. http://www.webanswers.com/misc/how-many-feet-equal...
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https://online.stat.psu.edu/stat415/book/export/html/822
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# 13.2 - The ANOVA Table
13.2 - The ANOVA Table
For the sake of concreteness here, let's recall one of the analysis of variance tables from the previous page:
One-way Analysis of Variance
Source DF SS MS F P
Factor 2 2510.5 1255.3 93.44 0.000
Error 12 161.2 13.4
Total 14 2671.7
In working to digest what is all contained in an ANOVA table, let's start with the column headings:
1. Source means "the source of the variation in the data." As we'll soon see, the possible choices for a one-factor study, such as the learning study, are Factor, Error, and Total. The factor is the characteristic that defines the populations being compared. In the tire study, the factor is the brand of tire. In the learning study, the factor is the learning method.
2. DF means "the degrees of freedom in the source."
3. SS means "the sum of squares due to the source."
4. MS means "the mean sum of squares due to the source."
5. F means "the F-statistic."
6. P means "the P-value."
Now, let's consider the row headings:
1. Factor means "the variability due to the factor of interest." In the tire example on the previous page, the factor was the brand of the tire. In the learning example on the previous page, the factor was the method of learning.
Sometimes, the factor is a treatment, and therefore the row heading is instead labeled as Treatment. And, sometimes the row heading is labeled as Between to make it clear that the row concerns the variation between the groups.
2. Error means "the variability within the groups" or "unexplained random error." Sometimes, the row heading is labeled as Within to make it clear that the row concerns the variation within the groups.
3. Total means "the total variation in the data from the grand mean" (that is, ignoring the factor of interest).
With the column headings and row headings now defined, let's take a look at the individual entries inside a general one-factor ANOVA table:
Hover over the lightbulb for further explanation.
One-way Analysis of Variance
Source DF SS MS P
Factor m-1 SS (Between) MSB MSB/MSE 0.000
Error n-m SS (Error) MSE
Total n-1 SS (Total)
Yikes, that looks overwhelming! Let's work our way through it entry by entry to see if we can make it all clear. Let's start with the degrees of freedom (DF) column:
1. If there are n total data points collected, then there are n−1 total degrees of freedom.
2. If there are m groups being compared, then there are m−1 degrees of freedom associated with the factor of interest.
3. If there are n total data points collected and m groups being compared, then there are nm error degrees of freedom.
Now, the sums of squares (SS) column:
1. As we'll soon formalize below, SS(Between) is the sum of squares between the group means and the grand mean. As the name suggests, it quantifies the variability between the groups of interest.
2. Again, as we'll formalize below, SS(Error) is the sum of squares between the data and the group means. It quantifies the variability within the groups of interest.
3. SS(Total) is the sum of squares between the n data points and the grand mean. As the name suggests, it quantifies the total variability in the observed data. We'll soon see that the total sum of squares, SS(Total), can be obtained by adding the between sum of squares, SS(Between), to the error sum of squares, SS(Error). That is:
SS(Total) = SS(Between) + SS(Error)
The mean squares (MS) column, as the name suggests, contains the "average" sum of squares for the Factor and the Error:
1. The Mean Sum of Squares between the groups, denoted MSB, is calculated by dividing the Sum of Squares between the groups by the between group degrees of freedom. That is, MSB = SS(Between)/(m−1).
2. The Error Mean Sum of Squares, denoted MSE, is calculated by dividing the Sum of Squares within the groups by the error degrees of freedom. That is, MSE = SS(Error)/(nm).
The F column, not surprisingly, contains the F-statistic. Because we want to compare the "average" variability between the groups to the "average" variability within the groups, we take the ratio of the Between Mean Sum of Squares to the Error Mean Sum of Squares. That is, the F-statistic is calculated as F = MSB/MSE.
When, on the next page, we delve into the theory behind the analysis of variance method, we'll see that the F-statistic follows an F-distribution with m−1 numerator degrees of freedom and nm denominator degrees of freedom. Therefore, we'll calculate the P-value, as it appears in the column labeled P, by comparing the F-statistic to an F-distribution with m−1 numerator degrees of freedom and nm denominator degrees of freedom.
Now, having defined the individual entries of a general ANOVA table, let's revisit and, in the process, dissect the ANOVA table for the first learning study on the previous page, in which n = 15 students were subjected to one of m = 3 methods of learning:
Hover over the lightbulb for further explanation.
One-way Analysis of Variance
Source DF SS MS F P
Factor 2 2510.5 1255.3 93.44 0.000
Error 12 161.2 13.4
Total 14 2671.7
1. Because n = 15, there are n−1 = 15−1 = 14 total degrees of freedom.
2. Because m = 3, there are m−1 = 3−1 = 2 degrees of freedom associated with the factor.
3. The degrees of freedom add up, so we can get the error degrees of freedom by subtracting the degrees of freedom associated with the factor from the total degrees of freedom. That is, the error degrees of freedom is 14−2 = 12. Alternatively, we can calculate the error degrees of freedom directly from nm = 15−3=12.
4. We'll learn how to calculate the sum of squares in a minute. For now, take note that the total sum of squares, SS(Total), can be obtained by adding the between sum of squares, SS(Between), to the error sum of squares, SS(Error). That is:
2671.7 = 2510.5 + 161.2
5. MSB is SS(Between) divided by the between group degrees of freedom. That is, 1255.3 = 2510.5 ÷ 2.
6. MSE is SS(Error) divided by the error degrees of freedom. That is, 13.4 = 161.2 ÷ 12.
7. The F-statistic is the ratio of MSB to MSE. That is, F = 1255.3 ÷ 13.4 = 93.44.
8. The P-value is P(F(2,12) ≥ 93.44) < 0.001.
Okay, we slowly, but surely, keep on adding bit by bit to our knowledge of an analysis of variance table. Let's now work a bit on the sums of squares.
## The Sums of Squares
In essence, we now know that we want to break down the TOTAL variation in the data into two components:
1. a component that is due to the TREATMENT (or FACTOR), and
2. a component that is due to just RANDOM ERROR.
Let's see what kind of formulas we can come up with for quantifying these components. But first, as always, we need to define some notation. Let's represent our data, the group means, and the grand mean as follows:
Group Data Means
1 $$X_{11}$$ $$X_{12}$$ . . . $$X_{1_{n_1}}$$ $$\bar{{X}}_{1.}$$
2 $$X_{21}$$ $$X_{22}$$ . . . $$X_{2_{n_2}}$$ $$\bar{{X}}_{2.}$$
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$$m$$ $$X_{m1}$$ $$X_{m2}$$ . . . $$X_{m_{n_m}}$$ $$\bar{{X}}_{m.}$$
Grand Mean $$\bar{{X}}_{..}$$
That is, we'll let:
1. m denotes the number of groups being compared
2. $$X_{ij}$$ denote the $$j_{th}$$ observation in the $$i_{th}$$ group, where $$i = 1, 2, \dots , m$$ and $$j = 1, 2, \dots, n_i$$. The important thing to note here... note that j goes from 1 to $$n_i$$, not to $$n$$. That is, the number of the data points in a group depends on the group i. That means that the number of data points in each group need not be the same. We could have 5 measurements in one group, and 6 measurements in another.
3. $$\bar{X}_{i.}=\dfrac{1}{n_i}\sum\limits_{j=1}^{n_i} X_{ij}$$ denote the sample mean of the observed data for group i, where $$i = 1, 2, \dots , m$$
4. $$\bar{X}_{..}=\dfrac{1}{n}\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n_i} X_{ij}$$ denote the grand mean of all n data observed data points
Okay, with the notation now defined, let's first consider the total sum of squares, which we'll denote here as SS(TO). Because we want the total sum of squares to quantify the variation in the data regardless of its source, it makes sense that SS(TO) would be the sum of the squared distances of the observations $$X_{ij}$$ to the grand mean $$\bar{X}_{..}$$. That is:
$$SS(TO)=\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n_i} (X_{ij}-\bar{X}_{..})^2$$
With just a little bit of algebraic work, the total sum of squares can be alternatively calculated as:
$$SS(TO)=\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n_i} X^2_{ij}-n\bar{X}_{..}^2$$
Can you do the algebra?
Now, let's consider the treatment sum of squares, which we'll denote SS(T). Because we want the treatment sum of squares to quantify the variation between the treatment groups, it makes sense that SS(T) would be the sum of the squared distances of the treatment means $$\bar{X}_{i.}$$ to the grand mean $$\bar{X}_{..}$$. That is:
$$SS(T)=\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n_i} (\bar{X}_{i.}-\bar{X}_{..})^2$$
Again, with just a little bit of algebraic work, the treatment sum of squares can be alternatively calculated as:
$$SS(T)=\sum\limits_{i=1}^{m}n_i\bar{X}^2_{i.}-n\bar{X}_{..}^2$$
Can you do the algebra?
Finally, let's consider the error sum of squares, which we'll denote SS(E). Because we want the error sum of squares to quantify the variation in the data, not otherwise explained by the treatment, it makes sense that SS(E) would be the sum of the squared distances of the observations $$X_{ij}$$ to the treatment means $$\bar{X}_{i.}$$. That is:
$$SS(E)=\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n_i} (X_{ij}-\bar{X}_{i.})^2$$
As we'll see in just one short minute why the easiest way to calculate the error sum of squares is by subtracting the treatment sum of squares from the total sum of squares. That is:
$$SS(E)=SS(TO)-SS(T)$$
Okay, now, do you remember that part about wanting to break down the total variation SS(TO) into a component due to the treatment SS(T) and a component due to random error SS(E)? Well, some simple algebra leads us to this:
$$SS(TO)=SS(T)+SS(E)$$
and hence why the simple way of calculating the error of the sum of squares. At any rate, here's the simple algebra:
Proof
Well, okay, so the proof does involve a little trick of adding 0 in a special way to the total sum of squares:
$$SS(TO) = \sum\limits_{i=1}^{m} \sum\limits_{i=j}^{n_{i}}((X_{ij}-\color{red}\overbrace{\color{black}\bar{X}_{i_\cdot})+(\bar{X}_{i_\cdot}}^{\text{Add to 0}}\color{black}-\bar{X}_{..}))^{2}$$
Then, squaring the term in parentheses, as well as distributing the summation signs, we get:
$$SS(TO)=\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n_i} (X_{ij}-\bar{X}_{i.})^2+2\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n_i} (X_{ij}-\bar{X}_{i.})(\bar{X}_{i.}-\bar{X}_{..})+\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n_i} (\bar{X}_{i.}-\bar{X}_{..})^2$$
Now, it's just a matter of recognizing each of the terms:
$$S S(T O)= \color{red}\overbrace{\color{black}\sum\limits_{i=1}^{m} \sum\limits_{j=1}^{n_{i}}\left(X_{i j}-\bar{X}_{i \cdot}\right)^{2}}^{\text{SSE}} \color{black}+2 \color{red}\overbrace{\color{black}\sum\limits_{i=1}^{m} \sum\limits_{j=1}^{n_{i}}\left(X_{i j}-\bar{X}_{i \cdot}\right)\left(\bar{X}_{i \cdot}-\bar{X}_{. .}\right)}^{\text{O}} \color{black}+ \color{red}\overbrace{\color{black}\left(\sum\limits_{i=1}^{m} \sum\limits_{j=1}^{n_{i}}\left(\bar{X}_{i \cdot}-\bar{X}_{* . *}\right)^{2}\right.}^{\text{SST}}$$
That is, we've shown that:
$$SS(TO)=SS(T)+SS(E)$$
as was to be proved.
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Question Video: Using Trigonometry to Find Lengths in Isosceles Triangles | Nagwa Question Video: Using Trigonometry to Find Lengths in Isosceles Triangles | Nagwa
# Question Video: Using Trigonometry to Find Lengths in Isosceles Triangles Mathematics
π΄π΅πΆ is an isosceles triangle where π΄π΅ = π΄πΆ = 10 cm and πβ πΆ = 52Β°20β²21β³. Find the length of π΅πΆ giving the answer to one decimal place.
05:11
### Video Transcript
π΄π΅πΆ is an isosceles triangle where π΄π΅ is equal to π΄πΆ which is equal to 10 centimeters and the angle πΆ is equal to 52 degrees 20 minutes and 21 seconds. Find the length of π΅πΆ giving the answer to one decimal place.
In any question like this, it is always worth trying to draw a diagram first. We were told in the question that triangle π΄π΅πΆ is isosceles. This means that two of the sides have equal length. The length of π΄π΅ and π΄πΆ are both equal to 10 centimeters. We were also told that angle πΆ was equal to 52 degrees 20 minutes and 21 seconds.
Our first step is to convert this angle into just degrees. One degree is equal to 60 minutes. This means that we can change 20 minutes into degrees by dividing 20 by 60. 20 minutes is equal to 0.333 and so on degrees, or 0.3 recurring degrees. One degree is also equal to 3600 seconds, as there are 60 seconds in a minute. 60 multiplied by 60 is equal to 3600. We can, therefore, convert 21 seconds into degrees by dividing 21 by 3600. This is equal to 0.00583 and so on degrees.
We can now add these two values together to calculate 20 minutes and 21 seconds in degrees. This is equal to 0.33916 and so on degrees. We can, therefore, say that angle πΆ is equal to 52.34 degrees to two decimal places. Whilst weβve rounded our answer here, it is important for accuracy to use the full answer on our calculator display in any further calculations.
Our next step is to create two identical right-angle triangles by drawing a vertical line from π΄ to the line π΅πΆ. We will call the point where this line meets π΅πΆ, π·. We were asked to calculate the length of π΅πΆ. Well, the length of π΅πΆ will be double the length of π·πΆ, which we have labelled π₯.
We can use right angle trigonometry, or SOHCAHTOA, to calculate the length π₯. The length π΄πΆ is the hypotenuse of the right-angled triangle, as it is the longest side. The length π΄π· is the opposite, as it is opposite the 52.34-degree angle. Finally, the length πΆπ·, labelled π₯, is the adjacent, as it is adjacent, or next to, the right angle and the 52.34-degree angle. We know the length of the hypotenuse. And we want to calculate the length of the adjacent. Therefore, we will use the cosine ratio.
This states that cos of π is equal to the adjacent divided by the hypotenuse. Substituting in our values gives us cos of 52.34 is equal to π₯ divided by 10. Multiplying both sides of this equation by 10 gives us π₯ is equal to 10 multiplied by cos of 52.34. Typing this into the calculator gives us a value of π₯ of 6.109 and so on. We can, therefore, say that π·πΆ is equal to 6.109 and so on centimeters.
As previously mentioned, π΅πΆ is double the length of π·πΆ. Multiplying 6.109 and so on by two gives us 12.219 and so on. We were asked to give our answer to one decimal place. This means our answer needs to have one number after the decimal point. The deciding number is the one in the hundredths column. As this is less than five, we will round down. We can, therefore, say that the length π΅πΆ in the isosceles triangle π΄π΅πΆ is 12.2 centimeters.
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# Math
Please illustrate how to rotate triangle A(2,0), B(7,0),C(7,2) 90 degrees.
1. 👍 0
2. 👎 0
3. 👁 29
1. About what point? Clockwise or counterclockwise?
Plot the first points on a graph and you will see many possibilities. The fact that this is a right triangle aligned with axis makes it simple. One possibility would be (7,0), (9,0) and (7,5) (For clockwise rotation about point B).
1. 👍 0
2. 👎 0
posted by drwls
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meebuigenzg
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Determine the polynomials p(x) satisfying $x\cdot p\left(x-1\right)=\left(x-26\right)\cdot p\left(x\right)$
Haylie Campbell
It is easy to see that the polynomial p has roots $0,1,...,25$ and therefore
$p\left(x\right)=x\left(x-1\right)...\left(x-25\right)g\left(x\right)$
where $g\left(x\right)$ is also polynomial.
Replacing the original equality we obtain that $g\left(x-1\right)=g\left(x\right)$. Now noting $g\left(x\right)-g\left(0\right)=h\left(x\right)$ polinomial with degree n, we find that h is roots $0,1,...,n$ ie having more root than degree n. Hence $g\left(x\right)=c$ constant. Finally
$p\left(x\right)=cx\left(x-1\right)...\left(x-25\right).$
Do you have a similar question?
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# Pentagon shape: area, perimeter, and lines of symmetry
An outstanding figure in geometry is the pentagon. It is usual that, for various reasons, we need to find the area of the pentagon.
Obtaining the area and perimeter of a pentagon is easy. First, however, we must know some basic characteristics, such as how many sides a pentagon has (five) or how many diagonals a pentagon have (five).
One of the most recognizable buildings in the world with this shape is The Pentagon. It is the headquarters of the United States Department of Defense. The Pentagon is a five-sided building with a perimeter of 2,611 feet (796 meters).
## What is a pentagon shape?
A pentagon shape is a polygon with five sides and five vertices.
A polygon is a two-dimensional geometric figure with a finite number of non-collinear consecutive line segments connected by their two ends, forming a closed space.
These five-sided-polygons can be classified into several categories, depending on their shape:
• Regular pentagon: It is the one that has all its sides of equal length, and all angles are equal.
• Irregular pentagon: It does not have all its sides and angles equal.
• Convex pentagon: All interior angles are less than 180°.
• Concave pentagon: At least one of the interior angles is greater than 180°.
## How many diagonals does a pentagon have?
A diagonal is a line or segment that joins one vertex to another within the same shape. The number of diagonals in a pentagon shape is five.
## How many lines of symmetry does a pentagon have
A pentagon has five axes of symmetry in a regular polygon.
In geometry, axes of symmetry are imaginary lines dividing the pentagon shape into equal parts so that each polygon has a certain number of axes of symmetry. Some geometric figures have only one axis, while others have two or even infinity, as with the circle or the oval.
The pentagon lines of symmetry are drawn from each middle point of its edges to the furthest vertex. As the number of sides of this regular polygon is equal to five, a pentagon has five symmetry lines.
## What are the pentagon properties?
Below we list the most significant characteristics of this type of flat shape:
• A pentagon is made up of five sides and five angles. The different types of pentagons are classified based on their sides and angles.
• All five interior angles add up to 540º.
• If it is regular, all interior angles measure 108°, and therefore all exterior angles measure 72°.
• The central angle of the regular pentagon 360° / 5 = 72º. The central angle is formed by two straight lines that join the two ends of a side to the center of the figure.
• A pentagon shape is composed of five diagonals.
• From each vertex, two diagonals can be drawn to another vertex.
• When the pentagon is regular, the diagonals at each vertex form three 36° angles and divide the regular pentagon into three isosceles triangles.
• The perimeter of a regular or irregular pentagon is the sum of the lengths of all its sides. If it is a regular polygon, it is the length of one side multiplied by five.
## How can you find the area of a pentagon?
There are several ways to obtain the area of a regular pentagon. An area pentagon formula is:
A = (5/2)·c·a
Where "c" refers to the length of a side, "a" refers to the size of the apothem of the pentagon.
If we know the perimeter: the formula becomes A = p·a / 2; where "p" is the perimeter.
If we know the apothem, we can use the formula:
A = (5 · (a2)) / (4 · tan(36))
In the formula, "a" refers to the apothem of the shape.
### Calculating the area of a triangle
The area of a pentagon can also be calculated by dividing it into triangles:
If we draw two diagonals, from vertex to vertex, without intersecting, three isosceles triangles are formed. The sum of the area of these triangles is equal to the area of the pentagon.
If we join each vertex to the center of the geometric figure, we will have five equilateral triangles left. We will also obtain the total surface with the sum of the surface of these five triangles.
Autor:
Data de publicació: October 1, 2022
Última revisió: October 1, 2022
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# 2007 AMC 12B Problems/Problem 3
## Problem
The point $O$ is the center of the circle circumscribed about triangle $ABC$, with $\angle BOC = 120^{\circ}$ and $\angle AOB = 140^{\circ}$, as shown. What is the degree measure of $\angle ABC$?
$\mathrm {(A)} 35 \qquad \mathrm {(B)} 40 \qquad \mathrm {(C)} 45 \qquad \mathrm {(D)} 50 \qquad \mathrm {(E)} 60$
## Solution
Since triangles $ABO$ and $BOC$ are isosceles, $\angle ABO=20^o$ and $\angle OBC=30^o$. Therefore, $\angle ABC=50^o$, or $\mathim{(D)}$ (Error compiling LaTeX. ! Undefined control sequence.).
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# finding position in triangle matrix
I have symmetric matrix stored in column major way. But I am storing only lower part of matrix to save space.
so my matrix looks like this:
1
2 6
3 7 10
4 8 11 13
5 9 12 14 15
I have to write code to find position of element in this matrix depending on index i(row),j(col) of that matrix.
I have written sth like this:
pos = (n*j) - j*j/2 + (i - j);
pos - position of my element in matrix - a[pos] n - size of matrix
Unfortunately It doesn't find good position always. I write program to test it and it prints:
1
2 6
3 7 11
4 8 12 14
5 9 13 15 17
6 10 14 16 18 18
I know it happens like that because when we divide j*j/2 we get int/int. But I have no idea what to do to make it work correctly.
-
if it is homework add homework tag – Pradeep Aug 24 '11 at 12:00
you appear to be printing out more rows than there are! Could you show code. – David Heffernan Aug 24 '11 at 12:04
@Pelsono: are you using 1D array to store the matrix? And, what's pos - position of my element in matrix - a[pos] n - size of matrix? – Eric Z Aug 24 '11 at 12:11
What are the i,j values for the entry "2"? – Beta Aug 24 '11 at 12:27
Let's examine the calculation one column at a time, assuming:
i <= j
1 <= i <= n
1 <= j <= n:
then:
i=1, pos=j
i=2, pos=n+j-1
i=3, pos=n+n-1+j-2
i=4, pos=n+n-1+n-2+j-3
etc...
we can deduce from this a general formula:
p=n*(i-1)+j-(i-1)*i/2
which can be tested using a simple bit a C#:
using System;
using System.IO;
namespace Stream
{
class Program
{
static void Main (string [] args)
{
for (int j = 1 ; j <= 5 ; ++j)
{
for (int i = 1 ; i <= 5 ; ++i)
{
Console.Write (GetIndex (i, j).ToString ("00 "));
}
Console.WriteLine ("");
}
}
static int GetIndex (int in_i, int in_j)
{
int
n = 5,
i = Math.Min (in_i, in_j),
j = Math.Max (in_i, in_j);
return n * (i - 1) + j - (i - 1) * i / 2;
}
}
}
-
it's C++ here ;) – Eric Z Aug 24 '11 at 12:34
I left it as an exercise for the OP to port it C++ - don't want to make life too easy now do we? – Skizz Aug 24 '11 at 12:40
Yup, just reminds me of last time when my answer gets downvoted here as its written in C, not C++;) – Eric Z Aug 24 '11 at 12:46
First of all, it would be much more simple to store the matrix by line (first line at the beginning of the array) to compute the indexes.
Your formula seems wrong, I would say it is (for j > 0):
n + (n-1) + ... + (n-j+1) + (i - j)
= (( sum [0 <= k <= j-1] (n-j+1) + k )) + (i - j)
= (n-j+1) * j + (( sum [0 <= k <= j-1] k )) + (i - j)
= (n-j+1) * j + (j * (j+1)) / 2 + (i - j)
if your indices (i,j) start at 0 and you store the first column (with n elements) at the first position in the array.
Edit: modified for indexes starting at 0.
-
#include <iostream>
using namespace std;
int main()
{
int i=0,k=1,j=0;
int a[5][5];
int n=0;
for (j=0; j<5;j++)
{
for (i=n; i<5;i++)
{
a[i][j]=k++;
cout<<i<<" "<<j<<" "<<(a[i][j])<<'\n';
}
n++;
}
return 0;
}
-
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# general mean value theorem
Can anyone give me the intuitive explanation of the general mean value theorem stated in my notes as under:
Let $f:U\rightarrow \mathbb R$ and $U\subseteq \mathbb R^n$ and let $f$ is differentiable at $K\subseteq U$ which is convex.. If $\gamma(t)=(1-t)a+t(b)$ is a line segment joining $a,b$ and $t\in[0,1]$ Then there is a point $c$ on the line segment s.t.
$$f(b)-f(a)=\nabla f(c)(b-a).$$
I'm facing problem how to interpret this theorem.How is it similar to mean value theorem in one-dimension. Please help....
• I don't think it would say "then $f$ is differentiable". Maybe it said "and let $f$ be differentiable". – Michael Hardy Oct 4 '14 at 18:01
As you stated it, it is exactly the one dimensional theorem applied to the function $g:[0,1]\to\mathbb{R}$ given by $g(t)=f(\gamma(t))$. Indeed, $$f(b)-f(a)=f(\gamma(1))-f(\gamma(0))=g(1)-g(0)=g'(s)$$ for some $s\in(0,1)$, but $$g'(t)=\nabla f(\gamma(t))\cdot \gamma'(t)=\nabla f(\gamma(t))\cdot(b-a)$$ where $\cdot$ is the scalar product. So $$f(b)-f(a)=g'(s)=\nabla f(\gamma(s))\cdot (b-a)$$ if you set $c=\gamma(s)$, you have your formula.
If you set $\hat{e}=\dfrac{b-a}{\|b-a\|}$, that is the unit vector pointing from $a$ to $b$, you can rewrite the statement as $$\dfrac{f(b)-f(a)}{\|b-a\|}=\nabla f( c)\cdot \hat{e}$$ which is maybe closer to the one-dimensional one and could help you to grasp the meaning of this.
• You have $\nabla f(\gamma(t))$ where you presumably meant $\nabla (f\circ\gamma)(t)$. A reasonable reader could think you meant $(\nabla f)(\gamma(t))$, which would not be correct. – Michael Hardy Oct 4 '14 at 18:04
• Uhm, no, I think I really mean $\nabla f$ calculated at the point $\gamma(t)$ … I already have a derivative of $\gamma$ outside. Moreover $f\circ \gamma$ is a real function of one real variable, so it does not really make any sense to write its gradient instead of simply writing its derivative… or am I missing something? – wisefool Oct 4 '14 at 22:10
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# Ravens
The tale of the Seven Ravens were seven brothers, each of whom was born exactly 2.5 years after the previous one. When the eldest of the brothers was 2-times older than the youngest, mother all curse. How old was seven ravens brothers when their mother cursed?
k1 (youngest) = 15
k2 = 17.5
k3 = 20
k4 = 22.5
k5 = 25
k6 = 27.5
k7 (oldest) = 30
### Step-by-step explanation:
${k}_{2}={k}_{1}+2.5=17.5$
${k}_{3}={k}_{2}+2.5=20$
${k}_{4}={k}_{3}+2.5=22.5$
${k}_{5}={k}_{4}+2.5=25$
${k}_{6}={k}_{5}+2.5=27.5$
${k}_{7}={k}_{6}+2.5=30$
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https://www.thestudentroom.co.uk/showthread.php?t=2656796
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# Rms voltageWatch
Announcements
Thread starter 5 years ago
#1
I know how to find it, but what exactly is it?
Why is it useful?
How does it apply to heating effect?
Thanks!
0
reply
5 years ago
#2
From my (basic) understanding, it is effectively the average voltage value when you've got an alternating voltage. An alternating voltage is one which goes from positive to negative again and again. Google "sine wave" if you don't know what I'm talking about. You can't just calculate the typical "average" value of this voltage wave, because that would be zero, so rms is a clever mathematical way of getting around that so you get a true measure of the average voltage.
1
reply
5 years ago
#3
(Original post by mintsponge)
From my (basic) understanding, it is effectively the average voltage value when you've got an alternating voltage. An alternating voltage is one which goes from positive to negative again and again. Google "sine wave" if you don't know what I'm talking about. You can't just calculate the typical "average" value of this voltage wave, because that would be zero, so rms is a clever mathematical way of getting around that so you get a true measure of the average voltage.
All good advice.
Just to elaborate a little more - what you're interested in is power (which tells you how much energy you're converting from electrical to heat form in a particular component over time). This energy will be dissipated regardless of which direction your current is moving in!.
In most circumstances, the voltage sine wave is useful because it tells us two things - the amplitude of our voltage, and its direction (backwards or forwards aka positive or negative) for every moment in time. But as discussed above, we don't care whether it's going backwards or forwards, only what amplitude it is. If you've done any statistics, you've probably learned how to calculate the standard deviation - we had the same problem there in that we only want to know how far a particular value is from the mean, and NOT whether it is above or below the mean. We calculate this by taking the SQUARE, and then taking the ROOT of the square. Sounds familiar now, right? It's the same idea
Hope that helps and good luck.
0
reply
5 years ago
#4
(Original post by Zenarthra)
I know how to find it, but what exactly is it?
Why is it useful?
How does it apply to heating effect?
Thanks!
The other posters have explained it well.
It's worth adding that rms voltage and current (both calculated the same way) are essentially a way of identifying the AC equivalent of a DC source that produces the same rate of conversion of electrical energy in a circuit.
So if you have a lamp in a circuit running on DC with a current, say, 3A and you swap this over to an AC supply resulting in the lamp having the same brightness (same power) then the AC supply was 3A rms.
An AC supply of Vrms = 5V and Irms = 2A supplies power at the same rate (10W) as a DC supply of V = 5V and I = 2A
In practice it's this aspect of rms values that is often important.
0
reply
5 years ago
#5
RMS of any function is basically the square root of the sum of all the values squared. i.e. sqrt( value1^2+vlue2^2+value3^2...) for a sine wave, this will always work out at the maximum height (peak amplitude) of the wave divided by root 2.
Wikipedia explains it quite well, albeit with Wiki's typical treatment of any maths subject by making it seem more complicated than it really is...
http://en.wikipedia.org/wiki/Root_mean_square
0
reply
Thread starter 5 years ago
#6
Thanks guys!
0
reply
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