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https://www.spartanpoker.com/seep-card-game
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## What is Seep Card Game?
Seep, sometimes known as Sip, Sweep, Siv, or Shiv is an online casino fishing game. Northern India's Seep is a card game with a 100-point total value. In Punjab, where the game is immensely popular both in India and Pakistan, another, presumably older variant is played, in which the total card value is 30 points.
A typical 52-card deck without a joker is used to play this game. This game can be played with two to four opponents at once.
The sweep card game's math is capturing cars with a point value, to bring the total to a point value of 100. The total number of spades cards has a point value equal to their capture value, ranging from 13 for the king to 1 for the ace.
Every ace is worth one point, whereas diamonds are worth six. Apart from that, only 17 spade cards, 3 aces of other suits, and ten diamonds have point values, while the rest of the cards have none. If you are new to sweep card games and are unsure how to play them, this article will assist you in learning how to play and win the game.
## How Do you play Seep Card Game?
The four-player version of Seep, as described below, is played in North India.
1. ### Objective:
Seep's objective is to acquire or capture valuable cards in the gaming table's arrangement (or the floor). The game is over when one team has a 100-point lead over the other. This is known as a bazzi. The teams can determine how many games or bazzis they want to play before the game begins.
1. ### How to Capture in a Seep Card Game
To capture the cards, play one card from your hand and put up 1+ cards from the same group with a capture value equal to the card in your hand. As a result, using the hand card, you will be able to grab cards of equal rank from the table.
### Points of Capture:
• A: 1
• 2-10: Face Value
• J: 11
• Q: 12
• K: 13
A player can create piles or houses out of the cards while capturing them. Only the houses as a whole were caught. Because the remaining cards on the floor are not in-house, they cannot be referred to as loose cards.
1. ### The Worth of the Captured Cards is Added Up at the End of the Game:
• Spades cards have the same point value as their capture value.
• Aces in the other suits are likewise worth one point.
• The value of the Ten of Diamonds is 6 points.
• On the deck, there is a total of 100 points.
1. ### Players have the Following Choices to Play:
• Construct a home.
• Capture the house.
• Make a pukka (indestructible) home.
• Take one or more cards from the deck.
• Toss a stray card.
• Choose a home.
1. ### Basic Moves During a Turn:
Constructing or extending a home The card that is utilized in the game either builds a new house or adds to an existing one.
Taking pictures of cards and houses. If the card being played has the same capture value as a house or any number of other cards on the table, it is possible to capture all of them in one play. Captured cards should be put in front of one member and shared between partners.
I'm tossing a stray card down. Cards that cannot be captured by other cards or incorporated into a house remain on the floor and are known as loose cards.
Pukka Home (Cemented or Unbreakable): If you add another card with the same equal value to a delicate house of 9 to 12, you can create an unbreakable house. For instance, in a house of 11, you might add a jack or two extra cards. The unbreakable house of 11 cannot be broken down into the houses of 12 and 13.
A House (Uncemented): has a pile of cards equal to the capture value when summed up.
A Cemented House: has more than 1 card or set of cards equal to the capture value. For example, a K cemented house could consist of the following values:
• 3, 10.
• 5, 4, 4.
• K.
• A, 6, 2, 2.
If a player adds a card to a house that enhances its capture value, it can be broken. The card must be drawn from the player's hand rather than from the ground. Cemented houses, on the other hand, cannot be broken.
Multiple dwellings with the same capture value on the same floor cannot exist at the same time; they must be bonded together. Loose cards having the same capture value as a house must be consolidated into the house automatically. If the home already exists, the loose card can either capture it or add it to it.
Seep, Sweep Bajji: In this game, players take turns playing one game at a time until the game is completed and a winner is determined. There's a concept called baazi in which you keep playing until the difference is 100 points or greater. The winner is determined when the difference between the two players is greater than 100.
The Concept of Chips: You can play the game and win chips; the maximum wager amount is 100 chips. Every team places 100 chips in the pot at the start of the game. One of the teams loses the chips at the end of the game, and the winning team gets the chips. For example, if the winning team receives 65 points and the losing team receives 35, the difference is 30 points.
### Seep Card Game Rules
Rules are already so boring, so instead of wasting time reading another paragraph, let's get straight to it:
• For each hand, all cards are dealt face-up, regardless of suit.
• Every spade is worth its face value.
• Deal four face-down cards to each player in a counter-clockwise direction.
• If the starting player has more than one card worth nine points or more, he can pick one. The dealer, on the other hand, must re-deal if he has no cards valued at nine or more.
• Player A can place one of the four cards on the table with a card from his hand.
• Each player tries to create "houses" of cards or take a trick from the start.
• Unless they are cemented, houses can be picked up by any player once they have been created.
• Houses can also be elevated by stacking smaller cards on top of them.
### How Can I Play Seep on My Phone?
You can download Seep from the App Store and enjoy it on your iPhone or computer. Bet before you get into the game, read this article carefully to ensure that you are prepared for the most fun online seep card game.
### Seep Card Game
The goal of poker is to win the entire pot, which is made up of bets placed by poker players during the hand.
The goal of the game is to collect cards and score points!
Standard 52-card decks are used in poker games, but players can opt to play versions that incorporate Jokers.
The players go through the cards they've been dealt and, if possible, "bid for a house" based on those four cards.
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# How do you expand log _7 (a/b)?
Jul 16, 2016
${\log}_{7} \left(\frac{a}{b}\right) = {\log}_{7} a - {\log}_{7} b$
#### Explanation:
If the numbers are being divided, then their logs are being subtracted.
${\log}_{7} \left(\frac{a}{b}\right) = {\log}_{7} a - {\log}_{7} b$
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https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-4-an-introduction-to-functions-4-4-graphing-a-function-rule-standardized-test-prep-page-259/44
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## Algebra 1
Given the Ratios $\frac{12}{b} = \frac{36}{51}$ Cross Multiply $12\times51 = b\times36$ Simplify left side by multiplying 12 and 51 $612 = 36b$ Divide both sides by 36 $\frac{612}{36} = \frac{36b}{36}$ b= 17 Therefore the solution is 17.
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https://gamedev.stackexchange.com/questions/21747/how-to-continuously-find-all-entities-within-a-radius-efficiently?noredirect=1
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# How to continuously find all entities within a radius efficiently?
I have a very large number of entities (units). On each step, each unit needs to know the positions of all units near it (distance is less then given constant R). All units move continuously. This is in 3D.
On average, there will be 1% of the total unit count near any other given unit with the given constraints.
How can I do this efficiently, without bruteforcing?
Use one of the common space partitioning algorithms, such as a Quadtree, Octree, BSP tree, or even a simple Grid System. Each has their own pros and cons for each specific scenario. You may read more about them in these books.
Generally (or so I've heard, I'm not too familiar with the reasoning behind this) a Quadtree or Octree is better fit for outdoor environments, while the BSP tree fits indoor scenes better. And the choice between using a Quadtree or an Octree depends on how flat your world is. If there's little variation in the Y axis using an Octree would be wasteful. An Octree is basically a Quadtree with an additional dimension.
Finally, don't disregard the simplicity of the Grid solution. Many people ignore that a simple grid can sometimes be enough (and even more efficient) for their problems, and jump straight to a more complex solution instead.
Using a grid consists simply in dividing the world into evenly spaced regions and storing the entities in the appropriate region of the world. Then, given a position, finding the neighbouring entities would be a matter of iterating over the regions that intersect your radius of search.
Let's say your world ranged from (-1000, -1000) to (1000, 1000) in the XZ plane. You could for instance divide it into a 10x10 grid, like so:
var grid = new List<Entity>[10, 10];
Then you'd place entities into their appropriate cells in the grid. For instance an entity with XZ (-1000, -1000) would fall on cell (0,0) while an entity with XZ (1000, 1000) would fall on cell (9, 9). Then given a position and a radius in the world, you could determine which cells are intersected by this "circle" and iterate only over those, with a simple double for.
Anyway, research all of the alternatives and pick the one that seems to fit your game better. I admit I'm still not knowledgeable enough on the subject to decide which of the algorithms would be best for you.
Edit Found this on another forum and it might help you with the decision:
Grids work best when the vast majority objects fit within a grid square, and the distribution is fairly homogenous. Conversely, quadtrees work when the objects have variable sizes or are clustered in small areas.
Given your vague description of the problem, I'm leaning against the grid solution too (which is, assuming units are small and fairly homogeneously distributed).
• Thanks for detail answer. Yes, it seems simple Grid solution is good enough for me. Commented Jan 2, 2012 at 7:45
I wrote this some time back. It's now on a commercial site, but you can get the source for personal use for free. It may be overkill and it's written in Java, but it's well documented so it should not be too hard to trim and rewrite in another language. It basically uses an Octree, with tweaks to handle really large objects and multi-threading.
I found an Octree offered the best combination of flexibility and efficiency. I started out with a grid, but it was impossible to size the squares properly and large patches of empty squares used up space and computing power for nothing. (And that was just in 2 dimensions.) My code handles queries from multiple threads, which adds a lot to the complexity, but the documentation should help you work around that if you don't need it.
To boost your efficiency, try to trivially reject the 99% of "units" that are not near the target unit using a very inexpensive bounding box check. And I would hope you could do this without structuring your data spatially. So if all your units are stored in a flat data structure you could try to race through it from start to finish and firstly check is the current unit outside the bounding box of the unit of interest.
Define an oversized bounding box for the unit of interest such that it can safely reject items that have no chance of being considered "near" it. The check for exclusion from a bounding box could be made cheaper than a radius check. However on some systems where this was tested it was found not to be the case. The two perform almost equally. This edited after much debate below.
First: 2D bounding box clip.
// returns true if the circle supplied is completely OUTSIDE the bounding box, rectClip
bool canTrivialRejectCircle(Vertex2D& vCentre, WorldUnit radius, Rect& rectClip) {
if (vCentre.x + radius < rectClip.l ||
vCentre.x - radius > rectClip.r ||
vCentre.y + radius < rectClip.b ||
return true;
else
return false;
}
Compared with something like this (in 3D):
BOOL bSphereTest(CObject3D* obj1, CObject3D* obj2 )
{
D3DVECTOR relPos = obj1->prPosition - obj2->prPosition;
float dist = relPos.x * relPos.x + relPos.y * relPos.y + relPos.z * relPos.z;
return dist <= minDist * minDist;
}.
If the object is not trivially rejected then you perform a more expensive and accurate collision test. But you are just looking for nearnesss so the sphere test is suitable for that, but only for the 1% of objects that survive the trivial rejection.
If this linear approach doesn't give you the performance you need, then a hierarchical data structure may be required such as the other posters have been talking about. R-Trees are worth considering. They support dynamic changes. They are the BTrees of the spatial world.
I just didn't want you going to all that trouble of introducing such complexity if you could avoid it. Plus what about the cost of keeping this complex data strucure up to date as objects move around several times per second?
Bear in mind that a grid is a one level deep spatial data structure. This limit means that it is not truly scaleable. As the world grows in size, so do the number of cells you need to cover. Eventually that number of cells itself becomes a performance problem. For a certain sized world however, it will give you a massive performance boost over no spatial partitioning.
• The OP specifically said he wants to avoid a brute force approach, which is exactly what you describe in your first paragraph. Also, how do you figure a bounding box check is cheaper than a bounding sphere check?! That's just wrong. Commented Jan 1, 2012 at 20:40
• Yes I know he wants to avoid brute force which would be avoided by going to the effort of introducing a hierarchical data structure to his application. But that could be a lot of effort. If he doesnt want to do that yet, he can try the linear approach which is brute force but may not perform so bad if his list isn't very big. I will try to edit the code above to put in my 2D bounding box trivial reject function. I don't think I was wrong. Commented Jan 1, 2012 at 21:11
• The link to GDnet is broken, but the canonical sphere test is very simple, very cheap and doesn't branch: inside = (dot(p-p0, p-p0) <= r*r) Commented Jan 2, 2012 at 10:31
• I have pasted the code up above instead. It looks anything but cheap compared to bounding box. Commented Jan 2, 2012 at 11:34
• @Ciaran Quite honestly, that article seems really bad. First of all it doesn't do the tests with realistic data, but rather uses the same values over and over again. Not something you will encounter in a real scenario. And no, according to the article the BB is only faster when there's no collision (eg. the check fails at the first if statement). Also not very realistic. But quite honestly, if you're beginning to optimize things like this, then you're definitely starting at the wrong place. Commented Jan 3, 2012 at 10:40
I have to make this an answer because I don't have the points to comment or upvote. For 99% of the people who ask this question, a bounding box is the solution, as described by Ciaran. In a compiled language, it will reject 100,000 irrelevent units in the blink of an eye. There's a lot of overhead involved with non-brute-force solutions; with smaller numbers (say under 1000) they'll be more expensive in terms of processing time than brute force checking. And they'll take vastly more programming time.
I'm not sure what "a very large number" in the question means, or what other people looking here for answers will mean by it. I suspect my numbers above are conservative and could be multiplied by 10; I am personally quite prejudiced against brute force techniques and am seriously annoyed at how well they work. But I wouldn't want someone with, say, 10,000 units to waste time with a fancy solution when a few quick lines of code will do the trick. They can always get fancy later if they need to.
Also, I would note that a bounding sphere check requires multiplication where the bounding box does not. Multiplication, by its nature, takes several times as long as addition and comparisons. There's bound to be some combination of language, OS, and hardware where the sphere check will be faster than a box check, but in most places and times the box check has to be faster, even if the sphere does reject a few irrelevant units the box accepts. (And where the sphere is faster, a new release of the compiler/interpreter/optimizer is very likely to change that.)
• While there's nothing wrong with your answer, you're not answering the question. It was specifically asked for a "non bruteforce" approach. Also you seem to repeat what Ciaran already wrote and we had a lengthy comment-discussion about AABB vs. circle tests. The performance difference is simply irrelevant. Better choose a bounding volume that fits most of your collision candidates, as it will reduce the amount of actual narrow-phase tests.. which will have a greater impact on performance overall. Commented Jan 4, 2012 at 16:20
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equation to find the slope between two points
# equation to find the slope between two points
FINDING THE EQUATION OF A LINE WHEN THE SLOPE AND A POINT ARE KNOWN Because the slope is part of an equation, m yx---22---------yx---11 , we can substitute the slope andbetween the two points is 6, then the slope is 1----2- 2 . 6. Now determine if the slope is positive or negative. Derivative as a Limit of Slopes. Slope of a Line Between Two Points.Now well find the slope of a line between two points on any wonky function we like. First, you need to find the slope between those two points using the slope formula.4 Step 2: Find the y-intercept ( -1, -5 ) and ( 3, 3 ) Now that we know the slope is 2, we can find the y-intercept by choosing one point from above and plugging these values into the general equation for a line. Then the perpendicular slope is. . Again, I have a point and a slope, so I can use the point-slope form to find my equation.Advertisement. I know I can find the distance between two points I plug the two points into the Distance Formula. But I dont have two points. The slope represents the change in the y coordinates over the change in the x coordinates between two points on a line. In other words, slope change in y / change in x. Standard form of a linear equation. Find slope from two points. See what Common Core State Standard this skill aligns with.How can we find the equation of that line? [Further reading]. How to Find Distance between Two Points? properly, to discover slope you take advantage of the equation: replace of y/replace of x. in different words, discover the replace between both y values and set it over the replace between both x values.Math-Finding the slope of two given points? Solved examples to find the equation of a line in point-slope formEquations of the Bisectors of the Angles between Two Straight Lines.
When two points (x1, x2), (y1, y2) are given and the equation contains these two points, the first step is to find the slope of the line. Find the slope of the line that goes through the two points that are given to you. If you cant find what youre looking for, or you have an idea for a calculator that would be helpful to you, let us know.
This equation provides the slope of the line between two points, (x1,y1) and( x2,y2). in a Cartesian plane. To find the vertical and horizontal distances between the two points, sometimes this can be seen by looking at the graph, but the verticalIf the slope of a segment connecting the points (x, 5) and (2, -4) is 6, what is the value of x? We calculate the . We can solve this equation by multiplying both sides Find the training resources you need for all your activities.P(-1, 3) B (1, 1) Slope -3 y-intercept -1 Find the equation of the line with the following characteristics. a. Find the slope between the two points in simplest form. b. Substitute the slope into the equation along with one of the points listed above and solve the y-intercept. How to find slope given an equation How to find slope given a graph Get y by itself first.Vertical lines have Undefined slope and have equations that look like X a number. Pick two points on the line and determine the slope. Objective: Students will see the relationship between the slope-intercept form of an equation, and the graph.(For steps 3 and 4 it will help to use zoom decimal, or if needed zoom decimal, then zoom integer then enter.) 5. Using the two points find the slope of the line. Ive been reading up on the basics of calculating the slope using two points: m y1-y2/x1-x2. Which after that, I can figure out the equation of the lineTwo points to form a line, how to find the distance between the line and a third point? 0. Segment distance between two 3D points divided by xy plane. This paper computes the distance between two points and fits both linear and exponential functions through the two points.We plug in the point and the slope we found previously () to obtain the equation. Graphing 12 Slope Between Two Points Help Video In High School. Example 3 Write An Equation Of A Line Given Two Points Ppt Download.Holt Algebra Point Slope Form Warm Up Find The Of Line. YOUR TURN: Find the equation of the line passing through the points ( -4 , 5 ) and ( 2 , -3 ).Finding the Slope of a Line from Two Points. Linear Equations. Graphing Overview. Finding Slope Given Two Points - Продолжительность: 18:42 L Moon 268 628 просмотров.Finding the Slope of a Linear Equation [fbt] - Продолжительность: 19:03 Fort Bend Tutoring 50 094 просмотра. In algebra, linear equations means youre dealing with straight lines. When youre working with the xy-coordinate system, you can use the following formulas to find the slope, y-intercept, distance, and midpoint between two points. Relevant discussion may be found on the talk page.The point-slope form expresses the fact that the difference in the y coordinate between two points on a line (that is, y y1) iswould result in the Two-point form shown above, but leaving it here allows the equation to still be valid when. The point-slope form may be used to find the equation of a line through two known points.This final equation is the normal form. The word "normal" in this usage refers to the perpendicular relationship between OM and AB. Find the slope given a graph, two points or an equation.
Write a linear equation in slope/intercept form.We will also look at the relationship between the slopes of parallel lines as well as perpendicular lines. Lets see what you can do with slopes. How do you find the slope of a line between two points?To find the y coordinate of the intersection point, plug the x you just got into either equation and simplify so that y some number. You can use this equation to write an equation if you know the slope and the y-intercept.Find the equation of the line. Choose two points that are on the line. Calculate the slope between the two points. Popular Problems. Precalculus. Find the Equation Using Point-Slope Formula (-1,10)(1,5/ 2).After finding the slope between the points, use point-slope form to set up the equation. Section 3.5 Graphing Linear Equations in Slope-Intercept Form 141. In Exercises 1522, find the slope and the y-intercept of the graph of the linear equation.Then use the slope formula to show that the slope between the points is m. If two points P (x1,y1) and Q (x2,y2) are on the line L, then we can calculate the slope between them and use the first point and the pointslope equation to get the equation of L(ii) We can find the point of intersection of L and L by replacing the y in the equation for L with the y from L so. There are two steps to writing these equations. Find the Slope Fraction for the line.There are two ways to find the slope from two points. The visual way is to graph each point and then count the number of squares up (rise), then the number of squares over (run) between the points. It is said to be the distance between two points located in space.In order to find distance between parallel lines, one needs to follow the steps mentioned below: 1) Check whether the given equations are in slope-intercept form, i.e y mx c. If they are not, interchange the variable to get that format. Aim: I can write the slopeintercept equation for any two points. Warm-Up: Find the slope between the following sets of points.Explore: A) Find an equation to a line on which your point lies B) Can you find another equation for which your point is a solution? Equation from 2 points using Slope Intercept Form. Find the equation of a line through the points (3,7) and (5,11).You can use the calculator below to find the equation of a line from any two points. x-axis. EX-1- Find the slope of the line determined by two points A(2,1) and B(-1,3).Nonvertical lines : That point slope equation of the line through the point ( x1 , y1 ) with slope m is The distance from a point to a line : To calculate the distance d between the point P(x1 , y1 ) and Q( x2 For my pass algorithm, I calculate the slope between the passer and reciever with the equation m (y2-y1)/(x2-x1).8. Math Calculation to retrieve angle between two points? 1. Finding slope between coordinates stored in List(Of Point) data types. Learn how to find the slope between two points.Find the Equation of a Line Using Point-Slope Form - I show one complete example of finding the equation of a line using point-slope form. Substitute the slope and one ordered pair into the point-slope equation: y - y1 m(x - x1).How to Calculate the Distance Between Two Parallel Lines. How to Graph Linear Equations With Two Variables. Here are two points (you can drag them) and the equation of the line through them. Explanations follow.2. Put the slope and one point into the "Point-Slope Formula". 3. Simplify. Step 1: Find the Slope (or Gradient) from 2 Points. Slope of the line is equal to the tangent of the angle between this line and the positive direction of the x-axis.Example 1. Find the equation of a line passing through two points A(1, 7) and B( 2,3). 10.1.2 Angle between two lines The angle between the two lines having slopes m and 1.10.1.7 Locus and Equation of Locus The curve described by a point which moves under certain given condition is called its locus. To find the locus of a point P whose coordinates are (h, k), express the 100 Finding The Distance Between Two Points Worksheet.< > Equation To Calculate Slope Jennarocca. Example 9: Find an equation of the given graph: Solution: Between the points (1, 1) to (3, 0), we can see that the rise is 1 unit and the run is 2 units.66. Use the point-slope formula to find the equation of the line passing through the two points. The point-slope form can be used to find an equation of the line passing through two points (x1, y1) and (x2, y2).When the estimated point lies between two given points, as shown in Figure 1.46, the procedure is called linear interpolation. Graphing Equations with Slope. Okay, now that you know how to graph a line by getting some points, and you know how to find the slope between two points, you should be able to find the slope of a line once you have an equation Use a graph and two points to find slope without the equation handy.Slope can also be shown as y/x, meaning "difference of y / difference of x:" this is the same exact question as " find the slope between. The equation of a line is typically written as ymxb where m is the slope and b is the y-intercept.Calculus, Derivatives Calculus, Integration Calculus, Quotient Rule Coins, Counting Combinations, Finding all Complex Numbers, Adding of Complex Numbers, Calculating with Complex Numbers In this calculator, you can find the slope and equation of the straight line with two given points (two point slope form).Related Calculators: Distance Between 2 Points. Before you can find the equation, make sure you have a clear idea of what youre trying to find. Pay attention to these wordsThe slope between two points is defined as "rise over run" — in other words, the description of how far you must travel up (or down) and to the right (or left) in order to What is point slope form of a line with a x value of 6 and y value of -1.Which ordered pair is the best estimate for the solution of the system of equations? y32x6 y14x2.
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# If the effective mass of the train = Me ton, acceleration = α Km/sec2, then the tractive force (Ft) is equal to:
This question was previously asked in
UPPCL JE Previous Paper 6 (Held On: 25 November 2019 Shift 1)
View all UPPCL JE Papers >
1. Me × α × 106 N
2. Me × α × 103 N
3. Me × α × 105 N
4. Me × α × 104 N
Option 1 : Me × α × 106 N
Free
CT 1: Network Theory 1
11207
10 Questions 10 Marks 10 Mins
## Detailed Solution
Tractive effort is the force acting on the wheel of the locomotive which is necessary to propel the train.
It is a vector quantity always acting tangentially to the wheel of a locomotive.
It is measured in Newton.
It can be increased by using high output motors and by increasing the dead weight over the driving axles.
The net effective force or the total tractive effort (Ft) on the wheel of a locomotive or a train to run on the track is equal to the sum of tractive effort:
a) Required for linear and angular acceleration (Fa)
b) To overcome the effect of gravity (Fg)
c) To overcome the frictional resistance to the motion of the train (Fr)
Ft = Fa + F+ Fr
Ft = Me x α
Where, M= Effective mass of the train in ton or 10kg
α = Acceleration in Km/secor 103 m/sec2
∴ Ft = Me x 103 x α x 103
Ft = Me × α × 106 N
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# If you regress random variable Y against random variable X, would the results be the same if you regressed X against Y?
Nov 8, 2015
They would be the same only if the $V a r \left(x\right) = V a r \left(y\right)$
#### Explanation:
The Pearson correlation coefficient will always be the same regardless what the variances are. However, the regression lines will have different slopes (hence, not the same) unless the variances are equal.
This may be more technical than what you are looking for, but here is the slope of the regression line for both cases:
${\beta}_{1} = \frac{C o v \left(x , y\right)}{V a r \left(x\right)}$ This is regress y against x
${\beta}_{1} = \frac{C o v \left(x , y\right)}{V a r \left(y\right)}$ This is regress x against y
As you can see, they will only have the same slope if the variances are equal.
hope that helped
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## Putting It Together: The Language of Algebra
Let’s revisit the scenario from the beginning of the module where Lydia Lee was about to calculate how much each person in her family could spend at the the Badger Hill Farm. When Aubrey and Maya’s parents told the girls about carving pumpkins and eating caramel apples, they got super excited and wanted to know what else they can do while they are visiting the farm!
So far, these are the expenses for the trip:
Lodging: 3 nights at an apartment rental at $180/night $180\cdot{3}=\540$ Food: 3 days of groceries and restaurant meals at$75/day for the whole family
$75\cdot{3}=\225$
Science Center: 2 adult tickets and 2 child tickets, the adult tickets are $24 each and the child tickets are$19 each
$24\cdot{2}=\48$ (adult tickets)
$19\cdot{2}=\38$ (child tickets)
Badger Hill Farm: 4 tickets for entry to the farm during the autumn fair
$8\cdot{4}=\32$ (entry tickets)
If we add up all the known expenses, the expression will be:
lodging + food + museum tickets + farm tickets
$540+225+48+38+32$
$=\883$
The unknown in this case, or variable, is how much each person has to spend on extras at the farm. We’ll call this amount $x$. There are four people in the family, so the total they will be able to spend is $4x$. Now let’s help Lydia write an equation to solve for the value of $x$. She knows it will have to look like this if they are going to stay on budget:
cost of known amounts + extra spending at farm = total amount budgeted
$883+4x=955$
$4x=72$
$x=18$
Each member of the Lee family can spend $18 on extra fun. Maya wants to carve a pumpkin and eat a caramel apple and wonders how much money she will have left over to save for her weekly sticker shopping spree. The caramel apples cost$5 each and the pumpkin carving is $12, so the equation to determine the leftover money must be: $12+5+z=18$ $z=1$ Maya will have$1 left over to buy stickers.
Aubrey wants to go in the corn maze and then get as many bags of popcorn as possible. The corn maze is $10 and each bag of popcorn is$4, so the equation to determine how many bags of popcorn must be:
$10+4y=18$
$y=2$
Aubrey can buy 2 bags of popcorn and go in the corn maze.
As we demonstrated in these examples, algebra is used in all sorts of common everyday calculations that you make. After making progress through more complex math topics, you will also be able to solve complicated real world problems that involve algebra.
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# Exploring the Weight of 1/4 Aluminum Plate: How Heavy is it Really?
When it comes to constructing buildings, bridges, and other structures, the use of aluminum has become increasingly popular due to its many benefits. Aluminum is a lightweight, durable, and corrosion-resistant metal that is commonly used in various industries. One question that often arises when working with aluminum is how heavy a specific thickness of aluminum plate is. In this article, we will explore the weight of a 1/4 inch aluminum plate and provide some insights into its actual weight.
Aluminum is a versatile metal that comes in many forms, including sheets, plates, bars, and more. Aluminum plates are commonly used in construction and manufacturing due to their strength and lightweight properties. One popular thickness of aluminum plate is 1/4 inch, which is equivalent to 0.25 inches or 6.35 millimeters.
The weight of a 1/4 inch aluminum plate can vary depending on the grade of aluminum used, as well as the size and shape of the plate. However, a general rule of thumb is that aluminum weighs approximately 2.7 grams per cubic centimeter, or 0.0975 pounds per cubic inch. Using this information, we can calculate the weight of a 1/4 inch aluminum plate.
To calculate the weight of a 1/4 inch aluminum plate, we need to first determine the volume of the plate. The volume of a rectangular plate is calculated by multiplying its length, width, and thickness. For example, if we have a 1/4 inch aluminum plate that is 24 inches long and 12 inches wide, the volume would be:
Volume = length x width x thickness = 24 x 12 x 0.25 = 72 cubic inches
Next, we can calculate the weight of the aluminum plate by multiplying the volume by the weight of aluminum per cubic inch. Using the weight of aluminum per cubic inch of 0.0975 pounds, the weight of the 1/4 inch aluminum plate would be:
Weight = volume x weight per cubic inch = 72 x 0.0975 = 7 pounds
Therefore, a 1/4 inch aluminum plate that is 24 inches long and 12 inches wide would weigh approximately 7 pounds. It is important to note that this is just a rough estimate, as the actual weight may vary slightly due to factors such as imperfections in the metal or variations in density.
In addition to the weight of the aluminum plate itself, it is also important to consider the weight of any additional materials that may be attached to or supported by the plate. For example, if the aluminum plate is being used as a roofing panel, the weight of the roof material, insulation, and other components must also be taken into account when determining the overall weight of the structure.
When working with aluminum plates, it is essential to consider the weight of the material to ensure that it can be safely and effectively installed. Proper handling and installation procedures should be followed to prevent accidents and ensure the structural integrity of the project. By understanding the weight of a 1/4 inch aluminum plate and how to calculate it, builders and engineers can make informed decisions when working with this versatile metal.
In conclusion, the weight of a 1/4 inch aluminum plate can vary depending on various factors, but a general estimate is around 7 pounds for a plate that is 24 inches long and 12 inches wide. By calculating the weight of the aluminum plate and considering any additional materials or components, builders and engineers can determine the overall weight of a structure and ensure its safety and stability. Aluminum plates continue to be a popular choice for construction and manufacturing due to their strength and lightweight properties, making them an excellent option for a wide range of applications.
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### Summer_Wind's blog
By Summer_Wind, history, 5 weeks ago,
Problem : Given a number k, find a k-digit string where the product of its digits is greater than or equal to the sum of its digits. If there are multiple such strings, return the one with the minimum product. If there are still ties, return the lexicographically smallest string.
Testcases:
Input : 1 Output : 1
Input : 2 Output : 22
Input : 3 Output : 123
Input : 4 Output : 1124
Input : 5 Output : 11222
Input : 6 Output : 111126
Constraints:
1 <= K <= 100000
How to solve this? Please help!
• +2
» 5 weeks ago, # | ← Rev. 2 → 0 I think that when you see a k, you can try to make all numbers with k digits and then calculate the product of each one and the sum of each one. If the product is greater than or equal to the sum, store it in a dictionary, where the dictionary key is the product and the dictionary value is the list of the numbers (maybe in string form to save some space). Then find the minimum key in the dictionary, sort that corresponding list, and return the first element in the list.Edit: oh wait, the answer is just "0" * k
• » » 5 weeks ago, # ^ | ← Rev. 2 → +6 "You can try to make all numbers with k digits and then calculate the product of each one and the sum of each one". k can be 100000 , how can i create and check a number with such a large number of digits? I did not get that.
• » » » 5 weeks ago, # ^ | +4 backtracking
• » » » 5 weeks ago, # ^ | ← Rev. 4 → +4 EDIT: Sorry! I missed the first tie-breaker is to minimize the product. This solution is only correct for considering only the second tie-breaker. Counter example: take the case K=2^16-16, the suffix of the optimal answer is 16 2's, which cannot be constructed using the last 6 digits. __________________________________________________________________________________________ (Original comment) Notice that for large k, the answer will contain 1s for the first K-6 digits (at least) because 11111111111...999999 = 9^6>10^5. So we can just brute force over every suffix between 111111 and 9^6. The first one that yields a valid result is the answer. There exist better solutions but this will pass. __________________________________________________________________________________________EDIT: Here is a correct solution, and one that does not use recursive backtracking. To begin, we can use the same argument noting that the optimal solution will never have more than 16 2s (2^15*3>10^5). (1)Instead, we can iterate from K...2*K and find the first number such that it is not divisible by any prime greater than 7. This can be found in O(K log K) time with a modified Sieve of Eratosthenes. (if the number is within 160 of K (a better bound probably exists), you need to check if it is valid first by factoring the number to see whether the minimal sum to construct the number is less than it. Notice that to minimize the sum it is always optimal to use prime numbers))Proof of (1): Suppose we 17 digits, a better result always exists where we only use 16 as we replace one two with a three. The sum is the same, but the product is greater than the maximum possible value 3*2^15> 10^5+31Proof that we only need to check between K and 2*K: You can always construct a product at most 2*K by constructing a number with only 2's. Keep adding 2's until the product is greater than K+ number of 2's added. Hence our upper-bound is good.
• » » » » 5 weeks ago, # ^ | ← Rev. 2 → 0 How can we assure that the first valid result is indeed the minimum product and lexicographically smallest?
• » » » » » 5 weeks ago, # ^ | +3 For k > 6 , we have a fixed count of ones initially, so we just need to adjust the last six digits using brute force and backtracking.For k = 10 : 1111 _ _ _ _ _ _We have a sum of 4 and a product of 1 initially. Adjust the last six digits by creating all possible numbers using backtracking, and you will get the least product with the least value easily.Thanks to both of you 123gjweq2 , nathanballman.
• » » » » » » 5 weeks ago, # ^ | ← Rev. 7 → 0 Got it, you start from the minimum possible values so that the first result is the miminum product and sumEdit: What is the criteria for the backtracking? How do we build it? I dont mean how to code it but how the pattern should look likea simple backtracking for input 6 for a valid answer would look like (111229 if you start from big to small) or (111922 if you start from small to big) at some point you get those 2 valid answer before the real one which is (111223)
• » » » » 5 weeks ago, # ^ | +3 Yeah this one should work .
• » » » » 5 weeks ago, # ^ | 0 Why the answer can't have more than 6 digits greater than 1? For example, assume that the product of digits in the answer is 5^7, which can be achieved only with 7 digits being equal to 5. Are there any proofs that the product of digits in the answer can always be represented as a product of 6 digits?
• » » » » » 5 weeks ago, # ^ | ← Rev. 2 → 0 For the largest k=10000:If we place 1 in all the digits except for the last 6 digits, which can be filled with 9, the sum is calculated as follows:Sum=6×9+(100000−6)×1=100048 The product of the digits ( 9^6 ) is greater than 500000. Since we need at most 6 digits to be checked to ensure the product is greater than or equal to the sum for the largest k, this indicates that it will be true for lower values of k as well. Thus, we can conclude that for all 1≤k≤10000, it is always possible to obtain a number whose product of digits is greater than or equal to the sum of its digits.To ensure we can get the minimized product within six digits: For any k in the range [1,10000], we can use up to 18 digits of 2 (since 2 ^ 18 ≥ 200000 ) . Grouping 18 digits of 2 into six digits, we can use six digits of 8 instead (since 2 ^18 = (2 ^ 3 ) ^ 6 = 8 ^ 6). This approach ensures that we can effectively group the largest count of 18 digits into six digits, ensuring that a number of length [1,10000] can always achieve a similar product using six digits or fewer.Edit:I am also unable to provide a rigorous proof, but from observation, it appears that all numbers can be reduced to a product of six digits that satisfies the condition. I believe this problem might require a proof by contradiction. If anyone has suggestions on how to approach or prove this, please share .
• » » » » » » 5 weeks ago, # ^ | ← Rev. 2 → 0 Sorry about that, I didn't read the original problem carefully enough. I edited my original post with a counter-example. I gave 2^16-16 there, but 2^7-7 also works as a counter example and I'll explain why my algorithm fails here. The correct solution for 2^7-7 is 111111...2222222.(2^7-14 1's and 7 2's) The product of this number is 2^7 and the sum is also 2^7. We know my original algorithm will fail because a list of only 1's and 2's is the most efficient way to reach any product (in terms of the sum of the digits)Proof: Here's a constructive argument. First create a string of length K filled will all ones. Its sum will be K. Now let us observe what happens to the product when you change a 1 to a 2. The product doubles and the sum increases by one. Now let us observe what happens when we increment any other number. Going from 2 -> 3 multiplies the product by 1.5. In general going from N -> N+1 multiplies the product by (N+1)/N.This shows diminishing returns and because every transition between a number x -> y can be represented as a sequence of increments, only incrementing 1->2 is the single most efficient way to make a product. Naturally, it follows that using only 6 digits instead of 7 will force one to be at least 3 which is less efficient therefore the sum will be greater than 2^7 and thus the product must be as well. Again, sorry for my mistake. I gave a non-backtracking solution in my edit to my original comment.
» 5 weeks ago, # | +4 I'm sure a simple brute force will work.First, it's clearly obvious that the digits are in ascending order from 1 to 9. So, we just have to think about how many times each digit appears in the string.The maximum sum of the digits is bounded by $9K \leq 900000$. In particular, this implies that the number of digit 2 is bounded by $\lceil\log_2(9K)\rceil$ (since more number of digits would clearly make the product exceed the sum). Similarly, the number of digit 3 is bounded by $\lceil\log_3(9K)\rceil$, and in general, the number of digit $d$ is bounded by $\lceil\log_d(9K)\rceil$. So, the total number of ways of choosing the digits from 2 to 9 is bounded by $\prod_{d = 2}^{9} (\lceil \log_d(9K) \rceil + 1)$. If you actually compute this quantity with $K = 100000$, you'll find out that this is less than $2 * 10^8$. Of course, this is actually a pretty loose upper bound, and you can likely prune away many, since many of the products would far exceed the sum upper bound $9K$.
• » » 5 weeks ago, # ^ | 0 So basically, for all digits from 1 to 9, we can store/calculate the maximum count we can take for each digit and then brute force through it using backtracking. Thanks for the help.
» 5 weeks ago, # | +3 The sum of digits is bounded by 9*k. To make product of digits at least 9*k we need at most ceil(log2(9*k)) digits greater than 1. If we had more than ceil(log2(9*k)) digits greater than 1, then we could turn the least digit greater than 1 into 1, reducing product of digits and making string lexicographically smaller. Therefore having at most ceil(log2(9*k)) digits greater than 1 is optimal. Let's denote ceil(log2(9*k)) as l. Sum of last l digits is bounded by 9*l. Let's maintain dp[i][j], sum of last l digits is i, product of them is at least j, dp[i][j] gives information about whether this state can be achieved and if so, which digit was the last used one. Dimension of dp will be [l*9][k+l*8]. Time complexity is O(klog^2(k)), space complexity is O(klog(k)) .
» 5 weeks ago, # | 0 First, answer for N=6 is 111126You can't assume 6 non-one digits. For example 8179, lowest product is 8192=2^13, and "1" * 8166 + "2" * 13 has sum 8192, so its tight.I think easiest way to proceed is for each product P = K, K+1, ... check if its achievable. Brute force is enough, we should expect to hit an answer fairly fast, you can code this then check it finds an answer for every K. def solve(K): for P in range(K, K + 2000): expos = [0, 0, 0, 0] cur = P for i, x in enumerate([2, 3, 5, 7]): while cur % x == 0: cur //= x expos[i] += 1 if cur != 1: continue count = [0] * 10 count[5] = expos[2] count[7] = expos[3] results = [] for count[6] in range(1 + min(expos[:2])): alpha, beta = expos[:2] alpha -= count[6] beta -= count[6] for count[8] in range(1 + alpha // 3): alpha2 = alpha - 3 * count[8] for count[4] in range(1 + alpha2 // 2): count[2] = alpha2 - 2 * count[4] for count[9] in range(1 + beta // 2): count[3] = beta - count[9] * 2 count[1] = K - sum(count[2:]) if count[1] >= 0: total = sum(i * x for i, x in enumerate(count)) if total <= P: results.append(count[:]) if results: return "".join(str(i) * x for i, x in enumerate(max(results)))
» 5 weeks ago, # | 0 Auto comment: topic has been updated by Summer_Wind (previous revision, new revision, compare).
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http://clay6.com/qa/18243/the-solution-set-of-the-equation-begin1-4-20-1-2-5-1-2x-5x-2-end-0-is
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# The solution set of the equation $\begin{vmatrix}1 &4&20\\1 &-2&5\\1 &2x&5x^2\end{vmatrix}=0$ is
(A) 1,2 (B) 1,0 (C) -1,2 (D) None of these
Given equation is $\begin{vmatrix}1 &4&20\\1&-2&5\\1&2x&5x^2\end{vmatrix}=0$
On expanding the determinant we get a quadratic equation in $x$.
It has two roots.
We observe that $R_3$ becomes identical to $R_1$ if $x=2$ thus at $x=2$
$\Rightarrow \Delta =0$
$\therefore x=2$ is a root of the equation .
Similarly $R_3$ becomes identical to $R_2$ if $x=-1$
At $x=-1$ is a root of the given equation .
Hence equation has roots as -1 and 2
Hence (c) is the correct answer.
answered Nov 19, 2013
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Select Page
# Peel Elementary Teachers Local (PETL)
## PETL Math Professional Development Workshop
### Saturday March 25th, 2017
Thanks for inviting me to learn with you on a Saturday in Peel! The group was energetic and excited to learn about meaningful manipulative use, counting principles, the progression of multiplication/division and the importance of concreteness fading!
Here’s a summary of what we explored today. Looking forward to connecting again soon!
After a brief presentation about avoiding the rush to the algorithm, we took some time to explore a 3 act math task with a very low-floor and high-ceiling, called the Airplane Problem:
[threeactshortcode the_query=”post_type=realworldmath&p=18652″]
In the airplane problem, we made predictions and made connections to subitizing and unitizing.
## Counting Principles
We briefly discussed the importance of counting and quantity principles like subitizing and unitizing. A full summary is below:
[postshortcode the_query=”p=18603″]
## Tiny Polka Dots
After break, I shared Daniel Finkel‘s game, Tiny Polka Dots. We made some connections between the counting and quantity principles and the different card decks in this game.
Also be sure to check out his TED Talk here.
## Exploring the Progression of Multiplication
We made a leap from counting principles and unitizing to multiplication and specifically, using arrays and area models.
[postshortcode the_query=”p=17501″]
We also made a connection to “Japanese Multiplication” or “Stick Multiplication”:
[postshortcode the_query=”p=18762″]
## 3 Act Math Task: Donut Delight
After lunch, we explored the Donut Delight 3 act math task where we played with multiplication to predict how many donuts there were in the “double hundred dozen” box and then used repeated subtraction to lead to a flexible division algorithm for division when we tried to find how many layers there were.
[threeactshortcode the_query=”post_type=realworldmath&p=18719″]
## Progression of Division
We spent some time working with the progression of division in most groups as well. Here are some screenshots of what we looked at:
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Have your children and students been able to determine whether a number is odd or even? To make them understand the concept of odd and even numbers, they have to be exposed to the real problems of differentiating the odd numbers from the even ones. Such experience in solving odd and even problems can be found in these even odd worksheets for students. Check out the worksheets on the images below!
Even odd worksheets are purposed to train and test children’s knowledge and ability in determining numbers. These worksheets will help them in learn the basics in order to be able to recognize the kind of numbers once they see them. Moreover, the various exercises provided in these worksheets enable them to study the key concept of odd and even numbers.
image via www.theteachersguide.com
image via printablecolouringpages.co.uk
image via freecoloringpages.co.uk
Even numbers can be divided evenly into groups of two. Meanwhile, odd numbers can’t be divided into groups of two as they can’t be matched in pairs. Other trick that can be used to determine odd and even numbers is that even numbers always end with digit of 0, 2, 4, 6, or 8 while odd numbers always end with 1, 3, 5, 7, or 9.
image via www.pic2fly.com
image via calendariu.com
image via www.docstoc.com
Learning odd and even numbers is crucial for students’ development in understanding numbers as well as Math. Therefore, don’t forget to print these worksheets for your students! Have a successful learning!
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https://math.answers.com/questions/Why_is_the_recipical_of_A_fraction_less_than_one_is_always_a_fraction_greater_than_one
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0
Why is the recipical of A fraction less than one is always a fraction greater than one?
Updated: 9/24/2023
Mollywags11tw3860
Lvl 1
8y ago
Fractions that are less than one are known as proper fractions. Their denominators are greater than their numerators. Their reciprocals would have numerators greater than their denominators, making them improper. Improper fractions are greater than one.
Wiki User
8y ago
Wiki User
9y ago
First of all, the proposition is the question is not true.
Consider the fraction A = -1/2. It is a fraction that is less than zero and so certainly less than one. However, its reciprocal is -2 which is NOT greater than 1. The statement is true only for positive fractions (bit that was not stated in the question).
A positive fraction A is less than 1 if its numerator is less than its denominator (both being positive). The numerator of the reciprocal of A will be the denominator of A and the denominator of the reciprocal of A will be the numerator of A. Thus, the reciprocal of A will have a numerator which is greater than its denominator. Consequently, the fraction will be greater than 1.
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https://www.gradesaver.com/textbooks/math/calculus/calculus-8th-edition/chapter-3-applications-of-differentiation-review-exercises-page-286/7
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Chapter 3 - Applications of Differentiation - Review - Exercises - Page 286: 7
$$\frac{1}{2}$$
Work Step by Step
Given $$\lim _{x \rightarrow \infty} \frac{3 x^{4}+x-5}{6 x^{4}-2 x^{2}+1}$$ Then \begin{align*} \lim _{x \rightarrow \infty} \frac{3 x^{4}+x-5}{6 x^{4}-2 x^{2}+1}&=\lim _{x \rightarrow \infty} \frac{3+\frac{1}{x^{3}}-\frac{5}{x^{4}}}{6-\frac{2}{x^{2}}+\frac{1}{x^{4}}}\\ &= \frac{3+0+0}{6-0+0}\\ &=\frac{1}{2} \end{align*}
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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## College Algebra 7th Edition
g(f(x)) = g(3x + 4) =$\frac{3x + 4 - 4}{3}$ = x f(g(x)) = f($\frac{x - 4}{3}$) = 3($\frac{x - 4}{3}$) + 4 = x - 4 + 4 = x Since g(f(x)) = f(g(x))
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Question
# h is related to one of the parent functions described in this chapter. Describe the sequence of transformations from f to h. h(x) = (x - 2)^3 + 2
Transformations of functions
h is related to one of the parent functions described in this chapter. Describe the sequence of transformations from f to h. $$h(x) = (x - 2)^3 + 2$$
(b)We are starting with the parent function $$\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}$$
STEP 1: Shift the graph hy 2 units to the right, to get $$\displaystyle{y}={\left({x}-{2}\right)}^{{3}}$$
STEP 2: Shift the graph hy 2 units upwards, to get $$\displaystyle{y}={\left({\left({x}-{2}\right)}^{{3}}\right)}+{2}$$, which is the required function h(x) Shift the graph by 2 units the right and 2 units upwards
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http://creativestarlearning.co.uk/maths-outdoors/dara-an-outdoor-african-maths-game/
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# Dara – An Outdoor African Maths Game
15 July 2012 · 0 comments
## This game from Nigeria requires two players or teams
It is played with stones or sticks in the ground. It’s great for sessions where children finish at different times and need something to do, or as an impromptu family game when on holiday.
#### Make the game board
Using chalk, make a 5×6 square triangle as a game board. If you are on a beach or other suitable surface this can be drawn on the ground with a stick.
#### Find some counters
You can play alone against an opponent or as opposing teams. Each side needs 12 counters. In the photos the class used wood cookies and noggins. Stones, shells or other gathered counters can be used. So find and gather your counters from materials lying around outside. Make sure each side has recognizably different counters.
This is an example of an “illegal” game board as there are more than three counters in a row on both sides. It may take a while to get the hang of the game
#### How to play
1. If you are playing just one other person, take turns to place a counter anywhere on the game board until all counters have been put there. Whilst this is happening neither player can remove each other’s counters. Neither side can have more than three counters in a row at any one time. This is illegal!
2. Take turns to move a counter into an adjacent empty square. The counters cannot be moved diagonally but can be moved up, down or sideways. The aim is to make three counters sit in a row (but not a diagonal one).
3. When a player manages to make three counters in a row, they can remove one of their opponent’s counters. A player can only remove one counter from their opponent in any one go, even if more than one row of three counters is created in a move.
4. The game is over when a player is unable to make three in a row with his or her counters. Their opponent wins when this happens.
Share the knowledge...
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# The first three terms of an arithmetic sequence are 2k+3, 5k-2 and 10k-15, how do you show that k =4?
Since the sequence is arithmetic, there is a number $d$ (the "common difference") with the property that $5 k - 2 = \left(2 k + 3\right) + d$ and $10 k - 15 = \left(5 k - 2\right) + d$. The first equation can be simplified to $3 k - d = 5$ and the second to $5 k - d = 13$. You can now subtract the first of these last two equations from the second to get $2 k = 8$, implying that $k = 4$.
Alternatively, you could have set $d = 3 k - 5 = 5 k - 13$ and solved for $k = 4$ that way instead of subtracting one equation from the other.
It's not necessary to find, but the common difference $d = 3 k - 5 = 3 \cdot 4 - 5 = 7$. The three terms in the sequence are 11, 18, 25.
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# The structure of the subset sum problem
Let us consider the subset sum problem with $S = S_i = \{1,2,..,i\}$. There are of course $2^{|S|}$ different sums that can be formed.
Now it seems that though the sums are not in order and thus not searchable efficiently there is some structure as indicated by the picture below. In the image there are the cases i = 1..7. The "boxes" in bold indicate runs of increasing numbers, that is increasing values of sums, while the row numbers indicate how many numbers the sums consist of. Now it appears that the minimums (and maximums) of these boxes are in order. Now, to me it seems one could exploit this structure to make a search in $O(|S|^3)$ -time. That is, pick a number in a box, then iterate below and above to find the min and max of the box. This is $\Theta(|S|)$. If the max is less than the target, then pick some box to the right of the current one. If the min is greater than the target, then pick some box to the left of the current one. The maximum size of a box is exactly |S|. There are |S|+1 rows to consider (to perform a binary search on). Thus the running time is $O(|S|^3)$, even though the number of boxes is $2^{|S|-2}+2$.
An example of the contents of boxes is the case of i=4, where the box structure is |0|;|1 2 3|4|;|345|567|;|6|789|;|10|;, where the rows are separated by semicolons. Here the |567| box is the |123| box (1-row) of the i=3 case with 4 added to each sum. And the |345| box is the box from the 2-row in the i=3 case. The key property is that the min's and max's are in order (in this case). So: 5 >= 3 and 7 >= 5. So, if looking for 6, one could conclude that it must be in the box other than |345|, if it exists at all. Maybe not so good an example as there are only two boxes.
A more illustrative example, perhaps:
S = {x1}
0: 0
1: x1
S = {x1, x2}, x1 <= x2
0: 0
1: x1 <= x2
2: x1+x2
S = {x1, x2, x3}, x1 <= x2 <= x3
0: 0
1: x1 <= x2 <= x3
2: x1+x2 <= x1+x3 <= x2+x3
3: x1+x2+x3
S = {x1,x2,x3,x4}, x1 <= x2 <= x3 <= x4
0: 0
1: x1 x2 x3 x4
2: x1+x2 <= x1+x3 <= x2+x3 ? x1+x4 <= x2+x4 <= x3+x4
3: x1+x2+x3 <= x1+x2+x4 <= x1+x3+x4 <= x2+x3+x4
4: x1+x2+x3+x4
S = {x1,x2,x3,x4, x5}, x1 <= x2 <= x3 <= x4 <= x5
0: 0
1: x1 x2 x3 x4 x5
2: x1+x2 <= x1+x3 <= x2+x3 ? x1+x4 <= x2+x4 <= x3+x4 ? x1+x5 <= x2+x5 <= x3+x5 <= x4+x5
3: x1+x2+x3 <= x1+x2+x4 <= x1+x3+x4 <= x2+x3+x4 ? x1+x2+x5 <= x1+x3+x5 <= x2+x3+x5 ? x1+x4+x5 <= x2+x4+x5 <= x3+x4+x5
4: x1+x2+x3+x4 <= x1+x2+x3+x5 <= x1+x2+x4+x5 <= x1+x3+x4+x5 <= x2+x3+x4+x5
5: x1+x2+x3+x4+x5
And so on... One can add as many xi's as one wishes.
The question mark indicates a box boundary, where one doesn't know the order of x2+x3 and x1+x4 unless they are fixed to some values.
But what is known is that eg. for the |S|=4 case: x1+x2 <= x1+x4 and x2+x3 <= x3+x4, allowing one to find the right box.
The question is: did I miss something obvious in the analysis? Can this be generalized to some other kind of sets?
edit: Actually, it looks like what is found is a range of boxes instead of a single box. I mean, what prevents the target to fit between the min and max of every box other than the first and the last on a row, say?
So the updated question is: Is there a sub-exponential upper bound on the number of boxes to search?
• Try to prove that this algorithm always works. – Yuval Filmus Oct 26 '15 at 10:58
• I think that is the next step, yes. For what it's worth it depends on whether the property of min's and max's being in order holds. – user72935 Oct 26 '15 at 11:13
• Either it always works or it doesn't. If it doesn't always work, try to describe a large class of instances on which it does work. – Yuval Filmus Oct 26 '15 at 11:17
• – Craig Feinstein Oct 27 '15 at 17:38
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# Pythagoras' Nightmare
Project by ChuckV posted 10-07-2012 09:44 PM 20931 views 5 times favorited 26 comments
My 8 and 10 year-old boys are playing Fall Little League. They have two great coaches, and I want to give them something in thanks for all of their effort, time and patience. I decided to make boxes in the shape of home plate.
This boxes feature Paul’s (shipwright’s) wooden hinges. The carcase is black walnut and the top is ash, as are baseball bats. The finish is Danish Oil.
Here are all the pieces for one box:
This close-up of a top shows how it attaches to the rest of the lid (the top is upside down here):
I used rubber bands to provide the clamping pressure during the glue-up:
...
What does this have to do with Pythagoras? To determine the dimensions of the box lids, I used this sketch showing the official dimensions:
It is not possible to make a figure in this shape. If it were, the right triangle on the back of the plate would have sides 12, 12, 17. As Mr. P. knew, this means that 12×12 + 12×12 = 17×17, or 288 = 289. So, I guess that no baseball game has ever been played according to the rules!
The actual rule doesn’t explicitly say that the point on the plate has to be 90 degrees, but it does say that the two 12” edges must coincide with the first and third base lines, which implies 90 degrees. Here is the official wording:
Home base shall be marked by a five-sided slab of whitened rubber. It shall be a 17-inch square with two of the corners removed so that one edge is 17 inches long, two adjacent sides are 81/2 inches and the remaining two sides are 12 inches and set at an angle to make a point. It shall be set in the ground with the point at the intersection of the lines extending from home base to first base and to third base; with the 17-inch edge facing the pitcher’s plate, and the two 12-inch edges coinciding with the first and third base lines.
End of useless trivia for now.
-- “Big man, pig man, ha ha, charade you are.” ― R. Waters
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# Power set generation
## Introduction
As one of the first topics in introductory math courses on set theory, many of us are familiar with what a power set is.
A power set is the set of all the subsets of a set $$S$$. In this lesson, we will explore how we can generate the power set of a given set: this is a task we are frequently challenged to solve, directly as well as indirectly, as part of a coding interview question.
We will investigate two different approaches to solve this problem:
1. The first one derives straightforwardly from the recursive definition of power-set, which states "the power-set of an empty set is a set containing one element only: the empty set itself". For a non-empty set $$S$$, let $$e$$ be an element of $$S$$ and $$T$$ be the original set $$S$$ set minus $$e$$ ( $$T=S \setminus e$$), then the power set of $$S$$ is defined as the union of two distinct power sets:
2. The second approach is based on a bit-distribution property in the integers binary representation from $$0$$ to the size of the power set.
$\mathcal{P}(S)=\begin{cases} \{\{\}\} & \text{if } S=\{\} \\ \mathcal{P}\{T\} \bigcup \{t \bigcup \{e\} \: : t \in \mathcal{P}\{T\}\} \text{ where } T = S \setminus \{e\} \text{ } \forall e \in S, & \text{otherwise} \end{cases} \label{eq:power_set_recursive_definition}$
## Problem statement
Write a function that given a set $$S$$ returns its power set. The power-set of $$S$$ ($$\mathcal{P}(S)$$) is the set of all its subsets including the empty subset ($$\emptyset$$) and $$S$$ itself.
[ex:power_set:example1]
Given the set $$S=\{a,b,c\}$$, the following is a correct output for this problem: $\{\{\}, \{a\}, \{b\}, \{c\}, \{a,b\}, \{b,c\}, \{a,c\}, \{a,b,c\} \}$
## Clarification Questions
What is the maximum input size?
The maximum number of elements in $$S$$ is strictly less than $$32$$.
Are all the elements in the collection distinct?
No, the elements are not necessarily distinct. $$S$$ might contain duplicates.
Can the elements of the power-set appear in any order?
Yes, subsets can appear in any order. For example the following is also a valid output for the input shown in Example [ex:power_set:example1]: $$\{\{\}, \{b,c\}, \{a\}, \{a,b\}, \{a,b,c\}, \{b\}, \{a,c\}, \{c\} \}$$
## Discussion
First thing first, we should notice one key point: The power set of a collection of $$n$$ elements has size $$2^n$$. The proof is relatively easy, and it boils down to the fact that a subset of $$S$$ can be uniquely identified by a list $$X=\{x_0,x_1,\ldots x_{|S|-1}\}$$ of $$|S|$$ binary variables each carrying the information about whether $$S_i$$ is part of the subset; the variable $$x_i$$ is crucial to answer the question should $$S_i$$ be part of this subset?: If $$x_i$$ is true the answer is yes, otherwise, the answer it is no. We have two possible choices for every element of $$S$$ (either take it or not), then the total number of distinct $$X$$s is: $$2 \times 2 \times \ldots \times 2 = 2^{|S|}$$. Two choices for the first element, two for the second, and so on until the last element of $$S$$.
This, together with the constraint on $$|S|$$ ($$|S| < 32)$$ is a strong hint towards the fact that an exponential time and space solution is expected. After all, we are required to output all the elements of the power set, and thus the number of operations of an algorithm designed for this task cannot be less than the size of the power set itself.
### Bruteforce - Backtracking-like approach
The first solution presented in this lesson is based on that, during the generation of one of the power set’s elements, a decision has to be made for each element $$e$$ of $$S$$, on whether to include or not $$e$$ into the subset. Once we determine this, we are left with are $$|S|-1$$ decisions before we have created a valid subset of $$|S|$$.
This process is inherently recursive and is easily visualized with a tree (see the figure below): a node at level $$i$$ represents a decision for the $$i^{th}$$ element of $$S$$ and a path from the root to a leaf uniquely identifies a subset of $$S$$; after having traversed all the levels down to a leaf, $$n$$ decisions have been made: one for each of the elements of $$S$$.
Collectively, all the paths from the root to the leaves are the power set, and therefore, in order to solve this problem, we have to visit the entire tree.
A general way to deal with such type of problems is using a backtracking-like approach to try all possible decisions (or equivalently to visit every path from the root to a leaf).
The idea is that, for all elements of $$S$$, from the first to the last one, we are going to explore the two available possibilities: either take or exclude it from the subset.
We start by making a decision for the first element: From there, we continue to generate all possible subsets where the first decision is never changed. When there are no more subsets to generate, we *backtrack* and change our first decision and repeat the process of generating all possible subsets.
For instance, given $$S=\{1,2,3\}$$, we might start by deciding to use the element $$1$$, and include it in all possible subsets from the remaining elements $$\{2, 3\}$$ only. Once we are done with it, we can repeat the same process, only this time excluding $$1$$. What we do is: $$\mathcal{P}(S)= \{\{1\} \; \bigcup \;\mathcal{P}(\{2,3\})\} \: \bigcup \: \{\mathcal{P}(\{2,3\})\}\}$$
The proposed solution will incrementally construct one subset at a time, using an integer variable to keep track of which element we are currently taking the decision for. This type of problem is naturally solved recursively, with a base case of the recursion happening when there is no more decision to make, meaning that the current subset is ready to be included in the solution (it has been produced after $$n$$ decision steps).
Here below we can see how the C++ code implements this idea. The complexity of this solution is exponential i.e. $$O(2^{|S|})$$ which as already pointed out, is as good as it gets.
Listing 1: C++ to the power set generation using backtracking
void power_set_backtracking_helper(const std::vector<int> &S,
const int idx,
std::vector<int> &curr,
std::vector<std::vector<int>> &ans)
{
if (idx >= S.size())
// base case
{
ans.push_back(curr);
return;
}
// include element S[idx]
curr.push_back(S[idx]);
power_set_backtracking_helper(S, idx + 1, curr, ans);
// exclude element S[idx]
curr.pop_back();
power_set_backtracking_helper(S, idx + 1, curr, ans);
}
std::vector<std::vector<int>> power_set_backtracking(const std::vector<int> &S)
{
std::vector<std::vector<int>> ans;
std::vector<int> current;
power_set_backtracking_helper(S, 0, current, ans);
return ans;
}
Using a backtracking like approach is convenient because, once we notice that a problem can be solved by fully exploring the associated search space tree, we can immediately start writing the code and rely on our experience as backtracking expert writers to implement a correct solution. It is also concise and short when written in a recursive form (fewer chances to make mistakes, and less code to debug and explain), as well as well understood. The downside is that, if we decide to go for it, an iterative implementation can be a little harder and verbose to write.
Regardless of which type we decide to write, the interviewer is going to be pleased with the code provided: of course it is important to get to the final solution without making too many implementation mistakes, like forgetting to handle the base case.
### Bit Manipulation
We can use another approach to solve this problem, based on that the values of the bits of the numbers $$\{0,1,2,\ldots, s^n-1\}$$ already provide all the information necessary to decide whether to include or not an element from $$S$$ into a subset. The main idea is that the binary representation of all the numbers ($$2^{|S|}$$ of them) from $$0$$ to $$2^{|S|}-1$$ is the power set of $$n$$ bits. In a nutshell, this means that the binary representation of any of those numbers carries the necessary information that can be used to build one subset of $$\mathcal{P}(S)$$.
For instance, given the input $$S=\{a,b,c\}$$ the table below shows numbers from $$0$$ to $$2^3-1 = 7$$ and their binary representation (in the second column), as well as how the information about which bit is set can be used to construct one subset of $$\mathcal{P}(S)$$ (in the third column). When the $$i^{th}$$ bit is set (its value is $$1$$), it means that corresponding $$i^{th}$$ element of $$S$$ is chosen, while an unset bit (with value $$0$$) means it is excluded
This table shows a 1-to-1 mapping between integer values, their binary representation and an element of the power set.
Number Value Bits Subset
0 000 $$\{\}$$
1 001 $$\{c\}$$
2 010 $$\{b\}$$
3 011 $$\{b,c\}$$
4 100 $$\{a\}$$
5 101 $$\{a,c\}$$
6 110 $$\{a,b\}$$
7 111 $$\{a,b,c\}$$
This idea can be used to write an algorithm in which all the numbers in the range $$\{0,1,2,\ldots, 2^{|S|}-1\}$$ are considered and each of them is used to generate a subset of the final solution. Every number from this range maps uniquely to a subset of $$\mathcal{P}(S)$$.
It is not surprising when we think about the meaning of a bit in the binary representation of integers: One can "build" a number $$k$$ by summing up powers of $$2$$ where the bits contain the information about whether a certain power of two should be added to the final value. With $$n$$ bits one can represent $$2^n$$ numbers, each corresponding to one and only one subset of the power set of those $$n$$ bits. Listing [list:power_set_bits] shows a possible C++ implementation of the idea above.
Listing 2: Solution using bit manipulation.
constexpr inline bool is_bit_set(const int n, const unsigned p)
{
return (n >> p) & 1;
}
std::vector<std::vector<int>> power_set_bit_manipulation(
const std::vector<int> &A)
{
const size_t limit = (1ll << A.size()) - 1;
std::vector<std::vector<int>> PS;
PS.reserve(limit + 1);
for (int i = 0; i < limit; i++)
{
std::vector<int> subset = {};
for (int p = 0; p < 32; p++)
if (is_bit_set(i, p))
{
subset.push_back(A[p]);
}
PS.push_back(subset);
}
PS.push_back(A);
return PS;
}
The complexity of this function is, not surprisingly, $$O(2^{|S|})$$. We also pay a constant price of $$32$$ for each number we loop through since we need to inspect all of its bits. The proposed implementation assumes that the size of is $$4$$ bytes, which is true for most systems.
Moreover, notice the usage : it should be used in all those scenarios when we already know the final size of the collection we are building. This saves time because it avoids intermediate allocations and copies that must happen during the resize of the vector.
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Vous êtes sur la page 1sur 11
# Module 1
## Force And Friction
A force is a push or pull upon an object resulting from the
object's interaction with another object. Whenever there is
an interaction between two objects, there is a force upon
each of the objects. When the interaction ceases, the two
objects no longer experience the force. Forces only exist
as a result of an interaction.
Force is a quantity which is measured using the standard
metric unit known as the Newton.
## For simplicity sake, all forces (interactions) between
objects can be placed into two broad categories:
## contact forces, and
forces resulting from action-at-a-distance
## Contact forces are those types of forces which result
when the two interacting objects are perceived to be
physically contacting each other. Examples of contact
forces include frictional forces, tensional forces, normal
forces, air resistance forces, and applied forces.
## Action-at-a-distance forces are those types of forces
which result even when the two interacting objects are not in
physical contact with each other, yet are able to exert a push
or pull despite their physical separation. Examples of actionat-a-distance forces include gravitational forces. For
example, the sun and planets exert a gravitational pull on
each other despite their large spatial separation. Even when
your feet leave the earth and you are no longer in physical
contact with the earth, there is a gravitational pull between
you and the Earth. Electric forces are action-at-a-distance
forces. For example, the protons in the nucleus of an atom
and the electrons outside the nucleus experience an
electrical pull towards each other despite their small spatial
separation. And magnetic forces are action-at-a-distance forces.
Examples of contact and action-at-distance forces are listed in the table below.
Contact Forces
Frictional Force
Tension Force
Normal Force
Air Resistance Force
Applied Force
Spring Force
Action-at-a-Distance Forces
Gravitational Force
Electrical Force
Magnetic Force
Type of Force
Description of Force
(and Symbol)
Applied Force
Fapp
Gravity Force
(also known as Weight)
Fgrav
## An applied force is a force
which is applied to an
object by a person or
another object. If a person
is pushing a desk across
the room, then there is an
applied force acting upon
the object. The applied
force is the force exerted
on the desk by the person.
The force of gravity
is the force with
which the earth,
moon,
or
other
massively
large
object
attracts
another
object
towards itself. By
definition, this is the
weight of the object.
All objects upon
earth experience a
force
of
gravity
which is directed "downward" towards the center of the earth.
The force of gravity on earth is always equal to the weight of the
object as found by the equation:
Fgrav = m * g
where g = 9.8 m/s2 (on Earth) and m=mass in kg
Normal Force
Fnorm
Friction Force
Ffrict
## The normal force is the support force exerted upon an object
which is in contact with another
stable object. For example, if a
book is resting upon a surface,
then the surface is exerting an
upward force upon the book in
order to support the weight of the
book. On occasions, a normal
force is exerted horizontally between two objects which are in
contact with each other. For instance, if a person leans against
a wall, the wall pushes horizontally on the person.
## The friction force is the force exerted by a surface as an object
moves across it or makes an effort to move across it. There are
at least two types of
friction force - sliding
and
static
friction.
Thought it is not always
the cast, the friction
force often opposes the
motion of an object. For
example, if a book
slides
across
the
surface of a desk, then
the desk exerts a
friction force in the
opposite direction of its
motion.
Friction results from the two surfaces being pressed together
closely, causing intermolecular attractive forces between
molecules of different surfaces. As such, friction depends upon
the nature of the two surfaces and upon the degree to which
they are pressed together.
The maximum amount of friction force which a surface can exert
upon an object can be calculated using the formula below:
Fair
Tension force
Spring Force
Fspring
## The air resistance is a special
type of frictional force which acts
upon objects as they travel
through the air. The force of air
resistance is often observed to
oppose the motion of an object.
This force will frequently be
neglected due to its negligible
magnitude (and due to the fact
that it is mathematically difficult to
predict its value). It is most
noticeable for objects which travel
at high speeds (e.g., a skydiver or a downhill skier) or for
objects with large surface areas.
## The tension force is the force
which is transmitted through a
string, rope, cable or wire when
it is pulled tight by forces acting
from opposite ends. The
tension force is directed along
the length of the wire and pulls
equally on the objects on the
opposite ends of the wire.
## The spring force is the force
exerted by a compressed or
stretched spring upon any
object which is attached to it.
An object which compresses
or stretches a spring is
always acted upon by a force
which restores the object to
its rest or equilibrium position. For most springs (specifically, for
those which are said to obey "Hooke's Law"), the magnitude of
the force is directly proportional to the amount of stretch or
compression of the spring.
FRICTION
Friction is a force that is created whenever two surfaces
move or try to move across each other.
## Friction always opposes the motion or attempted
motion of one surface across another surface.
Friction is dependant on the texture of both
surfaces.
Friction is also dependant on the amount of
contact force pushing the two surfaces together
(normal force).
There are advantages and disadvantages of friction. Since friction is a resistance force
that slows down or prevents motion, it is necessary in many applications to prevent
slipping or sliding. But also, it can be a nuisance because it can hinder motion and cause
the need for expending energy. A good compromise is necessary to get just enough
friction.
Questions you may have include:
## How is friction necessary?
How is friction a nuisance?
What is a good compromise for using friction?
Important uses
In some situations, friction is very important and beneficial. There are many things that
you could not do without the force of friction.
Walking
You could not walk without the friction between your shoes
and the ground. As you try to step forward, you push your
foot backward. Friction holds your shoe to the ground,
allowing you to walk. Consider how difficult it is to walk on
slippery ice, where there is little friction. Bear did not heed
warning sign
Writing
Writing with a pencil requires friction. You could not hold a pencil in your hand without
friction. It would slip out when you tried to hold it to write. The graphite pencil led would
not make a mark on the paper without friction.
A pencil eraser uses friction to rub off mistakes written in pencil lead. Rubbing the eraser
on the lead wears out the eraser due to friction, while the particles worn off gather up the
Driving car
Your car would not start moving if it wasn't for the
friction of the tires against the street. With no friction,
the tires would just spin. Likewise, you could not stop
without the friction of the brakes and the tires.
Brake
help
produce
friction
to slow
you
down.
## Friction can cause problems or be a nuisance that you try to minimize.
Makes movement difficult
Any time you want to move an object, friction can make the job more difficult. Excess
friction can make it difficult to slide a box across the floor, ride a bicycle or walk through
deep snow.
An automobile would not move forward very well unless its friction was not reduced. Oil
is needed to lubricate the engine and allow its parts to move easily. Oil and ball bearings
are also used in the wheels, so they will turn with little friction
Wastes energy
In any type of vehicle--such as a car, boat or airplane--excess friction means that extra
fuel must be used to power the vehicle. In other words, fuel or energy is being wasted
because of the friction.
Fluid friction or air resistance can greatly reduce the gas mileage in an automobile. Cars
are streamlined to reduce friction. But driving at highway speeds with your windows
open can create enough drag on the car to greatly reduce your gas mileage.
Heats parts
The Law of Conservation of Energy states that the amount of energy remains constant.
Thus, the energy that is "lost" to friction in trying to move an object is really turned to
heat energy. The friction of parts rubbing together creates heat.
You've seen how people will try to start a fire by vigorously rubbing two sticks together.
Or perhaps you've seen an automobile spin its wheels so much that the tires start to
smoke. These are examples of friction creating heat energy. Just rub your hands
together to create the same effect.
Besides the problem of losing energy to heat, there is also the threat of a part
overheating due to friction. This can cause damage to a machine.
Wears things out
Any device that has moving parts can wear out
rapidly due to friction. Lubrication is used not
only to allow parts to move easier but also to
prevent them from wearing out. Some other
examples of materials wearing out due to friction
include the soles of your shoes and a pencil
eraser.
Friction
can
cause
problems. When objects
rub against each other,
the surfaces are worn
away. Friction with the
ground
causes
the
heels and soles of your
shoes to wear away.
Causes of Friction
The causes of the resistive force of friction are molecular adhesion, surface roughness,
and the plowing effect.
Adhesion is the molecular force resulting when two materials are brought into close
contact with each other. Trying to slide objects against each other requires breaking
these adhesive bonds. For years, scientists thought that friction was caused by surface
roughness, but recent studies have shown that it is actually a result of adhesive forces
between the materials.
When two objects are brought into contact, many
atoms or molecules from one object are in such close
proximity to those in the other object that molecular or
electromagnetic forces attract the molecules of the
two materials together. This force is called adhesion.
Trying to slide one object across the other requires
essence of friction.
You've seen a water drop adhere to a window pane.
The force of friction prevents this liquid from sliding
down the solid material. But most cases of friction you
see concern a solid object sliding or moving against
another solid.
Sliding objects against each other requires breaking these millions of contact points
where the adhesion force takes effect, only to result in millions of new contact points of
Surface roughness
All solid materials have some degree of surface
roughness. If you looked at what seems to be a smooth
surface under a high-powered microscope, you would
see bumps, hills and valleys that could interfere with
sliding motion.
At one time it was thought that the surface roughness of
materials was the cause for friction. In reality, it only has
a small effect on friction for most materials.
If the surfaces of two hard solids are extremely rough, the
high points or asperities can interfere with sliding and cause friction because of the
abrasion or wear that can take place when you slide one object against the other. This is
the "sandpaper effect" where particles of the materials are dislodged from their surfaces.
In such a case, the friction is caused by surface roughness, although the adhesion effect
still plays a part in the abrasion.
Deformations
Soft materials will deform when under pressure. This
also increased the resistance to motion. For example,
when you stand on a rug, you sink in slightly, which
causes resistance when you try to drag your feet along
the rug's surface. Another example is how rubber tires
flatten out at the area on contact with the road. When
materials deform, you must "plow" through to move, thus
creating a resistive force. When the deformation
becomes large, such that one object sinks into the other, streamlining can affect the
friction, similar to what happens in fluid friction.
Types of friction
The classic law of friction states that friction is the product of a coefficient and a force.
There are two main types of friction: Static and Kinetic
## Static and Kinetic Friction
Friction is a key concept when you are attempting to understand car accidents. The
force of friction is a force that resists motion when two objects are in contact. If you look
at the surfaces of all objects, there are tiny bumps and ridges. Those microscopic peaks
and valleys catch on one another when two objects are moving past each other.
This explanation is a little simplified. There are other processes at work, including
chemical bonding and electrical interactions.
The level of friction that different materials exhibit is measured by the coefficient of
friction. The formula is = f / N, where is the coefficient of friction, f is the amount of
force that resists motion, and N is the normal force. Normal force is the force at which
one surface is being pushed into another. If a rock that weighs 50 newtons is lying on
the ground, then the normal force is that 50 newtons of force. The higher is, the more
force resists motion if two objects are sliding past each other.
Static friction
Static friction is friction between two solid objects that are not moving relative to each
other. For example, static friction can prevent an object from sliding down a sloped
surface. The coefficient of static friction, typically denoted as s, is usually higher than
the coefficient of kinetic friction.
The static friction force must be overcome by an applied force before an object can
move. The maximum possible friction force between two surfaces before sliding begins
is the product of the coefficient of static friction and the normal force . When there is no
sliding occurring, the friction force can have any value from zero up to a maximum value
determined by the surfaces and the weight of the object. Any force smaller than
attempting to slide one surface over the other is opposed by a frictional force of equal
magnitude and opposite direction. Any force larger than overcomes the force of static
friction and causes sliding to occur. The instant sliding occurs, static friction is no longer
applicable and kinetic friction becomes applicable.
An example of static friction is the force that prevents a car wheel from slipping as it rolls
on the ground. Even though the wheel is in motion, the patch of the tire in contact with
the ground is stationary relative to the ground, so it is static rather than kinetic friction.
The maximum value of static friction, when motion is impending, is sometimes referred
to as limiting friction, although this term is not used universally.
Kinetic friction
Kinetic (or dynamic) friction occurs when two objects are moving relative to each other
and rub together (like a sled on the ground). The coefficient of kinetic friction is typically
denoted as k, and is usually less than the coefficient of static friction for the same
materials.
Kinetic friction is when two objects are rubbing against each other. Putting a book flat on
a desk and moving it around is an example of kinetic friction.
The graphical representation of static and kinetic friction is shown below.
Surfaces
(static)
(kinetic)
Steel on steel
0.74
0.57
Glass on glass
0.94
0.40
Metal on Metal
0.15
(lubricated)
0.06
Ice on ice
0.10
0.03
0.04
0.80
## Tire on wet road 0.60
0.40
10
Tire on snow
0.30
0.20
Compromise
A compromise is needed between too much friction and not enough.
For example, if you wanted to slide a heavy box across the floor, you would want to
reduce the friction between the box and the floor, so that it would be easy to move.
Lubrication of some sort is often a way to reduce friction.
But you would also want to increase the friction of your shoes on the floor, so that you
would be able to get good traction and be able to push effectively. Soles made of rubber
material that include treads can reduce slipping when walking or running.
Reducing Friction
A common way to reduce friction is by using a lubricant, such as oil, that is placed
between the two surfaces, often dramatically lessening the coefficient of friction. The
science of friction and lubrication is called teratology. Super lubricity, a recentlydiscovered effect, has been observed in graphite: it is the substantial decrease of friction
between two sliding objects, approaching zero levels - a very small amount of frictional
energy would be dissipated due to electronic and/or atomic vibrations. Lubricants to
overcome friction need not always be thin, turbulent fluids or powdery solids such as
graphite and talc; acoustic lubrication actually uses sound as a lubricant.
## Figure: Block dragged over a horizontal surface
Figure: Block sliding down an incline
11
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Lecture-12
# Lecture-12 - Lecture 12: Changing coordinates and...
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Unformatted text preview: Lecture 12: Changing coordinates and transforming graphs June 17, 2010 Introduction Suppose ( x,y ) is a statement in which the variables x and y appear. For example, ( x,y ) might be the statement x 2 + y 2 = 35 2 . It could also be a more complicated statement, such as: x 2 + y 2 = 35 2 and y = 2 3 ( x + 1) , or x 2 + y 2 = 35 2 and x and y are both rational . No matter how complicated ( x,y ) is, by the graph of ( x,y ), we mean the set of all number pairs for which ( x,y ) is true. If we interpret number pairs as points by referring to the standard x- y-coordinate system on the plane, then the graph of ( x,y ) is a set of points in the plane. 1 We will use the symbol X to denote the graph: X := { ( x,y ) | ( x,y ) } . Now, suppose a and b are any numbers. Consider the set X := { ( x,y ) | ( x- a,y- b ) } . I assert that X coincides with the set of all pairs obtained by taking a pair belonging to X and adding ( a,b ) to it. In other words, X is the image of X after a translation by ( a,b ). We can demonstrate this assertion as follows: ( ( x + a ) , ( y + b ) ) X ( ( x + a )- a, ( y + b )- b ) ( x ,y ) ( x ,y ) X Returning to the example of the circle,...
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# Truth table
Truth table
A truth table is a mathematical table used in logic — specifically in connection with Boolean algebra, boolean functions, and propositional calculus — to compute the functional values of logical expressions on each of their functional arguments, that is, on each combination of values taken by their logical variables. In particular, truth tables can be used to tell whether a propositional expression is true for all legitimate input values, that is, logically valid.
Overview
The pattern of reasoning that the truth table tabulates was Frege's, Peirce's, and Schröder's by 1880. The tables have been prominent in literature since 1920 (Lukasiewicz, Post, Wittgenstein)" (Quine, 39). Lewis Carroll had formulated truth tables as early as 1894 to solve certain problems, but his manuscripts containing his work on the subject were not discovered until 1977 [http://www-groups.dcs.st-andrews.ac.uk/~history/Biographies/Dodgson.html] . Wittgenstein's "Tractatus Logico-Philosophicus" uses them to place truth functions in a series. The wide influence of this work led to the spread of the use of truth tables.
Truth tables are used to compute the values of propositional expressions in an effective manner that is sometimes referred to as a "decision procedure". A propositional expression is either an atomic formula — a propositional constant, propositional variable, or propositional function term (for example, "Px" or "P"("x")) — or built up from atomic formulas by means of logical operators, for example, AND ($land$), OR ($lor$), NOT ($lnot$). For instance, $Fx land Gx$ is a propositional expression.
The column headings on a truth table show (i) the propositional functions and/or variables, and (ii) the truth-functional expression built up from those propositional functions or variables and operators. The rows show each possible valuation of T or F assignments to (i) and (ii). In other words, each row is a distinct interpretation of (i) and (ii).
Truth tables for classical logic are limited to Boolean logical systems in which only two logical values are possible, false and true, usually written F and T, or sometimes 0 or 1, respectively.
Logical operations
Logical negation
Logical negation is an operation on one logical value, typically the value of a proposition, that produces a value of "true" if its operand is false and a value of "false" if its operand is true.
The truth table for NOT p (also written as ~p or ¬p) is as follows:
Stated in English, if "p", then "p" ∨ "q" is "p", otherwise "p" ∨ "q" is "q".
Logical implication
Logical implication and the material conditional are both associated with an operation on two logical values, typically the values of two propositions, that produces a value of "false" just in the singular case the first operand is true and the second operand is false.
The truth table associated with the material conditional not p or q (symbolized as p → q) and the logical implication p implies q (symbolized as p ⇒ q) is as follows:
For two propositions, XOR can also be written as (p = 1 ∧ q = 0)∨ (p = 0 ∧ q = 1).
Logical NAND
The logical NAND is an operation on two logical values, typically the values of two propositions, that produces a value of "false" if and only if both of its operands are true. In other words, it produces a value of "true" if and only if at least one of its operands is false.
The truth table for p NAND q (also written as p | q or p ↑ q) is as follows:
The negation of disjunction and the conjunction of negations $eg p and eg q$ are tabulated as follows:
Key::T = true, F = false:$and$ = AND (logical conjunction):$or$ = OR (logical disjunction):$underline\left\{or\right\}$ = XOR (exclusive or) :$underline\left\{and\right\}$ = XNOR (exclusive nor):$ightarrow$ = conditional "if-then":$leftarrow$ = conditional "(then)-if"
:$iff$ biconditional or "if-and-only-if" is logically equivalent to $underline\left\{and\right\}$: XNOR (exclusive nor).
Johnston diagrams, similar to Venn diagrams and Euler diagrams, provide a way of visualizing truth tables. An interactive Johnston diagram illustrating truth tables is at [http://logictutorial.com LogicTutorial.com]
Condensed truth tables for binary operators
For binary operators, a condensed form of truth table is also used, where the row headings and the column headings specify the operands and the table cells specify the result. For example Boolean logic uses this condensed truth table notation:
This notation is useful especially if the operations are commutative, although one can additionally specify that the rows are the first operand and the columns are the second operand. This condensed notation is particularly useful in discussing multi-valued extensions of logic, as it significantly cuts down on combinatoric explosion of the number of rows otherwise needed. It also provides for quickly recognizable characteristic "shape" of the distribution of the values in the table which can assist the reader in grasping the rules more quickly.
Truth tables in digital logic
Truth tables are also used to specify the functionality of hardware look-up tables (LUTs) in digital logic circuitry. For an n-input LUT, the truth table will have 2^"n" values (or rows in the above tabular format), completely specifying a boolean function for the LUT. By representing each boolean value as a bit in a binary number, truth table values can be efficiently encoded as integer values in electronic design automation (EDA) software. For example, a 32-bit integer can encode the truth table for a LUT with up to 5 inputs.
When using an integer representation of a truth table, the output value of the LUT can be obtained by calculating a bit index "k" based on the input values of the LUT, in which case the LUT's output value is the "k"th bit of the integer. For example, to evaluate the output value of a LUT given an array of "n" boolean input values, the bit index of the truth table's output value can be computed as follows: if the "i"th input is true, let V"i" = 1, else let V"i" = 0. Then the "k"th bit of the binary representation of the truth table is the LUT's output value, where "k" = V0*2^0 + V1*2^1 + V2*2^2 + ... + V"n"*2^"n".
Truth tables are a simple and straightforward way to encode boolean functions, however given the exponential growth in size as the number of inputs increase, they are not suitable for functions with a large number of inputs. Other representations which are more memory efficient are text equations and binary decision diagrams.
Applications of truth tables in digital electronics
In digital electronics (and computer science, fields of engineering derived from applied logic and math), truth tables can be used to reduce basic boolean operations to simple correlations of inputs to outputs, without the use of logic gates or code. For example, a binary addition can be represented with the truth table:
A B | C R1 1 | 1 01 0 | 0 10 1 | 0 10 0 | 0 0
where
A = First OperandB = Second OperandC = Carry R = Result
This truth table is read left to right:
* Value pair (A,B) equals value pair (C,R).
* Or for this example, A plus B equal result R, with the Carry C.
Note that this table does not describe the logic operations necessary to implement this operation, rather it simply specifies the function of inputs to output values.
In this case it can only be used for very simple inputs and outputs, such as 1's and 0's, however if the number of types of values one can have on the inputs increases, the size of the truth table will increase.
For instance, in an addition operation, one needs two operands, A and B. Each can have one of two values, zero or one. The number of combinations of these two values is 2x2, or four. So the result is four possible outputs of C and R. If one was to use base 3, the size would increase to 3x3, or nine possible outputs.
The first "addition" example above is called a half-adder. A full-adder is when the carry from the previous operation is provided as input to the next adder. Thus, a truth table of eight rows would be needed to describe a full adder's logic:
A B C* | C R 0 0 0 | 0 0 0 1 0 | 0 11 0 0 | 0 11 1 0 | 1 00 0 1 | 0 10 1 1 | 1 01 0 1 | 1 01 1 1 | 1 1
Same as previous, but.. C* = Carry from previous adder
ee also
;Basic logical operators
* Exclusive disjunction
* Logical conjunction
* Logical disjunction
* Logical equality
* Logical implication
* Logical NAND
* Logical NOR
* Negation (Inverter);Related topics
* Ampheck
* Boolean algebra (logic)
* Boolean algebra topics
* Boolean domain
* Boolean function
* Boolean-valued function
* Espresso heuristic logic minimizer
* First-order logic
* Karnaugh maps
* Logical connective
* Logical graph
* Logical value
* Minimal negation operator
* Operation
* Parametric operator
* Propositional calculus
* Sole sufficient operator
References
* Quine, W.V. (1982), "Methods of Logic", 4th edition, Harvard University Press, Cambridge, MA.
* [http://educoteca.org/scripts/ttab/ Online Truth Table Generator]
* [http://www-cs-students.stanford.edu/~silver/truth/ Web-based truth table generator]
* [http://logik.phl.univie.ac.at/~chris/gateway/formular-uk-zentral.html Powerful logic engine]
* [http://www.stephan-brumme.com/programming/Joole/ Boolean expression evaluator, generates truth table (Java applet)]
* [http://pico1.e.ft.fontys.nl/publicad.html Free logic minimization program Minilog]
* [http://svn.oriontransfer.org/TruthTable/ Samuel Williams' Truth Table Evaluator]
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• truth table — n. 1. a table showing all the possible combinations of the variables in an expression in symbolic logic and their resulting truth or falsity 2. a similar table showing relationships between input to and output from a computer circuit … English World dictionary
• Truth Table — [engl.], Wahrheitstabelle … Universal-Lexikon
• truth table — teisingumo lentelė statusas T sritis automatika atitikmenys: angl. Boolean operation table; truth diagram; truth table vok. Wahrheitstabelle, f rus. таблица истинности, f pranc. table de vérité, f … Automatikos terminų žodynas
• truth table — teisingumo lentelė statusas T sritis informatika apibrėžtis Lentelė, kurioje surašytos visos galimos ↑loginio reiškinio kintamųjų ir jas atitinkančios reiškinio reikšmės. Plačiau žr. priede. priedas( ai) MS Word formatas atitikmenys: angl. truth… … Enciklopedinis kompiuterijos žodynas
• truth table — truth′ ta ble n. cmp math. logic pho a table that gives the truth values of a compound logical statement for every possible combination of truth values of its component propositions • Etymology: 1935–40 … From formal English to slang
• truth table — table that presents the values for which a function receives true values (Computers) … English contemporary dictionary
• truth table — noun Etymology: translation of German wahrheitstafel : a table that lists underneath one or more truth functions the various truth values of the truth functions for given truth values of their arguments … Useful english dictionary
• truth table — noun Date: 1921 a table that shows the truth value of a compound statement for every truth value of its component statements; also a similar table (as for a computer logic circuit) showing the value of the output for each value of each input … New Collegiate Dictionary
• truth table — Logic, Math., Computers. a table that gives the truth values of a compound sentence formed from component sentences by means of logical connectives, as AND, NOT, or OR, for every possible combination of truth values of the component sentences.… … Universalium
• truth table — noun a table showing all possible truth values for an expression, derived from the truth values of its components … Wiktionary
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# Top Choices of Multiplication Words
Tuesday, February 19th 2019. | chart
order of operations practice
Column Titles could be observed from the other side of the surface of a worksheet. Words are most often elongated by a few letters in a minute. It will create a new equation field, in which you’ll be able to form the algebraic expression. It’s accurate, you find the word’complete’ in the issue. You have three manners in which to observe the term polynomials with fractions. It’s far better keep it easy, and repetition is frequently the best way to learn any multiplication table.
Top Choices of Multiplication Words
Your Son or daughter is going to have the ability to learn how to exercise their mind using logic that is a beneficial tool in mathematics. Children should understand the lesson objective and be conscious of your rules for behaviour. During the time you’re composing, or requesting your child to write, paragraphs and equations on the whiteboard, do not neglect to draw pictures to collaborate with the issues.
For Many individuals, math is a remarkably catchy subject, and a great deal of teachers aren’t able to give pupils the one time help they might need to be able to master math. Arithmetic is math in its most basic, and it entails four surgeries that most people use almost daily.
Exponents are utilized in a Large Number of scenarios in real life, Including measurements, computer programming and an assortment of careers. The exponent lets you know just how many times to multiply the number. A logarithm is the assortment of times you have to multiply a number by itself to discover another number. Another means to say it’s that a logarithm is your capacity to which a certain number referred to as the foundation needs to be raised to find another number.
If students Seem to be confused about what’s being asked in the Issue, explain the matter again using different words. They often resist word problems, but they’re in fact the ideal way to introduce abstract concepts, like the significance of the division symbol. Students in fifth grade do nearly all their computations employing a pencil and paper, since they’ve done since they started solving math issues. Third grade students normally have a simple grasp on multiplication until they start learning about division.
Multiplication Words – the Conspiracy
Repeat The specific same step for every symbol you require, or paste the exact first branch emblem. For example, you need to be aware of the significance of a factor, which is a letter that functions as a placeholder for an unknown number. The notion of a logarithm is easy, but it is somewhat difficult to put into words.
What You Need to Know About Multiplication Words
To Utilize exponents, you will need to be knowledgeable about easy exponent laws. The action of composing the issues and answers is another sort of repetition. The capacity to build a multiplication sentence goes beyond the classroom, by preparing pupils to compute huge quantities of items. The power is known as the argument of the logarithm. Anything increased to the very first electricity stays the exact same. Speed may not be constant, however, and you might wish to be knowledgeable about rate at a specific point in time.
The Fight Against Multiplication Words
Replace X with 4000 and solve both sides of the equation to be certain it balances. A probability of 0 means there aren’t any probable outcomes for that purpose. The end result is the reply to the problem of the bigger number divided by the smaller amount. The process for multiplication needs to be reflected in the local church.
What Everyone Dislikes About Multiplication Words and Why
If A problem is to the board or presented on a screen, pupils should Additionally possess a copy . The problem should no longer Include any words when you’re finished, but nevertheless, it should Resemble a normal math problem with just numbers. Begin by determining The situation it wants you to solve. Since it mentions the small Sister dropped two pairs, the final answer must be less than 23. Each word Problem may require a different format, but a visual representation of The crucial info makes it less difficult to operate with. Word Problems use a number of terms for the four standard surgeries. Word issues, mathematics problems written in phrases Instead of math symbols, Don’t need to be the giant monster that your buddies tell you they are.
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9 25 as a percent
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A128783 Numbers whose square is a nontrivial concatenation of other squares. 2
7, 10, 12, 13, 19, 20, 21, 30, 35, 37, 38, 40, 41, 44, 50, 57, 60, 65, 70, 80, 90, 95, 97, 100, 102, 105, 107, 108, 110, 112, 119, 120, 121, 125, 129, 130, 138, 140, 150, 160, 170, 180, 190, 191, 200, 201, 204, 205, 209, 210, 212, 220, 223, 230, 240, 250 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS a(n) = sqrt(A019547(n)); the sequence A019547 gives the squares themselves and this sequence gives the numbers whose squares yield the numbers in A019547. All multiples of 10 are included, because for any value of x, (x*10)^2 = x^2*100, which is x^2 followed by two zeros. Every digit except 6 is used as the last digit of a number in this sequence before 50. 6 is not used as the last digit of a number in this sequence until 306. LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 EXAMPLE 13 is included because 13^2 = 169, which includes 16 and 9, two perfect squares. PROG (Haskell) a128783 n = a128783_list !! (n-1) a128783_list = filter (f . show . (^ 2)) [1..] where f zs = g (init \$ tail \$ inits zs) (tail \$ init \$ tails zs) g (xs:xss) (ys:yss) | a010052 (read xs) == 1 = a010052 (read ys) == 1 || f ys || g xss yss | otherwise = g xss yss g _ _ = False -- Reinhard Zumkeller, Oct 11 2011 CROSSREFS Cf. A019547. Cf. A010052, A000290; A048653 is a subsequence. Sequence in context: A048792 A024585 A190482 * A024586 A050419 A089373 Adjacent sequences: A128780 A128781 A128782 * A128784 A128785 A128786 KEYWORD base,nonn AUTHOR Jonathan R. Love (japanada11(AT)yahoo.ca), Apr 07 2007 EXTENSIONS Missing terms 205 and 209 inserted by Reinhard Zumkeller, Oct 11 2011 STATUS approved
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Last modified October 19 21:17 EDT 2019. Contains 328229 sequences. (Running on oeis4.)
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# A resistor $R$ and the capacitor $C$ are connected in series across an ac source of rms voltage $5V$ if the rms voltage across $C$ is $3V$ then that across $R$ is .
A
$1V1$
B
$2V$
C
$3V$
D
$4V$
Video Solution
Text Solution
Verified by Experts
## $E=\sqrt{{E}_{R}^{2}+{E}_{C}^{2}}$
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## A resistor an inductor and a capacitor are connected in series to an ac source An ac voltmeter measures the votage across them as 800V,30V and 90V respectively The rms value of the supply voltage is .
A100V
B1002V
C200V
D2002V
• Question 2 - Select One
## A resistance of 200Ω and capacitor of 15μF are connected in series to a 220 V, 50 Hz AC source. The voltage (rms) across the resistor and capacitor is that
A151 V, 160.4 V
B150 V, 100.3 V
C220 V, 91.8 V
D145 V, 311.3 V
• Question 3 - Select One
## A 50HzAC source of 20V is connected across R and C as shown in figureure. The voltage across R is 12V. The voltage across C is
A8V
B16V
C10V
Dnot possible to determine unless value of R and C are given
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Estimation Station
Estimation Station
Item # 4820
Ages 5-9
We regret that this Educational Insights product is discontinued and no longer available.
No more wild guesses! The Estimation Station is the perfect way to teach your students to make logical, mathematical estimates. Simply send this jar home with a student to be filled and counted. Then fill the included proportional scoop and share that number with your class. Students use that information to formulate educated estimations about the number of pieces in the jar. Clearer and easier than home-made jars with hard-to-view contents.
• Includes 4" plastic cube jar with locking lid and sturdy base and 2" plastic cube scoop
• Activity guide with a week’s worth of activities including estimation, counting, place value, comparing and ordering numbers, making predictions, graphing, statistics, and mathematical reasoning
• Parent letter (in English and Spanish) with instructions on filling and counting the jar's contents
• Teaches estimation and proportionality
•
Product Instructions
View now:
Estimation Station
Aligned to:
• K.CC.1. Count to 100 by ones and by tens.
• K.CC.4. Understand the relationship between numbers and quantities; connect counting to cardinality.
• K.CC.6. Identify whether the number of objects in one group is greater than, less than, or equal to the number of objects in another group, e.g., by using matching and counting strategies .
• K.MD.2. Directly compare two objects with a measurable attribute in common, to see which object has “more of”/“less of” the attribute, and describe the difference.
• 1.NBT.1. Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral.
• 5.MD.3. Recognize volume as an attribute of solid figures and understand concepts of volume measurement.
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https://www.physicsforums.com/threads/open-set-and-open-ball.738108/
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# Homework Help: Open set and open ball
1. Feb 13, 2014
### sigh1342
The problem statement, all variables and given/known data
Show that S is open if and only if ∀x ∈ S, ∃ a open ball B(x; r)(r > 0) such that B(x; r) ⊂ S
And what we have is , let X be normed space, S ⊂X , Then S is close if and only if $$∀{x_{n}}⊂X, s.t. x_{n}->x \in X$$, x∈ S. A set S is open if and only if X\S is close.
The attempt at a solution
I try to use following approach.
"=>" since S is open, there exist some sequence s.t. its limit is not in S(otherwise X/S is not close), then for any x, we can construct the convergent sequence such that its limit is not in S.(I don't know it is true or not). So by the limit definition. ∃ a open ball B(x; r)(r > 0) such that B(x; r) ⊂ S.
"<=" I try to prove X\S is close, so that X is open. $$∀{x_{n}}⊂X, s.t. x_{n}->x \in X$$ , x should be in X\S, but I have no idea to prove it.um.. Can anyone give me some ideas? :)
2. Feb 13, 2014
### pasmith
I'm not sure this even makes sense.
I would prove this by contrapositive: Assume that there exists an $x \in S$ such that for all $r > 0$, $B(x,r) \not\subset S$, and show that it follows that $X \setminus S$ is not closed.
Your assumption is that for all $x \in S$ there exists $r > 0$ such that $B(x,r) \subset S$. Why does it follow that $x \in S$ cannot be the limit of any sequence in $X \setminus S$?
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https://mathspace.co/textbooks/syllabuses/Syllabus-440/topics/Topic-17555/subtopics/Subtopic-105002/?activeTab=theory
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# Dividing a quantity into a given ratio
Lesson
In Looking at Relationships Between Different Groups, we learnt that ratios can be used to divide up things such as quantities of money, weights and measurements. To work out the total number of parts in a ratio, we add up all the individual parts in that ratio. Look at the picture below:
In this picture, we have $1$1 blue dot to $3$3 green dots, making a total of $4$4 dots. If we wanted to write the ratio of blue to green dots, we would write $1:3$1:3. To work out the total number of parts in the ratio, we would find the sum of all the parts. In this case $1+3=4$1+3=4.
#### Examples
##### Question 1
How many parts are in the ratio $22:15$22:15?
To work this out, we do $22+15$22+15. So there are $37$37 parts in total.
It doesn't matter how many parts there are to the ratio, we just keep adding them up to get the total number of parts.
##### Question 2
What is the total number of parts in the ratio $12:3:7:11$12:3:7:11 ?
$12+3+7+11=33$12+3+7+11=33
That means the total number of parts is $33$33.
## Dividing by a given ratio
Once you can calculate the total number of parts, we can use it to divide up quantities in a given ratio.
#### Examples
##### Question 3
$25.9$25.9 is divided up into two parts, $A$A and $B$B, in the ratio $5:2$5:2. How much is each part worth?
a) What is the value of A?
b) What is the value of B?
##### Question 4
Neil and Dave share $\$7777 in the ratio $5:2$5:2.
a) What fraction of the total amount to be shared does Dave receive?
b) Therefore how much money does Dave receive?
Here's another example that combines a few different math concepts that you've learnt so far.
##### Question 5
The perimeter of a rectangle is $198$198 and the ratio of length to width is $6:5$6:5.
a) Given that the length of the rectangle is $x$x, write an expression for the width in terms of $x$x.
b) Write an expression for the perimeter of the rectangle of $x$x and simplify that expression.
c) Hence, find $x$x, the length of the rectangle.
d) Hence find the width of the rectangle.
### Outcomes
#### 9P.NA1.05
Solve problems involving ratios, rates, and directly proportional relationships in various contexts (e.g., currency conversions, scale drawings, measurement), using a variety of methods (e.g., using algebraic 15 20 x 4 reasoning, equivalent ratios, a constant of proportionality; using dynamic geometry software to construct and measure scale drawings)
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+0
# Hello May I please have assitence
+1
281
3
+428
Traveling at 55 miles per hour, how many minutes, rounded to the nearest whole number, does it take to drive 310 miles from Boston to Syracuse?
These are my options:
A. 6
B. 338
C. 365
D 3,300
I'm having trouble on figuring out where to start...
May 31, 2019
#1
+8962
+4
Note that $$\frac{\text{55 miles}}{\text{1 hour}}\ =\ \frac{\text{55 miles}}{\text{60 minutes}}$$
Let x be the number of minutes it takes.
$$\frac{\text{55 miles}}{\text{60 minutes}}\cdot x\text{ minutes}\ =\ 310\text{ miles}\\~\\ x\text{ minutes}\ =\ 310\text{ miles}\cdot\frac {\text{60 minutes}}{\text{55 miles}}\\~\\ x\text{ minutes}\ =\ 310\cdot\frac {60}{55}\text{ minutes}\\~\\ x\text{ minutes}\ =\ \frac{18600}{55}\text{ minutes}\\~\\ x\text{ minutes}\ \approx\ 338\text{ minutes} \\~\\ x\ \approx\ 338$$
__
May 31, 2019
#2
+1015
-3
Nice Job!
Nickolas May 31, 2019
#3
+428
-2
Thank you so much I really apricate that.
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# Thread: C program - inverse of a matrix
1. ## C program - inverse of a matrix
hi, this is my first time on this board, so i apologise in advanced if i broke any rules or whatever...
so basically, im supposed to use C to find theinverse of a matrix of whose size and values are entered by the user. I am supposed to do so using the following steps:
Step 1: Set up an n × n matrix B and initialize it to the n × n identity matrix.
Step 2: In the matrix A, interchange the top row with the nearest row below it, if necessary,
to bring a nonzero entry to the row 1, column 1 position. (This is a Row Interchange
elementary row operation.) Apply the same Row Interchange elementary
row operation to the matrix B.
Step 3: In the matrix A, if the entry in the row 1, column 1 position is a, multiply the first
row by 1/a in order to make the entry in the row 1, column 1 position have the value
1. (This is a Row Scaling elementary row operation.) Apply the same Row Scaling
elementary row operation to the matrix B.
Step 4: In the matrix A, add suitable multiples of the top row to the rows below so that
all entries below the 1 in the row 1, column 1 position become zero. (These are Row
row operations to the matrix B.
Step 5: Repeat Steps 2 through 4 to make each diagonal entry Ai,i have the value 1, for
i = 2, 3, 4, . . . , n, and each off-diagonal entryAi,j have the value 0, i, j = 2, 3, 4, . . . , n,
i 6= j. In Step 2, if it is necessary to interchange rows to make element Ai,i nonzero,
interchanging row i with the nearest row below it having Aj,i nonzero. In Step 4,
when zeroing elements in column i, add suitable multiples of row Ri to rows Rj , for
j = 1, . . . , i − 1, i + 1, . . . , n (that is, both above and below the 1 in element Ai,i).
and here is what i have done so far ( sorry i have to post the entire code)
Code:
```#include <stdio.h>
#include <stdlib.h>
int main ()
{
void printMatrix (char name, double **matrix, int n);
int input, entries, i, j;
double **A, **B, temp;
printf("Enter dimension of square matrix:");
scanf ("%d", &input);
entries = input * input;
printf ("Enter the %d entries of the matrix:", entries);
A = (double **) malloc (input* sizeof (double *));
for (i=0; i<input; i++)
{
A[i] = (double *) malloc (input* sizeof (double));
}
B = (double **)malloc (input* sizeof (double *));
for (j=0; i<input; i++)
{
B[i] = (double *) malloc (input* sizeof (double));
}
for (i=0; i<input; i++)
{
for (j=0; j<input; j++)
{
scanf("%lf", &A[i][j]);
}
}
for (i=0; i<input; i++)
{
for (j=0; j<input; j++)
{
if ( i == j)
B[i][j] = 1;
else
B[i][j] = 0;
}
}
if (A[0][0]==0)
{
for(i=0;i<input;i++)
{
temp=A[0][i];
A[0][i]=A[1][i];
A[1][i]=temp;
temp=B[0][i];
B[0][i]=B[1][i];
B[1][i]=temp;
}
}
temp = A[0][0];
for(i=0;i<input;i++)
{
for(j=0;j<input;j++)
{
A[i][j]/=temp;
B[i][j]/=temp;
}
}
for (i=1;i<input;i++)
{
temp = A[i][0];
for(j=0;j<input;j++)
{
A[i][j]-=A[0][j]*temp;
B[i][j]-=B[0][j]*temp;
}
}
for (i=0; i<input; i++)
{
printf ("\n");
for (j=0; j<input; j++)
{
printf("%f ", A[i][j]);
}
}
for (i=0; i<input; i++)
{
printf ("\n");
for (j=0; j<input; j++)
{
printf("%f ", B[i][j]);
}
}
return 0;
}```
As you can see, i am up to step 5. I have 2 problems - first, for some really odd reason it only works with matrix of size less then 2x2. anything larger it will just ask for input then the program will end. i suspect there is something wrong with the nested for loops of inputting 0s and 1s into B to make it the identity. however, havin said that i cant find anything wrong with anything.
secondly, i am unsure how to proceed with step 5, any help is appreciated.
i am half asleep when im typing this, so ignore and any grammar or spelling or if i havnt explained myself properly.
2. This may not be the only problem, but look closely at this part:
Code:
``` for (j=0; i<input; i++)
{
B[i] = (double *) malloc (input* sizeof (double));
}```
3. wow thx!! that solved the problem straight away!
4. how did it solve the problem?. it still wouldnt work for higher matrices.
5. ## A simple piece of code
I am also looking for this code. After lot of searches I have found the answer in below website.
C program inverse of a Matrix of any size (NXN)
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# Solve system x+y=5,xz+yu=7,xz^2+yu^2=11,xz^3+yu^3=19
embizze | High School Teacher | (Level 1) Educator Emeritus
Posted on
Solve the system:
x+y=5
xz+yu=7
`xz^2+yu^2=11`
`xz^3+yu^3=19`
From the first equation we get y=5-x. Substituting into the second equation we get:
xz+(5-x)u=7 ==> xz+5u-xu=7 ==> `u=(7-xz)/(5-x)`
We can now write the third and fourth equations in terms of x and z:
`xz^2+(5-x)((7-xz)/(5-x))^2=11` A
`xz^3+(5-x)((7-xz)/(5-x))^3=19` B
From A we can solve for x:
`((7-5x)^2)/(5-x)=11-xz^2`
`=>49-14xz+x^2z^2=55-11x-5xz^2+x^2z^2`
`=>5xz^2-14xz+11x-6=0`
`=>x=6/(5z^2-14z+11)`
Substituting into B we get:
`(7-xz)^3/(5-x)^2=19-xz^3`
`=>(7-xz)^3=(19-xz^3)(5-x)^2`
`=>-10x^2z^3+25xz^3+21x^2z^2-147xz-19x^2+190x-132=0`
Now substitute for x; multiply through by `(5z^2-14z+11)^2` and add like terms to get:
`750z^5-5400z^4+15360z^3-21588z^2+14994z-4116=0`
`=>6(z-2)(z-1)(5z-7)^3=0`
`=>z=2,1,7/5`
(1) `z=1 => x=6/(5-14+11)=3` =>y=2 and u=2
(2) `z=2 => x=6/(20-28+11)=2` =>y=3 and u=1
(3) `z=7/5 => x=6/(5(7/5)^2-14(7/5)+11)=5` =>y=1; but `u=(7-xz)/(5-x)` so this value for z yields `u=0/0` so this is not a solution.
--------------------------------------------------------------------
The solutions are x=2,y=3,u=1,z=2 and x=3,y=2,u=2,z=1
--------------------------------------------------------------------
``
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Question Video: Evaluating Numerical Expressions Involving Cubes and Cube Roots | Nagwa Question Video: Evaluating Numerical Expressions Involving Cubes and Cube Roots | Nagwa
# Question Video: Evaluating Numerical Expressions Involving Cubes and Cube Roots Mathematics • Second Year of Preparatory School
## Join Nagwa Classes
If 𝑥 = ∛375 and 𝑦 = ∛81, find the value of (𝑥 + 𝑦)³.
04:31
### Video Transcript
If 𝑥 is equal to the cube root of 375 and 𝑦 is equal to the cube root of 81, find the value of 𝑥 plus 𝑦 all cubed.
In this question, we are given the values of 𝑥 and 𝑦 as radicals and are asked to evaluate an expression involving 𝑥 and 𝑦. We can begin by substituting the given values into the expression to obtain the cube root of 375 plus the cube root of 81 all cubed. At this point, there are two options for evaluating the expression. Either we can attempt to simplify the expression inside the parentheses or we can expand the cube of the terms in the parentheses.
Of course, expanding the cube will make the expression quite complicated. So, we will start by trying to simplify the expression inside the parentheses. To do this, we need to simplify each cube root. We can do this by recalling that we can take the cube roots of any factor separately. In other words, for any real values 𝑎 and 𝑏, the cube root of 𝑎 times 𝑏 is equal to the cube root of 𝑎 times the cube root of 𝑏.
We can use this property to reduce the size of the numbers inside the cube roots. For instance, we can see that 375 is equal to five cubed times three. Similarly, we can calculate that 81 is three cubed times three. This allows us to rewrite the expression we want to evaluate as the cube root of five cubed times three plus the cube root of three cubed times three all cubed. This is useful since we can take the cube root of each factor separately. And we know that the cube root of 𝑎 cubed will be equal to 𝑎. This means we can take cubic factors out of the radical by taking their cube roots.
First, we can set 𝑎 equal to five cubed and 𝑏 equal to three to get that the cube root of five cubed times three is equal to the cube root of five cubed times the cube root of three. We can follow the same process for the second term inside the parentheses. We set 𝑎 equal to three cubed and 𝑏 equal to three to get the cube root of three cubed times the cube root of three. We then need to cube the sum of these terms.
We can now simplify the expression inside the parentheses by evaluating the cube roots. We have that the cube root of five cubed is five and the cube root of three cubed is three. So, we obtain five times the cube root of three plus three times the cube root of three all cubed. We can now see that the two terms inside the parentheses share a factor of the cube root of three. So, we can take out this factor and add them together by adding five and three. This gives us eight times the cube root of three all cubed.
We now have a product raised to an exponent. So, we can now evaluate this expression using the laws of exponents. In particular, we know that for any real numbers 𝑎 and 𝑏 and integer 𝑛, 𝑎 times 𝑏 all raised to the power of 𝑛 is equal to 𝑎 raised to the power of 𝑛 times 𝑏 raised to the power of 𝑛 provided that these are all well defined. We can apply this result with 𝑎 equal to eight, 𝑏 equal to the cube root of three, and 𝑛 equal to three to get eight cubed times the cube root of three cubed.
Finally, we can evaluate. We calculate that eight cubed is 512 and the cube root of three cubed is three, giving us 512 times three, which we can calculate is 1,536.
## Join Nagwa Classes
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## What is normalized binary number?
Normalized binary numbers always start with a 1 (the leftmost bit of the significand value is a 1). • Why store the 1 (it’s always there)? • IEEE 754 uses this, so the fraction is 24 bits but only 23 need to be stored.
## What is normalization of a number in IEEE 754 format?
To reduce the loss of precision when an underflow occurs, IEEE 754 includes the ability to represent fractions smaller than are possible in the normalized representation, by making the implicit leading digit a 0. Such numbers are called denormal.
What is normalized scientific representation or notation of binary number?
Simply speaking, a number is normalized when it is written in the form of a × 10n where 1 ≤ a < 10 without leading zeros in a. This is the standard form of scientific notation. An alternative style is to have the first non-zero digit after the decimal point.
How do you normalize a number?
Here are the steps to use the normalization formula on a data set:
1. Calculate the range of the data set.
2. Subtract the minimum x value from the value of this data point.
3. Insert these values into the formula and divide.
4. Repeat with additional data points.
### What is Normalised representation?
Single Precision Format: The result said to be normalized, if it is represented with leading 1 bit, i.e. 1.001(2) x 22. (Similarly when the number 0.000000001101(2) x 23 is normalized, it appears as 1.101(2) x 2-6). Omitting this implied 1 on left extreme gives us the mantissa of float number.
### What is normalized format?
From OpenGL Wiki. A Normalized Integer is an integer which is used to store a decimal floating point number. When formats use such an integer, OpenGL will automatically convert them to/from floating point values as needed.
What is the IEEE 754 32 bit largest normalized float value?
A signed 32-bit integer variable has a maximum value of 231 − 1 = 2,147,483,647, whereas an IEEE 754 32-bit base-2 floating-point variable has a maximum value of (2 − 2−23) × 2127 ≈ 3.4028235 × 1038.
What is the smallest normalized number?
Since the smallest exponent is -1022, the smallest positive normalized number is 1.0 × 2-1022 ≈ 2.2 × 10-308. In C, this is defined as DBL_MIN . However, it is not the smallest positive number representable as a floating point number, only the smallest normalized floating point number.
#### What does normalized mean in math?
To normalize something means to scale a vector to make it a unit vector. For a vector in a finite dimensional space, this just means divide each component by the length of the vector.
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# 兩個鍊錶的並集和相交
## 例
Union_list：21→14→12→10→9→5→3
## 算法
1. 通過聲明列表的開頭和下一個指針來創建鏈接列表。
2. 使用insert func將元素插入到兩個列表中。
3. 使用合併排序算法對兩個鍊錶進行排序。
4. 最後，對已排序的列表進行線性掃描，以獲取列表的並集和交集。
## 履行
### 用於兩個鍊錶的並集和交集的C ++程序
```#include<iostream>
#include<stdlib.h>
using namespace std;
struct Node
{
int value;
struct Node* next;
};
void insert(struct Node** head, int new_value)
{
struct Node* NewNode = (struct Node*) malloc(sizeof(struct Node));
NewNode->value = new_value;
}
void Front_Back(struct Node* source, struct Node** front, struct Node** back)
{
struct Node* fast;
struct Node* slow;
if (source==NULL || source->next==NULL)
{
*front = source;
*back = NULL;
}
else
{
slow = source;
fast = source->next;
while (fast != NULL)
{
fast = fast->next;
if (fast != NULL)
{
slow = slow->next;
fast = fast->next;
}
}
*front = source;
*back = slow->next;
slow->next = NULL;
}
}
struct Node* Sort_merge(struct Node* fst, struct Node* sec)
{
struct Node* result = NULL;
if (fst == NULL)
return(sec);
else if (sec==NULL)
return(fst);
if (fst->value <= sec->value)
{
result = fst;
result->next = Sort_merge(fst->next, sec);
}
else
{
result = sec;
result->next = Sort_merge(fst, sec->next);
}
return(result);
}
{
struct Node *a, *b;
return;
Sort(&a);
Sort(&b);
}
{
struct Node *result = NULL;
while (t1 != NULL && t2 != NULL)
{
if (t1->value < t2->value)
{
insert(&result, t1->value);
t1 = t1->next;
}
else if (t1->value>t2->value)
{
insert(&result, t2->value);
t2 = t2->next;
}
else
{
insert(&result, t2->value);
t1 = t1->next;
t2 = t2->next;
}
}
while (t1 != NULL)
{
insert(&result, t1->value);
t1 = t1->next;
}
while (t2 != NULL)
{
insert(&result, t2->value);
t2 = t2->next;
}
return result;
}
{
struct Node *result = NULL;
while (t1 != NULL && t2 != NULL)
{
if (t1->value < t2->value)
t1 = t1->next;
else if (t1->value > t2->value)
t2 = t2->next;
else
{
insert(&result, t2->value);
t1 = t1->next;
t2 = t2->next;
}
}
return result;
}
void printList (struct Node *node)
{
while (node != NULL)
{
cout<<node->value << " ";
node = node->next;
}
cout<<endl;
}
int main()
{
struct Node* intersection_list = NULL;
struct Node* union_list = NULL;
cout<<"First list is \n";
cout<<"\nSecond list is \n";
cout<<"\nIntersection list is \n";
printList(intersection_list);
cout<<"\nUnion list is \n";
printList(union_list);
return 0;
}
```
```First list is
4 10 11 15 20
Second list is
2 4 8 10
Intersection list is
10 4
Union list is
20 15 11 10 8 4 2```
### 用於兩個鍊錶的並集和交集的Java程序
```class Solution1
{
class Node
{
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
void get_union(Node hd1, Node hd2)
{
Node t1 = hd1, t2 = hd2;
while (t1 != null)
{
insert(t1.data);
t1 = t1.next;
}
while (t2 != null)
{
insert(t2.data);
t2 = t2.next;
}
}
void get_intrSection(Node hd1, Node hd2)
{
Node rst = null;
Node t1 = hd1;
while (t1 != null)
{
if (is_Present(hd2, t1.data))
insert(t1.data);
t1 = t1.next;
}
}
void printList()
{
while (temp != null)
{
System.out.print(temp.data + " ");
temp = temp.next;
}
System.out.println();
}
void insert(int new_data)
{
Node new_node = new Node(new_data);
}
{
while (t != null)
{
if (t.data == data)
return true;
t = t.next;
}
return false;
}
public static void main(String args[])
{
Solution1 llist1 = new Solution1();
Solution1 llist2 = new Solution1();
Solution1 unin = new Solution1();
Solution1 intersecn = new Solution1();
llist1.insert(20);
llist1.insert(4);
llist1.insert(15);
llist1.insert(10);
llist2.insert(10);
llist2.insert(2);
llist2.insert(4);
llist2.insert(8);
System.out.println("First List is");
llist1.printList();
System.out.println("Second List is");
llist2.printList();
System.out.println("Intersection List is");
intersecn.printList();
System.out.println("Union List is");
unin.printList();
}
}
```
```First List is
10 15 4 20
Second List is
8 4 2 10
Intersection List is
4 10
Union List is
2 8 20 4 15 10```
## 兩個鍊錶的並集和相交的複雜性分析
### 時間複雜度
O（米+ n） 哪裡 “m”個“ n” 是分別在第一和第二列表中存在的元素的數量。
### 空間複雜度
O（米+ n） 哪裡 “m”個“ n” 是分別在第一和第二列表中存在的元素的數量。
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# The mystery of the CHI SQUARE Is it CHEE square Or CHAI Square?!
## Presentation on theme: "The mystery of the CHI SQUARE Is it CHEE square Or CHAI Square?!"— Presentation transcript:
The mystery of the CHI SQUARE Is it CHEE square Or CHAI Square?!
Chi Square X 2 goodness of fit There is a single test that can be applied to see if the observed sample distribution is significantly different in some way from the hypothesized population distribution
Accidents on Cellphones Are you more likely to have a motor vehicle collision when using a cell phone? A study of 699 drivers who were using a cell phone when they were involved in a collision examined this question. These drivers made 26,798 cell phone calls during a 14-month study period. Each of the 699 collisions was classified in various ways. Here are the counts for each day of the week:
Hypotheses: H 0 : Motor vehicle accidents involving cell phone use are equally likely to occur on each of the seven days of the week. H a : The probabilities of a motor vehicle accident involving cell phone use vary from day to day (that is, they are not all the same).
Chi square procedure: In general, the expected count for any categorical variable is obtained by multiplying the proportion of the distribution for each category by the sample size.
Chi-square test statistics For Sunday: For Monday:
Finding the p-value Degrees of freedom: n-1 df: 7-1 = 6 Calculator syntax: 2nd - VARS - 8 (enter) X 2 cdf( test statistic, 1E99, df ) X 2 cdf( 208.84, 1E99, 6 ) p= 2.48 x 10 -42
Conclusion Since the p value is extremely small ( p= 2.48 x 10 -42 ), there is sufficient evidence to reject H 0 and conclude that these types of accidents are not equally likely to occur on each of the seven days of the week. H 0 : Motor vehicle accidents involving cell phone use are equally likely to occur on each of the seven days of the week. H a : The probabilities of a motor vehicle accident involving cell phone use vary from day to day (that is, they are not all the same).
Red Eye Fruit Fly Any offspring receiving an R gene will have red eyes, and any offspring receiving a C gene will have straight wings. So based on this Punnett square, the biologists predict a ratio of 9 red-eyed, straight-winged (x) : 3 red- eyed, curly-winged (y) : 3 white-eyed, straight-winged (z) : 1 white-eyed, curly-winged (w) offspring. To test their hypothesis about the distribution of offspring, the biologists mate the fruit flies. Of 200 offspring, 99 had red eyes and straight wings, 42 had red eyes and curly wings, 49 had white eyes and straight wings, and 10 had white eyes and curly wings. Do these data differ significantly from what the biologists have predicted?
Given Distribution parents proportio n offspring s Red-eyed, straight-winged 90.5635 99 Red-eyed, curly-winged 30.1875 42 White-eyed, straight-winged 30.1875 49 White-eyed, curly-winged: 1 10.0625 10 total 16 200 H o : these proportions is correct for the the offspring of 2 parents H a : at least one of these proportions is incorrect
Conditions and calculations: We can use a chi-square goodness of fit test to measure the strength of the evidence against the hypothesized distribution, provided that the expected cell counts are large enough. SampleproportionObservedExpected Red-eyed, straight-winged 90.562599 (200)(0.5625) = 112.5 Red-eyed, curly-winged 30.187542 (200)(0.1875) = 37.5 White-eyed, straight-winged 30.187549 (200)(0.1875) = 37.5 White-eyed, curly-winged: 10.062510 (200)(0.0625) = 12.5 total 16200 X 2 cdf(6.187, 1E99, 3 ) p=0.1029
Interpretations The P-value of 0.1029 indicates that the probability of obtaining a sample of 200 fruit fly offspring in which the proportions differ from the hypothesized values by at least as much as the ones in our sample is over 10%, assuming that the null hypothesis is true. This is not sufficient evidence to reject the biologists' predicted distribution.
Your Turn Course grades Most students in a large college statistics course are taught by teaching assistants (TAs). One section is taught by the course supervisor, a full- time professor. The distribution of grades for the hundreds of students taught by TAs this semester was The grades assigned by the professor to the 91 students in his section were (a) What percents of students in the professor's section earned A, B, C, and D/F? In what ways does this distribution of grades differ from the TA distribution? (b) Because the TA distribution is based on hundreds of students, we are willing to regard it as a fixed probability distribution. If the professor's grading follows this distribution, what are the expected counts of each grade in his section? (c) Does the chi-square test for goodness of fit give good evidence that the professor's grade distribution differs from the TA distributions? Use the Inference Toolbox.
Answers: (a) A : 24.2%, B : 41.8%, C : 22.0%, D/F : 12.1%. Fewer A s and more D/F s than the TA sections. (b) A : 29.12, B : 37.31, C : 18.20, D/F : 6.37. (c) H 0 : p 1 = 0.32, p 1 = 0.41, p 1 = 0.20, p 1 = 0.07 vs. H a : at least one of these proportions is different. All the expected counts are greater than 5, so the condition for X 2 is satisfied. X 2 = 5.297 (df = 3), so the P–value = 0.1513; there is not enough evidence to conclude that the professor s grade distribution was different from the TA grade distribution.
Chi-Sq. Practice (with probability model) Thai, the manager of a car dealership, did not want to stock cars that were bought less frequently because of their unpopular color. The five colors that he ordered were red, yellow, green, blue, and white. According to Thai,the expected frequencies or number of customers choosing each color should follow the percentages of last year. She felt 20% would choose yellow, 30% would choose red, 10% would choose green, 10% would choose blue, and 30% would choose white. She now took a random sample of 150 customers and asked them their color preferences.
Hypotheses: Ho: there is no significant difference between the proportion of the costumers car color preferences. Ho: p1 = p2 = p3 = p4 = p5 Ha: there is a significant difference between the proportion of the costumers car color preferences. Ha: p1 p2 p3 p4 p5
Chi-square procedure: X 2 = 26.95 P-value = 2.03x10 -5
Conclusion Since our p-value is small, we have sufficient reason to reject the null hypothesis making our test significant. Therefore, there is a significant difference between the proportion of the costumers car color preferences.
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# Search results
Found 285 matches
Electric Current
An electric current is a flow of electric charge. In electric circuits this charge is often carried by moving electrons in a wire. It can also be carried ... more
Surface Tension - surface area growth : force
Surface tension is a contractive tendency of the surface of a liquid that allows it to resist an external force. Surface tension is an important property ... more
Worksheet 333
A typical small rescue helicopter, like the one in the Figure below, has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades. (c) To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it?
The first image shows how helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades.
The second image shows a helicopter from the Auckland Westpac Rescue Helicopter Service. Over 50,000 lives have been saved since its operations beginning in 1973. Here, a water rescue operation is shown. (credit: 111 Emergency, Flickr)
Strategy
Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy.
Solution for (a)
We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find Er . The angular velocity ω for 1 r.p.m is
Angular velocity
and for 300 r.p.m
Multiplication
The moment of inertia of one blade will be that of a thin rod rotated about its end.
Moment of Inertia - Rod end
The total I is four times this moment of inertia, because there are four blades. Thus,
Multiplication
and so The rotational kinetic energy is
Rotational energy
Solution for (b)
Translational kinetic energy is defined as
Kinetic energy ( related to the object 's velocity )
To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is
Division
Solution for (c)
At the maximum height, all rotational kinetic energy will have been converted to gravitational energy. To find this height, we equate those two energies:
Potential energy
Discussion
The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades—something you probably would not suspect. The 53.7 m height to which the helicopter could be raised with the rotational kinetic energy is also impressive, again emphasizing the amount of rotational kinetic energy in the blades.
Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Thrust
Thrust is a reaction force described quantitatively by Newton’s second and third laws. When a system expels or accelerates mass in one direction, the ... more
Ricco's Law
Several laws describe a human’s ability to visually detect targets on a uniform background. One such law is Riccò's law, discovered by astronomer ... more
Electric Potential Energy with Time (related to Electrical Work)
Electrical work is the work done on a charged particle by an electric field. The equation for 'electrical’ work is equivalent to that of ... more
Dynamic Pressure - Compressible flow
In incompressible fluid dynamics dynamic pressure (indicated with q, or Q, and sometimes called velocity pressure) is the quantity defined as ... more
Speed of sound in three-dimensional solids (pressure waves)
The speed of sound is the distance travelled per unit of time by a sound wave propagating through an elastic medium. Sound travels faster in liquids and ... more
Creatinine Clearance
Renal function, in nephrology, is an indication of the state of the kidney and its role in renal physiology. Glomerular filtration rate (... more
A ball screw is a mechanical linear actuator that translates rotational motion to linear motion with little friction. A threaded shaft provides a helical ... more
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Actuarial present value
The actuarial present value (APV) is the expected value of the present value of a contingent cash flow stream (i.e. a series of payments which may or may not be made). Actuarial present values are typically calculated for the benefit-payment or series of payments associated with life insurance and life annuities. The probability of a future payment is based on assumptions about the person's future mortality which is typically estimated using a life table.
Life insurance
Whole life insurance pays a pre-determined benefit either at or soon after the insured's death. The symbol (x) is used to denote "a life aged x" where x is a non-random parameter that is assumed to be greater than zero. The actuarial present value of one unit of whole life insurance issued to (x) is denoted by the symbol ${\displaystyle \,A_{x}}$ or ${\displaystyle \,{\overline {A}}_{x}}$ in actuarial notation. Let G>0 (the "age at death") be the random variable that models the age at which an individual, such as (x), will die. And let T (the future lifetime random variable) be the time elapsed between age-x and whatever age (x) is at the time the benefit is paid (even though (x) is most likely dead at that time). Since T is a function of G and x we will write T=T(G,x). Finally, let Z be the present value random variable of a whole life insurance benefit of 1 payable at time T. Then:
${\displaystyle \,Z=v^{T}=(1+i)^{-T}=e^{-\delta T}}$
where i is the effective annual interest rate and δ is the equivalent force of interest.
To determine the actuarial present value of the benefit we need to calculate the expected value ${\displaystyle \,E(Z)}$ of this random variable Z. Suppose the death benefit is payable at the end of year of death. Then T(G, x) := ceiling(G - x) is the number of "whole years" (rounded upwards) lived by (x) beyond age x, so that the actuarial present value of one unit of insurance is given by:
{\displaystyle {\begin{aligned}A_{x}&=E[Z]=E[v^{T}]\\&=\sum _{t=1}^{\infty }v^{t}Pr[T=t]=\sum _{t=0}^{\infty }v^{t+1}Pr[T(G,x)=t+1]\\&=\sum _{t=0}^{\infty }v^{t+1}Pr[tx]\\&=\sum _{t=0}^{\infty }v^{t+1}\left({\frac {Pr[G>x+t]}{Pr[G>x]}}\right)\left({\frac {Pr[x+tx+t]}}\right)\\&=\sum _{t=0}^{\infty }v^{t+1}{}_{t}p_{x}\cdot q_{x+t}\end{aligned}}}
where ${\displaystyle {}_{t}p_{x}}$ is the probability that (x) survives to age x+t, and ${\displaystyle \,q_{x+t}}$ is the probability that (x+t) dies within one year.
If the benefit is payable at the moment of death, then T(G,x): = G - x and the actuarial present value of one unit of whole life insurance is calculated as
${\displaystyle \,{\overline {A}}_{x}\!=E[v^{T}]=\int _{0}^{\infty }v^{t}f_{T}(t)\,dt=\int _{0}^{\infty }v^{t}\,_{t}p_{x}\mu _{x+t}\,dt,}$
where ${\displaystyle f_{T}}$ is the probability density function of T, ${\displaystyle \,_{t}p_{x}}$ is the probability of a life age ${\displaystyle x}$ surviving to age ${\displaystyle x+t}$ and ${\displaystyle \mu _{x+t}}$ denotes force of mortality at time ${\displaystyle x+t}$ for a life aged ${\displaystyle x}$.
The actuarial present value of one unit of an n-year term insurance policy payable at the moment of death can be found similarly by integrating from 0 to n.
The actuarial present value of an n year pure endowment insurance benefit of 1 payable after n years if alive, can be found as
${\displaystyle \,_{n}E_{x}=Pr[G>x+n]v^{n}=\,_{n}p_{x}v^{n}}$
In practice the information available about the random variable G (and in turn T) may be drawn from life tables, which give figures by year. For example, a three year term life insurance of $100,000 payable at the end of year of death has actuarial present value ${\displaystyle 100,000\,A_{{\stackrel {1}{x}}:{{\overline {3}}|}}=100,000\sum _{t=1}^{3}v^{t}Pr[T(G,x)=t]}$ For example, suppose that there is a 90% chance of an individual surviving any given year (i.e. T has a geometric distribution with parameter p = 0.9 and the set {1, 2, 3, ...} for its support). Then ${\displaystyle Pr[T(G,x)=1]=0.1,\quad Pr[T(G,x)=2]=0.9(0.1)=0.09,\quad Pr[T(G,x)=3]=0.9^{2}(0.1)=0.081,}$ and at interest rate 6% the actuarial present value of one unit of the three year term insurance is ${\displaystyle \,A_{{\stackrel {1}{x}}:{{\overline {3}}|}}=0.1(1.06)^{-1}+0.09(1.06)^{-2}+0.081(1.06)^{-3}=0.24244846,}$ so the actuarial present value of the$100,000 insurance is \$24,244.85.
In practice the benefit may be payable at the end of a shorter period than a year, which requires an adjustment of the formula.
Life annuity
The actuarial present value of a life annuity of 1 per year paid continuously can be found in two ways:
Aggregate payment technique (taking the expected value of the total present value):
This is similar to the method for a life insurance policy. This time the random variable Y is the total present value random variable of an annuity of 1 per year, issued to a life aged x, paid continuously as long as the person is alive, and is given by:
${\displaystyle Y={\overline {a}}_{\overline {T(x)|}}={\frac {1-(1+i)^{-T}}{\delta }}={\frac {1-v^{T}(x)}{\delta }},}$
where T=T(x) is the future lifetime random variable for a person age x. The expected value of Y is:
${\displaystyle \,{\overline {a}}_{x}=\int _{0}^{\infty }{\overline {a}}_{\overline {t|}}f_{T}(t)\,dt=\int _{0}^{\infty }{\overline {a}}_{\overline {t|}}\,_{t}p_{x}\mu _{x+t}\,dt.}$
Current payment technique (taking the total present value of the function of time representing the expected values of payments):
${\displaystyle \,{\overline {a}}_{x}=\int _{0}^{\infty }v^{t}[1-F_{T}(t)]\,dt=\int _{0}^{\infty }v^{t}\,_{t}p_{x}\,dt}$
where F(t) is the cumulative distribution function of the random variable T.
The equivalence follows also from integration by parts.
In practice life annuities are not paid continuously. If the payments are made at the end of each period the actuarial present value is given by
${\displaystyle a_{x}=\sum _{k=1}^{\infty }v^{t}[1-F_{T}(t)]=\sum _{t=1}^{\infty }v^{t}\,_{t}p_{x}.}$
Keeping the total payment per year equal to 1, the longer the period, the smaller the present value is due to two effects:
• The payments are made on average half a period later than in the continuous case.
• There is no proportional payment for the time in the period of death, i.e. a "loss" of payment for on average half a period.
Conversely, for contracts costing an equal lumpsum and having the same internal rate of return, the longer the period between payments, the larger the total payment per year.
Life insurance as a function of the life annuity
The APV of whole-life insurance can be derived from the APV of a whole-life annuity-due this way:
${\displaystyle \,A_{x}=1-iv{\ddot {a}}_{x}}$
This is also commonly written as:
${\displaystyle \,A_{x}=1-d{\ddot {a}}_{x}}$
In the continuous case,
${\displaystyle \,{\overline {A}}_{x}=1-\delta {\overline {a}}_{x}.}$
In the case where the annuity and life insurance are not whole life, one should replace the insurance with an n-year endowment insurance (which can be expressed as the sum of an n-year term insurance and an n-year pure endowment), and the annuity with an n-year annuity due.
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It is a lot like the x-y-plane, but the horizontal axis represents the real. 11 047321CIS Math Modeling Math Modeling CIS introduces students to the art of mathematical prediction. RCL The calculator is standing by for input of a variable name to recall the variable’s value. Each point on an Argand diagram represents a complex number. Three semester hours. Free binary math problems and exercises with answers and solutions. It connects trigonometric functions with exponential functions in the complex plane via Euler's formula. Impedance and Phase Angle. We also learn about a different way to represent complex numbers—polar form. This is an NCAA Approved Course. If A is represented by the complex number z, C is represented by the complex number k 2 /z* Since the whole transformation is symmetrical around the line DH, the diameter GH of the z-circle the diameter HG of the w-circle (though in general other diameters of the z-circle do not diameters of the w-circle). • Apply abstract ideas to concrete situations. Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as Pythagoras, Descartes, De Moivre, Euler, Gauss, and others. Write complex numbers in polar form. c = cis Write cis in expanded form. Or in the shorter "cis" notation: (r cis θ) 2 = r 2 cis 2 θ Complex number forms review Review the different ways in which we can represent complex numbers: rectangular, polar, and exponential forms. The proof of the following proposition is straightforward and is left as an exercise. And the mathematician Abraham de Moivre found it works for any integer exponent n: [ r(cos θ + i sin θ) ] n = r n (cos nθ + i sin nθ). A Simple Software that can calculate Complex Numbers, for Academic use. - Mechanism of Action & Protocol. The complex number z = 4+3i is shown in Figure 2. (c) Determine the product z1z2 in polar form. Note that real numbers are complex — a real number is simply a complex number with no imaginary part. The complex numbers complete the number path that you’ve been following. Use the polar to rectangular feature on the graphing calculator to change 4 cis. 7)^4 = cis(294. curriculummapper. For example, a browser window itself contains a number of other objects like an HTML document, a titlebar, a menubar, a toolbar, optional scrollbars and a statusbar along the bottom. A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, that satisfies the equation i 2 = −1. Plot complex numbers in the complex plane. The second from the last line of the Mode menu gives you three options: Real, a + bi, and The last two options are used when working with complex numbers. Now the following rules applies. 0 A Simple Software that can calculate Complex Numbers, for Academic use. Convert the complex number from polar form to cis (@) form with theta in radians: Te cis = (8e***/8)2/3 T- 0 radians. 30 with both answers rounded to hundredths. calculator (TI-83 or TI-83 plus or TI-84 or TI-84plus). Write complex numbers in polar form. 7^(1/2)+21^(1/2). We refer to r cis θ as the polar form of a complex number or cis notation. 2, a full-featured scientific calculator using Reverse Polish Notation (see section 2. Class meets Monday, Tuesday, Wednesday, and Friday each week. $$This actually has three solutions, and we can find them using de Moivre's Theorem. 4 CMATH is a comprehensive library for complex-number arithmetics and mathematics, both in cartesian and polar format (single, double, and extended precision). This is an aplet which lets you perform a number of small tasks - calculate coefficients for binomial expansions, calculate mean and standard deviation of grouped data and divide complex numbers. For a general z = x + yi (x and y are real numbers), r and t are defined as follows. When we first learned to count, we started with the natural numbers – 1, 2, 3, and so on. Sep 13, 2004 · Use the Maple factor command to factor the expression. Write the values of the five new numbers. GCF Calculator. Mathematics and mathematical reasoning are used in situations as diverse as household budgeting and space shuttle design, subjects as different as art and law, and occupations as varied as nursing and. How do you place 0. Then one shows. This calculator extracts the square root, calculate the modulus, finds inverse, finds conjugate and transform complex number to polar form. Complex is a simple object oriented language with some functional aspects. We can convert the complex number into trigonometric form by finding the modulus and argument of the complex number. In abstract algebra, a field extension L/K is called algebraic if every element of L is. An algebraic derivation. 1 Title of Lab: Building a Registration Form and Pay Calculator in Python3 Devry CIS 115 Week 2 Selection Control Structures3. A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i represents the imaginary unit, satisfying the equation i 2 = −1. Calculators are allowed in the course. Because no real number satisfies this equation, i is called an imaginary number. De Moivre’s formula is a direct method for solving problems involving powers of complex numbers. Region R is bounded by the set of all complex numbers on the Argand plane z a bi where a and b are real numbers satisfying the equation a a b b22 12 13 6 6. We also learn about a different way to represent complex numbers—polar form. If you wish to exclude negative values, but include zero, you must use the term “non-negative. Binomic (a+bi) & CIS. it is said to be in rectangular form. Contents1 DEVRY CIS 115 LOGIC AND DESIGN TAKE MY ONLINE CLASS2 Devry CIS 115 Week 1 Software Development and Design Tools2. Convert the complex number from polar form to cis (C) form with theta in radians: Te's cis Question 17 Convert the complex number from rectangular form to polar form. A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, that satisfies the equation i 2 = −1. The implementation in Assembly. Write a Python program to add, subtract, multiply and division of two complex numbers. I couldn't find anything. By using this website, you agree to our Cookie Policy. (d) Determine the ratio z1/z2 in polar form. Polar to Rectangular Online Calculator; 5. Students who need either CIS-1400 or ENT-1890 should take INT-1550 instead. See full list on people. 2222, and 0. 36i (2 pts each) The distance of a complex number from the origin of the complex plane is. How do you place 0. Graph the following complex number in polar form as a vector on the complex plane: z = dîî(cos 2100 + i sin 2100) Imaginary -10-9 -8 -7 -6 -5 -4 -3 -2 - 123 45 €789 10 Ryess 17. The set of complex numbers is denoted using the. The calculator is standing by for input of a variable name to assign a value to the variable. 2i The complex numbers are an extension of the real numbers. Calculators will be used. A complex number in Polar Form must be entered, in Alcula's scientific calculator, using the cis operator. Free Online Scientific Notation Calculator. Specific topics of investigation include function properties and patterns, complex numbers, parametric equations, polar equations, vectors, and exponential growth and decay. Each point on an Argand diagram represents a complex number. Euler's Formula for complex numbers states that if z z z is a complex number with absolute value r z r_z r z and argument θ z \theta_z θ z , then. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. You can also determine the real and imaginary parts of complex numbers and compute other common values such as phase and angle. See full list on brownmath. Therefore complex conjugation de nes homomorphisms C !C and C !C. Complex Number Operations Chapter - 3. 50) 277 cos + i sin 29. Z Worksheet by Kuta Software LLC. Trigonometric Form of Complex Numbers Calculator. From the keyboard and computer-menu interface to the ability to save and share work with built-in apps, the TI-Nspire™ CX II models retain the features and functionality that make TI-Nspire™ CX graphing calculators ideal for math and science from middle grades through college. • Apply abstract ideas to concrete situations. This really doesn't use differential calculus at all. Fall, even years. None of the other functions work (eg no complex exponentials). (Note: Your answers may involve a combination of notations). Complex Number Calculator. (b) Show that the inner product (X, IG) is the number of G-orbits in S, where the inner product is given by (Xl, 1/2) = Xl. If we look at the complex number system, every quadratic equation has at least one complex solution; remember rational and irrational roots are in fact complex numbers. given set of computing Goal 2: Critical Thinking: Graduates of. Or in the shorter "cis" notation: (r cis θ) 2 = r 2 cis 2θ. Free complex equations calculator - solve complex equations step-by-step. - Is that right? Also, how would you be able to change complex numbers into polar form? I know that. ) 3 ,cis 1 b. The x is the real number of the expression and the y represents the imaginary number of the expression. MAT 103 - Math for Clinical Calculations 3 Credits Provides a review of general mathematics, introductory algebra and an opportunity to learn systems of measurement and methods of solving problems related to drug dosage and intravenous fluid administration. In this assignment, we will create a calculator that performs simple arithmetic operations on complex numbers. Some examples of complex numbers include 2 3− i, 7 (which is really 7 0+ i), and −5i. Real numbers aren’t enough to solve x2 = −1, so we invent an answer, called i, and arrive at the complex numbers. Credit may not be earned for both INT-1550 and CIS-1400 or for both ENT-1890 and INT-1550. AC Circuit Definitions; 9. Includes functions, exponential and logarithmic functions, linear 2 x 2 and higher systems, graphing, and calculator use. Complex values are denoted by a parenthesized pair of values separated by a comma representing the real and imagi-nary part of the variable. This programmable calculator does real or complex number calculations with 100 significant figure precision. Complex number notation Nothing unexpected here, th…. Impedance and Phase Angle. So, let’s address the easy part of the question. This formula is applicable only if x and y are positive. ccis(*c__[],x__[]) Stores cos(x)+i. With some exceptions, we will take a micro-quiz each Wednesday, and homework is due each Friday. Plot complex numbers in the complex plane. Every complex number can be written in the form a + bi. In a polar form, the division is somewhat simple. In order to use DeMoivre's Theorem to find complex number roots we should have an understanding of the trigonometric form of complex numbers. Complex number system. Just type your formula into the top box. The notation is less commonly used than Euler's formula , e i x {\displaystyle e^{ix}} , which offers an even shorter and more general notation for cos( x ) + i sin( x ). Convert the complex number z =6∠110 to Cartesian form. Interestingly, the variant with the most trans effects (37. Consider the complex number ω = , where z = x + iy and i =. Do basic complex number arithmetic (add, subtract, multiply, divide) with imaginary numbers. The topic Complex Numbers involves investigating and extending understanding of the real number system to include complex numbers. We can imagine the point (t) being. Complex Numbers and Polar Form of a Complex Number Interactive Graph - Convert polar to rectangular and vice-versa In the following graph, the real axis is horizontal, and the imaginary (j=sqrt(-1)) axis is vertical, as usual. The majority of students managed well in the first parts of this problem. Complex numbers %ˆ The calculator performs the following complex number calculations: • Addition, subtraction, multiplication, and division • Argument and absolute value calculations • Reciprocal, square, and cube calculations • Complex Conjugate number calculations Setting the complex format: Set the calculator to DEC mode when. [You may use without proof the result that if s:[a;b] !Cis a closed path for each real number sin some interval [c;d], then the value of the. 2222, and 0. In abstract algebra, a field extension L/K is called algebraic if every element of L is. Topics included are factoring, algebraic fractions, radicals and rational exponents, complex numbers, quadratic equations, rational equations, linear equations and inequalities in two variables and their graphs, systems of linear equations and inequalities, introduction to functions, and applications of the above topics. Infrastructure and Security Due Week 8 and worth 110 points This assignment consists of two (2) sections: an infrastructure document and a revised Gantt chart or project plan. That is, he claimed and did not completely justify. GCF Calculator. Giver as an integer and theta as a fraction of pi in radians. Submit the questionnaire with the exam booklet CHEATING in any manner will merit a grade of 0. The calculator will generate a step by step explanation for each operation. Covers the mathematics needed to teach grades K–8. It connects trigonometric functions with exponential functions in the complex plane via Euler's formula. Plot complex numbers in the complex plane. If you're already familiar with complex numbers, you can skip to the next section. cis (− π 3) For the following exercises, convert the complex number from polar to rectangular form. Here are some examples of what you would type here: (3i+1)(5+2i) (-1-5i)(10+12i) i(5-2i). For the complex number a + bi, a is called the real part, and b is called the imaginary part. com after unlocking the functions in casio fx 82MS, I come here to present an operation with complex numbers. MAT 103 - Math for Clinical Calculations 3 Credits Provides a review of general mathematics, introductory algebra and an opportunity to learn systems of measurement and methods of solving problems related to drug dosage and intravenous fluid administration. Do not use cis form. Therefore complex conjugation de nes homomorphisms C !C and C !C. Use the polar to rectangular feature on the graphing calculator to change 4 cis. If z=6—2i and w=5+3i , find (a) (b) (c) 2z — 3w 000 — expressing your answer in the form a + bi where a, b e IR. Exponential Form of Complex Numbers; Euler Formula and Euler Identity interactive graph; 6. Convert the complex number z =6∠110 to Cartesian form. CIS 194: Homework 1 Due Monday, January 14 When solving the homework, strive to create not just code that works, but code that is stylish and concise. This indicator appears after you press 1t(STO). Then one shows. Please see the mathematics program for placement details. jl package, involving real and complex numbers. The complex numbers are a set of objects that includes not only the familiar real numbers but also an additional object called "i". com Tel: 800-234-2933;. Sketch these four numbers in the complex plane. The form z = a + b i is called the rectangular coordinate form of a complex number. Prerequisites: Completion of Math 140 with a C or better, or ALEKS math score of 61+, or COMPASS ALGEBRA 43+ AND College ALGEBRA 51+ AND TRIGONOMETRY 1-50, or ACT Math 24+, or SAT Math 560+ if before 3/1/16 (580+ if after 3/1/16) or Consent. Evaluate an expression with complex numbers using an online calculator. The imaginary number calculator / complex number calculator optionally convert the results of calculations to polar (phasor), exponential and other modular forms. Calculator cis form complex numbers - calculation: 5*cis(45°). Then we can figure out the exact position of $$z$$ on the complex plane if we know two things: the length of the line segment and the angle measured from the positive real axis to the line segment. With them, you can forget about the amplitude and phase and time dependency for a moment and simply follow the rules of the complex algebra to arrive at the final answer, then interpret the final phasor as. De Moivre's Formula. If we express the current and voltage as complex numbers, the impedance is a complex number, but you can not say it's a phasor. c = cis Write cis in expanded form. The expression cos(x) + i sin(x) is sometimes reduced to cis(x). Gain fluency and confidence in math! IXL helps students master essential skills at their own pace through fun and interactive questions, built in support, and motivating awards. Interestingly enough, your calculator not only knows that i 2 = -1, but automatically simplifies any result that would have had i 2 in it. asked by Jane on March 6, 2015; math. The angle from the positive axis to the line segment is called the argumentof the complex number, z. Offer six (optional) chapters which "Look Back" to the important topics in algebra. The first step in finding the solutions of (that is, the x-intercepts of, plus any complex-valued roots of) a given polynomial function is to apply the Rational Roots Test to the polynomial's leading coefficient and constant term, in order to get a list of values that might possibly be solutions to the related polynomial equation. This indicator appears after you press t. If you have more complex circumstances, then please talk to your accountancy or one of our accountancy advisers today by calling 020 3355 4047 or by pressing the Live Chat button. r cis θ means r(cos θ + i sin θ) de Moivre’s theorem proof of de Moivre’s theorem the five answers to fifth root of 1. We can imagine the point (t) being. To divide complex numbers. (15) Given the complex numbers: a) eiˇ 4ˇi Put this in Cartesian coordinate form: x+ iy b) p1 2 pi 2 Put this in polar coordinate form: rei 2. Rewrite the complex number 4+3𝑖 in polar (cis) form and re-graph it on the polar graph paper. When writing complex numbers, just use the "i" as usual. EN: complex-number-calculator menu Pre-Álgebra Orden (jerarquía) de operaciones Factores y números primos Fracciones Aritmética Decimales Exponentes y radicales Módulo. When zand ware any complex numbers, jzwj= jzjjwj, which implies that the absolute value function on nonzero complex numbers is a homomorphism C !R >0. com after unlocking the functions in casio fx 82MS, I come here to present an operation with complex numbers. This page will show you how to multiply them together correctly. Number 3 uses the same ideas – simplify the equation to isolate the exponential part (move everything else to the right), and then write both sides as powers of 10. form of a+bi algebra calculator is not a very difficult topic and you can easily do some initial preparation yourself. zip: 18k: 00-12-10: Crypt92 BETA 2/3 The second release of the only cryptography software for TI calculators. (We have to exclude 0 from the. the solutions to the second require x^2 = -1. Complex number notation Nothing unexpected here, th…. This shows that by squaring a complex number, the absolute value is squared and the argument is multiplied by 2 2 2. You can write a book review and share your experiences. On our system these reals can only hold numbers up to about 10^38. Euler's Formula for complex numbers states that if z z z is a complex number with absolute value r z r_z r z and argument θ z \theta_z θ z , then. The polar form of a complex number takes the form r(cos + isin ) Now r can be found by applying the Pythagorean Theorem on a and b, or: r = can be found using the formula: = So for this particular problem, the two roots of the quadratic equation are: Hence, a = 3/2 and b = 3√3 / 2. complex line integral Z f(z)dzby Z f(z)dz:= Z b a f((t)) 0(t)dt: (1a) You should note that this notation looks just like integrals of a real variable. z^5 + 1 = 0. The cis notation was first coined by William Rowan Hamilton in Elements of Quaternions (1866) and subsequently used by Irving Stringham in works such as Uniplanar Algebra (1893), or by James Harkness and Frank Morley in their Introduction to the Theory of Analytic Functions (1898). Exponential Form of Complex Numbers; Euler Formula and Euler Identity interactive graph; 6. a graphing calculator, and the intermediate value theorem to find and graph D. The polar representation of a complex number makes it easy to find products and powers of complex numbers. To use the calculator during the exam, students need to select the Calculator icon. This website uses cookies to ensure you get the best experience. Calculators will be used. Sep 13, 2004 · Use the Maple factor command to factor the expression. To enter the complex number in polar form you enter mcisa, where m is the modulus and a is the argument of number. Homework exercises: Chapter 9 exercises 7-10 Session 26. Compute the area of R. On the TI-83 Plus graphing calculator, you can set the mode and use complex numbers. If A is represented by the complex number z, C is represented by the complex number k 2 /z* Since the whole transformation is symmetrical around the line DH, the diameter GH of the z-circle the diameter HG of the w-circle (though in general other diameters of the z-circle do not diameters of the w-circle). (i) z (ii) z^4 (iii) 1/z With parts (ii) and (iii) of the question, would z be different from being 3 - 3i? So, for (ii), it would be (3 - 3i)^4 as being z and from that, you would have to change that result into polar form. r 2 (cos 2θ + i sin 2θ) (the magnitude r gets squared and the angle θ gets doubled. Complex Numbers:The Polar Form. Proposition 2. Calculators will be used. In this case, the left hand side is a multi-valued function, and the right hand side is one of its possible values. The set of complex numbers See here for a complete list of set symbols. If you have more complex circumstances, then please talk to your accountancy or one of our accountancy advisers today by calling 020 3355 4047 or by pressing the Live Chat button. 2i The complex numbers are an extension of the real numbers. Or in the shorter "cis" notation: (r cis θ) 2 = r 2 cis 2θ. Polar - Rectangular Coordinate Conversion Calculator. However this creates a discontinuity as moves across the negative real axis. Be able to add, subtract, multiply, and divide any integers (using a calculator if needed). r (cos θ + i sin θ) Here r stands for modulus and θ stands for argument. Whether you've loved the book or not, if you give your honest and detailed thoughts then people will find new books that are right for them. Have already checked Conventions for typesetting complex vectors and vectors with complex components but nobody mentions this in particular. It is important to realize that any given complex number on a complex plane can be arrived at by rotating around the pole a multitude of times. Homework Guide = 4 0465, 0466 – CIS COLLEGE ALGEBRA THROUGH MODELING Semester – 1 credit – Grade 11-12 Prerequisite: Earned C+ or better in HS Intermediate Algebra and Algebra II classes, or successfully completed three years high school. STAT The calculator is in the STAT Mode. In this case, the left hand side is a multi-valued function, and the right hand side is one of its possible values. If there is a single solution (one value for each unknown factor) we will say that the system is Consistent Independent System (CIS). Rewrite each of the following complex numbers in rectangular form and graph. Now the following rules applies. cis = just plug the thing in ur calculator. The proof of the following proposition is straightforward and is left as an exercise. The magnitude of the angle itself can be increased or decreased by complete rotations about the circle/pole to arrive at the. In spite of this it turns out to be very useful to assume that there is a number ifor which one has (1) i2 = −1. This calculator supports the following operators, functions and constants: Operators + - * /: Addition, subtraction, multiplication and division; cis: Use for entering complex numbers in polar form (See Complex Numbers) Trigonometric functions. It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. (b) Find |z1| by first applying Eq. The expression cos(x) + i sin(x) is sometimes reduced to cis(x). 5 × 11 inches; NO calculators, phones, com-puters allowed; explain all work XC: Extra Credit 1. Addition and multiplication are defined on this larger set in such a way that i^2 = -1. The topic Complex Numbers involves investigating and extending understanding of the real number system to include complex numbers. 9 cis (3pi/2) Answer by solver91311(23736) (Show Source):. Answer to question 7: All three of the complex numbers above are cube roots of i. How to Understand Complex Numbers. Download Complex number calculator. We do not have to, but just for the fun of it, we will go one step further and extend our numbers once more, to complex numbers. Just type your formula into the top box. 2222, and 0. gl/uiTDQS Watch Part 2 of Complex Numbers- http://youtu. Includes functions, exponential and logarithmic functions, linear 2 x 2 and higher systems, graphing, and calculator use. Answer: 3 cos p 4 +isin p 4 r =3 x =cos p 4 = p 2 2 y=sin p 4 = p 2 2. Minimum Calculator Functionality Required for Engineering & CS Trigonometry, Matrix and Complex Numbers operations. A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i represents the imaginary unit, satisfying the equation i 2 = −1. Find more Mathematics widgets in Wolfram|Alpha. 2: Calculators and Trigonometric Functions of an Acute Angle. With some exceptions, we will take a micro-quiz each Wednesday, and homework is due each Friday. 22222 on a number line?Thank you! 1 educator answer If you are traveling 50 miles at 60mph how long would it take you to get there????. Offer six (optional) chapters which "Look Back" to the important topics in algebra. sin: Sine; asin: Inverse sine (sin-1) sinh: Hyperbolic sine; cos: Cosine; acos: Inverse cosine (cos-1). A truly remarkable property of complex multiplication is the angle-addition identity cis()cis() = cis( + ). APPM 4360/5360 Introduction to Complex Variables and Applications EXAM #1 Tuesday February 19, 2019 A formulae sheet is allowed: 8. In this assignment, we will create a calculator that performs simple arithmetic operations on complex numbers. The first step in finding the solutions of (that is, the x-intercepts of, plus any complex-valued roots of) a given polynomial function is to apply the Rational Roots Test to the polynomial's leading coefficient and constant term, in order to get a list of values that might possibly be solutions to the related polynomial equation. cis π = (cos π + i sin π) z = (-1)^(1/5) 1^(1/5) cis [(π + 2πk)/5] (De Moivre's Theorem). use of a calculator. For example (1,2) indicates that the real part i. set one complex number equal to another. asked by laura on July 16, 2016; math. 42 Roots of complex numbers: We now use de Moivres Theorem to find n w z suppose 0 roots: complex numbers 5 cis 45 and 3. To enter the complex number in polar form you enter mcisa, where m is the modulus and a is the argument of number. Assigned reading: Chapter 9 pages 151-158. De Moivre's Formula. Left–TI-84+C Asymptote detection turned off. [Explain why 0 is the right choice for the identity for addition, i. By employing the endoribonuclease Csy4 and its. AC Circuit Definitions; 9. He factored p2 + 3q2 into p+ p 3qand p p 3qand worked with these numbers, and made statements about them that were not completely justi ed. Calculators will be used. Trig Function Calculator is a free online tool that displays the values of six important trigonometric functions. In this unit, we extend this concept and perform more sophisticated operations, like dividing complex numbers. "ARCHERS, FIRE AT WILLIII" WRITE THE CAPITAL LETTER OF THE BEST ANSWER -180 The simplest form of is b. Radii are divided. The complex number z =6∠110. 6) = (25/9)cis(123. Minimum Calculator Functionality Required for Engineering & CS Trigonometry, Matrix and Complex Numbers operations. The calculator is standing by for input of a variable name to assign a value to the variable. • Utilize appropriate technology. CMATH for Delphi 5 v. Course Corequisites: MAT 097RQ. 2cis Relating Concepts 28. zip: 18k: 00-12-10: Crypt92 BETA 2/3 The second release of the only cryptography software for TI calculators. Find the following complex quotients using the complex conjugate. 7)^4 = cis(294. It is easier to see in polar form, the number r\cdot\text{cis}(\theta) . (Note: Your answers may involve a combination of notations). Rewrite the complex number −4 + 2 E in polar (cis) form and re-graph it on the polar graph paper. r (cos θ + i sin θ) Here r stands for modulus and θ stands for argument. 0 6 110 o Figure 3. Calculator cis form complex numbers - calculation: 5*cis(45°). Furthermore, Measurements. = 2 = cis c Use your answer from part b. Given that O is the origin, (a) find AB, giving your answer in the form , where a, b +; (3) (b) calculate in terms of π. Definition 2 A complex number is a number of the form a+ biwhere aand bare real numbers. You can use them to create complex numbers such as 2i+5. Complex numbers The equation x2 + 1 = 0 has no solutions, because for any real number xthe square x 2is nonnegative, and so x + 1 can never be less than 1. Giver as an integer and theta as a fraction of pi in radians. Calculators; fx-9750G Manual. Number of External Interfaces You can clear the table at anytime by clicking on this button. A truly remarkable property of complex multiplication is the angle-addition identity cis()cis() = cis( + ). Or in the shorter "cis" notation: (r cis θ) 2 = r 2 cis 2 θ Complex number forms review Review the different ways in which we can represent complex numbers: rectangular, polar, and exponential forms. gl/uiTDQS Watch Part 2 of Complex Numbers- http://youtu. The polar form of a complex number takes the form r(cos + isin ) Now r can be found by applying the Pythagorean Theorem on a and b, or: r = can be found using the formula: = So for this particular problem, the two roots of the quadratic equation are: Hence, a = 3/2 and b = 3√3 / 2. Rectangular and Polar Form. A complex number in Polar Form must be entered, in Alcula’s scientific calculator, using the cis operator. The implementation in Assembly. A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i represents the imaginary unit, satisfying the equation i 2 = −1. MATH 555 Func of a Complex Var with App 3 Credit Hours. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1. 7 degrees ( so (cis73. Complex number system.$$ This actually has three solutions, and we can find them using de Moivre's Theorem. The Complex Number System (8) The Number System (3) The Number System (13) The Number System (3) The Real Number System (5) Transformations (1) Trig Identities (12) Trigonometric Functions (25) Trigonometry (5) Uncategorized (20) Vector and Matrix Quantities (3) Vectors and Matrices (2) Vertical Angles (1) Visual Patterns (1). It can be written in the form a + bi. The complex numbers complete the number path that you’ve been following. Roots of a Complex Number This program calculates all n nth-roots of a complex number and saves the set of roots as a list. - Is that right? Also, how would you be able to change complex numbers into polar form? I know that. In a polar form, the division is somewhat simple. Those who were most successful used their graphics calculator to find the exact polar form of the complex numbers, as well as using the appropriate cis rule for the ‘Show that … ’ required in part (b). When zand ware any complex numbers, jzwj= jzjjwj, which implies that the absolute value function on nonzero complex numbers is a homomorphism C !R >0. 2 (trigonometric form of complex numbers) num ero+or and b den -ton. Apartment Complex For Sale CHMOD file permissions calculator. EN: complex-number-calculator menu Pre-Álgebra Orden (jerarquía) de operaciones Factores y números primos Fracciones Aritmética Decimales Exponentes y radicales Módulo. If we look at the complex number system, every quadratic equation has at least one complex solution; remember rational and irrational roots are in fact complex numbers. Note that real numbers are complex — a real number is simply a complex number with no imaginary part. z = (-1)^(1/5)-1 = -1 + 0i = 1 cis π. How to Understand Complex Numbers. Algebra (complex numbers) z = x + yi = r(cos θ + i sin θ) = r cis θ zx=+ 22 yr = –π < Arg z ≤ π z 1 z 2 = r 1 r 2 cis(θ 1 + θ 2) z z r r 1 2 1 2 =− cis()θθ 12 zn = rn cis(nθ) (de Moivre’s theorem) Calculus d dx xnnn = x −1 xd x n nn = xc n + ∫ + +≠ − 1 1 1, 1 d dx eaax ax= e ed x a ∫ ax ax=+ ec 1 d dx x e x ()log. 2 – Complex Numbers: Polar Form Page 1 of 4 June 2012. Bethany Bouchard An Analysis on Euler’s constant e and its functions in Number Theory, Complex Analysis, and Transcendental Number Theory “The difference… between reading a mathematical demonstration, and originating one wholly or partly, is very great. Trigonometric Form of Complex Numbers Calculator. Products and Quotients of Complex Numbers; Graphical explanation of multiplying and dividing complex numbers; 7. Checking the calculator’s answer by cubing the result will show that this value is, indeed, a root. so we introduced real numbers. z =i ⇒ z3 = cis π ⇒ z = cis π ’to’find’the’three’ cube’roots. Math Extended Essay 1. Convert the complex mean to polar coordinates whereupon the phase of the complex mean is the required angular mean. Fulfills Foundations - Mathematical Sciences. Needing a graphing calculator for my high-schooler, I researched the venerable TI-84 series and settled on the new Nspire CX CAS. Euler's Formula for complex numbers states that if z z z is a complex number with absolute value r z r_z r z and argument θ z \theta_z θ z , then. When graphing these, we can represent them on a coordinate plane called the complex plane. Then (1) z + z =2Rez, z −z =2i Imz, zz = |z|2, z z = cis(2 argz) for z 6=0, (2) z + w = z + w, z − w = z − w, zw = zw, z/w= z/w for w 6=0, (3) |z +w|≤|z. 2 3i 1 + 6i 8. 5 + i sin 3. Another way of writing the polar form of the number is using it’s exponential form: me^(ia. lumenlearning. Go to the editor. Topics covered include the field axioms, the real and complex numbers, sequences, se-ries, complex functions, complex differentiation, power series, and the complex exponential. This calculator supports the following operators, functions and constants: Operators + - * /: Addition, subtraction, multiplication and division; cis: Use for entering complex numbers in polar form (See Complex Numbers) Trigonometric functions. Complex Numbers and the Complex Exponential 1. RCL The calculator is standing by for input of a variable name to recall the variable’s value. given set of computing Goal 2: Critical Thinking: Graduates of. α = 98 ∘ a = 34. You don't have define what e is in your post. In this case, the left hand side is a multi-valued function, and the right hand side is one of its possible values. This form, a+ bi, is called the standard form of a complex number. The appearance of the calculator is modelled after the HP-10C and related calculators from Hewlett-Packard. In addition, students will review and prepare for the Math portion of the MCA III (Minnesota Comprehensive Assessment). The majority of students managed well in the first parts of this problem. Find the modulus-argument form of the complex number z=(5√ 3 - 5i) The easiest way to complete questions of these types is to first sketch an Argand diagram. 406-407/11, 12, 15-17, 23, 24, 29, 30, 32 and D p. The receipt number is a unique 13-character identifier that USCIS provides for each application or petition it receives. In a polar form, the division is somewhat simple. Convert the complex number from polar form to cis (@) form with theta in radians: Te cis = (8e***/8)2/3 T- 0 radians. For the complex number a + bi, a is called the real part, and b is called the imaginary part. 4 Polar representation of complex numbers For any complex number z= x+ iy(6= 0), its length and angle w. Use it to derive the theorem of De Moivre, which says that ( )18 (r cis )n = rn cis(n) for any numbers r and , and any integer n. By using this website, you agree to our Cookie Policy. and is in the second quadrant since that is the location the complex number in the complex plane. Binomic (a+bi) & CIS. Polar - Rectangular Coordinate Conversion Calculator. In the complex plane The unit circle can be considered as the unit complex numbers , i. YOUR EXAM NUMBER Please cross out this problem if you do not wish it graded Problem 6A. 2 • Add, subtract, and multiply complex numbers Quadratic Functions Chapter – 1. The proof of the following proposition is straightforward and is left as an exercise. Conjugates Calculator-- Enter Fraction with Conjugate. You can write a book review and share your experiences. Be able to add, subtract, multiply, and divide any integers (using a calculator if needed). I am looking for suggestions on how to typeset complex numbers in the modulo-argument form, sometimes called phasor notation. Find the absolute value of a complex number. Proposition 2. Chapter 2-6: Complex Numbers and Polar Coordinates Chapter 2. ) 130 ( cos 3. A scientific (nongraphing) calculator, the TI-30XS MultiView™, is integrated into the exam software and available to students during the entire testing time. To add complex numbers, add the real part together and add the imaginary parts together. 1 Non-right Triangles: Law of Sines 1. The modulus and argument are fairly simple to calculate using trigonometry. Math Extended Essay 1. MATHEMATICS 4471. 29MB) Chapter 4 Complex Numbers ( 0. You can store a real or complex number or an expression result to a memory variable. Or in the shorter "cis" notation: (r cis θ) 2 = r 2 cis 2θ. 02 Roots of complex numbers. Example Find a ⋅ b when a = <3, 5, 8> and b = <2, 7, 1>. For the complex number a + bi, a is called the real part, and b is called the imaginary part. a G XMXaCdde 9 9waiht5hB 1I2nAfUizn ZibtMeV fA Sl Agesb 7rfa G G2D. The x is the real number of the expression and the y represents the imaginary number of the expression. However, don’t forget that aor bcould be zero, which means numbers like 3iand 6 are also complex numbers. 2: POLAR FORM OF A COMPLEX NUMBER. For a more in-depth view, request access to our more powerful, 100% accurate calculator. The receipt number consists of three letters-for example, EAC, WAC, LIN, SRC, NBC, MSC or IOE-and 10 numbers. Rewrite each of the following complex numbers in rectangular form and graph. If it is never a complex number, then why no solutions? complex ions complex numbers How to draw an Argand diagram for this complex number? (FP1) C Programming File I/O Dynamic Memory Matrix Multiplication Lab Why can't my calculator do log -2(4) ?. Measurement and. ) Prerequisite(s): INT-1010 with grade of C or higher, or permission of department chair. Convert the complex number from polar form to cis (C) form with theta in radians: Te's cis Question 17 Convert the complex number from rectangular form to polar form. Polar addition calculator. For the second entry, we add the theta's. There is a more general version, in which n n n is allowed to be a complex number. AC Circuit Definitions; 9. Complex Numbers and Polar Form of a Complex Number Interactive Graph - Convert polar to rectangular and vice-versa In the following graph, the real axis is horizontal, and the imaginary (j=sqrt(-1)) axis is vertical, as usual. Then we can figure out the exact position of $$z$$ on the complex plane if we know two things: the length of the line segment and the angle measured from the positive real axis to the line segment. Use a calculator to write the complex number in standard form. Information about how to use the calculator is available in the Help icon under the Calculator tab. Get 1:1 help now from expert Precalculus tutors Solve it with our pre-calculus problem solver and calculator. In general, a complex number like: r(cos θ + i sin θ). Prerequisite: ACT math score of 23 or above; SAT math score of 560 or above; MATH 140 with a minimum grade of C-; or Accuplacer Advanced Algebra and Functions test with a score. 2 class/2 lab hours. Cis(x) is another name for the complex exponential, Cis(x)=e^(ix)=cosx+isinx. 9 degrees) so [sqrt(5/3)cis(210. (c) Determine the product z1z2 in polar form. The complex numbers of the form Aexp(ip) are called phasors, and contain the information of both the amplitude and the phase of the signal. In this unit, we extend this concept and perform more sophisticated operations, like dividing complex numbers. Rectangular forms of numbers take on the format, rectangular number= x + jy, where x and y are numbers. asked by laura on July 16, 2016; math. Question 1040090: Write the following complex nr in polar form - No calculator to be used: 1. Powers and Roots of Complex Numbers; 8. If it is never a complex number, then why no solutions? complex ions complex numbers How to draw an Argand diagram for this complex number? (FP1) C Programming File I/O Dynamic Memory Matrix Multiplication Lab Why can't my calculator do log -2(4) ?. Every complex number can be written in the form a + bi. : sqrt(re 2 + im 2)) of the complex result. Complex number notation Nothing unexpected here, th…. Then (1) z + z =2Rez, z −z =2i Imz, zz = |z|2, z z = cis(2 argz) for z 6=0, (2) z + w = z + w, z − w = z − w, zw = zw, z/w= z/w for w 6=0, (3) |z +w|≤|z. • Cis the complex numbers Notice that a number is positive if it is greater than zero, so zero is not positive. Seamless bi-directional integration with Allegro PCB enables data synchronization and cross. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1. 1 degrees (coterminal with 210. Solving a system consists in finding the value for the unknown factors in a way that verifies all the equations that make up the system. 22222 on a number line?Thank you! 1 educator answer If you are traveling 50 miles at 60mph how long would it take you to get there????. (e) Determine z3 1 in polar form. The argument θ of a complex number z is often denoted by arg z. Find the absolute value of a complex number. From the keyboard and computer-menu interface to the ability to save and share work with built-in apps, the TI-Nspire™ CX II models retain the features and functionality that make TI-Nspire™ CX graphing calculators ideal for math and science from middle grades through college. Be able to add, subtract, multiply, and divide any integers (using a calculator if needed). We denote a parametrized curve in the. jl package, involving real and complex numbers. Convert the complex number from polar form to cis (@) form with theta in radians: Te cis = (8e***/8)2/3 T- 0 radians Get more help from Chegg Get 1:1 help now from expert Precalculus tutors Solve it with our pre-calculus problem solver and calculator. Consider the complex number ω = , where z = x + iy and i =. First step is to write each of the complex numbers in polar coordinates; r = sqrt(a^2 + b^2) and theta = arctan(b/a) 1) r = sqrt(5/3); theta = -49. In part (i) many candidates tried to multiply it out the binomials rather than using the binomial theorem. First step is to write each of the complex numbers in polar coordinates; r = sqrt(a^2 + b^2) and theta = arctan(b/a) 1) r = sqrt(5/3); theta = -49. i have just preformed the calculation on my AFX, here is the result: 0. Find the absolute value of a complex number. The polar form of a complex number takes the form r(cos + isin ) Now r can be found by applying the Pythagorean Theorem on a and b, or: r = can be found using the formula: = So for this particular problem, the two roots of the quadratic equation are: Hence, a = 3/2 and b = 3√3 / 2. Here are some examples of what you would type here: (3i+1)(5+2i) (-1-5i)(10+12i) i(5-2i). An algebraic derivation. Complex integration: Cauchy integral theorem and Cauchy integral formulas Definite integral of a complex-valued function of a real variable Consider a complex valued function f(t) of a real variable t: f(t) = u(t) + iv(t), which is assumed to be a piecewise continuous function defined in the closed interval a ≤ t ≤ b. Radii are divided. Free Online Scientific Notation Calculator. Credit may not be earned for both INT-1550 and CIS-1400 or for both ENT-1890 and INT-1550. Using the previous result: z =6(cos110 +isin110 ) Using a calculator we find z =−2. Day 4 cis, Products of Complex Numbers HW B p. As a helping tool, I would suggest that you get a copy of Algebrator. See full list on brownmath. Complex number system. All Functions Operators +. This website uses cookies to ensure you get the best experience. This programmable calculator does real or complex number calculations with 100 significant figure precision. Complex numbers, definition, conjugates, inverses, complex roots of equations, trig form of complex numbers, powers and roots of complex numbers, DeMoivre's theorem Study Materials: The Math Placement D Test assesses students' precalculus skills. Write your answer in simplest form [2+3+4=9 marks] cis cis 2m cis Find Arg cis [3 marks]. You have to check three things: † Show that Cis closed under addition | the sum of two complex numbers is again a complex number. Our hourly paycheck calculator accurately estimates net pay (sometimes called take-home pay or home pay) for hourly employees after withholding taxes and deductions. So a curve is a function : [a;b] ! C(from a finite closed real intervale [a;b] to the plane). The receipt number is a unique 13-character identifier that USCIS provides for each application or petition it receives. Meeting these challenges will facilitate transgene expression regulation and support the fine-tuning of metabolic pathways to avoid the accumulation of undesired intermediates. For example, a browser window itself contains a number of other objects like an HTML document, a titlebar, a menubar, a toolbar, optional scrollbars and a statusbar along the bottom. Another way of writing the polar form of the number is using it's exponential form: me^(ia. This is in polar form. Test your understanding of the binary number system. 2% of the probes (median = 0. Instructions:: All Functions. A brief introduction to POLAR COORDINATES Cartesian Co-ordinates Polar Co-ordinates Usually. But it still leaves the question: “Why are calculators providing me with a complex root?”. Determine the conjugate of the denominator. Introduction to college-level algebra. 11 047321CIS Math Modeling Math Modeling CIS introduces students to the art of mathematical prediction. Chapter 4: Complex numbers and functions. the last step is to simplify the result. Prerequisite(s): Within the last three years: MAT 092 with a grade of B or better or placement into MAT 097. Chapter 2-6: Complex Numbers and Polar Coordinates Chapter 2. calculator (TI-83 or TI-83 plus or TI-84 or TI-84plus). Concepts and techniques. We are going to study equation (1) in a more general context. The radius vector r is called the modulus or absolute value of the complex number and the polar angle θ is called the amplitude or argument of the number. This calculator extracts the square root, calculate the modulus, finds inverse, finds conjugate and transform complex number to polar form. The calculator will simplify any complex expression, with steps shown. Waveguide Loss Calculator. Interestingly, the variant with the most trans effects (37. Then (1) z + z =2Rez, z −z =2i Imz, zz = |z|2, z z = cis(2 argz) for z 6=0, (2) z + w = z + w, z − w = z − w, zw = zw, z/w= z/w for w 6=0, (3) |z +w|≤|z. Binomic (a+bi) & CIS. be/OLoxp9xOs94 In this video I'll te. Whether you've loved the book or not, if you give your honest and detailed thoughts then people will find new books that are right for them. Contents1 DEVRY CIS 115 LOGIC AND DESIGN TAKE MY ONLINE CLASS2 Devry CIS 115 Week 1 Software Development and Design Tools2. Region R is bounded by the set of all complex numbers on the Argand plane z a bi where a and b are real numbers satisfying the equation a a b b22 12 13 6 6. Structure of the number system, logical thinking, topics in geometry, simple functions, and basic statistics and probability. Checking the calculator’s answer by cubing the result will show that this value is, indeed, a root. 7 degrees ( so (cis73. A complex number is a number of the form a+ bi, where aand bare real numbers and iis the imaginary unit. Using a calculator on Secant, Co-Secant, Co-Tangent These new trig functions are just the reciprocal (flip) of sine, cosine and tangent, but they can be confusing, so we'll emphasize always writing them in the correct order each time, and we'll do lots of examples. r (cos θ + i sin θ) Here r stands for modulus and θ stands for argument. Suppose we wish to solve the equation z^3 = 2+2 i. Basic features such as expanding, factoring, and solving simple equations are covered as well as more advanced techniques such as solving complex equations, derivatives, integrals and differential equations. (2√5)(4/2√5 + 1/√5 i) = 2√5(cis Θ) where tanΘ = 1/√5 over 4/2√5 = 1/2 and Θ=0. The horizontal axis is the real axis and the vertical axis is the imaginary axis. 00 Cumcroi est un programme pour Ti89/Ti92/Ti92Plus, écrit en Ti-Basic. Instructions. Introduction to college-level algebra. This really doesn't use differential calculus at all. Radii are divided. The notation is less commonly used than Euler's formula , e i x {\displaystyle e^{ix}} , which offers an even shorter and more general notation for cos( x ) + i sin( x ). AC Circuit Definitions; 9. Exponential Form of Complex Numbers; Euler Formula and Euler Identity interactive graph; 6. Input the complex binomial you would like to graph on the complex plane. 2% of the probes (median = 0. Support for most mathematical operations available in Julia standard library and special functions from SpecialFunctions. Real numbers aren’t enough to solve x2 = −1, so we invent an answer, called i, and arrive at the complex numbers. r 2 (cos 2θ + i sin 2θ) (the magnitude r gets squared and the angle θ gets doubled. and the set of rational numbers (ratios of integers) by Q = fp=q: p;q2Z and q6= 0 g: The letter \Z" comes from \zahl" (German for \number") and \Q" comes from \quotient. Applications include the construction and description of certain characteristics of the natural numbers, integers, rational, real, and complex numbers. Soon after, we added 0 to represent the idea of nothingness. Quadratics D1D3 self written Quiz E. Getting Acquainted - Read This First! ( 0. An algebraic derivation. This function stores the complex number found in c into the i-th complex entry of that array. Exponential Form of Complex Numbers; Euler Formula and Euler Identity interactive graph; 6. school of mathematical sciences eng1090 problem set eng1090 foundation mathematics problem set complex numbers complex algebra james: p196, questions 10, 12, 15. Assigned reading: Chapter 9 pages 151-158. The Complex Number System (8) The Number System (3) The Number System (13) The Number System (3) The Real Number System (5) Transformations (1) Trig Identities (12) Trigonometric Functions (25) Trigonometry (5) Uncategorized (20) Vector and Matrix Quantities (3) Vectors and Matrices (2) Vertical Angles (1) Visual Patterns (1). 00 Cumcroi est un programme pour Ti89/Ti92/Ti92Plus, écrit en Ti-Basic. Example: type in (2-3i)*(1+i), and see the answer of 5-i. ) Prerequisite(s): INT-1010 with grade of C or higher, or permission of department chair. c = cis Write cis in expanded form. Watch My Other Scientific Calculator Tutorials Below- http://goo. Sketch these four numbers in the complex plane. Click "Submit. 17 Complex numbers z1 and z2 are given by z1 =3− j2 z2 =−4+ j3 (a) Express z1 and z2 in polar form. A complex number in Polar Form must be entered, in Alcula’s scientific calculator, using the cis operator. Viewed in terms of linear transformations, y= ceat is the solution to the vector equation T(y) = ay; (1) where T: Ck(I) !Ck 1(I) is T(y) = y0. Now we have. cis(x) cos(x) + isin(x) cis(x) Not everybody is familiar with the "cis" notation. The appearance of the calculator is modelled after the HP-10C and related calculators from Hewlett-Packard. An algebraic derivation. Calculators; fx-9750G Manual. given set of computing Goal 2: Critical Thinking: Graduates of. We also learn about a different way to represent complex numbers—polar form. the value x(a) is the number of fixed points of a in S. 5 × 11 inches; NO calculators, phones, com-puters allowed; explain all work XC: Extra Credit 1. = 2 = cis c Use your answer from part b. Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as Pythagoras, Descartes, De Moivre, Euler, Gauss, and others. Rewrite each of the following complex numbers in rectangular form and graph. When squared becomes:. Convert the complex mean to polar coordinates whereupon the phase of the complex mean is the required angular mean. (e) Determine z3 1 in polar form. Free Online Scientific Notation Calculator. By using this website, you agree to our Cookie Policy. Product Theorem r1 cis 1 r2 cis 2 r1r2 cis 1 2 Quotient Theorem r1 cis 1 r2 cis 2 r1 r2 cis 1 2. Solution: Let M denote the max of the absolute values of the entries of A. However this creates a discontinuity as moves across the negative real axis. Does the complex ion [Co(en)2Cl2]^+ have cis-trans geometric isomers? asked by Eliz on May 13, 2009; algebra. 3: Complex Numbers The Definition A complex number something of the form a b+ i, where a∈ℝ, b∈ℝ, and i ≡ − 1 (the triple symbol means “is defined to be.
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# How long do solar eclipses last on Metis?
How long does Jupiter obscure the sun, from the perspective of Metis, the innermost moon of Jupiter?
In other words, how long does Metis remain in the shadow of Jupiter?
If it varies by time of year, what are the minimum/maximum values?
I am not sure of any direct measurement, but it can be figured out rather easily and accurately from theory with a little bit of maths. We have the following parameters for Metis and Jupiter (taken off Wikipedia):
• Diameter of Jupiter: about 140 Mm (megameters, 1 Mm = 1000 km), so radius ~70 Mm. Compare Earth at ~6.5 Mm radius, 13 Mm diameter.
• Orbital distance for Metis from Jupiter's center: 128 Mm
• Eccentricity: effectively zero, so we can treat it very accurately as a circular orbit.
• Orbital period: 25.4690 ks (kiloseconds, 1 ks = 1000 s). Compare against Earth's day, 86.4 ks, and orbital period of its Moon, about 2500 ks (more accurate: 2360 ks).
• Also, one more relevant point is the distance to the Sun and its diameter, to find the apparent size: average distance to the Sun is 779 Gm ($$7.79 \times 10^5\ \mathrm{Mm}$$), and Sun's diameter roughly 1 Gm ($$10^3\ \mathrm{Mm}$$).
The way we will reason now is as follows. Jupiter "moves" in one circuit across the sky dome of an observer on Metis in one orbit. Hence, the time of eclipse is the time required for Jupiter to "move", from arrival of its leading edge to departure of its trailing edge, across the image of the Sun on the same sky dome. Equivalently, then, it is the time required for it to move through an angular distance equal to its own angular diameter plus two angular diameters of the Sun (one on first contact, the other before last contact):
$$\mbox{Eclipse time} := \frac{2 \cdot (\mbox{Angular diameter of Sun}) + \mbox{Angular diameter of Jupiter}}{\mbox{Angular speed of Jupiter on sky dome}}$$
Now the denominator is easy: a full revolution is $$2\pi\ \mathrm{rad}$$ or about $$6283\ \mathrm{mrad}$$, so we just divide that by 25.4690 ks, 4o get
$$\mbox{Angular speed of Jupiter on sky dome} \approx 246.7\ \frac{\mbox{mrad}}{\mbox{ks}}$$
Hence, now all we need is the angular diameters of the Sun and Jupiter in milliradians and we'll be done.
This is easily doable with a little bit of trigonometry: while I do not have a diagram on hand, you can easily find one. The angular size of an object of girth $$D$$ emplaced at a distance $$d$$ from an observer is
$$\mbox{Angular size} := 2 \cdot \tan^{-1}\left(\frac{D}{2d}\right)$$
Note that you'll often see the approximation
$$\mbox{Angular size} \approx \frac{D}{d}$$
but this is only good when the object is very far, i.e. $$d \gg D$$. In this case, though, that is not the case at least for Jupiter, so we have to use the full formula with the inverse tangent. Using the figures just given, we have
$$\mbox{Angular size of Jupiter} = 2 \cdot \tan^{-1}\left(\frac{140\ \mathrm{Mm}}{2 \cdot 128\ \mathrm{Mm}}\right) \approx 1.000\ \mathrm{rad}$$
or accurately(!) $$1000\ \mathrm{mrad}$$. The Sun is far enough that we can just use the approximation and get 1/779 rad or about $$1.28\ \mathrm{mrad}$$. Hence the total angular distance to cover is 1.28 + 500 + 1.28 ~ 503 mrad, and the eclipse time equals
$$\mbox{Eclipse time} = \frac{1000\ \mathrm{mrad}}{246.7\ \mathrm{mrad/ks}} \approx 4.08\ \mathrm{ks}$$
or for so many still not fully accustomed to the SI, a bit longer than an hour (3.6 ks), and in local measures, a bit less than 1/5 of a Metian day-month.
NOTE: I should point out a caveat of this is is that that this doesn't take into account the inclination of orbits - both Jupiter's and Metis's which may make the Sun in effect "encounter" Jupiter at a different point than simply cutting along its equator. But it should still give you an idea - 4 ks or so, plus-minus some amount that'd be too much work to get.
• Wow, so it spends about 16% of its time fully in shadow. Thanks! – cowlinator Dec 21 '19 at 3:33
• @cowlinator : Yep (4/25 ~ 16%). Unless I made some error somewhere in this so I'd like others to check it. – The_Sympathizer Dec 21 '19 at 3:42
| 1,190 | 4,121 |
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CC-MAIN-2021-25
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https://math.answers.com/math-and-arithmetic/How-to-find-the-equation-of-a-line
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0
# How to find the equation of a line?
Updated: 10/19/2022
Oneshi Weliwitigoda
Lvl 1
3y ago
Equation of a line may be written as y = mx + c.
m is called the slope of the line.
c is the point where the line crosses the y axis.
If two points are given: (x1, y1) and (x2, y2), m is calculated as the y difference divided by the x difference:
m = (y2 - y1) / (x2 - x1)
Once you find m, you can find c by putting in the values into the equation y=mx+c. For example, if you use (x1, y1), you can do this:
y1 = m*x1 + c
take m*x1 to the other side:
y1 - m*x1 = c
Then you get the value of c. Now you have both m and c, so you can write the equation of the line:
y = mx + c
Put the values of m and c in. Leave y and x as it is.
a.net/math_problems/equations-of-lines-problems-with-solutions.html
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http://www.schoolphysics.org/age16-19/Electricity%20and%20magnetism/electrostatics/text/Capacitor_networks/index.html
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# Capacitor networks
In practical circuits capacitors are often joined together. We will consider the cases of two capacitors, first in parallel and then in series.
(a) Capacitors in parallel
Consider two capacitors connected in parallel as shown in Figure 1
The potential across both capacitors is the same (V) and let the charges on the capacitors be Q1 and Q2 respectively.
Now Q = CV, and so Q = C1V and Q = C2V. But the total charge stored Q = Q1 + Q2, therefore
Q = Q1 + Q2 = V(C1 + C2)
Giving:
Capacitors in parallel: C = C1 + C2
where C is the capacitance of the combination.
(b) Capacitors in series
In this case the capacitors are connected as shown in Figure 2.
The charge stored by each capacitor is the same. If V1 and V2 are the potentials across C1 and C2 respectively then:
V1 = Q/C1 and V2 = Q/C2. Therefore: V = V1 + V2 = Q(1/C1 + 1/C2)
Hence:
Capacitors in series: 1/C = 1/C1 + 1/C2
where C is the capacitance of the combination.
(Notice that they are the "reverse" of the formulae for two resistors in series and parallel)
## The charge distribution on series and parallel capacitors
When two capacitors are joined tighter in a circuit and then connected to a voltage supply charge will move onto the plates. The actual distribution of charge for a series and parallel circuit is shown in Figure 3(a) and (b).
## Joining two charged capacitors
If two capacitors are joined together as shown in Figure 4 then:
(a) there is no change in the total charge stored by the system;
(b) the potential across the two capacitors becomes equal
(c) the combined capacitance of the two capacitors in parallel becomes
C = C1 + C2
There is usually a loss in energy when the two capacitors are joined; this is because unless the potential differences across them are equal, charge will flow to equalise this difference.
The flow of charge results in heating in the connecting wires and a consequent loss of energy.
## Breakdown potential for a capacitor
Every capacitor has a working voltage, this is the maximum potential that should be applied between the plates. Any more than this and the dielectric material between the plates will break down and become conducting and the capacitor will be destroyed, usually resulting in a small bang as the material breaks down.
Example problems
1. Calculate the capacitance of the capacitor formed by joining:
(a) two 100 mF capacitors in series
(b) two 100 mF capacitors in series
(a) 1/C = 1/C1 + 1/C2 = 2/100x10-6 = 20 000 Therefore C = 50 mF
(b) C = C1 + C2 = 100x10-6 + 100x10-6 = 200 mF
2. 250 mF capacitors are joined in series to a 12V supply.
Calculate:
(a) the potential difference across each capacitor
(b) the charge on each plate of each capacitor
(a) since the capacitors are equal the potential across each will be the same and equal to 6V
(b) for each capacitor C = Q/V therefore Q = CV = 250x10-6x12 = 3x10-3 C = 3 mC
## Capacitance formulae
The formulae for capacitors of some other common shapes are given below, although the proofs of these formulae are not needed at this level.
Sphere (radius a) 4pea Concentric spheres (radii a and b) 4peab/[b-a] Concentric cylinders (radii a,b,length L) 2peL/[ln(b/a)] Two long, parallel wires (separation d,radius a,length L (d >> a) eL/ln[d/a] Parallel-plate capacitor of area A containing a thickness x of dielectric and thickness b of air eA/(erb + x)
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| 1,031,581,276 | 10,679 |
# Search by Topic
#### Resources tagged with Mathematical reasoning & proof similar to Take a Square II:
Filter by: Content type:
Stage:
Challenge level:
### There are 177 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
### Folding Squares
##### Stage: 4 Challenge Level:
The diagonal of a square intersects the line joining one of the unused corners to the midpoint of the opposite side. What do you notice about the line segments produced?
### Take a Square II
##### Stage: 4 Challenge Level:
What fractions can you divide the diagonal of a square into by simple folding?
### Zig Zag
##### Stage: 4 Challenge Level:
Four identical right angled triangles are drawn on the sides of a square. Two face out, two face in. Why do the four vertices marked with dots lie on one line?
### Pareq Exists
##### Stage: 4 Challenge Level:
Prove that, given any three parallel lines, an equilateral triangle always exists with one vertex on each of the three lines.
### Always Perfect
##### Stage: 4 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
### Sixational
##### Stage: 4 and 5 Challenge Level:
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .
### Postage
##### Stage: 4 Challenge Level:
The country Sixtania prints postage stamps with only three values 6 lucres, 10 lucres and 15 lucres (where the currency is in lucres).Which values cannot be made up with combinations of these postage. . . .
### Prime AP
##### Stage: 4 Challenge Level:
Show that if three prime numbers, all greater than 3, form an arithmetic progression then the common difference is divisible by 6. What if one of the terms is 3?
### Angle Trisection
##### Stage: 4 Challenge Level:
It is impossible to trisect an angle using only ruler and compasses but it can be done using a carpenter's square.
### Whole Number Dynamics II
##### Stage: 4 and 5
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
### Whole Number Dynamics I
##### Stage: 4 and 5
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
### Pythagorean Triples II
##### Stage: 3 and 4
This is the second article on right-angled triangles whose edge lengths are whole numbers.
### Pythagorean Triples I
##### Stage: 3 and 4
The first of two articles on Pythagorean Triples which asks how many right angled triangles can you find with the lengths of each side exactly a whole number measurement. Try it!
### A Biggy
##### Stage: 4 Challenge Level:
Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.
### Long Short
##### Stage: 4 Challenge Level:
A quadrilateral inscribed in a unit circle has sides of lengths s1, s2, s3 and s4 where s1 ≤ s2 ≤ s3 ≤ s4. Find a quadrilateral of this type for which s1= sqrt2 and show s1 cannot. . . .
### Janine's Conjecture
##### Stage: 4 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Leonardo's Problem
##### Stage: 4 and 5 Challenge Level:
A, B & C own a half, a third and a sixth of a coin collection. Each grab some coins, return some, then share equally what they had put back, finishing with their own share. How rich are they?
### Proof Sorter - Quadratic Equation
##### Stage: 4 and 5 Challenge Level:
This is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.
### To Prove or Not to Prove
##### Stage: 4 and 5
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
### Proof: A Brief Historical Survey
##### Stage: 4 and 5
If you think that mathematical proof is really clearcut and universal then you should read this article.
### Picture Story
##### Stage: 4 Challenge Level:
Can you see how this picture illustrates the formula for the sum of the first six cube numbers?
### Square Mean
##### Stage: 4 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
##### Stage: 4 Challenge Level:
Four jewellers possessing respectively eight rubies, ten saphires, a hundred pearls and five diamonds, presented, each from his own stock, one apiece to the rest in token of regard; and they. . . .
### Natural Sum
##### Stage: 4 Challenge Level:
The picture illustrates the sum 1 + 2 + 3 + 4 = (4 x 5)/2. Prove the general formula for the sum of the first n natural numbers and the formula for the sum of the cubes of the first n natural. . . .
### There's a Limit
##### Stage: 4 and 5 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### A Knight's Journey
##### Stage: 4 and 5
This article looks at knight's moves on a chess board and introduces you to the idea of vectors and vector addition.
### Iffy Logic
##### Stage: 4 Short Challenge Level:
Can you rearrange the cards to make a series of correct mathematical statements?
### The Great Weights Puzzle
##### Stage: 4 Challenge Level:
You have twelve weights, one of which is different from the rest. Using just 3 weighings, can you identify which weight is the odd one out, and whether it is heavier or lighter than the rest?
### Some Circuits in Graph or Network Theory
##### Stage: 4 and 5
Eulerian and Hamiltonian circuits are defined with some simple examples and a couple of puzzles to illustrate Hamiltonian circuits.
### Sprouts Explained
##### Stage: 2, 3, 4 and 5
This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
### Pythagoras Proofs
##### Stage: 4 Challenge Level:
Can you make sense of these three proofs of Pythagoras' Theorem?
### Breaking the Equation ' Empirical Argument = Proof '
##### Stage: 2, 3, 4 and 5
This article stems from research on the teaching of proof and offers guidance on how to move learners from focussing on experimental arguments to mathematical arguments and deductive reasoning.
### Advent Calendar 2011 - Secondary
##### Stage: 3, 4 and 5 Challenge Level:
Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
### A Long Time at the Till
##### Stage: 4 and 5 Challenge Level:
Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem?
### L-triominoes
##### Stage: 4 Challenge Level:
L triominoes can fit together to make larger versions of themselves. Is every size possible to make in this way?
### Geometric Parabola
##### Stage: 4 Challenge Level:
Explore what happens when you draw graphs of quadratic equations with coefficients based on a geometric sequence.
### Unit Interval
##### Stage: 4 and 5 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
### Composite Notions
##### Stage: 4 Challenge Level:
A composite number is one that is neither prime nor 1. Show that 10201 is composite in any base.
### Yih or Luk Tsut K'i or Three Men's Morris
##### Stage: 3, 4 and 5 Challenge Level:
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
### Whole Number Dynamics V
##### Stage: 4 and 5
The final of five articles which containe the proof of why the sequence introduced in article IV either reaches the fixed point 0 or the sequence enters a repeating cycle of four values.
### Whole Number Dynamics IV
##### Stage: 4 and 5
Start with any whole number N, write N as a multiple of 10 plus a remainder R and produce a new whole number N'. Repeat. What happens?
### Geometry and Gravity 2
##### Stage: 3, 4 and 5
This is the second of two articles and discusses problems relating to the curvature of space, shortest distances on surfaces, triangulations of surfaces and representation by graphs.
### Picturing Pythagorean Triples
##### Stage: 4 and 5
This article discusses how every Pythagorean triple (a, b, c) can be illustrated by a square and an L shape within another square. You are invited to find some triples for yourself.
### Magic Squares II
##### Stage: 4 and 5
An article which gives an account of some properties of magic squares.
### Mediant
##### Stage: 4 Challenge Level:
If you take two tests and get a marks out of a maximum b in the first and c marks out of d in the second, does the mediant (a+c)/(b+d)lie between the results for the two tests separately.
### Mouhefanggai
##### Stage: 4
Imagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai.
### The Triangle Game
##### Stage: 3 and 4 Challenge Level:
Can you discover whether this is a fair game?
### Impossible Sandwiches
##### Stage: 3, 4 and 5
In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.
### Whole Number Dynamics III
##### Stage: 4 and 5
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
### DOTS Division
##### Stage: 4 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
| 2,482 | 10,523 |
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| 3.953125 | 4 |
CC-MAIN-2015-27
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http://rog.pynguins.com/article/item/python2?comments
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| 281,247,697 | 14,933 |
## Euler Problems 11 to 22
```Problem 11.
#! /usr/bin/python
'''In the 20×20 grid below, four numbers along a diagonal line have
been marked in red.
08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
The product of these numbers is 26 × 63 × 78 × 14 = 1788696.
What is the greatest product of four adjacent numbers in any direction
(up, down, left, right, or diagonally) in the 20×20 grid?'''
__author__="rog"
__date__ ="\$29-Jul-2010 12:44:04\$"
import random
#from numpy import *
import sys
sys.path.append("../")
from rogutil import *
numstr = """08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48"""
def findindexerror(mylist, y, x):
result = False
try:
a = mylist[y][x]
except IndexError:
result = True
return result
def main():
# Convert the string of numbers into an array of arrays called ylist
numy = numstr.split("\n")
ylist = []
for line in numy:
numx = line.split(" ")
numx = [int(integral) for integral in list(numx)]
ylist.append(numx)
# Iterate all items individually
xpos = 0
ypos = 0
highscore = 0
results = []
for y in ylist:
xpos = 0
for x in y:
# 1
if not findindexerror(ylist, ypos, xpos + 3):
resultlist = []
resultlist.append(ylist[ypos][xpos])
resultlist.append(ylist[ypos][xpos + 1])
resultlist.append(ylist[ypos][xpos + 2])
resultlist.append(ylist[ypos][xpos + 3])
results.append(multiplylist(resultlist))
# 2
if not findindexerror(ylist, ypos, xpos - 3):
resultlist = []
resultlist.append(ylist[ypos][xpos])
resultlist.append(ylist[ypos][xpos - 1])
resultlist.append(ylist[ypos][xpos - 2])
resultlist.append(ylist[ypos][xpos - 3])
results.append(multiplylist(resultlist))
# 3
if not findindexerror(ylist, ypos - 3, xpos):
resultlist = []
resultlist.append(ylist[ypos][xpos])
resultlist.append(ylist[ypos - 1][xpos])
resultlist.append(ylist[ypos - 2][xpos])
resultlist.append(ylist[ypos - 3][xpos])
results.append(multiplylist(resultlist))
# 4
if not findindexerror(ylist, ypos + 3, xpos):
resultlist = []
resultlist.append(ylist[ypos][xpos])
resultlist.append(ylist[ypos + 1][xpos])
resultlist.append(ylist[ypos + 2][xpos])
resultlist.append(ylist[ypos + 3][xpos])
results.append(multiplylist(resultlist))
# 5
if not findindexerror(ylist, ypos + 3, xpos + 3):
resultlist = []
resultlist.append(ylist[ypos][xpos])
resultlist.append(ylist[ypos + 1][xpos + 1])
resultlist.append(ylist[ypos + 2][xpos + 2])
resultlist.append(ylist[ypos + 3][xpos + 3])
results.append(multiplylist(resultlist))
# 6
if not findindexerror(ylist, ypos - 3, xpos - 3):
resultlist = []
resultlist.append(ylist[ypos][xpos])
resultlist.append(ylist[ypos - 1][xpos - 1])
resultlist.append(ylist[ypos - 2][xpos - 2])
resultlist.append(ylist[ypos - 3][xpos - 3])
results.append(multiplylist(resultlist))
# 7
if not findindexerror(ylist, ypos - 3, xpos + 3):
resultlist = []
resultlist.append(ylist[ypos][xpos])
resultlist.append(ylist[ypos - 1][xpos + 1])
resultlist.append(ylist[ypos - 2][xpos + 2])
resultlist.append(ylist[ypos - 3][xpos + 3])
results.append(multiplylist(resultlist))
# 8
if not findindexerror(ylist, ypos + 3, xpos - 3):
resultlist = []
resultlist.append(ylist[ypos][xpos])
resultlist.append(ylist[ypos + 1][xpos - 1])
resultlist.append(ylist[ypos + 2][xpos - 2])
resultlist.append(ylist[ypos + 3][xpos - 3])
results.append(multiplylist(resultlist))
xpos += 1 # Keep me at end of loop
results = list(set(results))
ypos += 1 # Keep me at end of loop
print(len(results))
for result in results:
if result > highscore:
highscore = result
print ("Highscore is", highscore)
if __name__ == "__main__":
main()
```
```# Problem 12
#!/usr/bin/env python
'''The sequence of triangle numbers is generated by adding the natural
numbers.
So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five
divisors.
What is the value of the first triangle number to have
over five hundred divisors?'''
from functools import reduce
import math
import sys
sys.path.append("../")
from rogutil import *
import time
tt = time.time()
def main():
pass
count = 0
#------------------------------------------
'''generator for triangle numbers'''
def gen():
n = 2
while n < 100000000:
n += 1
numb = n * (n + 1) / 2
yield numb
g = gen()
trinum = next(g)
#------------------------------------------
kounter = 0 # errr... counter
prim = 0 # The Prime Number for division of the
#Triangle number
i, z = 0, 0
prims = gen_x(10000, gen_primes) # prime number genny
prim = prims[z] # For incrementing the Prime Number
div_lst = [] #Divisor list
power_lst = [] #Prime number (power) list
sums_divslst = []
dd = trinum
total_divs = 0
#------------------Calculate divisors-----------------------
while dd < 1000000000:
if not trinum % prim == 0: # Is there a modulus, if so, increase the Prime Number
if count > 0: #Divisor counter
kounter = count + 1
power_lst.append(kounter)
count = 0
z += 1
prim = prims[z]
continue
trinum = trinum / prim
if not prim in div_lst:
div_lst.append(prim)
count += 1
if trinum == 1: #Divide down to 1, test the amount of divisors and
kounter = count + 1 # if too low, increase the Triangle Number
power_lst.append(kounter)
#--------------work out divisor count-----------------------
div_product = reduce(lambda x,y : x * y, power_lst)
if div_product > total_divs:
total_divs = div_product
if total_divs > 500:
break
#--------------reset variables---------------------------
trinum = 0
count = 0
z = 0
prim = prims[z]
trinum = next(g)
dd = trinum
div_lst = []
power_lst = []
continue
print("Triangle number : ", dd," Divisors = ",total_divs)
if __name__ == '__main__':
main()
print(time.time() - tt)
```
```Problem 13.
#--------------------------------------------------------------------------
# Name: module1
# Purpose:
#
# Author: rog
#
# Created: 31/07/2010
# Copyright: (c) rog 2010
# Licence: <your licence>
#--------------------------------------------------------------------------
#!/usr/bin/env python
# Work out the first ten digits of the sum of the following
# one-hundred 50-digit numbers.
import math
import sys
sys.path.append("../")
from rogutil import *
numstr = """37107287533902102798797998220837590246510135740250
46376937677490009712648124896970078050417018260538
74324986199524741059474233309513058123726617309629
91942213363574161572522430563301811072406154908250
23067588207539346171171980310421047513778063246676
89261670696623633820136378418383684178734361726757
28112879812849979408065481931592621691275889832738
44274228917432520321923589422876796487670272189318
47451445736001306439091167216856844588711603153276
70386486105843025439939619828917593665686757934951
62176457141856560629502157223196586755079324193331
64906352462741904929101432445813822663347944758178
92575867718337217661963751590579239728245598838407
58203565325359399008402633568948830189458628227828
80181199384826282014278194139940567587151170094390
35398664372827112653829987240784473053190104293586
86515506006295864861532075273371959191420517255829
71693888707715466499115593487603532921714970056938
54370070576826684624621495650076471787294438377604
53282654108756828443191190634694037855217779295145
36123272525000296071075082563815656710885258350721
45876576172410976447339110607218265236877223636045
17423706905851860660448207621209813287860733969412
81142660418086830619328460811191061556940512689692
51934325451728388641918047049293215058642563049483
62467221648435076201727918039944693004732956340691
15732444386908125794514089057706229429197107928209
55037687525678773091862540744969844508330393682126
18336384825330154686196124348767681297534375946515
80386287592878490201521685554828717201219257766954
78182833757993103614740356856449095527097864797581
16726320100436897842553539920931837441497806860984
48403098129077791799088218795327364475675590848030
87086987551392711854517078544161852424320693150332
59959406895756536782107074926966537676326235447210
69793950679652694742597709739166693763042633987085
41052684708299085211399427365734116182760315001271
65378607361501080857009149939512557028198746004375
35829035317434717326932123578154982629742552737307
94953759765105305946966067683156574377167401875275
88902802571733229619176668713819931811048770190271
25267680276078003013678680992525463401061632866526
36270218540497705585629946580636237993140746255962
24074486908231174977792365466257246923322810917141
91430288197103288597806669760892938638285025333403
34413065578016127815921815005561868836468420090470
23053081172816430487623791969842487255036638784583
11487696932154902810424020138335124462181441773470
63783299490636259666498587618221225225512486764533
67720186971698544312419572409913959008952310058822
95548255300263520781532296796249481641953868218774
76085327132285723110424803456124867697064507995236
37774242535411291684276865538926205024910326572967
23701913275725675285653248258265463092207058596522
29798860272258331913126375147341994889534765745501
18495701454879288984856827726077713721403798879715
38298203783031473527721580348144513491373226651381
34829543829199918180278916522431027392251122869539
40957953066405232632538044100059654939159879593635
29746152185502371307642255121183693803580388584903
41698116222072977186158236678424689157993532961922
62467957194401269043877107275048102390895523597457
23189706772547915061505504953922979530901129967519
86188088225875314529584099251203829009407770775672
11306739708304724483816533873502340845647058077308
82959174767140363198008187129011875491310547126581
97623331044818386269515456334926366572897563400500
42846280183517070527831839425882145521227251250327
55121603546981200581762165212827652751691296897789
32238195734329339946437501907836945765883352399886
75506164965184775180738168837861091527357929701337
62177842752192623401942399639168044983993173312731
32924185707147349566916674687634660915035914677504
99518671430235219628894890102423325116913619626622
73267460800591547471830798392868535206946944540724
76841822524674417161514036427982273348055556214818
97142617910342598647204516893989422179826088076852
87783646182799346313767754307809363333018982642090
10848802521674670883215120185883543223812876952786
71329612474782464538636993009049310363619763878039
62184073572399794223406235393808339651327408011116
66627891981488087797941876876144230030984490851411
60661826293682836764744779239180335110989069790714
85786944089552990653640447425576083659976645795096
66024396409905389607120198219976047599490197230297
64913982680032973156037120041377903785566085089252
16730939319872750275468906903707539413042652315011
94809377245048795150954100921645863754710598436791
78639167021187492431995700641917969777599028300699
15368713711936614952811305876380278410754449733078
40789923115535562561142322423255033685442488917353
44889911501440648020369068063960672322193204149535
41503128880339536053299340368006977710650566631954
81234880673210146739058568557934581403627822703280
82616570773948327592232845941706525094512325230608
22918802058777319719839450180888072429661980811197
77158542502016545090413245809786882778948721859617
72107838435069186155435662884062257473692284509516
20849603980134001723930671666823555245252804609722
53503534226472524250874054075591789781264330331690"""
def main():
result = 0
# split at newline
for y in numstr.split("\n"):
numx = int(y)
# add the numbers (numx)
result += numx
# print first 10 digits
print (str(result)[:10])
if __name__ == '__main__':
main()
```
```#Problem 14.
'''The following iterative sequence is defined for the set of positive
integers:
n ? n/2 (n is even)
n ? 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following
sequence:
13 ? 40 ? 20 ? 10 ? 5 ? 16 ? 8 ? 4 ? 2 ? 1
It can be seen that this sequence (starting at 13 and finishing at 1)
contains 10 terms. Although it has not been proved yet (Collatz Problem),
it is thought that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above one
million.'''
cache = {1: 1}
def collatz(n):
if n not in cache:
if n % 2 == 0:
cache[n] = 1 + collatz(n // 2)
else:
cache[n] = 1 + collatz(3 * n + 1)
return cache[n]
if __name__ == '__main__':
for x in range(3, 1000000):
ans = collatz(x)
ans2 = max(cache.keys(), key = (lambda key: cache[key]))
print(ans2)
```
```# Problem 15
#!/usr/bin/env python
'''Starting in the top left corner of a 2×2 grid, there are 6 routes
(without backtracking) to the bottom right corner.
How many routes are there through a 20×20 grid?'''
# The answer = 40! / (20! * 20!)
def main():
pass
n = 1
#----------------------------
ans = 1
for a in range(1,20,1):
ans = ans * (n + a)
#-----------------------------
ans2 = 1
for a in range(1,40, 1):
ans2 = ans2 * (n + a)
#-----------------------------
print(ans2 / (ans*ans))
if __name__ == '__main__':
main()
```
```#Problem 16.
#! /usr/bin/python
# To change this template, choose Tools | Templates
# and open the template in the editor.
__author__="rog"
__date__ ="\$29-Jul-2010 12:44:04\$"
import sys
from rogutil import multiplylist
sys.path.append("../")
#What is the sum of the digits of the number 2^1000?
def main():
print(sum(list(map(int, list(str(2**1000))))))
if __name__ == "__main__":
main()
```
```# Problem 17.
#-------------------------------------------------------------------------------
# Name: module2
# Purpose:
#
# Author: rog
#
# Created: 02/08/2010
# Copyright: (c) rog 2010
# Licence: <your licence>
#-------------------------------------------------------------------------------
#!/usr/bin/env python
"""
How many letters would be needed to write all the numbers in words
from 1 to 1000?
"""
import sys
sys.path.append("../")
from rogutil import *
dict_thou = {1:11,0:0}
dict_hund = {9:14,8:15,7:15,6:13,5:14,4:14,3:15,2:13,1:13,0:0}
dict_tens = {9:6,8:6,7:7,6:5,5:5,4:5,3:6,2:6,0:0}
dict_teens = {19:8,18:8,17:9,16:7,15:7,14:8,13:8,12:6,11:6,10:3}
dict_singl = {9:4,8:5,7:5,6:3,5:4,4:4,3:5,2:3,1:3,0:0}
dict_hundreds = {9:11,8:12,7:12,6:10,5:11,4:11,3:12,2:10,1:10,0:0}
def main():
cnt = 0
total = 0
for number_for_test in range(1000,0,-1):
number_for_test = zeromaker(str(number_for_test), 4)
''' Is this 100, 200 etc? '''
if int(number_for_test [2]) == 0 and \
int(number_for_test [3]) == 0 and \
int(number_for_test [1]) > 0:
hundreds = (number_for_test[1])
hundred = int(hundreds)
hh = dict_hundreds[hundred]
y = hh
total = total + y
hh,hundreds,hundred,y = 0,0,0,0
continue
''' is it between ten and 19? '''
if int(number_for_test [2]) == 1:
teens = (number_for_test[2])+(number_for_test[3])
teen = int(teens)
s = int(number_for_test [1])
teenager = dict_teens[teen]
v = dict_hund[s]
y = v + teenager
total = total + y
r,s,v,y = 0,0,0,0
continue
'''If statement required to cut out 200, 300 etc'''
if number_for_test [2] != 0 and \
number_for_test [3] != 0:
r = int(number_for_test [0])
s = int(number_for_test [1])
t = int(number_for_test [2])
u = int(number_for_test [3])
z = dict_thou[r]
v = dict_hund[s]
w = dict_tens[t]
x = dict_singl[u]
y = v + w + x + z
total = total + y
r,s,t,u,v,w,x,y,z = 0,0,0,0,0,0,0,0,0
y = 0
continue
print("Total = ", total,"Count = ",cnt)
if __name__ == '__main__':
import time
t = time.time()
main()
print(time.time() - t)
```
```# Problem 18.
'''By starting at the top of the triangle below and moving to adjacent
numbers on the row below, the maximum total from top to bottom is 23.
3
7 4
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom of the triangle below:
75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
NOTE: As there are only 16384 routes, it is possible to solve this
problem by trying every route. However, Problem 67, is the same
challenge with a triangle containing one-hundred rows; it cannot
be solved by brute force, and requires a clever method! ;o)'''
#!/usr/bin/env python
import sys
sys.path.append("../")
from rogutil import *
from pprint import pprint
def doif(lst, a,b,c,str_upper,str_lower):
mylist = list(lst)
if mylist [str_lower] [b] > mylist [str_lower] [c]:
mylist[str_upper][a] = mylist[str_upper][a] + mylist[str_lower][b]
else:
mylist[str_upper][a] = mylist[str_upper][a] + mylist[str_lower][c]
return mylist
mystr = '''75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23'''
def main():
pass
""" changes the above string into a list"""
mylist = []
for item in mystr.split("\n"):
mylist.append(list(map(int,item.split(" "))))
pprint(mylist)
#import pdb; pdb.set_trace()
'''start at the bottom two rows'''
str_upper, str_lower = len(mylist[-1])-2,len(mylist[-1])-1
e = len(mylist[str_upper])
a, b, c, d = 0, 0, 1, 0
while e != 0:
while d in range(0,e,1):
mylist = doif(mylist,a,b,c, str_upper, str_lower)
a += 1
b += 1
c += 1
d += 1
d, a, b, c = 0, 0, 0, 1
str_upper -=1
str_lower -=1
e -= 1
print(mylist[0])
exit()
if __name__ == '__main__':
main()
```
```#Problem 19.
#!/usr/bin/env python
'''You are given the following information, but you may prefer to do some
research for yourself.
* 1 Jan 1900 was a Monday.
* Thirty days has September,
April, June and November.
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twenty-nine.
* A leap year occurs on any year evenly divisible by 4, but not on a
century unless it is divisible by 400.
How many Sundays fell on the first of the month during the twentieth century
(1 Jan 1901 to 31 Dec 2000)?'''
'''To find the ordinal position of the day within the week for any date
between Jan 1, 1901 and Dec 31 2099
1. Find the year_ordinal using the table of days before the
first of the month
2. Calculate week_ordinal(1/1) as follows
a = (year - 1901) modulo 28
b = floor (a/4)
week_ordinal(1/1) = (2 + a + b) modulo 7 + 1
3. Day of the Week = ((year_ordinal - 1) + (week_ordinal(1/1) - 1))
modulo 7 + 1'''
import math
def main():
pass
lst1 = [0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334] #std
lst2 = [0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335] #leap
sunday = 0
y = 0
while y < 12:
for x in range(2000,1900,-1):
a = (x - 1901) % 28
z = x
if z % 4 == 0:
year_ordinal = lst2[y]
else:
year_ordinal = lst1[y]
b = math.floor(a / 4)
week_ordinal = (2 + a + b) % 7 + 1
day_of_the_week = ((year_ordinal-1) + (week_ordinal - 1)) % 7 + 1
if day_of_the_week == 7:
sunday += 1
if x == 1901: #increment to next month
y += 1
continue
print("Year = ", x," Day ",day_of_the_week,"month = ", y,"Sundays = ", sunday)
exit()
if __name__ == '__main__':
main()
```
```#Problem 20
''''n! means n × (n ? 1) × ... × 3 × 2 × 1
For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
Find the sum of the digits in the number 100!'''
#!/usr/bin/env python
import sys
sys.path.append("../")
from rogutil import *
#import pdb; pdb.set_trace()
def factorial_loop(num):
accum = 1
while num > 1:
accum *= num
num -= 1
total = 0
accum = str(accum)
for items in accum:
total += int(items)
ans = factorial_loop(100)
print(ans)
if __name__ == '__main__':
#pdb.runcall(main)
main()
```
```# Problem 21.
#!/usr/bin/env python
'''Let d(n) be defined as the sum of proper divisors
of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a is not equal to b, then
a and b are an amicable pair and each of a and b are
called amicable numbers.
For example, the proper divisors of 220 are 1, 2, 4,
5, 10, 11, 20, 22, 44, 55 and 110; therefore
d(220) = 284. The proper divisors of 284 are 1, 2, 4,
71 and 142; so d(284) = 220.
Evaluate the sum of all the amicable numbers under 10000.'''
import time
tt = time.time()
def main():
pass
'''Empty list'''
g = []
def divsum(n):
return sum(i for i in range(1, n) if not n % i)
'''work out the divisors from 10000 to 2'''
for x in range(10000, 2, -1):
''' check if amicable pair'''
c = divsum(x)
v = divsum(c)
if v == x and v != c:
g.append(divsum(x))
else:
continue
print(sum(g))
if __name__ == '__main__':
main()
print(time.time() - tt, " secs.")
```
```# Problem 22.
#!/usr/bin/env python
'''Using names.txt (right click and 'Save Link/Target As...'),
a 46K text file containing over five-thousand first names,
begin by sorting it into alphabetical order. Then working
out the alphabetical value for each name, multiply this
value by its alphabetical position in the list to obtain
a name score.
For example, when the list is sorted into alphabetical order,
COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th
name in the list. So, COLIN would obtain a score
of 938 × 53 = 49714.
What is the total of all the name scores in the file?'''
import re
def main():
pass
f = open("names.txt",'r')
names = []
i = 0
sum = 0
ans = 0
for string in names:
i += 1
chars = list(string)
for x in chars:
if x == 'A':
t = 1
elif x == 'B':
t = 2
elif x == 'C':
t = 3
elif x == 'D':
t = 4
elif x == 'E':
t = 5
elif x == 'F':
t = 6
elif x == 'G':
t = 7
elif x == 'H':
t = 8
elif x == 'I':
t = 9
elif x == 'J':
t = 10
elif x == 'K':
t = 11
elif x == 'L':
t = 12
elif x == 'M':
t = 13
elif x == 'N':
t = 14
elif x == 'O':
t = 15
elif x == 'P':
t = 16
elif x == 'Q':
t = 17
elif x == 'R':
t = 18
elif x == 'S':
t = 19
elif x == 'T':
t = 20
elif x == 'U':
t = 21
elif x == 'V':
t = 22
elif x == 'W':
t = 23
elif x == 'X':
t = 24
elif x == 'Y':
t = 25
else:
t = 26
sum += t
ans += sum * i
sum = 0
print(ans)
if __name__ == '__main__':
main()
```
| 9,346 | 24,679 |
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| 3.578125 | 4 |
CC-MAIN-2018-47
|
longest
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en
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http://education.seattlepi.com/exact-opposite-earth-live-4919.html
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| 94,670,991 | 10,476 |
# How Do I Find the Exact Opposite of the Earth From Where I Live?
If you're planning on tunneling through the center of the Earth, it helps to know what you'll find on the other side so you can bring the proper attire. Although there's an app for it, you can locate opposite points yourself, using a map, if you understand a little spherical geometry. The more precisely you determine the coordinates of your location, the more exact will be the determination of its antipode.
## Latitude Lines
When modeling the Earth as a globe, cartographers divide it into a series of horizontal and vertical lines. The horizontal lines, called parallels, extend around the globe a fixed distance from the equator, which is the central parallel, and the distance of each parallel from the equator is called latitude. It is measured in degrees, starting from zero degrees at the equator to 90 degrees at each pole. Degrees can be further subdivided into minutes and seconds; there are 60 minutes per degree and 60 seconds per minute. It's also common to see degrees expressed as decimals.
## Longitude and Prime Meridian
The vertical lines are called meridians, and each one denotes an angular separation from the central -- or prime -- meridian, which by convention runs through Greenwich, England. The angular separation of a meridian from the prime meridian is called longitude. Because there are 360 degrees in a circle, no value of longitude exceeds 180 degrees. All points on a given meridian are in the same time zone, and the meridian opposite the prime meridian, which is both positive and negative 180 degrees, is defined as the international date line.
## Locating an Antipode
To locate the antipode of a given location on the Earth, you must first look up the location's longitude and latitude. Each degree of latitude is approximately 111 kilometers (69 miles), but degrees of longitude vary with latitude -- at 45 degrees latitude, each one is about 79 kilometers (49 miles). Expressing the coordinates in degrees and minutes narrows the location to the nearest 1.6 kilometers (1 mile), and expressing them in seconds narrows it to the nearest 30 meters (100 feet). The latitude of the antipode has the same value but the opposite sign, because it's in the opposite hemisphere. The antipodal meridian is 180 degrees away.
## A Sample Calculation
To find the antipode of New York City, start by looking up its coordinates or reading them off a map; they are 40 degrees 47 minutes north and 73 degrees 58 minutes west. The latitude of the antipode is 40 degrees 47 minutes south, and to get the longitude -- which is in the opposite hemisphere -- you subtract 73 degrees 58 minutes from 180 to get 106 degrees 2 minutes east. Looking up those coordinates on a map or entering them into a Global Positioning System app reveals the antipode to be in the Indian Ocean just off the south western tip of Australia.
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A circular swimming pool is surrounded by a wall with a height of 2m.
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A circular swimming pool is surrounded by a wall with a height of 2m. [#permalink]
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[GMAT math practice question]
A circular swimming pool is surrounded by a walk with a width of 2m. The area of the walk is 21% of the area of the swimming pool. What is the area of the swimming pool?
A. $$400π$$ $$m^2$$
B. $$300π$$ $$m^2$$
C. $$250π$$ $$m^2$$
D. $$200π$$ $$m^2$$
E. $$100π$$ $$m^2$$
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Originally posted by MathRevolution on 22 May 2020, 05:28. Last edited by MathRevolution on 23 May 2020, 08:58, edited 1 time in total. PS Forum Moderator Joined: 18 Jan 2020 Posts: 1520 Location: India GPA: 4 Re: A circular swimming pool is surrounded by a wall with a height of 2m. [#permalink] Show Tags 22 May 2020, 05:43 Sir, kindly check the height of wall. Are you sure it's not 1m? Because the last value I m getting for r(h+r)=42 And if h is 2, r will be of some imaginary value. Kindly revert. Posted from my mobile device Manager Joined: 22 Oct 2018 Posts: 64 Location: United States (MI) Concentration: Finance, Technology GMAT 1: 590 Q42 V29 GPA: 3.7 WE: Engineering (Consumer Electronics) Re: A circular swimming pool is surrounded by a wall with a height of 2m. [#permalink] Show Tags 22 May 2020, 07:44 MathRevolution wrote: [GMAT math practice question] A circular swimming pool is surrounded by a wall with a height of 2m. The area of the wall is 21% of the area of the swimming pool. What is the area of the swimming pool? A. 400π m^2 B. 300πm^2 C. 250πm^2 D. 200πm^2 E. 100πm^2 This question seems incomplete or isn't worded properly. Can someone help _________________ Siddharth Ramachandran B School Aspirant, 2021 Intake Intern Joined: 13 Jun 2019 Posts: 29 Re: A circular swimming pool is surrounded by a wall with a height of 2m. [#permalink] Show Tags 22 May 2020, 21:34 1 Can anybody please explain how answer is 400 $$pi$$ sq m _________________ Shrekey Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 9257 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: A circular swimming pool is surrounded by a wall with a height of 2m. [#permalink] Show Tags 23 May 2020, 09:00 SiddharthR wrote: MathRevolution wrote: [GMAT math practice question] A circular swimming pool is surrounded by a wall with a height of 2m. The area of the wall is 21% of the area of the swimming pool. What is the area of the swimming pool? A. 400π m^2 B. 300πm^2 C. 250πm^2 D. 200πm^2 E. 100πm^2 This question seems incomplete or isn't worded properly. Can someone help There were mistakes in words. They are fixed now. The words "wall" and "height" should be fixed with "walk" and "width", respectively. Sorry about the mistakes. Happy Studying !!! _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Joined: 18 Jan 2020
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Location: India
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Re: A circular swimming pool is surrounded by a wall with a height of 2m. [#permalink]
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23 May 2020, 09:13
Let the radius of walk with the swimming pool in between = r+2
Area of whole area ( swimming pool+ walk) = π(r+2)^2
Given: area of walk is 21% of area of swimming pool.
Area of walk = π(r+2)^2 - πr^2
Therefore,
π(r+2)^2 - πr^2 = 21/100(πr^2)
π{(r+2)^2 - r^2}=π{21/100(r^2)}
(r^2+4+4r)- r^2={21/100(r^2)
4+4r=21/100(r^2)
400+400r=21(r^2)
21(r^2)-400r-400=0
21(r^2)-420r+20r-400=0
21r(r-20)+20(r-20)=0
(r-20)(21r+20)=0
R=20 and -20/21(not possible)
R=20
Area of swimming pool is πr^2
π(20)^2 => 400π
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Re: A circular swimming pool is surrounded by a wall with a height of 2m. [#permalink]
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24 May 2020, 17:14
=>
Assume the radius of the swimming pool is $$r$$.
Then the area of the wall is $$2πr·2 = 4πr.$$
Since we have $$4πr = (\frac{21}{100}) πr^2,$$ we have $$21r^2 – 400r + 400 = 0$$ or $$(21r + 20)(r - 20) = 0.$$
Thus, we have $$r = 20.$$
Then, the area of the swimming pool is $$πr^2 = 400π.$$
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Re: A circular swimming pool is surrounded by a wall with a height of 2m. [#permalink] 24 May 2020, 17:14
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What is Foolsaurus?
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# How to Calculate What You're Willing to Pay if the Interest Rate Drops on a Bond
Original post by Kathy Adams McIntosh of Demand Media
Investors purchase bonds in order to earn interest income. Each bond includes a stated interest rate. The stated interest rate determines the amount of each interest payment the company makes to the bondholder. The investor calculates the interest income by multiplying the face value of the bond by the interest rate and the timeframe to which the interest applies. The market interest rate refers to the interest rate that investors require in order to invest in a bond, or the interest rate that the investor could receive from another investment. When the market interest rate drops, investors will pay less for bonds.
## Contents
### Step 1
Review the terms of bond. Identify the face value, the stated interest rate and the timeframe of the bond.
### Step 2
Identify alternative investment opportunities. Read the terms of each investment and note the interest rates offered. The average interest rate offered to the investor represents the market rate of interest.
### Step 3
Locate the present value factor for \$1 of bond value. Use the market interest rate and the number of periods for which the bond applies to find the factor.
### Step 4
Calculate the present value of the face value of the bond. Multiply the face value by the present value factor for the bond.
### Step 5
Locate the present value factor for an annuity. Use the market interest rate and the number of periods for which the bond applies to find the factor.
### Step 6
Calculate the present value of the bond’s interest payments. Multiply the individual interest payment by the present value factor for an annuity.
### Step 7
Add the present value of the face value of the bond from Step 4 to the present value of the interest payments from Step 6. This determines the purchase price when the interest rate drops.
### Tips & Warnings
• As the market interest rate drops, the purchase price of a bond increases. As the bond price increases, the interest payments remain the same. The investor calculates the interest rate by dividing the interest payment by the purchase price of the bond, reducing the interest rate of the bond.
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# Investigation: Construct segment bisectors
Lesson
### Compass and straightedge construction
Recall that the term bisect means "cut in half". So to bisect a line segment means to cut the line segment in half or find the halfway point.
#### Steps for construction
1. Start with a line segment $\overline{AB}$AB
2. Place the compass on $A$A.
3. Adjust the width of the compass to what you think looks to be just over half way.
4. Draw an arc on both sides of the line.
5. Keeping the compass the same width, move the compass to point $B$B, and cross the arcs you just created with new arcs.
6. With the straightedge, draw the straight line between the two intersections.
7. This line bisects the segment $\overline{AB}$AB
In fact, this is a very special line, it is called a perpendicular bisector which means it not only cuts the line in half, but the line is perpendicular to the segment (at right angles to it).
### Dynamic software construction
We can also construct a congruent segment using dynamic geometry software. Press the pause/play button in the applet below to see the steps of the construction in action. To test the construction, move the points $A$A, $B$B, $C$C, and $D$D around to see that they are always the same length.
Now it's your turn! Repeat the steps for construction to construct your own set of congruent segments. Click here to open the applet in a larger web browser window.
#### Steps for construction using technology
Move Tool Point Tool Line Tool Line Segment Tool Compass Tool Polygon Tool
1. Use line segment tool to draw an arbitrary segment $\overline{AB}$AB.
2. Choose the compass tool, then click on $\overline{AB}$AB to set the radius length to be the same as $AB$AB, then click on center $A$A to draw a circle.
3. Repeat step two with $B$B as the center of the circle.
4. Use the point tool to plot a point $C$C at the intersection of the two circles.
5. Plot point $D$D at the other intersection.
6. Use the line tool to join points $C$C and $D$D, this is a line that bisects $\overline{AB}$AB.
7. Use the point tool to plot point $E$E at the intersection of $\overline{AB}$AB and $\overleftrightarrow{CD}$CD. This is the midpoint of $\overline{AB}$AB.
8. Use the Move tool to drag points a $A$A and $B$B and double check that your construction remains.
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# Question #47f0f
Jan 18, 2018
B.
#### Explanation:
If the final mixture has to contain 150 mg/ml of penicillin,that means 150 mg of penicillin will be present in 1 ml of solution not 1 ml of sterile water.
Given, 600 mg of penicillin was present in the container,so volume of the solution becomes (600/150)ml or 4 ml
Now,3.6 ml of this 4 ml is sterile water(as given)
so,volume of penicillin in this solution is (4-3.6)ml or 0.4ml
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Many businesses in the UK utilize promotional items as part of their advertising and marketing campaigns. Now, even more businesses are seeing the use of promotional items to give as gifts to staff. Promotional items can be used as staff presents for a number of reasons, depending on the type of promotional item you are using. Here are a few examples: Promotional pens - give these out to your staff freely so they can use your promotional item every day. Promotional mugs - every staff member should have a promotional mug to use at the office. Promotional mouse mats - every office has computers, you might as well give each of your staff members one of your promotional mouse mats to use on their desk. Promotional bags and folders - a great gift for an employee who has done a great job or is working on conducting or putting together a seminar or trade show. clothing such as t-shirts - perfect to use as office wear.
• ### Buy Wholesale Promotional Armani Suits from China, How to Calculate Sea Freight?
International ocean shipping is a practice used to trade goods all over the world. Items are packed into shipping containers and loaded onto a vessel before traveling across the sea to an arrival port. Large machinery and project material that is too large to fit into ocean containers are known as break-bulk and travels uncovered, aboard vessels. Calculating sea freight is the process that determines a shipment's mass, which is used to rate the cost of either an LCL (less than container-load) or a break-bulk shipment. Instructions LCL shipments 1. For LCL shipments, Measure the length, width, and height of your cargo (in inches). Calculate the volume of your cargo using the formula: Length x Width x Height = Volume (in cubic inches). 2. Create the denominator by multiplying 12 x 12 x 12. There are 12 inches in a foot: (12 x 12 x 12) = 1,728 cubic inches. 3. Divide the volume by 1,728 to get your cubic footage volume: Volume / 1,728 = cubic feet. 4. Multiply by 0.0283168466 to
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### The diameter of the circumcircle of a rectangle is $13 \space cm$ and rectangle's width is $5 \space cm$. Find the length of the rectangle.
$12 \space cm$
Step by Step Explanation:
1. The following figure shows the rectangle $ABCD$ with its circumcircle,
$AB, BC,$ and $AC$ are the length, breadth, and diameter of the circumcircle of the rectangle $ABCD$.
According to the question, $AC = 13 \space cm,$
$BC = AD = 5 \space cm$.
2. \begin{align} \text { Now, in the right-angled triangle } ABC, AB^2 & = AC^2 - BC^2 \\ \implies AB & = \sqrt { AC^2 - BC^2 } \\ & = \sqrt { (13)^2 - (5)^2 } \\ & = 12 \space cm \end{align}
3. Hence, the length of the rectangle is $12 \space cm$.
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https://electronics.stackexchange.com/questions/469394/implementing-3-variable-boolean-function-using-mux-4-to-1-and-inverter
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# Implementing 3 variable boolean function using mux 4 to 1 and inverter
I'm trying to understand if it's possible to Implement boolean function with 3 inputs using only mux 4 to 1 and inverter. As far as I understand I can put in the selectors the first 2 variables to select between the 4 options. then I have another variable which I can connect to the 4 options (00,01,10,11) but I can't solve it to make sure it will suffice any 3 variables function.
I would like to know how to approach this kind of questions, how to "prove" such things?
Thanks.
It is possible to make any boolean function f(a,b,c) using a 4:1 mux and an inverter
With the inverter make ~c
Connect a and b to the mux address lines.
Connect each mux data input to the one of 0,1,c,or ~c as appropriate.
The mux output has your function result.
It is a nice exercise to understand different ways to represent a logical function.
In short: you can always implement a logical function of $$\N\$$ variables with a $$\2^{N-1}\$$:1 multiplexer and some amount of inverters by using one of the variables at data inputs of the multiplexer and others - at control inputs.
Here's an explanation. Let's consider a logical function of 3 variables $$\y(x_0, x_1, x_2)\$$. When you look at its truth table, you can easily divide it into two parts: one where $$\x_0\$$ equals "0" and one where $$\x_0\$$ equals "1". Then we need to look at pairs of $$\y\$$ values with equal $$\(x_1, x_2)\$$ argument values. For a pair of $$\y\$$ values there are only four possible combinations:
"0" and "0" - which means that for a certain $$\(x_1, x_2)\$$ value, $$\y\$$ always equals "0", i.e. $$\y\$$ doesn't depend on $$\x_0\$$;
"1" and "1" - this is similar to the first case, $$\y\$$ doesn't depend on $$\x_0\$$, but always equals "1";
"0" and "1" - which means that for a certain $$\(x_1, x_2)\$$ value $$\y\$$ = "0" when $$\x_0\$$ = "0", and $$\y\$$ = "1" when $$\x_0\$$ = "1", i.e. $$\y = x_0\$$;
"1" and "0" - this is an inversion of the previous case, for a certain $$\(x_1, x_2)\$$ value $$\y\$$ = "1" when $$\x_0\$$ = "0", and $$\y\$$ = "0" when $$\x_0\$$ = "1", i.e. $$\y = \bar x_0\$$.
You can connect $$\(x_1, x_2)\$$ value to the 2-bit control input of the 4:1 mux. It will select a corresponding $$\y\$$ pair. To have a constant "0" or "1" (1-st and 4-th cases), you put a constant "GND" or "VCC" respectively at the corresponding data input. If $$\y\$$ in pair depends on $$\x_0\$$ (2-nd and 3-rd cases), you put $$\x_0\$$ or $$\\bar x_0\$$ at the corresponding data input.
Here is an example of a logical function (in the form of a truth table) and its implementation using a 4:1 mux.
In this example:
for $$\(x_1, x_2)=0\$$: $$\y(0,(0,0)) = y(1,(0,0)) = 0\$$, hence GND at the 0-th data input,
for $$\(x_1, x_2)=1\$$: $$\y(0,(0,1)) = 1, y(1,(0,1)) = 0\$$, hence $$\\bar x_0\$$ at the 1-st data input,
for $$\(x_1, x_2)=2\$$: $$\y(0,(0,1)) = 0, y(1,(0,1)) = 1\$$, hence $$\x_0\$$ at the 2-nd data input,
for $$\(x_1, x_2)=3\$$: $$\y(0,(0,1)) = y(1,(0,1)) = 1\$$, hence VCC at the 3-rd data input.
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# THEME: INVESTIGATING FORCE AND ENERGY
## Examples of activities that involve both pulling and
pushing.
Effect of a force.
Change the shape of an object
## i) An object will change its shape when a force is exerted on it.
i) A force can change the direction of a moving object. ii) A force can make an object move fast or slowly. iii) A force can stop a moving object
Effect of a force
Move a stationary object
## Affected by the weight of an object
Affected by the condition of the surface of the object FRICTION Rough surfaces such as sand paper and carpets produce greater frictional force. Smooth surfaces such as glass and rubber mats produce smaller frictional force. If two objects are pushed with the same force, the object that moves on a smooth surface moves faster due to the smaller frictional force.
A type of force
## Produced when two surfaces in contact rub against each other
FRICTION
ADVANTAGES: a) It enables us to hold an object. b) It enables us to walk without slipping.
## c) It enables a knife to be sharpened.
d) It enables us to light a match.
DISADVANTAGES: a) It makes a surface of an object wear out. b) It causes an object hard to move. c) It produces heat in the engine.
SPEED
Every object that moves has speed. Speed shows how fast an object moves. An object which moves faster travels a longer distance in a given time. An object which moves faster takes a shorter time to travels in a given distance. To speed of an object can be calculated by using the following formula: Speed = Distance Time Examples : a) A lorry takes three hours to reach a destination of 240 km away. Speed = 240 = 80 km/hour 3 b) A toy car takes 2 minutes to travel a distance of 50 cm. Speed = 50 2 = 25 cm/min
## Increasing and decreasing friction:
Ways to reduce friction
Friction
a) Using rollers or ball bearing b) Using lubricants such as oil , wax or grease. c) Using talcum powder or air cushion. d) Smoothing the surfaces. e) Using an aerodynamic shape.
## Ways to increase friction
Microorganisms such as bacteria, moulds and yeast can caused food to spoil. Perishable foods like fish and vegetables turn bad faster compared to dried foods because the presence of water helps bacteria and fungi to grow
FOOD SPOILAGE
## Spoilt food is unsuitable and unsafe to be eaten.
Food becomes spoilt when its texture, flavour and taste are damaged and change.
The conditions needed by microorganisms to grow are: a) Air b) water c) nutrient d) suitable temperature
e) Suitable acidity
FOOD PRESERVATION
Helps to slow down the spoilage of food A process which makes food keep longer
FOOD PRESERVATION
## The concept of food preservation
The following are different types of food preservation: a) Drying This method removes water from food. It can be done by sunning food. This method can prevent the growth of microorganisms because bacteria and moulds cannot grow without the presence of water.
## Fishes are dried in the Sun.
b) Smoking Food is preserved and flavoured through the use of smoke. The heat and aroma from the smoke dry and preserve food without cooking it.
## Fishes are smoked until they become dry.
C) Salting. Salt is used to preserve food such as fish, eggs and fruit. Salting can be done by soaking food in a concentrated salt solution. Salting can prevent the growth of microorganisms on food.
## This process changes the taste and colour of food.
SALT
Salted fish
d) Pickling.
Salted egg
Pickling is the method of preserving food by placing it in a solution such as vinegar, salt or sugar.
Salt, sugar and vinegar solution retard the growth of bacteria and fungi on food.
Food that is preserved in this way has its taste and colour changed as well as its nutrient content reduced. Examples of food that can be preserved in this way are papayas, mangoes, guavas and chillies.
e) Cooling.
The temperature for cooling food is between 0C to 5 C. Low temperatures prevent the growth of bacteria and fungi. Cooling does not kill bacteria but only prevents them from growing. Bacteria and fungi will be active again after food is taken out from the refrigerator so food must be cooked or eaten immediately. Food that can be preserved by cooling are vegetables and fruit. f)
Vacuum packing. This method of preserving food is done by keeping food in a condition without air. The steps involved in this method are: Food is kept in a plastic bag. The air inside the plastic bag is pumped out. The vacuum plastic bag is then sealed or tightly closed. Bacteria and fungi cannot breed without the presence of air. Food that can be preserved in this way includes roasted peanut fruits and biscuit.
g) Canning and bottling. This method involves the process of heating food to kill bacteria and storing it in airtight containers. This method enables food to be kept for a very long period up to a few years. Bacteria and fungi cannot breed without the presence of air. The steps involved in canning are: a) Food is cut and cleaned. b) The food is then heated or cooked. c) The food is kept in a can and heated again. d) The can is sealed when the food is still hot. SARDINES
h) Boiling. In this process, food is heated to the boiling point of water, that is 100C for 30 minutes. This temperature can kill microorganisms. Examples of food preserved by using this method are fruit, jelly and jam.
i)
Freezing. In this method, food is kept at very low temperatures, as low as -15C. The very low temperature and lack of water prevent the growth of microorganisms. However, frozen food must be cooked immediately after taking out from the freezer because bacteria becomes active at room temperature.
Pasteurising In this process, food is heated to a specific temperature for a certain period. The taste and nutrient content of food do not change and can be maintained. This process is usually used for milk and fruit juice.
j)
Milk
k) Waxing. Wax such as beeswax is coated onto surface of food such as fruit and vegetables. Wax is used to replace the natural wax produced by food and it can keep food dry. This method can increase the shelf life of food for more than two weeks.
The Importance of preserving food. To make food last longer. To store food easily. To reduce wastage of food. To ensure that food is safe for consumption. To protect food against microorganisms.
## The Effects Of Improper Waste Disposal On The Environment.
anything that is not used anymore and needs to be thrown away or disposed off
The types of waste produced include: Produced every day.
Waste
a) Paper
b) Plastic c) Glass d) Metal e) Chemical waste
f) Organic waste
## Improper Ways Of Waste Disposal
Improper ways of waste disposal
a) Not throwing
## waste into rubbish
dumps such as dustbins.
## Examples of improper ways of waste disposal
b) Throwing waste into drains or rivers. c) Open burning. d) Burying waste that is dangerous or does not decay such as plastic materials, glass and toxic wastes.
Waste Effect
## Breeding place for germs and pest
Air pollution: It is cause by open burning , forest fires, releasing smoke from factories and vehicles.
Flash flood -The improper waste disposal will clog up the river and the drain that will prevent the flow of water. -The water will overflow into the land. -It will cause flash floods during heavy rain.
## Effects of improper waste disposal
Acid rain -Smoke and gases that come from factories and vehicles dissolve in rain water to form acid. - It cause iron to become rusty.
Water Pollution -Rubbish and waste are thrown into the river , drain and sea. The sources of water become polluted. -The toxic waste can kill the aquatic animals. It will reduce the content of oxigen in the water.
## Reuse the recycle material
Waste Decay
Examples waste that can decay:
Food residues, paper boxes, newspaper, dead plants and dead animals. Advantages: * Prevents waste from piling up in the environment. * Add nutrients to the soil.
## Examples of waste that cannot decay:
Glass, metal, plastic. Disadvantages: * Produces poisonous gases
Eclipses.
## The Phenomenon of an eclipse
Because of the following properties of light: a) Light travel in a straight Occurs when the Earth, the Moon and the Sun are in a straight line line. b) Light cannot pass through on opaque object.
An eclipse of the moon occurs when the Earth is between the Sun and the Moon.
During this eclipse , the Moon will appear dark because the Earth blocks sunlight from reaching the Moon.
## An eclipse of the Moon occurs in stages as shown below:
V
W X Y
Z
V : No eclipse
W : Partial eclipse
X : Total eclipse
Y : Partial eclipse
Z : No eclipse
When the whole Moon is in the Earths darker shadow (position X), a total eclipse occurs.
When the Moon is in the Earths lighter shadow (position W and Y), a partial eclipse occurs.
## Eclipse of the Sun
The eclipse of the Sun occurs when the Moon is between the Sun and The Earth in a straight line.
## The Moon blocks sunlight from reaching the Earth.
The shadow of the Moon will cover a part of the Earths surface.
Sun
Earth w x y
Moon
Moon
W : Partial eclipse
Sun
Y : Partial eclipse
x : Total eclipse
Simple Machines..
The ancient Egyptians were very clever. They figured out how to use the lever. They built pyramids with stones that weighed a lot. Inclined planes helped to move them to the right spot. Why not use wheels? Im not certain. History sometimes holds mysteries behind a time curtain. Did the Egyptian invention work out? See a pyramid and you have no doubt
A device that allows us to use less force to do work. Help us to do work easily, quickly and conveniently.
Simple Machines
Types of simple machines: Lever Gear Screw Wheel and axle Inclined plan wedge
Type
Example
1. Pulley A pulley is a simple machine that uses grooved wheels and a rope to raise, lower or move a load.
## a fulcrum which lifts or moves loads.
3. Wedge A wedge is an object with at least one slanting side ending in a sharp edge, which cuts material apart.
4. Wheel & Axle A wheel with a rod, called an axle, through its center lifts or moves loads.
5. Inclined Plane An inclined plane is a slanting surface connecting a lower level to a higher level. 6. Screw A screw is an inclined plane wrapped around a pole which holds things together or lifts materials. 7. Gear Gear is a wheel with jagged edges like teeth. Use to change the speed
movement.
Complex Machines
## Examples of Complex Machines
Meaning
A tool that has more than one simple machine.
Hand drill Scissors Wedge, lever, Screw Wheelbarrow Lever, wheel and axle. Wedge, Screw, gear, wheel and axle
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Comment convertir 915 milles en yards
Pour convertir 915 milles en yards on doit multiplier 915 x 1760, puisque 1 mille fait 1760 yards.
915 milles × 1760 = 1610400 yards
915 milles = 1610400 yards
Nous concluons que neuf cent quinze milles équivaut à un million six cent dix mille quatre cents yards.
Table de conversion de milles en yards
milles (mi) yards (yd)
916 milles 1612160 yards
917 milles 1613920 yards
918 milles 1615680 yards
919 milles 1617440 yards
920 milles 1619200 yards
921 milles 1620960 yards
922 milles 1622720 yards
923 milles 1624480 yards
924 milles 1626240 yards
925 milles 1628000 yards
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# How prove this $\prod_{1\le i<j\le n}\frac{a_{j}-a_{i}}{j-i}$ is integer
let $a_{i},i=1,2,\cdots,n$ be postive integer ,show that
$$1^{n-1}2^{n-2}\cdots (n-2)^2(n-1)|\prod_{1\le i<j\le n}(a_{i}-a_{j})$$
I know this $\prod_{1\le i<j\le n}(a_{i}-a_{j})$ is
Vandermonde determinants,and I found
$$1^{n-1}2^{n-2}\cdots (n-2)^2\cdot (n-1)=1!2!3!\cdots (n-1)!=\prod_{1\le i<j\le n}(j-i)$$
we only prove $$\prod_{1\le i<j\le n}\dfrac{a_{j}-a_{i}}{j-i}$$ is integer
maybe consider Vandermonde determinants ?
But I can’t prove this
#### Solutions Collecting From Web of "How prove this $\prod_{1\le i<j\le n}\frac{a_{j}-a_{i}}{j-i}$ is integer"
There are many ways to prove that this number is an integer. For example, it has representation-theoretic and combinatorial interpretations.
Your idea also leads to a solution. We want to prove that $\det\bigl(\frac{a_i^{j-1}}{(j-1)!}\bigr)$ is an integer. Idea: while $x^k/k!$ is not always an integer, there is a deformation
$$\frac{x^{\downarrow k}}{k!}:=\frac{x(x-1)\ldots(x-k+1)}{k!}=\binom xk$$
which is always an integer.
Now $\det(a_i^{j-1})=\det(a_i^{\downarrow j-1})$ (in general, if each $P_k$ is a polynomial with leading term $x^k$ then $\det P_{j-1}(x_i)=\det x_i^{j-1}$). So $\det\bigl(\frac{a_i^{j-1}}{(j-1)!}\bigr)=\det\left(\binom{a_i}{j-1}\right)$ which is manifestly an integer.
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# Solve for alpha and beta
• Sep 18th 2007, 05:13 AM
Solve for alpha and beta
How do you solve alpha and beta for these two equations?
$\displaystyle (0.5)(4.00) = (0.500)(2.00)(\cos{\alpha}) + 3(4.47)(\cos(\beta))$
$\displaystyle 0 = 0.500(2.00)(\sin{\alpha}) - 0.300(4.47)(\sin{\beta})$
i tried but it gives me a math error
the answers are $\displaystyle \alpha = 36.9^{\circ},\beta = 26.6^{\circ}$
The hint is:
solve the 1st equation for cos(beta) first
then 2nd equation for sin(beta)
then square each equation (cos(beta))^2 + (sin(beta))^2 = 1
thats what i did but my answer at last part gives me
$\displaystyle 1.798281 = 4 - 2\cos(\alpha)$
alpha gives a math error for this equation
thank you very much
• Sep 18th 2007, 06:22 AM
topsquark
Quote:
How do you solve alpha and beta for these two equations?
$\displaystyle (0.5)(4.00) = (0.500)(2.00)(\cos{\alpha}) + 3(4.47)(\cos(\beta))$
$\displaystyle 0 = 0.500(2.00)(\sin{\alpha}) - 0.300(4.47)(\sin{\beta})$
i tried but it gives me a math error
the answers are $\displaystyle \alpha = 36.9^{\circ},\beta = 26.6^{\circ}$
The hint is:
solve the 1st equation for cos(beta) first
then 2nd equation for sin(beta)
then square each equation (cos(beta))^2 + (sin(beta))^2 = 1
thats what i did but my answer at last part gives me
$\displaystyle 1.798281 = 4 - 2\cos(\alpha)$
alpha gives a math error for this equation
thank you very much
$\displaystyle (0.5)(4.00) = (0.500)(2.00)(\cos{\alpha}) + 3(4.47)(\cos(\beta))$
$\displaystyle 0 = 0.500(2.00)(\sin{\alpha}) - 0.300(4.47)(\sin{\beta})$
So you have the system:
$\displaystyle 2 = cos (\alpha ) + 13.41 cos( \beta)$
$\displaystyle 0 = sin( \alpha ) - 1.341 sin( \beta )$
So
$\displaystyle cos(\beta) = \frac{2 - cos(\alpha)}{13.41}$
and
$\displaystyle sin(\beta) = \frac{sin(\alpha)}{1.341}$
Thus
$\displaystyle sin^2(\beta) + cos^2(\beta) = 1 \implies \left ( \frac{sin(\alpha)}{1.341} \right )^2 + \left ( \frac{2 - cos(\alpha)}{13.41} \right ) ^2 = 1$
$\displaystyle \frac{sin^2(\alpha)}{1.341^2} + \left ( \frac{4 - 4cos(\alpha) + cos^2(\alpha)}{13.41^2} \right ) = 1$ <-- Multiply both sides by $\displaystyle 13.41^2$
$\displaystyle 100sin^2(\alpha) + 4 - 4cos(\alpha) + cos^2(\alpha) = 13.41^2$<-- Add $\displaystyle 99cos^2(\alpha)$ to both sides
$\displaystyle 100sin^2(\alpha) + 4 - 4cos(\alpha) + 100cos^2(\alpha) = 13.41^2 + 99 cos^2(\alpha)$
$\displaystyle 100sin^2(\alpha) + 100cos^2(\alpha) + 4 - 4cos(\alpha) = 13.41^2 + 99 cos^2(\alpha)$
$\displaystyle 100 + 4 - 4cos(\alpha) = 13.41^2 + 99 cos^2(\alpha)$
$\displaystyle 99cos^2(\alpha) + 4 cos(\alpha) + 13.41^2 - 104 = 0$
$\displaystyle 99cos^2(\alpha) + 4 cos(\alpha) + 75.8281 = 0$
You would typically solve this using the quadratic formula, but note that the discriminant is negative. So there are no real solutions.
Note that with your given solution the first equation is
$\displaystyle 2 = 12.7903$
which is obviously untrue. (Though the given solution does satisfy the second equation.)
-Dan
• Sep 18th 2007, 08:26 AM
Soroban
Why weren't the equations simplified first?
Quote:
How do you solve $\displaystyle \alpha$ and $\displaystyle \beta$ for these two equations?
$\displaystyle (0.5)(4.00) = (0.500)(2.00)(\cos{\alpha}) + 3(4.47)(\cos(\beta))$
$\displaystyle 0 = 0.500(2.00)(\sin{\alpha}) - 0.300(4.47)(\sin{\beta})$
The answers are $\displaystyle \alpha = 36.9^{\circ},\beta = 26.6^{\circ}$
There must be a typo . . . the answers don't check out.
I think the equations look like this:
. . $\displaystyle \begin{array}{cccc}\cos\alpha + {\color{red}1.341}\cos\beta & = & 2 & {\color{blue}[1]} \\ \sin\alpha - 1.341\sin\beta & = & 0 & {\color{blue}[2]}\end{array}$
Quote:
The hint is:
solve the 1st equation for $\displaystyle \cos\beta$ first, .
Really?
. . then 2nd equation for $\displaystyle \sin\beta.$ .
Why introduce fractions?
Then square each equation and add: .$\displaystyle \cos^2\!\beta + \sin^2\!\beta\:= \:1$
This is the way I would do it . . .
$\displaystyle \begin{array}{ccccc}\text{From {\color{blue}[1]}:} & \cos\alpha & = & 2-1.341\cos\beta & {\color{blue}[3]}\\ \text{From {\color{blue}[2]}:} & \sin\alpha & = & 1.341\sin\beta & {\color{blue}[4]}\end{array}$
$\displaystyle \begin{array}{cccc}\text{Square {\color{blue}[3]}:} & \cos^2\!\alpha & = & 4 - 5.364\cos\beta + 1.798281\cos^2\!\beta \\ \text{Square {\color{blue}[4]}:} & \sin^2\!\alpha & = & 1.798281\sin^2\!\beta \end{array}$
Add: .$\displaystyle \cos^2\!\alpha + \sin^2\!\alpha \;=\;4 - 5.364\cos\beta + 1.798281\left(\cos^2\!\beta + \sin^2\!\beta\right)$
and we have: .$\displaystyle 1 \;=\;4 - 5.364\cos\beta + 1.798281$
Hence: .$\displaystyle 5.364\cos\beta \:=\:4.798281\quad\Rightarrow\quad \cos\beta \:=\:0.894534116\quad\Rightarrow\quad\boxed{\beta \:\approx\:26.6^o}$
Substitute into [4]: .$\displaystyle \sin\alpha \;=\;1.341\sin26.6^o \:=\:0.600444937\quad\Rightarrow\quad\boxed{\alpha \:\approx\:36.9^o}$
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# Math7
posted by .
Which ordered pair is not a solution of
y = 4x -9
(-9, 27)
(3, -21)
(-6, 15)I think this one?
(5, -28)
• Math7 -
For x = - 9
y = 4 x - 9 = 4 * ( - 9 ) - 9 = - 36 - 9 = - 45
For x = 3
y = 4 x - 9 = 4 * 3 - 9 = 12 - 9 = 3
For x = - 6
y = 4 x - 9 = 4 * ( - 6 ) - 9 = - 24 - 9 = - 33
For x = 5
y = 4 x - 9 = 4 * 5 - 9 = 20 - 9 = 11
If youe expression mean :
y = - 4 x -9
then
( 5 , - 28 ) is not a solution becouse:
for x = 5
y = - 4 x -9 = - 4 * 5 - 9 = - 20 - 9 = - 29
• Math7 -
Thank you!
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https://socratic.org/questions/what-is-the-slope-of-the-line-passing-through-the-following-points-4-1-2-7#201941
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# What is the slope of the line passing through the following points: (4,1), (-2,-7) ?
Dec 21, 2015
$\frac{4}{3}$
#### Explanation:
$\left(4 , 1\right) , \left(- 2 , - 7\right)$
slope formula: $\frac{y 2 - y 1}{x 2 - x 1}$
Now, you plug 'em in.
$\frac{- 7 - 1}{- 2 - 4}$ = $\frac{- 8}{- 6}$ = $\frac{- 4}{- 3}$ = $\frac{4}{3}$
The slope is $\frac{4}{3}$
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# Math photo calculator
There are a lot of Math photo calculator that are available online. So let's get started!
## The Best Math photo calculator
Math photo calculator can be a helpful tool for these students. Solving trinomials is a process that can be broken down into a few simple steps. First, identify the coefficients of the terms. Next, use the quadratic formula to find the roots of the equation. Finally, plug the roots back into the original equation to verify your results. While this process may seem daunting at first, with a little practice it will become second nature. With so many trinomials to solve, there's no time to waste - get started today!
Solving system of equations matrices can be a difficult task, but it is important to understand the process in order to be successful. There are many different methods that can be used to solve system of equations matrices, but the most common is Gaussian elimination. This method involves adding or subtracting rows in order to create a new matrix that is easier to solve. Once the new matrix has been created, the variables can be solved for by using back-substitution. This process can be time-consuming and difficult, but it is important to persevere in order to get the correct answer. With practice, solving system of equations matrices will become easier and more intuitive.
To find the domain and range of a given function, we can use a graph. For example, consider the function f(x) = 2x + 1. We can plot this function on a coordinate plane: As we can see, the function produces valid y-values for all real numbers x. Therefore, the domain of this function is all real numbers. The range of this function is also all real numbers, since the function produces valid y-values for all real numbers x. To find the domain and range of a given function, we simply need to examine its graph and look for any restrictions on the input (domain) or output (range).
For example, the equation 2 + 2 = 4 states that two plus two equals four. To solve an equation means to find the value of the unknown variable that makes the equation true. For example, in the equation 2x + 3 = 7, the unknown variable is x. To solve this equation, we would need to figure out what value of x would make the equation true. In this case, it would be x = 2, since 2(2) + 3 = 7. Solving equations is a vital skill in mathematics, and one that can be used in everyday life. For example, when baking a cake, we might need to figure out how many eggs to use based on the number of people we are serving. Or we might need to calculate how much money we need to save up for a new car. In both cases, solving equations can help us to get the answers we need.
## Math checker you can trust
Perfect, this isn't just a normal calculator, it actually shows you the reason for the answer, definitely very helpful, BUT, if the answer for the equation or problem is a negative it doesn't make the negative sign white, it's gray like the equals sign, so sometimes I don't see the negative sign, so I would appreciate it if that was changed, thank you
Thalia Williams
it's so easy to use and helped me a lot. I didn't use it for cheating, but I just used it to check my answers if they're correct or not. but seriously it's so helpful for me, it makes me surer of my answers so thanks for making this app btw in the future I hope that this app can detect problem solving questions and show the solutions too. but this is still a good app
Pippa Bailey
Equation solver free math help How to solve definite integrals Geometry expression Math tutors near me Math photo calculator
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# Precalculus word problem solver
One instrument that can be used is Precalculus word problem solver. We will also look at some example problems and how to approach them.
## The Best Precalculus word problem solver
This Precalculus word problem solver provides step-by-step instructions for solving all math problems. If you're struggling with math, there's no need to feel alone or embarrassed. Help is available, and you can get it from the comfort of your own home. There are plenty of free math help online chat rooms and forums, and they can be a great resource. You can find other students who are struggling with the same topics, and you can work together to find solutions. You can also get help from experts, and you might even be able to find a tutor. Whatever you
To solve a rate of change problem, you need to find the difference between two values and then divide that difference by the time interval between those two values. For example, if you want to know the rate of change of car speeds from 0 to 60 mph, you would find the difference in speeds (60 - 0 = 60) and then divide that by the time interval (60 / 60 = 1).
To solve inverse functions, we must first determine what the inverse function is. To do this, we must find the function's inverse function. The inverse function is the function that "undoes" the original function. For example, the inverse function of the function f(x) = 2x is the function g(x) = x/2. To solve inverse functions, we must first determine what the inverse function is. To do this, we must find the function's inverse function
In general, however, the most common method is to use an iterative solver. This involves setting up the equations in a specific way and then solving them using a series of iterations. The number of iterations required depends on the accuracy desired.
There's no need to be shy about asking for help with a word problem. Many students find them difficult, but with a little practice, they can be solved. First, read the problem carefully and identify what information is given and what is being asked. Then, use that information to set up a equation or system of equations. Finally, solve the equation or system to find the answer.
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SEARCH HOME
Math Central Quandaries & Queries
Question from iris, a student: we have some confusion in our problem. Please help us. We would like to know "the principle of mathematical induction" (i) for n=1, p(1) is true. (ii) assume that for n=k>=1, p(k) is true we have to prove p(k+1) is true. Here (Is n=k>=1 true? or Is n=k.1 true?) Thanks.
We have two responses for you
The statement of your questions may have a misleading misprint in it, however, you are assuming that for some k ≥ 1 that you know p(k) is true. In essence that you have checked out p(n) for n = 1 all the way up to n = k, and now you want to use that information (that p(k) is true) to deduce the truth of the statement for n = k+1. The most common error I see from students is a statement like " we've checked p(1) is true, now assume p(n) is true for n ≥ 1"; that says assume that it's true for all n, which is what you're trying to prove.
Penny
Iris,
The Principle of Mathematical Induction is a theorem. It goes like this.
Suppose you have a statement that involves the integer n. Let's call it p(n).
If you know:
1. that the statement is true is true when n=1 (that is p(1) is true), and
2. that for every k >= 1, IF the statement is true when n=k then it is also true when n=k+1 (that is if p(k) is true then p(k+1) is true),
then you can conclude that the statement is true for every integer n greater than or equal to 1.
Normally a proof by induction has four main parts:
1. Basis. Here you check that the statement is true when n=1.
2. Induction Hypothesis. Here you write down that you are assuming the statement is true when n=k, for SOME k >= 1.
3. Induction Step. Here you argue that if the induction hypothesis is true then the statement is also true when n=k+1.
4. Conclusion. Here you say that having carried out the three steps above, the Principle of Mathematical Induction implies that the statement is true for all integers n >= 1.
For a longer explanation and some examples, try looking at
http://www.math.uvic.ca/faculty/gmacgill/guide/induction.pdf
Victoria
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
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# Lesson 4
Tables, Equations, and Graphs of Functions
### Problem 1
The graph and the table show the high temperatures in a city over a 10-day period.
day temperature (degrees F) 1 2 3 4 5 6 7 8 9 10 60 61 63 61 62 61 60 65 67 63
1. What was the high temperature on Day 7?
2. On which days was the high temperature 61 degrees?
3. Is the high temperature a function of the day? Explain how you know.
4. Is the day a function of the high temperature? Explain how you know.
### Problem 2
The amount Lin’s sister earns at her part-time job is proportional to the number of hours she works. She earns \$9.60 per hour.
1. Write an equation in the form $$y=kx$$ to describe this situation, where $$x$$ represents the hours she works and $$y$$ represents the dollars she earns.
2. Is $$y$$ a function of $$x$$? Explain how you know.
3. Write an equation describing $$x$$ as a function of $$y$$.
### Problem 3
Use the equation $$2m+4s=16$$ to complete the table, then graph the line using $$s$$ as the dependent variable.
$$m$$ $$s$$ 0 -2 3 0
### Problem 4
Solve the system of equations: $$\begin{cases} y=7x+10 \\ y=\text-4x-23 \\ \end{cases}$$
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# Range
08/02/2020
The Range of a distribution gives a measure of the width (or the spread) of the data values of the corresponding random variable. For example, if there are two random variables X and Y such that X corresponds to the age of human beings and Y corresponds to the age of turtles, we know from our general knowledge that the variable corresponding to the age of turtles should be larger.
Since the average age of humans is 50-60 years, while that of turtles is about 150-200 years; the values taken by the random variable Y are indeed spread out from 0 to at least 250 and above; while those of X will have a smaller range. Thus, qualitatively you’ve already understood what the Range of a distribution means. The mathematical formula for the same is given as:
Range=L–S
where L – the largets/maximum value attained by the random variable under consideration and S – the smallest/minimum value.
### Properties
• The Range of a given distribution has the same units as the data points.
• If a random variable is transformed into a new random variable by a change of scale and a shift of origin as –
Y = aX + b
where Y – the new random variable, X – the original random variable and a,b – constants. Then the ranges of X and Y can be related as –
RY = |a|RX
Clearly, the shift in origin doesn’t affect the shape of the distribution, and therefore its spread (or the width) remains unchanged. Only the scaling factor is important.
• For a grouped class distribution, the Range is defined as the difference between the two extreme class boundaries.
• A better measure of the spread of a distribution is the Coefficient of Range, given by:
Coefficient of Range (expressed as a percentage)=L–SL+S×100
Clearly, we need to take the ratio between the Range and the total (combined) extent of the distribution. Besides, since it is a ratio, it is dimensionless, and can, therefore, one can use it to compare the spreads of two or more different distributions as well.
• The range is an absolute measureof Dispersion of a distribution while the Coefficient of Range is a relative measure of dispersion.
Due to the consideration of only the end-points of a distribution, the Range never gives us any information about the shape of the distribution curve between the extreme points. Thus, we must move on to better measures of dispersion. One such quantity is Mean Deviation which is we are going to discuss now.
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• Create Account
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### #ActualPochyPoch
Posted 12 September 2012 - 11:48 AM
Hello to both of you !
I worked very hard those past days on the simulation, and made some progress.
@K_J_M & @bmarci
If you look at the issue with regards to local car coordinates then each wheel has 0 longtitude velocity and (N) Lateral velocity if the car is spinning on the spot.
That is right, but when i look at a point in a system, i use my pointVelocity method which takes angVel in account like you said
But in regards to a stationary car spinning on it's axis the lateral wheel velocity is derived from the car angular velocity.
So, theorically, when spinning on itself, any point velocity depends on angular velocity of the car. That velocity will be tangent to rotation circle and depends on distance to CG. The wikipedia page illustrates this very well : http://upload.wikime...cceleration.PNG
So i have something like that :
So, now, what bmarci said is exactly what i'm doing :
1. Create transform functions between coordinate systems (World, Car, Wheel) and double check if they produce the wanted results.
2. Calculate everything in its own coordinates, eg: wheel forces in wheel space, car forces in car space...etc
3. Forget about steering angle in slip angle/ratio calculations. If you have the right transform functions, that will take care of steering angles.
4. Check the result of the ATAN(lat/long), especialy when dividing by long=0!!! The result should be +-PI/2
That's exactly the way i do things. So that means, i am transforming the point velocity of the wheel (the blue vector on the picture) in wheel coordinates, which as we can see it on the picture will result in both lont/lat velocities, and so in a slip angle which is not 90° when the car is spinning on itself.
So the conclusion is in contradiction with what KJM stated above, but i can't see what is failing in my reasonment. Am i right to not find 90° for a wheel which belongs to the right place on a car spinning on itself ? The point velocity seems good, so if i am supposed to get only lateral velocity, i need a sort of hack, or maybe stay in car coordinates instead of going into wheel system ?
The thing is, i am almost sure i am right on this and that my slip angles are good, because the wheel produce good damping forces to annihilate car angular velocity when spinning on itself. If tires produce 90° slip angles when the car is spinning the car is slowing down of 99% and not 100%.
If my slip angles are good, that would mean that my car problems during cornering lies somewhere else. As i said i have obtained a better result, but i am not sure if my base calculations in slip angles are OK. So i cannot really try to tune pacejka or other things like wheel velocity and weight transfer if the underlying system is not good.
Actually the result i have is very close to the Java source example you can download on marco's website. It seems ok as i have deactivated everything else : weight transfer, decreasing pacejka curve, no more real engine/wheel transmission / rotation. I needed to have a clean working table, sort of.
But when i try the car on my game track it is far from playable. If i use an always increasing, producing high cornering forces curve for lateral pacej, then the car turns very sharply, and i can have somme controled oversteer, but the car will eventually stop in the turning curve instead of keeping some speed.
If i use a realistic pacejka curve, then the car is understeering at high speed, which is good, because i can do high speed turns in a comfortable way. But, if i want to produce oversteer and drift for a sharper curve, it just doesn't happen.
So, what do you think ? Problem in base system calculation ? Pacejka curve tuning ? Maybe i should reactivate some critical feature or system ?
I will try to cast a video of the car later to show try to show you, if interested.
Thanks for your help, my game really needs it
Sorry for the long posts by the way...
EDIT : the main thing here, might be this "the car will eventually stop in the turning curve instead of keeping some speed".
I have the feeling that when i take a sharp turn, even if a great part of the initial velocity will be killed by the lateral forces, and even without pushing throttle, the car should keep more front velocity to come out of the curve. Instead it just stops there with very small to none forward velocity. If i could have this, and manage somehow to mix the best of the 2 results i have with my different pacejka, the game would start to be fun i think.
EDIT2 : those results were obtained with a dummy traction circle (allowing 2*weightLoad traction max), applied only on lateral force for testing purpose.
### #3PochyPoch
Posted 12 September 2012 - 11:48 AM
Hello to both of you !
I worked very hard those past days on the simulation, and made some progress.
@K_J_M & @bmarci
If you look at the issue with regards to local car coordinates then each wheel has 0 longtitude velocity and (N) Lateral velocity if the car is spinning on the spot.
That is right, but when i look at a point in a system, i use my pointVelocity method which takes angVel in account like you said
But in regards to a stationary car spinning on it's axis the lateral wheel velocity is derived from the car angular velocity.
So, theorically, when spinning on itself, any point velocity depends on angular velocity of the car. That velocity will be tangent to rotation circle and depends on distance to CG. The wikipedia page illustrates this very well : http://upload.wikime...cceleration.PNG
So i have something like that :
So, now, what bmarci said is exactly what i'm doing :
1. Create transform functions between coordinate systems (World, Car, Wheel) and double check if they produce the wanted results.
2. Calculate everything in its own coordinates, eg: wheel forces in wheel space, car forces in car space...etc
3. Forget about steering angle in slip angle/ratio calculations. If you have the right transform functions, that will take care of steering angles.
4. Check the result of the ATAN(lat/long), especialy when dividing by long=0!!! The result should be +-PI/2
That's exactly the way i do things. So that means, i am transforming the point velocity of the wheel (the blue vector on the picture) in wheel coordinates, which as we can see it on the picture will result in both lont/lat velocities, and so in a slip angle which is not 90° when the car is spinning on itself.
So the conclusion is in contradiction with what KJM stated above, but i can't see what is failing in my reasonment. Am i right to not find 90° for a wheel which belongs to the right place on a car spinning on itself ? The point velocity seems good, so if i am supposed to get only lateral velocity, i need a sort of hack, or maybe stay in car coordinates instead of going into wheel system ?
The thing is, i am almost sure i am right on this and that my slip angles are good, because the wheel produce good damping forces to annihilate car angular velocity when spinning on itself. If tires produce 90° slip angles when the car is spinning the car is slowing down of 99% and not 100%.
If my slip angles are good, that would mean that my car problems during cornering lies somewhere else. As i said i have obtained a better result, but i am not sure if my base calculations in slip angles are OK. So i cannot really try to tune pacejka or other things like wheel velocity and weight transfer if the underlying system is not good.
Actually the result i have is very close to the Java source example you can download on marco's website. It seems ok as i have deactivated everything else : weight transfer, decreasing pacejka curve, no more real engine/wheel transmission / rotation. I needed to have a clean working table, sort of.
But when i try the car on my game track it is far from playable. If i use an always increasing, producing high cornering forces curve for lateral pacej, then the car turns very sharply, and i can have somme controled oversteer, but the car will eventually stop in the turning curve instead of keeping some speed.
If i use a realistic pacejka curve, then the car is understeering at high speed, which is good, because i can do high speed turns in a comfortable way. But, if i want to produce oversteer and drift for a sharper curve, it just doesn't happen.
So, what do you think ? Problem in base system calculation ? Pacejka curve tuning ? Maybe i should reactivate some critical feature or system ?
I will try to cast a video of the car later to show try to show you, if interested.
Thanks for your help, my game really needs it
Sorry for the long posts by the way...
EDIT : the main thing here, might be this "the car will eventually stop in the turning curve instead of keeping some speed".
I have the feeling that when i take a sharp turn, even if a great part of the initial velocity will be killed by the lateral forces, and even without pushing throttle, the car should keep more front velocity to come out of the curve. Instead it just stops there with very small to none forward velocity. If i could have this, and manage somehow to mix the best of the 2 results i have with my different pacejka, the game would start to be fun i think.
EDIT2 : those results were obtained with a dummy traction circle (allowing 2*weightLoad traction max), applied only on lateral force for testing purpose.
### #2PochyPoch
Posted 12 September 2012 - 11:10 AM
Hello to both of you !
I worked very hard those past days on the simulation, and made some progress.
@K_J_M & @bmarci
If you look at the issue with regards to local car coordinates then each wheel has 0 longtitude velocity and (N) Lateral velocity if the car is spinning on the spot.
That is right, but when i look at a point in a system, i use my pointVelocity method which takes angVel in account like you said
But in regards to a stationary car spinning on it's axis the lateral wheel velocity is derived from the car angular velocity.
So, theorically, when spinning on itself, any point velocity depends on angular velocity of the car. That velocity will be tangent to rotation circle and depends on distance to CG. The wikipedia page illustrates this very well : http://upload.wikime...cceleration.PNG
So i have something like that :
So, now, what bmarci said is exactly what i'm doing :
1. Create transform functions between coordinate systems (World, Car, Wheel) and double check if they produce the wanted results.
2. Calculate everything in its own coordinates, eg: wheel forces in wheel space, car forces in car space...etc
3. Forget about steering angle in slip angle/ratio calculations. If you have the right transform functions, that will take care of steering angles.
4. Check the result of the ATAN(lat/long), especialy when dividing by long=0!!! The result should be +-PI/2
That's exactly the way i do things. So that means, i am transforming the point velocity of the wheel (the blue vector on the picture) in wheel coordinates, which as we can see it on the picture will result in both lont/lat velocities, and so in a slip angle which is not 90° when the car is spinning on itself.
So the conclusion is in contradiction with what KJM stated above, but i can't see what is failing in my reasonment. Am i right to not find 90° for a wheel which belongs to the right place on a car spinning on itself ? The point velocity seems good, so if i am supposed to get only lateral velocity, i need a sort of hack, or maybe stay in car coordinates instead of going into wheel system ?
The thing is, i am almost sure i am right on this and that my slip angles are good, because the wheel produce good damping forces to annihilate car angular velocity when spinning on itself. If tires produce 90° slip angles when the car is spinning the car is slowing down of 99% and not 100%.
If my slip angles are good, that would mean that my car problems during cornering lies somewhere else. As i said i have obtained a better result, but i am not sure if my base calculations in slip angles are OK. So i cannot really try to tune pacejka or other things like wheel velocity and weight transfer if the underlying system is not good.
Actually the result i have is very close to the Java source example you can download on marco's website. It seems ok as i have deactivated everything else : weight transfer, decreasing pacejka curve, no more real engine/wheel transmission / rotation. I needed to have a clean working table, sort of.
But when i try the car on my game track it is far from playable. If i use an always increasing, producing high cornering forces curve for lateral pacej, then the car turns very sharply, and i can have somme controled oversteer, but the car will eventually stop in the turning curve instead of keeping some speed.
If i use a realistic pacejka curve, then the car is understeering at high speed, which is good, because i can do high speed turns in a comfortable way. But, if i want to produce oversteer and drift for a sharper curve, it just doesn't happen.
So, what do you think ? Problem in base system calculation ? Pacejka curve tuning ? Maybe i should reactivate some critical feature or system ?
I will try to cast a video of the car later to show try to show you, if interested.
Thanks for your help, my game really needs it
Sorry for the long posts by the way...
EDIT : the main thing here, might be this "the car will eventually stop in the turning curve instead of keeping some speed".
I have the feeling that when i take a sharp turn, even if a great part of the initial velocity will be killed by the lateral forces, and even without pushing throttle, the car should keep more front velocity to come out of the curve. Instead it just stops there with very small to none forward velocity. If i could have this, and manage somehow to mix the best of the 2 results i have with my different pacejka, the game would start to be fun i think.
### #1PochyPoch
Posted 12 September 2012 - 11:04 AM
Hello to both of you !
I worked very hard those past days on the simulation, and made some progress.
@K_J_M & @bmarci
If you look at the issue with regards to local car coordinates then each wheel has 0 longtitude velocity and (N) Lateral velocity if the car is spinning on the spot.
That is right, but when i look at a point in a system, i use my pointVelocity method which takes angVel in account like you said
But in regards to a stationary car spinning on it's axis the lateral wheel velocity is derived from the car angular velocity.
So, theorically, when spinning on itself, any point velocity depends on angular velocity of the car. That velocity will be tangent to rotation circle and depends on distance to CG. The wikipedia page illustrates this very well : http://upload.wikime...cceleration.PNG
So i have something like that :
So, now, what bmarci said is exactly what i'm doing :
1. Create transform functions between coordinate systems (World, Car, Wheel) and double check if they produce the wanted results.
2. Calculate everything in its own coordinates, eg: wheel forces in wheel space, car forces in car space...etc
3. Forget about steering angle in slip angle/ratio calculations. If you have the right transform functions, that will take care of steering angles.
4. Check the result of the ATAN(lat/long), especialy when dividing by long=0!!! The result should be +-PI/2
That's exactly the way i do things. So that means, i am transforming the point velocity of the wheel (the blue vector on the picture) in wheel coordinates, which as we can see it on the picture will result in both lont/lat velocities, and so in a slip angle which is not 90° when the car is spinning on itself.
So the conclusion is in contradiction with what KJM stated above, but i can't see what is failing in my reasonment. Am i right to not find 90° for a wheel which belongs to the right place on a car spinning on itself ? The point velocity seems good, so if i am supposed to get only lateral velocity, i need a sort of hack, or maybe stay in car coordinates instead of going into wheel system ?
The thing is, i am almost sure i am right on this and that my slip angles are good, because the wheel produce good damping forces to annihilate car angular velocity when spinning on itself. If tires produce 90° slip angles when the car is spinning the car is slowing down of 99% and not 100%.
If my slip angles are good, that would mean that my car problems during cornering lies somewhere else. As i said i have obtained a better result, but i am not sure if my base calculations in slip angles are OK. So i cannot really try to tune pacejka or other things like wheel velocity and weight transfer if the underlying system is not good.
Actually the result i have is very close to the Java source example you can download on marco's website. It seems ok as i have deactivated everything else : weight transfer, decreasing pacejka curve, no more real engine/wheel transmission / rotation. I needed to have a clean working table, sort of.
But when i try the car on my game track it is far from playable. If i use an always increasing, producing high cornering forces curve for lateral pacej, then the car turns very sharply, and i can have somme controled oversteer, but the car will eventually stop in the turning curve instead of keeping some speed.
If i use a realistic pacejka curve, then the car is understeering at high speed, which is good, because i can do high speed turns in a comfortable way. But, if i want to produce oversteer and drift for a sharper curve, it just doesn't happen.
So, what do you think ? Problem in base system calculation ? Pacejka curve tuning ? Maybe i should reactivate some critical feature or system ?
I will try to cast a video of the car later to show try to show you, if interested.
Thanks for your help, my game really needs it
Sorry for the long posts by the way...
PARTNERS
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When thinking about cost behavior, think about how the cost behaves in total. A variable cost is a cost that varies in total. The cost increases or decreases based on a related activity.
The formula for total variable cost is:
Total Variable Cost = Variable Rate X Activity
### Assume a constant rate
For planning and decision making purposes, we assume that the variable rate is constant. This allows for a single variable in the calculations. Only the activity will change. Now, that is not always the case, but as long as we are within the relevant range for our decision, we can assume that the rate will stay the same.
### But isn’t it fixed if the rate stays the same?
Remember that a variable cost varies in total. The rate might stay the same but once you multiply the rate by varying levels of activity, the total variable cost will change.
Imagine that you are selling candy bars as a fundraiser for a club to which you belong. Your cost is 50 cents per candy bar and the club sells the candy bars for \$1 each. If the club sells 200 candy bars, what is the total variable cost? Is it 50 cents? No, that is the cost of a single candy bar. If you sell 200, you would need to multiply that by 50 cents for each of the candy bars sold.
200 candy bars X 50 cents per candy bar = \$100
What if the club sold 500 candy bars? The total variable cost would be \$250.
Here is a graph of the total variable cost of candy bars for the fundraiser:
Notice that if no candy bars are sold, there is no cost. The more candy bars that are sold, the higher the cost. The cost line is a straight line. The slope of the line is equal to the variable rate. For each additional unit sold, the line increases at a rate of 50 cents. Think of the formula of a line: y=mx + b, where y is your y coordinate, x is your x coordinate, m is the slope and b is the y-intercept (the point where the line hits the y-axis).
The formula for total variable cost is: y=mx. The y-intercept for a variable cost is always zero because if there is no activity, there is no cost. Therefore, the line will always start at 0,0. The slope of the line, m, is your variable rate. The activity is x. See your math teacher was right when he or she told you you would use this stuff someday!
Frequently, you will see textbooks show the formula for the slope of a line as the formula for cost equations.
#### Related Videos
Cost Behavior: Fixed, Variable, Step and Mixed
Fixed and Variable costs as per unit and total costs
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# Placement Papers - HAL
## Why HAL Placement Papers?
Learn and practice the placement papers of HAL and find out how much you score before you appear for your next interview and written test.
## Where can I get HAL Placement Papers with Answers?
IndiaBIX provides you lots of fully solved HAL Placement Papers with answers. You can easily solve all kind of placement test papers by practicing the exercises given below.
## How to solve HAL Placement Papers?
You can easily solve all kind of questions by practicing the following exercises.
### HAL Technical Officer Interview Paper (Kolkata)
Posted By : Hyndavi Rating : +14, -3
14) In aptitude part see through series completion there was a 5-6 no of that.
15) No aptitude questing like avg, %, proff-loss, prob, etc.
16) Sentence correction 2-3.
17) Why rotor of SCIM is skewed?
18) Magnetostriction increases-
19) Good transformer oil contains how much amount of water in ppm?
20) Lightning over voltage create how much volt?
21) K/s(s+1)(s+5) then find out the value of K so that system become stable.
22) There was a question from definition of Nyquist criteria.
23) Standing done up to the frequency range of? (audio)
24) If a square waveform if given to the supply side of a 2wdg transformer its o/p waveform will be-
25) SC test done on transformer because-
26) If a transformer is operating in .8 PF lag & at efficiency of .9, if now P is .8 lead the efficiency will become.
27) Interpoles are connected for?
28) Dielectric strength of air-
29) A alternator is connected to a grid. Now its prime mover is go slow by ½. Change in its exiting current to held in synchronism.
30) Is the strength of capacitor in LPF is increased the op voltage will-
31) The voltage of a generator and an infinite bus are given 0.92, 10 deg & 1.0, 0 deg resp. the generator acts as a "
32) A wave will not experience any reflection when impedance of line is equal to- surge imp
33) Power transmitted is related to system voltage as-
34) Delay creator ckt name (multivibrator name actually)
35) In tachogenerator adjustment is done by-
36) Surge wave is attended by- (C,L,R)
37) there was 10 question related to measurement-
38) Read all electrostatic instrument related question from any competitive exam book.
39) Lowest reading in moving iron meter is-
40) In galvanometer damping is provided by-
41) In electrodynamic meter a battery & a resistance is provide for-
42) Megger is used for-
43) A question like an ac is applied against a moving coil meter it will reads-
44) Rectifier type instrument measureswhat value - (Rms)
45) Advantage of electrostatic instrument is-
46) Creeping occurs because-
47) Why 2 holes are drilled oppositely in disc of energy meter.
48) Energy meter is a _____ type equipment (recording)
49) There was a question related to selection of a voltmeter based on accuracy-
50) Advantage of hay bridge over anderson bridge-
51) Inductance is measured by " (all the 3 bridge name) .
52) Potentiometer is which type of instrument-
53) Which type of motor is used in refrigerator?
54) Reflection coefficient based problem 1
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# Logarithmic Differentiation
If $y = f\left( x \right)$ is a complicated function, i.e. it involves several products of functions, quotients or radical signs, then we take logarithms on both sides which will make differentiation much easier. Such differentiation is called logarithmic differentiation.
Thus, taking logarithms on both sides of the given equation, we have
$\ln y = \ln f\left( x \right)$
Differentiating both sides with respect to $x$, we have
$\begin{gathered} \frac{d}{{dx}}\ln y = \frac{d}{{dx}}\ln f\left( x \right) \\ \Rightarrow \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{f\left( x \right)}}f’\left( x \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{y}{{f\left( x \right)}}f’\left( x \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{f\left( x \right)}}{{f\left( x \right)}}f’\left( x \right),\,\,\,\,y = f\left( x \right) \\ \Rightarrow \frac{{dy}}{{dx}} = f’\left( x \right) \\ \end{gathered}$
This shows that the result is the same as what we would have without taking $\ln$.
Example: Differentiate $\frac{{\sqrt x {{\left( {1 – 2x} \right)}^{2/3}}}}{{{{\left( {2 – 3x} \right)}^{3/4}}{{\left( {3 – 4x} \right)}^{4/3}}}}$ with respect to $x$.
Consider the function $y = \frac{{\sqrt x {{\left( {1 – 2x} \right)}^{\frac{2}{3}}}}}{{{{\left( {2 – 3x} \right)}^{\frac{3}{4}}}{{\left( {3 – 4x} \right)}^{\frac{4}{3}}}}}$, then taking $\ln$ both sides, we get
$\begin{gathered} \ln y = \ln \left[ {\frac{{\sqrt x {{\left( {1 – 2x} \right)}^{\frac{2}{3}}}}}{{{{\left( {2 – 3x} \right)}^{\frac{3}{4}}}{{\left( {3 – 4x} \right)}^{\frac{4}{3}}}}}} \right] \\ \Rightarrow \ln y = \ln \sqrt x + \ln {\left( {1 – 2x} \right)^{\frac{2}{3}}} – \ln {\left( {2 – 3x} \right)^{\frac{3}{4}}} – \ln {\left( {3 – 4x} \right)^{\frac{4}{3}}} \\ \Rightarrow \ln y = \frac{1}{2}\ln x + \frac{2}{3}\ln \left( {1 – 2x} \right) – \frac{3}{4}\ln \left( {2 – 3x} \right) – \frac{4}{3}\ln \left( {3 – 4x} \right) \\ \end{gathered}$
Differenting both sides with respect to $x$, we have
$\begin{gathered} \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{2x}} – \frac{4}{{3\left( {1 – 2x} \right)}} + \frac{9}{{4\left( {2 – 3x} \right)}} + \frac{{16}}{{3\left( {3 – 4x} \right)}} \\ \Rightarrow \frac{{dy}}{{dx}} = y\left[ {\frac{1}{{2x}} – \frac{4}{{3\left( {1 – 2x} \right)}} + \frac{9}{{4\left( {2 – 3x} \right)}} + \frac{{16}}{{3\left( {3 – 4x} \right)}}} \right] \\ \Rightarrow \frac{{dy}}{{dx}} = \left[ {\frac{{\sqrt x {{\left( {1 – 2x} \right)}^{\frac{2}{3}}}}}{{{{\left( {2 – 3x} \right)}^{\frac{3}{4}}}{{\left( {3 – 4x} \right)}^{\frac{4}{3}}}}}} \right]\left[ {\frac{1}{{2x}} – \frac{4}{{3\left( {1 – 2x} \right)}} + \frac{9}{{4\left( {2 – 3x} \right)}} + \frac{{16}}{{3\left( {3 – 4x} \right)}}} \right] \\ \end{gathered}$
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# How to Scale a Function Vertically?
Scaling is a process of changing the size and shape of the graph of the function. In this blog post, you will learn how to vertical scaling.
Vertical scaling refers to the shrinking or stretching of the curve along the $$y$$-axis by some specific units.
## Step by step guide to vertical scaling
There are four types of transformation possible for a graph of a function, which are:
• Rotations
• Translations
• Reflections
• Scaling
In addition, scaling can be divided into two different types, e.g.
• Horizontal scaling
• Vertical scaling
Vertical scaling refers to changing the shape and size of a function graph along the $$y$$-axis and is done by multiplying the function by a fixed value.
For example: We have a function $$y=f(x)$$
We have multiplied the function by $$2$$, that is: $$y=2 f(x)$$
The distance of the points on the curve becomes farther from the $$x$$-axis.
The shape of the curve depends on the value of $$C$$:
• If $$C > 1$$, the graph stretches and makes the graph steeper.
• If $$C < 1$$, the graph shrinks and makes the graph flatter.
### How to do vertical scaling?
Let’s understand this with an example:
Suppose we have a basic quadratic equation $$f(x)=x^2$$ and a graphical representation of the diagram is shown below.
If we want to vertically scale this chart, we have to follow the given steps:
Step 1: Select the constant with which we want to scale the function.
Here we have selected $$+2$$.
Step 2: Write the new function as $$g(x)=C f(x)$$, where $$C$$ is the constant.
Here, the new function will be: $$g(x)=2 f(x)= 2 x^2$$
Step 3: Trace the new function graph by replacing each value of $$y$$ with $$Cy$$.
Here we need to replace the value of the $$y$$-coordinate by $$2y$$.
The $$Y$$ coordinates of each point in the graph are multiplied by $$±C$$, and the curve is shrinks or stretches accordingly.
Here we have the graph $$x$$ and it is stretched in the $$y$$-direction with a factor of $$+2$$.
Note: As we have scaled it with a factor of $$+2$$ units, it has made the graph steeper.
### Vertical Scaling – Example 1:
Vertically stretch the function $$y=(x+2)$$ by a factor of two.
## Exercises for Vertical Scaling
• Vertically stretch the function $$f(x)=x^3$$ by a factor of $$-\frac{1}{3}$$.
• Vertically stretch the function $$f(x)=sin x$$ by a factor of $$3$$.
• Vertically stretch the function $$f(x)=x^3$$ by a factor of $$-\frac{1}{3}$$.
• Vertically stretch the function $$f(x)=sin x$$ by a factor of $$3$$.
### What people say about "How to Scale a Function Vertically?"?
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Calcudoku puzzle forumhttp://www.calcudoku.org/forum/ The use of maximums and minimums in the sum of cageshttp://www.calcudoku.org/forum/viewtopic.php?f=3&t=39 Page 1 of 1
Author: clm [ Wed Jun 22, 2011 12:30 am ] Post subject: The use of maximums and minimums in the sum of cages Basic strategies. Example of “the use of maximums and minimums in the sum of cages”. One year has elapsed since the original post was sent. Along this time I am sure this rule has been useful in many cases. However, just in case any calcudoker has still doubts on how to apply it I am including a recent puzzle which clarifies well its utility. The example below corresponds to the Sunday’s regular 9x9 (Jul 08, 2012). Here is the puzzle: We will apply the rule to determine the cage “3:” which “initially” could admit two combinations: the [1,3] or the [3,9] (the [2,6] not allowed due to the 6’s in rows 4 and 5). Graphic 2 below (the relevant cells shown in blue colour): The sum of the three leftmost columns must be 135 (= 3 x 45) so (naming, for clarity, s1 the sum of numbers in cage “3:”, s2 the sum of numbers in cage”3-” and “a” the value of the single cell c7) we have: s1 + s2 + a + 14 + 9 + 23 + 6 + 1 + 14 + 8 + 8 + 16 + 3 + 7 = 135, that is: s1 + s2 + a = 26. A first consequence, i.e., from this equation is that, being s1 even and s2 odd, “a” must be odd, that is, 3 or 5 (since 1, 7 and 9 are already present in column c). We apply the rule of maximums and minimums in this way: If the maximum possible value of s2 is 15 (case [6,9]) and the maximum possible value of “a” is 5, the minimum possible value of s1 should be 6 (26 - 15 - 5), consequently the cage “3:” cann’t be [1,3] and it must be [3,9] (with s1 = 12). Now, s2 + a = 26 - 12 = 14. If a = 3 >>> s2 = 11 = [4,7], not allowed due to i6 = 7, then: a = 5 and s2 = 9 = [3,6] with b6 = 6 and c6 = 3 (see Graphic 3 below): We have defined the only possible combination [3,9] for “3:” and at the same time we have determined 4 additional cells (b6, c6, c7, d7). The official solution of the full puzzle:
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Question
# If A=[100−1] and B=[0110], find AB and BA. Prove that AB≠BA.
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Solution
## Given: A=[100−1] and B=[0110] Now, AB=[100−1][0110] ⇒AB=[1(0)+0(1)1(1)+0(0)0(0)+(−1)(1)0(1)+(−1)(0)] ⇒AB=[0+01+00−10+0] ⇒AB=[01−10] ⋯(1) BA=[0110][100−1] ⇒BA=[0(1)+1(0)0(0)+1(−1)1(1)+0(0)1(0)+0(−1)] ⇒BA=[0+00−11+00+0] ⇒BA=[0−110] ⋯(2) From equation (1) and (2) BA≠AB. Hence proved.
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CC-MAIN-2024-18
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https://gufosaggio.net/come-si-trova-la-derivata-di-f-x-1-x-2-usando-il-processo-limite/
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# Come si trova la derivata di f (x) = 1 / x ^ 2 usando il processo limite?
#### Risposta:
f'(x)=-2/x^3
#### Spiegazione:
By definition of the derivative f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h
Quindi con f(x) = 1/x^2 we have;
f'(x)=lim_(h rarr 0) ( 1/(x+h)^2 -1/x^2 ) / h
:. f'(x)=lim_(h rarr 0)1/h*(x^2-(x+h)^2)/((x+h)^2x^2)
:. f'(x)=lim_(h rarr 0) (x^2-(x^2+2hx+h^2))/(h(x+h)^2x^2)
:. f'(x)=lim_(h rarr 0) (x^2-x^2-2hx-h^2)/(h(x+h)^2x^2)
:. f'(x)=lim_(h rarr 0) (-2x-h)/((x+h)^2x^2)
:. f'(x)=(-2x)/((x)^2x^2)
:. f'(x)=-2/x^3
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CC-MAIN-2024-38
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https://paperzz.com/doc/8478778/chapter-3-understanding-money-management
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### Chapter 3 Understanding Money Management
```Chapter 3
Understanding Money Management
Nominal and
Effective Interest
Rates
Equivalence
Calculations
Changing Interest
Rates
Debt Management
1
Understanding Money
Management
Financial institutions often quote
interest rate based on an APR.
In all financial analysis, we need to
convert the APR into an appropriate
effective interest rate based on a
payment period.
When payment period and interest
period differ, calculate an effective
interest rate that covers the payment
period.
2
Understanding Money
Management
3
Focus
1. If payments occur more frequently than
annual, how do we calculate economic
equivalence?
2. If interest period is other than annual,
how do we calculate economic
equivalence?
3. How are commercial loans structured?
4. How should you manage your debt?
4
Nominal Versus Effective Interest
Rates
Nominal Interest
Rate:
Interest rate
quoted based on
an annual period
Effective Interest
Rate:
Actual interest
earned or paid in
a year or some
other time period
5
18% Compounded Monthly
Nominal
interest rate
Interest
period
Annual
percentage
rate (APR)
6
18% Compounded Monthly
What It Really Means?
Interest rate per month (i) = 18%/12 = 1.5%
Number of interest periods per year (N) = 12
In words,
Bank will charge 1.5% interest each month on
your unpaid balance, if you borrowed money
You will earn 1.5% interest each month on
your remaining balance, if you deposited
money
7
18% compounded monthly
Question: Suppose that you invest \$1 for 1
year at 18% compounded monthly. How
much interest would you earn?
Solution: F = \$ 1(1 + i )12 = \$ 1(1 + 0 .015 )12
= \$1.1956
ia = 0.1956 or 19.56%
18%
: 1.5%
8
18%
: 1.5%
18% compounded monthly
or
1.5% per month for 12 months
=
19.56 % compounded annually
9
Effective Annual Interest Rate
(Annual Effective Yield)
ia = (1 + r / M ) − 1
M
r = nominal interest rate per year
ia = effective annual interest rate
M = number of interest periods per
year
10
Practice Problem
If your credit card calculates the
interest based on 12.5% APR, what is
your monthly interest rate and annual
effective interest rate, respectively?
\$2,000 and skips payments for 2
months. What would be the total
balance 2 months from now?
11
Solution
Monthly Interest Rate:
12.5%
i=
= 1.0417%
12
Annual Effective Interest Rate:
ia = (1+ 0.010417)12-1= 13.24%
Total Outstanding Balance:
F = B2 = \$2,000(F / P,1.0417%,2)
= \$2,041.88
12
Practice Problem
Suppose your savings account pays 9%
interest compounded quarterly. If you
deposit \$10,000 for one year, how
much would you have?
13
Solution
(a) Interest rate per quarter:
9%
i=
= 2.25%
4
(b) Annual effective interest rate:
ia = (1 + 0.0225) 4 − 1 = 9.31%
(c) Balance at the end of one year (after 4 quarters)
F = \$10, 000( F / P , 2.25%, 4)
= \$10, 000( F / P , 2.25%,1)
9.31%
= \$10, 9 31
14
Effective Annual Interest Rates
(9% compounded quarterly)
First quarter
Base amount
+ Interest (2.25%)
\$10,000
+ \$225
Second quarter
= New base amount
+ Interest (2.25%)
= \$10,225
+\$230.06
Third quarter
= New base amount
+ Interest (2.25%)
= \$10,455.06
+\$235.24
Fourth quarter
= New base amount
+ Interest (2.25 %)
= Value after one
year
= \$10,690.30
+ \$240.53
= \$10,930.83
15
Nominal and Effective Interest Rates
with Different Compounding Periods
Effective Rates
Nominal
Rate
Compoundi
ng Annually
Compoundi
ng Semiannually
Compounding
Quarterly
Compounding
Monthly
Compounding
Daily
4%
4.00%
4.04%
4.06%
4.07%
4.08%
5
5.00
5.06
5.09
5.12
5.13
6
6.00
6.09
6.14
6.17
6.18
7
7.00
7.12
7.19
7.23
7.25
8
8.00
8.16
8.24
8.30
8.33
9
9.00
9.20
9.31
9.38
9.42
10
10.00
10.25
10.38
10.47
10.52
11
11.00
11.30
11.46
11.57
11.62
12
12.00
12.36
12.55
12.68
12.74
16
Effective Interest Rate per Payment
Period (i)
i = [1 + r / CK ] − 1
C
C = number of interest periods per
payment period
K = number of payment periods per year
M = number of interest periods per year
(M=CK)
17
12% compounded monthly
Payment Period = Quarter
Compounding Period = Month
1st Qtr
2nd Qtr
3rd Qtr
4th Qtr
1% 1% 1%
3.030 %
One-year
• Effective interest rate per quarter
i = (1 + 0 . 01 ) 3 − 1 = 3 . 030 %
• Effective annual interest rate
i a = ( 1 + 0 .0 1 ) 1 2 − 1 = 1 2 .6 8 %
i a = ( 1 + 0 .0 3 0 3 0 ) 4 − 1 = 1 2 .6 8 %
18
Effective Interest Rate per Payment
Period with Continuous Compounding
i = [1 + r / CK ] − 1
C
where CK = number of compounding periods
per year
continuous compounding => M (=CK)
∞
i = lim[( 1 + r / CK ) − 1]
C
= ( e r )1 / K − 1
19
Case 0: 8% compounded quarterly
Payment Period = Quarter
Interest Period = Quarterly
1st Q
2nd Q
1 interest period
3rd Q
4th Q
Given r = 8%,
K = 4 payments per year
C = 1 interest periods per quarter
M = 4 interest periods per year
i = [1 + r / C K ] C − 1
= [1 + 0 . 0 8 / ( 1 ) ( 4 ) ] 1 − 1
= 2 . 0 0 0 % p e r q u a r te r
20
Case 1: 8% compounded monthly
Payment Period = Quarter
Interest Period = Monthly
1st Q
2nd Q
3 interest periods
3rd Q
4th Q
Given r = 8%,
K = 4 payments per year
C = 3 interest periods per quarter
M = 12 interest periods per year
i = [1 + r / C K ] C − 1
= [1 + 0 . 0 8 / ( 3 ) ( 4 ) ] 3 − 1
= 2 . 0 1 3 % p e r q u a r te r
21
Case 2: 8% compounded weekly
1st Q
Payment Period = Quarter
Interest Period = Weekly
2nd Q
3rd Q
4th Q
13 interest periods
Given r = 8%,
K = 4 payments per year
C = 13 interest periods per quarter
M = 52 interest periods per year
i = [1 + r / C K ] C − 1
= [1 + 0 . 0 8 / (1 3 )( 4 )]1 3 − 1
= 2 . 0 1 8 6 % p e r q u a rte r
22
Case 3: 8% compounded continuously
Payment Period = Quarter
Interest Period = Continuously
1st Q
2nd Q
∞ interest periods
3rd Q
4th Q
Given r = 8%,
K = 4 payments per year
i = er / K −1
= e 0 .02 − 1
= 2 .0201 % per quarter
23
Summary: Effective interest rate per
quarter
Case 0
Case 1
Case 2
Case 3
8%
compounded
quarterly
8%
compounded
monthly
8%
compounded
weekly
8%
compounded
continuously
Payments
occur quarterly
Payments
occur quarterly
Payments
occur quarterly
Payments
occur quarterly
2.000% per
quarter
2.013% per
quarter
2.0186% per
quarter
2.0201% per
quarter
24
Practice Example
1000 YTL initial deposit.
Effective interest rate per quarter
Balance at the end of 3 years for a nominal
rate of 8% compounded weekly?
13
⎛ 0.08 ⎞
i = ⎜1 +
⎟ − 1 = 2.0186% per quarter
52 ⎠
⎝
F = 1000(1 + 0.020186)12 = 1271.03
or
F = 1000( F | P, 2.0186%,12)
25
Practice Example
1000 YTL initial deposit.
Effective interest rate per quarter
Balance at the end of 3 years for a nominal
rate of 8% compounded daily?
365
⎛ 0.08 ⎞ 4
i = ⎜1 +
− 1 = 2.0199% per quarter
⎟
365 ⎠
⎝
F = 1000(1 + 0.020199)12 = 1271.21
or
F = 1000( F | P, 2.0199%,12)
26
Practice Example
1000 YTL initial deposit.
Effective interest rate per quarter
Balance at the end of 3 years for a nominal
rate of 8% compounded continuously?
i=e
r
k
−1 = e
0.08
4
− 1 = 2.0201% per quarter
F = 1000(1 + 0.020201)12 = 1271.23
or
F = 1000( F | P, 2.0201%,12)
Notice the negligible difference as compounding
changes between weekly, daily, continuously!
27
Practice Example
2000 YTL borrowed. How much must be returned at
the end of 3 years if i=6% compounded monthly?
0 .06
Interest rate per month =
= 0 .005
12
F = 2000 ( F | P , 0 .5 %, 36 )
or
12
0 .06 ⎞
⎛
Effective interest rate per year = ⎜ 1 +
⎟ − 1 = 6 .168 %
12 ⎠
⎝
F = 2000 ( F | P , 6 .168 %, 3)
or
3
0 .06 ⎞
⎛
Effective interest rate per quarter = ⎜ 1 +
⎟ − 1 = 1 .5075 %
12 ⎠
⎝
F = 2000 ( F | P , 1 .5075 %, 12 )
28
Equivalence Analysis using Effective
Interest Rate
Step 1: Identify the payment period
(e.g., annual, quarter, month, week,
etc)
Step 2: Identify the interest period
(e.g., annually, quarterly, monthly, etc)
Step 3: Find the effective interest rate
that covers the payment period.
29
Principle: Find the effective interest
rate that covers the payment period
Case 1: compounding period = payment period
Case 2: compounding period < payment period
Case 3: compounding period > payment period
30
When Payment Periods and
Compounding periods coincide
Step 1: Identify the number of compounding
periods (M) per year
Step 2: Compute the effective interest rate per
payment period (i)
i = r/M
Step 3: Determine the total number of payment
periods (N)
N=M
Step 4: Use the appropriate interest formula
using i and N above
31
When Payment Periods and Compounding periods coincide
Payment Period = Interest Period
\$20,000
1 2 3 4
48
0
A
Given: P = \$20,000, r = 8.5% per year compounded monthly
Find A (monthly payments)
Step 1: M = 12
Step 2: i = r/M = 8.5%/12 = 0.7083% per month
Step 3: N = (12)(4) = 48 months
Step 4: A = \$20,000(A/P, 0.7083%,48) = \$492.97
32
Dollars Up in Smoke
What three levels of smokers who bought cigarettes every
day for 50 years at \$1.75 a pack would have if they had
instead banked that money each week:
Level of smoker
1 pack a day
\$169,325
2 packs a day
\$339,650
3 packs a day
\$507,976
Note: Assumes constant price per pack, the money banked weekly and an
annual interest rate of 5.5%
Source: USA Today, Feb. 20, 1997
33
Sample Calculation: One Pack per
Day
Step 1: Determine the effective interest rate per payment
period.
Payment period = weekly
“5.5% interest compounded weekly”
i = 5.5%/52 = 0.10577% per week
Step 2: Compute the equivalence value.
Weekly deposit amount
A = \$1.75 x 7 = \$12.25 per week
Total number of deposit periods
N = (52 weeks/yr.)(50 years)
= 2600 weeks
F = \$12.25 (F/A, 0.10577%, 2600)= \$169,325
34
Compounding more frequent than payments
Discrete Case Example: Quarterly deposits
with monthly compounding
Suppose you make equal quarterly
deposits of \$1000 into a fund that pays
interest at a rate of 12% compounded
monthly. Find the balance at the end of
year three.
35
Compounding more frequent than payments
Discrete Case Example: Quarterly deposits
with monthly compounding
Year 1
0
1
2
3
Year 2
4
5
6
7
F=?
Year 3
8
9 10 11
12
Quarters
A = \$1,000
Step 1: M = 12 compounding periods/year
K = 4 payment periods/year
C = 3 interest periods per quarter
Step 2: i = [1 + 0 .12 /( 3)( 4 )] 3 − 1
= 3 .030 %
Step 3: N = 4(3) = 12
Step 4: F = \$1,000 (F/A, 3.030%, 12)
= \$14,216.24
36
Continuous Case: Quarterly deposits with
Continuous compounding
Suppose you make equal quarterly
deposits of \$1000 into a fund that pays
interest at a rate of 12% compounded
continuously. Find the balance at the
end of year three.
37
Continuous Case: Quarterly deposits with
Continuous compounding
Year 1
0
1
2
3
Year 2
4
5
6
7
F=?
Year 3
8
9 10 11
12
Quarters
A = \$1,000
Step 1:
K = 4 payment periods/year
Step 2:
i = e0.12/ 4 − 1
C = ∞ interest periods per quarter
= 3.045% per quarter
Step 3:
Step 4:
N = 4(3) = 12
F = \$1,000 (F/A, 3.045%, 12)
= \$14,228.37
38
Compounding less frequent than payments
Suppose you make \$500 monthly deposits to
an account that pays interest at a rate of
10%, compounded quarterly. Compute the
balance at the end of 10 years.
i = (1 + 0.10 / 4)1/ 3 − 1 = 0.826% per month
F = 500( F | A,0.826%,120) = 101907.89
F = 1500( F | A,2.5%,40) = 101103.83
Whenever a deposit is made, it
starts to earn interest.
Money deposited during a
quarter does not earn interest
39
Credit Card Debt
Annual fees
Annual
percentage
rate
Grace period
Minimum
payment
Finance
charge
40
Methods of Calculating Interests on your
Credit Card
Method
Description
Interest You Owe
The bank subtracts the amount of
balance and charges you interest on
the remainder. This method costs
you the least.
With the \$1,000 payment, your
new balance will be \$2,000. You
pay 1.5% on this new balance,
which will be \$30.
Average Daily Balance
The bank charges you interest on the
average of the amount you owe each
day during the period. So the larger
the payment you make, the lower the
interest you pay.
With your \$1,000 payment at the
15th day, your balance will be
reduced to \$2,000. Therefore,
(1.5%)(\$3,000+\$2,000)/2=\$37.50.
Previous Balance
The bank does not subtract any
payments you make from your
previous balance. You pay interest on
the total amount you owe at the
beginning of the period. This method
costs you the most.
the bank will charge 1.5% on your
beginning balance \$3,000:
(1.5%)(\$3,000)=\$45.
41
Commercial Loans
Amortized Loans
• Effective interest rate specified
• Paid off in installments over time (equal periodic amounts)
• What is the cost of borrowing? NOT necessarily the loan
with lowest payments or lowest interest rate. Have to look
at the total cost of borrowing (interest rate and fees, length
of time it takes you to repay (term))
42
Auto Loan
\$20,000
1
2
24 25
48
0
Given: APR = 8.5%, N = 48 months, and
P = \$20,000
Find: A
A = \$20,000(A/P,8.5%/12,48)
= \$492.97
43
Suppose you want to pay off the remaining loan in
lump sum right after making the 25th payment.
How much would this lump be?
\$20,000
1
2
48
24 25
0
\$492.97
25 payments that were
\$492.97
23 payments that are
still outstanding
P = \$492.97 (P/A, 0.7083%, 23)
= \$10,428.96
44
For the 33rd payment, what is the interest
payment and principal payment?
492.97( P | A, 8.5% / 12,16) = 7431.12
7431.12(0.0071) = 52.76
492.97 − 52.76 = 440.21
Remaining
balance after
32 payments
Interest
component of the
33rd payment
Principal payment
component of the
33rd payment
45
Option 1
Option 2
Debt Financing Lease Financing
Price
\$14,695
\$14,695
Down payment
\$2,000
0
APR (%)
Monthly payment
Length
3.6%
\$372.55
\$236.45
36 months
36 months
Fees
\$495
Cash due at lease end
\$300
Purchase option at lease
end
Cash due at signing
\$8.673.10
\$2,000
\$731.45
46
6% compounded monthly
47
Which Option is Better?
Debt Financing:
Pdebt = \$2,000 + \$372.55(P/A, 0.5%, 36)
- \$8,673.10(P/F, 0.5%, 36)
= \$6,998.47
Lease Financing:
Please = \$495 + \$236.45 + \$236.45(P/A, 0.5%, 35)
+ \$300(P/F, 0.5%, 36)
= \$8,556.90
48
Example
Suppose you borrowed \$10000 at an
interest rate of 12% compounded
monthly over 36 months. At the end of
the first year (after 12 payments), you
want to negotiate with the bank to pay
off the remainder of the loan in 8 equal
quarterly payments. What is the
amount of this quarterly payment, if the
interest rate and compounding
frequency remain the same?
49
Example Solution
A = 10000( A | P,1%,36) = 332.14
Remaining debt (end of 1st) = 332.14( P | A,1%,24) = 7055.77
Effective rate per quarter = (1 + 0.01)3 − 1 = 3.03%
A = 7055.77( A | P,3.03%,8) = 1006.41 per quarter
50
Example
An individual wants to make equal monthly
deposits into an account for 15 years in order
to then make equal monthly withdrawals of
\$1500 for the next 20 years, reducing the
balance to zero. How much should be
deposited each month for the first 15 years if
the interest rate is 8% compounding weekly?
What is the total interest earned during this
35 year process?
51
Example
A family has a \$30000, 20 year mortgage at
15% compounded monthly.
Find the monthly payment and the total interest
paid.
Suppose the family decides to add an extra \$100
to its mortgage payment each month starting from
the first payment of the 6th year. How long will it
take the family to pay off the mortgage? How
much interest will the family save?
52
Example
A man borrows a loan of \$10000 from a bank. Aaccording to the
agreement between the bank and the man, the man will pay
nothing during the first year and will pay equal amounts of A
every month for the next 4 years. If the interest rate is 12%
compounded weekly, find
the payment amount A
The interest payment and principal payment for the 20th
payment.
Suppose you want to pay off the remaining loan in lump sum
right after making the 25th payment. How much would this
lump be?
53
Example
A man’s current salary is \$60000 per year and he is planning to
retire 25 years from now. He anticipates that his annual salary
will increase by \$3000 each year and he plans to deposit 5% of
his yearly salary into a retirement fund that earns 7% interest
compounded daily. What will be the amount accumulated at the
time of his retirement.
54
Example
A series of equal quarterly payments of \$1000 extends over a
period of 5 years. What is the present worth of this quarterlypayment series at 9.75% interest compounded continuously?
55
Example
A lender requires that monthly mortgage payments be no more
than 25% of gross monthly income, with a maximum term of 30
years. If you can make only a 15% down payment, what is the
minimum monthly income needed in order to purchase a
\$200,000 house when the interest rate is 9% compounded
monthly?
56
Example
Alice wanted to purchase a new car for \$18,400. A dealer offered her
financing through a local bank at an interest rate of 13.5%
compounded monthly. The dealer’s financing required a 10% down
payment and 48 equal monthly payments. Because the interest rate is
rather high, Alice checked with her credit union for other possible
financing options. The loan officer at the credit union quoted her
10.5% interest for a new-car loan and 12.25% for a used-car loan. But
to be eligible for the loan, Alice had to have been a member of the
credit union for at least six months. Since she joined the credit union
two months ago, she has to wait four more months to apply for the
loan. Alice decides to go ahead with the dealer’s financing and 4
months later refinances the balance through the credit union at an
interest rate of 12.25%(because the car is no longer new)
a)Compute the monthly payment to the dealer
b)Compute the monthly payment to the credit union
c)What is the total interest payment for each loan transaction
57
Example
A loan of \$10000 is to be financed over a period of 24 months.
The agency quotes a nominal interest rate of 8%for the first 12
months and a nominal interest rate of 9% for any remaining
unpaid balance after 12 months, with both rates compounded
monthly. Based on these rates, what equal end-of-month
payments for 24 months would be required in order to repay
the loan?
58
Example
Suppose you are in the market for a new car worth \$18000. You are
offered a deal to make a \$1800 down payment now and to pay the
balance in equal end-of-month payments of \$421.85 over a 48
month period. Consider the following situations:
a)
b)
Instead of going through the dealer’s financing, you want to make a
down payment of \$1800 and take out an auto loan from a bank at
11.75% compounded monthly. What would be your monthly
payments to pay off the loan in 4 years?
If you were to accept the dealer’s offer, what would be the effective
rate of interest per month charged by the dealer on your financing?
59
Example
A man borrowed money from a bank to finance a small
fishing boat. The bank’s loan terms allowed him to defer
payments for six months and then to make 36 equal end-ofmonth payments thereafter. The original bank note was for
\$4800 with an interest rate of 12% compounded weekly.
After 16 monthly payments, David found himself in a financial
bind and went to a loan company for assistance in lowering
his monthy payments. Fortunately, the loan company offered
to pay his debts in one lump sum, provided that he pays the
company \$104 per month for the next 36 months. What
monthly rate of interest is the loan company charging on this
transaction?
60
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Bloomberg Anywhere Remote Login Bloomberg Terminal Demo Request
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# SAT Tip: Solving Quadratic Equations
Photograph by Sean De Burca/Corbis
This tip on improving your SAT score was provided by Vivian Kerr at Veritas Prep.
To factor means to break down a number into two smaller parts. For example, we know 12 = 3 x 4, so we can say that 3 and 4 are factors of 12, because they divide evenly into 12. We’d call 12 a multiple of both 3 and 4.
A polynomial is a math expression with multiple terms, such as z + 2y. It can consist of numbers and variables. To factor a polynomial, we just divide out what is common to all terms. We couldn’t factor the polynomial z + 2y because the two terms have nothing in common. But we could factor 6x + 7xy since both terms contain an“x.” If we divided the “x” out from both terms, we’d be left with 6 + 7y. The factors would look like this: (x)(6 + 7y).
If you need to check your work and are not sure if you factored correctly, you can always multiply the factors to see if you arrive at the original polynomial.
A quadratic equation is a special polynomial with three terms in the form ax² + bx + c. Why do quadratics look like this? Because of FOIL!
FOIL stands for First-Outer-Inner-Last. When we multiply the first two terms of the factors, the outer two terms, the inner two terms, and the last two terms, then sum them together, we’ll always arrive at the form ax² + bx + c! Let’s try it!
Say (x + 1)(x – 5) are the factors. The first two terms are (x)(x) = x2. The outer two terms are (x)(-5) = -5x. The inner two terms are (1)(x) = x. The last two terms are (1)(-5) = -5. Summed together we get: x2 + x -5x – 5 = x2 -4x -5. We can use this knowledge to help us factor.
1. Factor x² – 9x + 18.
Let’s start by setting up two parentheses: ( )( ).
The first term in each parentheses will always be x, since x*x = x². Then we look at the coefficient of the second term, -9. The second terms in the factors must add together to equal the middle term’s coefficient.
This means we need two numbers that sum to -9. There are quite a few combinations that would work: -1 and -8, -2 and -7, etc. But we know the third term in a quadratic will equal the product of the second terms in its factors, so the only two numbers that work are -6 and -3. Therefore the factors must be: (x – 3) (x – 6).
To find the roots or solutions of a quadratic equation, set the two factors each equal to zero and solve.
(x – 3) = 0 (x – 6) = 0
+3 +3 +6 +6
x = 3 x = 6
The “solutions” to this quadratic are 3 and 6. Remember on SAT test day that the “solution” to a question asking for the factors will be the factors, but the “solutions” to a quadratic are different.
Vivian Kerr has been teaching and tutoring in the Los Angeles area since 2005. She graduated from the University of Southern California, studied abroad in London, and has worked for several test-prep giants tutoring, writing content, and blogging about all things SAT, ACT, GRE, and GMAT.
For more SAT advice from Veritas Prep watch “SAT Tip: What Should You Do When You’re Stuck On a Question?”
LIMITED-TIME OFFER SUBSCRIBE NOW
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The monthly incomes of Aryan and Babban are in the ratio $3:4$ and their monthly expenditures are in the ratio $5:7$. If each saves ${\text{Rs}}.15000$per month, find their monthly incomes using matrix method. This problem reflects which value?
Last updated date: 21st Jul 2024
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Hint: Convert the rations into a system of equations of linear equations and then solve using Gaussian Elimination method.
Let the incomes of Aryan and Babban be $3x$ and $4x$ respectively.
In the same way, their expenditures would be $5y$ and $7y$ respectively.
Since Aryan and Babban saves ${\text{Rs}}.15000$ , we get
$3x - 5y = 15000.........................................\left( 1 \right) \\ 4x - 7y = 15000........................................\left( 2 \right) \\$
These are the systems of linear equations in the form $AX = B$, which can be solved by using Gaussian Elimination method.
So, $A = \left[ {\begin{array}{*{20}{c}} 3&{ - 5} \\ 4&{ - 7} \end{array}} \right],B = \left[ {\begin{array}{*{20}{c}} {15000} \\ {15000} \end{array}} \right]{\text{ and }}X = \left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right]$
This can be written in matrix form as
$\left[ {\begin{array}{*{20}{c}} 3&{ - 5} \\ 4&{ - 7} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {15000} \\ {15000} \end{array}} \right]$
Applying the row operation ${R_2} \to {R_2} - \dfrac{4}{3}{R_1}$
We get $\left[ {\begin{array}{*{20}{c}} 3&{ - 5} \\ 0&{ - \dfrac{1}{3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {15000} \\ { - 5000} \end{array}} \right]$
Applying the row operation ${R_2} \to {R_2} \times - 3$
We get $\left[ {\begin{array}{*{20}{c}} 3&{ - 5} \\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {15000} \\ {15000} \end{array}} \right]$
Applying the row operation ${R_1} \to {R_1} + 5{R_2}$
We get $\left[ {\begin{array}{*{20}{c}} 3&0 \\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {90000} \\ {15000} \end{array}} \right]$
Applying the row operation ${R_1} \to {R_1} \times \dfrac{1}{3}$
We get $\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {30000} \\ {15000} \end{array}} \right]$
Converting back into system of equations we have
$x = 30000 \\ y = 15000 \\$
Thus, the income of Aryan is $3x = 3 \times 30000 = 90000$ and the income of Babban is $4x = 4 \times 30000 = 120000$.
Hence incomes of Aryan and Babban are ${\text{Rs}}.90,000{\text{ and Rs}}.1,20,000$ respectively.
The value of the problem is Aryan is more interested in saving money than Babban. So, one must save money, no matter how much one earns.
Note: In this problem the matrix method is used to solve the system of $n$ linear equations in $n$ unknowns. Some types of matrix methods to solve the system of linear equations are Gaussian Elimination method, Inverse matrix method and Cramer's rule. Here we have used the Gaussian Elimination method.
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stest1-2005
# stest1-2005 - t u(2-t(a Determine whether x t is a power...
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Name BME 171, Spring 2005 - TEST #1 1. Consider an LTI system with delay, whose output to an input x ( t ) = 6 u ( t ) is y ( t ) = 3(1 - e - 5( t - 2) ) u ( t - 2) . (a) What is the impulse response of this system? Compute and make a sketch. (b) Without using convolution, compute the response of this system to the input x 1 ( t ) = - 12 u ( t - 8). Sketch both the input and the response. Do not forget to label the axes and to indicate all pertinent features. 2. The filter is given by the block diagram shown below (left): n z -1 z -1 z -1 + + x[n] y[n] y[n] (a) Find and sketch impulse response h [ n ] of this filter. Determine the type and the order of this filter. (b) Consider an input x [ n ] = 2 u [ n + 2] u [2 - n ]. The response y [ n ] of the filter to this input is shown above (right) but the axes are incomplete. Show how you can determine the position of the vertical axis, the support of the output, and the scale of the vertical axis. Then complete the plot of y [ n ].
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3. An LTI system has an impulse response: h ( t ) = 3 e - t u ( t ) . The input to this system is
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Unformatted text preview: ( t ) u (2-t ). (a) Determine whether x ( t ) is a power signals, an energy signal, or neither. (b) Determine whether the system is BIBO stable. What can you conclude regard-ing the response of this system to x ( t )? (c) The response to input x ( t ) can be computed by evaluating the convolution integral: y ( t ) = Z ∞-∞ x ( λ ) h ( t-λ ) dλ. (1) Draw a sketch indicating the position of the two functions inside the convo-lution integral at time t = 0 . 5. Carefully label all pertinent features on the graph. 4. Consider a sinusoidal signal x ( t ):-8-6-4-2 2 4 6 8-4 4 t x(t) (a) Write this signal in the form of a rotating phasor, ¯ x ( t ) = X e jωt , where X is a complex number. (b) Assuming that x ( t ) was sampled with frequency f s = 2 Hz, determine a sinu-soidal signal x a ( t ) that will be represented by the same samples as x ( t ) due to aliasing. 2...
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# Simon Fraser University Spring 2014 Econ 302 D200 Midterm Exam Instructor: Songzi Du
```Simon Fraser University
Spring 2014
Econ 302 D200 Midterm Exam
Instructor: Songzi Du
Monday March 10, 2014, 8:30 – 10:20 AM
Write your name, SFU ID number, and tutorial section number both on the exam
booklets and on the questionnaire. Hand in both the exam booklets and this
questionnaire. But note that only the exam booklets are graded.
• Name:
• SFU ID number:
• Tutorial section number:
General instructions
1. This is a closed-book exam: no books, notes, computer, cellphone, internet, or other
aids. A scientific, non-graphing calculator is allowed.
2. If you use decimals in calculation, keep two decimal places.
3. Write clearly. Illegible answers will receive no credit.
4. Show your work! Partial credits are given. Answers without proper explanation/calculation
will be penalized.
5. A request for regrade can be accepted only if the exam is written with a pen.
Do not turn over this page until instruction to start is given.
1
1. (10 points) Consider the following game.
W
X
Y
Z
A
2, 1
0, 3
2, 0
2, 0
B
1, 2 3, 1
1, 1
2, 1
C
1, 4
1, 0
4, 3
3, 3
D
0, 1
2, 1
2, 3
4, 0
Part i: Find the strategies that survive iterative deletion of strictly dominated strategies
(ISD). For each strategy that you delete, write down the strategy that strictly dominates it.
Part ii: Find all Nash equilibria (pure and mixed) in this game, and for each Nash
equilibrium that you find, calculate the two players’ expected payoffs in the equilibrium.
2. (10 points) Find and describe the pure-strategy subgame perfect equilibria (SPE) in
the following game. Explain the steps that you use to find the SPE.
(This is a three-player game: the payoffs of (1, 3, 1) (given by actions a, d and f ) mean
that player 1 gets 1, player 2 gets 3, and player 3 gets 1.)
1
a
c
b
2
2
e
d
3
2
3
d0
e
d
3
3
e0
3
3
0
g
2
0
0
f
g
f
g
f
g
f
g
f
1
3
1
3
0
0
3
0
0
1
1
3
0
1
1
1
3
2
1
2
3
0
1
1
−1
1
1
0
f
00
g 00
0
1
0
2
0
1
3. (10 points) The market demand function for a monopolist’s product is P (q) = 20 − q.
The monopolist’s cost function is C(q) = 2q 2 , where C(q) is the cost to produce q units of
the product. Assume that the government imposes a sales tax of 50%, and the sales tax
does not impact the monopolist’s cost. Calculate the monopolist’s price and quantity, and
calculate the corresponding consumer’s price.
2
4. (10 points) Consider a market setting with three firms. Firm 2 and 3 are already
operating as monopolists in two different industries (they are not competitors). Firm 1 must
decide whether to enter firm 2’s industry and thus compete with firm 2, or enter firm 3’s
industry and thus compete with firm 3. Production in firm 2’s industry occurs at zero cost,
whereas the cost of production in firm 3’s industry is 2 per unit. Demand in firm 2’s industry
is given by p = 9 − Q, whereas demand in firm 3’s industry is given by p0 = 14 − Q0 , where
p and Q denote the price and total quantity in firm 2’s industry and p0 and Q0 denote the
price and total quantity in firm 3’s industry.
The game runs as follows: First, firm 1 chooses between E 2 and E 3 . (E 2 means “enter
firm 2’s industry” and E 3 means “enter firm 3’s industry.”) This choice is observed by firm
2 and 3. Then, if firm 1 chooses E 2 , firm 1 and 2 competes in quantity (simultaneously
setting their production quantities q1 and q2 ); in this case firm 3 is the monopolist in his
industry. On the other hand, if firm 1 chooses E 3 , then firm 1 and 3 competes in quantity
(simultaneously setting their production quantities q10 and q30 ); in this case firm 2 is the
monopolist in his industry.
Calculate and report the subgame perfect equilibrium (SPE) of this game. Note that the
SPE involves the strategies of all three firms.
5. (10 points) The following (stage) game is repeated twice:
L
M
R
U
8, 8
0, 9
0, 0
C
9, 0
0, 0
3, 1
D
0, 0
1, 3
3, 3
In each stage the two players move simultaneously, and in the second stage the players
observe the actions taken in the previous stage. Assume that the final payoff of a player is
the sum of his payoffs from the two stages.
Find and describe a subgame perfect equilibrium (SPE) in which (U, L) is played in the
first stage. Explain why it is a SPE.
6. (10 points) Player 1 and player 2 simultaneously make demands m1 and m2 , where
m1 and m2 are real numbers between 0 and 1. If m1 + m2 ≤ 1 (compatible demands, given
that the surplus to be divided equals 1), then player 1 obtains the payoff m1 and player 2
3
obtains m2 . On the other hand, if m1 + m2 > 1 (incompatible demands), then both players
get 0. Find and describe all pure-strategy Nash equilibria of this game. Explain your answer.
4
```
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# §14.8 Behavior at Singularities
## §14.8(i) $x\to 1-$ or $x\to-1+$
As $x\to 1-$,
14.8.1 $\displaystyle\mathsf{P}^{\mu}_{\nu}\left(x\right)$ $\displaystyle\sim\frac{1}{\Gamma\left(1-\mu\right)}\left(\frac{2}{1-x}\right)^% {\mu/2},$ $\mu\neq 1,2,3,\dots$, 14.8.2 $\displaystyle\mathsf{P}^{m}_{\nu}\left(x\right)$ $\displaystyle\sim(-1)^{m}\frac{{\left(\nu-m+1\right)_{2m}}}{m!}\left(\frac{1-x% }{2}\right)^{m/2},$ $m=1,2,3,\dots$, $\nu\neq m-1,m-2,\dots,-m$, 14.8.3 $\displaystyle\mathsf{Q}_{\nu}\left(x\right)$ $\displaystyle=\frac{1}{2}\ln\left(\frac{2}{1-x}\right)-\gamma-\psi\left(\nu+1% \right)+O\left(1-x\right),$ $\nu\neq-1,-2,-3,\dots$,
where $\gamma$ is Euler’s constant (§5.2(ii)). In the next three relations $\Re\mu>0$.
14.8.4 $\mathsf{Q}^{\mu}_{\nu}\left(x\right)\sim\frac{1}{2}\cos\left(\mu\pi\right)% \Gamma\left(\mu\right)\left(\frac{2}{1-x}\right)^{\mu/2},$ $\mu\neq\tfrac{1}{2},\tfrac{3}{2},\tfrac{5}{2},\dots$,
14.8.5 $\mathsf{Q}^{\mu}_{\nu}\left(x\right)\sim(-1)^{\mu+(1/2)}\frac{\pi\Gamma\left(% \nu+\mu+1\right)}{2\Gamma\left(\mu+1\right)\Gamma\left(\nu-\mu+1\right)}\left(% \frac{1-x}{2}\right)^{\mu/2},$ $\mu=\tfrac{1}{2},\tfrac{3}{2},\tfrac{5}{2},\dots$, $\nu\pm\mu\neq-1,-2,-3,\dots$,
14.8.6 $\mathsf{Q}^{-\mu}_{\nu}\left(x\right)\sim\frac{\Gamma\left(\mu\right)\Gamma% \left(\nu-\mu+1\right)}{2\Gamma\left(\nu+\mu+1\right)}\left(\frac{2}{1-x}% \right)^{\mu/2},$ $\nu\pm\mu\neq-1,-2,-3,\dots$.
The behavior of $\mathsf{P}^{\mu}_{\nu}\left(x\right)$ and $\mathsf{Q}^{\mu}_{\nu}\left(x\right)$ as $x\to-1+$ follows from the above results and the connection formulas (14.9.8) and (14.9.10).
## §14.8(ii) $x\to 1+$
14.8.7 $\displaystyle P^{\mu}_{\nu}\left(x\right)$ $\displaystyle\sim\frac{1}{\Gamma\left(1-\mu\right)}\left(\frac{2}{x-1}\right)^% {\mu/2},$ $\mu\neq 1,2,3,\dots$, 14.8.8 $\displaystyle P^{m}_{\nu}\left(x\right)$ $\displaystyle\sim\frac{\Gamma\left(\nu+m+1\right)}{m!\Gamma\left(\nu-m+1\right% )}\left(\frac{x-1}{2}\right)^{m/2},$ $m=1,2,3,\dots$, $\nu\pm m\neq-1,-2,-3,\dots$, 14.8.9 $\displaystyle\boldsymbol{Q}_{\nu}\left(x\right)$ $\displaystyle=-\frac{\ln\left(x-1\right)}{2\Gamma\left(\nu+1\right)}+\frac{% \frac{1}{2}\ln 2-\gamma-\psi\left(\nu+1\right)}{\Gamma\left(\nu+1\right)}+O% \left(x-1\right),$ $\nu\neq-1,-2,-3,\dots$,
14.8.10 $\boldsymbol{Q}_{-n}\left(x\right)\to(-1)^{n+1}(n-1)!,$ $n=1,2,3,\dots$,
14.8.11 $\boldsymbol{Q}^{\mu}_{\nu}\left(x\right)\sim\frac{\Gamma\left(\mu\right)}{2% \Gamma\left(\nu+\mu+1\right)}\left(\frac{2}{x-1}\right)^{\mu/2},$ $\Re\mu>0$, $\nu+\mu\neq-1,-2,-3,\dots$.
## §14.8(iii) $x\to\infty$
14.8.12 $\displaystyle P^{\mu}_{\nu}\left(x\right)$ $\displaystyle\sim\frac{\Gamma\left(\nu+\frac{1}{2}\right)}{\pi^{1/2}\Gamma% \left(\nu-\mu+1\right)}(2x)^{\nu},$ $\Re\nu>-\tfrac{1}{2}$, $\mu-\nu\neq 1,2,3,\dots$, 14.8.13 $\displaystyle P^{\mu}_{\nu}\left(x\right)$ $\displaystyle\sim\frac{\Gamma\left(-\nu-\frac{1}{2}\right)}{\pi^{1/2}\Gamma% \left(-\mu-\nu\right)(2x)^{\nu+1}},$ $\Re\nu<-\tfrac{1}{2}$, $\nu+\mu\neq 0,1,2,\dots$, 14.8.14 $\displaystyle P^{\mu}_{-1/2}\left(x\right)$ $\displaystyle\sim\frac{1}{\Gamma\left(\frac{1}{2}-\mu\right)}\left(\frac{2}{% \pi x}\right)^{1/2}\ln x,$ $\mu\neq\tfrac{1}{2},\tfrac{3}{2},\tfrac{5}{2},\dots$,
14.8.15 $\boldsymbol{Q}^{\mu}_{\nu}\left(x\right)\sim\frac{\pi^{1/2}}{\Gamma\left(\nu+% \frac{3}{2}\right)(2x)^{\nu+1}},$ $\nu\neq-\tfrac{3}{2},-\tfrac{5}{2},-\tfrac{7}{2},\dots$,
14.8.16 ${\boldsymbol{Q}^{\mu}_{-n-(1/2)}\left(x\right)\sim\frac{\pi^{1/2}\Gamma\left(% \mu+n+\frac{1}{2}\right)}{n!\Gamma\left(\mu-n+\frac{1}{2}\right)(2x)^{n+(1/2)}% }},$ $n=1,2,3,\dots$, $\mu-n+\frac{1}{2}\neq 0,-1,-2,\dots$.
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Study Materials
# NCERT Solutions for Class 10th Mathematics
Page 2 of 5
## Chapter 13. Surface Areas and Volumes
### Exercise 13.2
Exercise 13.2
1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
3. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 13.15).
4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5
cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig. 13.16).
5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius
0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use π = 3.14)
7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Page 2 of 5
Chapter Contents:
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# Algorithms for two and three dimensional Knapsack
I know that the 2D and 3D Knapsack problems are NPC, but is there any way to solve them in reasonable time if the instances are not very complicated? Would dynamic programming work?
By 2D (3D) Knapsack I mean I have a square (cube) and a I have list of objects, all data are in centimeters and are at most 20m.
• What forms do your objects have? How big is the surrounding area; has it bounded size? Apr 24, 2012 at 5:52
• Are you searching for an exact solver or are heuristics sufficient? Apr 24, 2012 at 19:01
• Could you be more specific? What are "sizes", and what is $m$? What precisely is your input, what precisely are your constraints, and what precisely are you trying to optimize? Apr 25, 2012 at 11:46
• Also, what have you already tried? Apr 25, 2012 at 11:47
• The problem you're talking about isn't generally referred to as a knapsack problem; it usually goes by the name bin-packing problem, and you should be able to find a lot more information about it under that name. May 14, 2012 at 20:36
Knapsack can be solved by dynamic programming in pseudo-polynomial time $O(nW)$ with $n$ the number of objects and $W$ the size of the knapsack. So, as long as your container is small (numerically), you can solve the problem efficiently. Note that you can adjust $W$ by changing resolution; no need to measure a shipping container to the µm, but meters are probably to coarse (depending on your objects).
Knapsack can also be approximated arbitrarily well in polynomial time (see polynomial-time approximation schemes).
However, Knapsack only considers fitting numbers into another number; it does not care about geometrics. If you need to "puzzle", you need another problem; considering Tetris, this is probably much harder than Knapsack.
GREEDY will always find a reasonable solution, but not necessarily the optimal one. Simply put the largest object which will fit each time in the knapsack. Stop when no more objects will fit.
• No, that is not true. Note that in the Knapsack problem, objects have values, too. Filling greedily by size can yield an arbitrarily bad solution. Jul 31, 2012 at 9:07
• @Raphael: Well, not arbitrarily bad, but I wouldn't consider the greedy solution a reasonable solution. The greedy approach gets worse for the higher dimensional analogues. Aug 2, 2012 at 9:08
• @A.Schulz Actually yes arbitrarily bad! The greedy heuristic for knapsack, using either size or bang-for-buck can easily be shown to not have any finite approximation guarantee to OPT. Oct 29, 2012 at 18:56
• People... please stop saying "Well, I don't know about that! But..." before doing your %#\$( homework on it! Dec 18, 2016 at 18:46
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# Excel’s Number Calculation: Understanding the Process
Excel’s Number Calculation: Understanding the Process – Hello Readers of RikudesignCom! We all know that Microsoft Excel is one of the best tools to manage and analyze data. But have you ever wondered how it calculates numbers? Well, in this post, we are going to explore how Excel performs its calculations, so buckle up and let’s dive into it!
Excel uses formulas and functions to do its calculations. Formulas are expressions that tell Excel which calculation to perform on specific numbers. Functions are predefined formulas that perform specific calculations in a particular order. Some examples of functions include SUM, AVERAGE, MAX, MIN, and COUNT. Once you input the formula or function in the appropriate cell, Excel will calculate the answer using the BODMAS rule (Brackets, Order, Division, Multiplication, Addition, Subtraction).
The target of how Excel calculates numbers is anyone who uses Excel to analyze their data. Whether you’re a business owner trying to keep track of your finances or an analyst working on complex projects, understanding how Excel does its calculations can help you make informed decisions when analyzing data.
In conclusion, we have learned that Excel performs calculations using formulas and functions, and applies BODMAS rule to solve them. Understanding how Excel calculates numbers is essential for anyone who uses Excel to analyze data. You can learn more about this topic by following the link below. Happy Excel-ing!
## Factors Influencing How Does Excel Calculate Numbers
Welcome to this article, where we will explore the different factors that influence how Excel calculates numbers. Excel is an incredibly powerful spreadsheet program used by millions of people worldwide. The features and functions, quality and reputation of the software, competition, development difficulty, development costs, target market, and platform are all factors that impact how Excel calculates numbers. We will explore each of these factors below.
### Features and Functions
Excel contains a vast array of features and functions that impact how it calculates numbers. For example, Excel allows users to perform basic arithmetic operations such as addition, subtraction, multiplication, and division, but it can also perform more complex calculations like statistical analysis, financial modeling, and data visualization. These features and functions determine what types of calculations Excel can perform, making it an ideal tool for specific needs.
### Quality and Reputation
Excel has an excellent reputation for producing accurate calculations, which is why it is the go-to tool for many individuals and businesses looking for reliable calculations. However, quality can fluctuate as Microsoft releases new versions or updates to Excel, making it important to stay up to date on the latest versions and patches to ensure accurate calculations. Reputation also plays a significant role in how people perceive Excel and whether they trust its calculations.
More: Creative Ways Writers Use Excel: Tips & Tricks
### Level of Competition
Excel’s market dominance also influences how it calculates numbers. As one of the most popular spreadsheet programs, Excel faces stiff competition from other programs offering similar features and functions. However, Excel’s widespread use gives it a competitive edge and provides a larger user base, allowing developers to access more data and improve algorithms used in calculations. Below is a table showing Excel’s market share.
Spreadsheet Program Market Share (%)
Excel 80
LibreOffice Calc 5
Others 8
### Development Difficulty
The complexity of developing a powerful and reliable spreadsheet program is substantial, affecting how Excel calculates numbers. Development difficulty determines how much time, resources, and expertise are required in creating Excel’s wide range of features and functions. As such, developers must understand the math and statistics behind various calculations before they can code them into the software.
### Development Costs
Development costs are the financial implications of creating and maintaining Excel over its lifespan. These long-term expenses affect how frequently Microsoft upgrades and improves Excel, and ultimately influences how the program calculates numbers. The development cost of a software application like Excel typically consists of salaries for developers, infrastructure costs, marketing expenses, and general overhead.
### Target Market
The choice of the target market can impact how users interact with and, consequently, how Excel calculates numbers. For instance, businesses that need advanced financial analysis and modeling capability are more likely to use Excel than individual consumers who may only require basic arithmetic calculation functions. By targeting a specific market, Excel can align its features and functions with an intended use case, thus influencing how it calculates numbers.
### Platform
Lastly, the platform on which Excel operates also impacts how it calculates numbers. Excel runs on various operating systems, including Windows, Mac, Android, and iOS. However, different platforms may have varying levels of compatibility or specific bugs, causing issues with calculation accuracy. For example, functions that work perfectly on Windows may not work as expected on a Mac or mobile device. Below is a table showcasing estimates of Excel usage based on the platform.
Operating System Estimated Usage (%)
Windows 60
MacOS 20
Android/iOS 20
As we have seen from the above discussion, multiple factors impact how Excel calculates numbers. From the features and functions within Excel to the development costs and target market, each factor has a unique influence on overall calculation accuracy.
## How Does Excel Calculate Numbers?
### Determination Strategy
Excel is a spreadsheet program that enables users to perform calculations and analysis of data in a table format. However, have you ever wondered how Excel calculates these numbers? Excel uses mathematical formulas to calculate the result of any data input into its cells. Once a mathematical formula is entered into a cell, Excel executes the formula above/below all other entries within that row/column, and the results are displayed in the active cell as an output value.
More: Boost Excel size: Tips to increase
Excel offers a vast library of functions that enable users to perform complicated calculations, including arithmetic operations such as addition, subtraction, multiplication, division, percentages, ratios, and much more. Additionally, users can create custom Excel functions to handle specific requirements. In summary, Excel calculates by using the formulas sequentially and, after completing the calculations, returns the result to the cell where the formula was entered.
## Changes and Reasons
### How Does Excel Calculate Numbers changes and reasons
Updates to Excel’s calculation engine primarily reflect critical data processing and auditing needs. Microsoft updates Excel with every new version or service pack released to improve its computational capabilities, fix bugs, and improve overall functionality. The latest versions of Excel have made significant adjustments to its calculation engine, making it faster and more accurate, improving user productivity significantly.
Another change to Excel’s calculation strategy includes Conditional Formula Evaluation, which enhances the software ability to process dependent and massive multi-stage logic expressions. Therefore, the new evaluation protocol only calculates formulas that have been recently identified. Consequently, conditional formula evaluation reduces Excel’s calculation time while preserving its performance.
In conclusion, every update to Excel brings about improvement in its calculation engine’s accuracy, speed, functionality, and reliability. This guarantees users a seamless and fast experience when performing calculations and data analysis in Excel.
## How Does Excel Calculate Numbers Determination Errors
### Round-off errors
When working with large sets of data, Excel may encounter round-off errors. These errors occur when Excel rounds a number to a certain number of decimal places, which can cause inaccuracies in calculations. Round-off errors occur because Excel only stores 15 significant digits for each number.
### Data entry mistakes
Another common error is data entry mistakes. Users may input incorrect data or formulas that do not accurately reflect the intended calculation. This can lead to incorrect results and further errors in subsequent calculations.
## How Does Excel Calculate Numbers Determination Solutions
### Using the ROUND function
To address round-off errors, users can utilize Excel’s built-in ROUND function. This function allows users to specify the number of decimal places to round to, reducing the likelihood of rounding errors. For example, =ROUND(A1,2) would round the value in cell A1 to two decimal places.
### Data validation
To minimize data entry mistakes, Excel offers several tools for data validation. Users can set rules to ensure that data is entered correctly and within a specified range. Additionally, Excel can flag potential errors, such as misspelled words or inconsistent data, to help users identify mistakes before they become a problem.
### Using absolute references
Another solution for accurate calculations is using absolute references in formulas. Absolute references lock a specific cell reference in a formula, preventing it from changing when the formula is copied or moved. This ensures that the correct cells are used in each calculation, reducing the risk of errors.
More: Stop Excel from Reducing Fractions
Function Description
SUM Adds a range of cells
AVERAGE Calculates the average of a range of cells
MAX Returns the highest value in a range of cells
MIN Returns the lowest value in a range of cells
In conclusion, Excel is a powerful tool for data analysis and calculations. However, accuracy can be compromised by round-off errors and data entry mistakes. By utilizing functions such as ROUND, data validation, and absolute references, users can ensure accurate calculations and avoid costly errors.
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# Cool Facts
## Famous Mathematician of the Day
23th November 2012 marks 408 years since the death of Francesco Barozzi, an Italian mathematician, astronomer and humanist. Barozzi was born on the island of Crete, at Candia (now Heraklion), at the time a Venetian possession, the son of Iacopo Barozzi, a Venetian nobleman. He was educated at Padua, and studied mathematics at the University …
## Why I never wanted to be an archaeologist
As you can see in the picture below, you can use all 3 shards from each column and each row to create a complete plate. That explains why no one bothered to dig them up for so many years! Credit: Lee Sallows
## Why does the shower curtain move towards the water?
The worst part about taking a shower is getting attacked by the wet shower curtain. Luckily, David Schmidt discovered a logic explanation for why this happens so often and you are about to hear it. To do the calculation, I drafted a model of a typical shower and divided the shower area into 50,000 minuscule …
## The fourth dimension: Tesseract 101
The Tesseract is the four dimensional analogue of a cube and it is also known as a hypercube. Because imagining a four dimensional object is impossible, you will have to try to take it all in from this picture. An even better example is a moving image: You can read everything you need to know …
## Tupper’s self-referential formula
Tupper’s self-referential formula is a self-referential formula defined by Jeff Tupper that, when graphed in two dimensions, can visually reproduce the formula itself. It is used in various math and computer science courses as an exercise in graphing formulae. The formula is an inequality defined by: where [.] denotes the floor function and mod is …
## Easy way to memorize the value of e
That’s a 20 dollar bill, so memorize “2″ and put down the decimal point. The picture on the bill is of Andrew Jackson. He was our seventh President, so put a “7″ after the decimal point to get 2.7. Jackson was elected in 1828, so put down “1828″ next. Since there’s a 2 in front …
## Mathematics art in The University of Manchester
While walking randomly, the guy behind http://www.walkingrandomly.com/?p=4203 discovered something interesting on the windows of an old building from The University of Manchester. This mathematics themed stained glass window is part of The Sackville Street Building and it seems that other mathematical windows can be found in buildings from around the world, including Caius College, Cambridge.
## How high can you count? – The powers of 10
Did you ever think about how far you can count? Here are some of the names of the powers of 10. English names of the first 10000 powers of 10 10 to a power (number of 0’s after the 1) Latin power (number of 000’s after the 1) prefix cardinal (determines letters before the illion) …
## Complex Roots Made Visible — Math Fun Facts
This is a neat way you can visualize complex roots of quadratics. What you do is take a quadratic like: 2x^2 – 8x + 10 If you tried to factor or find real roots of this quadratic, you will run into problems. In fact, this polynomial has two complex roots (but no real roots). According …
## Tautochrone curves
This is pretty neat! A tautochrone curve is one where the time taken by an object sliding down (without friction in uniform gravity) to the bottom is INDEPENDENT of the starting position. As seen above, all four points end up at the bottom at the same time! More info on wikipedia: http://en.wikipedia.org/wiki/Tautochrone_curve
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# 10+ Amazing Mass Vs Weight Worksheet Answers
Pages 1-2 features a reading exercise on gravity mass and weight including the following topics. A Your mass in kilograms.
Mass And Weight Worksheet A Science Measurement Resource Science Worksheets Mass
### Mass percent is also known as percent by weight or ww.
Mass vs weight worksheet answers. No conversions of units between the two systems are needed. To play this quiz please finish editing it. A The weight of the matter decreases.
Mass and Weight Mass is an important property of matter. If the mass of an object is 1000kg on Earth what. Students will reinforce the concept by answering questions about the differences between mass and weight and the final question will be for them to distinguish the difference in their own words between mass and weight.
Distinguish between mass and weight. Our quiz and worksheet contains a series of helpful questions on how to measure mass and weight with Newtons Laws. How much would a 10 kg suitcase weigh on the surface of.
Finding the mass of an object is easy to do with todays technology. Recall and use the equation W mg. A 60 kg person standing on the dwarf planet ceres would weigh 16 2 n.
State that weight is a gravitational force. An object consists of the same amount of matter wherever it may be in the universe. Demonstrate an understanding that mass is a property that resists change in motion.
Convert metric units from one to another. Ie an apply is likely to weigh 150 g or 1 Kg. On Earth Mars or even space.
He might want to stretch himself once an employee knows his efforts don t go. 1 The amount of matter in something determines its _____. A worksheet for students to compare the weight of objects on each of the planets using the gravitational pull of each planet.
Mass and Weight Use the following formula to solve for weight. Weight worksheet Concept Development Practice Page 3-1 from the Conceptual Physics textbook. – What is gravity -.
Try it risk-free for 30 days. A worksheet designed to highlight the differences between weight and mass the equation that links them together and an acitivity whereby pupils can work out how much they would weigh on various objects in. D The gas in the balloon will have no net weight since helium defies gravity and vertically rises.
About This Quiz Worksheet. Mass is a measure of how much matter an object contains while weight is a measure of the force of gravity on the The units of mass are typically grams. Mass and Weight – I.
Worksheet for Fourth Grade Math. 98 ms2 Weight is measured in Newtons 1 N 1 kg ms 2 Answer the following questions show ALL WORK and UNITS 1. B Use this mass to solve for your weight on these other planets.
3 Your weight will change depending on _____. Mass vs Weight Answers mass weight Is affected by the gravitational field strength of a planetary body and so can be different for the same object depending on which planet it is measured on. Mass weight gravity Mass is a measure of the amount of matter in an object.
Showing top 8 worksheets in the category mass or weight. The answer to this question lies within the difference between mass and weight. A 4-page worksheet that covers the nature of gravity and the difference between weight and massthrough a 2-page reading and 7 review questions.
Here is a worksheet to tie up introducing the differences between mass and weight. Mass vs Weight. Mass is not related to gravity.
Demonstrate understanding that weights and hence masses may be compared using a balance. PROBLEM PageIndex8 Determine the mass of each of the following. Weight W Mass m x gravity g W mg Mass is measured in kilograms kg Gravity on earth is a constant.
This resource was reviewed using the Curriki Review rubric and received an overall Curriki Review System rating of 170 as of 2016-07-27. E Weight will stay the same since only gases are filling it. The weight of an.
A 00146 mol KOH b 102 mol ethane C2H6 c 16 103 mol Na2SO4 d 6854 103 mol glucose C6H12O6 e 286 mol CoNH36Cl3 Answer a 0819 g Answer b 307 g Answer c 023 g Answer d 1235 106 g 1235 kg Answer e 765 g PROBLEM. 2 How strongly the planet youre on pulls on you is your _____. One worksheet is in customary units ounces lbs one worksheet uses metric units gm kg and one worksheet mixes the units.
B Use this mass to solve for your weight on these other planets. A Your mass in kilograms. Weight mass and gravity.
Not affected by the gravitational field strength and so remains unchanged whichever planet it. Define Weight 3. Planet Moon Your Mass Here kg Gravitational Acceleration Here ms2 Your Weight Here N Earth 98 Moon 16 Sun 274 Jupiter 259 Pluto 061 Mercury 373 Neptune 1128 Saturn 1119 Mimas 08 Saturns moon Mimas is on the left.
Mass is the measurement of the amount of matter in an object. Weight is a measure of the gravitational force between two objects. Mass and weight worksheet answers.
C The weight of the matter increases. Quiz Worksheet – Distinguishing Between Mass and Weight. Worksheet includes 54 MCQs with answers.
Anything that is made of matter has mass. Define Mass 2. Mass vs weight on different planets – worksheet.
Matthew clemens created date. Choose an answer and hit next. Measurement Customary and Metric Measurement Mass and Weight.
Guess the reasonable estimates of weights of various daily life objects. Free 27 joni7 home sweet habitat. Live worksheets English.
Below are 3 versions of our grade 5 word problems worksheet involving the measurement of mass or weight. Determine weight on other planets and answer questions about gravity. Planet Moon Your Mass Here kg Gravitational Acceleration Here ms2 Your Weight Here N Earth 98 Moon 16 Sun 274 Jupiter 259 Pluto 061 Mercury 373 Neptune 1128 Saturn 1119 Mimas 08 Saturns moon Mimas is on the left.
18 Questions Show answers. MASS WEIGHT PRACTICE Answers below Weight Mass x g g earth 981 Nkg 10 Nkg Strength of gravity g on the surface in Newtons per Kilogram Nkg Mercury Venus Moon Mars Jupiter Saturn Uranus Neptune Pluto 36 89 16 38 260 111 87 107 04 1. Mass is commonly measured in grams or kilograms.
You will receive your score and answers. The mass of an object does not change when it is moved from one place to another. Weight Practice Worksheet 15 minutes After students have learned the properties of mass and weight and the equation Weight mass x acceleration due to gravity they can use to convert between mass and weight I ask students to complete the Mass vs.
B The weight of the matter does not change. Displaying all worksheets related to mass vs weight.
Mass Weight And Gravity Worksheet Distance Learning Physical Science Middle School Science Lesson Plans Elementary Persuasive Writing Prompts
Mass And Weight Worksheet A Science Measurement Resource Worksheets Science Weight
Mass Vs Weight Worksheet Measurement Worksheets Science Worksheets Science Curriculum
Weight Grams And Kilograms Worksheets Measurement Worksheets Volume Worksheets Math Worksheet
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1. ## Average speed problem
A cyclist travels from Paris to the Louvre at an average speed of x km/h.
find the time taken in term of x ?
On his return journey from louvre to Paris, he decreases his average speed by 3 km/h.
Find the time taken in return journey in terms of x ?
if the difference between the times is one hour 20 minutes,
find the value of x ?
2. ## Re: Average speed problem
First of all, how far away is Paris from the Louvre?
3. ## Re: Average speed problem
I was under the impression that the Louver was in Paris!
4. ## Re: Average speed problem
Originally Posted by HallsofIvy
I was under the impression that the Louver was in Paris!
As was I, I just expected that the OP meant something like the distance between the Louvre to the centre of Paris...
5. ## Re: Average speed problem
let the distance = $\displaystyle d$
$\displaystyle x = \frac{d}{t_1} \implies t_1 = \frac{d}{x}$
$\displaystyle t_2 = \frac{d}{x-3}$
$\displaystyle t_1 - t_2 = \frac{d}{x} - \frac{d}{x-3} = \frac{4}{3}$
plug in $\displaystyle d$ (once you find out what it is) and solve for $\displaystyle x$
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# math
posted by on .
Estimate a 15% tip on a taxi ride that cost 23.25.
• math - ,
• math - ,
15/23.25= x/100 then you cross multiply?
• math - ,
No. Besides, your instructions tell you to estimate.
When I estimate I first figure 10%. Then I double the 10%, and tip some amount between.
10% of 23 = 2.30
2.30 * 2 = 4.60
Estimate a number between 2.30 and 4.60
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# How to do this problem fast
posted by .
Sara works as a data entry operator. her current wage is \$8.50 per hour. After her excellent performance rating, her new hourly rate is \$9.18. What is the percent of increase in her wages?
• How to do this problem fast -
8%.
Because -> percentage change = new-old/old x 100
so 0.68/8.50 x100
Hope this helps :)
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Septic Tank lab
Introduction.
Most houses outside of cities use septic tank for deposal of house hold waste. The design is controlled by state law. In New Mexico the state regulations are controlled by the Title 20: Environmental Protection Chapter 7: Wastewater and Water Supply Facilities Part 3: Liquid Waste Disposal.
Before starting lab go through the septic tank tutorial. Look at picture of leach line installation
Objectives:
Design a septic tank for the first house on Tellbrook Rd in Las Cruces NM. Use the terra server to locate and measure the size of the lot. Roughly 150 m by 150 m . Calculate the size of the lot in ac.
Look up the soil type and topography in the soil survey. Place the location of the septic tank downstream from the house. Design the size of the tank and the leach line.
Determine the size of the house from the aerial photo graph and that will determine the number of people living in it and the number of rooms. House size is roughly 4000 ft sq. and should support a family of 5.
Look at the topographic maps to determine the slope of the land. See the terra server. Capture the topographic map and draw in the house and location of septic tank parts. Use a graphic program to accomplish this.
Design criteria
The typical flow is 75 gallons per person per day and design for 2 people per bedroom. The leach field design equation:
A= (Q*T^0.5)/5 multiple by 1.5 for some states
A= ft^2
Q= gpd/person
T= soil percolation rate, minutes/inch
Minimum size leach field
A= Q/ LTAR
LTAR = long term acceptable rate gpd/FT^2
Soil type LTAR Design gpd/f^2 Medium sand 1.2 Sandy loam to loam 0.72 Loam 0.5 Silt loam 0.4 Clay loam 0.3 Silty clay loam 0.2
The outlet pipe needs to flow downhill at least 1/8 inch per foot downhill to the leach field where the septic tank effluent enters a manifold or distribution ("D") box.
Assume a 4000 ft sq house has six rooms. Max number of people 12. Max flow of wastewater= 900 gpd.
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https://phys.libretexts.org/Bookshelves/Mathematical_Physics_and_Pedagogy/Computational_Physics_(Chong)/07%3A_Finite-Difference_Equations/7.01%3A_Derivatives
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# 7.1: Derivatives
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Suppose we have discretized a function of one variable, obtaining a set of $$\psi_n \equiv \psi(x_n)$$ as described above. For simplicity, we assume that the discretization points are evenly-spaced and arranged in increasing order (this is the simplest and most common discretization scheme). The spacing between points is defined as
$h \equiv x_{n+1} - x_n.$
Let us discuss how the first and higher-order derivatives of $$\psi(x)$$ can be represented under discretization.
## 7.1.1 First Derivative
The most straightforward representation of the first derivative is the forward-difference formula:
$\psi'(x_n) \approx \frac{\psi_{n+1} - \psi_n}{h}$
This is inspired by the usual definition of the derivative of a function, and approaches the true derivative as $$h \rightarrow 0$$. However, it is not a very good approximation. To see why, let's analyze the error in the formula, which is defined as the absolute value of the difference between the formula and the exact value of the derivative:
$\mathcal{E} = \left|\psi'(x_n) - \frac{\psi_{n+1} - \psi_n}{h}\right|$
We can expand $$\psi_{n+1}$$ in a Taylor series around $$x_{n}$$:
$\psi_{n+1} = \psi_n + h\, \psi'(x_n) + \frac{h^2}{2}\psi''(x_n) + \frac{h^3}{6}\psi'''(x_n) + O(h^4)$
Plugging this into the error formula, we find that the error decreases linearly with the spacing:
$\mathcal{E} = \left| \frac{h}{2}\psi''(x_n) + O(h^2)\right| \sim O(h).$
There is a better alternative, called the mid-point formula. This approximates the first derivative by sampling the points to the left and right of the desired position:
$\psi'(x_n) \approx \frac{\psi_{n+1} - \psi_{n-1}}{2h}.$
To see why this is better, let us write down the Taylor series for $$\psi_{n\pm1}$$:
$\psi_{n+1} = \psi_n \,+\, h\, \psi'(x_n) \,+\, \frac{h^2}{2}\psi''(x_n) \,+\, \frac{h^3}{6}\psi'''(x_n) \,+\, \frac{h^4}{24}\psi''''(x_n) + O(h^5)$
$\psi_{n-1} = \psi_n \,-\, h\, \psi'(x_n) \,+\, \frac{h^2}{2}\psi''(x_n) \,-\, \frac{h^3}{6}\psi'''(x_n) \,+\, \frac{h^4}{24}\psi''''(x_n) \,+\, O(h^5)$
Note that the two series have the same terms involving even powers of $$h$$, whereas the terms involving odd powers of $$h$$ have opposite signs. Hence, if we subtract the second series from the first, the result is
$\psi_{n+1} - \psi_{n-1} = 2 h\, \psi'(x_n) + 2 \frac{h^3}{6}\psi'''(x_n) + O(h^5)$
Because the $$O(h^{2})$$ terms are equal in the two series, they cancel out under subtraction, and only the $$O(h^{3})$$ and higher terms survive. After re-arranging the above equation, we get
$\psi'(x_n) = \frac{\psi_{n+1} - \psi_{n-1}}{2 h} + O(h^2).$
Hence, the error of the mid-point formula scales as $$O(h^2)$$, which is a good improvement over the $$O(h)$$ error of the forward-difference formula. What's especially nice is that the mid-point formula requires the same number of arithmetic operations to calculate as the forward-difference formula, so this is a free lunch!
It is possible to come up with better approximation formulas for the first derivative by including terms involving $$\psi_{n\pm 2}$$ etc., with the goal of canceling the $$O(h^{3})$$ or higher terms in the Taylor series. For most practical purposes, however, the mid-point rule is sufficient.
## 7.1.2 Second Derivative
The discretization of the second derivative is easy to figure out too. We again write down the Taylor series for $$\psi_{n\pm1}$$:
$\psi_{n+1} = \psi_n \,+\, h\, \psi'(x_n) \,+\, \frac{h^2}{2}\psi''(x_n) \,+\, \frac{h^3}{6}\psi'''(x_n) \,+\, \frac{h^4}{24}\psi''''(x_n) \,+\, O(h^5)$
$\psi_{n-1} = \psi_n \,-\, h\, \psi'(x_n) \,+\, \frac{h^2}{2}\psi''(x_n) \,-\, \frac{h^3}{6}\psi'''(x_n) \,+\, \frac{h^4}{24}\psi''''(x_n) \,+\, O(h^5)$
When we add the two series together, the terms involving odd powers of $$h$$ cancel, and the result is
$\psi_{n+1} + \psi_{n-1} = 2\psi_n + h^2 \psi''(x_n) + \frac{h^4}{12}\psi''''(x_n) + O(h^5).$
A minor rearrangement of the equation then gives
$\psi''(x_n) \approx \frac{\psi_{n+1} - 2\psi_n + \psi_{n-1}}{h^2} + O(h^2).$
This is called the three-point rule for the second derivative, because it involves the value of the function at the three points $$x_{n+1}$$, $$x_{n}$$, and $$x_{n-1}$$.
This page titled 7.1: Derivatives is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Y. D. Chong via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# A “Geometric” probability problem
n points are chosen randomly from a circumference of a circle. Find the probability that all points are on the same half of the circle. My intuition was that the probability is $1/2^{n-1}$ since if the first point is placed somewhere on the circumference, for each point there is probability of $1/2$ to be placed on the half to the right of it and $1/2$ to the left..but i feel i might be counting some probabilitys twice. Any ideas?
• What you have done is assume a division of the circle into a right half and a left half before you distribute the points, and calculate the probability that all the points end up in the same half. Your answer is therefore too small, because you haven't taken into account that you're free to choose a division of the circle into halves after all the points have been placed. Specifically, for two points, the probability should be $1$. – Arthur Mar 2 '18 at 11:58
• You may distribute the points on a line segment of length $2\pi$. Then calculate the probability that distance between the largest and smallest value is less than $\pi$. – Michael Hoppe Mar 2 '18 at 12:43
• @MichaelHoppe That is not good enough. If the points concentrate at utmost left and utmost right then the value you mention is more than $\pi$ but still it is possible that a half-circle exists that contains them all. – drhab Mar 2 '18 at 12:55
• @drhab Yes, it’s not that easy. Then just find a tangent such that all projections of the points on the tangent lie on the same side of the point of contact (just kidding). – Michael Hoppe Mar 2 '18 at 13:21
If a half-circle exists containing all $n$ points then also a half-circle exists that starts from one of the points, goes in counter-clockwise direction and contains all points.
If the points are numbered with $1,\dots,n$ and $E_i$ denotes the event that the half-circle as described above and starting at point $i$ contains all points then $E_1,\dots E_n$ are mutually disjoint events.
By symmetry we have $P(E_1)=\cdots=P(E_n)$.
Actually it remains now to find $P(E_1)$.
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# Search by Topic
Filter by: Content type:
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Challenge level:
### There are 32 results
Broad Topics > Transformations and their Properties > Symmetry
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##### Stage: 2 Challenge Level:
Use the information on these cards to draw the shape that is being described.
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This problem explores the shapes and symmetries in some national flags.
### Poly Plug Pattern
##### Stage: 1 Challenge Level:
Create a pattern on the left-hand grid. How could you extend your pattern on the right-hand grid?
### Exploded Squares
##### Stage: 1 Challenge Level:
This practical activity challenges you to create symmetrical designs by cutting a square into strips.
### Coordinate Challenge
##### Stage: 2 Challenge Level:
Use the clues about the symmetrical properties of these letters to place them on the grid.
### Stringy Quads
##### Stage: 2 Challenge Level:
This practical problem challenges you to make quadrilaterals with a loop of string. You'll need some friends to help!
### Symmetry Challenge
##### Stage: 2 Challenge Level:
Systematically explore the range of symmetric designs that can be created by shading parts of the motif below. Use normal square lattice paper to record your results.
### A Cartesian Puzzle
##### Stage: 2 Challenge Level:
Find the missing coordinates which will form these eight quadrilaterals. These coordinates themselves will then form a shape with rotational and line symmetry.
### Colouring Triangles
##### Stage: 1 Challenge Level:
Explore ways of colouring this set of triangles. Can you make symmetrical patterns?
### Always, Sometimes or Never? Shape
##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
### Sorting Letters
##### Stage: 1 and 2 Challenge Level:
This interactivity allows you to sort letters of the alphabet into two groups according to different properties.
### Reflector ! Rotcelfer
##### Stage: 2 Challenge Level:
Can you place the blocks so that you see the relection in the picture?
### Topkapi Palace
##### Stage: 2 and 3 Challenge Level:
These images are taken from the Topkapi Palace in Istanbul, Turkey. Can you work out the basic unit that makes up each pattern? Can you continue the pattern? Can you see any similarities and. . . .
### Tiles in a Public Building
##### Stage: 2 Challenge Level:
What is the same and what is different about these tiling patterns and how do they contribute to the floor as a whole?
### Animated Triangles
##### Stage: 1 Challenge Level:
Watch this "Notes on a Triangle" film. Can you recreate parts of the film using cut-out triangles?
### Triangle Shapes
##### Stage: 1 and 2 Challenge Level:
This practical problem challenges you to create shapes and patterns with two different types of triangle. You could even try overlapping them.
### Dancing with Maths
##### Stage: 2, 3 and 4
An article for students and teachers on symmetry and square dancing. What do the symmetries of the square have to do with a dos-e-dos or a swing? Find out more?
### Beat the Drum Beat!
##### Stage: 2 Challenge Level:
Use the interactivity to create some steady rhythms. How could you create a rhythm which sounds the same forwards as it does backwards?
### Making Maths: Five-point Snowflake
##### Stage: 2 Challenge Level:
Follow these instructions to make a five-pointed snowflake from a square of paper.
### Making Maths: Indian Window Screen
##### Stage: 2 Challenge Level:
Can you recreate this Indian screen pattern? Can you make up similar patterns of your own?
### Making Maths: Snowflakes
##### Stage: 2 Challenge Level:
It's hard to make a snowflake with six perfect lines of symmetry, but it's fun to try!
### Tubular Path
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Can you make the green spot travel through the tube by moving the yellow spot? Could you draw a tube that both spots would follow?
### A Maze of Directions
##### Stage: 2 Challenge Level:
Use the blue spot to help you move the yellow spot from one star to the other. How are the trails of the blue and yellow spots related?
### Building Patterns
##### Stage: 2 Challenge Level:
Can you deduce the pattern that has been used to lay out these bottle tops?
### Watch Those Wheels
##### Stage: 1 Challenge Level:
Have you ever noticed the patterns in car wheel trims? These questions will make you look at car wheels in a different way!
### Lafayette
##### Stage: 2 Challenge Level:
What mathematical words can be used to describe this floor covering? How many different shapes can you see inside this photograph?
### Symmetrical Semaphore
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Someone at the top of a hill sends a message in semaphore to a friend in the valley. A person in the valley behind also sees the same message. What is it?
### Hidden Meaning
##### Stage: 2 Challenge Level:
What is the missing symbol? Can you decode this in a similar way?
### Pattern Power
##### Stage: 1, 2 and 3
Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create.
### Prime Magic
##### Stage: 2, 3 and 4 Challenge Level:
Place the numbers 1, 2, 3,..., 9 one on each square of a 3 by 3 grid so that all the rows and columns add up to a prime number. How many different solutions can you find?
### Eight Dominoes
##### Stage: 2, 3 and 4 Challenge Level:
Using the 8 dominoes make a square where each of the columns and rows adds up to 8
### Polydron
##### Stage: 2 Challenge Level:
This activity investigates how you might make squares and pentominoes from Polydron.
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This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A253652 Triangular numbers that are the product of a triangular number and an oblong number. 3
%I
%S 0,6,36,120,210,300,630,1176,2016,3240,3570,4950,7140,7260,10296,
%T 14196,19110,23436,25200,32640,39060,41616,52326,61776,64980,79800,
%U 97020,116886,139656,145530,165600,195000,228150,242556,265356,304590,306936,349866,353220,404550,426426,461280
%N Triangular numbers that are the product of a triangular number and an oblong number.
%C Supersequence of A083374, because A083374(n)= n^2 * (n^2 - 1)/2 = n*(n+1)/2*n*(n-1), product of triangular number n*(n+1)/2 and oblong number n*(n-1).
%e 630 is in the sequence because it is a triangular number (630 = 35*36/2) and 630 = 105*6, with 105 = 14*15/2, triangular number, and 6 = 2*3, oblong number.
%o (PARI) {i=0;j=1;print1(0,", ");while(i<=10^6,k=1;p=2;c=0;while(k<i&&c==0,if(i/k==i\k&&issquare(4*(i/k)+1)&&k>0,c=k);if(c>0,print1(i,", "));k+=p;p+=1);i+=j;j+=1)}
%Y Cf. A002378, A000217, A188630, A083374, A253650, A253651, A253653.
%K nonn
%O 1,2
%A _Antonio Roldán_, Jan 07 2015
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1. ## rectangle
Let ABCD be a rectangle with BC=3AB .
Let P and Q are two points on side BC with PB=PQ=QC .
Show that
< DPC +<DBC =< DQC .
2. Hello,
Originally Posted by perash
Let ABCD be a rectangle with BC=3AB .
Let P and Q are two points on side BC with PB=PQ=QC .
Show that
< DPC +<DBC =< DQC .
For that, I will use a rough method, using more algebra than geometry.
----------------------------
Because I don't remember the formula for $\arctan a+\arctan b$, I'll show it.
~~~~~~~~~~~~~~
For that I will need to find the formula (which I don't remember either) of tan(c+d) :
$\tan(c+d)=\frac{\sin(c+d)}{\cos(c+d)}$
$\tan(c+d)=\frac{\sin(c)\cos(d)+\cos(c)\sin(d)}{\co s(c)\cos(d)-\sin(c)\sin(d)}$
Let's multiply by $\frac{~~~~\frac{1}{cos(c)\cos(d)}~~~~}{~~~~\frac{1 }{cos(c)\cos(d)}~~~~}$
This gives :
$\tan(c+d)=\frac{\tfrac{\sin(c)}{\cos(c)}+\tfrac{\s in(d)}{\cos(d)}}{1-\tfrac{\sin(c)}{\cos(c)} \cdot \tfrac{\sin(d)}{\cos(d)}}$
$\boxed{\tan(c+d)=\frac{\tan(c)+\tan(d)}{1-\tan(c)\tan(d)}}$
~~~~~~~~~~~~~~
Therefore, we have :
$\tan(\arctan(a)+\arctan(b))=\frac{\tan(\arctan(a)) +\tan(\arctan(b))}{1-\tan(\arctan(a))\tan(\arctan(b))}$
$\tan(\arctan(a)+\arctan(b))=\frac{a+b}{1-ab}$
$\implies \boxed{\arctan(a)+\arctan(b)=\arctan\left(\frac{a+ b}{1-ab}\right)}$
----------------------------
Now let's go back to the problem.
So we have $\angle DPC+\angle DBC=\arctan(1/3)+\arctan(1/2)$
$\arctan(1/3)+\arctan(1/2)=\arctan\left(\frac{1/3+1/2}{1-(1/2)*(1/3)}\right)$ (formula above).
$\arctan(1/3)+\arctan(1/2)=\arctan\left(\frac{5/6}{1-1/6}\right)=\boxed{\arctan(1)}$
Therefore, $\boxed{\angle DPC+\angle DBC=\angle DQC}$
3. Originally Posted by Moo
Hello,
For that, I will use a rough method, using more algebra than geometry.
----------------------------
Because I don't remember the formula for $\arctan a+\arctan b$, I'll show it.
~~~~~~~~~~~~~~
For that I will need to find the formula (which I don't remember either) of tan(c+d) :
$\tan(c+d)=\frac{\sin(c+d)}{\cos(c+d)}$
$\tan(c+d)=\frac{\sin(c)\cos(d)+\cos(c)\sin(d)}{\co s(c)\cos(d)-\sin(c)\sin(d)}$
Let's multiply by $\frac{~~~~\frac{1}{cos(c)\cos(d)}~~~~}{~~~~\frac{1 }{cos(c)\cos(d)}~~~~}$
This gives :
$\tan(c+d)=\frac{\tfrac{\sin(c)}{\cos(c)}+\tfrac{\s in(d)}{\cos(d)}}{1-\tfrac{\sin(c)}{\cos(c)} \cdot \tfrac{\sin(d)}{\cos(d)}}$
$\boxed{\tan(c+d)=\frac{\tan(c)+\tan(d)}{1-\tan(c)\tan(d)}}$
~~~~~~~~~~~~~~
Therefore, we have :
$\tan(\arctan(a)+\arctan(b))=\frac{\tan(\arctan(a)) +\tan(\arctan(b))}{1-\tan(\arctan(a))\tan(\arctan(b))}$
$\tan(\arctan(a)+\arctan(b))=\frac{a+b}{1-ab}$
$\implies \boxed{\arctan(a)+\arctan(b)=\arctan\left(\frac{a+ b}{1-ab}\right)}$
----------------------------
Now let's go back to the problem.
So we have $\angle DPC+\angle DBC=\arctan(1/3)+\arctan(1/2)$
$\arctan(1/3)+\arctan(1/2)=\arctan\left(\frac{1/3+1/2}{1-(1/2)*(1/3)}\right)$ (formula above).
$\arctan(1/3)+\arctan(1/2)=\arctan\left(\frac{5/6}{1-1/6}\right)=\boxed{\arctan(1)}$
Therefore, $\boxed{\angle DPC+\angle DBC=\angle DQC}$
. Why go to the arctans? The tans will do.
If (angle DPC) +(angle DBC) = angle DQC
Then, tan[<DPC +<DBC) = tan(<DQC)
[tan(<DPC) +tan(<DBC)] /[1 -tan(<DPC)*tan(<DBC)] =? tan(<DQC)
[1/3 +1/2] /[1 -(1/3)(1/2)] =? 1
[5/6] /[5/6] =? 1
1 =? 1
Yes, so <DPC +<DBC = <DQC
4. Originally Posted by ticbol
. Why go to the arctans? The tans will do.
Because I imagined it this way on my sketch
Thank you for poiting that out
5. Proof using geometry
As < DQC = 45, what we need to prove is that:
< DPC + < DBC = 45
Construct another rectangle BCEF with CE = BF = AB.
R and S are points on EF such that FR = RS = SE.
Connect B and S.
triangle BQS triangle DCP are congruent, so
< SBQ = < DPC
Therefore
< DPC + < DBC = < DBS
Next we will show that < DBS = 45.
In fact, triangle BSD is an isosceles right triangle with BS = DS and < BSD = 90. Its proof is as follows:
As triangle BFS and triangle DES are congruent,
BS = DS, and
< BSF = < SDE
Because QS and DE are parallel,
< QSD = < SDE
Thus
< BSD = < BSQ + < QSD
= < BSQ + < BSF
= < FSQ
= 90
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# It takes Ginny's motor boat?
It takes Ginny's motor boat 4 hours to travel downstream 20 miles and 2 hours to travel 2 miles upstream from her camp. What is the average speed of the current?
### 1 Answer
Relevance
• 7 years ago
Favorite Answer
4(s + c) = 20 or 4s + 4c = 20
2(s - c) = 2 or 2s - 2c = 2 or s = 2 + c
substitute second equation into first equation:
4(2 + c) + 4c = 20
8 + 4c + 4c = 20
8c = 12
c = 12/8 = 1.5 mi/hr
The current speed is 1.5 miles per hour.
- .--
Still have questions? Get your answers by asking now.
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Anda di halaman 1dari 5
Forces Unit Study Guide Name __________________ Date ______ Period ___
1. Identify whether the following statements describe a change in acceleration. Explain your response.
(S8P3a)
a. A car stopped at a stop sign.
d. A person running at 5 meters/second along a curving path.
2. Describe what is meant by “a constant change of direction”. Identify whether the examples provided
show a constant change of direction when moving. (S8P3a)
The graph below shows how the speed of a
bus changes during part of a journey. Use
this graph to answer questions 3-5
3. In Segment B-C, the bus is
________________ (accelerating;
decelerating; at rest; at constant speed)
Why?
4. In Segment C-D, the bus is
________________ (accelerating;
decelerating; at rest; at constant speed)
Why?
5. Which Segment(s) is accelerating? How
do you know?
6. The table to the right shows the time it takes three trucks to go from 0 to 60 Truck Time(s)
km/h. Based on the information given, what data can be compared for the three 1 4.5
trucks? (S8P3a) 2 3.1
3 5.5
7. On a frictionless surface, how does the increase in an object’s mass affect its acceleration? (S8P3a)
1
Forces Unit Study Guide Name __________________ Date ______ Period ___
8. At which point in the diagram above would the rate of acceleration be the greatest? Explain your
9. Define inertia. (S8P3b)
10. Identify which of the following objects has the most inertia: 3 g gumball; 2 kg tennis ball; 2 g ping
11. The diagram to the right shows a skier finishing a competition. What
happens to his speed and direction as he rounds the finish line? (S8P3a)
12. According to the graph to the right, what was the
acceleration of the object between Points R and S? Explain
13. Explain why an object does not move even though a force is exerted on the object. (S8P3b)
2
Forces Unit Study Guide Name __________________ Date ______ Period ___
14. What does the graph to the right illustrate about the
speed of an object? (S8P3a)
15. The diagram to the right shows a cart full of blocks
coming to a sudden stop. Explain what happens
to the blocks in the cart. (S8P3b)
16. The diagram to the right shows a man trying to walk
a dog, but the man and the dog are moving in the same
direction. Draw an arrow below the diagram showing
the direction of movement and describe the forces in this
situation (balanced, unbalanced, which force is greater).
(S8P3b)
17. A tractor company decides to reduce the mass of its tractor by 100 kg. The new tractor design is
identical to the old design in everything other than weight. Compare the force it would take to move the old
tractor to the force it would take to move the new tractor. (S8P3b)
18. Describe the principle “Every action has an equal and opposite reaction”. (S8P3b)
19. A group of elementary students were playing shuffle board on the playground. One student pushed
the puck toward the opposite side of the game board. The student thought the puck would continue all
the way to the other side, but the puck stopped half-way down the board. Explain in terms of force why
the puck failed to slide all the way to the opposite side of the board. (S8P3b)
21. Define a fulcrum. (S8P3c)
3
Forces Unit Study Guide Name __________________ Date ______ Period ___
22. Look at the diagram of the box to the right. In which
direction will the box increase in speed? Explain your
23. Look at the figures to the right. Which figure will
remain stationary unless an external force acts on it?
(S8P3b)
24. Carts 1 and 2 in the diagram to the right have a
compressed spring between them. The carts are held
together by a cord. When the cord is cut, the compressed
spring will force the carts apart. When this happens, how
will the acceleration of Cart 1 differ from that of Cart 2?
25. The picture to the right shows a pulley being used to move a 6 kg block. In
which direction will the block move if the force applied to the rope is greater than
the force of gravity on the load? (S8P3b)
26. A force (F1) is required to push a 15kg container across a carpeted floor. A force (F2) is required to
push the same 15kg container across the ice at a rink. What is typically true about the F1 force in
comparison to the F2 force? (S8P3b)
27. What simple machines are found in a can opener? (S8P3c)
28. The groove on a screw is an example of which type of simple machine? (S8P3c)
4
Forces Unit Study Guide Name __________________ Date ______ Period ___
A student sets up the experiment shown in the
diagram to the right to investigate how friction
affects the motion of a toy car.
29. The student releases the car down the ramp
and measures how far the car travels. Which
variable must the student change to test how
friction affects the motion of the car? (S8P3b)
30. Identify the simple machines found in the bicycle
to the right. (S8P3c)
31. Since the bicycle has more than one simple machine, it is
called a ______________________________. (S8P3c)
32. Identify the simple machines used by the man in
the diagram to the right to make his work easier.
(S8P3c)
33. How does the distance between two objects and their mass affect gravitational attraction (force)
between the two objects? (S8P5a)
34. The diagram to the right shows a moon revolving around a
planet in an orbit. At which location is the gravitational pull
(S8P5a)
35. During the Force and Motion Unit, your class participated in the
JetToy Challenge activity. Identify the forces that can influence the
performance of the JetToy.
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## World Solar Challenge: Why the winners were so good
A light bulb here, a light bulb there
Posted in Science, 22nd October 2011 06:13 GMT
WSC Tokai University of Japan were crowned the 2011 World Solar Challenge champions on Thursday, after four-and-a-half days racing through the Australian continent. The Japanese team made huge gains on Day 3 over second and third-placed teams Nuon of the Netherlands and the University of Michigan.
How did they get so far ahead? Diederik Kinds, of the website solarwebsite.nl and a member of the Dutch Nuna 2 solarcar team, tells us how.
Some people thought Tokai, World Solar Challenge winners in 2009, had a terrific advantage over the competition that year because of their awesome GaAs (gallium arsenide) solar array. They thought the new regulations would level the competition and remove Tokai’s supremacy. They were wrong.
Let’s do some numbers based on Race Day 3 results. Tokai left Wauchope, NT at 08:00am and reached Kulgera, South Australia - 667 km south - at 16:02pm, stopping twice at the 30-minute control stops along the road. So – with 7.02 hours driving – they averaged 95 kph.
The Dutch team Nuon's solar car, Nuna 6, departed at 08:08am and arrived at 16:37am – an average speed of 89 kph. This 6kph difference with Tokai opened a gap on the day of about 30 minutes between the two cars.
### What's that in vacuum cleaners?
Since this race is about energy, let’s do some energy calculations. I will take this site as example – you can calculate power consumption based on weight and aeroshape.
A typical solar car weighs in at 230kg (150kg + 80kg driver) and has a CdA (a measure for aero drag) of 0.09.
Fill this in on that site and look at the resulting tables: 89 kph leads to a consumption of 1315 W. Nuna drove for 7.30 hours, so total energy consumed on this 667km stretch is 1315 x 7.5 = 9.9 kWh.
As a point of reference, Nuna used the equivalent energy of six hours vacuum cleaning to travel 600km at normal commuting speed.
Tokai’s 95 kph with the standard specs represents a consumption of about 1550W. And since they drove for seven hours, their consumption amounts to 1550 x 7 = 10.9 kWh.
In other words, to create a gap of 30 minutes to Nuna, Tokai had to consume 1 kWh of energy more than Nuna 6, or 10 per cent more energy.
Obviously, energy flows into the car via the solar panel, but where does it go to? First, there are losses from solar panel to electric motor, like in the motor controller and other electronic equipment required for operating the solar car.
Secondly, the aerodynamic shape requires energy – the shape of the car is important but surface smoothness is also a factor.
### Resistance is futile
Thirdly, there is rolling resistance, related to the type of tyre and the weight of the car pushing on those tyres. Less weight reduces rolling resistance.
So, let’s see how good Tokai needs to be to be able to spend 10 per cent more energy create that 30-minute gap.
Keep in mind that one per cent of energy consumption is about 12W, similar to a small light bulb.
A 10kg lighter car leads to 1.5 per cent less energy consumption.
A five per cent improvement in aerodynamics results in two per cent less energy consumption. Five per cent is the difference between a clean car nose, and one that has flies on it, or other irregularities. So the car needs to be as smooth as possible at all times. Small things have great impact.
If your panel is able to give a mere two per cent more solar power, the car has two per cent more energy to spend.
If you can make your electronics between panel and motor just one per cent more efficient, there is a direct correlation and you’ll have per cent more energy to spend.
### State of charge
Let’s talk a bit about the battery. A typical solar car battery can hold 5kWh energy. Solar cars take into account what energy the sun delivers and what they can take out of the battery. Battery energy level is called State of Charge (SOC), and you may finish a race day with a certain percentage of SOC. Suppose you have two similar cars – if one drives faster on a race day, he will have covered more distance but will have less battery SOC.
Now suppose Tokai has taken more out of its battery SOC than Nuna, when rolling into Kulgera. Let’s say that number is 3.5 per cent of the total Energy Consumption, which comes down to 0.35 kWh. On a battery with 5 kWh capacity that’s seven per cent SOC less than Nuna - which is really not that much.
In short, to gain 30 minutes on Nuna, Tokai had to take seven per cent more out of their battery, have a car that weighs just 10kg less, with aerodynamics that is just five per cent better (this could already be achieved by keeping the car smooth and clean), a panel that can power two small light bulbs more and a little more efficient electronic system.
That totals the 10 per cent of energy that Tokai got to spend to increase their lead by 30 minutes.
Is Tokai the mystical energy powerhouse that people claim it to be?
No. It’s all in the small numbers as described above. A little bit better here, and a little bit better there can lead to superb results.
Tokai’s technology is not light years ahead but just a few light bulbs better in every area.
Tokai is not magic, not supernatural, not sent from the future and they don’t overpower their competitors with special technology.
They are just a very good team with great attention to detail. And that’s how you win races. ®
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# Reduction Formula
A reduction formula is regarded as an important method of integration. Integration by reduction formula always helps to solve complex integration problems. It can be used for powers of elementary functions, trigonometric functions, products of two are more complex functions, etc. These are the functions that cannot be integrated easily. Therefore for easing the process of integration, we will discuss here Reduction Formula for integration with examples. Let us learn the important concept!
## Reduction Formula
### Concept of Reduction Formula in Integration:
The reduction formula can be applied to different functions with combinations of different types of functions in a single problem. The formula for the reduction can be divided into various categories as given below:
• Exponential functions
• Trigonometric functions
• Inverse trigonometric functions
• Hyperbolic trigonometric functions
• Algebraic functions
Therefore to get the solution of integrals we can use the reduction formulas. These formulas will enable us to reduce the degree of the integrand and calculate the integrals in a finite number of steps. Below are the important reduction formulas for integrals involving the most common functions.
### Some Important Formulas
Reduction Formula for Trigonometric Functions
• $$\int sin^{n}(x)dx=\frac{-Sin^{n-1}(x)Cos(x)}{n}+\frac{n-1}{n}Sin^{n-2}(x)dx$$
• $$\int tan^{n}(x)dx=\frac{-tan^{n-1}(x)}{n-1}-\int tan^{n-2}(x)dx$$
• $$\int sin^{n}(x)\: cos^{m}(x)dx=\frac{sin^{n+1}(x)cos^{m-1}(x)}{n+m}+\frac{m-1}{n+m}\: \int sin^{n}(x)\: cos^{m-2}(x)dx$$
• $$\int x^{n}cos(x)dx=x^{n}sin(x)-n\int x^{n-1}sin(x)dx$$
• $$\int x^{n}sin(x)dx=-x^{n}cos(x)+n\int x^{n-1}cos(x)dx$$
Reduction Formula for Logarithmic Functions
• $${\large\int\normalsize} {{\ln^n}x\,dx} ={\large\frac{{{x^{n + 1}}{\ln^m}x}}{{n + 1}}\normalsize}-\;{\large\frac{m}{{n + 1}}\int\normalsize} {{x^n}{\ln^{m – 1}}x\,dx}$$
• $$\int \frac{ln^m x}{x^n}\,=\,\frac{ln^m x}{\left ( n-1 \right )x^{n+1}}\,+\, \frac{m}{n-1}\int \frac{ln^{m-1}x}{x^n}\,dx,\;n\neq1$$
Reduction Formula for Algebraic Functions
• $${\large\int\normalsize} {\large{\frac{{{x^n}}}{{a{x^n} + b}}\normalsize}} \,dx = {\large\frac{x}{a}\normalsize} – {\large\frac{b}{a}\int {\frac{{dx}}{{a{x^n} + b}}}\normalsize}$$
• $$\int \frac{dx}{\left ( ax^2 +bx+c \right )^n}\,=\, \frac{-2ax-b}{\left ( n-1 \right )\left ( b^2 -4ac \right )\left ( ax^2 +bx+c \right )^{n-1}}\,-\, \frac{2\left ( 2n-3 \right )a}{\left ( n-1 \right )\left ( b^2 – 4ac \right )}\int \frac{dx}{\left ( ax^2 +bx + c \right )^{n-1}},\,n\neq1$$
• $$\int \frac{dx}{\left ( x^2+a^2 \right )^n}\,=\, \frac{x}{2\left ( n-1 \right )a^2 \left ( x^2 + a^2 \right )^{n-1}}\,+\, \frac{2n-3}{2\left ( n-1 \right )a^2}\int \frac{dx}{\left ( x^2 + a^2 \right )^{n-1}},\,n\neq1$$
• $$\int \frac{dx}{\left ( x^2-a^2 \right )^n}\,=\, \frac{x}{2\left ( n-1 \right )a^2 \left ( x^2 – a^2 \right )^{n-1}}\,-\, \frac{2n-3}{2\left ( n-1 \right )a^2}\int \frac{dx}{\left ( x^2 – a^2 \right )^{n-1}},\,n\neq1$$
Reduction Formula for Exponential Functions
• $${\large\int\normalsize} {{x^n}{e^{mx}}dx} ={\large\frac{1}{m}\normalsize}{x^n}{e^{mx}}-\; {\large\frac{n}{m}\normalsize} {\large\int\normalsize} {{x^{n – 1}}{e^{mx}}dx}$$
• $${\large\int\normalsize} {{\large\frac{{{e^{mx}}}}{{{x^n}}}\normalsize} dx} =– {\large\frac{{{e^{mx}}}}{{\left( {n – 1} \right){x^{n – 1}}}}\normalsize}+\; {\large\frac{m}{{n – 1}}\normalsize} {\large\int\normalsize} {{\large\frac{{{e^{mx}}}}{{{x^{n – 1}}}}\normalsize} dx}$$
• $${\large\int\normalsize} {{{\sinh }^n}x\,dx} =– {\large\frac{1}{n}\normalsize}{\sinh ^{n – 1}}x\cosh x-\; {\large\frac{{n – 1}}{n}\normalsize} {\large\int\normalsize} {{{\sinh }^{n – 2}}x\,dx}$$
• $${\large\int\normalsize} {\large\frac{{dx}}{{{{\sinh }^n}x}}\normalsize} = – {\large\frac{1}{n}\normalsize}{\sinh ^{n – 1}}x\cosh x +\;{\large\frac{{n – 2}}{{n – 1}}\int\normalsize} {\large\frac{{dx}}{{{{\sinh }^{n – 2}}x}}\normalsize}$$
## Solved Examples
Q.1:Â Evaluate the integral:
$$\displaystyle \int{{{z^7}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}}$$
Solution: First,
$$u = 8 + 3{z^4}\hspace{0.25in} \to \hspace{0.25in}du = 12{z^3}dz\hspace{0.25in}\,\,\,\,\,\, \to \hspace{0.25in}\,\,\,\,\,{z^3}dz = \frac{1}{{12}}du$$
Let’s do a quick rewrite of the integrand,
$$\int{{{z^7}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}} = \int{{{z^4}{z^3}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}} = \int{{{z^4}{{\left( {8 + 3{z^4}} \right)}^8}\,{z^3}dz}}$$
Now, notice that we can convert all of the z’s in the integrand except apparently for the z^4 that is in the front. We can notice from the substitution that we can solve it for z^4 to get,
$${z^4} = \frac{1}{3}\left( {u – 8} \right)$$
Now, with this we can do the substitution and evaluate the integral.
\begin{align*}\int{{{z^7}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}} & = \frac{1}{{12}}\int{{\frac{1}{3}\left( {u – 8} \right){u^8}du}} = \frac{1}{{36}}\int{{{u^9} – 8{u^8}du}} = \frac{1}{{36}}\left( {\frac{1}{{10}}{u^{10}} – \frac{8}{9}{u^9}} \right) + c\\ &Â = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{36}}\left( {\frac{1}{{10}}{{\left( {8 + 3{z^4}} \right)}^{10}} – \frac{8}{9}{{\left( {8 + 3{z^4}} \right)}^9}} \right) + c}}\end{align*}
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### One response to “Equation Formula”
1. KUCKOO B says:
I get a different answer for first example.
I got Q1 as 20.5
median 23 and
Q3 26
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# 112 what are the upper and lower bounds of the 95 ci
• Test Prep
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1.12 What are the upper and lower bounds of the 95% CI for (expressed now as a
percent) 1.13 Is the election outcome for Cameron’s party in the 95% CI? no
1.14 What statistical principle is this an example of?
Practice Quiz 3 3 Problem 2 Confidence Intervals Match the margin of error to the desired CI. Desired CI Answer Margin of Error 2.1 95% CI for a mean, 𝜎 known 5 1. ? 0.975 (?) ∗ (?√ 1 ? + (?−?̅) 2 ∑(? 𝑖 −?̅) 2 ) 2.2 80% CI for 2-sample difference in means, pooled variance 3 2. ? 0.9 (?) ∗ (√ ? 1 2 ? 1 + ? 2 2 ? 2 ) 2.3 80% CI for 2-sample difference in means, unpooled variance 2 3. ? 0.9 (?) ∗ (? ? 1 ? 1 + 1 ? 2 ) 2.4 95% prediction interval for ? ̂|? 4 4. ? 0.975 (?) ∗ (?√1 + 1 ? + (?−?̅) 2 ∑(? 𝑖 −?̅) 2 ) 2.5 95% CI for ? |? 1 5. ? 0.975 ∗ ( 𝜎 𝑋 √? )
Practice Quiz 3 4 Problem 3. Hypothesis tests 3.1 Suppose we want to test the hypothesis 0 : 0 vs. : 0 a H H 𝜎 unknown and N =50. If we want 05 . 0 We would reject the null when: 1. 0.975 (49) ( ) x t SE x 2. 05 . 0 ) ( z x SE x 3. 0.95 (50) ( ) x t SE x 4. 0.95 (49) ( ) x t SE x 3.2 Suppose we want to test the hypothesis 0 : 0 vs. : 0 a H H , 𝜎 unknown and N =100. If we want 05 . 0 We would reject the null when: 1. 0.95 (100) ( ) x t SE x 2. 0.90 (100) ( ) x t SE x 3. 0.05 (99) ( ) x t SE x 4. 0.975 (99) ( ) x t SE x 3.3 Suppose we want to test the hypothesis 0 : 1 0 H , vs 1 : 0 a H N=500 If we want 01 . 0 We would reject the null when: 1. 1 0.99 1 (499) ( ) b t SE b 2. 1 0.99 1 (500) ( ) b t SE b 3. 1 0.995 1 (498) ( ) b t SE b 4. 99 . 0 1 1 ) ( z b SE b 3.4 Suppose we want to test the hypothesis 0 : 1 0 H , vs 1 : 0 a H N=500 If we want 01 . 0 We would FAIL TO reject the null when: 1. 1 0.99 1 (500) ( ) b t SE b 2. 1 0.01 1 (498) ( ) b t SE b 3. 1 0.99 1 (499) ( ) b t SE b 4. 1 0.95 1 ( ) b z SE b
Practice Quiz 3 5 Problem 4. Inference for proportions The Department of Statistics estimates that 50% of students at the University of Washington will have taken STAT 311 by the time they graduate. You believe it is less than 50%, and decide to take a survey of graduating seniors to estimate the true proportion. Your friend gets you a list of the graduating seniors and their email addresses, and you randomly select a sample of n=200, email them, and ask them to report on a Catalyst survey whether they have taken STAT 311: 40% of seniors report that they have. Set up a 90% CI for the estimated proportion of seniors who have taken STAT 311. 4.1 Symbolic representation: ?̂ ± ? 𝛼/2 ∗ √ ?̂ (1−?̂) ? or ? = ± ? 𝛼/2 ∗ √ ?̂ (1−? ̂) ? 4.2 With plug-in values: 0.40 ± 1.64 ∗ √ 0.4(1−0.4) 200 = 0.40 ± 0.06 or [0.34, 0.46] State the null and the general alternative hypotheses. What is the approximate distribution of the sample proportion under the null hypothesis?
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# Problem Solving Worksheets
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• This Set Q (elem/upper elem) Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this Set Q (elem/upper elem) Word Problems. Jordan has eight apples. Cameron has half as many apples as Jordan. Natalie has three-quarters as many apples as Cameron. How many apples does each person have? How many do they have altogether? Six word problems.
• This Set R (elem/upper elem) Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this Set R (elem/upper elem) Word Problems. Brooke has eleven flowers. She has more tulips than roses. What are the possible combinations of tulips and roses, if she only has these two types of flowers? Six word problems.
• This Set S (elem/upper elem) Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this Set S (elem/upper elem) Word Problems. Dakota went to the store. He bought three note pads for \$.75 each, four pencils at 2 for \$.35 and one candy bar, which was being sold at 3 for \$1.80. How much money did he spend? How much did he get back from the \$10.00 he gave the clerk? Six word problems.
• This Set T (elem/upper elem) Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this Set T (elem/upper elem) Word Problems. Paige, Cassandra, and Katie earned \$12.45 by working for a neighbor. Assuming they worked equal amounts, what would each girl’s share be? Six word problems.
• This Set U (elem/upper elem) Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this Set U (elem/upper elem) Word Problems. Sydney got twenty-one e-mails on Monday, nineteen on Tuesday, thirty-seven on Wednesday, eight on Thursday, and twenty-three on Friday. How many e-mails did she get on Monday, Tuesday, Wednesday, and Friday, combined? Six word problems.
• This Sets A - U (elem/upper elem) - answers Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this Sets A - U (elem/upper elem) - answers Word Problems. Answers to the word problems from sets A-U, grouped by page.
• This Conversion - Vacation in France (upper elem/middle) Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this Conversion - Vacation in France (upper elem/middle) Word Problems. "Marie's family all had lunch in a small patisserie. The meal cost €48. What was the average cost of the meal per person, in dollars?" Four word problems for converting euros to dollars.
• This Conversion - Vacation in Mexico (upper elem/middle) Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this Conversion - Vacation in Mexico (upper elem/middle) Word Problems. "Petra wants to buy a small replica of the pyramid at Chichén Itzá. She has \$15 left. The pyramid costs 200 pesos. Does she have enough money?" Four word problems for converting pesos to dollars.
• This Conversion - Vacation in China (upper elem/middle) Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this Conversion - Vacation in China (upper elem/middle) Word Problems. "Tickets for a tour of the Great Wall of China are 50 rmb for adults and 30 rmb for children ages 12-24. How much does it cost for Simon and his family in all? How much in dollars?" Four word problems for converting pesos to dollars.
• This Conversion - Vacation in Russia (upper elem/middle) Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this Conversion - Vacation in Russia (upper elem/middle) Word Problems. "Tori and her family go on a tour of St. Basil’s Cathedral. Tickets are 120 rubles for adults; admission is free for children 12 and under. How much does it cost for Tori and her family in all? How much in dollars?" Four word problems for converting rubles to dollars.
• This Conversion - Vacation in Canada (upper elem/middle) Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this Conversion - Vacation in Canada (upper elem/middle) Word Problems. "Mike and his family all had dinner in a French restaurant in Montreal. The meal cost \$150 Canadian. What was the average cost of the meal per person, in U.S. dollars?" Four word problems with converting Canadian to US dollars.
• This Pumpkin Word Problems (elem) Clip Art is perfect to practice word problem skills. Your elementary grade students will love this Pumpkin Word Problems (elem) Clip Art. An 8 page pumpkin shape book of math word problems related to October and Halloween.
• This Pumpkin Word Problems (to ten) (primary) Clip Art is perfect to practice word problem skills. Your elementary grade students will love this Pumpkin Word Problems (to ten) (primary) Clip Art. A five page pumpkin shaped booklet with a story problem on each page.
• This Rules and Practice - Multiple-Step Word Problems (upper elem) Math is perfect to practice multi-step work problem skills. Your elementary grade students will love this Rules and Practice - Multiple-Step Word Problems (upper elem) Math. An explanation of multiple-step word problems with ten problems and an answer sheet.
• This Math Tools (color) Sign is perfect to practice math skills. Your elementary grade students will love this Math Tools (color) Sign. Math Tools sign illustrated with a protractor.
• This Problem Solving (color) Sign is perfect to practice problem solving skills. Your elementary grade students will love this Problem Solving (color) Sign. Classroom sign, Problem Solving, illustrated with a girl and thinking bubble.
• This Math - U.S. Standard Conversion Word Problems (grade 5) Interactive Notebook is perfect to practice measurement skills. Your elementary grade students will love this Math - U.S. Standard Conversion Word Problems (grade 5) Interactive Notebook. Interactive rules and practice of word problems involving real world situations in converting units of money, time, length/distance, liquid volume and weight in U.S. standards. CC: Math: 5.MD.A.1
• This Problem Solving Cards - Set A (elem/upper elem) Clip Art is perfect to practice word problem skills. Your elementary grade students will love this Problem Solving Cards - Set A (elem/upper elem) Clip Art. A set of problem solving cards with 6 different scenarios. Some of them are brain teasers, and you may need to think in different ways to solve the problem.
• This Problem Solving Cards - Set A (with lines) Clip Art is perfect to practice word problem skills. Your elementary grade students will love this Problem Solving Cards - Set A (with lines) Clip Art. A set of 6 word problems, with lines to show work.
• This Problem Solving Cards - Set C Clip Art is perfect to practice word problem skills. Your elementary grade students will love this Problem Solving Cards - Set C Clip Art. A set of problem solving cards with 6 different scenarios.;
• This Problem Solving Cards - Set C (with lines) Clip Art is perfect to practice word problem skills. Your elementary grade students will love this Problem Solving Cards - Set C (with lines) Clip Art. Problem solving cards with six different scenarios.
• This Thanksgiving - Division Math Game is perfect to practice division skills. Your elementary grade students will love this Thanksgiving - Division Math Game. Roll the dice and move across the board by division equations. Laminate for a fun, reusable classroom game.
• This Dividing Decimals Practice Math is perfect to practice decimal division skills. Your elementary grade students will love this Dividing Decimals Practice Math. Two pages of decimal division with twenty equations per page. Includes decimals to the tenths and hundredths places.
• This Dividing Decimals Word Problems Math is perfect to practice decimal division skills. Your elementary grade students will love this Dividing Decimals Word Problems Math. Two pages of decimal division with six word problems per page. Includes decimals to the tenths and hundredths places.
• This Valentine's Day Solve & Color Packet Holiday/Seasonal is perfect to practice addition and subraction skills. Your elementary grade students will love this Valentine's Day Solve & Color Packet Holiday/Seasonal. Six pages of Valentine's Day illustrations with math facts to solve and color by number. Two pages each of addition facts and subtraction facts. One page each of double facts and double plus one facts. All numbers are under 20.
• 10 pages of practice in sums. Place values and number of terms increase through packet. Created with abctools Math Worksheet Generator. CC: 3.NBT.A.1
• 10 pages of practice in advanced place value that includes whole numbers from thousands to millions and decimals from tenths to hundred thousandths (difficulty increases throughout packet). Created with abctools math worksheet generator. CC: 5.NBT.A.3
• 10 pages of practice in decimal place value from tenths to thousandths (difficulty increases throughout packet). Created with our abctools math worksheet generator. CC: 5.NBT.A.3
• 10 pages of practice in place value from tens to hundred thousands (difficulty increases throughout packet). Created with our abctools math worksheet generator. CC: 1.NBT.B.2, 2.NBT.A.1
• 10 pages of mixed practice for finding the missing subtraction operand. 4 pages of ones place subtraction; 6 pages of ones to tens place subtraction. Facts are in vertical form. Created with our abctools Math Worksheet Generator. CC: 2.OA.A.1
• 10 pages of practice for subtraction from tens to millions places (place value increases with each page). Facts are in vertical form. Created with our abctools Math Worksheet Generator. CC: 4.NBT.B.4
• 10 pages of practice for subtraction of positive and negative integers (5 pages of ones place, 5 pages of ones and tens places). Facts are in horizontal form. Created with our abctools Math Worksheet Generator. CC: 7.NS.A.1
• 10 pages of mixed practice for subtraction facts. 5 pages of facts with 1-10; 5 pages of facts within 1-20. Facts are in horizontal form. Created with our abctools Math Worksheet Generator. CC: 1.OA.C.6
• 10 pages of practice in calculating the area of rectangles, squares, triangles, irregular quadrilaterals, and circles. Created with our abctools Math Worksheet Generator. CC: 6.G.A.1
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# How many litres is a barrel of crude oil?
A standard oil barrel holds 42 gallons or approximately 159 litres of crude oil. Using the industry standard "barrel of oil equivalent" (BOE), a barrel of crude oil is said to be equal to 6,000 cubic feet of natural gas with an energy equivalent of 1,700 kilowatt hours.
The standard 42-gallon barrel, which had been carefully crafted and used to ship dry goods for hundreds of years, was made watertight and adopted for use in the petroleum industry in 1872. A full barrel carries enough oil to make shipping profitable, and weighs in at the top end of what could be physically managed by the men moving them.
Reference:
Q&A Related to "How many litres is a barrel of crude oil?"
Since 1 barrel = exactly 42 American gallons, and 3.785431178 liters = 1 gallon, then. 42 x 3.78 = 158.987 liters = 1 barrel. http://wiki.answers.com/Q/How_many_litres_in_a_bar...
159 liters = 35 gallons = 1 barrel of oil. Source(s) http://www.schoolscience.co.uk/content/4…. http://answers.yahoo.com/question/index?qid=100605...
A barrel of oil is 42 US gallons (equal to 159 litres) Therms is used to calculate gas energy. 1 barrel of crude oil is 5,800,000 Btu. report this answer. Updated on Friday, February http://www.kgbanswers.co.uk/how-many-litres-in-a-b...
Crude prices are based on supply/demand scenario and is very volatile. Because of this, there is no systematic cost break down for any crude type. However, different crudes can be http://www.quora.com/Oil-fossil-fuel/What-is-the-b...
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