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http://schoolbag.info/mathematics/sat_1/86.html | 1,537,867,191,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161350.69/warc/CC-MAIN-20180925083639-20180925104039-00021.warc.gz | 218,255,105 | 4,509 | IMAGINARY NUMBERS - Imaginary and Complex Numbers - MISCELLANEOUS TOPICS - SAT SUBJECT TEST MATH LEVEL 1
CHAPTER 17Imaginary and Complex Numbers
• Imaginary Numbers
• Complex Numbers
• Exercises
Approximately 47 of the 50 questions on the Math 1 test are on topics covered in Chapters 2–16. However, as indicated on the chart on WHAT TOPICS ARE COVERED?, there are generally three questions on miscellaneous topics not covered in those chapters. In Chapters 17–19, you will read about the three topics that occur most often among the miscellaneous questions: imaginary numbers, sequences, and logic.
If your study time is limited, you may want to skip this material and concentrate on the chapters about algebra and geometry, which are the sources of the overwhelming majority of the questions on the Math 1 test. If, however, you have the time, you should study this chapter carefully. If you review the material in Chapters 17, 18, and 19, you should be able to answer any question that comes up on these topics.
IMAGINARY NUMBERS
By KEY FACT A1, if x is a real number, then x is positive, negative, or zero. If x = 0, then x 2 = 0; and if x is either positive or negative, then x 2 is positive. So if x is a real number, x 2 CANNOT be negative.
In the set of real numbers, the equation x 2 = –1 has no solution. In order to solve such an equation, mathematicians defined a new number, i, called the imaginary unit, with the property that i 2 = –1. This number is often referred to as the square root of –1: i. Note that i is not a real number and does not correspond to any point on the number line.
All of the normal operations of mathematics—addition, subtraction, multiplication, and division—can be applied to the number i.
Addition: i + i = 2i 2i + 5i = 7i Subtraction: i – i =0 2i – 5i = –3i Multiplication: (i)(i) = i2 = –1 (2i)(5i) = 10i2 = – 10 Division:
Key Fact P1
If x is a positive number, then .
EXAMPLE 1:
Don’t Get Fooled
= (2i)(2i) = 4i2 = –4
is not equal to
EXAMPLE 2: What is
For the Math 1 test, you must be able to raise i to any positive integer power. In particular, note:
i1 = i (a1 = a for any number) i 2 = –1 (by definition) i 3 = –i i.i i = (i.i )(i ) = i2.i = –1(i ) = –i i 4 = 1 i 4 = i.i.i.i = (i.i)(i.i ) = (–1)(–1) = 1 i 5 =i i5 = i4.i = 1.i = i i 6 = –1 i 6 = i5.i = i.i = i2 = –1 i 7 = –i i7 = i6.i = (–1)i = –i i 8 = 1 i 8 = i7.i = (–i)(i ) = –i2 = –(–1) = 1
Note that the powers of i form a repeating sequence in which the four terms, i, –1, –i, 1 repeat in that order indefinitely.
As you will see in KEY FACT P2, this means that to find the value of in for any positive integer n, you should divide n by 4 and calculate the remainder.
Key Fact P2
For any positive integer n:
• If n is a multiple of 4, n1
• If n is not a multiple of 4, nr, where r is the remainder when n is divided by 4.
EXAMPLE 3: To evaluate i 375, use your calculator to divide 375 by 4:
375 4 = 93.75 the quotient is 93
Then multiply 93 by 4:
93 × 4 = 372 the remainder is 375 – 372 = 3
So i 375 = i 3 = –i.
The concepts of positive and negative apply only to real numbers. If a is positive, a is to the right of 0 on the number line. Since imaginary numbers do not lie on the number line, you cannot compare them. It is meaningless even to ask whether i is positive or negative or whether 12i is greater than or less than 7i.
| 1,155 | 3,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2018-39 | latest | en | 0.825788 |
http://activityvillage.co.uk/counting-to-20 | 1,722,969,009,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640497907.29/warc/CC-MAIN-20240806161854-20240806191854-00799.warc.gz | 830,916 | 13,545 | # Counting to 20
Help your children learn the numbers from 11 to 20 by repetition, fun games and challenges, and this collection of videos! This first stage is more about recognising the number words and being able to say them in the right order rather than associating them with a particular written number or number of objects. (Skip ahead to Counting on from 10 if that is what you are looking for). Remember, practice makes perfect!
## Hands-On Ideas
• Counting to 20. Count fast, count slow, chant, sing!
• Making it active. Jump and count. Hop and count. Throw a ball and count. Count as you bounce up, and count as you bounce down!
• Making it fun. Count in turns - between pairs, groups or 3, or round in a bigger circle.
• Count backwards too. When you have counting to 20 under your belt, try counting backwards too!
• Start from any number. Make sure that kids don't have to start at 1 to count up to 20! Start from 5, or 10, or 12, or 16...
• Play Snap Challenge. At random times during the day, call out "Snap Challenge! Count to 20 starting from 6!" Your child can snap challenge you, too.
## Use Our Resources
Here are some printables that you can use with the kids to help learn the numbers 11 to 20.
## Autumn Number Peg Cards To 20
Give the kids a collection of pegs and ask them to count the number of autumn objects on each card, then "clip" the correct number. A fun way to practise counting skills! There are 20 peg cards in this set.
## Butterfly Count the Spots 2
Here's a simple activity for children learning to count up to 20. Count the spots! We have both colour and black and white versions of this page.
## Christmas Number Peg Cards
Count the Christmas pictures and use a clothes peg to peg the correct number, up to 20. There are 5 pages in this pdf file, so 20 peg cards to cut out in total.
## Count the Petals Folding or Matching Cards
Count the petals! Our pretty flowers with from 4 petals to 20 petals challenge the kids to count. Use this set as double-sided flashcards by folding, or cut them all apart for some matching activities. Start with just a few cards with lower numbers and then work up.
## Counting Cubes Large Ten Colours
This is a ten page file with ten different colours of large counting cubes - perfect for when you need something for cutting and sticking into maths notebooks. They might also be useful when you want to do a "hands-on" activity but don't have any maths manipulatives to hand.
## Counting Cubes Ten of Each Colour
This set of printable counting cubes includes 10 of each colour, including a "top" cube. We've spaced them so that you can cut them out as quickly as possible.
## Counting Cubes Towers of Ten All Colours
This 2-page printable contains 10 ready-stacked towers of our counting cubes for you to cut out and use with other same-size printables in the collection, in 10 different colours.
## Flowers Number Match Up
Practise counting with these pretty flower number match up cards! Print out the five pages in the PDF file onto some card, cut each pair in half then mix up and give to the kids to reassemble.
## Guest Post - Count the Petals
Shelly brightens up learning maths skills by using the cheerful Count the Petals printout from Activity Village in this guest blog post.
## Number Lines to 20 - Bordered - Numbers Above
Here are some colourful, bordered number lines to 20 to print for children learning to count, count on, add and subtract.
## Number Lines to 20 - Bordered - Numbers Below
Print this colourful bordered number lines to 20 in 5 different colours - or just choose the colour that you prefer and print that page as many times as required. Each page gives you 3 bordered number lines from 0 to 20.
## Number Lines to 20 - Kids
These fun printable number lines to 20 have a parade of children ready to encourage the kids in their counting! They print 2 to a page and are simply sliced apart.
## Number Lines to 20 - Kids - Dotty
Kids will love learning to count to 20 with these fun printable number lines! There are two on the page, ready to be sliced apart.
## Number Lines to 20 - Simple
Choose from 2 versions of our simple number line strips to 20 - one iwth the numbers below (pictured) and one with the numbers above. Simply slice the paper for 3 quick-to-print number lines from 0 to 20.
## Ten Frame Posters 11 to 20
Use this set of 10 printables as individual posters, or line them up as a frieze or banner. Alternatively, laminate them and use them as mats in the classroom, to help children visualise and understand how the numbers are made up of 10 plus another number.
## Next Steps
Number Symbols 11 to 20
## Explore Activity Village
Become a Member to access 39,215 printables! | 1,103 | 4,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-33 | latest | en | 0.930379 |
https://programmingpraxis.com/2012/08/21/two-more-random-exercises/?like=1&source=post_flair&_wpnonce=7f81158ee4 | 1,500,849,490,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424623.68/warc/CC-MAIN-20170723222438-20170724002438-00275.warc.gz | 697,909,792 | 35,139 | ## Two More Random Exercises
### August 21, 2012
We did two tasks related to random numbers in the most recent exercise, and we have looked at high-quality random number generators in several previous exercises. In today’s exercise we look at two very low-quality random number generators, which should not be used for any production application.
The first, invented by John von Neumann in 1949, occasioned his famous quip “Any one who considers arithmetical methods of producing random digits is, of course, in a state of sin.” The middle-square method takes a number with an even number of digits, squares it, and extracts the middle digits for the next iteration; for instance, if the seed is 675248, the square is 455959861504, and the middle digits are 959861.
The second, invented by IBM in the early 1960s, caused Donald Knuth to claim “its very name RANDU is enough to bring dismay into the eyes and stomachs of many computer scientists!”. RANDU is based on the recursion xn+1 = 65539 · xn (mod 231), with x0 odd.
Your task is to write functions that generate random numbers by the middle-square and RANDU methods. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Pages: 1 2
### 7 Responses to “Two More Random Exercises”
Needs -std=c++0x if compiling with g++.
```#include <iostream>
constexpr int ipow( int v, int e )
{
return e == 0 ? 1 : ipow(v, e-1) * v;
}
template<int digits>
class RandMidSqr
{
public:
RandMidSqr( int seed = 675248 ) : value( seed ){}
int operator()()
{
return value = ((long long)value * value)
/ ipow(10,digits/2)
% ipow(10,digits);
}
private:
int value;
};
struct Randu
{
public:
Randu( unsigned int seed = 1 ) : lastVal( seed ){}
unsigned int operator()()
{
return lastVal = (lastVal * 65539) % ipow(2,31);
}
private:
int lastVal;
};
int main( int argc, char**args )
{
RandMidSqr<6> rms;
for( int i = 0; i < 100; ++i )
{
std::cout << rms() << std::endl;
}
Randu ru;
for( int i = 0; i < 100; ++i )
{
std::cout << ru() << std::endl;
}
return 0;
}
```
2. JP said
I went ahead and coded up both of them in Scheme. This is the sort of thing where a language that has functions as first class citizens can really shine. You can write a function that generates a function that generates random numbers!
3. […] Praxis put out another two “random” exercises, this time about making psuedorandom number generators (where the previous had us composing already […]
4. A Python solution:
```import math
def rand_middle_square(seed):
n = seed
seed_len = int(round(math.log(seed, 10)))
while True:
yield n
n = (n * n) / (10 ** (seed_len / 2)) % (10 ** seed_len)
def randu(seed):
n = seed
while True:
yield n
n = (65539 * n) % 2147483648
def random(count, seed, rand_fn):
nums = []
random_gen = rand_fn(seed)
for _ in xrange(count):
nums.append(random_gen.next())
return nums
print(random(5, 675248, rand_middle_square))
print(random(5, 7, randu))
```
5. edza said
Hey! I think I have bug, or I implemented it wrongly. Any tips anyone?
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace NeumannRandNumGen
{
class Program
{
//The middle-square method takes a number with an even number of digits, squares
//it, and extracts the middle digits for the next iteration; for instance, if the
//seed is 675248, the square is 455959861504, and the middle digits are 959861
static void Main(string[] args)
{
int seed = 45505986; // seed
while(true) {
int randomNumber = NeumannRandom0To100(ref seed);
Console.WriteLine(randomNumber.ToString());
if (randomNumber == -1)
break;
}
}
static SortedSet alreadySeen = new SortedSet();
static int NeumannRandom0To100(ref int seed)
{
if (seed == 0 || seed == -1)
return -1;
// jabūt pāra skaitlim ciparu
if (seed.ToString().Length % 2 != 0)
seed /= 10;
unchecked
{
seed *= seed;
}
string newSeed = seed.ToString();
int length = newSeed.Length;
int middleFirst = length / 2; // int truncate division
int generatedNumber = int.Parse(newSeed[middleFirst].ToString() + newSeed[middleFirst + 1].ToString());
return -1;
return generatedNumber;
}
}
}
6. […] built several random number generators: [1], [2], [3], [4], [5], [6], [7], [8], [9] (I didn’t realize it was so many until I went back and looked). In […] | 1,183 | 4,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-30 | longest | en | 0.795249 |
https://community.fabric.microsoft.com/t5/Desktop/Need-help-with-lat-long-coordinates/td-p/1689803 | 1,720,868,883,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514493.69/warc/CC-MAIN-20240713083241-20240713113241-00641.warc.gz | 144,257,039 | 53,524 | cancel
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Anonymous
Not applicable
## Need help with lat/long coordinates
Hi!
I am hoping you can assist me with an issue I am having.
I want to be able to get a number value, such as distance, using longitude and latitude values. So I want to gain insight on driving distance, using calculations based on longitude and latitude. Therefore, I am looking to create an expression that gives me a numeric value (such as distance for example), based on comparing how far the driving distance between two points represented by lat & long are.
Hope this makes sense! I basically want to see driving distance using values from longitude and latitude.
Really hoping you can assist me with this
6 REPLIES 6
Community Support
Hi @Anonymous,
You need Latitude and Longitude to calculate the distance between two locations with following formula:
=acos(sin(lat1)*sin(lat2)+cos(lat1)*cos(lat2)*cos(lon2-lon1))*6371
ps: 6371 is Earth radius in km.
You can refer:
http://www.girlswithpowertools.com/2014/05/distance/
Best Regards,
If this post helps then please consider Accept it as the solution to help the other members find it more quickly.
Anonymous
Not applicable
How can I determine lat1 and lat2 values? The lat values are all in the same column, and Power BI does not allow for chooing individual cells. So how does power bi know which exact values (such as lat1) to work with? So I do not just the whole column..
Community Support
Hi @Anonymous,
The two coordinates to be calculated need to be in two different columns in order to facilitate the calculation of the distance between the coordinates.
Best Regards,
Anonymous
Not applicable
I see, however my column with latitude and column with longitude are 10,455.. so I would have to have separate columns for all of these values?
Community Support
Hi @Anonymous
Sorry, could you provide sample data without sensitive information?
Best Regards,
Super User
Hi @Anonymous ,
You can look into the ArcGIS Map visual which helps in calculating driving distances and driving times.
Do refer the following documentation on this:
https://doc.arcgis.com/en/power-bi/design/find-nearby-locations.htm
Thanks,
Pragati
Best Regards,
Pragati Jain
Did I answer your question? Mark my post as a solution! This will help others on the forum!
Proud to be a Super User!! | 552 | 2,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-30 | latest | en | 0.900039 |
https://jstechroad.com/category/algorithms/tree-traversal/ | 1,723,042,411,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694594.35/warc/CC-MAIN-20240807143134-20240807173134-00025.warc.gz | 261,325,765 | 16,525 | Posts published in “Tree Traversal”
You are given the `root` of a binary tree containing digits from `0` to `9` only.
Each root-to-leaf path in the tree represents a number.
• For example, the root-to-leaf path `1 -> 2 -> 3` represents the number `123`.
Return the total sum of all root-to-leaf numbers.
leaf node is a node with no children.
Example 1:
```Input: root = [1,2,3]
Output: 25
Explanation:
The root-to-leaf path `1->2` represents the number `12`.
The root-to-leaf path `1->3` represents the number `13`.
Therefore, sum = 12 + 13 = `25`.
```
Example 2:
```Input: root = [4,9,0,5,1]
Output: 1026
Explanation:
The root-to-leaf path `4->9->5` represents the number 495.
The root-to-leaf path `4->9->1` represents the number 491.
The root-to-leaf path `4->0` represents the number 40.
Therefore, sum = 495 + 491 + 40 = `1026`.
```
Constraints:
• The number of nodes in the tree is in the range `[1, 1000]`.
• `0 <= Node.val <= 9`
• The depth of the tree will not exceed `10`.
Idea:
Use DFS Template
Solution 1:
``````/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var sumNumbers = function(root) {
if (root === null) return;
let res = [];
let cur = [];
dfs(root, res, cur);
let sum = 0;
for (let num of res) {
sum += num;
}
return sum;
};
function dfs(root, res, cur) {
// exit recursion:
if (root === null) return;
// found leaf, save the number
if (root.left === null && root.right === null) {
cur.push(root.val);
res.push(parseInt(cur.join('')));
cur.pop();
return;
}
// possible solution
if (root !== null) {
cur.push(root.val);
dfs(root.left, res, cur);
dfs(root.right, res, cur);
cur.pop();
}
return;
}``````
Solution 2: (Version 2)
``````var sumNumbers = function(root) {
if (root === null) return;
let res = [0];
let cur = [0];
dfs(root, res, cur);
return res[0];
};
function dfs(root, res, cur) {
// exit recursion:
if (root === null) return;
// found leaf, save the number
if (root.left === null && root.right === null) {
cur[0] = cur[0] * 10 + root.val;
res[0] += cur[0];
cur[0] = (cur[0] - root.val) / 10;
return;
}
// possible solution
if (root !== null) {
cur[0] = cur[0] * 10 + root.val;
dfs(root.left, res, cur);
dfs(root.right, res, cur);
cur[0] = (cur[0] - root.val) / 10;
}
return;
}``````
Given the `root` of a binary tree and an integer `targetSum`, return `true` if the tree has a root-to-leaf path such that adding up all the values along the path equals `targetSum`.
leaf is a node with no children.
Example 1:
```Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
```
Example 2:
```Input: root = [1,2,3], targetSum = 5
Output: false
```
Example 3:
```Input: root = [1,2], targetSum = 0
Output: false
```
Constraints:
• The number of nodes in the tree is in the range `[0, 5000]`.
• `-1000 <= Node.val <= 1000`
• `-1000 <= targetSum <= 1000`
Solution: (Recursion)
• Time complexity: O(n)
• Space complexity: O(n)
``````/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @param {number} targetSum
* @return {boolean}
*/
var hasPathSum = function(root, targetSum) {
if (root === null) return false;
if (root.val === targetSum && root.left === null && root.right === null) return true;
return hasPathSum(root.left, targetSum - root.val) ||
hasPathSum(root.right, targetSum - root.val);
};``````
Given the `root` of a binary tree, return the postorder traversal of its nodes’ values.
Example 1:
```Input: root = [1,null,2,3]
Output: [3,2,1]
```
Example 2:
```Input: root = []
Output: []
```
Example 3:
```Input: root = [1]
Output: [1]
```
Example 4:
```Input: root = [1,2]
Output: [2,1]
```
Example 5:
```Input: root = [1,null,2]
Output: [2,1]
```
Constraints:
• The number of the nodes in the tree is in the range `[0, 100]`.
• `-100 <= Node.val <= 100`
Idea:
Use two stacks:
```1. Push root to first stack.
2. Loop while first stack is not empty
2.1 Pop a node from first stack and push it to second stack(res[])
2.2 Push left and right children of the popped node to first stack
3. Pop all elements from second stack(res.reverse())```
Solution:
``````/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var postorderTraversal = function(root) {
let res = []; // use res as another stack
// empty tree case:
if (root === null) return res;
let stack = [];
// postorder visit: left, right, root
stack.push(root);
while (stack.length !== 0) {
let cur = stack.pop();
// treat it as stack, store them in reverse order:
// i.e root, left, right
res.push(cur.val);
if (cur.left) stack.push(cur.left);
if (cur.right) stack.push(cur.right);
}
// we can pop all elements one by one, or just reverse them
return res.reverse();
};``````
Given the `root` of a binary tree, return the preorder traversal of its nodes’ values.
Example 1:
```Input: root = [1,null,2,3]
Output: [1,2,3]
```
Example 2:
```Input: root = []
Output: []
```
Example 3:
```Input: root = [1]
Output: [1]
```
Example 4:
```Input: root = [1,2]
Output: [1,2]
```
Example 5:
```Input: root = [1,null,2]
Output: [1,2]
```
Constraints:
• The number of nodes in the tree is in the range `[0, 100]`.
• `-100 <= Node.val <= 100`
Idea:
• Create an empty stack and push root node to stack
• Do the following while stack is not empty:
1. Pop an item from the stack and store to result array
2. Push right child of a popped item to stack
3. Push left child of a popped item to stack
Right child is pushed first: The right child is pushed before the left child to make sure that the left subtree is processed first.
Solution:
``````/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var preorderTraversal = function(root) {
let res = [];
// empty tree case:
if (root === null) return res;
let stack = [];
// preorder visit: root -> left -> right
// so the right child is pushed before the left child
// to make sure that the left subtree is processed first.
stack.push(root);
while (stack.length !== 0) {
let cur = stack.pop();
res.push(cur.val);
if (cur.right) stack.push(cur.right);
if (cur.left) stack.push(cur.left);
}
return res;
};``````
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as:
a binary tree in which the left and right subtrees of every node differ in height by no more than 1.
Example 1:
```Input: root = [3,9,20,null,null,15,7]
Output: true
```
Example 2:
```Input: root = [1,2,2,3,3,null,null,4,4]
Output: false
```
Example 3:
```Input: root = []
Output: true
```
Constraints:
• The number of nodes in the tree is in the range `[0, 5000]`.
• `-104 <= Node.val <= 104`
Solution:
``````/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
var isBalanced = function(root) {
if (root === null) return true;
const res = [true];
height(root, res);
return res[0];
};
function height(root, res) {
if (root === null) return 0;
const l = height(root.left, res);
const r = height(root.right, res);
if (Math.abs(l - r) > 1) res[0] = false;
return Math.max(l, r) + 1;
}``````
Given the `root` of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level).
Example 1:
```Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]
```
Example 2:
```Input: root = [1]
Output: [[1]]
```
Example 3:
```Input: root = []
Output: []
```
Constraints:
• The number of nodes in the tree is in the range `[0, 2000]`.
• `-1000 <= Node.val <= 1000`
Idea:
Use BFS Template
Note: Don’t forget root === null case
Solution:
``````/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[][]}
*/
var levelOrder = function(root) {
const res = [];
if (root === null) return res;
const queue = [];
queue.push(root);
while(queue.length !== 0) {
let size = queue.length;
let level = []; // store same level nodes
while(size--) {
let cur = queue.shift();
level.push(cur.val);
if (cur.left) queue.push(cur.left);
if (cur.right) queue.push(cur.right);
}
res.push(level.concat());
}
return res;
};``````
Given the `root` of a binary tree, return the inorder traversal of its nodes’ values.
Example 1:
```Input: root = [1,null,2,3]
Output: [1,3,2]
```
Example 2:
```Input: root = []
Output: []
```
Example 3:
```Input: root = [1]
Output: [1]
```
Example 4:
```Input: root = [1,2]
Output: [2,1]
```
Example 5:
```Input: root = [1,null,2]
Output: [1,2]
```
Constraints:
• The number of nodes in the tree is in the range `[0, 100]`.
• `-100 <= Node.val <= 100`
Idea:
Iterating method using Stack
Solution:
``````/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var inorderTraversal = function(root) {
let stack = [];
let res = [];
while(root !== null || stack.length !== 0) {
while(root !== null) {
stack.push(root);
root = root.left;
}
root = stack.pop();
res.push(root.val);
root = root.right;
}
return res;
}`````` | 3,137 | 10,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-33 | latest | en | 0.464561 |
https://arstechnica.com/science/2020/02/flat-surfaces-surf-past-each-other-on-the-peak-of-a-wave/?comments=1 | 1,618,501,414,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038085599.55/warc/CC-MAIN-20210415125840-20210415155840-00253.warc.gz | 194,092,769 | 27,329 | # Flat surfaces surf past each other on the peak of a wave
## Simple model describes the transition from static to dynamic friction.
A cook's reaction to cockroaches in the kitchen is a good approximation of a physicist's reaction to friction: not only is it undesirable, it is hard to get rid of. OK, our analogy falls at a last hurdle: unlike cooks, most physicists don't really want to understand friction.
That said, there are a few strong-stomached physicists who are trying to understand friction. One of the problems they face is that friction is so… individual. Every experiment is different. Even copies of the same experiment are slightly different. Despite that, one group has now managed to come up with a general model that replicates many of the main features of friction.
## Friction, how I stab at thee
Actually, part of that last sentence is a lie. The model focuses on the transition between two different types of friction. I'll get to that in a moment.
When we look at friction in detail, the reasons we find describing it difficult become clear. Imagine two plates being slid across each other. At one level, we can say that friction comes from the force of attraction between the two plates due to local imbalances in electron density. In other words, the positively charged bits of one plate are attracted to the negatively charged bits of the other plate. So far, so mind-numbing.
But the strength of the attraction between the plates depends on all sorts of details about the two surfaces: the materials, the roughness of the surface, the conductivity of the surface, the separation between the surfaces, and many more things. To model this in detail digs a big painful hole of code and buries the physics at the bottom of it.
Experimentally, it is even worse. If we were to repeat the experiment twice using the same plates, we are likely to get a different result. Why? Because the first experiment would have modified the surface, so the second experiment is conducted under different conditions. It is enough to make you despair.
## Friction is dynamic
Now, let's make friction even trickier by starting our experiment from rest. The plates aren't moving, and we apply a force… and more force… and more force. Suddenly, the plates break free and start to slide. The initial friction is much larger than the friction while the plate is moving. These two frictions are called static and dynamic friction, and the transition between the two is rather more complex than I have just described. You would think that any model that seeks to replicate even the most basic features would be complicated. And you would be wrong.
To understand the transition between static and dynamic friction, a group of researchers created a breathtakingly simple model. One surface is perfectly flat. A series of wheels is placed on the surface so that all the wheels are in contact with each other. The wheels can slide due to an applied force, and they can rotate due to a torque. If one wheel does anything, it is in contact with the next, so the forces and torques travel down the chain.
The wheels are held to the surface by a force that is perpendicular to the surface. Friction is described by a single number. No complicated physics for us, thank you very much. The game the researchers play is to push on one of the wheels and apply Newton's laws repeatedly until the wheels' motion stabilizes. Then they do it again for a different force… and again.
## Setting the wheels in motion
As with reality, the researchers observe a change from static (high) friction to dynamic friction once the force gets above a certain threshold. Examining the dynamics of the wheels more closely, they show that the transition to dynamic friction occurs because the applied force creates a wave that propagates stably along the wheels. That wave allows motion along the surface with far less force than if there is no wave.
The appearance of the wave is presaged by behavior observed for forces below the threshold. For smaller forces, similar waves are generated, but they quickly die out as they propagate, preventing the transition to easy motion. The researchers show all this with a beautiful numerical analysis of the dynamical system.
The model does a remarkably good job of showing the generally observed characteristics of the transition from static to dynamic friction. The model has, as the authors point out, the advantage that you can dig right into the workings and see exactly what is going on. That's because the model is not something that you can use to calculate real frictional forces for realistic situations; instead, it is a tool for discovery.
Physical Review Journals, 2020, DOI: 10.1103/PhysRevLett.124.030602 (About DOIs)
## 51 Reader Comments
1. Very cool. I always thought the ability to "wring" exceedingly flat metallic surfaces like gauge blocks together and have them stick as though they were glued bordered on the realm of magic. Surface interactions at the micro-scale are one of those mind-blowing things that we blithely use everyday, blissfully unaware of the amazing and barely understood stuff going on out of sight.
2. So in this model, are the experimenters measuring differences in rotation (and change in rotation) across the wheels?
3. Do we get small enough gaps between the two surfaces that in places the Casimir effect is of importance? It would be cool to think that part of my car's gas consumption is due to quantum effects of virtual particle formation in the vacuum. I said cool, not necessarily useful.
4. I'm still not clear on the execution of the experiment.
"One surface is perfectly flat. A series of wheels is placed on the surface so that all the wheels are in contact with each other. The wheels can slide due to an applied force, and they can rotate due to a torque. If one wheel does anything, it is contact with the next, so the forces and torques travel down the chain.
The wheels are held to the surface by a force that is perpendicular to the surface. Friction is described by a single number. No complicated physics for us, thank you very much. The game the researchers play is to push on one of the wheels and apply Newton's laws repeatedly until the wheels' motion stabilizes. Then they do it again for a different force… and again."
How are the wheels kept in contact with each other?
How and where is the force applied to the <first> wheel?
Is it in the direction of potential rotation or perpendicular so it can't really rotate?
Is the force applied directly to the wheel in the direction of potential rotation at the exact mid point of the wheel between top and bottom or below that midpoint where it is likely to push the wheel so that it skids instead of rotating?
I get that the intent here is not to provide the necessary steps to recreate the experiment but I'm having a hard time visualizing how it was constructed.
5. Quote:
The model has, as the authors point out, the advantage that you can dig right into the workings and see exactly what is going on.
But the inner workings of the model don't match the actual situation of two rough (at some level) surfaces on contact, so there's no reason to expect any new detailed behavior to match.
Plus, they still have to model friction in their model as just some number they put in. I'm not seeing how modeling a bunch of wheels plus a single friction coefficient is better than just using two coefficients.
6. ZenBeam wrote:
Plus, they still have to model friction in their model as just some number they put in. I'm not seeing how modeling a bunch of wheels plus a single friction coefficient is better than just using two coefficients.
This has something to do with Occam's razor.
A model using one variable to describe a set of behaviours is better than two models each have one variable to describe two subsets of the same behaviours. For example the better one more adequately describes those sets of behaviours which are outside of the scope of the two other models.
7. Quote:
A series of wheels is placed on the surface so that all the wheels are in contact with each other. The wheels can slide due to an applied force, and they can rotate due to a torque. If one wheel does anything, it is contact with the next, so the forces and torques travel down the chain.
Something is missing in this picture that makes it very hard to visualize. A series of wheels placed in contact with each other cannot rotate, period (try drawing a picture and seeing how the wheels would rotate), since the touching points on the wheels would have to move linearly in opposite directions for the wheels to rotate in the same direction. Unless somehow a friction between the wheels at their contact point is competing against the friction between the wheels and the surface, but then that just seems to devolve the modeling of friction down a level, essentially circularly (pun intended).
So it's not clear exactly how the wheels are coupled to each other.
edit: fixed "must cannot"! If only that was possible...
8. See the diagrams on how the balls are arranged and the forces. This really should be in the article!
https://doi.org/10.1103/PhysRevLett.124.030602
9. Quote:
A series of wheels is placed on the surface so that all the wheels are in contact with each other. The wheels can slide due to an applied force, and they can rotate due to a torque. If one wheel does anything, it is contact with the next, so the forces and torques travel down the chain.
Something is missing in this picture that makes it very hard to visualize. A series of wheels placed in contact with each other cannot rotate, period (try drawing a picture and seeing how the wheels would rotate), since the touching points on the wheels would have to move linearly in opposite directions for the wheels to rotate in the same direction. Unless somehow a friction between the wheels at their contact point is competing against the friction between the wheels and the surface, but then that just seems to devolve the modeling of friction down a level, essentially circularly (pun intended).
So it's not clear exactly how the wheels are coupled to each other.
edit: fixed "must cannot"! If only that was possible...
This reminds me of one of Ponder Stibbons's remarks in one of the Pratchett books - something like "that is an analogy which is very illuminating but is completely wrong."
10. I love Ars... and articles like this from Chris are among the best examples of why I read it.
11. Quote:
A series of wheels is placed on the surface so that all the wheels are in contact with each other. The wheels can slide due to an applied force, and they can rotate due to a torque. If one wheel does anything, it is contact with the next, so the forces and torques travel down the chain.
Something is missing in this picture that makes it very hard to visualize. A series of wheels placed in contact with each other cannot rotate, period (try drawing a picture and seeing how the wheels would rotate), since the touching points on the wheels would have to move linearly in opposite directions for the wheels to rotate in the same direction. Unless somehow a friction between the wheels at their contact point is competing against the friction between the wheels and the surface, but then that just seems to devolve the modeling of friction down a level, essentially circularly (pun intended).
So it's not clear exactly how the wheels are coupled to each other.
edit: fixed "must cannot"! If only that was possible...
This reminds me of one of Ponder Stibbons's remarks in one of the Pratchett books - something like "that is an analogy which is very illuminating but is completely wrong."
I think that quote has force for the same reason that Pauli is still quoted many years later for sadly saying that a theory was so bad that 'It is not even wrong' (and hence the name of the Not Even Wrong blog). Because a wrong theory can have some redeeming value in terms of analogies or illumination of the constraints on a correct theory, but if a theory is so bad it doesn't even do that....
BTW I'm not saying that applies to this case, it may well be I simply don't understand it.
12. I haven't (yet) read the PRL article, but I did write a thesis on friction -- lo these many years ago.
My thesis adviser was always dead set against these kinds of models. He disparaged them as "Friction as a component of friction." In the present model, that would correspond to the wheels being locked together.
For my interests (earthquakes), the only behavior of interest would be, if you coupled this friction model to a block attached to a spring, and loaded the spring, would the "stick slip" naturally repeat itself, or would it be a "one and done" phenomenon. If the latter, it is not a good model for seismicity.
Any geophysicists looking for a MS thesis out there? Put this into such a spring-block model and run it...
13. So... cooks want to understand cockroaches?
14. R4sterMan wrote:
See the diagrams on how the balls are arranged and the forces. This really should be in the article!
https://doi.org/10.1103/PhysRevLett.124.030602
This is the one thing Chris' articles could use, more diagrams. (And puns, we could always use more puns)
15. Quote:
As with reality, the researchers observe a change from static (high) friction to dynamic friction once the force gets above a certain threshold.
IIRC, that threshold is called the coefficient of friction; the necessary amount of force needed to go from static friction to dynamic friction.
f = \mu N
f = friction force
\mu = coefficient of friction
N = normal force
For folks who may not get friction, imagine driving. Your tires against the road are in a STATIC state. Yes, the tires are spinning, but the parts that touch the road remain in place against the road. The spinning part is an illusion of dynamic motion. Like the treads on a tank, the vehicle is propelled forward by the motion of the tires going backward on the ground relative to the direction of travel.
Slam on the brakes (and no anti-lock here for the purposes of demonstration) and all of a sudden, the tires stop spinning, but the car keeps going. The inertia of the car is applying the force that changed the state from a static to a dynamic friction. Once the normal force from inertia was lost, the tires return to a static state, albeit not spinning.
Mash the accelerator, and all of a sudden, your engine provides normal power in excess of the coefficient of friction and induces another dynamic friction state as the rear tires spin against the pavement. The front ones, being still in a static state, begin to roll. Eventually, the rear tires stop spinning relative to their position on the pavement, the smoke ends and the rear tires enter the same static friction state as the front tires, albeit with them moving along with the car, but not relative to where the tread meets the road.
The force came from inertia to stop, and from the engine to go. In both cases, rubber was lost. The force was applied over the amount of time it took to make the tires stop skidding (in either direction)
The coefficient of friction for rubber on dry pavement is actually fairly high. Throw in some water, and that coefficient drops like a stone because the water molecules begin to separate the rubber from the pavement, and water has an exceptionally low coefficient of friction.
Oddly, friction was one of the areas I got in the few physics classes I took. Then I switched to chemistry/biochemistry, because I thought THAT was easier (It was, for me). It's good to see Chris in his element, after he took an interesting stab at biochemistry in another Ars article today.
Thanks for the article(s), Chris! Friction was fun for me, but the physics department underwent a radical change in management/curricula which, unknown to me, messed up the reqs for all the classes. So, I decided that if real life was that messy, I may as well study the mess, than make one of my own life pursuing a subject my school screwed up at a critical time for me. Turns out, I was pretty good at it back in the day. But I do wonder where I'd have gone if the school hadn't been such a dick about the changes.
16. m4gw4s wrote:
ZenBeam wrote:
Plus, they still have to model friction in their model as just some number they put in. I'm not seeing how modeling a bunch of wheels plus a single friction coefficient is better than just using two coefficients.
This has something to do with Occam's razor.
A model using one variable to describe a set of behaviours is better than two models each have one variable to describe two subsets of the same behaviours. For example the better one more adequately describes those sets of behaviours which are outside of the scope of the two other models.
They still have more than one variable, though. Otherwise they would never be able to get different ratios of static and dynamic friction. Maybe they choose the wheel size or equivalently how many wheels per unit length to use to get them both to match. But somehow, they still have as many variables to select, so it's not a simpler model.
17. More research like this and spherical cows may become an endangered species.
18. More research like this and spherical cows may become an endangered species.
I don't know, we could end up with models saying "Approximate a walking cow as a series of wheels contacting each other". Interesting, but similarly unrealistic.
19. R4sterMan wrote:
See the diagrams on how the balls are arranged and the forces. This really should be in the article!
https://doi.org/10.1103/PhysRevLett.124.030602
This is the one thing Chris' articles could use, more diagrams. (And puns, we could always use more puns)
No, that would spoil it. Unless of course they were hidden under spoilers. I treat the articles and their DOI links like a cryptic crossword which provides an evening's entertainment.
You may read this as a joke but it is not intended as such.
20. Quote:
A series of wheels is placed on the surface so that all the wheels are in contact with each other. The wheels can slide due to an applied force, and they can rotate due to a torque. If one wheel does anything, it is contact with the next, so the forces and torques travel down the chain.
Something is missing in this picture that makes it very hard to visualize. A series of wheels placed in contact with each other cannot rotate, period (try drawing a picture and seeing how the wheels would rotate), since the touching points on the wheels would have to move linearly in opposite directions for the wheels to rotate in the same direction. Unless somehow a friction between the wheels at their contact point is competing against the friction between the wheels and the surface, but then that just seems to devolve the modeling of friction down a level, essentially circularly (pun intended).
So it's not clear exactly how the wheels are coupled to each other.
edit: fixed "must cannot"! If only that was possible...
This reminds me of one of Ponder Stibbons's remarks in one of the Pratchett books - something like "that is an analogy which is very illuminating but is completely wrong."
I think that quote has force for the same reason that Pauli is still quoted many years later for sadly saying that a theory was so bad that 'It is not even wrong' (and hence the name of the Not Even Wrong blog). Because a wrong theory can have some redeeming value in terms of analogies or illumination of the constraints on a correct theory, but if a theory is so bad it doesn't even do that....
BTW I'm not saying that applies to this case, it may well be I simply don't understand it.
I took Pauli's meaning as being that there was so little content (or so little connection to observations) that one could not point to an actual error. "42" is an example (as the ultimate solution).
21. Well, here's the image from the paper showing how the wheels are laid out:
This still leaves me with more questions. So if the wheels skid, then there are 10 points of friction? Or 20 counting the force from above (no axles are shown)? If the wheels rotate, then are there 9 slipping points between the 10 wheels, maybe as well as additional sources of friction from where force is exerted (from above and the side)?
22. Fatesrider wrote:
For folks who may not get friction, imagine driving. Your tires against the road are in a STATIC state. Yes, the tires are spinning, but the parts that touch the road remain in place against the road. The spinning part is an illusion of dynamic motion. Like the treads on a tank, the vehicle is propelled forward by the motion of the tires going backward on the ground relative to the direction of travel.
Actually I believe there is always a degree of slip between a tyre and the road when the vehicle is moving. This is because the weight of the vehicle is being supported by the contact patches, the rubber is flexible, and so as a tiny piece of the tyre reaches the end of the contact patch it is subjected to both a reduced upward force and a reduced longitudinal force, rather than the pure up/down force that would imply no slip.
Given that tyres have internal friction as well as friction between them and the road, which is why they get warm in use, a tyre is perhaps the most complicated example of frictional effects that one can readily think of*.
Spoiler: show
*well, I can think of an even more complicated one that a lot of people are familiar with, but discussion of it is probably unsuitable for a family publication.
23. Quote:
A series of wheels is placed on the surface so that all the wheels are in contact with each other. The wheels can slide due to an applied force, and they can rotate due to a torque. If one wheel does anything, it is contact with the next, so the forces and torques travel down the chain.
Something is missing in this picture that makes it very hard to visualize. A series of wheels placed in contact with each other cannot rotate, period (try drawing a picture and seeing how the wheels would rotate), since the touching points on the wheels would have to move linearly in opposite directions for the wheels to rotate in the same direction. Unless somehow a friction between the wheels at their contact point is competing against the friction between the wheels and the surface, but then that just seems to devolve the modeling of friction down a level, essentially circularly (pun intended).
So it's not clear exactly how the wheels are coupled to each other.
edit: fixed "must cannot"! If only that was possible...
It's not a lab experiment; it's a model.
The wheels are coupled "mathematically".
24. Quote:
A series of wheels is placed on the surface so that all the wheels are in contact with each other. The wheels can slide due to an applied force, and they can rotate due to a torque. If one wheel does anything, it is contact with the next, so the forces and torques travel down the chain.
Something is missing in this picture that makes it very hard to visualize. A series of wheels placed in contact with each other cannot rotate, period (try drawing a picture and seeing how the wheels would rotate), since the touching points on the wheels would have to move linearly in opposite directions for the wheels to rotate in the same direction. Unless somehow a friction between the wheels at their contact point is competing against the friction between the wheels and the surface, but then that just seems to devolve the modeling of friction down a level, essentially circularly (pun intended).
So it's not clear exactly how the wheels are coupled to each other.
edit: fixed "must cannot"! If only that was possible...
It's not a lab experiment; it's a model.
The wheels are coupled "mathematically".
Mathematical coupling, the cleanest type.....
25. Fatesrider wrote:
For folks who may not get friction, imagine driving. Your tires against the road are in a STATIC state.
It's not static because the rubber stretches continuously as the weight of the vehicle shifts over the section of tyre. It's basically sliding all the time and that's why tyres wear out.
Also strangely tyres have the most grip when they their rotation speed is offset by around 10% compared to the speed of the vehicle. In other words — there is more friction when they're sliding. Until they slide more than about 10%, and then there's less friction. Why? There are many theories, but I don't think I've ever heard one that sounds "right".
And some things are just bizarre, for example UV has an affect. Race cars have more cornering grip whenever a cloud covers the sun. And the extra grip vanishes instantly when the sun emerges from behind the cloud. The obvious answer would be heat, but the temperature change from cloud cover is very slow and not large enough to explain the drastic change in friction that occurs whenever UV is applied or removed from the track. It doesn't make sense, but it's clearly a thing and can make the difference between first or last.
Billions of dollars are poured into researching wheel friction, and that's not enough for anyone to understand the fundamentals. All they have to work with are observations, of you do X you get Y. Therefore we do X even though we don't understand it.
26. Fatesrider wrote:
For folks who may not get friction, imagine driving. Your tires against the road are in a STATIC state. Yes, the tires are spinning, but the parts that touch the road remain in place against the road. The spinning part is an illusion of dynamic motion. Like the treads on a tank, the vehicle is propelled forward by the motion of the tires going backward on the ground relative to the direction of travel.
Actually I believe there is always a degree of slip between a tyre and the road when the vehicle is moving. This is because the weight of the vehicle is being supported by the contact patches, the rubber is flexible, and so as a tiny piece of the tyre reaches the end of the contact patch it is subjected to both a reduced upward force and a reduced longitudinal force, rather than the pure up/down force that would imply no slip.
Given that tyres have internal friction as well as friction between them and the road, which is why they get warm in use, a tyre is perhaps the most complicated example of frictional effects that one can readily think of*.
Spoiler: show
*well, I can think of an even more complicated one that a lot of people are familiar with, but discussion of it is probably unsuitable for a family publication.
rug burn?
27. I thought friction was caused by small imperfections stopping things from moving.
Apply enough force an a combination of the following happens:
a) The imperfections are broken off
b) The surfaces are forced apart enough so that the imperfections are no longer in contact.
With the result that they can move parallel to each other.That is why adding a liquid helps - it fills in the imperfections.
Maybe I am just unusually dense today. Am I missing something? Because all this talk of wheels has me thoroughly confused.
28. I love Ars... and articles like this from Chris are among the best examples of why I read it.
I have to disagree here... The recent article on the Apollo lander computer was a great Ars article, but this one frustrated me by coming so close to being scientifically interesting, or providing insight, while not actually being illuminating or helpful in describing the work on friction. Several not terribly useful analogies about cockroaches and wheels, but frustratingly little interpretive insight.
Chris, I challenge you to do better here... The last three paragraphs get into what the paper is trying to describe, but I think there's room for improvement in the narrative...
29. It's great to see an article on friction here on Ars. It's a major theme of mine at work (automotive assembly, particularly threaded bolts and fittings) and has been a key factor in most of my main successes and failures. I work closely with the institute of tribology of a nearby university, where the team, right up to the lead professor, is really focussing on experimentation rather than modelling. They haven't completely given up on modelling, but it appears to be massively tricky, particularly once you start adding coatings to the dry surfaces modelled in the paper described in this article.
The balance of tribology and rheology (how things flow) looks to be a true wicked problem - which of course the lab doesn't mind, as it keeps them in a job ;-)
30. Fatesrider wrote:
For folks who may not get friction, imagine driving. Your tires against the road are in a STATIC state.
It's not static because the rubber stretches continuously as the weight of the vehicle shifts over the section of tyre. It's basically sliding all the time and that's why tyres wear out.
Also strangely tyres have the most grip when they their rotation speed is offset by around 10% compared to the speed of the vehicle. In other words — there is more friction when they're sliding. Until they slide more than about 10%, and then there's less friction. Why? There are many theories, but I don't think I've ever heard one that sounds "right".
And some things are just bizarre, for example UV has an affect. Race cars have more cornering grip whenever a cloud covers the sun. And the extra grip vanishes instantly when the sun emerges from behind the cloud. The obvious answer would be heat, but the temperature change from cloud cover is very slow and not large enough to explain the drastic change in friction that occurs whenever UV is applied or removed from the track. It doesn't make sense, but it's clearly a thing and can make the difference between first or last.
Billions of dollars are poured into researching wheel friction, and that's not enough for anyone to understand the fundamentals. All they have to work with are observations, of you do X you get Y. Therefore we do X even though we don't understand it.
Time to put UV lights under the race cars?
31. "Nice? Disk? Oh, very!"
Even though, or especially, since lots of real world details are slipped over ...
Last edited by Torbjörn Larsson, OM on Mon Feb 03, 2020 7:26 am
32. I love Ars... and articles like this from Chris are among the best examples of why I read it.
I don’t mean to be harsh. Generally and even in the article Lee does a good job of simplifying conplexity. But the first few paragraphs of this article are a mess. Do cooks want to understand cockroaches or was that a joke? Wait what part of what sentence was a lie? Of course you will clarify later. That’s how all articles work.
Or maybe that is a style folks like and I am the outlier.
33. I love Ars... and articles like this from Chris are among the best examples of why I read it.
I don’t mean to be harsh. Generally and even in the article Lee does a good job of simplifying conplexity. But the first few paragraphs of this article are a mess. Do cooks want to understand cockroaches or was that a joke? Wait what part of what sentence was a lie? Of course you will clarify later. That’s how all articles work.
Or maybe that is a style folks like and I am the outlier.
[Insert joke about the problems of modeling humor here.]
34. I love Ars... and articles like this from Chris are among the best examples of why I read it.
I don’t mean to be harsh. Generally and even in the article Lee does a good job of simplifying conplexity. But the first few paragraphs of this article are a mess. Do cooks want to understand cockroaches or was that a joke? Wait what part of what sentence was a lie? Of course you will clarify later. That’s how all articles work.
Or maybe that is a style folks like and I am the outlier.
[Insert joke about the problems of modeling humor here.]
Perhaps humor could be modeled as a series of wheels.
35. And this, folks, is why racing drivers are paid the big bucks to wring every last newton of grip out...
This has caught out several game developers, too - I can remember the developers of Live For Speed disappearing into a big black hole for months/years figuring out a more realistic tire model. Actually, I don't know if they ever emerged from that black hole - at last check they were improving the rendering of lighting and other easier fixes.
36. And some things are just bizarre, for example UV has an affect. Race cars have more cornering grip whenever a cloud covers the sun. And the extra grip vanishes instantly when the sun emerges from behind the cloud.
From the article I take that friction has something to do with unevenly distributed electrons in the surfaces. As UV light can "free" electrons (photoelectric effect), the connection seems to be right there?!
37. I didn't know letting my foot off the brake and stepping on the accelerator involved debatable physics. Now, I'm afraid to drive.
38. Fatesrider wrote:
For folks who may not get friction, imagine driving. Your tires against the road are in a STATIC state.
It's not static because the rubber stretches continuously as the weight of the vehicle shifts over the section of tyre. It's basically sliding all the time and that's why tyres wear out.
Also strangely tyres have the most grip when they their rotation speed is offset by around 10% compared to the speed of the vehicle. In other words — there is more friction when they're sliding. Until they slide more than about 10%, and then there's less friction. Why? There are many theories, but I don't think I've ever heard one that sounds "right".
And some things are just bizarre, for example UV has an affect. Race cars have more cornering grip whenever a cloud covers the sun. And the extra grip vanishes instantly when the sun emerges from behind the cloud. The obvious answer would be heat, but the temperature change from cloud cover is very slow and not large enough to explain the drastic change in friction that occurs whenever UV is applied or removed from the track. It doesn't make sense, but it's clearly a thing and can make the difference between first or last.
Billions of dollars are poured into researching wheel friction, and that's not enough for anyone to understand the fundamentals. All they have to work with are observations, of you do X you get Y. Therefore we do X even though we don't understand it.
So what you're saying is that those street racers with blacklight underglow were actually getting a real performance advantage?
39. The Article wrote:
A cook's reaction to cockroaches in the kitchen is a good approximation of a physicist's reaction to friction: not only is it undesirable, it is hard to get rid of. OK, our analogy falls at a last hurdle: unlike cooks, most physicists don't really want to understand friction.
What on earth is this first paragraph? What purpose or value does it add to the article?
Even more confounding when it’s lacking a sensible description of the experiment - thankfully other commenters have taken it upon themselves to add the diagram.
You must to comment. | 7,391 | 35,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-17 | latest | en | 0.964883 |
https://www.grc.nasa.gov/www/K-12/airplane/sound.html | 1,603,142,303,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107866404.1/warc/CC-MAIN-20201019203523-20201019233523-00116.warc.gz | 733,570,898 | 6,458 | + Text Only Site
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This page is intended for college, high school, or middle school students. For younger students, a simpler explanation of the information on this page is available on the Kid's Page.
Air is a gas, and a very important property of any gas is the speed of sound through the gas. Why are we interested in the speed of sound? The speed of "sound" is actually the speed of transmission of a small disturbance through a medium. Sound itself is a sensation created in the human brain in response to sensory inputs from the inner ear. (We won't comment on the old "tree falling in a forest" discussion!)
Disturbances are transmitted through a gas as a result of collisions between the randomly moving molecules in the gas. The transmission of a small disturbance through a gas is an isentropic process. The conditions in the gas are the same before and after the disturbance passes through. Because the speed of transmission depends on molecular collisions, the speed of sound depends on the state of the gas. The speed of sound is a constant within a given gas and the value of the constant depends on the type of gas (air, pure oxygen, carbon dioxide, etc.) and the temperature of the gas. An analysis based on conservation of mass and momentum shows that the speed of sound a is equal to the square root of the ratio of specific heats g times the gas constant R times the temperature T.
a = sqrt [g * R * T]
Notice that the temperature must be specified on an absolute scale (Kelvin or Rankine). The dependence on the type of gas is included in the gas constant R. which equals the universal gas constant divided by the molecular weight of the gas, and the ratio of specific heats.
The speed of sound in air depends on the type of gas and the temperature of the gas. On Earth, the atmosphere is composed of mostly diatomic nitrogen and oxygen, and the temperature depends on the altitude in a rather complex way. Scientists and engineers have created a mathematical model of the atmosphere to help them account for the changing effects of temperature with altitude. Mars also has an atmosphere composed of mostly carbon dioxide. There is a similar mathematical model of the Martian atmosphere. We have created an atmospheric calculator to let you study the variation of sound speed with planet and altitude.
Here's a JavaScript program to calculate speed of sound and Mach number for different planets, altitudes, and speed. You can use this calculator to determine the Mach number of a rocket at a given speed and altitude on Earth or Mars.
# Mach and Speed of Sound Calculator
Input
Press->
Output
Speed
Speed of Sound
Mach
To change input values, click on the input box (black on white), backspace over the input value, type in your new value, and hit the Enter key on the keyboard (this sends your new value to the program). You will see the output boxes (yellow on black) change value. You can use either English or Metric units and you can input either the Mach number or the speed by using the menu buttons. Just click on the menu button and click on your selection. There is a sleek version of this program for experienced users who do not need these instructions.
You can also download your own copy of this program to run off-line by clicking on this button:
As an object moves through the atmosphere, the air is disturbed and the disturbances are transmitted through the air at the speed of sound. You can study how the disturbances are transmitted with an interactive sound wave simulator. If we consider the atmosphere on a standard day at sea level static conditions, the speed of sound is about 761 mph, or 1100 feet/second. We can use this knowledge to approximately determine how far away a lightning strike has occurred.
The speed of sound in the atmosphere is a constant that depends on the altitude, but an aircraft can move through the air at any desired speed. The ratio of the aircraft's speed to the speed of sound affects the forces on the aircraft. Aeronautical engineers call the ratio of the aircraft's speed to the speed of sound the Mach number, M. If the aircraft moves much slower than the speed of sound, conditions are said to be subsonic, 0 < M << 1, and compressibility effects are small and can be neglected. If the aircraft moves near the speed of sound, conditions are said to be transonic, M ~ 1, and compressibility effects like flow choking become very important. For aircraft speeds greater than the speed of sound, conditions are said to be supersonic, 1 < M < 3, and compressibility effects are important. Depending on the specific shape and speed of the aircraft, shock waves may be produced in the supersonic flow of a gas. For high supersonic speeds, 3 < M < 5, aerodynamic heating becomes very important. If the aircraft moves more than five times the speed of sound, conditions are said to be hypersonic, M > 5, and the high energy involved under these conditions has significant effects on the air itself. The Space Shuttle re-enters the atmosphere at high hypersonic speeds, M ~ 25. Under these conditions, the heated air becomes an ionized plasma of gas and the spacecraft must be insulated from the high temperatures.
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https://hubpages.com/education/Microwave-Hyper-Quanta-Faster-Than-Light | 1,582,692,595,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146186.62/warc/CC-MAIN-20200226023658-20200226053658-00500.warc.gz | 411,174,896 | 35,117 | Microwave Hyper Quanta: Faster Than Light
Updated on April 12, 2015
Can You Go Faster Than Speed of Light?
Microwave Hyper Quanta: Faster Than Light!
Just as a photon is the packet of information (frequency) received at the end of a ray of variable distance (wavelength), so too is a tachyon, tachyon - A hypothetical particle that travels faster than the speed of light (and therefore also travels back in time). The existence of tachyons is allowed by the equations of Einstein's special theory of relativity. that point of extension of an electromagnetic microwave moving faster than a photon. To call a tachyon electromagnetic would be similar to calling a photon particle the wavelength of a light ray. The spin of the photon is equal to the information of the wavelength light ray, and the spin of a tachyon is equal to the information of the microwave length.
The electromagnetivity of the microwave length itself is contingent upon the concept of it bearing a charge arising as a result of its frequency’s fluctuation, and therefore a magnetic field arranged perpendicularly to the flow of the electrical current. Thus to call the microwaves electromagnetic energy depends on whether or not they carry an electrical charge that can produce magnetic polarity, which is indicated by their structure as a wavelength, similar to the known wavelengths, waveforms and fields of the electromagnetic spectrum.
The microwave lengths are smaller than the point of charge of the electron when it is singularized, because they comprise the geometric vibrational framework underlying the electron’s different possible orbital shell diameters. By establishing orbital shells for the charge to occupy as potential, microwaves act like gravity. Therefore, while the microwaves look electromagnetic, because of their very short wavelengths and very high frequency, they do not so much act like electrons as act on electrons, and thus in the apparent form of gravity.
The prevalent modern theory is that tachyon particles do possess aethyreal energy, or what I have called elsewhere, astral light. The reason for this is that particles, such as photons, when converted into phased waves of pulsed energy through solid media can exhibit faster than light characteristic velocities, and therefore it is assumed that tachyons, which share this class of velocity, might also possess the electromagnetic charge characteristic to the accelerated subliminal particles. It should be remembered however, that tachyons are of a quantifiably different dimension altogether, and should therefore be expected to possess qualitatively different traits.
While modern theory does not argue over the notion of tachyons being commutative with subliminal electromagnetic and atomic radiation, it has postulated mathematically that tachyons accelerate their speed as they lose energy, which is the elegant contra-positive to the constant speed of light in a vacuum rule of relativity, which states that as subliminal matter increases it gains mass. This expectation of tachyons will likely be proven according to countless mathematical applications of the relativistic equations, and during all that time delay the true nature of tachyons being discovered by experimental physics.
It should suffice to say that tachyons are visibly radiates as Cerenkov light emitted by photons. Since the light from one photon alone would not illuminate the entire diameter of the universe, however, it is also safe to say that this illuminating energy wears out. In this regard, one could say that the energy of tachyons which is positively charged when emitted from photons is negatively charged when absorbed into atomic nuclei, moving from a dynamic to a static state. The modern theory speculates that when a tachyon has burned off its Cerenkov radiation, it loses all of its energy and achieves infinite velocity, occupying every point on its gravitational microwave length superstring history simultaneously. Such a tachyon is called a “transcendental” tachyon and, having no energy, is referred to as “zero point energy” — a category shared with certain scalar phased waves.
The microwaves do not just travel through the electron cloud without disrupting it, they also move faster than light. Ordinarily they are invisible as a result of this, and though the waves themselves are harmless, they are still believed by theoretical physicists to be associated with the explosive conclusion of the lesser, similar trajectories of antimatter because breaking the light barrier goes against the observable laws of physics simply by being considered improbable.
The only time it is possible to see, or rather, to measure these microwaves is when they are projected through a solid medium. Here it is seen that energy beginning as an electromagnetic charge can travel through the material substance by quantum tunneling faster than the speed of a photon traveling the same distance unimpeded. Here we might predict that the photon would win because it is larger, about the size of the electron cloud. However because the microwave is generated from an originally electromagnetic source, it carries a charge that interacts with the electron. It does not compress the orbital cloud into a singular point of charge, as does the photon, but instead the microwave length and the electron field inter-phase, and the microwave exits the opposite side of the orbital shell at the same time as it is entering it from the opposite side.
The particles at the tip of these microwaves, whose history forms the fractal waveform, involute. Tachyons turn inside out, and they do this continuously. They are doing this right now. When one contacts the surface of the electron’s potential energy orbital cloud, the tachyon absorbs the pattern of the shell, and moves through this geometry according to its own trajectory. In other words, when the tachyon and electron inter-phase, they combine to form a dimensional warp in the probabilistic continuum of space time.
This is how the tachyons travel faster than photons. Because of this they are considered as moving counter to the flow of entropy, or the increase of disorder. This comparison, again, implies a false similarity between tachyons and antimatter, since the particles and waves that have been classified antimatter were all discovered as being generated by the decomposition of larger particles at speeds approaching the speed of light. However the difference between the world below the speed of light and the world above it is like the difference between the world under the surface of the water in a fish pond as opposed to the vast expanse of the world above the water.
Related Science Topics of Interest:
http://hubpages.com/hub/Macroverse-VS-Microverse
http://hubpages.com/hub/The-Concept-of-the-Mind-Directing-Energy-in-Physics-Zero-Point-Energy-Field
http://hubpages.com/hub/What-are-Fractals-Why-Important
http://hubpages.com/hub/Retrocausality-Reverse-Causality-Today-Effects-the-Past
http://hubpages.com/hub/newbook
http://hubpages.com/hub/Why-Is-Noetics-Important
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• Are Flood Defence Schemes Any Good?
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• AUTHOR
Dallas W Thompson
9 years ago from Bakersfield, CA
CMHypno, Nobody completely understands it! It is a process of theories and more research... Currently quantum physics defies "common sense..."
• CMHypno
9 years ago from Other Side of the Sun
I think that I will have to set the dial on my brain to 'faster than the speed of light' before I really understand this, but it is fascinating information. There is still so much for us to learn about quantum physics and the universe
• AUTHOR
Dallas W Thompson
9 years ago from Bakersfield, CA
MartieCoetser, It is a process. Nobody "really knows" what-the-hell-is-going-on in quantum physics.. It is a process of observing, measuring, questioning, "guessing," and trial and error... It is a challenging frontier of science... Surprisingly quantum physics is everywhere in nature. Photosynthesis for example (sunlight into food)... No one is "on top of quantum physics!" Enjoy the process of learning: we are all learning...
• Martie Coetser
9 years ago from South Africa
Dallas, I am laughing myself to tears over here. This is the utmost time I read a topic about quantum physics, and really, I am trying my utmost best to understand what I’m reading. Always! For I realize there is something in quantum I ought to know. So when I was done, and I read Nellieanna’s comment, I just burst out laughing, for there it was, already written by her, the exact words in my mind. I appreciate your explanation to her – that is really easy to understand. I promise you, if you keep on explaining quantum in this manner, I will for sure get the picture, and I guess Nellieanna too. I’ve bookmarked this one too, for I’m soon going to make time to get on top of this quantum stuff for once and for all.
• AUTHOR
Dallas W Thompson
9 years ago from Bakersfield, CA
FitnezzJim, Thanks for stopping by. Our perspective determines our relative...
• FitnezzJim
9 years ago from Fredericksburg, Virginia | 1,920 | 9,153 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-10 | longest | en | 0.927441 |
easter.oremus.org | 1,723,470,060,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641039579.74/warc/CC-MAIN-20240812124217-20240812154217-00007.warc.gz | 11,281,328 | 1,584 | Modulo Arithmetic
Modulo arithmetic is concerned with the division of whole numbers (or integers) into a quotient and a remainder.
For example, 7 divided by 3 is 2 (the quotient), remainder 1. 12 divided by 3 is 4 remainder 0.
In this paper we refer to the quotient of two numbers, a and b as
a div b
and the remainder as
a mod b
To use the above examples, 7 div 3 is 2, and 7 mod 3 is 1.
For a more complicated example 1996 div 19 is 105, and 1996 mod 19 is 1 (because 1996 divided 19 is 105, remainder 1).
A subtle point is that the remainder is always a positive number. So -7 divided by 3 is -3, remainder 2 (rather than -2, remainder -1). This is important for the date of Easter because we must often calculate the remainder when dividing into a negative number. Since in that case we are not interested in the quotient it is sufficient to calculate the negative remainder and then simply add to it the number we were dividing by. So -11 mod 3 may be calculated as: -11 divided by 3 is -3, remainder -2, so the number required is -2 + 3 = 1.
Simon Kershaw <simon@oremus.org>
10 February 2004 | 309 | 1,105 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-33 | latest | en | 0.921497 |
https://prepinsta.com/leetcode-top-100-liked-questions-with-solution/subarray-sum-equals-k/ | 1,709,516,311,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476409.38/warc/CC-MAIN-20240304002142-20240304032142-00357.warc.gz | 456,238,728 | 32,153 | # Subarray Sum Equals K Leetcode Solution
## Subarray Sum Equals K Leetcode Problem :
Given an array of integers nums and an integer k, return the total number of subarrays whose sum equals to k.
A subarray is a contiguous non-empty sequence of elements within an array.
## Subarray Sum Equals K Leetcode Solution :
### Constraints :
• 1 <= nums.length <= 2 * 104
• -1000 <= nums[i] <= 1000
• -107 <= k <= 107
### Example 1:
• Input: nums = [1,2,3], k = 3
• Output: 2
Approach :
The solution uses a hash table to store the cumulative sum of the array up to each index, along with the number of times that sum has occurred. The algorithm works as follows:
1. Initialize a hash table map to store the cumulative sum of the array up to each index, along with the number of times that sum has occurred.
2. Initialize a variable count to store the number of subarrays with sum k.
3. Iterate over the array, starting at index 0.
4. Add the current element to the cumulative sum.
5. If the cumulative sum is equal to k, increment count.
6. If the cumulative sum minus k is present in the hash table, increment count by the number of times that sum has occurred.
7. Add the current cumulative sum to the hash table, along with the number of times that sum has occurred.
8. Return count.
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https://alltopicall.com/tag/output/ | 1,540,172,293,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514443.85/warc/CC-MAIN-20181022005000-20181022030500-00336.warc.gz | 596,830,328 | 13,372 | ## Wages, capital: Substitution and Output Effects
Consider a CES production function $$Y=f(K,L)$$ with elasticity of factor substitution $$sigma>0$$.
The substitution effect of higher real wages naturally implies a shift along the isoquant to more $$K$$. The rise in costs will reduce output. The total effect of $$w uparrow$$ on $$Y$$ and $$K$$ is the sum of this substitution and output effect.
I’m interested in the link between the value of $$sigma$$ and the strength of these two effects. Does anyone know a good reference, or lecture notes on the algebra of these effects in the CES context?
## Python Dictionary multiple line keys input and output line by line of their subsequent keys
i want to generate a squence of token of each key into their values
i have made a dictionary in python it is taking a single key and showing the value of the key only once but
i want to make a multiple keys scanner(input) sperated by whitespace which will show values of the given key(string) line by line
CODE:
table = {‘BEGIN’:’301′, ‘DO’:’302′,’ELSE’:’303′,’END’:’304′,’FLOAT’:’305′,’FOR’:’306′,’IF’:’307′,’INTEGER’:’308′,’THEN’:’309′,’PRINT’:’310′,’PROGRAM’:’311′,’UNTIL’:’312′,’WHILE’:’313′,’AND’:’314′,’OR’:’315′,’RETURN’:’316′,'(‘:’201’, ‘)’:’202′, ‘;’:’203′,’,’:’204′,'[‘:’205′,’]’:’206′,’:=’:’101′,’-‘:’102′,’+’:’103′,’*’:’104′,’/’:’105′,’%’:’106′,’<>‘:’107′,’>’:’108′,’<':'109','<=':'110','>=’:’111′,’=’:’112′,’1′:’25’,’2′:’25’,’3′:’25’,’4′:’25’,’5′:’25’,’0′: ’25’,’1′: ’25’,’2′: ’25’,’3′: ’25’,’4′: ’25’,’5′: ’25’,’6′: ’25’,’7′ : ’25’,’8′: ’25’,’9′: ’25’,’10’: ’25’,’11’: ’25’,’12’: ’25’,’13’: ’25’,’14’: ’25’,’15’: ’25’,’16’: ’25’,’17’: ’25’,’18’: ’25’,’19’: ’25’,’20’: ’25’,’21’: ’25’,’22’: ’25’,’23’: ’25’,’24’: ’25’,’25’: ’25’,’26’: ’25’,’27’: ’25’,’28’: ’25’,’29’: ’25’,’30’: ’25’,’31’: ’25’,’32’: ’25’,’33’: ’25’,’34’: ’25’,’35’: ’25’,’36’: ’25’,’37’: ’25’,’38’: ’25’,’39’: ’25’,’40’: ’25’,’41’: ’25’,’42’: ’25’,’43’: ’25’,’44’: ’25’,’45’: ’25’,’46’: ’25’,’47’: ’25’,’48’: ’25’,’49’: ’25’,’50’: ’25’,’51’: ’25’,’52’: ’25’,’53’: ’25’,’54’: ’25’,’55’: ’25’,’56’: ’25’,’57’: ’25’,’58’: ’25’,’59’: ’25’,’60’: ’25’,’61’: ’25’,’62’: ’25’,’63’: ’25’,’64’: ’25’,’65’: ’25’,’66’: ’25’,’67’: ’25’,’68’: ’25’,’69’: ’25’,’70’: ’25’,’71’: ’25’,’72’: ’25’,’73’: ’25’,’74’: ’25’,’75’: ’25’,’76’: ’25’,’77’: ’25’,’78’: ’25’,’79’: ’25’,’80’: ’25’,’81’: ’25’,’82’: ’25’,’83’: ’25’,’84’: ’25’,’85’: ’25’,’86’: ’25’,’87’: ’25’,’88’: ’25’,’89’: ’25’,’90’: ’25’,’91’: ’25’,’92’: ’25’,’93’: ’25’,’94’: ’25’,’95’: ’25’,’96’: ’25’,’97’: ’25’,’98’: ’25’,’99’: ’25’,’100′:’25’,'” “‘:’75’}
x=input(‘token to see code: ‘)
print(table[x])
right now its working like this
OUTPUT:
token to see code: IF
307
what i want to make
I WANT TO MAKE IT WORK SOMETHING LIKE THIS
Example Run
The following program:
PROGRAM
INTEGER A;
FLOAT B;
BEGIN
IF A < 50 THEN
B := 2.3;
END IF;
PRINT(“Hello world”);
END; ##
produces the following
output:
311
308
50
203
305
50
203
301
307
50
109
25
309
50
101
25
203
304
307
203
310
201
75
202
203
304
203
555
555
i want to make it more by adding these logics too but dont know how
variables A variable begins with a letter followed by zero or more letters, digits, or underscores. A variable cannot be a reserved word. Output a token value of 50 for each variable encountered.
strings Strings are a group of 0 or more characters enclosed within double quotes “. Strings can contain any character, except newlines. Two double quotes in a row (“”), constitue a double quote in the string, and not the end of one string and the start of another. Output a token of 75 for each string encountered.
THANKS IN ADVANCE ITS KINDA MY CLASS PROJECT AND THAT HAS TO SUBMITTED TOMORROW IT WOULD BE HIGHLY APPRECIATED IF SOMEONE MAKE CODE FOR ME OF THIS
## Mr-ct Regression Model Output
I ran inference with the mr-ct regression model with the pre-trained weights from the model zoo, but the output it produced was an indistinct, low-contrast image that looked nothing like the test CT. Could you clarify what the output should look like? Does the model not come pre-trained?
We would also like to retrain the model on a dataset of our own 3D brain images, but we’re unsure where or how to specify a different dataset to train on. We can’t find the current training dataset in the model zoo so we’re wondering if it includes a training set or just the pre-trained weights and parameters.
## How could I sort the output of awk disregard of certain character?
I am trying to print out the Chinese font I have in my laptop, and sort the output with respect to their name that could be referred to in LaTeX. This could be done with the following command
fc-list :lang=zh-cn | awk -F": " '{$1=""; print}' | sort It finds the font list, then after first occurence of : and a single space, which is where the names of the font show up. Then we of course sort it. The problem here is though, there are fonts named like .PingFang SC, which will be sorted to the front and is not that consistent with the rest. Is there a way to sort disregarding the period? Also, it may not be a good choice to remove the period, which will change the actual name of the font and thus make the reference hard. By the way, there are one : and two spaces in front of the name of the font, however, if I added in the delimiter marker the second space, the output will be blank, could anyone help on that too? Thanks in advance! ## lspci output of NVMe SSD in M.2 slot I am developing some system management scripts and was wondering if a M.2 NVMe slot on the motherboard shows up in lspci output? Unfortunately, I don’t have the hardware currently to test this (I have a NVMe SSD in an HBA sitting in a PCIe slot, but not an M.2 slot built on to the motherboard). If anyone reading this has the needed hardware (ie., populated M.2 slot on motherboard having an NVMe SSD), could they help a fellow developer by providing the lspci output showing the NVMe drive in the M.2 slot? ## Why does the camera output an image with multi-colors not shown by the naked eye under the projector? I was shooting in a gallery that used what appears to be a multi-colored projector and had difficulty shooting in that type of scene as the color was not even across the scene. The image shown below is what I mean, you see traces of green, blue, and yellow colors in the picture when it’s supposed to be all white/gray as seen by the naked eye. Why does this occur to the camera’s output (DSLR or mirrorless) when shooting a scene that’s lit by a projector? Is it the type of projector used? Is there a way to fix this or camera settings wise? ## Why is Uniform Cost Search not output the optimal solution in this case? According to the AI lecture and textbook, the Uniform Cost Search algorithm should always output the optimal solution if the cost is greater than 0. Consider the search graph below, where S is the start node and G1, G2, and G3 are goal states. Arcs are labelled with the cost of traversing them and the heuristic cost to a goal is shown inside the nodes. By using the Uniform Cost Search algorithms, the path should be: S -> B -> F -> D -> G2, the total cost is 12. However, the optimal solution should be S -> D -> G2, the total cost is 9 Am I misunderstanding the UCS algorithm? Thanks for any comments! ## Error headers: ap_headers_output_filter() after putting cache header in htaccess file Receiving error: [debug] mod_headers.c(663): headers: ap_headers_output_filter() after I included this within the htaccess file: # 6 DAYS Header set Cache-Control "max-age=518400, public" # 2 DAYS Header set Cache-Control "max-age=172800, public, must-revalidate" # 2 HOURS Header set Cache-Control "max-age=7200, must-revalidate" Any help is appreciated as to what I could do to fix this? ## how does echo recognize, when it should output list of files or pure text? I encountered this question in my test: Make echo output all directories that start with a vowel (a,e,i,o,u,e), end with a number and have at least 3 characters. I thought this would be fairly easy, but soon I ended up confused and evetually failed the test. My first thought was simply: echo {a,e,i,o,u,e}*{0..9} Then I tried to use square brackets echo [a,e,i,o,u,e]*[0..9] and at the end i tried something like this: echo$(ls {a,e,i,o,u,e}*{0..9})
which gave me the required output, but with some error directory missing messages, and Im not even sure if it wouldnt be considered cheating as Im using another function to do it.
Could anyone clarify for me, how do I do it and when do I use which brackets?
And I would also like to know, how does echo recognize when it should output list of directories instead of my exact words.
## R code and Sweave output – EDIT
UPDATED Question 1 In the lattest update of KnitR the output is clean as what I expected. Maybe an update was made.
The new output :
Before update
I’m trying to get a particular output with Sweave and R code.
For now I have this code :
documentclass[11pt,french]{report}
usepackage{babel} %%french
usepackage{amsmath,amsfonts,amssymb} %%maths
usepackage[utf8]{inputenc} % LaTeX
usepackage[T1]{fontenc} % LaTeX
usepackage[dvipsnames,table,xcdraw]{xcolor}
%% Background code chunk
usepackage{listings}
usepackage{color}
definecolor{codegreen}{rgb}{0,0.6,0}
definecolor{codegray}{rgb}{0.5,0.5,0.5}
definecolor{codepurple}{rgb}{0.58,0,0.82}
definecolor{backcolour}{rgb}{0.95,0.95,0.92}
%% background code chunk listing
lstset{
language=R}
lstdefinestyle{mystyle}{
backgroundcolor=color{backcolour},
keywordstyle=color{magenta},
numberstyle=tinycolor{codegray},
stringstyle=color{codepurple},
basicstyle=footnotesize,
breakatwhitespace=false,
breaklines=true,
captionpos=b,
keepspaces=true,
numbers=left,
numbersep=5pt,
showspaces=false,
showstringspaces=false,
showtabs=false,
tabsize=2
}
lstset{style=mystyle}
witch is coming from this site.
The R code I’m trying to implement is :
%% script R
begin{lstlisting}[linerange=\begin{Sinput}-\end{Sinput},includerangemarker=false, caption = Code source en R pour l'exemple]
<>=
# dataset
x <- c(2,3,6,9,12); y <- c(2,5,3,6,5)
# Estimations des parametres
reg <- lm(y ~ x)
# Resume de l'estimation
summary(reg)
# Valeurs de Yt
fitted(reg)
# Residus
residuals(reg)
@
end{lstlisting}
The output :
My questions :
1- How can I get a better output for the ~ symbol ?
2- Is there a way to eval the code without getting and output like this ? | 3,100 | 10,475 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-43 | latest | en | 0.347983 |
https://convert-dates.com/days-before/1225/2024/04/04 | 1,713,222,457,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00324.warc.gz | 168,496,118 | 4,346 | ## 1225 Days Before April 4, 2024
Want to figure out the date that is exactly one thousand two hundred twenty five days before Apr 4, 2024 without counting?
Your starting date is April 4, 2024 so that means that 1225 days earlier would be November 26, 2020.
You can check this by using the date difference calculator to measure the number of days before Nov 26, 2020 to Apr 4, 2024.
November 2020
• Sunday
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November 26, 2020 is a Thursday. It is the 331st day of the year, and in the 48th week of the year (assuming each week starts on a Sunday), or the 4th quarter of the year. There are 30 days in this month. 2020 is a leap year, so there are 366 days in this year. The short form for this date used in the United States is 11/26/2020, and almost everywhere else in the world it's 26/11/2020.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 1225 weekdays before Apr 4, 2024, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Apr 4, 2024, which falls on a Thursday. Counting forward, the next day would be a Friday.
To get exactly one thousand two hundred twenty five weekdays before Apr 4, 2024, you actually need to count 1715 total days (including weekend days). That means that 1225 weekdays before Apr 4, 2024 would be July 25, 2019.
If you're counting business days, don't forget to adjust this date for any holidays.
July 2019
• Sunday
• Monday
• Tuesday
• Wednesday
• Thursday
• Friday
• Saturday
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July 25, 2019 is a Thursday. It is the 206th day of the year, and in the 206th week of the year (assuming each week starts on a Sunday), or the 3rd quarter of the year. There are 31 days in this month. 2019 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 07/25/2019, and almost everywhere else in the world it's 25/07/2019.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date. | 935 | 2,733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-18 | latest | en | 0.918091 |
https://www.itcsdaixie.com/daixie-cse514-spring-2023-programming-assignment-1/ | 1,720,973,399,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00269.warc.gz | 747,892,344 | 16,622 | # CS代写｜CSE514 – Spring 2023 Programming Assignment 1
## 这是一篇来自美国的关于为了增强对目标函数、回归模型和优化的梯度下降算法理解的CS代写。它包括一个编程分配（具有可选的奖励积分扩展）和一个报告。这个项目是个人的工作，没有代码共享请，但你可以发布bug问题到广场寻求帮助。
Topic
Design and implement a gradient descent algorithm or algorithms for regression.
Programming work
A) Data pre-processing
Pre-process the attribute values of your data by normalizing or standardizing each variable. Make sure to keep a copy that was not pre-processed, so you can analyze the effect that pre-processing the data has on the optimization.
B) Univariate linear regression
In lecture, we discussed univariate linear regression y = f(x) = mx+b, where there is only a single independent variable x, using MSE as the loss function.
Your program must specify the objective function of mean squared error and be able to apply the gradient descent algorithm for optimizing a univariate linear regression model.
C) Multivariate linear regression
In practice, we typically have multi-dimensional (or multi-variate) data, i.e., the input x is a vector of features with length p. Assigning a parameter to each of these features, plus the b parameter, results in p+1 model parameters. Multi-variate linear models can be succinctly represented as:
y = f(x) = (m · x)
(i.e., dot product between m and x),
where m = (m0, m1, …, mp) T and x = (1, x1, …, xp) T, with m0 in place of b in the model.
Your program must be able to apply the gradient descent algorithm for optimizing a multivariate linear regression model using the mean squared error objective function.
D) Optional extension 1 – Mean Absolute Error as the loss function
For bonus points, include the option of optimizing for the MAE instead of MSE. Calculating MAE as your error is insufficient! You must define a new gradient calculation to be used for the gradient descent optimization process.
E) Optional extension 2 – Ridge Regression
For bonus points, include the option of optimizing an l2 penalty as part of your loss function.
Calculating MSE + l2 as your error is insufficient! You must define a new gradient calculation to be used for the gradient descent optimization process. You must tune the ? hyperparameter value for minimizing test error.IMPORTANT: Regression is basic, so there are many implementations available, but you MUST implement your method yourself. This means that you cannot use an embedded function for regression or gradient descent from a software package. You may use other basic functions like matrix math, but the gradient descent and regression algorithm must be implemented by yourself.
Data to be used
We will use the Concrete Compressive Strength dataset in the UCI repository at UCI Machine Learning Repository: Concrete Compressive Strength Data Set
(https://archive.ics.uci.edu/ml/datasets/Concrete+Compressive+Strength)
Note that the last column of the dataset is the response variable (i.e., y).
There are 1030 instances in this dataset.
Use 900 instances for training and 130 instances for testing, randomly selected. This means that you should learn parameter values for your regression models using the training data, and then use the trained models to predict the testing data’s response values without ever training on the testing dataset.
What to submit – follow the instructions here to earn full points
• (80 pts total + 17 bonus points) The report as a pdf
o Introduction (15 pts + 5 bonus points)
• (4 pts) Your description/formulation of the problem (what’s the data and what practical application could there be for your work with it, beyond just “this is my homework” or “I want to optimize this equation”),
• (3 pts) a description of how you normalized or standardized your data. Include some figures that illustrate how the distribution of feature values changed because of your pre-processing
• (5 pts) the details of your algorithm (e.g., stopping criterion, is this stochastic gradient descent or not, how you chose your learning rate, etc),
• (3 pts) pseudo-code of your algorithm (see Canvas for an example)
• (+2 bonus pts) if you include a description of how you implemented MAE
• (+3 bonus pts) if you include a description of how you implemented Ridge Regression
o Results (52 pts + 8 bonus points)
• To report the performance of your models, calculate the variance explained (eg.
R-squared) for the response variable, which is:
1-MSE/Variance(observed)
In other words, calculate the average squared error between predicted responses and actual responses (MSE). Then calculate the average squared difference between actual responses and mean actual response (Variance). Divide the former by the latter, then subtract from 1.§ (26 pts) Variance explained of your models on the training dataset when using only one of the predictor variables (univariate regression) and when using all eight (multivariate regression).
You should have a total of nine values from optimizing on the raw data, and nine values from optimizing on the pre-processed data.
At least two of your models optimized on raw data must achieve a positive variance explained on the training data.
At least two of your models optimized on pre-processed data must achieve a positive variance explained on the training data
• (10 pts) Variance explained of your models on the testing data.
You should have a total of nine values from optimizing on the raw data, and nine values from optimizing on the pre-processed data.
• (16 pts) Plots of your univariate models on top of scatterplots of the training data used. Please plot the data using the x-axis for the predictor variable and the y-axis for the response variable.
• (+4 bonus points) if you include results from using the MAE loss function on the pre-processed data
• (+4 bonus points) if you include results from using Ridge Regression on the preprocessed data
o Discussion (13 pts + 4 bonus points)
• (8 pts) Compare and contrast your models.
• Did the same models that accurately predicted the training data also accurately predict the testing data?
• Did different models take longer to train or require different hyperparameter values?
• How did pre-processing change your results or optimization approach?
• (5 pts) Draw some conclusions about what factors predict concrete compressive strength. What would you recommend for making the hardest possible concrete?
• (+2 bonus points) if you include comparisons from using MAE
• (+2 bonus points) if you include comparisons from using Ridge Regression
Cement (component 1)(kg in a m^3 mixture)
0.0805381Note: We won’t be grading for good writing practices, but you may have points taken off if you don’t write in full sentences and paragraphs, or if you fail to correct spelling and grammar that a simple spell-check tool would alert you of. Results may be presented as a table, but you must label the rows/columns with enough detail for a reader to interpret it without searching your text, and the figures must be labeled as well.
• (20 pts total + 8 bonus points) Your program (in a language you choose) including:
o (15 pts) The code itself
o (5 pts) Sufficient instructions/documentation on how to run your program (input/output plus execution environment and compilation if needed)
o (+4 bonus points) if you include code for using MAE as the loss function
o (+4 bonus points) if you include code for Ridge Regression
Note: We won’t grade your program’s code for good coding practices or documentation.
However, if we find your code difficult to understand or run, we may ask you to run your program to show it works on a new dataset.
Due date
Monday, March 6 (midnight, STL time). Submission to Gradescope via course Canvas.
A one-week late extension is available in exchange for a 20% penalty on your final score.
About the extra credit:
The bonus point values listed are the upper limit of extra credit you can earn for each extension. How many points you get will depend on how well you completed each task. Feel free to include partially completed extensions for partial extra credit!
In total, you can earn up to 25 bonus points on this assignment, which means you can get a 125% as your score if you submit it on time, or you can submit this assignment late with the 20% penalty and still get a 100% as your score. It’s up to you how you would prefer to manage your time and effort. | 1,817 | 8,324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-30 | latest | en | 0.785285 |
https://www.studysmarter.co.uk/explanations/chemistry/physical-chemistry/rate-equations/ | 1,701,185,093,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099892.46/warc/CC-MAIN-20231128151412-20231128181412-00695.warc.gz | 1,132,925,388 | 78,202 | ### Select your language
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Rate Equations
Chemical Reactions are processes in which a set of reactants are converted into products as a result of changes to their structures. These structural changes can happen at different speeds, similar to how race cars can travel at different speeds. Just like how it's important to understand how the speed of…
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# Rate Equations
Lerne mit deinen Freunden und bleibe auf dem richtigen Kurs mit deinen persönlichen Lernstatistiken
Nie wieder prokastinieren mit unseren Lernerinnerungen.
Chemical Reactions are processes in which a set of reactants are converted into products as a result of changes to their structures. These structural changes can happen at different speeds, similar to how race cars can travel at different speeds. Just like how it's important to understand how the speed of a race car can be affected, understanding how the speed of chemical changes can be affected is an important part of Physical Chemistry.
The rate equation is an expression that links the rate of a reaction to the Concentration of the species involved.
• We will be looking at the rate equation.
• We'll see what it tells us about the rate of a reaction.
• Finally, we'll explore the Rate Constant and reaction orders, and we'll briefly touch on methods used to determine the rate equation.
## Rate equation chemistry
The rate of a reaction is how quickly a reaction occurs. But what does that mean and what does it tell us? Well, one way of looking at it is to think about how much product is made in a period of time, which will depend on how much reactant is used up. In essence, we can say that the rate of a reaction is the speed at which reactants are converted into products.
The rate of reaction is the change in Concentration of reactants or products over time. It is typically measured in mol dm-3 s-1.
### Measuring rate of reaction
To calculate the rate of a reaction, we need to measure the change in the amount of reactant/product from the start of the reaction to the end.
We define the rate of a reaction as a measure of how much product is formed, or how much reactant is used, over a period of time.
We can measure this by observing things like colour change, pH change, volume of gas produced, or change in mass of a solid reactant. You should then be able to convert your data values into figures for concentration. This data is plotted on a line graph, with time on the x-axis and concentration on the y-axis. From the graph, we can find a value for the rate of reaction by working out the line's gradient. We either calculate an overall rate of reaction or an instantaneous rate of reaction. Both use the following equation:
$\mathrm{rate}\mathrm{of}\mathrm{reaction}=\frac{\mathrm{change}\mathrm{in}\mathrm{concentration}}{\mathrm{time}\mathrm{taken}}$
#### Overall rate of reaction
Calculating an overall rate of reaction is fairly straightforward. You divide the overall change in concentration of a reactant or product by the time taken. For a graph of concentration against time, this means dividing the change in y-values by the change in x-values. Here's an example.
Calculate the overall rate of reaction for the following graph.
Fig. 1 - A concentration-time graph used to calculate rate of reaction
To find the overall rate of reaction, we divide the change in concentration by the time taken.
It doesn't matter whether you measure the concentration of a product or reactant - both will give you a valid answer.
Fig. 2 - A concentration-time graph used to calculate rate of reaction
Here, the concentration starts at 40 mol dm-3 and ends at 8 mol dm-3. This is a change of 40 - 8 = 32 mol dm-3. The reaction takes 200 seconds. The rate of reaction is therefore $\frac{32}{200}=0.16\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$.
#### Instantaneous rate of reaction
Sometimes, finding an overall rate of reaction isn't that useful. You might instead want to know how the rate of reaction changes over time. To do this, you calculate instantaneous rates of reaction. This involves drawing a tangent to the curve at a particular point and finding its gradient. Again, this is given by the change in concentration divided by the time taken - in other words, the change in y-values divided by the change in x-values.
Calculate the instantaneous rate of reaction for the following graph at 60 seconds.
Fig. 3 - A concentration-time graph used to calculate rate of reaction
We first need to find the 60-second mark on the curve. We draw a tangent to the curve at this point.
Remember that a tangent is a straight line that just touches the curve at a specified point.
Next, we calculate the gradient of this tangent by dividing the change in concentration by the time taken. You do this by turning the tangent into a right-angled triangle.
Fig. 4 - A concentration-time graph used to calculate rate of reaction
Here, we can see from our right-angled triangle that the concentration starts at 22 mol dm-3 and ends at 6 mol dm-3. This is an overall change of 16 mol dm-3. This change in concentration takes place between 20 and 120 seconds, meaning it takes 100 seconds in total. The instantaneous rate of reaction is therefore $\frac{16}{100}=0.16\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$
## Rate of reaction equation
Let's look at something different: the rate equation. The rate equation in chemistry is a formula that we can use to find the rate of a reaction using the concentration of species involved in the reaction. Here's what it looks like:
Fig. 5 - Rate equation
At first glance it certainly looks confusing, but once you understand what's going on it isn't all that bad.
• k is the Rate Constant.
• The letters A and B in the rate equation are used to represent species involved in the reaction. These could be reactants or Catalysts.
• The square brackets around the letters represent concentration. So, [A] is used to show the concentration of species A.
• The letters m and n represent the order of the reaction with respect to a certain species. They show the power that the concentration of that species is raised to in the rate equation. Overall, [A]m represents the concentration of A, raised to the power of m. This means that A has the order m.
### The rate constant
k is the rate constant. It is used in the rate equation to link the concentrations of certain species to the rate of that reaction. The value of k changes depending on the reaction and reaction conditions. However, k is always constant for a certain reaction at a particular temperature. If you were to carry out the exact same reaction at different temperatures, k would change, but if you carried it out at the same temperature, k would stay the same: after all, it is a constant!
To learn more about how the rate constant relates to temperature, read The Arrhenius Equation. And if you want to find out how to calculate the rate constant, alongside its units, head over to Determining Rate Constant.
### Orders of reaction
In Chemical Reactions, reactants and Catalysts (if there are any) have an order of reaction. The sum of the individual orders of species in a reaction equals the overall order of the equation.
In the rate equation, the order of a reaction with respect to a species is shown using a power. For example, in the rate equation we looked at above, the order of A is represented by the letter m. The order of a reaction with respect to a species tells us how the concentration of that particular species affects the reaction rate. Some species don't affect the rate whatsoever, while other species affect it dramatically.
Any non-negative number can be an order, and species can also have fractional orders like 5/2. But for the purpose of your exams, you only need to know about zero, first and second-order reactants.
#### Zero-order reactants
The concentration of a zero-order reactant doesn't affect the rate of reaction. If you double its concentration, the rate stays the same. This is because 20 = 1. Because they have no effect on the rate of reaction, zero-order reactants don't appear in the rate equation.
#### First-order reactants
The concentration of first-order reactants is directly proportional to the rate of reaction. If you double the concentration of a first-order reactant, the rate of reaction also doubles. This is because 21 = 2.
If a reactant is first-order, it appears in the rate equation raised to the power of 1. However, we don't tend to write the number 1 because raising something to the power of 1 has no effect on its value. You'll see first-order reactants in the rate equation as [A], where A represents the species.
#### Second-order reactants
The concentration of second-order reactants has an exponential effect on the rate of reaction. Doubling the concentration of a second-order reactant causes the rate of reaction to quadruple. This is because 22 = 4.
If a reactant is second-order then we put it in the rate equation raised to the power of 2; in other words, squared. You'll see it in the rate equation as [A]2.
#### Order of a reaction
The overall order of a reaction is the sum of all the individual reactant orders. Remember that the order of a reactant is the power that it is raised to in the rate equation. If you're ever asked to find the overall order of reaction, simply add together all of the powers present in the equation and you'll reach your final answer.
Understanding orders of reaction is a bit tricky, so let's look at an example to help you understand.
A reaction has the chemical equation and rate equation shown below.
$\mathrm{A}+\mathrm{B}+\mathrm{C}\to \mathrm{D}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{rate}=\mathrm{k}\left[\mathrm{A}\right]{\left[\mathrm{B}\right]}^{2}$
Describe the effect of doubling the concentrations of A, B, and C.
First of all, looking at the rate equation, we can see that the only species present are A and B. C does not appear at all. It must therefore be zero-order. Hence, doubling the concentration of C will have no effect on the rate of reaction.
On the other hand, A does appear in the rate equation. It looks like [A] isn't raised to any power, but as we learnt above, [A] is the same as saying [A]1. A is therefore first-order. Doubling the concentration of A will cause the rate of reaction to double, because 21 = 2.
B also appears in the rate equation. It is raised to the power of 2, meaning that it is second-order. Doubling the concentration of B will cause the rate of reaction to quadruple, because 22 = 4.
## Determining the rate equation
There are a few different methods we can use to determine the rate equation for a reaction. The basic principles come down to determining the species involved in the rate equation and then finding each of their orders. The main methods for doing this are:
• The initial rates method.
• Using rate-concentration graphs.
• Finding first-order reactants from their half-life.
• Inspecting the Reaction Mechanism.
We cover these methods in much more detail in Determining Reaction Order, but we'll explore them briefly now.
### Initial rates
The initial rates method involves measuring the rate of the same reaction over several experiments, each with different starting concentrations of a particular reactant. This method allows us to see numerically how the concentration of the reactant affects the rate of the reaction. We do this for each reactant, and can use the information to determine the reactant's order.
### Rate-concentration graphs
Earlier in the article, we looked at how you use graphs showing concentration of a species against time to calculate rate of reaction at a specific instant. You can then take the values for instantaneous rate of reaction and plot them against concentration to make a rate-concentration graph. These take specific shapes, depending on the order of the species involved.
• A horizontal straight line shows that the rate of reaction is unaffected by the concentration of the species. The species is therefore zero-order.
• A sloping straight line through the origin shows that the rate is directly proportional to the concentration of the species. The species is therefore first-order.
• A curved line through the origin shows that the rate is exponentially proportional to the concentration of the species. The species is second-order or higher.
Fig. 6 - Rate-concentration graphs for reactants with different orders
### Half-life equations
The half-life, ${\mathrm{t}}_{1/2}$, of a reactant is the time it takes for the concentration of that reactant to become half of what it was where you started measuring from. There's an interesting feature of first-order reactants: they have a constant half-life. This means that it takes the same amount of time to get from, say, a concentration of 1.0 to a concentration of 0.5 mol dm-3, as it does to get from a concentration of 0.8 to 0.4 mol dm-3. In both cases, the concentration has halved.
You can measure half-life using concentration-time graphs. Pick any point on the graph and look at the concentration for that time value. Then, see how long it takes to halve the concentration. Repeat this again to find multiple half-lives for a species. If all of the half-lives are the same, the species is first-order.
Fig. 7 - A concentration-time graph used to show half-life
#### Half-life and rate constant
The half-life of a first-order reactant relates to the rate constant, k, using the following equation:
$\mathrm{k}=\frac{\mathrm{ln}\left(2\right)}{{\mathrm{t}}_{1/2}}$
This means that once you know the half-life of a first-order reactant, you can easily find k.
### Reaction mechanism
Reactions can have mechanisms with one step or multiple steps. Each step happens at a different speed, and the rate of reaction is determined by the slowest step. We call the slowest step in a reaction the rate-determining step, and it gives us an idea of what the rate equation is likely to look like. This is because the rate equation is only made up of reacting species found in the steps up to and including the rate-determining step. The number of moles of each species relates to its order.
If you know the Reaction Mechanism and the rate-determining step of a reaction, you can predict the rate equation!
## Rate Equations - Key takeaways
• The rate of a chemical reaction is the change in concentration of reactants or products over time.
• Rate of reaction can be represented by a rate equation. Rate equations are composed of a rate constant (k), and reactant concentrations raised to the power of their respective order.
• Rate constants are constant for a particular reaction at a certain temperature.
• The order of a reaction with respect to a species tells us how the rate of reaction depends on the concentration of that species.
• The concentration of zero-order reactants has no effect on the rate of reaction.
• The concentration of first-order reactants is directly proportional to the rate of reaction.
• The concentration of second-order reactants has an exponential effect on the rate of reaction.
• The rate equation can be determined using the initial rates method, by identifying the shapes of graphs, by calculating half-lives, and by inspecting the reaction mechanism.
## Frequently Asked Questions about Rate Equations
To calculate the rate equation, you need to find out the order of reaction with respect to each species involved in the reaction. You also need to find the rate constant, k. You can do this experimentally. Once you've formed a rate equation, you can substitute in known concentration values and find the rate of reaction at a particular instant.
Rate equations are written in the form rate = k [A]m [B]n. The rate constant, k, is a value that is always constant for a particular reaction at a particular temperature. [A] represents the concentration of A, whilst the letter m represents the order of the reaction with respect to A. Overall, [A]m means the concentration of A, raised to the power of m. To write a rate equation, you work out the rate constant and the orders of reaction with respect to each species involved, and write them in the form given above.
The rate equation tells us the rate of a reaction. This means that it tells us the rate of change of reactant or product concentration during a reaction. So, by calculating the rate equation, you can find the rate of change.
You can find the rate of a reaction by using the rate equation. The rate equation is a formula that tells us the rate of any reaction from the concentration of its reactants.
Some factors that affect the rate of a reaction include reactant concentration, surface area, temperature, and activation energy.
## Rate Equations Quiz - Teste dein Wissen
Question
Why is the Arrhenius equation useful in chemistry?
Show answer
Answer
It allows us to relate the temperature of a reaction with its rate
Show question
Question
What does the letter A represent in the Arrhenius equation?
Show answer
Answer
A is the Arrhenius constant
Show question
Question
What information about a reaction can the value of ex give us in the Arrhenius equation?
Show answer
Answer
the number of reacting particles that have enough energy to react
Show question
Question
Rearrange the Arrhenius equation into its logarithmic form.
Show answer
Answer
ln(k) = ln(A) - Ea/RT
Show question
Question
Given a graph showing an Arrhenius plot for a chemical reaction, how could you use the plotted data to determine activation energy and the rate constant for that reaction?
Show answer
Answer
The activation energy would be equal to the gradient of the line.
The rate constant would be equal to the y-intercept.
Show question
Question
When drawing an Arrhenius plot, what would you label your axis?
Show answer
Answer
The x-axis would be 1/T
The y-axis would be ln(K)
Show question
Question
What does the Arrhenius equation show us about the effect of temperature on the rate of a reaction?
Show answer
Answer
The rate of a reaction will increase if the temperature at which the reaction occurs increases.
Show question
Question
What is a rate equation?
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Answer
A mathematical equation that links the rate of reaction with the concentration of species involved in the reaction.
Show question
Question
What does the letter k represent in the rate equation?
Show answer
Answer
The rate constant
Show question
Question
What do the letters m and n represent in the rate equation?
Show answer
Answer
The order of the reaction with respect to a certain species.
Show question
Question
Give four ways of determining the rate equation.
Show answer
Answer
• Initial rates method
• Rate-concentration graphs
• Half-life
• Reaction mechanism
Show question
Question
When using the initial rates method to determine the rate equation, you _____.
Show answer
Answer
Keep the external conditions the same.
Show question
Question
How can we use half-lives to help determine the rate equation?
Show answer
Answer
We can use half-lives to identify first-order reactants. First-order reactants have a constant half-life. You can see this by plotting a concentration-time graph.
Show question
Question
What is half-life?
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Answer
The half-life (t1/2) of a species is the time it takes for half of the species to be used in the reaction. In other words, it is the time it takes for its concentration to halve.
Show question
Question
True or false? Second-order reactants have a constant half-life.
Show answer
Answer
False
Show question
Question
Only species in _____ feature in the rate equation.
Show answer
Answer
The steps up to and including the rate-determining step
Show question
Question
Which step in a chemical reaction is the rate-determining step?
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Answer
The slowest step
Show question
Question
What is rate of reaction?
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Answer
Change in concentration of reactants or products over time.
Show question
Question
What are the units for rate of reaction?
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Answer
mol dm-3 s-1
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Question
Give three ways of measuring rate of reaction.
Show answer
Answer
• pH change
• Change in mass
• Volume of gas produced
• Colour change
Show question
Question
What does the letter k represent in the rate equation?
Show answer
Answer
The rate constant
Show question
Question
What does [A] represent in the rate equation?
Show answer
Answer
The concentration of species A, typically in mol dm-3
Show question
Question
What do the letters m and n represent in the rate equation?
Show answer
Answer
The order of reaction with respect to a particular species. They show the power that the species is raised to in the rate equation.
Show question
Question
rate = k [A]2 [B]
What is the order of reaction with respect to A?
Show answer
Answer
2
Show question
Question
rate = k [A]2 [B]
What is the overall order of the reaction?
Show answer
Answer
3
Show question
Question
'k' is constant at different temperatures. True or false?
Show answer
Answer
False
Show question
Question
What are the units of the rate constant k?
Show answer
Answer
It depends
Show question
Question
rate = k [A] [B]
What are the units of k for this reaction?
Show answer
Answer
mol-1 dm3 s-1
Show question
Question
How does the concentration of zero-order species affect the rate of reaction?
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Answer
It has no effect on the rate of reaction.
Show question
Question
How does the concentration of first-order species affect the rate of reaction?
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Answer
It is directly proportional to the rate of reaction.
Show question
Question
How does the concentration of second-order species affect the rate of reaction?
Show answer
Answer
It has an exponential effect on the rate of reaction.
Show question
Question
The concentration of a second-order species doubles. Predict the effect on the rate of reaction.
Show answer
Answer
The rate of reaction quadruples.
Show question
Question
The concentration of a first-order species increases by a factor of 3. Predict the effect on the rate of reaction.
Show answer
Answer
The rate of reaction increases by a factor of 3.
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Question
Which of these do not appear in the Arrhenius equation?
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Answer
ΔT
Show question
Question
What are the units of Ea in the Arrhenius equation?
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Answer
J mol-1
Show question
Question
What are the units of k in the Arrhenius equation?
Show answer
Answer
mol-1 dm3 s-1
Show question
Question
What does k represent in the Arrhenius equation?
Show answer
Answer
The rate constant
Show question
Question
What does T represent in the Arrhenius equation and what are its units?
Show answer
Answer
Temperature, K
Show question
Question
According to the Arrhenius equation, increasing the temperature _____ the rate of reaction.
Show answer
Answer
Increases
Show question
Question
According to the Arrhenius equation, decreasing the activation energy _____ the rate of reaction.
Show answer
Answer
Increases
Show question
Question
What is the significance of the y-value at the point where x = 0 on an Arrhenius plot?
Show answer
Answer
The y-value at the point where x = 0 is equal to ln(A)
Show question
Question
What is the Arrhenius equation?
Show answer
Answer
The Arrhenius equation is a mathematical formula that relates the rate constant of a reaction with the activation energy and temperature of that reaction.
Show question
Question
What goes on the y-axis on an Arrhenius plot?
Show answer
Answer
ln(k)
Show question
Question
What information does the y-coordinate of the point at which x = 0 give you?
Show answer
Answer
ln(A)
Show question
Question
A reaction takes place at 600K. The Arrhenius constant equals 3.8 x 1011 s-1 and the activation energy of this reaction is 240 kJ mol-1. Calculate the rate constant, k.
Show answer
Answer
4.73 x 10-10 s-1
Show question
Question
A reaction takes place at 650K. The Arrhenius constant equals 5.617 x 1012 mol-1 dm3 s-1 and the rate constant equals 0.53 mol-1 dm3 s-1. Calculate the activation energy, giving your answer to three significant figures.
Show answer
Answer
162000 J mol-1 (162 kJ mol-1)
Show question
Question
True or false? For reactions with an activation energy of around 50 kJ mol-1, increasing the temperature from 290K to 200K will triple the rate constant.
Show answer
Answer
False
Show question
Question
What is the rate constant?
Show answer
Answer
The rate constant is a proportionality constant that links the concentrations of certain species to the rate of a chemical reaction.
Show question
Question
What is the symbol for the rate constant?
Show answer
Answer
k
Show question
Question
Which of the following statements is always true about the rate constant?
Show answer
Answer
0
Show question
## Test your knowledge with multiple choice flashcards
When using the initial rates method to determine the rate equation, you _____.
True or false? Second-order reactants have a constant half-life.
Only species in _____ feature in the rate equation.
Flashcards in Rate Equations50+
Start learning
Why is the Arrhenius equation useful in chemistry?
It allows us to relate the temperature of a reaction with its rate
What does the letter A represent in the Arrhenius equation?
A is the Arrhenius constant
What information about a reaction can the value of ex give us in the Arrhenius equation?
the number of reacting particles that have enough energy to react
Rearrange the Arrhenius equation into its logarithmic form.
ln(k) = ln(A) - Ea/RT
Given a graph showing an Arrhenius plot for a chemical reaction, how could you use the plotted data to determine activation energy and the rate constant for that reaction?
The activation energy would be equal to the gradient of the line.
The rate constant would be equal to the y-intercept.
When drawing an Arrhenius plot, what would you label your axis?
The x-axis would be 1/T
The y-axis would be ln(K)
More about Rate Equations
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I’ve mentioned multiple times so far that my ‘true’ bra size is 30F. But really, how did I come up with this result? I used an online calculator and then tried on bras in store, having a good idea of how bras are supposed to fit. However, when I went in store and told the fitter I measured myself as 30F or 30FF, she told me we’d need to try on some bras to confirm that, and there’s a good reason to this: online bra size calculators are VERY frequently misleading, even extremely misleading.
Let’s take the time to measure exactly how misleading these calculators can be.
My measurements are 30″ loose underbust, 29″ snug underbust, 28″ tight underbust, 36″ standing bust, 38″ leaning bust, 36″ standing bust. Why did I take that many measurements? Because the most precise calculators can use all those 6 measurements, although most of them use only 2 measurements: snug underbust and standing bust. However, some weirder places, like Victoria’s Secret, also use another measurement instead of the underbust, which I took just for fun: the perimeter passing at the height of the band in the back and over the bust, which comes up at 35″ for me.
First, let’s calculate using the basic math I talked about in this post. My snug underbust is 29″, which puts my band size at 30, and my standing bust is 36″, 36 – 29 = 7″, which comes out as an F cup in UK sizing, or a G cup in US sizing. The other 5 measurements become useful if, for example, there was a big difference between my snug and tight underbust measurements, which could mean I’d want to get a number between the two instead of rounding up the snug measurements. It could also be useful to consider if my leaning bust was 3″ or more bigger than my standing or if my lying measurement was bigger than my standing, which could mean I’d need a FF or a G instead of F. But, since it’s not the case, 30F usually does the trick for me, so let’s go with that as my reference mark!
So, let’s start with stores that are familiar to me! La Vie en Rose is a Canadian lingerie company dedicated to the well-being of women. Their price tag is mid-range, they usually come up between CA\$40 and CA\$50, which is more expensive than La Senza and Victoria’s Secret. They carry sizes 32 A-C, 34-38 A-DD and 40 D-DD. Yeah, that’s right, they don’t go above C in 32 or under D in 40 (they may have 32-36 AA and 32D in some styles). They don’t have a size calculator, but here’s how their size chart works:
So… According to this chart, I should wear a 34C. Fairly normal, right? But 2 whole band sizes off and 2 cup sizes off too (my sister size in 34 is DD). While their grading would put people with 39″ underbusts in 38, which may actually be the right size, and people with 40-42″ underbusts in 40, their band size chart doesn’t get accurate until you are at least 32″ under the bust. Under that, they will fit you in bands that are too big for you. The cup size chart doesn’t make much sense either, with the increase between cup sizes being 3/4 of an inch instead of a full inch, which doesn’t make much sense. But it gets a bit weirder:
That’s right! La Vie en Rose knows about sister sizes! But they are totally oblivious to the fact that a bra should always be tried on on the loosest hook, not the middle one, and the fact that the small space between the hooks will not make a band that is 2″ too big to begin with a better fit, or a band that is 2″ too small less constricting (but that’s only if you were wearing the right band size to begin with). It can help someone stuck between band sizes or on a model that runs particularly tight or loose. But it’s not something to do “when your size is not available in the desired style”, as they put it, because wearing a band smaller than you need can quickly become uncomfortable, while wearing a band bigger than you need on the tightest hooks means the bra will stretch out too much to be wearable within less than a few months.
Now on to La Senza:
La Senza are so nice that they even did the math for me! I’m a 34C according to their handy size chart, again 2 band sizes too big and 2 cup sizes too small! If I was a normal person, I’d be very inclined to think that, since the two major bra stores in Canada tell me the same size, then they must be right. That’s to tell you how common my measurements are (enough that they take them for example), and how many women may be wearing the wrong bra size. However, as opposed to La Vie en Rose, La Senza consistently fits people in bands that are 2 sizes too big.
The last but not the least of mainstream stores is, of course, Victoria’s Secret:
Yeah, they’re the one using the weird measurement I took earlier on. They tell me I’m a 34B, so still 2 band sizes off, but this time 3 cup sizes too small. The weird thing is La Senza and Victoria’s Secret are owned by the same company, L Brands, which also owns Bath & Body Works, so I’d figured out they’d be using similar bra sizing techniques, but apparently not. La Senza was originally created in Canada but has always used similar marketing techniques to VS, so it was only natural that they’d be owned by the same head company at some point.
After these 3 companies, which are total disasters, I figured I would give a try to more specialized lingerie online stores and went with the 2 most popular online stores in North America: Herroom and Bare Necessities.
I’ll start with Herroom, since they make a big deal out of helping women finding their proper bra size:
I strongly suspect many people would feel overwhelmed by the amount of links they can clic and read through. It effectively seems like this website contains “Everything You’ll Ever Need to Know” about bra fitting. But the bra size calculator, what’s it worth?
Yup, they’re a band size off and 2 cup sizes too small. I have some old Passionata bras in 32D… And they’re awful: gore floating, band riding up with no support, straps digging in, quad-boob syndrome, etc. I’ve already dared to contact them about this matter and was met with a rather cold response indicating that it works fine for other women and they can’t make it perfect for everyone. Here’s their manual instructions:
It’s funny, but their advice concerning band sizes is only accurate after 38. They also inaccurately give the same cup size to people with different underbust measurements: someone with a 35″ underbust will be, according to their calculator, the same cup size as someone with a 38″ underbust if both are, let’s say, 40″ in the bust. In reality, those 2 people would need 2 very different sizes, the person with a 35″ underbust needing a 34DD/E or a 36D/DD, while the one with a 38″ underbust would need a 36C or a 38B (depending on their tight underbust measurement), as they’ll have very different breast volumes.
The fact that they mention that their calculator/calculation method is inaccurate after 4″ of difference between the bust and the band size and if people have pendulous breasts is quite curious for a website that stocks 974 different “Full Figure DD+” models and 590 “Plus-size” models, meaning that a good portion of the people shopping on their website fall out of their calculator’s accuracy range, which makes their size calculator quite nonsensical. It’s also quite curious considering that pendulousness doesn’t really influence the measurements (apart from the leaning bust measurement) and sizes as much as the shapes of bras, which they claim to have good knowledge about. I’m not going to take the time to break down the advice they give concerning bra styles as they give a lot of flawed advice, but I’ll let it be known that their categories are pretty much made-up. There is no such a thing as “petite” bras, they are conflating cropped bustiers with longline bras, they’re describing sports bras as “bras without underwires” (?), and they’re lumping together any and all 3-part cup bras, while 3-part cups are just a design feature and not a specific cup shape.
Now, let’s move on to Bare Necessities.
Yeah well, Bare Necessities doesn’t want you to measure. They consider that if you have fitting issues with your current bras, analyzing those issues should suffice to readjust your size. But the problem, as I have highlighted above, is that people who are feeling frustrated with their bra sizes are often multiple sizes apart from their true size; just being told to go down a cup/up a cup or down a band/up a band may not do much more than directing women to bras that they believe are the ideal fit, but are not, because they have no idea how a bra is truly supposed to fit. This approach also negates the many ways in which cup shapes can influence fit, or how wearing the wrong cup size can actually make you believe the band is tight enough/too small. Their “Check your Fit” page or “troubleshooting guide” don’t actually give much accurate advice about how to take care of these problems, and they are pretty oblivious about how shapes may influence fit. They still do give some calculation method to get a “starting point”, which is really what calculators are about anyway.
So, first thing: do NOT measure over your bust for the band size. It’s a handy way to overestimate band size that’s been used by many mainstream companies. It’s especially worse for people who have tall breast roots and/or V-shaped ribcages and/or muscular torsos. Everyone should measure for the band size where the band is supposed to pass (doing anything else is pretty illogical as, well, you’d want to know how large is the spot where the band is passing, not how large other spots are). As for the cup size, Bare Necessities also makes the mistake of advising to calculate cup size from the band size instead of the ribcage measurement, which doesn’t make sense but would still leave me one shy cup size away from my true size, with the right band size, which would be a 30E.
So, if all of these calculators are inaccurate, do I have a better option for you? Somewhat.
The most accurate calculator that I know of is maintained by the A Bra That Fits community of Reddit. Their calculator is not 100% perfect, but it gives a much closer idea of band and cup size that can actually be used as a base to figure out your best fitting size.
The pros of this calculator compared to others are numerous:
• Taking into account 5 measurements and having a more complex calculation algorithm means being able to suggest bra sizes fairly accurately for people down to 24 band size, up to 46 band size (but bras exist up to 52 and beyond) and up to MM cup size (but a few sizes exist beyond that too).
• Calculating things like the leaning bust gives a potential hint about breasts being pendulous or having softer tissue (which can bring their own fitting issues).
• Calculating breast size when naked means your breasts can’t be distorted by a too small or too big bra, which makes for more natural and accurate measurements.
• It suggests sister sizes, which are sizes in a different band but same cup volume, and puts up a warning when it’s likely that one may want to sister-size up for comfort or down for support.
But even if this calculator is more accurate, as a general rule, than others, it still suggests UK 30G for me, while my true size is UK 30F, because it can overestimate cup size in pendulous breasts that have a tinier difference between the leaning and standing/lying measurements. So even when you get the best calculation method out there, the results are still nothing more than a starting point, as bra fitting is not a science but an art. For what it’s worth, I would definitely recommend this calculator over the others shown in this post. Some indie bra shops also have fairly accurate size calculators, but from experience I can say that you should not follow what a size calculator tells you just because it’s a brand’s size calculator (be especially wary of calculators that put you in bigger band sizes than the usual). On contrary to what they may tell you, bra sizing is rather standardized, and they should not, and will not as a general rule run 2 band sizes smaller than other brands. Use what you know of your true size or contact the brand owner before ordering if need be, but not all bra designers actually know much about bra sizing, so don’t just trust people with your measurements!
Anyways, this is just to show you how the discourse about bra sizes can be confusing, even in places that claim to be specialists about bra sizing. So be careful, don’t buy things because someone tells you to buy them and trust your gut feeling when it comes to bras. Every little annoying thing about a bra you are trying on will get worse as you wear it (except in most cases fabric texture and band tightness).
Stay supported,
Justy
Note: This post is not meant to bash mentioned (or not mentioned) brands or stores, bra fitters, or anything or anyone else. It is also not endorsed by or supported in any way by any of the cited brands/stores. I do not mean that you should boycott those stores or brands, or that you are necessarily wearing the “wrong” size if you wear bra sizes that you found using these places’ calculators or following advice from bra fitters in these places. There are no right or wrong sizes, only best fitting sizes and less-than-best fitting sizes. This is simply meant as an illustration of how bra sizes are actually not well understood even within the industry, so it is normal that it can get very frustrating for people to try and find their best fitting size. | 3,071 | 13,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2018-43 | longest | en | 0.960856 |
https://www.ruby-forum.com/t/gmsk-gsm-initial-phase-for-a-burst/107173 | 1,722,871,064,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00700.warc.gz | 760,346,620 | 6,132 | # GMSK: GSM initial phase for a burst
Hello everyone.
I am looking at the frequency correction burst of a GSM signal.
According to spec the original data defines 148 bits of 0 that are
differentially encoded and GMSK modulated. After the data goes
through the differential encoder it ends up as [-1,1,1, (142 1s),
1,1,1]. I have taken this burst and plotted I vs Q (which is
attached). I was expecting to see the amplitude ramp up followed by a
negative pi/2 phase shift, followed by 147 positive pi/2 phase shifts,
and a ramp down. I never see the initial negative phase shift though.
Am I missing something in my sanity? Any thoughts or ideas?
Thank you for your time
Jeff
Jeffrey K. wrote:
Am I missing something in my sanity? Any thoughts or ideas?
Its hard to read the graph like that. Plot angle vs time or even
better, derivative of angle vs. time.
Matt
Great suggestion!
I have attached a couple of things:
1. The IQ plot (1e6 sample rate)
2. The burst data (MATLAB array)
3. The da vs time plot. (1 symbol = 3.69us)
Another quick question. I can now see that after the amplitude ramp
up that there is a short section of negative phase movement (.7
radians clockwise) prior to the remaining counterclockwise rotations.
Should I expect a full (-pi/2) to start off the burst? Is my sampling
rate just not high enough to catch the remaining rotation? Any other
suggestions?
Thank you for your time
Jeff
On 6/18/07, Jeffrey K. [email protected] wrote:
Another quick question. I can now see that after the amplitude ramp
up that there is a short section of negative phase movement (.7
radians clockwise) prior to the remaining counterclockwise rotations.
Should I expect a full (-pi/2) to start off the burst? Is my sampling
rate just not high enough to catch the remaining rotation? Any other
suggestions?
GMSK has a certain level of ISI built into the waveform. 0.7 radians
is about a pi/2 shift (0.2228 versus 0.25?) which would be pretty darn
close.
You can look at the inherent ISI within GMSK in a nice pretty graph in
this PDF:
http://www.ieee802.org/16/tg1/phy/contrib/802161pc-00_11.pdf
You may want to filter your samples down to 1 sample per symbol and
see if you get 3 dots close to each of your 4 constellation points (if
you have already compensated for any phase rotations).
Brian
On 6/18/07, Brian P. [email protected] wrote:
GMSK has a certain level of ISI built into the waveform. 0.7 radians
is about a pi/2 shift (0.2228 versus 0.25?) which would be pretty darn
close.
Whoops - looks like I’ve got QPSK on the brain today. Sorry about that.
Not really sure why you couldn’t see at least something close to a pi/2
shift.
Brian
Jeffrey K. wrote:
Should I expect a full (-pi/2) to start off the burst? Is my sampling
rate just not high enough to catch the remaining rotation? Any other
suggestions?
I don’t see anything at all wrong with this data.
Matt
Is there an initial condition to the differential encoder? Seems odd
that
they wouldn’t just send the 148 pi/2 phase shifts. -Clark
differentially encoded and GMSK modulated. After the data goes
through the differential encoder it ends up as [-1,1,1, (142 1s),
1,1,1]. I have taken this burst and plotted I vs Q (which is
attached). I was expecting to see the amplitude ramp up followed by a
negative pi/2 phase shift, followed by 147 positive pi/2 phase shifts,
and a ramp down. I never see the initial negative phase shift though.
Am I missing something in my sanity? Any thoughts or ideas?
Thank you for your time
Jeff
<< fcch2.png >>
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# IPS6eCh05_2bb - Sampling Distributions for Sample Means IPS...
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Sampling Distributions for Sample Means IPS Chapter 5.2 © 2009 W.H. Freeman and Company
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Objectives (IPS Chapter 5.2) Sampling distribution of a sample mean The mean and standard deviation of For normally distributed populations The central limit theorem Weibull distributions x
Reminder: What is a sampling distribution? The sampling distribution of a statistic is the distribution of all possible values taken by the statistic when all possible samples of a fixed size n are taken from the population. It is a theoretical idea — we do not actually build it. The sampling distribution of a statistic is the probability distribution of that statistic.
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Sampling distribution of the sample mean We take many random samples of a given size n from a population with mean μ and standard deviation σ . Some sample means will be above the population mean μ and some will be below, making up the sampling distribution. Sampling distribution of “ x bar” Histogram of some sample averages
Sampling distribution of x bar μ σ / n For any population with mean μ and standard deviation σ : The mean , or center of the sampling distribution of , is equal to x
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Mean of a sampling distribution of There is no tendency for a sample mean to fall systematically above or below μ , even if the distribution of the raw data is skewed. Thus, the mean of the sampling distribution is an unbiased estimate of the population mean μ — it will be “correct on average” in many samples. Standard deviation of a sampling distribution of The standard deviation of the sampling distribution measures how much the sample statistic varies from sample to sample. It is smaller than the standard deviation of the population by a factor of √ n . Averages are less variable than individual observations .
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# Beginner Programming Problem- Encryption/Decryption
Brandon Cahrenger
Greenhorn
Posts: 4
Hello, I'm new here but I've been browsing and there is a lot of very excellent information here. I'm also new to programming, having just started this semester. I like it a lot thus far, and although I'm still grasping the basics, I think I have a decent understanding of it all.
Anyways, here is my problem. I've written most of the code (along with another classmate) to get this to work, but I can't seem to return the encrypted value. I'm also trying to figure out how to separate the digits (or if I'm supposed to) to make the numbers encrypted. The main kicker is that we have to be using methods to return the final values. Everything compiles, but any help is very appreciated. Thanks!
Here is the question
A company wants to transmit data over the telephone, but they are concerned that their phones may be tapped. All of their data is transmitted as four-digit integers. They have asked you to write a program that will encrypt their data so that it may be transmitted more securely. Your program should read a four-digit integer number and encrypt it as follows: Replace each digit by the remainder after (the sum of that digit plus 7) is divided by 10. Then, swap the first digit with the third, and swap the second digit with the fourth. Then print the encrypted integer. Do the encryption in a method and send the encrypted number back to main. Here are the method headers: public static int encrypt(int num) // Takes num as a parameter and returns the encrypted value public static int getnum() // gets one number from the user Example: If the user enters 1234 they should see: 0189
Campbell Ritchie
Marshal
Posts: 56536
172
Welcome to the Ranch
I have added code tags to your post. Always use them: doesn't it look better.
You have written a+b+c+d. What is that supposed to do? Will you get a 4‑digit number from that?
Why have you marked all your methods static?
Campbell Ritchie
Marshal
Posts: 56536
172
You also have repeated code with adding 7 and taking the remainder. That should be separated into a method of its own.
Brandon Cahrenger
Greenhorn
Posts: 4
Thank you for the response Campbell Ritchie.
Thanks for the Code Tags, it does look much better.
The a+b+c+d is probably an error on my part (or it is an error on my part). That's definitely part of the problem and I'm trying to figure it out.
As for the static, that's what I've been taught so far, so that's all I know (although I'm researching it all as the info comes to me).
Campbell Ritchie
Marshal
Posts: 56536
172
What about 1000 * a + 100 * b + 10 * c + d?
Shiv Tattva
Greenhorn
Posts: 13
Try this
Tyson Lindner
Ranch Hand
Posts: 211
Basically you have to add the digits together as strings (not ints) to get a string, then convert your string back to an int before returning. It might be useful to use the Integer class from the start in order to do this. | 769 | 3,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-51 | latest | en | 0.937174 |
https://www.physicsforums.com/threads/determining-the-position-from-the-velocity.238167/ | 1,656,124,545,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033925.2/warc/CC-MAIN-20220625004242-20220625034242-00391.warc.gz | 1,035,056,386 | 15,911 | # Determining the position from the velocity
## Homework Statement
An object (masse m) is placed on a slope with an angle a at a height of h above the ground. There is no friction on the inclined plane. The inclined plane leads to a horizontal plane 2 meters long that ends with a spring with a constant of k. There is friction on the horizontal plane (but not the inclined plane). Where is the object when its velocity is equal to 1m/s?
m=1kg
h=0.5m
k=1000N/m
angle a= 30 degrees
F(friction)=1N
## Homework Equations
E(kin)=0.5*mv²
W = E(cin final) - E(cin initial)
## The Attempt at a Solution
W = 0.5*mv² - 0.5*m(0)²
=0.5*mv² = F*d
d=(mv²)/(2F)
=.5m
but I don't know how to add in the energie given by the spring and the energy lost from friction.
Any help would be appreciated thanks
arildno
Homework Helper
Gold Member
Dearly Missed
Was the force of friction given as 1N, or is that your assumption?
Insofar as it was given, remember the work-energy formula.
Yes, it's given. And sorry my attempt is a bit poor, I just need a bit of help starting off in the right direction. Thanks
I seriously doubt that you need all the info (but maybe I'm doing something wrong):
$$\frac{1}{2}mv^2 =mgh$$
solving for v:
$$v = \sqrt{2gh}$$
Inserting data I get 5 cm lower w.r.t initial height?
tiny-tim
Homework Helper
… but I don't know how to add in the energie given by the spring and the energy lost from friction.
Hi zakare!
The energy absorbed by the spring is 1/2*kx², where x is the decrease in length of the spring;
and the energy lost from friction equals the work done against friction.
It will also reach 1 m/s when its a bit in the spring
edit: first post and im late :(
tiny-tim
Homework Helper
Welcome to PF!
Hi Multicol ! Welcome to PF!
It will also reach 1 m/s when its a bit in the spring
edit: first post and im late :(
that's ok … spring was early this year! | 535 | 1,895 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2022-27 | latest | en | 0.933498 |
https://calciosfondo.jdevcloud.com/fact-family-worksheets-kindergarten/19134/ | 1,675,777,953,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500619.96/warc/CC-MAIN-20230207134453-20230207164453-00678.warc.gz | 177,255,579 | 9,034 | # Fact Family Worksheets Kindergarten 2021
Fact Family Worksheets Kindergarten. Ad the most comprehensive library of free printable worksheets & digital games for kids. Ad the most comprehensive library of free printable worksheets & digital games for kids.
Add and subtract (sums to 10) complete the number family shown by each ten frame. Add and subtract (sums to 20) complete the number family shown by each pair of ten frames.
### Addition Fact Family Activity Fact Families Activities
Addition fact family worksheet for kindergarten if you are looking for a way to teach your children and have not had much success in the past, why not try one of the new “fact family workbooks.” these books, based on the popular television show “family matters,” are designed to be fun, educational, and also fun for your children! After that, the factors include only 2 and up.
### Fact Family Worksheets Kindergarten
Cats are also included, and dogs are also included for the older students.Check out the first worksheet below!Children cut and paste the equations to the matching sum on the windows.Constant practice with these tools strengthens their understanding of.
Download 20 first grade math worksheets and printables.Each page has 9 different math houses.Fact family house worksheets free printable fact family worksheets for first grade, free printable fact family worksheets for second grade, free printable multiplication division fact family worksheets, free printable math fact family worksheets, free printable first grade math fact family worksheets, , image source:Fact family worksheets help your students to begin to.
Family worksheets challenge your students by numbers up to 20.additional fact family worksheets &;For example, in the fact families to 100 worksheets, the factors.Free printable fact family worksheets.Free watermelon fact family worksheet.
In the first set with the yellow triangles, the student is given the numbers and must then show the addition and subtraction relationships between them.Kindergarten fact families are used to compose and decompose numbers.Kindergarten math springtime houses for fact family printables is an 8 page, 16 fact family resource.Let’s talk about it in the fact family trees lesson!
Math (springtime) houses are a fun way to learn and practice the fact family concept with addition and subtraction patterns.Our free fact family worksheets are fantastic for learning addition, subtraction, multiplication, and division facts.Printable fact family worksheets for addition and subtraction facts are also.Save and print this collection of printable fact family worksheets for testing children in their ability to build fact families based on their understanding of number relationship.
Save and print this collection of printable fact family worksheets for testing children in their ability to build fact families based on their understanding of number relationships.See more ideas about fact families, math classroom, math facts.Simply download pdf file with fact family worksheets and you are ready to play and learn with cute watermelon activities.Students will use the three numbers in the roof triangle to make four different addition and subtraction equations.
Teachers are provided with worksheets that break down addition and subtraction fact families and are encouraged to ask students how equations in fact.The first worksheets only have single digit numbers;The later worksheets use slightly larger numbers.The number 1 as a factor is only used in the fact families to 49 and 64 worksheets.
The picture quality will help to make it easy for them to find the fact they need.The story that is based on the popular family of animals, birds, mammals, birds, reptiles, and amphibians can be used.The structure of fact family worksheets allows students to complete them within a couple of minutes.The window flaps allow children to practice the facts and reinforce the fact families.
There are two basic types of papers.These 3 numbers make a set of related math facts.These fact family practice worksheets introduce your students to addition fact families, subtraction fact families, multiplication fact families, and division fact families.These fact family worksheets are perfect for practicing how to build the fact family sets for the given numbers for addition and subtraction as well as multiplication and division in the future.
These fact family worksheets for first grade are perfect for practicing how to build the fact family sets for the given numbers for addition and subtraction as well as multiplication and division.These fact family worksheets should help make those relationships very clear to students and perhaps help them to understand division a little better.These fact family worksheets will help you with many numbers relation exercises!These first grade fact family worksheets are perfect for practicing how to build the fact family sets for the given numbers for addition and subtraction as well as multiplication and division in the future.
These super cute, free printable watermelon worksheet pages are a fun way for kindergarten and first graders practice with hands on fact family worksheets.These worksheets are great for any math curriculum including singapore math that we use ( they.These worksheets ask students to write out the 4 different addition or subtraction equations representing each fact family.These worksheets provide children with an interactive way to see the relationships in a fact family.
This concept is just the same with how people in a family are related.This is great for seeing the relationship between them.This math lesson plan is perfect for first graders who are learning about subtraction within 20, addition within 20, and inverse operations.This method of counting saves them time and will effectively help them to move up to higher math skill.
Use the math worksheets below to build upon the concepts explained above.When 3 numbers are related, a fact family is built.You’ll notice a questions mark in some of the families. | 1,100 | 6,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-06 | latest | en | 0.907834 |
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Posted by4 months ago
## Could someone walk through what this code does? I'm new
:1→Y :16→A :1→B :5→E :0→Z :Lbl G :getKey→K :A-1→A :randInt(2,7)→D :ClrHome :If not(A:16→A :If not(A=16:0→B :If A=16:1→B :If B=1:D=E :If K=25:Y-1→Y :If not(K=25:Y+1→Y :Output(Y,3,"0 :For(I,1,8) :Output(I,A,"H :End :Output(E,A," "//1 space :Output(E+1,A," "//1 space :If A=3 :Then :If not(Y=E) and not(Y=E+1 :Output(10,10,"P :Output(1,8,Z :Z+1→Z :End :Goto G
1
Posted by1 year ago
Archived
## fractions
I want to program the calculator to keep the answer in a fraction even when there's a square root. Is there anyway to do that?
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the ti83-84 calculator programming language | 298 | 805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-30 | latest | en | 0.846218 |
https://www.cnblogs.com/UnderSilenceee/p/7295994.html | 1,563,838,927,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528290.72/warc/CC-MAIN-20190722221756-20190723003756-00388.warc.gz | 661,229,429 | 4,974 | hdu6035[dfs+思维] 2017多校1
/*hdu6035[dfs+思维] 2017多校1*/
//合并色块, 妙啊妙啊
#include<bits/stdc++.h>
using namespace std;
const double eps=1e-8;
const int inf=0x3f3f3f3f;
typedef long long LL;
vector<int>G[200005];
LL sum[200005];
int c[200005],son[200005],mark[200005];
int n,u,v,cnt=0,kase=1;
LL ans,res;
void dfs(int u,int fa){
son[u]=1;
LL y=sum[c[u]];
LL x=sum[c[u]];
for(int i=0;i<G[u].size();++i){
int v=G[u][i];
if(v!=fa){
dfs(v,u);
son[u]+=son[v];
LL temp=son[v]-sum[c[u]]+x;
res+=(temp-1LL)*temp/2LL;
x=sum[c[u]];
}
}
sum[c[u]]+=son[u]-(sum[c[u]]-y);
//printf("Node #%d, Color:%d, Sum[%d]:%lld, Son:%d\n",u,c[u],c[u],sum[c[u]],son[u]);
}
void solve(){
res=ans=0;
dfs(1,-1);
//cout<<"Res: "<<res<<endl;
ans=cnt*1LL*n*(n-1)/2LL-res;
for(int i=1;i<=n;i++){
if(mark[i]){
ans-=(n-sum[i])*(n-sum[i]-1)/2LL;
//printf("Color:%d, Sum[%d]:%d, Ans:%d\n",i,i,sum[i],ans);
}
}
printf("Case #%d: %lld\n",kase++,ans);
}
int main(){
while(~scanf("%d",&n)){
cnt=0;
memset(sum,0,sizeof(sum));
memset(son,0,sizeof(son));
memset(mark,0,sizeof(mark));
for(int i=0;i<=n+1;i++)
G[i].clear();
for(int i=1;i<=n;i++){
scanf("%d",&c[i]);
if(!mark[c[i]]){
mark[c[i]]++;
cnt++;
}
}
for(int i=1;i<=n-1;i++){
scanf("%d%d",&u,&v);
G[u].push_back(v);
G[v].push_back(u);
}
solve();
}
return 0;
}
/*
8
1 2 1 1 2 1 2 1
1 2
1 3
2 4
2 5
3 6
4 7
5 8
12
1 2 1 1 2 1 2 2 1 2 2 2
1 2
1 3
2 4
2 5
3 6
3 7
4 8
5 9
5 10
7 11
7 12
13
1 2 1 1 2 1 2 2 1 1 2 2 2
1 2
1 3
2 4
2 5
3 6
3 7
4 8
4 9
5 10
5 11
7 12
7 13
*/
posted @ 2017-08-06 21:07 UnderSilence 阅读(...) 评论(...) 编辑 收藏 | 746 | 1,529 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-30 | latest | en | 0.116664 |
https://www.jiskha.com/display.cgi?id=1320431979 | 1,516,774,548,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893397.98/warc/CC-MAIN-20180124050449-20180124070449-00128.warc.gz | 929,509,188 | 4,068 | # statistics
posted by .
A sample of n = 9 has a mean of M = 72 with SS = 162. Use the sample to make a point estimate of ì and to construct the 90% confidence interval for ì. (Round all answers to two decimal places
ì = ? ± ? ( ? )
The 90% confidence interval for ì:
( ? , ? )
• statistics -
Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.
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More Similar Questions | 661 | 2,851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-05 | latest | en | 0.89223 |
https://fr.slideserve.com/bach/static-force-analysis-powerpoint-ppt-presentation | 1,721,778,356,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00138.warc.gz | 214,851,644 | 21,103 | 1 / 13
# STATIC FORCE ANALYSIS
STATIC FORCE ANALYSIS. WEEK 1. COUPLE MOMENT OF COUPLE. Couple : Two equal and opposite forces ( F & F ’) Moment of Couple : A vector normal to the plane of the couple ( M = R BA X F ). Conditions of Equilibrium. A system of bodies is in equilibrium, if, and only if:.
Télécharger la présentation
## STATIC FORCE ANALYSIS
E N D
### Presentation Transcript
1. STATIC FORCE ANALYSIS WEEK 1
2. COUPLE MOMENT OF COUPLE Couple: Two equal and opposite forces (F & F’) Moment of Couple: A vector normal to the plane of the couple (M=RBAX F)
3. Conditions of Equilibrium A system of bodies is in equilibrium, if, and only if: In 2-D (planar) systems:
4. Two-Force Members Not in equilibrium F0, M0 Not in equilibrium F=0, M0 In equilibrium F=0, M=0 Condition of equilibrium, for any two-force member with no applied torque: The forces are equal, opposite and have the same line of action.
5. Three-Force Members O Not in equilibrium M0 In equilibrium F=0, M=0 Condition of equilibrium, for any three-force member with no applied torque: Forces should be coplanar Four-Force Members:The problem is reduced to one of three-force member. Then the approach above is applied.
6. Example (Graphical Solution) a) Link 3 is a two-force member C
7. 1 N=0.225 lb b) Link 4 is a three-force member c )1 st approach: OR c’ ) 2 nd approach: Since link 4 is is a three-force member; lines of action of forces P, F34 should intersect at a point . Therefore direction of F14 F14 is obtained. d) Force triangle is used. If 1st approach was used, this triangle gives direction and magnitude of If 2nd approach was used, this triangle gives magnitudes of F34, F14 Since, graphically F14 e) Link 3// Action and reaction forces are equal: and f)Link 2// F=0
8. (Analytical Solution) 4// = 5(cos68.4i+sin68.4j) X 120(-cos40i-sin40j) +12(cos68.4i+sin68.4j) X F34(cos 22.4i+sin 22.4j) = 0 F34 = 33.1 lb 2//
9. Problem (AnalyticalSolution) (withoutfriction) O2A=75 mm P=0.9 kN AB=350 mm M12=? 2 3 4 4// F=0 F34+P+F14=0 F34(cos11,95i-sin11,95j)-900i+F14j=0 i: 0.978 F34 = F34x= 900 F34 = 920.25 N j: 0.207 F34 = F34y= F14F14 = 190.5 N F34= 920,25 /-11,95 N = 900i-190.5j N F14= 190.5j N For the moment balance, all of the force vectors should pass through point B.
10. 3// A two-force member F=0 F23=-F43=F34 2// F=0 F12=-F32=F23=F34 MO2=0 M12 + O2A x F32 = 0 M12k + 0.075(cos105i+sin105j) x (-900i+190.5j) = 0 M12k –3.69k + 65.2k = 0 M12 = -61.51 N.m M12 = -61.51k N.m
11. Problem (AnalyticalSolution) (withfriction) =0.2 (Between piston and cylinder) 3 2 4 4// F=0F34+P+F14=0 F34+P+(0.2Ni+Nj)=0 F34(cos11.95i-sin11.95j)-900i+0.2Ni+Nj=0 i: 0.978 F34 -900+0.2N = 0 j: 0.207 F34 + N= 0 N= 0.207 F34 0.978 F34 -900+0.2(0.207F34)= 0 F34 = 882.61 Newton N=182.7 NewtonF14 = 36.54i+182.7j Newton F34= 882.61 /-11,95 Newton = 863.48i-182.75j Newton F=0F12=-F32=F23=F34 MO2=0M12 + O2A x F32 = 0 M12k + 0.075(cos105i+sin105j) x (-863.48i+182.75j) = 0 M12k –3.547k + 62.55k = 0 M12 = -59.003 N.mM12 = -59.003k N.mCONCLUSION ? 0.2N 2//
More Related | 1,221 | 3,086 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-30 | latest | en | 0.820015 |
https://nrich.maths.org/public/topic.php?code=17&cl=4&cldcmpid=336 | 1,601,531,478,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402130615.94/warc/CC-MAIN-20201001030529-20201001060529-00171.warc.gz | 484,295,201 | 8,138 | # Resources tagged with: Powers & roots
Filter by: Content type:
Age range:
Challenge level:
### There are 41 results
Broad Topics > Properties of Numbers > Powers & roots
### Rational Roots
##### Age 16 to 18 Challenge Level:
Given that a, b and c are natural numbers show that if sqrt a+sqrt b is rational then it is a natural number. Extend this to 3 variables.
### Absurdity Again
##### Age 16 to 18 Challenge Level:
What is the value of the integers a and b where sqrt(8-4sqrt3) = sqrt a - sqrt b?
### Plus or Minus
##### Age 16 to 18 Challenge Level:
Make and prove a conjecture about the value of the product of the Fibonacci numbers $F_{n+1}F_{n-1}$.
### Golden Eggs
##### Age 16 to 18 Challenge Level:
Find a connection between the shape of a special ellipse and an infinite string of nested square roots.
### In Between
##### Age 16 to 18 Challenge Level:
Can you find the solution to this algebraic inequality?
### Ab Surd Ity
##### Age 16 to 18 Challenge Level:
Find the value of sqrt(2+sqrt3)-sqrt(2-sqrt3)and then of cuberoot(2+sqrt5)+cuberoot(2-sqrt5).
### How Many Solutions?
##### Age 16 to 18 Challenge Level:
Find all the solutions to the this equation.
### Square Pair Circles
##### Age 16 to 18 Challenge Level:
Investigate the number of points with integer coordinates on circles with centres at the origin for which the square of the radius is a power of 5.
### Irrational Arithmagons
##### Age 16 to 18 Challenge Level:
Can you work out the irrational numbers that belong in the circles to make the multiplication arithmagon correct?
### Googol
##### Age 16 to 18 Short Challenge Level:
Find the smallest value for which a particular sequence is greater than a googol.
### Pythagorean Fibs
##### Age 16 to 18 Challenge Level:
What have Fibonacci numbers got to do with Pythagorean triples?
### Fibonacci Fashion
##### Age 16 to 18 Challenge Level:
What have Fibonacci numbers to do with solutions of the quadratic equation x^2 - x - 1 = 0 ?
### Giants
##### Age 16 to 18 Challenge Level:
Which is the bigger, 9^10 or 10^9 ? Which is the bigger, 99^100 or 100^99 ?
### The Root of the Problem
##### Age 14 to 18 Challenge Level:
Find the sum of this series of surds.
### Rationals Between...
##### Age 14 to 16 Challenge Level:
What fractions can you find between the square roots of 65 and 67?
### Bina-ring
##### Age 16 to 18 Challenge Level:
Investigate powers of numbers of the form (1 + sqrt 2).
### Em'power'ed
##### Age 16 to 18 Challenge Level:
Find the smallest numbers a, b, and c such that: a^2 = 2b^3 = 3c^5 What can you say about other solutions to this problem?
### Staircase
##### Age 16 to 18 Challenge Level:
Solving the equation x^3 = 3 is easy but what about solving equations with a 'staircase' of powers?
### Cube Roots
##### Age 16 to 18 Challenge Level:
Evaluate without a calculator: (5 sqrt2 + 7)^{1/3} - (5 sqrt2 - 7)^1/3}.
### Guesswork
##### Age 14 to 16 Challenge Level:
Ask a friend to choose a number between 1 and 63. By identifying which of the six cards contains the number they are thinking of it is easy to tell them what the number is.
### Roots Near 9
##### Age 14 to 16 Short Challenge Level:
For how many integers 𝑛 is the difference between √𝑛 and 9 is less than 1?
### Perfectly Square
##### Age 14 to 16 Challenge Level:
The sums of the squares of three related numbers is also a perfect square - can you explain why?
### Lost in Space
##### Age 14 to 16 Challenge Level:
How many ways are there to count 1 - 2 - 3 in the array of triangular numbers? What happens with larger arrays? Can you predict for any size array?
### Surds
##### Age 14 to 16 Challenge Level:
Find the exact values of x, y and a satisfying the following system of equations: 1/(a+1) = a - 1 x + y = 2a x = ay
### Enriching Experience
##### Age 14 to 16 Challenge Level:
Find the five distinct digits N, R, I, C and H in the following nomogram
### Unusual Long Division - Square Roots Before Calculators
##### Age 14 to 16 Challenge Level:
However did we manage before calculators? Is there an efficient way to do a square root if you have to do the work yourself?
### Function Pyramids
##### Age 16 to 18 Challenge Level:
A function pyramid is a structure where each entry in the pyramid is determined by the two entries below it. Can you figure out how the pyramid is generated?
### Root to Poly
##### Age 14 to 16 Challenge Level:
Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.
### Fit for Photocopying
##### Age 14 to 16 Challenge Level:
Explore the relationships between different paper sizes.
### Route to Root
##### Age 16 to 18 Challenge Level:
A sequence of numbers x1, x2, x3, ... starts with x1 = 2, and, if you know any term xn, you can find the next term xn+1 using the formula: xn+1 = (xn + 3/xn)/2 . Calculate the first six terms of this. . . .
### Archimedes Numerical Roots
##### Age 16 to 18 Challenge Level:
How did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?
### Mod 7
##### Age 16 to 18 Challenge Level:
Find the remainder when 3^{2001} is divided by 7.
### Negative Power
##### Age 14 to 16 Challenge Level:
What does this number mean ? Which order of 1, 2, 3 and 4 makes the highest value ? Which makes the lowest ?
### Rachel's Problem
##### Age 14 to 16 Challenge Level:
Is it true that $99^n$ has 2n digits and $999^n$ has 3n digits? Investigate!
### Archimedes and Numerical Roots
##### Age 14 to 16 Challenge Level:
The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?
### Consecutive Squares
##### Age 14 to 16 Challenge Level:
The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false?
### Number Rules - OK
##### Age 14 to 16 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Power Countdown
##### Age 14 to 16 Challenge Level:
In this twist on the well-known Countdown numbers game, use your knowledge of Powers and Roots to make a target.
### Deep Roots
##### Age 14 to 16 Challenge Level:
Find integer solutions to: $\sqrt{a+b\sqrt{x}} + \sqrt{c+d.\sqrt{x}}=1$
### Equal Temperament
##### Age 14 to 16 Challenge Level:
The scale on a piano does something clever : the ratio (interval) between any adjacent points on the scale is equal. If you play any note, twelve points higher will be exactly an octave on. | 1,725 | 6,709 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-40 | latest | en | 0.751797 |
https://ask.sagemath.org/question/49634/how-to-better-plot-elliptic-curves-over-finite-fields/ | 1,716,012,993,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057327.58/warc/CC-MAIN-20240518054725-20240518084725-00657.warc.gz | 89,710,640 | 15,356 | # How to better plot elliptic curves over finite fields?
I'm trying to get an elliptic curve plot, but the points are too thick and the resolution too low. This makes the points stick together in a mess. The following code
E = EllipticCurve(GF(next_prime(20000)),[0,1])
E.plot()
results in this image(https:// imgur.com/a/aMFpFXN). How can i make the points less thick/the resolution bigger?
edit retag close merge delete
1
From E.plot?:
• "args, *kwds" - all other options are passed to the circle graphing primitive.
So did you try circle? and circle.options?
( 2020-01-22 06:23:57 +0200 )edit
Sort by ยป oldest newest most voted
I don't think it will be instructive, for learning purposes, to use a prime that big and plot it (assuming this is what you want to do) and I suspect the problem is the size of the field.
Here's the curve I use as an example, with cofactor 1, for "ecc" curves that are like NIST ones.
E=EllipticCurve(GF(17),[3,5])
E.order()
E.is_supersingular()
E.plot()
As you can see, it's a prime order group (so cofactor 1) and not supersingular, although it is obviously way too small for real world use. The plot however is nicer. Can't upload it to show you, but you can reproduce from this.
more
Let us understand first the called plot routine, asking for "help" and "full help". Using
E = EllipticCurve(GF(next_prime(20000)),[0,1])
E.plot?
we obtain also the information where the code is located:
Init docstring: Initialize self. See help(type(self)) for accurate signature.
File: /usr/lib/python3.8/site-packages/sage/schemes/elliptic_curves/ell_finite_field.py
Type: method
Usually, E.plot?? would be enough to get the way the method works,
Source:
def plot(self, *args, **kwds):
"""
!!! ... long doc string ... !!! PLEASE CHECK IT
"""
R = self.base_ring()
if not R.is_prime_field():
raise NotImplementedError
G = plot.Graphics()
G += plot.points([P[0:2] for P in self.points() if not P.is_zero()], *args, **kwds)
return G
File: /usr/lib/python3.8/site-packages/sage/schemes/elliptic_curves/ell_finite_field.py
but in this case we need more, since the plot in the line G = plot.Graphics() is not the global plot, but the one in the lines...
import sage.plot.all as plot
class EllipticCurve_finite_field(EllipticCurve_field, HyperellipticCurve_finite_field):
at the beginning of the file. OK, we know the origin of the plot method, can also ask for its documentation:
sage: import sage.plot.all as xplot
sage: ?xplot.points
and the last points method sends us to point2d, where we finally get the meaning of the options we can use, for instance
sage: xplot.point2d?
also contains examples using size and pointsize. Defaults are
sage: xplot.point2d.options
{'alpha': 1,
'aspect_ratio': 'automatic',
'faceted': False,
'legend_color': None,
'legend_label': None,
'marker': 'o',
'markeredgecolor': None,
'rgbcolor': (0, 0, 1),
'size': 10}
So in order to get smaller points, one can try:
sage: E.plot(pointsize=1)
and note that the option pointsize=1 is passed to the plot through the line in the code of E.plot:
Source:
def plot(self, *args, **kwds):
"""etc
"""
R = self.base_ring()
if not R.is_prime_field():
raise NotImplementedError
G = plot.Graphics()
G += plot.points([P[0:2] for P in self.points() if not P.is_zero()], *args, **kwds)
return G
in the line with G += ... where we get the *args, **kwds from the E.plot caller, and insert them to the plot.points. To have access to the size of the graphics, we need to have access to G, so it is maybe a good idea to construct the G object with bare hands and use its G.show method with the right / needed figsize. See the example in xplot.Graphics?
more | 993 | 3,708 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-22 | latest | en | 0.895432 |
https://www.jiskha.com/display.cgi?id=1286227767 | 1,503,462,317,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117519.92/warc/CC-MAIN-20170823035753-20170823055753-00071.warc.gz | 917,468,971 | 4,224 | # Physics
posted by .
A bicycle wheel that initially spins at 3 revolutions per second is braked uniformly to a stop in 1 second. A spot is painted at the outer edge of the wheel. How many revolutions does the spot go during the time it takes the wheel to stop?
The answer I got was 0.5 but was incorrect so can someone show me how to do this problem.
• Physics -
I like moving things to radians, not revs.
angularaccleration=changevleociyt/time
finalangulvelocity^2=intialangularvelocity^2+2*angularacceleration*displacement
0=(3*2PI)^2+2(-2*3PI)*displacment
solve for displacement (in radians) ; divide by 2PI to get revs.
• Physics -
Ok thanks.
• Physics -
1.5 revolutions.
If it starts at 3 rev/sec and ends at 0 rev/sec and decelerates uniformly, then the avg is 1.5 rev/sec. It spent 1 second, so 1.5 rev/sec * 1 sec = 1.5 rev
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More Similar Questions | 772 | 3,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-34 | latest | en | 0.88336 |
https://www.physicsforums.com/threads/conservation-of-motion.345994/ | 1,532,237,465,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593010.88/warc/CC-MAIN-20180722041752-20180722061752-00109.warc.gz | 989,590,835 | 13,488 | # Homework Help: Conservation of motion
1. Oct 15, 2009
### Chuck 86
1. The problem statement, all variables and given/known data
boy in a wheelchair (total mass, 48.7 kg) wins a race with a skateboarder. He has a speed of 1.23 m/s at the crest of a slope 2.37 m high and 14.0 m long. At the bottom of the slope, his speed is 6.27 m/s. If air resistance and rolling resistance can be modeled as a constant frictional force of 41.8 N, calculate the work he did in pushing forward on his wheels during the downhill ride.
2. Relevant equations
Ei-Ef=Wfriction+Woutside forces
1/2mv^2, mgh
3. The attempt at a solution
cant seem to set this up right. dont know the equations for the 2 work
2. Oct 15, 2009
### Donaldos
Can't you think of a general expression for work that you could use here?
3. Oct 15, 2009
### Chuck 86
Not of the outside forces
4. Oct 15, 2009
### Chuck 86
that would be cool to get some help instead of pointing out that im not smart, thats why im asking for help
5. Oct 16, 2009
### Donaldos
What are the outside forces?
In particular, what is the work done by gravity?
6. Oct 16, 2009
### Chuck 86
MgH=(48.7)(9.8)(14)
7. Oct 16, 2009
### Donaldos
OK but H=2.37 m here, not 14m.
Now you just need to calculate the work done by friction. | 389 | 1,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2018-30 | latest | en | 0.920576 |
https://www.cpalms.org/PreviewStandard/Preview/5363 | 1,656,949,200,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104432674.76/warc/CC-MAIN-20220704141714-20220704171714-00512.warc.gz | 770,605,078 | 19,768 | # MAFS.3.OA.4.9
Identify arithmetic patterns (including patterns in the addition table or multiplication table), and explain them using properties of operations. For example, observe that 4 times a number is always even, and explain why 4 times a number can be decomposed into two equal addends.
General Information
Subject Area: Mathematics
Domain-Subdomain: Operations and Algebraic Thinking
Cluster: Level 3: Strategic Thinking & Complex Reasoning
Cluster: Solve problems involving the four operations, and identify and explain patterns in arithmetic. (Major Cluster) -
Clusters should not be sorted from Major to Supporting and then taught in that order. To do so would strip the coherence of the mathematical ideas and miss the opportunity to enhance the major work of the grade with the supporting clusters.
Date of Last Rating: 02/14
Status: State Board Approved
Assessed: Yes
Test Item Specifications
• Assessment Limits :
Adding and subtracting is limited to whole numbers within 1,000. All values in items may not exceed whole number multiplication facts of 10 x 10 or the related division facts.
• Calculator :
No
• Context :
No context
Sample Test Items (1)
• Test Item #: Sample Item 1
• Question:
A multiplication table is shown.
Which statement correctly describes how to find the multiples of 6 in the multiplication table?
• Difficulty: N/A
• Type: MC: Multiple Choice
## Related Courses
This benchmark is part of these courses.
5012050: Grade Three Mathematics (Specifically in versions: 2014 - 2015, 2015 - 2022 (current), 2022 and beyond)
7712040: Access Mathematics Grade 3 (Specifically in versions: 2014 - 2015, 2015 - 2018, 2018 - 2022 (current), 2022 and beyond)
5012055: Grade 3 Accelerated Mathematics (Specifically in versions: 2019 - 2022 (current), 2022 and beyond)
5012015: Foundational Skills in Mathematics 3-5 (Specifically in versions: 2019 - 2022 (current), 2022 and beyond)
## Related Access Points
Alternate version of this benchmark for students with significant cognitive disabilities.
MAFS.3.OA.4.AP.9a: Identify and describe the rule for a numerical pattern where numbers increase by 2, 5 or 10.
MAFS.3.OA.4.AP.9b: Select or name the three next terms in a numeral pattern where numbers increase by 2, 5, or 10.
MAFS.3.OA.4.AP.9c:
Identify multiplication patterns in a real-world setting.
## Related Resources
Vetted resources educators can use to teach the concepts and skills in this benchmark.
## Assessments
Sample 3 - Third Grade Math State Interim Assessment:
This is a State Interim Assessment for third grade.
Type: Assessment
Sample 1 - Third Grade Math State Interim Assessment:
This is a State Interim Assessment for third grade.
Type: Assessment
## Educational Game
Number Facts Bingo:
This Flash applet generates number fact questions for the game of Bingo. Each of the six levels focuses on a different range of number facts (addition, subtraction, and multiplication), which are displayed one at a time in a variety of question formats. The applet is intended for use in a class/group setting with a projector or interactive whiteboard. Downloadable cards for each level are available from the menu page. At any time in a game the "number facts so far" feature will reveal all the questions presented in the current round to facilitate review or verification of a winning board.
Type: Educational Game
## Formative Assessments
Students are presented with an equation and asked to find a pattern within the equation and to determine if the equation is true or not.
Type: Formative Assessment
Students are asked to consider what type of number results when adding two odd numbers and when adding three odd numbers.
Type: Formative Assessment
Students are asked to consider the parity of the sums of two even numbers, two odd numbers, and an even and an odd.
Type: Formative Assessment
Patterns Within the Multiplication Table:
Students are asked to find the missing numbers in a column of a multiplication table by using a pattern found within the table.
Type: Formative Assessment
Multiplication of Even Numbers:
Students are asked to determine if the total number of students in five classes will be even or odd.
Type: Formative Assessment
## Lesson Plans
Tricky Rice Math Patterns MEA:
This is a 3rd grade MEA that requires students to use mathematical patterns to solve the problem, along with the analysis of data. After reading One Grain of Rice by Demi, students will look for ways to help Rani's relative find a new pattern so she can secure a large supply of rice to feed the people of her province in India. The twist is likely to cause controversy, so prepare for some strong debates.
Type: Lesson Plan
The Power of Patterns:
Students will work a real world problem to discover similarities and differences between the patterns of adding 2 to a number and doubling a number. The problem is set in the real world context of twin brothers who choose different patterning strategies given by their grandma to save for buying a car.
Type: Lesson Plan
## Lesson Study Resource Kit
Operations and Algebraic Thinking Lesson Study Resource Kit – Third Grade:
This lesson study resource kit can be used to guide and support teams of third grade teachers as they engage in lesson study focused on the academic standards in the Operations and Algebraic Thinking domain.
Type: Lesson Study Resource Kit
## Original Student Tutorials
Party Patterns: Odds and Evens in Addition - Part 3:
Determine if the sum of three odd or three even numbers will be odd or even as Lilly prepares for a math celebration in this interactive tutorial.
This is part 3 in a 3-part series. Click below to explore the other tutorials in the series.
Type: Original Student Tutorial
Party Patterns: Evens and Odds in Addition – Part 2:
Explore addition patterns to find if the sum of an odd and an even number will be odd or even in this interactive tutorial.
This is part 2 in a 3-part series. Click below to explore the other tutorials in the series.
Type: Original Student Tutorial
Party Patterns: Evens and Odds in Addition – Part 1:
Determine whether the sum of two odd numbers is odd or even and whether the sum of two even numbers is odd or even by helping Lilly prepare for a math celebration in this interactive tutorial.
This is part 1 in a 3-part series. Click below to explore the other tutorials in the series.
• Part 2: Party Patterns: Evens and Odds in Addition (COMING SOON)
• Part 3: Party Patterns: Evens and Odds in Addition (COMING SOON)
Type: Original Student Tutorial
The goal of this task is to help students understand the commutative property of addition by examining the addition facts for single digit numbers. This is important as it gives students a chance, at a young age, to do more than memorize these arithmetic facts which they will use throughout their education.
Making a ten:
This task asks students to study more carefully the make-a-ten strategy that they should already know and use intuitively. In this strategy, knowledge of which sums make a ten, together with some of the properties of addition and subtraction, are used to evaluate sums which are larger than 10. This task is intended for instruction purposes as it takes time to identify the patterns involved and understand the steps in the procedures.
The purpose of this task is to study some patterns in a small addition table. Each pattern identified persists for a larger table and if more time is available for this activity students should be encouraged to explore these patterns in larger tables.
Patterns in the multiplication table:
The goal is to look for structure and identify patterns and then try to find the mathematical explanation for this. This problem examines the ''checkerboard'' pattern of even and odd numbers in a single digit multiplication table. The even numbers in the table are examined in depth using a grade appropriate notion of even, namely the possibility of reaching the number counting by 2's or expressing the number as a whole number of pairs.
## Student Center Activity
Students can practice answering mathematics questions on a variety of topics. With an account, students can save their work and send it to their teacher when complete.
Type: Student Center Activity
## STEM Lessons - Model Eliciting Activity
Tricky Rice Math Patterns MEA:
This is a 3rd grade MEA that requires students to use mathematical patterns to solve the problem, along with the analysis of data. After reading One Grain of Rice by Demi, students will look for ways to help Rani's relative find a new pattern so she can secure a large supply of rice to feed the people of her province in India. The twist is likely to cause controversy, so prepare for some strong debates.
## MFAS Formative Assessments
Students are asked to consider what type of number results when adding two odd numbers and when adding three odd numbers.
Students are asked to consider the parity of the sums of two even numbers, two odd numbers, and an even and an odd.
Students are presented with an equation and asked to find a pattern within the equation and to determine if the equation is true or not.
Multiplication of Even Numbers:
Students are asked to determine if the total number of students in five classes will be even or odd.
Patterns Within the Multiplication Table:
Students are asked to find the missing numbers in a column of a multiplication table by using a pattern found within the table.
## Original Student Tutorials Mathematics - Grades K-5
Party Patterns: Evens and Odds in Addition – Part 1:
Determine whether the sum of two odd numbers is odd or even and whether the sum of two even numbers is odd or even by helping Lilly prepare for a math celebration in this interactive tutorial.
This is part 1 in a 3-part series. Click below to explore the other tutorials in the series.
• Part 2: Party Patterns: Evens and Odds in Addition (COMING SOON)
• Part 3: Party Patterns: Evens and Odds in Addition (COMING SOON)
Party Patterns: Evens and Odds in Addition – Part 2:
Explore addition patterns to find if the sum of an odd and an even number will be odd or even in this interactive tutorial.
This is part 2 in a 3-part series. Click below to explore the other tutorials in the series.
Party Patterns: Odds and Evens in Addition - Part 3:
Determine if the sum of three odd or three even numbers will be odd or even as Lilly prepares for a math celebration in this interactive tutorial.
This is part 3 in a 3-part series. Click below to explore the other tutorials in the series.
## Student Resources
Vetted resources students can use to learn the concepts and skills in this benchmark.
## Original Student Tutorials
Party Patterns: Odds and Evens in Addition - Part 3:
Determine if the sum of three odd or three even numbers will be odd or even as Lilly prepares for a math celebration in this interactive tutorial.
This is part 3 in a 3-part series. Click below to explore the other tutorials in the series.
Type: Original Student Tutorial
Party Patterns: Evens and Odds in Addition – Part 2:
Explore addition patterns to find if the sum of an odd and an even number will be odd or even in this interactive tutorial.
This is part 2 in a 3-part series. Click below to explore the other tutorials in the series.
Type: Original Student Tutorial
Party Patterns: Evens and Odds in Addition – Part 1:
Determine whether the sum of two odd numbers is odd or even and whether the sum of two even numbers is odd or even by helping Lilly prepare for a math celebration in this interactive tutorial.
This is part 1 in a 3-part series. Click below to explore the other tutorials in the series.
• Part 2: Party Patterns: Evens and Odds in Addition (COMING SOON)
• Part 3: Party Patterns: Evens and Odds in Addition (COMING SOON)
Type: Original Student Tutorial
The goal of this task is to help students understand the commutative property of addition by examining the addition facts for single digit numbers. This is important as it gives students a chance, at a young age, to do more than memorize these arithmetic facts which they will use throughout their education.
Making a ten:
This task asks students to study more carefully the make-a-ten strategy that they should already know and use intuitively. In this strategy, knowledge of which sums make a ten, together with some of the properties of addition and subtraction, are used to evaluate sums which are larger than 10. This task is intended for instruction purposes as it takes time to identify the patterns involved and understand the steps in the procedures.
The purpose of this task is to study some patterns in a small addition table. Each pattern identified persists for a larger table and if more time is available for this activity students should be encouraged to explore these patterns in larger tables.
Patterns in the multiplication table:
The goal is to look for structure and identify patterns and then try to find the mathematical explanation for this. This problem examines the ''checkerboard'' pattern of even and odd numbers in a single digit multiplication table. The even numbers in the table are examined in depth using a grade appropriate notion of even, namely the possibility of reaching the number counting by 2's or expressing the number as a whole number of pairs.
## Student Center Activity
Students can practice answering mathematics questions on a variety of topics. With an account, students can save their work and send it to their teacher when complete.
Type: Student Center Activity
## Parent Resources
Vetted resources caregivers can use to help students learn the concepts and skills in this benchmark.
The goal of this task is to help students understand the commutative property of addition by examining the addition facts for single digit numbers. This is important as it gives students a chance, at a young age, to do more than memorize these arithmetic facts which they will use throughout their education.
Making a ten:
This task asks students to study more carefully the make-a-ten strategy that they should already know and use intuitively. In this strategy, knowledge of which sums make a ten, together with some of the properties of addition and subtraction, are used to evaluate sums which are larger than 10. This task is intended for instruction purposes as it takes time to identify the patterns involved and understand the steps in the procedures. | 3,056 | 14,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-27 | latest | en | 0.821704 |
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1 AutoCAD 2008 Tutorial First Level: 2D Fundamentals by Randy H. Shih Oregon Institute of Technology MultiMedia CD by Jack Zecher Indiana University Purdue University Indianapolis SDC PUBLICATIONS Schroff Development Corporation
2 AutoCAD 2008 Tutorial 1-1 Chapter 1 AutoCAD Fundamentals Create and Save AutoCAD drawing files Use the AutoCAD visual reference commands Draw, using the LINE and CIRCLE commands Use the ERASE command Define Positions using the Basic Entry methods Use the AutoCAD Pan Realtime option
4 AutoCAD Fundamentals 1-3 Note that AutoCAD automatically assigns generic name, Drawing X, as new drawings are created. In our example, AutoCAD opened the graphics window using the default system units and assigned the drawing name Drawing1. Drawing Units Setup Every object we construct in a CAD system is measured in units. We should determine the value of the units within the CAD system before creating the first geometric entities. 1. In the pull-down menus, select: [Format] [Units]
5 1-4 AutoCAD 2008 Tutorial 2. Click on the Length Type option to display the different types of length units available. Confirm the Length Type is set to Decimal. 3. On your own, examine the other settings that are available. 4. In the Drawing Units dialog box, set the Length Type to Decimal. This will set the measurement to the default English units, inches. 5. Set the Precision to two digits after the decimal point as shown in the above figure. 6. Pick OK to exit the Drawing Units dialog box.
6 AutoCAD Fundamentals 1-5 Drawing Area Setup Next, we will set up the Drawing Limits; setting the Drawing Limits controls the extents of the display of the grid. It also serves as a visual reference that marks the working area. It can also be used to prevent construction outside the grid limits and as a plot option that defines an area to be plotted/printed. Note that this setting does not limit the region for geometry construction. 1. In the pull-down menus, select: [Format] [Drawing Limits] 2. In the command prompt area, near the bottom of the AutoCAD drawing screen, the message Reset Model Space Limits: Specify lower left corner or [On/Off] <0.00,0.00>: is displayed. Press the ENTER key once to accept the default coordinates <0.00,0.00>. 3. In the command prompt area, the message Specify upper right corner <0.00,0.00>: is displayed. Press the ENTER key once to accept the default coordinates <12.00,9.00>. 4. On your own, move the graphic cursor near the upper-right comer inside the drawing area and note that the drawing area is unchanged. (The Drawing Limits command is used to set the drawing area, but the display will not be adjusted until a display command is used.)
8 AutoCAD Fundamentals 1-7 Drawing lines with the LINE command 1. Move the graphics cursor to the first icon in the 2D Draw and Modify panel. This icon is the Line icon. A help-tip box appears next to the cursor. 2. Select the icon by clicking once with the leftmouse-button, which will activate the Line command. 3. In the command prompt area, near the bottom of the AutoCAD drawing screen, the message _line Specify first point: is displayed. AutoCAD expects us to identify the starting location of a straight line. Move the graphics cursor inside the graphics window and watch the display of the coordinates of the graphics cursor at the bottom of the AutoCAD drawing screen. The three numbers represent the location of the cursor in the X, Y, and Z directions. We can treat the graphics window as if it was a piece of paper and we are using the graphics cursor as if it were a pencil with which to draw. Coordinates of the location of the graphics cursor We will create a freehand sketch of a fivepoint star using the Line command. Do not be overly concerned with the actual size or the accuracy of your freehand sketch. This AutoCAD 2008 user interface. 1 4 exercise is to give you a feel for the
9 1-8 AutoCAD 2008 Tutorial 4. We will start at a location about one-third from the bottom of the graphics window. Left-click once to position the starting point of our first line. This will be point 1 of our sketch. Next move the cursor upward and toward the right side of point 1. Notice the rubber-band line that follows the graphics cursor in the graphics window. Left-click again (point 2) and we have created the first line of our sketch. 5. Move the cursor to the left of point 2 and create a horizontal line about the same length as the first line on the screen Repeat the above steps and complete the freehand sketch by adding three more lines (from point 3 to point 4, point 4 to point 5, and then connect to point 5 back to point 1) Notice that the Line command remains activated even after we connected the last segment of the line to the starting point (point 1) of our sketch. Inside the graphics window, click once with the right-mouse-button and a popup menu appears on the screen Select Enter with the left-mouse-button to end the Line command. (This is equivalent to hitting the [ENTER] key on the keyboard.) 9. On your own, move the cursor near point 2 and point 3, and estimate the length of the horizontal line by watching the displayed coordinates for each point.
10 AutoCAD Fundamentals 1-9 Visual reference The method we just used to create the freehand sketch is known as the interactive method, where we use the cursor to specify locations on the screen. This method is perhaps the fastest way to specify locations on the screen. However, it is rather difficult to try to create a line of a specific length by watching the displayed coordinates. It would be helpful to know what one inch or one meter looks like on the screen while we are creating entities. AutoCAD 2008 provides us with many tools to aid the construction of our designs. We will use the GRID and SNAP options to get a visual reference as to the size of objects and learn to restrict the movement of the cursor to a set increment on the screen. The Status Bar area is located at the bottom of the AutoCAD drawing screen. The words SNAP, GRID, ORTHO, POLAR, OSNAP, OTRACK, DUCS, LWT and MODEL appearing to the right of the coordinates are buttons that we can left-click to turn these special options ON and OFF. When the corresponding button is highlighted, the specific option is turned ON. These buttons act as toggle switches; each click of the button will toggle the option ON or OFF. Using the buttons is a quick and easy way to make changes to these drawing aid options. We can toggle the options on and off in the middle of another command. Option Buttons GRID ON 1. Left-click the GRID button in the Status Bar to turn ON the GRID option. (Notice in the command prompt area, the message <Grid on> is also displayed.) 2. Move the cursor inside the graphics window, and estimate the distance in between the grid points by watching the coordinates display at the bottom of the screen.
11 1-10 AutoCAD 2008 Tutorial The GRID option creates a pattern of dots that extends over an area on the screen. Using the grid is similar to placing a sheet of grid paper under a drawing. The grid helps you align objects and visualize the distance between them. The grid is not displayed in the plotted drawing. The default grid spacing, which means the distance in between two dots on the screen, is 0.5 inches. We can see that the sketched horizontal line in the sketch is about 4.0 inches long. SNAP ON 1. Left-click the SNAP button in the Status Bar to turn ON the SNAP option. 2. Move the cursor inside the graphics window, and move the cursor diagonally on the screen. Observe the movement of the cursor and watch the coordinates display at the bottom of the screen. The SNAP option controls an invisible rectangular grid that restricts cursor movement to specified intervals. When SNAP mode is on, the screen cursor and all input coordinates are snapped to the nearest point on the grid. The default snap interval is 0.5 inches, and aligned to the grid points on the screen. 3. Click on the Line icon in the 2D Draw and displayed. Modify toolbar. In the command prompt area, the message _line Specify first point: is
12 AutoCAD Fundamentals On your own, create another sketch of the five-point star with the GRID and SNAP options switched ON. 5. Use the right-mouse-button and select Enter in the popup menu to end the Line command if you have not done so. Using the ERASER command One of the advantages of using a CAD system is the ability to remove entities without leaving any marks. We will erase two of the lines using the Erase command. 1. Pick Erase in the 2D Draw and Modify toolbar. (The icon is the first icon in the Modify toolbar. The command prompt area and AutoCAD awaits us to select the objects to erase. icon is a picture of an eraser at the end of a pencil.) The message Select objects is displayed in the
13 1-12 AutoCAD 2008 Tutorial 2. Left-click the SNAP button on the Status Bar to turn OFF the SNAP option so that we can more easily move the cursor on top of objects. We can toggle the Status Bar options ON or OFF in the middle of another command. 3. Select any two lines on the screen; the selected lines are displayed as dashed lines as shown in the figure below. 4. Right-mouse-click once to accept the selections. The selected two lines are erased. Repeat the last command 1. Inside the graphics window, click once with the rightmouse-button to bring up the popup option menu. 2. Pick Repeat Erase, with the left-mouse-button, in the popup menu to repeat the last command. Notice the other options available in the popup menu. AutoCAD 2008 offers many options to accomplish the same task. Throughout this text, we will emphasize the use of the AutoCAD Heads-up Design TM interface, which means we focus on the screen, not on the keyboard.
14 AutoCAD Fundamentals Move the cursor to a location that is above and toward the left side of the entities on the screen. Left-mouse-click once to start a corner of a rubber-band window. First corner Second corner 4. Move the cursor toward the right and below the entities, and then left-mouseclick to enclose all the entities inside the selection window. Notice all entities that are inside the window are selected. 5. Inside the graphics window, right-mouse-click once to proceed with erasing the selected entities. On your own, create a free-hand sketch of your choice using the Line command. Experiment with using the different commands we have discussed so far. Also try switching the GRID and SNAP options ON and OFF in the middle of a command.
16 AutoCAD Fundamentals 1-15 Changing to the 2D UCS icon Display In AutoCAD 2008, the UCS icon is displayed in various ways to help us visualize the orientation of the drawing plane. 1. In the pull-down menus, select: [View] [Display] [UCS Icon] [Properties] 2. In the UCS icon style section, switch to the 2D option as shown. 3. Click OK to accept the settings. Note the W symbol in the UCS icon indicates the UCS is aligned to the world coordinate system.
17 1-16 AutoCAD 2008 Tutorial Cartesian and Polar Coordinate Systems In a two-dimensional space, a point can be represented using different coordinate systems. The point can be located, using a Cartesian coordinate system, as X and Y units away from the origin. The same point can also be located using the polar coordinate system, as r and θ units away from the origin. For planar geometry, the polar coordinate system is very useful for certain applications. In the polar coordinate system, points are defined in terms of a radial distance, r, from the origin and an angle θ between the direction of r and the positive X axis. The default system for measuring angles in AutoCAD 2008 defines positive angular values as counter-clockwise from the positive X-axis. Absolute and Relative Coordinates AutoCAD 2008 also allows us to use absolute and relative coordinates to quickly construct objects. Absolute coordinate values are measured from the current coordinate system's origin point. Relative coordinate values are specified in relation to previous coordinates. Note that the coordinate display area can also be used as a toggle switch; each left-mouse-click will toggle the coordinate display on or off. In AutoCAD 2008, the absolute coordinates and the relative coordinates can be used in current coordinate system's origin point. We can switch to using the relative coordinates by using symbol. symbol is used as the relative coordinates specifier, conjunction with the Cartesian and polar coordinate systems. By default, AutoCAD expects us to enter values in absolute Cartesian coordinates, distances measured from the which means that we can specify the position of a point in relation to the previous point.
18 AutoCAD Fundamentals 1-17 Defining Positions In AutoCAD, there are five methods for specifying the locations of points when we create planar geometric entities. Interactive method: Use the cursor to select on the screen. Absolute coordinates (Format: X,Y): Type the X and Y coordinates to locate the point on the current coordinate system relative to the origin. Relative rectangular coordinates Type the X and Y coordinates relative to the last point. Relative polar coordinates Type a distance and angle relative to the last point. Direct Distance entry technique: Specify a second point by first moving the cursor to indicate direction and then entering a distance. The GuidePlate We will next create a mechanical design using the different coordinate entry methods. Use the Erase command and erase all entities on the screen before proceeding to the next section.
19 1-18 AutoCAD 2008 Tutorial The rule for creating CAD designs and drawings is that they should be created at full size using real-world units. The CAD database contains all the definitions of the geometric entities and the design is considered as a virtual, full-sized object. Only when a printer or plotter transfers the CAD design to paper is the design scaled to fit on a sheet. The tedious task of determining a scale factor so that the design will fit on a sheet of paper is taken care of by the CAD system. This allows the designers and CAD operators to concentrate their attention on the more important issues the design. 1. Select the Line command icon in the 2D Draw and Modify toolbar. In the command prompt area, near the bottom of the AutoCAD graphics window, the message _line Specify first point: is displayed. AutoCAD expects us to identify the starting location of a straight line. 2. In the command prompt area, we will locate the starting point of our design at the origin of the world coordinate system. Command: _line Specify first point: 0,0 [ENTER] (Type 0,0 and press the [ENTER] key once.) 3. We will create a horizontal line by entering the absolute coordinates of the second point. Specify next point or [Undo]: 5.5,0 [ENTER] (0,0) (5.5,0)
20 AutoCAD Fundamentals 1-19 The line we created is aligned to the bottom edge of the drawing window. Let us adjust the view of the line by using the Pan Realtime command. 4. Click on the Pan Realtime icon in the 2D Navigate toolbar panel. The icon is the picture of a hand with four arrows. The Pan command enables us to move the view to a different position. This function acts as if you are using a video camera. 5. Move the cursor, which appears as a hand inside the graphics window, near the center of the drawing window, then push down the left-mouse-button and drag the display toward the right and top side until we can see the sketched line. (Notice the scroll bars can also be used to adjust viewing of the display.) 6. Press the [Esc] key to exit the Pan command. Notice that AutoCAD goes back to the Line command. 7. We will create a vertical line by using the relative rectangular coordinates entry method, relative to the last point we specified: Specify next point or [ENTER] 8. We can mix any of the entry methods in positioning the locations of the endpoints. Move the cursor to the Status Bar area, and turn ON the GRID and SNAP options. SNAP & GRID ON 9. Create the next line by picking the location, world coordinates (8,2.5), on the screen. 10. We will next use the relative polar coordinates entry method, relative to the last point we specified: Specify next point or [ENTER] (Distance is 3 inches with an angle of 90 degrees.)
21 1-20 AutoCAD 2008 Tutorial 11. Using the relative rectangular Reference Coordinate System aligned at the previous point along the two reference axes. 12. Move the cursor directly to the left of the last point 14. Select Close with the left-mouse-button to connect back to the starting point and end the Line command. coordinates entry method to create the next line, we can imagine a reference coordinate system aligned at the previous point. Coordinates are measured Specify next point or [ENTER] (-1.5 and 1 inches are measured relative to the reference point.) and use the direct distance entry technique by entering 6.5 [ENTER]. 13. For the last segment of the sketch, we can use the Close option to connect back to the starting point. Inside the graphics window, right-mouse-click and a popup menu appears on the screen.
22 AutoCAD Fundamentals 1-21 Creating Circles The menus and toolbars in AutoCAD 2008 are designed to allow the CAD operators to select the different Draw commands through the pull-down menus. quickly activate the desired commands. Besides using the Draw toolbar, we can also 1. In the pull-down menus, select: [Draw] [Circle] [Center, Diameter] Notice the different options available under the circle submenu: Center Point: Draws a circle based on a center point and a diameter or a radius. 2 Points: Draws a circle based on two endpoints of the diameter. 3 Points: Draws a circle based on three points on the circumference. TTR Tangent, Tangent, Radius: Draws a circle with a specified radius tangent to two objects. TTT Tangent, Tangent, Tangent: Draws a circle tangent to three objects.
23 1-22 AutoCAD 2008 Tutorial 2. In the command prompt area, the message Specify center point for circle or [3P/2P/Ttr (tan tan radius)]: is displayed. AutoCAD expects us to identify world coordinates (2.5,3) as the center point for the first circle. the location of a point or enter an option. We can use any of the four coordinate entry methods to identify the desired location. We will enter the Specify center point for circle or [3P/2P/Ttr (tan tan radius)]: 2.5,3 [ENTER] 3. In the command prompt area, the message Specify diameter of circle: is displayed. Specify diameter of circle: 2.5 [ENTER] 4. Inside the graphics window, right-mouseclick to bring up the popup option menu. 5. Pick Repeat Center, Diameter with the left-mouse-button in the popup menu to repeat the last command. location as (2.5,2). 6. Using the relative rectangular coordinates entry method, relative to the center-point coordinates of the first circle, we specify the Specify center point for circle or [3P/2P/Ttr (tan tan [ENTER]
24 AutoCAD Fundamentals In the command prompt area, the message Specify Diameter of circle: <2.50> is displayed. The default option for the Circle command in AutoCAD is to specify the radius and the last radius used is also displayed in brackets. Specify Diameter of circle<2.50>: 1.5 [ENTER] Saving the CAD Design 1. In the pull-down menus, select: [File] [Save As] Note the command can also be activated with quick-key combination of [Ctrl]+[Shift]+[S].
25 1-24 AutoCAD 2008 Tutorial 2. In the Save Drawing As dialog box, select the folder in which you want to store the CAD file and enter GuidePlate in the File name box. Select the folder to store the file. Enter GuidePlate 3. Pick Save in the Save Drawing As dialog box to accept the selections and save the file. Exit AutoCAD 2008 To exit AutoCAD 2008, select File then choose Exit from the pull-down menu or type QUIT the command prompt. Note the command can also be activated with quick-key combination of [Ctrl]+[Q].
26 AutoCAD Fundamentals 1-25 Questions: 1. What are the advantages and disadvantages of using CAD systems to create engineering drawings? 2. How do the GRID and SNAP options assist us in sketching? 3. List and describe the different coordinate entry methods available in AutoCAD? 4. List and describe two types of coordinate systems commonly used for planar geometry. 5. Identify the following commands: (a) (b) (c) (d) Tan, Tan, Radius
27 1-26 AutoCAD 2008 Tutorial Exercises: (All dimensions are in inches.) 1. 2.
29 1-28 AutoCAD 2008 Tutorial NOTES:
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# Shapes Math Activities
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### Description
This is a collection of SHAPE math activities.
1. I can find the shapes with no points/vertices. Students take their color coded mat (4) to find the shapes with no points. All mats are similar but not exactly the same. There is a printable for students to color the shapes without vertices. There are separate I can and printables using the terms vertices instead of points.
2. I can find the shape (triangle, square, circles, ovals, rectangles, diamonds, rhombus.) Students take their color coded mat (4) to find the shape asked for. All mats are similar but not exactly the same. There are separate activities using the term diamond and rhombus to meet your needs. There are 3 coloring printables to go along with these areas.
3. I can play a shape identification game. There are two different game boards to choose from. One will take longer to play than the other. Students pull a shape card (use any of the shapes you have taught or that you choose) and if they know that shape, they get to roll the die. They can move a marker that many places.
4. I can count the points and lines in shapes. Students take their cards and count the shapes points and lines in the given shape. They write down the number on the top, in the given boxes. There are separate I can statements, cards, and printables which use the term vertices instead of points.
5. I can find the points and straight lines in a shape. Students take their cards and place dots of the points of the shapes. Then they write a number on the straight line of a shape or letters to show the curved line of a shape. Again there is a printable option. There is also a separate option which uses the term vertices instead of points.
6. I can draw flat shapes. Student use the mats provided to trace, then draw the shapes. Please note, although an oval and circle do not have points or vertices, I provided them points for drawing. To show these as “different” I made the points hallow. Use this as a teachable opportunity. I also made note when to draw a curved line to help their brains.
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by Mary Amoson
Sharing Kindergarten
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TPT empowers educators to teach at their best. | 643 | 3,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-14 | latest | en | 0.929893 |
https://www.convert-measurement-units.com/convert+Pascal+to+Micropascal.php | 1,721,892,444,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00114.warc.gz | 626,393,310 | 14,050 | Convert Pa to µPa (Pascal to Micropascal)
## Pascal into Micropascal
numbers in scientific notation
Direct link to this calculator:
https://www.convert-measurement-units.com/convert+Pascal+to+Micropascal.php
# Convert Pa to µPa (Pascal to Micropascal):
1. Choose the right category from the selection list, in this case 'Pressure'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Pascal [Pa]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Micropascal [µPa]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '323 Pascal'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Pascal' or 'Pa'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Pressure'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '8 Pa to µPa' or '75 Pa into µPa' or '61 Pascal -> Micropascal' or '15 Pa = µPa' or '68 Pascal to µPa' or '22 Pa to Micropascal' or '29 Pascal into Micropascal'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(96 * 50) Pa'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '89 Pascal + 43 Micropascal' or '4mm x 57cm x 11dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 5.352 459 951 292 6×1022. For this form of presentation, the number will be segmented into an exponent, here 22, and the actual number, here 5.352 459 951 292 6. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 5.352 459 951 292 6E+22. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 53 524 599 512 926 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 881 | 3,729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-30 | latest | en | 0.824573 |
https://id.scribd.com/document/284348359/Answers-to-EPSP-Study-Questions-1 | 1,566,747,051,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330750.45/warc/CC-MAIN-20190825151521-20190825173521-00333.warc.gz | 489,288,306 | 72,017 | Anda di halaman 1dari 3
# Study Questions on Post-synaptic Potentials
## Create a plot of the amplitude of an EPSP in mV (y-axis) as a function of
the membrane potential (x-axis), using a reversal potential of -10mV. On the
y-axis, depolarizing amplitudes should be plotted upward, as positive values.
This plot illustrates the original results of Takeuchi & Takeuchi for the
nicotinic ACh receptor, which is roughly equal in permeability to Na and K
ions. Now, draw additional plots on the same axis to illustrate the behavior
of the EPSP under each of the following conditions:
(a) Double the number of ACh receptors activated.
(b) Half the number of resting channels.
(c) Mutated ACh receptor permeable only to Na ions
200
V(EPSP)
150
a,b
c
100
Control
50
Vm
0
-100
-80
-60
-40
-20
20
40
60
-50
-100
-150
The control plot shows Vrev = -10 and is linear, with depolarizing EPSP
amplitudes negative to Vrev, hyperpolarizing positive to Vrev.
Doubling the number of ACh receptors activated or reducing by half the
number of resting channels each will double the EPSP amplitude at each
potential without changing Vrev.
Plot c, where we made the ACh receptor permeable only to Na ions,
shows Vrev = VNa, which I assumed to be +30mV in this cell. The slope
depends on your assumptions. If you assume the same total conductance,
the slope should stay the same, as I have drawn it above. Another
reasonable assumption would be that you removed the K conductance
leaving the Na conductance intact, in which case the slope would be reduced
by the fraction of total conductance originally contributed by Na. So, for
example, if the conductance ratio originally were 1.8:1 Na:K, then the slope
of line c should be 1.8/2.8 of the original slope (but still +30mV Vrev.
Now do the exact same set of plots, but this time plot the current
underlying the EPSP, as measured in voltage clamp. Plot inward current
down (negative) on the y-axis. Some of the conditions (a)-(c) change the
voltage plots in Question 1, but do not change the current plots in Question
2. Do you understand why?
I (EPSP)
150
100
50
0
-100
-80
-60
-40
-20
20
40
60
Vm
-50
Control, b
c
a
-100
-150
-200
## Voltage clamp plots of EPSP current are inverted, because current is
inward for a depolarizing EPSP, and inward currents are plotted as negative
number (convention). The control plot and plot (b) are the same, because
changing R (reducing resting channels) does not change EPSP current. Plot a
is double control, because twice as many receptors activated means twice as
much current at each potential. All have the same Vrev. The plot of c has
the same assumptions as in the first graph, and its slope would change with
a new assumption, just like in the first graph. | 736 | 2,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-35 | latest | en | 0.891965 |
https://www.physicsforums.com/threads/circuits-with-series-and-parallel-wiring.612509/ | 1,558,735,422,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257767.49/warc/CC-MAIN-20190524204559-20190524230559-00538.warc.gz | 910,326,207 | 22,302 | # Circuits with Series and Parallel Wiring
#### Snape1830
I attached the problem. I'm pretty confused. I remember the equations for parallel and series circuits, but I have no idea how to do this. I attempted, but failed. I know the two 10 ohm resistors are in parallel (so Rnet is 5), and the 20 ohm and the 6 ohm resistor are in series (So Rnet is 26). But then are those in parallel or series? And then i just don't know how to do the problem.
#### Attachments
• 5 KB Views: 267
Related Introductory Physics Homework News on Phys.org
#### vela
Staff Emeritus
Homework Helper
The 10-ohm resistors are in parallel, but the 20-ohm and 6-ohm resistor aren't in series.
#### Snape1830
The 10-ohm resistors are in parallel, but the 20-ohm and 6-ohm resistor aren't in series.
Is the 10, 10 and 20 ohm in parallel and the 6 ohm in series with those three? Otherwise, I can't see how the 20 and the 6 aren't in series.
#### vela
Staff Emeritus
Homework Helper
The 20-ohm and 6-ohm are not in series because there's something else connected to the node where the two resistors are connected. If two elements are in series, all of the current in one element has to go through the other element. For example, the battery and the 6-ohm resistor are in series. You should be able to see that all of the current that flows through the battery has to also go through the 6-ohm resistor. That's not the case with the 6-ohm and 20-ohm resistor. Some of the current that goes through the 6-ohm resistor will branch off to go through the 10-ohm resistors.
#### Snape1830
The 20-ohm and 6-ohm are not in series because there's something else connected to the node where the two resistors are connected. If two elements are in series, all of the current in one element has to go through the other element. For example, the battery and the 6-ohm resistor are in series. You should be able to see that all of the current that flows through the battery has to also go through the 6-ohm resistor. That's not the case with the 6-ohm and 20-ohm resistor. Some of the current that goes through the 6-ohm resistor will branch off to go through the 10-ohm resistors.
Ok....so how do I do the problem?
#### mrcheeses
10 ohm resistors aren't in a series and are parallel but the other two are in a series. I find it easy to see if they are parallel or series by redrawing the diagram to one you are comfortable with. At every split in the wire with resistors, just branch it off.
#### Snape1830
10 ohm resistors aren't in a series and are parallel but the other two are in a series. I find it easy to see if they are parallel or series by redrawing the diagram to one you are comfortable with. At every split in the wire with resistors, just branch it off.
So 20 and 6 are in series? That's what I thought. I'm not really sure how to redraw it. Can you can just help me get the answer?
Staff Emeritus
Homework Helper
Last edited:
#### ehild
Homework Helper
I redraw the circuit making the wires a bit shorter and moving the ammeter closer to the resistors, but the circuit is equivalent with the original one. Note that the ammeter can be replaced by a single piece of wire. Which resistors are connected in parallel?
ehild
#### Attachments
• 6.6 KB Views: 183
#### azizlwl
Yes, 10,10 and 20 ohms are in parallel.
#### Snape1830
I redraw the circuit making the wires a bit shorter and moving the ammeter closer to the resistors, but the circuit is equivalent with the original one. Note that the ammeter can be replaced by a single piece of wire. Which resistors are connected in parallel?
ehild
10, 10 and the 20 ohm resistor are in parallel. So the Rnet is 4 for those? So then do I just use V=RI to solve for current?
#### ehild
Homework Helper
Yes, the resultant of the parallel resistor is 4 ohm. And the resultant is connected to the 6 ohm resistor, and the whole is connected to the battery. You can use V=RI to get the total current.
What current does the ammeter read?
#### Snape1830
Yes, the resultant of the parallel resistor is 4 ohm. And the resultant is connected to the 6 ohm resistor, and the whole is connected to the battery. You can use V=RI to get the total current.
What current does the ammeter read?
Well, 4+6=10.
So...
20=10I
I=2 A
But 2 is wrong, so I don't know.
#### ehild
Homework Helper
The ammeter does not read the current flowing through the battery. It reads the current that flows through itself. It is connected in series with what ?
ehild
#### Snape1830
The ammeter does not read the current flowing through the battery. It reads the current that flows through itself. It is connected in series with what ?
ehild
The three resistors in parallel? But to find current you need to know the voltage.
#### ehild
Homework Helper
The three resistors in parallel? But to find current you need to know the voltage.
No it is not the three resistors in parallel, but?
And you can find the voltage.
ehild
#### Snape1830
No it is not the three resistors in parallel, but?
And you can find the voltage.
ehild
Just the two 10s? How do I find the voltage if I only have the resistors?
#### azizlwl
Well, 4+6=10.
So...
20=10I
I=2 A
But 2 is wrong, so I don't know.
Now you got 2 Amps flowing through the equilvalent parallel resistors and through 6 Ohms resistor.
Now what is the voltage across each of the parallel resistors?
From this voltage drop, you calculate the current flow.
What is required here is to find the current flowing through the equivalent parallel resistors of two 10 Ohms resistors.
Last edited:
#### Snape1830
Now you got 2 Amps flowing through the equilvalent parallel resistors and through 6 Ohms resistor.
Now what is the voltage across each of the parallel resistors?
From this voltage drop, you calculate the current flow.
V-RI
V=10(2)
V=20 V
V=20(2)
V=40 V
So across the 10 ohm resistors, the voltage drop is 20 V
Across the 20 ohm resistor the voltage drop is 40 V?
That 40 can't be right, though?
#### azizlwl
Wrong.
With equivalent parallel resistors, you have only 2 resistors. 6 ohms and 4 ohms.
What is the voltage drop across 4 ohms resistor.
This voltage drop is identical to all the resistors in parallel.
Remember the voltage across all resistore in parallel are equal.
#### Snape1830
Wrong.
With equivalent parallel resistors, you have only 2 resistors. 6 ohms and 4 ohms.
What is the voltage drop across 4 ohms resistor.
This voltage drop is identical to all the resistors in parallel.
Remember the voltage across all resistore in parallel are equal.
Oh, right, I misread the question.
V=2(4)
V=8 Volts
#### ehild
Homework Helper
Oh, right, I misread the question.
V=2(4)
V=8 Volts
So the voltage across the resultant of the parallel resistors is 8 V. You get the current through each of them if you divide that 8 V by the resistances.
What is the current that flows through the ammeter?
Remember Kirchhoff's nodal law.
ehild
#### Attachments
• 9.3 KB Views: 143
#### Snape1830
So the voltage across the resultant of the parallel resistors is 8 V. You get the current through each of them if you divide that 8 V by the resistances.
What is the current that flows through the ammeter?
Remember Kirchhoff's nodal law.
ehild
Kirchoff's loop rule? The current going in equals the current coming out.
So do I find the current through of the parallel resistors and then add them up...because that comes out to 2 and 2 is wrong.
#### ehild
Homework Helper
Kirchoff's loop rule? The current going in equals the current coming out.
So do I find the current through of the parallel resistors and then add them up...because that comes out to 2 and 2 is wrong.
I wrote Kirchhoff's node rule.
The ammeter measures the current that flows through it. Does I20 ,represented by the green arrow flow through the ammeter?
ehild
Last edited:
#### Snape1830
Wrote Kirchhoff's node rule.
The ammeter measures the current that flows through it. Does I20 ,represented by the green arrow flow through the ammeter?
ehild
No? So it would be 1.6 A?
And thanks for taking the time to edit the picture!
Last edited:
### Physics Forums Values
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• Solo and co-op problem solving | 2,173 | 8,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2019-22 | longest | en | 0.949417 |
http://www.chegg.com/homework-help/questions-and-answers/grindstone-mass-m-30-kg-radius-05-m-constantacceleration-30-rad-s2-2-seconds-angular-displ-q1367997 | 1,386,637,982,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164003799/warc/CC-MAIN-20131204133323-00073-ip-10-33-133-15.ec2.internal.warc.gz | 272,261,486 | 7,030 | 0 pts ended
A grindstone of mass M = 30 kg and radius 0.5 m has a constant
acceleration, ", of 3.0 rad/s2. After 2 seconds....
A. Find the angular displacement
B. Find the angular speed, T, after two seconds.
C. Find the distance traveled in meters of a point on the rim of the
grindstone after two seconds.
D. Find the linear (or tangential) speed, v, of a point on the rim after
two seconds.
E. Find the tangential acceleration of a point on the rim after two
seconds.
F. Find the centripetal acceleration of a point on the rim after two
seconds.
G. If the grindstone is considered a solid disc with a Moment of Inertia,
I, is given by the formula MR2/2, find the torque that is necessary for it to
have the give angular acceleration. | 196 | 737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2013-48 | latest | en | 0.913258 |
https://www.numbersaplenty.com/1037298080 | 1,713,434,623,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.28/warc/CC-MAIN-20240418093630-20240418123630-00646.warc.gz | 829,478,139 | 3,197 | Search a number
1037298080 = 255713191373
BaseRepresentation
bin111101110100111…
…110100110100000
32200021212012110102
4331310332212200
54111022014310
6250532521532
734463613440
oct7564764640
92607765412
101037298080
11492589445
1224b4802a8
13136b98c60
149ba9a120
15610ec8a5
hex3dd3e9a0
1037298080 has 192 divisors, whose sum is σ = 3040063488. Its totient is φ = 325693440.
The previous prime is 1037298079. The next prime is 1037298109. The reversal of 1037298080 is 808927301.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 2780774 + ... + 2781146.
It is an arithmetic number, because the mean of its divisors is an integer number (15833664).
Almost surely, 21037298080 is an apocalyptic number.
1037298080 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 1037298080, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (1520031744).
1037298080 is an abundant number, since it is smaller than the sum of its proper divisors (2002765408).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
1037298080 is a wasteful number, since it uses less digits than its factorization.
1037298080 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 599 (or 591 counting only the distinct ones).
The product of its (nonzero) digits is 24192, while the sum is 38.
The square root of 1037298080 is about 32207.1122580091. The cubic root of 1037298080 is about 1012.2812468529.
The spelling of 1037298080 in words is "one billion, thirty-seven million, two hundred ninety-eight thousand, eighty". | 576 | 1,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-18 | latest | en | 0.86489 |
https://forums.civfanatics.com/threads/civ-v-ideas-suggestions-summary.324159/page-27 | 1,571,076,272,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986654086.1/warc/CC-MAIN-20191014173924-20191014201424-00170.warc.gz | 500,905,029 | 18,425 | # Civ V Ideas & Suggestions Summary
Discussion in 'Civ - Ideas & Suggestions' started by Camikaze, Jun 13, 2009.
1. ### Dudu42Chieftain
Joined:
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Messages:
79
I dont get it.
Could someone explain me why hard caps are bad ideas?
Because it is... "arbitrary"?
And one per tile is better because it is... "less arbitrary"?
I dont see why we couldnt make something like small armies of 3 units. Maybe some units could be excludent, like sam and artillery, ranged and meelee. That sounds a good idea to me.
And the infinite workers sounds even better IMO. My point is to clear the tiles a bit. It's just nonsense when the movements of your unit are messed by your very units.
The workers stack is surely more urgent, I think.
2. ### rysmielChieftain
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One per tile isn't better. Any limit to number of units on a tile is bad.
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They limit the options available to the player, for starters. I don't think the player should be limited in what they can do, just penalised for doing the wrong thing. The other main reason hard caps are a bad idea is that they simply switch the number of units in a stack needed to as the optimal strategy. If you limit stacks at 5, then the dominant strategy is going to be to have stacks of 5 going around everywhere. It doesn't actually solve the problem, just limits it, at the same time as limiting the player. Exponential, or even linear penalties, on the other hand, allow the player the rule array of options, making you actually weigh up what will be advantageous in a given situation, varying the strategy, so you don't just have stacks of a certain number going around.
Another reason is that specifying a particular number simply makes no sense. Why 5? Why 6? Why any particular number?
It is the least arbitrary number that can be chosen, because it is the minimum endpoint of the hard cap spectrum.
Why 3?
Also, this just means groups of 3 will be going around everywhere, rather than singular units, or groups of 10 or 20.
Perhaps this is so. But not if involving a hard cap.
4. ### Dudu42Chieftain
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Im not complaining about the strategy. I really enjoyed the end of "Stack of Doom", one per tile is just better. But I think some battles become too crowded in choke points, needlessly. And they could low the unit cost a bit (just a bit, I like them a bit uncommom).
I chose 3, it's a sugestion. It's arbitrary, but the game is full of them, so that's not a good point to me. 3 is small, but more than enough to clean the tiles.
---------------------
Ok, enough of hard caps
I have another sugestion:
The return of book wonders!
I was thinking about the old Sun Tzu Art of War and Theory Evolution (could be represented as Origin of Species.)
There's a lot of books. Philosofiae Naturalis & Principia Matematica (Newton), Lusiadas (Camoes), The Prince (from Machiavelli).
No need to be only books. Just all kings of cultural Masterpieces with huge influence around the globe. Instead of using hammers for production, they could use culture... they also could be built by a great person (like Principia from a great scientist).
5. ### The Rusty GamerChieftain
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For me, the answer to this whole tile stack thing is the following points:
1) Each type of tile can have a different stack limit depending on what type of tile it is. eg: a plains tile might allow more room than a forest tile.
2) Each unit is given a size attribute. Therefore you can fit different amounts of different types of units on tiles. eg: A tank might have size=4 (say) and if the tile stack limit is 5 (say), then only 1 tank could fit on there but a unit of size 1 could join it.
3) Tile stack limits can change over time depending on techs, policies, buildings etc.
Implementation of the above ideas could create some flexability for modders.
egs:
a leader characteristic that adds 1 to tile stacks for a specific tile type such as a jungle
a civilization with a unique unit that has a smaller size than other civs with parallel unit
stack limit could be smaller when in enemy territory
a worker improvement such as a road could increase the stack limit of a tile, a mine could decrease it.
6. ### Dudu42Chieftain
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Interesting ideas, Rusty Gamer.
Also, putting all into a stack isnt necessarily the best strat.
We have the flanking ability, so putting 2 units on both sides of enemy might be better than stack them. Also, we could use some collateral damage againd...
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Even if the hard caps are different for different terrain types and change over time, it's still arbitrary, and still an unnecessary limitation on the player.
8. ### Dudu42Chieftain
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The science bonus is arbitrary, the food in a tile is arbitrary. The strengh of units is also arbitrary.
I dont see how this could be a issue exclusively for stacks, Camikaze.
9. ### The Rusty GamerChieftain
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You could say everything in the game is "arbitrary". Each stack level limit would allow for different game strategy. A one stack limt requires a certain strategy, a two-stack another and so on. Limitations are defined so that the game is acutally strategic, not just a free-for-all. Otherwise, why not just allow unlimited units that cost next to nothing with rediculously high amounts for attack, defense, movement and range ability and go for it? If you want to do that, you might as well play an RTS. But there's already a MOD for unlimited stack ability in any case so why the gripe?
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Those are not arbitrary limitations.
When there is no other way to solve a problem, then arbitrary limitations are acceptable. But in the case of stacks, there are other ways to solve the problem, namely by the introduction of a series of penalties that impact largely on stacks. A single exponential penalty based solely on the number of units in a stack would be one way to do it. A number of penalties that in conjunction created this effect would be another, and perhaps better way. These would firstly not actually prevent you from having stacks, and would allow for a greater variety of strategic options and considerations, dependent upon the individual situation.
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Good discussion. Camikaze, wouldn't the the penalty of stacking more than one unit also be an arbitrary value?
I like Rusty's suggestion of a puzzle like approach to it. Where tiles have 'capacity'
The current capacity is '1' warfighting unit and '1' worker/non fighting unit.
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Yes, but again, it wouldn't be an arbitrary limitation.
I'm just going to take a pretty straight forward example of what the penalty could be to illustrate my point.
x is the number of units in a stack.
y is the percentage penalty applied to the stack (0 ≤ y ≤1).
So, when x=1, you would want y=0, and as x → ∞, you would want y → 1. That much is fairly simple. A logarithmic function is the best way to realistically place firm penalties on stacks of size. To create such a function, that would simply have to choose one single point to fix. That single point is an arbitrary choice (although not a limitation in any way), but all other points are simply consequences of that. This is an arbitrary choice that is acceptable, as is the amount of happiness received from a colosseum, or the amount of research needed for a tech, and would be determined to fit in with game balance.
So, just to round off the example, if we wanted to place a penalty on a unit in a stack of 10, of 50%, then for x=10, y=0.5. So the equation here would be log100x=y (assuming my maths knowledge hasn't completed deserted me). This can then be translated to determine a penalty for a unit within a stack of any number. Seems a good way to solve the issue, to me, and requires only an arbitrary equation number (dependent on balance issues) rather than an arbitrary limitation on the player.
1 is, however, the least arbitrary number possible to be chosen for a hard cap, given it is a point of extremity.
13. ### The Rusty GamerChieftain
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For what is suggested here, this might be a little simplistic. You would also have to take into account the type of units such as ranged and/or armoured. You could end up with an incredibly complex formula which may never be perfect, tweaked and adjusted every time another unforseen combination instance arose exposing the equation as out of balance.
If we allow for stacking, do we also allow for units from different civs to be in the same tile?
I'm personally for the idea of increased stacking ability primarily based on some sort of logistics tech. I've already suggested having a size for units but have come up with an alternative idea. Instead of having a size attribute, the stack limit could be reached simply by adding up the units' attack and/or defense abilities. You could even have two stack limits - an attack stack limit and a defense stack limit, both of which could also be affected depending on the type of terrain. What type of logistics you choose to research could become a strategic decision, even an either or situation - do you want to emphasise attack or defense - and this choice could affect your entire game plan.
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I too think a singular penalty is a little simplistic, but that a few penalties should be implemented to have the same effect. None of these would have to be too complex, and in combination, would be very effective. One of the major points of this would be that you would be able to take advantage of the penalties and minimise them for certain strategic situations. For example, if there was a cost penalty also dependent on distance, you could afford a varying number of units in a stack depending on how far from your borders you are. But the overall effect of the penalties would still be one of generally discouraging stacking.
Remember, the aim shouldn't be to not allow stacking; a hard cap does that perfectly well, as does one unit per tile. The aim should be to disincentivise stacking.
And if you allow stacking, then yes, you do allow units from different civs on the one tile. That's a detail, not a key issue, though.
15. ### Dudu42Chieftain
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Some minor changes, I think, about transporting troops trough the water tiles.
I like the automatic transportation. Way better than building a transport, for sure.
But still, it's weird to see troops instantly getting into a boat in the middle of nowhere.
Shoudn't be better if they had to pass trough a friendly city?
(scouts should keep the automatic board, to keep exploring isolated isles).
16. ### The Rusty GamerChieftain
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A "bug?" appears to be when sending a unit more than one tile away. It doesn't always do it in the most efficient manner - for instance, it might go straight into a forest and have to wait one more turn when it could've got to the target tile in one turn had it gone via a clear piece of land - nor does it seem to know to avoid city-state borders. I've peeved them off accidentally at least a couple of time this way.
17. ### guspashoChieftain
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That has ALWAYS been a problem! I wish they would figure out how to fix it already!
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@Dudu42- Perhaps just having to pass through friendly territory should be necessary (i.e. not neutral territory), rather than friendly cities.
@The Rusty Gamer- Might be a good idea to post that in the bug reports forum. Sounds dodgy.
Guys, it's probably a good idea to move discussion from this thread to this thread. It's the replacement for this thread (which was designed as a Pre Civ5 discussion thread).
19. ### PestWulfChieftain
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*self delete*
20. ### Scilly_guyChieftain
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I'm not going to contribute anything useful, I'm just going to put it out there that I like 1upt, yes sometimes it annoys me maybe there could be some tweaks, especially with naval units and embarked units, workers and other non combat units, but apart from that I like it. | 3,004 | 12,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-43 | latest | en | 0.936435 |
https://www.arxiv-vanity.com/papers/2009.07072/ | 1,618,572,046,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038056325.1/warc/CC-MAIN-20210416100222-20210416130222-00379.warc.gz | 750,007,906 | 49,502 | # The linkedness of cubical polytopes: The cube
Hoa T. Bui Centre for Informatics and Applied Optimisation, Federation University Australia
Faculty of Science and Engineering, Curtin University, Australia
and Guillermo Pineda-Villavicencio & Julien Ugon Centre for Informatics and Applied Optimisation, Federation University Australia
School of Information Technology, Deakin University, Australia
April 10, 2021
###### Abstract.
The -dimensional cube is the convex hull in of the vectors . A cubical polytope is a polytope with all its facets being combinatorially equivalent to cubes. The paper is concerned with the linkedness of the graphs of cubical polytopes. A graph with at least vertices is -linked if, for every set of distinct vertices organised in arbitrary unordered pairs, there are vertex-disjoint paths joining the vertices in the pairs. We say that a polytope is -linked if its graph is -linked.
We establish that the -dimensional cube is -linked, for every ; this is the maximum possible linkedness of a -polytope. This result implies that, for every , a cubical -polytope is -linked, which answers a question of Wotzlaw [15]. We also found cubical -polytopes that are -linked, for every such that .
As intermediate results, we provide new proofs of a couple of known results: the characterisation of 2-linked 3-dimensional polytopes and the fact that 4-dimensional polytopes are 2-linked.
Finally, we introduce the notion of strong linkedness, which is slightly stronger than that of linkedness. A graph is strongly -linked if it has at least vetrtices and, for every set of exactly vertices organised in arbitrary unordered pairs, there are vertex-disjoint paths joining the vertices in the pairs and avoiding the vertices in not being paired. For even the properties of strongly -linkedness and -linkedness coincide, since every vertex in is paired; but they differ for odd . We show that 4-polytopes are strongly -linked and that -dimensional cubes are strongly -linked, for .
###### 2010 Mathematics Subject Classification:
Primary 52B05; Secondary 52B12
Julien Ugon’s research was partially supported by ARC discovery project DP180100602.
## 1. Introduction
A (convex) polytope is the convex hull of a finite set of points in ; the convex hull of is the smallest convex set containing . The dimension of a polytope in is one less than the maximum number of affinely independent points in the polytope. A polytope of dimension is referred to as a -polytope.
A face of a polytope in is itself, or the intersection of with a hyperplane in that contains in one of its closed halfspaces. Faces other than are polytopes of smaller dimension. A face of dimension 0, 1, and in a -polytope is a vertex, an edge, and a facet, respectively. The graph of a polytope is the undirected graph formed by the vertices and edges of the polytope.
This paper studies the the linkedness of cubical -polytopes, -dimensional polytopes with all their facets being cubes. A -dimensional cube is the convex hull in of the vectors . By a cube we mean any polytope that is combinatorially equivalent to a cube; that is, one whose face lattice is isomorphic to the face lattice of a cube.
Denote by the vertex set of a graph or a polytope . Given sets of vertices in a graph, a path from to , called an path, is a (vertex-edge) path in the graph such that and . We write path instead of path, and likewise, write path instead of .
Let be a graph and a subset of distinct vertices of . The elements of are called terminals. Let be an arbitrary labelling and (unordered) pairing of all the vertices in . We say that is linked in if we can find disjoint paths for , where denotes the interval . The set is linked in if every such pairing of its vertices is linked in . Throughout this paper, by a set of disjoint paths, we mean a set of vertex-disjoint paths. If has at least vertices and every set of exactly vertices is linked in , we say that is -linked. If the graph of a polytope is -linked we say that the polytope is also -linked.
Unless otherwise stated, the graph theoretical notation and terminology follow from [4] and the polytope theoretical notation and terminology from [16]. Moreover, when referring to graph-theoretical properties of a polytope such as minimum degree, linkedness and connectivity, we mean properties of its graph.
All the 2-linked graphs have been characterised [11, 13]. In the context of polytopes, one consequence is that, with the exception of simplicial 3-polytopes, no 3-polytope is 2-linked; a simplicial polytope is one in which every facet is a simplex. Another consequence is that every 4-polytope is 2-linked. We provide new proofs of these two results: Propositions 6 and 5.
There is a linear function such that every -connected graph is -linked, which follows from works of Bollobás and Thomason [2]; Kawarabayashi, Kostochka, and Yu [5]; and Thomas and Wollan [12]. In the case of polytopes, Larman and Mani [6, Thm. 2] proved that every -polytope is -linked, a result that was slightly improved to in [14, Thm. 2.2].
There are -polytopes that are -connected but not -connected, and so is an upper bound for the linkedness of -polytopes. Simplicial -polytopes are -linked, for every [6, Thm. 2]. In his PhD thesis [15, Question 5.4.12], Wotzlaw asked whether every cubical -polytope is -linked. In Theorem 18 we answer his question in the affirmative by establishing that a -cube is -linked, for every (see Theorem 16). We remark that the linkedness of the cube was first established in [7, Prop. 4.4] as part of a study of linkedness in Cartesian products of graphs, but not self-contained proof was available.
In a subsequent paper [3], we prove a stronger result: a cubical -polytope is -linked, for every ; which is best possible. In anticipation of this result, in Proposition 22 we proved that certain cubical -polytopes that are embedded in the -cube are -linked, for every .
Finally, we introduce the notion strong linkedness, a property marginally stronger than linkedness. We say that a -polytope is strongly -linked if, for every set of exactly vertices and every pairing with pairs from , the set is linked in and each path joining a pair in avoids the vertices in not being paired in . For odd the properties of strongly -linkedness and -linkedness coincide, since every vertex in is paired in ; but they differ for even . We show that 4-polytopes are strongly -linked and that -cubes are strongly -linked, for .
## 2. Preliminary results
This section groups a number of results that will be used in later sections of the paper.
Propositions 6 and 4 follow from the characterisation of 2-linked graphs carried out in [11, 13]. Both propositions also have proofs stemming from arguments in the form of Lemma 1; for the sake of completeness we give such proofs.
###### Lemma 1 ([10, Thm. 3.1]).
Let be a -polytope, and let be a linear function on satisfying for some . If and are vertices of with and , then there exists a path with and such that for each .
We state Balinski’s theorem on the connectivity of polytopes.
###### Theorem 2 (Balinski [1]).
For every , the graph of a -polytope is -connected.
Let be a set of vertices in a graph . Denote by the subgraph of induced by , the subgraph of that contains all the edges of with vertices in . Write for . A path in the graph is called -valid if no inner vertex of the path is in . The distance between two vertices and in a graph , denoted , is the length of a shortest path between the vertices.
###### Definition 3 (Configuration 3F).
Let be a set of at least four terminals in a 3-cube and let be a labelling and pairing of the vertices in . A terminal of , say , is in Configuration 3F if the following conditions are satisfied:
1. four vertices of appear in a 2-face of the cube;
2. the terminals in the pair are at distance two in ; and
3. the neighbours of in are all vertices of .
Configuration 3F is the only configuration in a 3-cube that prevents the linkedness of a pairing of four vertices, as Proposition 4 attests. A sequence of vertices in a cycle is in cyclic order if, while traversing the cycle, the sequence appears in clockwise or counterclockwise order. It follows that, if pairing of vertices in a 3-cube appears in cyclic order in a 2-face, then all the vertices in are in Configuration 3F.
###### Proposition 4.
Let be the graph of a 3-polytope and let be a set of four vertices of . The set is linked in if and only if there is no facet of the polytope containing all the vertices of .
###### Proof.
Let be a 3-polytope embedded in and let be an arbitrary set of four vertices in . We first establish the necessary condition by proving the contrapositive. Let be a 2-face containing the vertices of and consider a planar embedding of in which is the outer face. Label the vertices of so that they appear in the cyclic order . Then the paths and in must inevitably intersect, implying that is not linked.
Assume there is no 2-face of containing all the vertices of . Let be a (linear) hyperplane that contains , and , and let be a linear function that vanishes on (this may require a translation of the polytope). Without loss of generality, assume that for some and that .
First consider the case that is a supporting hyperplane of a 2-face . The subgraph is connected by Balinski’s theorem (Theorem 2), and so there is an -valid path on . Then, use Lemma 1 to find an path in which each inner vertex has positive -value. The paths and are clearly disjoint.
Now consider the case that intersects the interior of . Then there is a vertex in with -value greater than zero and a vertex with -value less than zero. Use Lemma 1 to find an path in which each inner vertex has negative -value and an path in which each inner vertex has positive -value. ∎
The subsequent corollary follows at once from Proposition 4.
###### Corollary 5.
The same reasoning employed in the proof of the sufficient condition of Proposition 4 settles Proposition 6.
###### Proof.
Let be the graph of a 4-polytope embedded in . Let be a given set of four vertices in and let a labelling and pairing of the vertices in .
Consider a linear function that vanishes on a linear hyperplane passing through . Consider the two cases in which either is a supporting hyperplane of a facet of or intersects the interior of .
Suppose is a supporting hyperplane of a facet . First, find an path in the subgraph , which is connected by Balinski’s theorem (Theorem 2). Second, use Lemma 1 to find an path that touches only at .
If instead intersects the interior of then there is a vertex in with -value greater than zero and a vertex with -value less than zero. Use Lemma 1 to find an path in which each inner vertex has negative -value and an path in which each inner vertex has positive -value. ∎
## 3. d-cube
Consider the -cube . Let be a vertex in and let denote the vertex at distance from , called the vertex opposite to . Besides, denote by the facet disjoint from a facet of ; we say that and is a pair of opposite facets.
###### Definition 7 (Projection π).
For a pair of opposite facets of , define a projection from to by sending a vertex to the unique neighbour of in , and a vertex to itself (that is, ); write to be precise, or write or if the cube and the facet are understood from the context.
We extend this projection to sets of vertices: given a pair of opposite facets and a set , the projection or of onto is the set of the projections of the vertices in onto . For an -face , the projection or of onto is the -face consisting of the projections of all the vertices of onto . For a pair of opposite facets in , the restrictions of the projection to and the projection to are bijections.
Let be a set of vertices in the graph of a -cube . If, for some pair of opposite facets , the set contains both a vertex and its projection , we say that the pair is associated with the set in and that is an associating pair. Note that an associating pair can associate only one pair of opposite facets.
The next lemma lies at the core of our methodology.
###### Lemma 8.
Let be a nonempty subset of . Then the number of pairs of opposite facets associated with is at most .
###### Proof.
Let and let with be given. Consider a pair of opposite facets. Define a direction in the cube as the set of the edges between and ; each direction corresponds to a pair of opposite facets. The directions partition the edges of the cube into sets of cardinality . (The notion of direction stems from thinking of the cube as a zonotope [16, Sec. 7.3])
A pair of facets is associated with the set if and only if the subgraph of induced by contains an edge from the corresponding direction.
If a direction is present in a cycle of , then the cycle contains at least two edges from this direction. Indeed, take an edge on that belongs to a direction between a pair of opposite facets. After traversing the edge from to , for the cycle to come back to the facet , it must contain another edge from the same direction. Hence, by repeatedly removing edges from cycles in we obtain a spanning forest of that contains an edge for every direction present in . As a consequence, the number of such directions is at most the number of edges in the forest, which is upper bounded by . (A forest is a graph with no cycles.) ∎
The relevance of the lemma stems from the fact that a pair of opposite facets not associated with a given set of vertices allows each vertex in to have “free projection”; that is, for every the projection is not in , and for the projection is not in .
## 4. Connectivity of the d-cube
We next unveil some further properties of the cube that will be used in subsequent sections.
Given sets of vertices in a graph , the set separates from if every path in the graph contains a vertex from . A set separates two vertices not in if it separates from . We call the set a separator of the graph. We will also require the following three assertions.
###### Proposition 9 ([9, Prop. 1]).
Any separator of cardinality in consists of the neighbours of some vertex in the cube and the subgraph has exactly two components, with one of them being the vertex itself.
A set of vertices in a graph is independent if no two of its elements are adjacent. Since there are no triangles in a -cube, Proposition 9 gives at once the following corollary.
###### Corollary 10.
A separator of cardinality in a -cube is an independent set.
###### Remark 11.
If and are vertices of a cube, then they share at most two neighbours. In other words, the complete bipartite graph is not a subgraph of the cube; in fact, it is not an induced subgraph of any simple polytope [8, Cor. 1.12(iii)].
## 5. Linkedness of the d-cube
In this section, we establish the linkedness of (Theorem 16). We make heavy use of Menger’s theorem [4, Thm. 3.3.1] henceforth, and so we remind the reader of the theorem and one of one of its consequences.
###### Theorem 12 (Menger’s theorem, [4, Sec. 3.3]).
Let be a graph, and let and be two subsets of its vertices. Then the minimum number of vertices separating from in equals the maximum number of disjoint paths in .
###### Theorem 13 (Consequence of Menger’s theorem).
Let be a -connected graph, and let and be two subsets of its vertices, each of cardinality at least . Then there are disjoint paths in .
Two vertex-edge paths are independent if they share no inner vertex.
###### Lemma 14.
Let P be a cubical -polytope with . Let be a set of vertices in , all contained in a facet . Let . Arbitrarily label and pair vertices in to obtain . Then, for at least of these pairs , there is an -valid path in .
###### Proof.
If, for each pair in there is an -valid path in connecting the pair, we are done. So assume there is a pair in , say , for which an -valid path does not exist in . Since is -connected, there are independent paths (Theorem 13), each containing a vertex from ; that is, the set , with cardinality , separates from in . By Proposition 9, the vertices in are the neighbours of or in , say of .
Take any pair in , say . If there was no -valid path in , then, by Proposition 9, the set would separate from and would consist of the neighbours of or in , say of . But in this case, a vertex in , which exists since , would form a triangle with and , a contradiction. See also Corollary 10. Since our choice of was arbitrary, we must have an -valid path in between any pair for . ∎
For a set of pairs of vertices in a graph, a -linkage is a set of disjoint paths with the path joining the pair for . For a path we often write for to denote the subpath . We are now ready to prove Theorem 16.
The definition of -linkedness gives the following lemma at once.
###### Lemma 15.
Let . Let be a set of distinct vertices of a -linked graph , let be a labelling and pairing of the vertices in , and let be a set of vertices in such that . Then there exists a -linkage in that avoids every vertex in .
###### Theorem 16 (Linkedness of the cube).
For every , a -cube is -linked.
###### Proof.
The cases of are trivially true. For the remaining values of , we proceed by induction, with given by Proposition 6.
Let , then . Let be any set of vertices, our terminals, in the graph of the -cube and let be a pairing and labelling of the vertices of . We aim to find a -linkage with joining the pair for . For a facet of , let denote the facet opposite to .
We consider three scenarios: (1) all the pairs in lie in some facet of , (2) a pair of lies in some facet of but not every vertex of is in , and (3) no pair of lies in a facet of , which amounts to saying that every pair in is at distance in . For the sake of readability, each scenario is highlighted in bold.
In the first scenario every vertex in lies in some facet of . Hence Lemma 14 gives an -valid path in joining a pair in , say . The projection in of every vertex in onto is not in . Define as the set of pairs of projections of the corresponding vertices in onto . By the induction hypothesis on , there is a -linkage with for . Since is disjoint from , each path can be extended with and to obtain a path for . And together, all the paths give the desired -linkage in the cube.
In the second scenario a pair of , say , lies in some facet of but not every vertex in is in . Let denote the set of neighbours of a vertex in a face of the cube and let denote the set of all the neighbours of in the cube.
In what follows, whenever we let . Let , and partition as follows.
X0 :={x∈XF:{x,y}∈Y and y∈NF(x)} X1 :={x∈XF∖X0:{x,xpFo}∈Y} X2 :={x∈XF∖(X0∪X1):xpFo∉X} X3 :={x∈XF∖(X0∪X1∪X2):xpFo∈X, and for {x,y}∈Y and y∈V(Fo) there is a unique X-% valid path xypFy} X4 :=XF∖(X0∪X1∪X2∪X3)
Let if and, or , and let if and, or .
###### Claim 1.
Let . For every vertex in , there is an -valid path of length at most two from to such that and the paths are pairwise disjoint.
Proof. We prove this claim by induction on the cardinality of .
In the base case , for , let . For , let be the unique -valid path with and . It is clear that these paths are pairwise disjoint and -valid.
Now suppose that the claim is true for any subset of of cardinality . Pick a vertex and let . By the induction hypothesis, there exist -valid and pairwise disjoint paths of length at most two from to for . To prove that the claim is true for , we only need to construct an -valid path disjoint from all these paths previously defined. We will construct it as for some . Define
Ox=⋃z∈X2∪X3∪X′′4(Mz∩NF(x))⋃(X∩NF(x)).
The set represents the set of vertices in that cannot be chosen as in the path . In other words, if then the claim is true for .
Excluding the path , there are exactly disjoint paths of length two in between and , each going through an element of . Thus, to show that there is a suitable vertex , it suffices to show an injection between and , which would imply . Observe that and .
For every vertex , map to . For every with , map to ; note that , since . For a vertex with there exists a unique vertex such that is the unique vertex in on the path . Since , it follows that . In this case, map to if , otherwise map to . Note that ; otherwise the vertices , and would all be pairwise neighbours but there are no triangles in . See Fig. 1(a)-(b) for a depiction of the different types of neighbours of the vertex and the injection from to .
The existence of an injection from to shows the existence of the vertex , and therefore, of the desired path . This concludes the proof of the claim.
We now finalise this second scenario. Let be the set of vertices for . Then and . The pairs of terminals in are already linked by -valid paths . We link the remaining pairs in thereafter.
Applying Claim 1 to , we get the paths from all the terminals in to . For every vertex , let . In this way, the paths have been defined for every vertex in . Denote by the set of vertices in for each in . Then
(*) |X′|+|X1|+|πFo(X1)|+|X3|+|Y3|≤2(k−1)≤d−1.
Let be the corresponding pairing of the vertices in : if with , then the corresponding pair in is .
The induction hypothesis ensures that is -linked. As a consequence, because of (*5) there is a -linkage that avoids every vertex in (Lemma 15). The -linkage gives the existence of paths in between and for . Each path is then extended with the paths and to obtain a path for .
It only remains to show the existence of a path in pairwise disjoint from the paths for . Suppose that we cannot find a path pairwise disjoint from the other paths with . Then there would be a set in separating from . The set would consist of terminal vertices in and nonterminal vertices on some path for . Each nonterminal vertex in amounts to the existence of a terminal vertex in , namely , since . Hence, the cardinality of would be at most . By the -connectivity of , which follows from Balinski’s theorem (Theorem 2), the set would have cardinality , which would imply that every terminal in and every nonterminal in that lies on a path for are in . By Proposition 9, the set would consist of the neighbours of or , say of . In this configuration all the vertices of would be in , which is a contradiction. Indeed, since there is no edge between any two vertices in (Corollary 10), no nonterminal on a path is in , and therefore, , or equivalently, , as desired. The existence of the path finally settles the second scenario. See Fig. 1(c).
It is instructive for the reader to convince herself that the proof of the second scenario works well by verifying the existence of the paths and the existence of the path for the cubes and .
Finally, let us move onto the third and final scenario: every pair in is at distance . From Lemma 8 it follows that there exists a pair of opposite facets of that is not associated with , since and there are pairs of the form . This means that for every , and that for every , . Without loss of generality, assume that and . We can further assume that for some with , say . Then . Let .
By the induction hypothesis, is -linked, and by Lemma 15, we can find disjoint paths in between and for , with each path avoiding . Let . Now using the -linkedness of , find disjoint paths and in , each avoiding the set (Lemma 15); there are vertices in and for . Let and .
The proof of the theorem is now complete. ∎
We are now in a position to answer Wotzlaw’s question ([15, Question 5.4.12]). We continue with a simple lemma from [14, Sec. 3].
###### Lemma 17 ([14, Sec. 3]).
Let be a -connected graph and let be a -linked subgraph of . Then is -linked.
Theorem 16 in conjunction with Lemma 17 gives the answer.
###### Theorem 18.
For every , a cubical -polytope is -linked.
###### Proof.
Let be a cubical -polytope. The results for are trivial. The case of follows from the connectivity of the graph of (Balinski’s theorem), while the case of follows from Proposition 6. For , since a facet of is a -cube with , by Theorem 16 it is -linked. So the -connectivity of the graph of , which follows from Balinski’s theorem (Theorem 2), together with Lemma 17 establishes the proposition. ∎
We improve Theorem 18 in a subsequent paper [3], where we establish the maximum possible linkedness of for a cubical -polytope with .
### 5.1. Strong linkedness of the cube
With Theorems 16, 14 and 6 at hand, it can be verified that 4-polytopes and -cubes for are strong linked. Theorem 19 shows that 4-polytopes are strongly -linked while Theorem 20 shows that -cubes for are strongly -linked.
###### Theorem 19 (Strong linkedness of 4-polytopes).
Every cubical 4-polytope is strongly -linked.
###### Proof.
Let denote the graph of a 4-polytope embedded in . Let be a set of five vertices in . Arbitrarily pair four vertices of to obtain . Let be the vertex of not being paired in . We aim to find two disjoint paths and such that each path avoids the vertex . The proof is very similar to that of Propositions 6 and 4.
Consider a linear function that vanishes on a linear hyperplane passing through . Assume that for some and that .
Suppose first that is a supporting hyperplane of a facet of . If , then find an -valid path in using the 3-connectivity of (Balinski’s theorem). Then use Lemma 1 to find an -valid path in which each inner vertex has positive -value. If instead , then and Lemma 14 ensures the existence of an -valid path in for some , say for . Then use Lemma 1 to find an -valid path in which each inner vertex has positive -value. So assume intersects the interior of . Then there is a vertex in with -value greater than zero and a vertex with -value less than zero. In this case, use Lemma 1 to find an -valid path in which each inner vertex has negative -value and an -valid path in which each inner vertex has positive -value. ∎
Not every 4-polytope is strongly -linked. Take a two-fold pyramid over a quadrangle . Then is a 4-polytope on six vertices, say . Let the sequence appears in in cyclic order, and let the vertex be in . To see that is not strongly 2-linked, observe that, for every two paths and in , either they intersect or one of them contains .
###### Theorem 20 (Strong linkedness of the cube).
For every , a -cube is strongly -linked.
###### Proof.
It suffices to prove the result for . Let be a set of vertices in the -cube for . Arbitrarily pair vertices in to obtain . Let be the vertex of not being paired in . We aim to find a -linkage where each path joins the pair and avoids the vertex .
The result for is given by Theorem 19. So assume .
From Lemma 8 it follows that there exists a pair of opposite facets of | 6,404 | 26,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-17 | latest | en | 0.901172 |
https://isaaclo97.github.io/problems/Spoj/POTHOLE.html | 1,721,898,701,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00334.warc.gz | 267,713,369 | 5,175 | ## Solución al problema número POTHOLE de Spoj - POTHOLE.
Cualquier duda no dudes en contactar.
``````// Adjacency list implementation of FIFO push relabel maximum flow
// with the gap relabeling heuristic. This implementation is
// significantly faster than straight Ford-Fulkerson. It solves
// random problems with 10000 vertices and 1000000 edges in a few
// seconds, though it is possible to construct test cases that
// achieve the worst-case.
//
// Running time:
// O(|V|^3)
//
// INPUT:
// - graph, constructed using AddEdge()
// - source
// - sink
//
// OUTPUT:
// - maximum flow value
// - To obtain the actual flow values, look at all edges with
// capacity > 0 (zero capacity edges are residual edges).
#include <cmath>
#include <vector>
#include <iostream>
#include <queue>
using namespace std;
typedef long long LL;
const int INF = 1000000000;
struct Edge {
int from, to, cap, flow, index;
Edge(int from, int to, int cap, int flow, int index) :
from(from), to(to), cap(cap), flow(flow), index(index) {}
};
struct PushRelabel {
int N;
vector<vector<Edge> > G;
vector<LL> excess;
vector<int> dist, active, count;
queue<int> Q;
PushRelabel(int N) : N(N), G(N), excess(N), dist(N), active(N), count(2*N) {}
void AddEdge(int from, int to, int cap) {
G[from].push_back(Edge(from, to, cap, 0, G[to].size()));
if (from == to) G[from].back().index++;
G[to].push_back(Edge(to, from, 0, 0, G[from].size() - 1));
}
void Enqueue(int v) {
if (!active[v] && excess[v] > 0) { active[v] = true; Q.push(v); }
}
void Push(Edge &e) {
int amt = int(min(excess[e.from], LL(e.cap - e.flow)));
if (dist[e.from] <= dist[e.to] || amt == 0) return;
e.flow += amt;
G[e.to][e.index].flow -= amt;
excess[e.to] += amt;
excess[e.from] -= amt;
Enqueue(e.to);
}
void Gap(int k) {
for (int v = 0; v < N; v++) {
if (dist[v] < k) continue;
count[dist[v]]--;
dist[v] = max(dist[v], N+1);
count[dist[v]]++;
Enqueue(v);
}
}
void Relabel(int v) {
count[dist[v]]--;
dist[v] = 2*N;
for (int i = 0; i < G[v].size(); i++)
if (G[v][i].cap - G[v][i].flow > 0)
dist[v] = min(dist[v], dist[G[v][i].to] + 1);
count[dist[v]]++;
Enqueue(v);
}
void Discharge(int v) {
for (int i = 0; excess[v] > 0 && i < G[v].size(); i++) Push(G[v][i]);
if (excess[v] > 0) {
if (count[dist[v]] == 1)
Gap(dist[v]);
else
Relabel(v);
}
}
LL GetMaxFlow(int s, int t) {
count[0] = N-1;
count[N] = 1;
dist[s] = N;
active[s] = active[t] = true;
for (int i = 0; i < G[s].size(); i++) {
excess[s] += G[s][i].cap;
Push(G[s][i]);
}
while (!Q.empty()) {
int v = Q.front();
Q.pop();
active[v] = false;
Discharge(v);
}
LL totflow = 0;
for (int i = 0; i < G[s].size(); i++) totflow += G[s][i].flow;
}
};
int main() {
int t;
cin >> t;
for (int i = 0; i < t; i++) {
int n;
cin >> n;
PushRelabel pr(n);
for (int j = 0; j < n-1; j++) {
int m;
cin >> m;
for (int k = 0; k < m; k++) {
int p;
cin >> p;
p--;
int cap = (j == 0 || p == n-1) ? 1 : INF; | 992 | 2,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-30 | latest | en | 0.342176 |
https://www.physicsforums.com/threads/screams-in-anger-ok-divergence-theorem-problem.94637/ | 1,508,379,953,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823214.37/warc/CC-MAIN-20171019012514-20171019032514-00733.warc.gz | 980,922,150 | 16,415 | # *screams in anger* ok, divergence theorem problem
1. Oct 14, 2005
### schattenjaeger
question says answer in whichever would be easier, the surface integral or the triple integral, then gives me(I'm in a mad hurry, excuse the lack of formatting...stuff)
the triple integral of del F over the region x^2+y^2+z^2>=25
F=((x^2+y^2+z^2)(xi+yj+zk)), so del F would be 3(x^2+y^2+z^2)?
In this case, the region is a sphere, it'd be easier to do it with the triple integral over the volume, right? Well, regardless I tried it that way, converting to spherical coordinates(my book mixes up the traditional phi and theta placement, but whatever)
triple integral of 3r^4*sin(theta)drd(theta)d(phi), and the limits of integration, going from the right integral to the left, 0-5, 0-pi, 0-2pi?
And somewhere before there is where I messed up 'cuz I can do the integral I have there easily enough and I get like some huge square of 5 times pi, and the answer is 100pi
-_-
:'(
2. Oct 14, 2005
### rachmaninoff
is that actually a $<=$ ? (you say the region is a sphere...)
I don't understand your notation. If you what you meant was $$\left( x^3 \hat{i} + y^3 \hat{j} + z^3 \hat{k} \right)$$, then your $$\nabla \mathbf{\vec{F}}$$ is correct. But what you wrote is $$\left( x^2 + y^2 + z^2 \right)\left( x \hat{i} + y \hat{j} + z \hat{k} \right)$$
$$=\left( x \left( x^2 + y^2 + z^2 \right) \hat{i} + y \left( x^2 + y^2 + z^2 \right) \hat{j} + z \left( x^2 + y^2 + z^2 \right)\hat{k} \right)$$, which has a different divergence.
Last edited by a moderator: Oct 14, 2005
3. Oct 14, 2005
### schattenjaeger
Oh I love you, you're fast
yes I mean <=
but yah, the latter is what I wrote and meant, but I'm confused, isn't the (x^2+y^2+z^2) like...if you just had a constant A, times (xi+yj+zk)(or A<x,y,z>)it's be like <Ax,Ay,Az>, so can you not do that with the (x^2+y^2+z^2) out front?
am I doing something stupid in the midst of night?
4. Oct 14, 2005
### schattenjaeger
F=((x^2+y^2+z^2)(xi+yj+zk)), so del F would be 3(x^2+y^2+z^2)?
I'll throw in the intermediate step I did too,
so F=<x^3+y^2+z^2,x^2+y^3+z^3,similarforz>
and the divergence would be the same in that case as <x^3,y^3,z^3>
5. Oct 14, 2005
### rachmaninoff
Not quite:
$$\left(x^2+y^2+z^2 \right)(x \mathbf{\hat{i}} \right))=\left(x^3+\mathbf{x}y^2+\mathbf{x}z^2 \right) \mathbf{\hat{i}}$$
and similarly...
6. Oct 14, 2005
### schattenjaeger
Well that was a dumb mistake. Remember kids, sleep is gooood
7. Oct 14, 2005
### schattenjaeger
So with that in mind
how the devil is the answer what it is?
8. Oct 14, 2005
### schattenjaeger
because using the correct divergence and going through it I'm getting like 12500pi
9. Dec 19, 2005
### neutrino
schattenjaeger,
I hope you're awake after a long :zzz: . I was having trouble with the exact problem,from Boas' book I presume, and a search for "divergence theorem" brought me here. Anyway, I arrived at the answer given at the back of the book,viz. $$4\pi5^5$$, by solving the surface integral. You also get it when
$$\nabla.\vec{F} = 3(x^2+y^2+z^2)$$,in the case of a volume integral, which is obviously wrong. So did you eventually solve it using the triple integral?
Anyone's help will be appreciated.
Last edited: Dec 19, 2005 | 1,098 | 3,264 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-43 | longest | en | 0.86809 |
http://metamath.tirix.org/mpests/mpbirand | 1,716,198,474,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058254.21/warc/CC-MAIN-20240520080523-20240520110523-00409.warc.gz | 17,610,541 | 1,906 | # Metamath Proof Explorer
## Theorem mpbirand
Description: Detach truth from conjunction in biconditional. (Contributed by Glauco Siliprandi, 3-Mar-2021)
Ref Expression
Hypotheses mpbirand.1 ${⊢}{\phi }\to {\chi }$
mpbirand.2 ${⊢}{\phi }\to \left({\psi }↔\left({\chi }\wedge {\theta }\right)\right)$
Assertion mpbirand ${⊢}{\phi }\to \left({\psi }↔{\theta }\right)$
### Proof
Step Hyp Ref Expression
1 mpbirand.1 ${⊢}{\phi }\to {\chi }$
2 mpbirand.2 ${⊢}{\phi }\to \left({\psi }↔\left({\chi }\wedge {\theta }\right)\right)$
3 1 biantrurd ${⊢}{\phi }\to \left({\theta }↔\left({\chi }\wedge {\theta }\right)\right)$
4 2 3 bitr4d ${⊢}{\phi }\to \left({\psi }↔{\theta }\right)$ | 265 | 678 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-22 | latest | en | 0.447124 |
https://www.kaysonseducation.co.in/questions/p-span-sty_2527 | 1,642,864,559,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303864.86/warc/CC-MAIN-20220122134127-20220122164127-00488.warc.gz | 901,621,980 | 10,699 | ## Question
### Solution
Correct option is
bx – ag + bg – af = 0
Any lie parallel to ax + by + λ = 0
bx – ay + λ = 0
It passes through (–g, –f
So λ = bg – af equation of diameter
bx – ag + bg – af = 0
#### SIMILAR QUESTIONS
Q1
Find the equation of normal at the point (5, 6) to the circle;
x2 + y2 – 5x + 2y – 48 = 0
Q2
Find the equation of tangent to the circle x2 + y2 = 16 drawn from the point (1, 4).
Q3
The angle between a pair of tangents from a point P to the circle
x2 + y2 + 4x – 6y + 9 sin2α + 13 cos2α = 0 is 2α.
Find the equation of the lows of the point P.
Q4
Find the length of tangents drawn from the point (3, – 4) to the circle
2x2 + 2y2 – 7x – 9y – 30 = 0
Q5
Find the condition that chord of contact of any external point (hk) to the circle x2 + y2 = a2 should subtend right angle at the centre of the circle.
Q6
The chord of contact of tangents drawn from a point on the circle x2 +y2 = a2 to the circle x2 + y2 = b2 thouchese x2 + y2 = e2 find ab in.
Q7
Find the middle point of the chord intercepted on line lx + my + n = 0 by the circle x2 + y2 = a2.
Q8
Find the equation of the tangents from the point A(3, 2) to the circle x2 +y2 + 4x + 6y + 8 = 0.
Q9
If two tangents are drawn from a point on the circle x2 + y2 = 25 to the circle x2 + y2 = 25. Then find the angle between the tangents.
Q10
Examine if the two circle x2 + y2 – 2x – 4y = 0 and x2 + y2 – 8y – 4 = 0 touch each other externally or internally. Also the pointed contact. | 632 | 1,490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-05 | latest | en | 0.446902 |
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Area of parallelograms
randFromArray([ ["in", "inch"], ["ft", "foot"], ["m", "meter"], ["cm", "centimeter"], ["", "unit"] ]) randRange(1, 8) randRange(1, 8) randRangeNonZero(-2, 2) B * H
Suppose a parallelogram has base length B \text{ UNIT} and height H \text{ UNIT}.
What is the parallelogram's area?
K square plural(UNIT_TEXT)
graph.p = parallelogram(B, H, UNIT, SH); graph.p.drawBase(); graph.p.drawHeight();
By moving a slice of the parallelogram, we can see that its area is equal to that of a rectangle with the same base and height.
graph.p.sliceHint(); graph.p.animHint();
area of rectangle = b \times h
graph.p.drawArea();
A = B \times H = K
Something went wrong with that request. Please try again. | 242 | 848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2016-18 | latest | en | 0.793702 |
https://gmatclub.com/forum/how-long-did-it-take-betty-to-drive-nonstop-on-a-trip-from-110158.html | 1,540,171,709,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514443.85/warc/CC-MAIN-20181022005000-20181022030500-00289.warc.gz | 651,890,862 | 53,863 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# How long did it take Betty to drive nonstop on a trip from
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Joined: 26 Feb 2011
Posts: 32
How long did it take Betty to drive nonstop on a trip from [#permalink]
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28 Feb 2011, 12:25
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How long did it take Betty to drive nonstop on a trip from her home to Denver, Colorado?
(1) If Betty's average speed for the trip had been 3/2 times as fast, the trip would have taken 2 hours.
(2) Betty's average speed for the trip was 50 miles per hour.
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Status: I am Midnight's Child !
Joined: 04 Dec 2009
Posts: 109
WE 1: Software Design and Development
Re: How long did it take Betty to drive nonstop on a trip from [#permalink]
### Show Tags
28 Feb 2011, 14:18
How long did it take Betty to drive nonstop on a trip from her home to Denver,Colorado?
1) If Bettys average speed for the trip had been 1,5 times as fast, the trip would have taken 2 hours
2) Bettys average speed for the trip was 50 miles per hour
IMO A
From (1),
$$d/s$$ -$$d/1.5s$$ = $$2$$
$$d$$ = $$6s$$
We know d = t*s
Hence $$t$$ =$$6$$ So, Sufficient
from (2) .
s= 50 So, Insufficient
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28 Feb 2011, 16:28
Agree with the answer. It does not matter for DS, but its d/1.5s = 2. t = 3.
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28 Feb 2011, 19:15
Let x be the average speed and t be the time taken during trip, so from (1) we have :
1.5(x) * 2 = x*t
=> t = 3, sufficient.
From (2), all we have is x = 50, so not sufficient.
Regards,
Subhash
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28 Feb 2011, 21:17
Statement 1.since the distance is equal:
V1T1=V2T2
V1T1=1.5V1*(T1-2)
T1=1.5T1-3
T1=6
Statement 1 is sufficient
Statement 2:Ony average velocity is given so nothing can be concluded.Hence Ststement 2 is insufficient.
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Re: How long did it take Betty to drive nonstop on a trip from [#permalink]
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03 Oct 2011, 22:45
subhashghosh wrote:
Let x be the average speed and t be the time taken during trip, so from (1) we have :
1.5(x) * 2 = x*t
=> t = 3, sufficient.
From (2), all we have is x = 50, so not sufficient.
Regards,
Subhash
Statement 1 says that if Betty' speed has been 1 1/2 times as fast , the trip would have taken 2 hours...
Doesnt it mean Is 'S' was original speed new speed = s + 1 1/2 S = 5/2 S and not 1.5 S????
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Re: How long did it take Betty to drive nonstop on a trip from [#permalink]
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04 Oct 2011, 05:03
Statement 2 is not enough. Gives us only one variable.
Statement one gives us enough to solve with an if then scenario.
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Re: How long did it take Betty to drive nonstop on a trip from [#permalink]
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04 Oct 2011, 05:07
Statement 2 is not enough. Gives us only one variable.
Statement one gives us enough to solve with an if then scenario.
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26 Oct 2011, 23:34
Can someone please comment if this is right approach?
Let 'S' be Speed and 'D' be distance.
st1)Original Equation: S * (t+2) = D
New Equation(If betty's speed was 1 1/2 times as fast the trip would have taken 2 hours) => (S + 3/2 S) * 2 = D
Doesnt it mean Is 'S' was original speed new speed = s + 1 1/2 S = 5/2 S and not 1.5 S????
S * (t+2) = 5/2 S * 2
Thus 3S = St
Thus t = 3
Hence A
Is this correct? I see people write 1.5 S and not 5/2 S in the new equation .... So is my approach correct or wrong ??? PLease comment
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18 Nov 2011, 07:15
siddhans wrote:
Can someone please comment if this is right approach?
Let 'S' be Speed and 'D' be distance.
st1)Original Equation: S * (t+2) = D
New Equation(If betty's speed was 1 1/2 times as fast the trip would have taken 2 hours) => (S + 3/2 S) * 2 = D
Doesnt it mean Is 'S' was original speed new speed = s + 1 1/2 S = 5/2 S and not 1.5 S????
S * (t+2) = 5/2 S * 2
Thus 3S = St
Thus t = 3
Hence A
Is this correct? I see people write 1.5 S and not 5/2 S in the new equation .... So is my approach correct or wrong ??? PLease comment
Can someone please reply? How do we know "as fast" means s + 3/2 s or 1.5 s???
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23 Nov 2017, 07:48
How long did it take Betty to drive nonstop on a trip from her home to Denver, Colorado?
Time = Distance/Rate ?
(1) If Betty's average speed for the trip had been 3/2 times as fast, the trip would have taken 2 hours.
(3/2*Rate)*2 = Distance;
Distance/Rate = 3. Sufficient.
(2) Betty's average speed for the trip was 50 miles per hour. Only the rate is clearly not sufficient, to get the time. Not sufficient.
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In this blog post, we will be discussing about Substitution solver with steps. Our website will give you answers to homework.
## The Best Substitution solver with steps
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Full Version: Negate an equation:
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A program is presented a list of equations: L0:={"x+y<=3","y<=9","c=-x-y"};
It is desired to change the signs of each of the terms in the third (string) element: L0(3) ==> "c=-x-y" which needs to become "-c=x+y"
How would you go about it?
eqs:={"x+y<=3","y<=9","c=-x-y"};
expr("neg("+eqs[3]+")"); returns -( c == (-x-y) ) this is a bug?
I do not know why change = to ==, Alters the expression, and becomes a comparison, affecting the logic of an algorithm,
simplify(Ans); returns 0 (false)
should be
eqs:={"x+y<=3","y<=9","c=-x-y"};
expr("neg("+eqs[3]+")"); returns -( c = (-x-y) )
simplify(Ans);
-c = -(-x-y)
-c = (x+y)
compsystems, thanks for trying. Unfortunately, your procedure doesn't meet the objective: "-c=x+y"
In general, I was hoping to find a more direct way to convert between a string object version of a function, and the function object, (and retain the result as a string object). For example, this works as specified, but is quite cumbersome:
STRING(-CAS("left(CAS(L0(3)))")=simplify(-CAS("right(CAS(L0(3)))")));
Better ways and means probably exist, and I just haven't stumbled upon them so far. Somehow, it seems like there would be a built in command for interacting with functions as string objects. I would appreciate other ideas, since I seem to encounter this problem quite a bit. Working with a "string function" can be a way to get around problems resulting from undesired evaluation steps; in programs, for example.
Have you tried: STRING(-expr(L0(3))) ?
Edit: the command above works in CAS but not in Home, in Home you need to do: STRING(CAS("-expr(L0(3))"))
Well, I tried many similar variants, but not precisely that ... I was hoping you would set me on the right path! For one thing, I was using EXPR, not expr, and it doesn't work with the upper case version. I like your method much better than the long winded version I've shown.
There is one other thing, it turns out that the last function in the list is the only one needing change of signs. So, in general, I can use something like:
L0(SIZE(L0));
To reach the objective in any size list of string elements. However that approach doesn't work in the CAS world, frustrating something like this:
STRING(CAS("-expr(L0( SIZE(L0) ))"));
This WILL work, however:
N:=SIZE(L0);
STRING(CAS("-expr(L0(N))"));
Would you have any thoughts on how to adapt your excellent method to any size list, in general?
You can use L0(0) to access to the last element of L0.
Yes, I had forgotten about that. That is very helpful. Thank you!
Be careful when evaluating a symbolic object (represented as a string) within a program whether it be through CAS() or EXPR().
Code:
``` export localwarning() begin local x:=1; L0:={ "x^2+y^2=r^2" }; return(string(CAS("-expr(L0(x))"))); end;```
There is no bug, but you should check whether the answer is what you wanted. If not, you'll need to create your own myEXP() and myCAS() functions when dealing with indirection and undefined variables. The expressions stored as strings in L0 will be evaluated, and therefore inherit whatever values are present as local variables inside the calling function. Since x was used here in our calling program (set to 1), then any expression parsed via EXPR() and CAS() will be evaluate x to 1 -- including the instance where the string is parsed into a symbolic expression. The short solution is to not use x or y or c in your calling program. However, a generalized approach would stead call CAS() or EXPR() without ANY local variables defined, except for possibly the variable used to pass the string (which you hope and cross your fingers does not exist inside the string itself).
Code:
``` begin myCAS(casString) return(CAS(EVAL(casString)); // cross your fingers that casString is itself not a sequence of characters in the variable casString end;```
No local variables (other than 'casString') are used in the myCAS() calling program so your chances of having your string/expression modified by pre-existing local variables is next to none. The EVAL() is needed because CAS commands treat their arguments as being literal.
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# Control theory
The aim of this lab Is to simulate motor position control with an open loop system and a closed loop system for a harder using Mutual. And compare the experimental results to theoretically derived results. Results Figure – Armature Control Motor Figure 1 represents the mechanisms of the harder. This is represented by the block diagram In figure 4. Open loop Figure – Block Diagram Figure 2 represents the transfer function for the open loop system.
Figure – Open Loop Response Figure 3 represents an open loop step response for the block diagram In figure 2 for a step in put of magnitude 0. 1 radar. When Aka = 100 using the following parameters: Table – Parameters Parameter Symbol lap. Values Inertia IN m so / radar Friction 20 N m s / radar Amplifier Aka 10-1000 Resistance Motor Constant Km Inductance 1 mm The response is a ramp as expected. This happens because there is no feedback in the system. Once the arm in the harder is set to go too position it will not stop and will keep moving at the same rate thus creating a ramp response when simulated on Mutual.
Closed Loop Figure 4 represents the transfer function for the closed loop system. Figure – Closed Loop Response Figure 5 represents a closed loop step response for the block diagram in figure 4 for a step input of magnitude 0. 1 radar. When Aka = 100 using the same parameters in table 1. The response is undermanned as expected. The blue line on the graph represents the position of the arm head in the harder. The dash line at an amplitude of 1 represents where the arm head wants to be. The feedback in the loop cause the arm head (blue line) get closer and eventually settles were it wants to be.
Experimental results Table – Results Time to Peak tip 0. 1559 sec Settling Time TTS 0 sec Max Peak Pm 1. 2182 amplitude Theoretical results Pole Data: -1000. 5+0. 00001 1st -9. 7+20. 11 2nd -9. 7-20. 11 3rd Where tip is the time to peak, Pm is the Max peak, TTS is the settling time, writ is the angular frequency and ( is the damping ratio. Figure – Pole Diagram Figure 6 shows the relationship between the poles on the s-plane. Where x is the angular frequency. Using the 2nd and 3rd poles and the formulas 1-3 the time to peak, settling time and Max peak can be calculated.
As shown on figure 5 = 20. 1 Using this in formula 1 the time to peak can be calculated. As stated before Now tip and writ are known using formula 1 ( (the damping ratio) can be calculated. Ana writ ten Max pea p Ana ten settling t Mime TTS can De calculated Table – Results Comparison Theoretical Experimental % Error 0. 156298 sec 0. 255 0. 4103 sec 0. 3757 sec 9. 2 1. 2179 amplitude 0. 025 Conclusion The experimental results verify the theory the percentage error for the Time to peak and Max peak are negligible. The settling time is higher at 9. 2% this could be caused by experimental error.
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Is 0 a rational number
is 0 a rational number
Contents
Is zero number a rational number?
The Answer is “Yes” Number Zero is a rational number.
As We already aware that the integer 0 may be written as following forms:
For Instance, 0/1, 0/-1, 0/2, 0/-2, 0/3, 0/-3, 0/4, 0/-4 and so on …..
In some other words,we can say that 0 = 0/y, where y is any non-zero integer number.
Thus, number Zero 0 may be written in the form of, where x/y = 0, where x = 0 and y is any non-zero integer number.
Hence, this proved that number Zero 0 is a rational number.
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### is 0 a rational number
6. शक संवत की शुरुवात कब हुई थी?
A. 78 ई
B. 80 ई
C. 30 ई
D. 639 ई
Ans: A
7. जिन्हें 1925 में केंद्रीय विधान सभा के स्पीकर के रूप में चुना गया था ?
A. एन सी केलकर
B. विठ्ठलभाई पटेल
C. मदन मोहन मालवीय
D. लाला लाजपत राय
Ans: B
8. दिल्ली सल्तनत के किस सुल्तान को ‘दुनिया का खान’ कहा गया था?
A.. पृथ्वीराज
B. पोरस
C. सिकंदर
D. मुहम्मद बिन तुगलक
Ans: D
9. इनमे से किस वर्ष अकबर ने तीर्थ-यात्रा समाप्त की थी?
A. 1540
B. 1550
C. 1563
D. 1572
Ans: C
10. भारत के राजधानी दिल्ली में “हिन्दुस्तान सोशलिस्ट रिपब्लिकन एसोसिएशन’ की स्थापना किस वर्ष हुई थी?
A. 1958
B. 1931
C. 1915
D. 1896
Ans: B
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### is 0 a rational number Part 1
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https://www.physicsforums.com/threads/current-through-ballistic-2deg-channel.1000859/ | 1,717,072,373,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00834.warc.gz | 797,612,695 | 17,071 | # Current through Ballistic 2DEG Channel
• FermiFrustration
In summary, to solve this problem, you need to solve the two-dimensional Schrödinger equation with the given parameters and use the energy eigenvalues to calculate the number of available channels through the channel, which can then be used to calculate the current in the system.
FermiFrustration
Homework Statement
See attachment!
Find current through ballistic 2DEG channel assuming a parabolic potential in the channel
Relevant Equations
Schrödinger Equation, Airy's Equation, I = N (2e^2)/h V
So I am a bit uncertain what approach is best for solving this problem and how exactly I should approach it, but my strategy right now is:
1. Solve the time-independent Schrödinger Equation with the given Hamiltonian and find energy eigenvalues of system:
-Here I struggle a bit with actually solving it; if my approach is right this should be the crux of the problem
-Since the y and x-dependent parts of the Schrödinger equation are possible to separate I think it is possible to solve this as two differential equations with only one variable like:
## ( -\frac{h^2}{2m^{*}}(\frac{d^2}{dx^2} + \frac{d^2}{dy^2}) + \frac{1}{2}m^* \omega^2 y^2 - E )\Psi = 0 ##
## (-\frac{h^2}{2m^{*}}\frac{d^2}{dx^2} - E )\Psi = (\frac{h^2}{2m^{*}}\frac{d^2}{dy^2} - \frac{1}{2}m^* \omega^2 y^2)\Psi ##
## \implies ¨(-\frac{h^2}{2m^{*}}\frac{d^2}{dx^2} - E )\Psi = G ##
## \implies (\frac{h^2}{2m^{*}}\frac{d^2}{dy^2} - \frac{1}{2}m^* \omega^2 y^2)\Psi = G ##
-I am however unsure how to reassemble this into a complete solution.
2. Find energy eigenstates below the fermi energy - their number should be the number of available channels through the channel
3. Plug this new obtained N into I = N (2e^2)/h V with the given voltage from source to drainIs my approach right and how should I go about solving the differential equation?
#### Attachments
• NanoP1.png
60.3 KB · Views: 111
Your approach is close, but you need to solve the two-dimensional Schrödinger equation with the given parameters, rather than separating it into two equations. The two-dimensional version of the Schrödinger equation is:## ( -\frac{h^2}{2m^{*}}(\frac{d^2}{dx^2} + \frac{d^2}{dy^2}) + \frac{1}{2}m^* \omega^2 (x^2+y^2) - E )\Psi = 0 ##There are several methods for solving this equation, such as separation of variables, Green's function techniques, and numerical methods. Once you have solved the equation to obtain the energy eigenvalues, you can then use those values to calculate the number of available channels through the channel, which you can then plug into I = N(2e^2)/h V with the given voltage from source to drain.
## 1. What is a ballistic 2DEG channel?
A ballistic 2DEG (two-dimensional electron gas) channel is a type of electronic device that consists of a thin layer of electrons confined to a two-dimensional plane. This channel is typically created in a semiconductor material and is used to study the behavior of electrons in a controlled environment.
## 2. How is current generated in a ballistic 2DEG channel?
Current is generated in a ballistic 2DEG channel when a voltage is applied across the channel, causing the electrons to move from one end to the other. The movement of electrons creates a flow of current, which can be measured and studied to understand the behavior of electrons in this type of channel.
## 3. What factors affect the current through a ballistic 2DEG channel?
The current through a ballistic 2DEG channel is affected by several factors, including the applied voltage, the thickness of the channel, the temperature of the device, and the properties of the material used to create the channel. Additionally, any external magnetic or electric fields can also influence the current through the channel.
## 4. How is the current through a ballistic 2DEG channel measured?
The current through a ballistic 2DEG channel is typically measured using specialized equipment, such as a current probe or a Hall effect sensor. These tools are able to detect the flow of electrons through the channel and provide accurate measurements of the current at different points along the channel.
## 5. What is the significance of studying the current through a ballistic 2DEG channel?
Studying the current through a ballistic 2DEG channel can provide valuable insights into the behavior of electrons in a confined environment. This research can have practical applications in the development of new electronic devices and technologies, as well as advancing our understanding of fundamental principles in physics.
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1K | 1,232 | 4,738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-22 | latest | en | 0.832826 |
http://www.fiberoptics4sale.com/wordpress/what-is-fiber-bragg-grating/ | 1,454,741,794,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701146196.88/warc/CC-MAIN-20160205193906-00133-ip-10-236-182-209.ec2.internal.warc.gz | 407,941,816 | 18,992 | # What is Fiber Bragg Grating ?
March 19, 2011
### What is Fiber Bragg Grating – FBG
A Fiber Bragg Grating FBG Device from JDSU
Normal optical fibers are uniform along their lengths. In a simple fiber Bragg grating, the refractive index of the fiber core vary periodically along the length of the fiber, as shown in the following figure.
### How Fiber Bragg Grating Reflects and Transmits Light
As shown in the above figure, the refractive index of the fiber core is modulated with a period of
Λ. When a light with a broad spectrum is launched into one end of fiber containing a fiber Bragg grating, the part of the light with wavelength matching the Bragg grating wavelength will be reflected back to the input end, with the rest of the light passing through to the other end. This reflection phenomena is explained in the following figure.
From the momentum conservation requirement of the Bragg grating condition, the following equation can be obtained:
where neff is the effective refractive index of the fiber core, and
λB is the wavelength of the light reflected by the Bragg grating.
Therefore, the Bragg grating wavelength λB can be expressed as
Note that the Bragg grating wavelength is the function of the effective index and the period of the grating.
Fiber Bragg grating can be used as a MUX/DEMUX device in WDM systems for extracting a signal (channel) with a particular wavelength from a stream of signals (channels). This is shown in the following two pictures..
### Reflection and Transmission in Fiber Gratings
The fundamental principle behind the operation of fiber Bragg grating (FBG) is Fresnel reflection. Where light traveling between media of different refractive indices may both reflect and refract at the interface.
The fiber Bragg grating will typically have a sinusoidal refractive index variation over a defined length. We have seen the definition of Bragg wavelength λB from the previous section.
The wavelength spacing between the first minima, (as shown in above figure), or the bandwidth
Δλ is given by,
where
δn0 is the variation in the refractive index (n3-n2), and
η is the fraction of power in the core.
### Fiber Bragg Grating Structure
The structure of the FBG can vary via the refractive index, or the grating period. The grating period can be uniform or graded, and either localised or distributed in a superstructure.
The refractive index has two primary characteristics, the refractive index profile, and the offset. Typically, the refractive index profile can be uniform or apodized, and the refractive index offset is positive or zero.
There are six common structures for Fiber Bragg Gratings;
1. Uniform positive-only index change
2. Gaussian apodized
3. Raised-cosine apodized
4. Chirped
5. Discrete phase shift
6. Superstructure
Structure of the Refractive Index Change in a Uniform Fiber Bragg Grating
Refractive Index Profile in the Core
### How to Make a Fiber Bragg Grating Device
Fiber Bragg gratings, which operate at wavelengths other than near the writing wavelength, are fabricated by techniques that broadly fall into two categories: those that are holographic and those that are noninterferometric, based on simple exposure to UV radiation periodically along a piece of fiber.
The holographic techniques use a beam splitter to divide a single input UV beam into two, interfering them at the fiber.
The noninterferometric techniques depend on periodic exposure of a fiber to pulsed sources or through a spatially periodic amplitude mask.
Here we only introduce the holographic approach with a bulk interferometer.
The interferometer is one encountered in standard holography, with the UV beam divided into two at a beam splitter and then brought together at a mutual angle of
θ, by reflections from two UV mirrors. This method allows the Bragg wavelength to be chosen independently of the UV wavelength.
### Fiber Bragg Gratings Applications
The following figure shows a communication system carries 4 wavelengths: 1550, 1552, 1554, and 1556nm. The 3 shorter wavelengths (1550, 1552, 1554nm) need to go from Town A to Town C.
The system needs to to send signals at 1556nm from Town A to B, and from Town B to C. It drops 1556nm at Town B, and replaces it with another signal at the same wavelength 1556nm from Town B to C.
2) Chromatic Dispersion Compensation
Another use of fiber gratings is to compensate for chromatic dispersion in an optical fiber. The grating serves as a selective optical delay line, which adjusts the transit times of different wavelengths in a pulse so they are approximately equal.
For example, the longer wavelengths in a pulse arrived first, and the shorter wavelengths arrived last. As shown in the following figure, the Bragg grating is made so segments which reflect different wavelengths are in different positions along the length of the grating.
In this case, the longer wavelengths (λ4) arrive first and are transmitted through to the last part of the grating. The first part of the grating reflects the shortest wavelengths, which arrive last. The longer wavelengths have to travel a longer distance, so they are delayed, allowing the shorter wavelengths to catch up.
The following figure shows a typical dispersion compensation module design based on Fiber Bragg Grating and Circulator.
Layout of a Fiber Bragg Grating Based Dispersion Compensation Module
3) Fiber Bragg Grating Sensors
As well as being sensitive to strain, the Bragg wavelength λB is also sensitive to temperature. This means that fiber Bragg gratings can be used as sensing elements in optical fiber sensors.
In a FBG sensor, the measurand causes a shift in the Bragg wavelength, ΔλB. The relative shift in the Bragg wavelength, ΔλB / λB, due to an applied strain (ε) and a change in temperature (ΔT) is approximately given by,
Fiber Bragg gratings can then be used as direct sensing elements for strain and temperature.
The following picture shows a unpackaged Fiber Bragg Grating FBG sensor supplied in a single mode optical fiber for strain and temperature measurements. Suitable for very low intrusion surface mounting, FRP composite embedment, or packaging into an FBG transducer. | 1,347 | 6,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-07 | latest | en | 0.91365 |
https://rdrr.io/cran/samplesize/src/R/n.ttest.R | 1,519,532,091,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816094.78/warc/CC-MAIN-20180225031153-20180225051153-00596.warc.gz | 767,051,828 | 14,401 | # R/n.ttest.R In samplesize: Sample Size Calculation for Various t-Tests and Wilcoxon-Test
#### Documented in n.ttest
```n.ttest <-
function(power = 0.80, alpha = 0.05, mean.diff = 0.8, sd1 = 0.83, sd2 = sd1, k = 1, design = "unpaired", fraction = "balanced", variance = "equal")
{
if(variance == "equal" & sd1 != sd2){
warning("Variance is set to equal, but sd's are different. This makes no sense!")
}
if(fraction == "unbalanced" & k == 1){
warning("Groups are chosen unbalanced, but fraction argument k is 1")
}
if(design == "paired" & fraction == "unbalanced"){
warning("Argument -unbalanced- is not used. Paired design is balanced")
}
if(design == "paired" & k != 1){
warning("Argument -k- is set to 1. Paired design is balanced")
}
if(design == "paired" & variance == "unequal"){
warning("Paired design assumes and uses equal variances")
}
if(design == "paired"){
fraction = "balanced"
}
if(design == "paired"){
variance = "equal"
}
if(design == "unpaired" & variance == "unequal"){
warning("Arguments -fraction- and -k- are not used, when variances are unequal")
}
if(power > 1 | power < 0){
stop("Power must be between 0 and 1.0")
}
if(power < 0.5){
warning("Are you sure that Power should be lower than 50 % ?")
}
if(alpha > 1 | alpha < 0){
stop("Type-I-error must be between 0 and 1.0")
}
if(alpha > 0.1){
warning("Are you sure that the two-sided Type-I-Error should be larger than 10 % ?")
}
if(k < 0){
stop("Fraction k must be greater than zero")
}
conf.level <- 1 - alpha / 2
n.start <- 4
switch(variance,
"unequal" = {
k <- sd2/sd1
n1.pri <- n.start/(1 + k)
n2.pri <- (k * n.start)/(1 + k)
n1 <- max(n1.pri, 2)
n2 <- max(n2.pri, 2)
gamma <- sd1/(sd1 + sd2)
c <- mean.diff/(sd1 + sd2)
df_approx <- 1/ ((gamma)^2/ (n1 - 1) + (1 - gamma)^2/ (n2-1))
tkrit.alpha <- qt(conf.level, df = df_approx)
tkrit.beta <- qt(power, df = df_approx)
n.temp <- ((tkrit.alpha + tkrit.beta)^2)/(c^2)
while(n.start <= n.temp){
n.start <- n1 + n2 + 1
n1 <- n.start/(1 + k)
n2 <- (k * n.start)/(1 + k)
df_approx <- 1/ ((gamma)^2/ (n1 - 1) + (1 - gamma)^2/ (n2-1))
tkrit.alpha <- qt(conf.level, df = df_approx)
tkrit.beta <- qt(power, df = df_approx)
n.temp <- ((tkrit.alpha + tkrit.beta)^2)/(c^2)
}
output <- list("Total sample size" = ceiling(n1) + ceiling(k * n1), "Sample size group 1" = ceiling(n1), "sample size group 2" = ceiling(n2))
return(output)
},
"equal" = {
{
switch(
design,
"paired" = {
n.start <- 2
c <- mean.diff / sd1
tkrit.alpha <- qt(conf.level, df = n.start -1)
tkrit.beta <- qt(power, df = n.start - 1)
n.temp <- ((tkrit.alpha + tkrit.beta)^2)/(c^2)
while(n.start <= n.temp){
n.start <- n.start + 1
tkrit.alpha <- qt(conf.level, df = n.start - 1)
tkrit.beta <- qt(power, df = n.start - 1 )
n.temp <- ((tkrit.alpha + tkrit.beta)^2)/(c^2)
}
output <- list("Total sample size" = n.start)
return(output)
},
"unpaired" = {
switch(
fraction,
"balanced" = {
c <- mean.diff / (2*sd1)
tkrit.alpha <- qt(conf.level, df = n.start - 1)
tkrit.beta <- qt(power, df = n.start - 1)
n.temp <- ((tkrit.alpha + tkrit.beta)^2)/(c^2)
while(n.start <= n.temp){
n.start <- n.start + 1
tkrit.alpha <- qt(conf.level, df = n.start - 1)
tkrit.beta <- qt(power, df = n.start - 1)
n.temp <- ((tkrit.alpha + tkrit.beta)^2)/(c^2)
}
n1 <- ceiling(n.start / 2)
n2 <- ceiling(n.start / 2)
output <- list("Total sample size" = 2 * n1, "Sample size group 1" = n1, "Sample size group 2" = n2)
return(output)
},
"unbalanced" = {
df <- n.start - 2
c <- (mean.diff/sd1)*(sqrt(k)/(1+k))
tkrit.alpha <- qt(conf.level, df = df)
tkrit.beta <- qt(power, df = df)
n.temp <- ((tkrit.alpha + tkrit.beta)^2)/(c^2)
while(n.start <= n.temp){
n.start <- n.start + 1
tkrit.alpha <- qt(conf.level, df = n.start - 2)
tkrit.beta <- qt(power, df = n.start - 2)
n.temp <- ((tkrit.alpha + tkrit.beta)^2)/(c^2)
}
n1 <- n.start / (1 + k)
n2 <- k * n1
output <- list("Total sample size" = ceiling(n1) + ceiling(n2), "Sample size group 1" = ceiling(n1), "Sample size group 2" = ceiling(n2), "Fraction" = k)
return(output)
})}
)}
return(output)
}
)
}
```
## Try the samplesize package in your browser
Any scripts or data that you put into this service are public.
samplesize documentation built on May 29, 2017, 6:59 p.m. | 1,415 | 4,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-09 | longest | en | 0.471766 |
https://www.doubtnut.com/ncert-solutions/class-12-maths-chapter-11-three-dimensional-geometry-exercise-miscellaneous-exercise-1 | 1,620,523,752,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988953.13/warc/CC-MAIN-20210509002206-20210509032206-00040.warc.gz | 768,809,919 | 46,883 | # NCERT Class 12 Three Dimensional Geometry Miscellaneous Exercise Maths Solutions
## Class 12 Three Dimensional Geometry Miscellaneous Exercise Maths NCERT Solutions
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### NCERT Class 12 | THREE DIMENSIONAL GEOMETRY | Miscellaneous Exercise | Question No. 19
Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes -> rdot( hat i- hat j+2 hat k)=5 and -> rdot(3 hat i+ hat j+ hat k)=6 .
### NCERT Class 12 | THREE DIMENSIONAL GEOMETRY | Miscellaneous Exercise | Question No. 18
Find the distance of the point (" "1," "" "5," "" "10) from the point of intersection of the line -> r=2 hat i- hat j+2 hat k+lambda(3 hat i+4 hat j+2 hat k) and the plane -> r=( hat i- hat j+ hat k)=5 .
### NCERT Class 12 | THREE DIMENSIONAL GEOMETRY | Miscellaneous Exercise | Question No. 15
Find the equation of the plane passing through the line of intersection of the planes -> rdot( hat i+ hat j+ hat k)=1 and -> rdot(2 hat i+3 hat j- hat k)+4=0 and parallel to x-axis.
### NCERT Class 12 | THREE DIMENSIONAL GEOMETRY | Miscellaneous Exercise | Question No. 14
If the points (1,1,p) a n d (3, "0, "1) be equidistant from the plane vecr *(3 hat i+4 hat j-12 hat k)+13=0 , then find the value of p.
### NCERT Class 12 | THREE DIMENSIONAL GEOMETRY | Miscellaneous Exercise | Question No. 17
Find the equation of the plane which contains the line of intersection of the planes -> rdot( hat i+2 hat j+3 hat k)-4=0, -> rdot(2 hat i+ hat j- hat k)+5=0 and which is perpendicular to the plane | 506 | 1,619 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-21 | latest | en | 0.61318 |
https://numpy.org/devdocs/reference/generated/numpy.polynomial.laguerre.lagvander3d.html | 1,723,502,694,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641052535.77/warc/CC-MAIN-20240812221559-20240813011559-00521.warc.gz | 327,654,399 | 8,244 | # numpy.polynomial.laguerre.lagvander3d#
polynomial.laguerre.lagvander3d(x, y, z, deg)[source]#
Pseudo-Vandermonde matrix of given degrees.
Returns the pseudo-Vandermonde matrix of degrees deg and sample points (x, y, z). If l, m, n are the given degrees in x, y, z, then The pseudo-Vandermonde matrix is defined by
$V[..., (m+1)(n+1)i + (n+1)j + k] = L_i(x)*L_j(y)*L_k(z),$
where 0 <= i <= l, 0 <= j <= m, and 0 <= j <= n. The leading indices of V index the points (x, y, z) and the last index encodes the degrees of the Laguerre polynomials.
If V = lagvander3d(x, y, z, [xdeg, ydeg, zdeg]), then the columns of V correspond to the elements of a 3-D coefficient array c of shape (xdeg + 1, ydeg + 1, zdeg + 1) in the order
$c_{000}, c_{001}, c_{002},... , c_{010}, c_{011}, c_{012},...$
and np.dot(V, c.flat) and lagval3d(x, y, z, c) will be the same up to roundoff. This equivalence is useful both for least squares fitting and for the evaluation of a large number of 3-D Laguerre series of the same degrees and sample points.
Parameters:
x, y, zarray_like
Arrays of point coordinates, all of the same shape. The dtypes will be converted to either float64 or complex128 depending on whether any of the elements are complex. Scalars are converted to 1-D arrays.
deglist of ints
List of maximum degrees of the form [x_deg, y_deg, z_deg].
Returns:
vander3dndarray
The shape of the returned matrix is x.shape + (order,), where $$order = (deg[0]+1)*(deg[1]+1)*(deg[2]+1)$$. The dtype will be the same as the converted x, y, and z.
Notes
New in version 1.7.0.
Examples
>>> import numpy as np
>>> from numpy.polynomial.laguerre import lagvander3d
>>> x = np.array([0])
>>> y = np.array([2])
>>> z = np.array([0])
>>> lagvander3d(x, y, z, [2, 1, 3])
array([[ 1., 1., 1., 1., -1., -1., -1., -1., 1., 1., 1., 1., -1.,
-1., -1., -1., 1., 1., 1., 1., -1., -1., -1., -1.]]) | 651 | 1,891 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-33 | latest | en | 0.642165 |
https://ro.scribd.com/doc/272312849/Dezv-P-M-a-Copil-Norm03 | 1,643,375,629,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305494.6/warc/CC-MAIN-20220128104113-20220128134113-00188.warc.gz | 526,702,835 | 77,324 | Sunteți pe pagina 1din 1
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Drepturi de autor © 2022 Scribd Inc. | 1,263 | 2,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-05 | latest | en | 0.086708 |
https://51zuoyejun.com/caseDetail/3519.html | 1,618,113,148,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038060927.2/warc/CC-MAIN-20210411030031-20210411060031-00558.warc.gz | 196,509,206 | 7,133 | Math 108B, W 21 Review Outline for Second Midterm
** Only material studied since the first midterm is listed on this review sheet, but you
should understand and be ready to use the earlier material in combination with this.
Material: Sections 7A (skip 7.16), 7B (7.23-24), 7C (7.37-43), 8A, 8B (8.20-30), 8C.
Reminder: This midterm will take place on Monday 22 February from 5–6 PM (Pacific
Standard Time).
Definitions.
conjugate transpose of a matrix
orthonormally diagonalizable operator
normal operator
isometry
generalized eigenvector of an operator
generalized eigenspace of an operator
nilpotent operator
strictly upper triangular matrix
multiplicity of an eigenvalue
block diagonal matrix
characteristic polynomial
monic polynomial
minimal polynomial
one polynomial divides another
one polynomial is a multiple of another
Symbols and Abbreviations.
T ∗
TFAE
G(λ, T )
Theorems. This is a list of the major theorems (and corollaries, etc.) that we have
developed, with phrases labelling them. You don’t need to memorize the numbers of these
theorems, but you should know the statements. (The phrases listed here are not full
statements.) The numbers are listed here to help you find things in the book.
Note: Some of these results concern arbitrary vector spaces and some concern arbitrary
inner product spaces. However, certain results require the vector space or inner product
space to be finite dimensional. If “finite dimensional” is in the hypothesis of a result, it
cannot be applied to arbitrary vector spaces. Some results require the field to be C, and
some require the field to be R. Any of these kinds of hypotheses must be included when
you give a complete statement of a result.
Existence of adjoints (discussion under 7.2)
Null spaces and ranges of adjoints (7.7)
The matrix of an adjoint (7.10)
Self-adjoint T with Tv ⊥ v for all v implies T = 0 (7.14 & 7.16)
Normality versus norm conditions (7.20)
Normal operators and their adjoints have the same eigenvectors (7.21)
Eigenvalues for a normal operator and its adjoint are conjugates (in class)
Orthogonality of eigenvectors of a normal operator (7.22)
Normal operators and adjoints restricted to invariant subspaces (in class)
Complex Spectral Theorem (7.24)
Characterizations of isometries (7.42)
Isometries are normal (in class)
Description of isometries over C (7.43)
Null spaces of powers of operators (8.2 & 8.3)
Limit to growth of null spaces (8.4)
Direct sum of null space and range for dimV -th power of an operator (8.5)
Null space description of generalized eigenspaces (8.11)
Linear independence of generalized eigenvectors (8.13)
Generalized eigenspaces make a direct sum (in class)
The dimV -th power of a nilpotent operator is zero (8.18)
Matrices of nilpotent operators (8.19)
Invariance of null spaces and ranges of polynomials in an operator (8.20)
Description of operators on complex vector spaces (8.21)
Bases of generalized eigenvectors (8.23)
Sum of multiplicities of eigenvalues of an operator (8.26)
Block diagonal matrices for operators (8.29)
Degrees and roots of characteristic polynomials (8.36)
Cayley-Hamilton Theorem (8.37)
Existence and uniqueness of minimal polynomial (8.40)
Minimal polynomial of T divides q iff q(T ) = 0 (8.46)
Minimal polynomial divides characteristic polynomial (8.48)
Roots of minimal polynomial are eigenvalues (8.49)
Vocabulary, Grammar, Reading, Writing. Remember that mathematics is a lan-
guage as well as a subject. The vocabulary consists of words and symbols with precise
definitions. Some of these you already have in your background; the new ones appearing
in this course are listed under “Definitions” and “Symbols” above. The basic grammar of
mathematics is built up from logic and the expressions (both verbal and symbolic) used
to make mathematical statements. Vocabulary and grammar are both needed in order to
read or write a piece of mathematics.
All of these aspects of mathematical language will be tested. Questions of the following
types may appear:
Give a precise definition of the concept or symbol “· · · ”.
Give a precise statement of the theorem, lemma, proposition, or corollary “· · ·”.
Such a statement should include the complete hypotheses and the full conclu-
sion(s).
Some proofs will also be asked for – proofs of statements done in the book and/or in
class; proofs that appeared as homework problems; or proofs of new statements.
Finally, there may be questions involving small calculations.
Email:51zuoyejun
@gmail.com | 1,097 | 4,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2021-17 | latest | en | 0.875009 |
https://www.careerride.com/view/comparison-of-rankine-and-carnot-cycles-mcqs-with-answers-23084.aspx | 1,701,492,756,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100327.70/warc/CC-MAIN-20231202042052-20231202072052-00729.warc.gz | 788,475,874 | 6,224 | # Comparison of Rankine and Carnot Cycles - MCQs with Answers
## Comparison of Rankine and Carnot Cycles - MCQs with Answers
Q1. How can we differentiate Rankine cycle from Carnot cycle?
a. Heat addition process of Rankine cycle is reversible isothermal whereas heat addition process of Carnot cycle is reversible isobaric
b. Heat addition process of Rankine cycle is reversible isobaric whereas heat addition process of Carnot cycle is reversible isothermal
c. Heat addition process of Rankine cycle is reversible isentropic whereas heat addition process of Carnot cycle is reversible isothermal
d. both cycles are identical except the working fluid used
ANSWER: b. Heat addition process of Rankine cycle is reversible isobaric whereas heat addition process of Carnot cycle is reversible isothermal
Q2. What is the relation between efficiencies of Rankine cycle () and Carnot cycle for the same pressure ratio?
a. (ηRankine) = (ηCarnot)
b. (ηRankine) > (ηCarnot )
c. (ηRankine) < (ηCarnot )
d. none of the above
ANSWER: c. (ηRankine) < (ηCarnot )
Q3. If Tm be the mean temperature of heat addition in Rankine cycle as shown in diagram,
what will the formula for efficiency of Rankine cycle?
a. (ηRankine) = ( T3 / Tm)
b. (ηRankine) = 1 – (T3 / Tm)
c. (ηRankine) = 1 – (T2 / Tm)
d. (ηRankine) = (T2 / Tm)
ANSWER: b. (ηRankine) = 1 – (T3 / Tm)
Q4. The maximum efficiency of Rankine cycle (ηRankine) is the function of
a. the mean temperature of heat addition (Tm) only
b. the mean temperature of heat addition (Tm) and temperature of steam at the exit of the turbine
c. the mean temperature of heat addition (Tm) and temperature of steam at the entry of the turbine
d. the mean temperature of heat addition (Tm) and temperature of steam at exit of the condenser
Q5. What is the effect of superheated steam on efficiency of Rankine cycle?
a. efficiency of Rankine cycle decreases with increase in superheat of the steam
b. efficiency of Rankine cycle increases with increase in superheat of the steam
c. efficiency of Rankine cycle is not affected by change in superheat of the steam
d. none of the above
ANSWER: b. efficiency of Rankine cycle increases with increase in superheat of the steam
Q6. What is the effect of increase in pressure at which heat is added on the pump work in the Rankine cycle?
a. the pump work increases with increase in pressure of heat addition
b. the pump work decreases with increase in pressure of heat addition
c. the pump work does not change with increase in pressure of heat addition
d. the pump work either increases or decreases with increase in pressure of heat addition
ANSWER: a. the pump work increases with increase in pressure of heat addition
Q7. When the pressure at which heat is added in Rankine cycle increases, the moisture content at the turbine exhaust
a. increases
b. decreases
c. remains same
d. cannot say
Q8. What is the condition for increase the chances of corrosion of blades of turbine?
a. decrease in the pressure difference between which the Rankine cycle operates
b. increase in the pressure difference between which the Rankine cycle operates
c. both a. and b.
d. none of the above
ANSWER: b. increase in the pressure difference between which the Rankine cycle operates
Q9. What is the maximum content of moisture allowed at the turbine exhaust in the steam power plant?
a. 50 %
b. 60 %
c. 30 %
d. 15 % | 854 | 3,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-50 | longest | en | 0.859433 |
http://www.cram.com/flashcards/scientific-vocabulary-311493 | 1,503,525,433,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886124563.93/warc/CC-MAIN-20170823210143-20170823230143-00305.warc.gz | 529,361,548 | 24,012 | • Shuffle
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### 30 Cards in this Set
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Speed distance/time Average time total distance/total time velocity speed in a given direction distance speed X time Unbalanced forces forces that push an object in same or opposite direction make the object move, stop or change direction balanced forces Equal forces acting on an object in opposite direction Acceleration The reate at which velocity changes gravity the force that pulls objects toward each other Law The rule in science Manipulated variable The ONE factor that a scientist changes in an experiment Variable Any factor that can change in an experiment Phrenology A pseudoscience that was the study of teh structure if the skull to determine intelligence and character in a person. Alchemy a pseudoscientific ancestor of chemistry in the Middle Ages Data Table A table used to calculate a lot of "what-ifs" in one operation and to show all the results on one worksheet Air resistance the fluid friction experienced by objects falling through the air Speed distance/time Average time total distance/total time velocity speed in a given direction distance speed X time Unbalanced forces forces that push an object in same or opposite direction make the object move, stop or change direction balanced forces Equal forces acting on an object in opposite direction Acceleration The reate at which velocity changes gravity the force that pulls objects toward each other Law The rule in science Manipulated variable The ONE factor that a scientist changes in an experiment Variable Any factor that can change in an experiment Phrenology A pseudoscience that was the study of teh structure if the skull to determine intelligence and character in a person. Alchemy a pseudoscientific ancestor of chemistry in the Middle Ages Data Table A table used to calculate a lot of "what-ifs" in one operation and to show all the results on one worksheet Air resistance the fluid friction experienced by objects falling through the air | 500 | 2,456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-34 | longest | en | 0.835418 |
https://www.zrivo.com/gross-to-net-calculator | 1,726,357,289,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651601.85/warc/CC-MAIN-20240914225323-20240915015323-00750.warc.gz | 988,002,660 | 11,525 | # Gross to Net Calculator
The gross pay is the amount you get paid before any taxes. Your employer processes payroll and makes sure that you pay your fair share of taxes. Several taxes start with your gross pay and reduces it to the net pay. After your net pay, you have no obligation to anyone as you’ve already paid taxes.
To understand how this works, we need to take a look at payroll processing. Assume your paycheck for the week is \$2,000. You will pay Social Security and Medicare taxes based on gross pay. Then, you’ll pay income taxes. The federal income taxes are mandatory but your state might not tax wages or income at all. So there are at least three different taxes you are subject to and must pay through payroll processing. You won’t pay any taxes directly yourself as it’s your employer’s job to withhold income taxes and forward them to the Internal Revenue Service or the state tax department. Again, only if the state taxes wages.
## Social Security and Medicare tax calculator
The combined rate of Social Security and Medicare taxes is 7.65 percent. All employees are responsible to pay at this rate. These taxes are also known as FICA taxes which stand for the Federal Insurance Contributions Act.
Individually, the Social Security tax rate is 6.2 percent and Medicare tax is 1.45 percent. There is also additional Medicare withholding which will affect net pay. For income over \$200,000, employees are subject to 0.9 percent additional Medicare tax. This additional withholding combines the total Medicare withholding to 2.35 percent, but only for the income earned above the threshold. There is no additional withholding for Social Security taxes.
## Federal, state, and local income taxes
If you’re going to earn more than \$12,400 during the tax year, you will withhold federal income taxes. You may also pay state and local income taxes. Those who are expected to earn less than \$12,400 during the tax year, can fill out Form W-4 to claim an exclusion from federal income taxes. Your exempt status will be enough to let your employer not withhold any income taxes.
The income tax rates are different for everyone but the majority of it depends on the salary. The more you make, the higher you’ll pay in income taxes. It’s as simple as that. The easiest way to ensure your employer withholds the right amount of income taxes is to provide the payroll department with Form W-4 or the state equivalent of the form. If not furnished, the employer may withhold tax at a higher rate than you would because the Internal Revenue Service mandates employers to withhold income tax at the highest single rate in absence of the withholding form. | 547 | 2,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.961173 |
https://en.wikipedia.org/wiki/9_(number) | 1,720,925,979,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00210.warc.gz | 198,349,032 | 71,209 | # 9
(Redirected from 9 (number))
← 8 9 10 →
Cardinalnine
Ordinal9th
(ninth)
Numeral systemnonary
Factorization32
Divisors1,3,9
Greek numeralΘ´
Roman numeralIX, ix
Greek prefixennea-
Latin prefixnona-
Binary10012
Ternary1003
Senary136
Octal118
Duodecimal912
Amharic
Arabic, Kurdish, Persian, Sindhi, Urdu٩
Armenian numeralԹ
Bengali
Chinese numeral九, 玖
Devanāgarī
Greek numeralθ´
Hebrew numeralט
Tamil numerals
Khmer
Telugu numeral
Thai numeral
Malayalam
Babylonian numeral𒐝
Egyptian hieroglyph𓐂
Morse code____.
9 (nine) is the natural number following 8 and preceding 10.
## Evolution of the Hindu–Arabic digit
Circa 300 BC, as part of the Brahmi numerals, various Indians wrote a digit 9 similar in shape to the modern closing question mark without the bottom dot. The Kshatrapa, Andhra and Gupta started curving the bottom vertical line coming up with a 3-look-alike.[1] How the numbers got to their Gupta form is open to considerable debate. The Nagari continued the bottom stroke to make a circle and enclose the 3-look-alike, in much the same way that the sign @ encircles a lowercase a. As time went on, the enclosing circle became bigger and its line continued beyond the circle downwards, as the 3-look-alike became smaller. Soon, all that was left of the 3-look-alike was a squiggle. The Arabs simply connected that squiggle to the downward stroke at the middle and subsequent European change was purely cosmetic.
While the shape of the glyph for the digit 9 has an ascender in most modern typefaces, in typefaces with text figures the character usually has a descender, as, for example, in .
The form of the number nine (9) could possibly derived from the Arabic letter waw, in which its isolated form (و) resembles the number 9.
The modern digit resembles an inverted 6. To disambiguate the two on objects and labels that can be inverted, they are often underlined. It is sometimes handwritten with two strokes and a straight stem, resembling a raised lower-case letter q, which distinguishes it from the 6. Similarly, in seven-segment display, the number 9 can be constructed either with a hook at the end of its stem or without one. Most LCD calculators use the former, but some VFD models use the latter.
## Mathematics
Nine is the fourth composite number, and the first composite number that is odd. Nine is the third square number (32), and the second non-unitary square prime of the form p2, and, the first that is odd, with all subsequent squares of this form odd as well. Nine has the even aliquot sum of 4, and with a composite number sequence of two (9, 4, 3, 1, 0) within the 3-aliquot tree. It is the first member of the first cluster of two semiprimes (9, 10), preceding (14, 15).[2] Casting out nines is a quick way of testing the calculations of sums, differences, products, and quotients of integers in decimal, a method known as long ago as the 12th century.[3]
By Mihăilescu's theorem, 9 is the only positive perfect power that is one more than another positive perfect power, since the square of 3 is one more than the cube of 2.[4][5]
9 is the sum of the cubes of the first two non-zero positive integers ${\displaystyle 1^{3}+2^{3}}$ which makes it the first cube-sum number greater than one.[6]
It is also the sum of the first three nonzero factorials ${\displaystyle 1!+2!+3!}$, and equal to the third exponential factorial, since ${\displaystyle 9=3^{2^{1}}.}$[7]
Nine is the number of derangements of 4, or the number of permutations of four elements with no fixed points.[8]
9 is the fourth refactorable number, as it has exactly three positive divisors, and 3 is one of them.[9]
A number that is 4 or 5 modulo 9 cannot be represented as the sum of three cubes.[10]
If an odd perfect number exists, it will have at least nine distinct prime factors.[11]
9 is a Motzkin number, for the number of ways of drawing non-intersecting chords between four points on a circle.[12]
The first non-trivial magic square is a ${\displaystyle 3}$ x ${\displaystyle 3}$ magic square made of nine cells, with a magic constant of 15.[13] Meanwhile, a ${\displaystyle 9}$ x ${\displaystyle 9}$ magic square has a magic constant of 369.[14]
There are nine Heegner numbers, or square-free positive integers ${\displaystyle n}$ that yield an imaginary quadratic field ${\displaystyle \mathbb {Q} \left[{\sqrt {-n}}\right]}$ whose ring of integers has a unique factorization, or class number of 1.[15]
### Geometry
#### Polygons and tilings
The regular hexagon contains a total of nine diagonals, and is one of only four polytopes with radial equilateral symmetry such that its long radius (center to vertex length) is the same as the edge-length: (the hexagon), the cuboctahedron, the tesseract, and the 24-cell.
A polygon with nine sides is called a nonagon.[16] Since 9 can be written in the form ${\displaystyle 2^{m}3^{n}p}$, for any nonnegative natural integers ${\displaystyle m}$ and ${\displaystyle n}$ with ${\displaystyle p}$ a product of Pierpont primes, a regular nonagon is constructed with a regular compass, straightedge, and angle trisector.[17] Also an enneagon, a regular nonagon is able to fill a plane-vertex alongside an equilateral triangle and a regular 18-sided octadecagon (3.9.18), and as such, it is one of only nine polygons that are able to fill a plane-vertex without uniformly tiling the plane.[18] In total, there are a maximum of nine semiregular Archimedean tilings by convex regular polygons, when including chiral forms of the snub hexagonal tiling. More specifically, there are nine distinct uniform colorings to both the triangular tiling and the square tiling (the simplest regular tilings) while the hexagonal tiling, on the other hand, has three distinct uniform colorings.
The fewest number of squares needed for a perfect tiling of a rectangle is nine.[19]
#### Polyhedra
There are nine uniform edge-transitive convex polyhedra in three dimensions:
Nine distinct stellation's by Miller's rules are produced by the truncated tetrahedron.[20] It is the simplest Archimedean solid, with a total of four equilateral triangular and four hexagonal faces.
Collectively, there are nine regular polyhedra in the third dimension, when extending the convex Platonic solids to include the concave regular star polyhedra known as the Kepler-Poinsot polyhedra.[21][22]
#### Higher dimensions
In four-dimensional space, there are nine paracompact hyperbolic honeycomb Coxeter groups, as well as nine regular compact hyperbolic honeycombs from regular convex and star polychora.[23] There are also nine uniform demitesseractic (${\displaystyle \mathrm {D} _{4}}$) Euclidean honeycombs in the fourth dimension.
There are only three types of Coxeter groups of uniform figures in dimensions nine and thereafter, aside from the many families of prisms and proprisms: the ${\displaystyle \mathrm {A} _{n}}$ simplex groups, the ${\displaystyle \mathrm {B} _{n}}$ hypercube groups, and the ${\displaystyle \mathrm {D} _{n}}$ demihypercube groups. The ninth dimension is also the final dimension that contains Coxeter-Dynkin diagrams as uniform solutions in hyperbolic space. Inclusive of compact hyperbolic solutions, there are a total of 238 compact and paracompact Coxeter-Dynkin diagrams between dimensions two and nine, or equivalently between ranks three and ten. The most important of the last ${\displaystyle {\tilde {E}}_{9}}$ paracompact groups is the group ${\displaystyle {\tilde {T}}_{9}}$ with 1023 total honeycombs, the simplest of which is 621 whose vertex figure is the 521 honeycomb: the vertex arrangement of the densest-possible packing of spheres in 8 dimensions which forms the ${\displaystyle \mathbb {E} _{8}}$ lattice. The 621 honeycomb is made of 9-simplexes and 9-orthoplexes, with 1023 total polytope elements making up each 9-simplex. It is the final honeycomb figure with infinite facets and vertex figures in the k21 family of semiregular polytopes, first defined by Thorold Gosset in 1900.
### List of basic calculations
Multiplication 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 20 25 50 100 1000
9 × x 9 18 27 36 45 54 63 72 81 90 99 108 117 126 135 144 180 225 450 900 9000
Division 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
9 ÷ x 9 4.5 3 2.25 1.8 1.5 1.285714 1.125 1 0.9 0.81 0.75 0.692307 0.6428571 0.6
x ÷ 9 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 1 1.1 1.2 1.3 1.4 1.5 1.6
Exponentiation 1 2 3 4 5 6 7 8 9 10
9x 9 81 729 6561 59049 531441 4782969 43046721 387420489 3486784401
x9 1 512 19683 262144 1953125 10077696 40353607 134217728 387420489 1000000000
Radix 1 5 10 15 20 25 30 40 50 60 70 80 90 100
110 120 130 140 150 200 250 500 1000 10000 100000 1000000
x9 1 5 119 169 229 279 339 449 559 669 779 889 1109 1219
1329 1439 1549 1659 1769 2429 3079 6159 13319 146419 1621519 17836619
### In base 10
9 is the highest single-digit number in the decimal system.
#### Occurrence
9 is related to just 3, 6 and itself. When we do the doubling of all natural numbers we get, 1, 2, 4, 8, 16, 32, 64... Now if we try to get the digital roots of two digit numbers, then, 16= 1+6 =7. 32= 3+2 =5. Such way, only 1, 2, 4, 5, 7, 8 occur. Same thing happens when we start halving the numbers.
#### Nikola tesla
Nikola Tesla was a great scientist. Despite his work on free energy etc. he stated, " If the understand the numbers 3, 6, 9, you understand the universe."
#### Divisibility
A positive number is divisible by nine if and only if its digital root is nine:
• 9 × 2 = 18 (1 + 8 = 9)
• 9 × 3 = 27 (2 + 7 = 9)
• 9 × 9 = 81 (8 + 1 = 9)
• 9 × 121 = 1089 (1 + 0 + 8 + 9 = 18; 1 + 8 = 9)
• 9 × 234 = 2106 (2 + 1 + 0 + 6 = 9)
• 9 × 578329 = 5204961 (5 + 2 + 0 + 4 + 9 + 6 + 1 = 27; 2 + 7 = 9)
• 9 × 482729235601 = 4344563120409 (4 + 3 + 4 + 4 + 5 + 6 + 3 + 1 + 2 + 0 + 4 + 0 + 9 = 45; 4 + 5 = 9)
That is, if any natural number is multiplied by 9, and the digits of the answer are repeatedly added until it is just one digit, the sum will be nine.[24]
In base-${\displaystyle N}$, the divisors of ${\displaystyle N-1}$ have this property.
#### Multiples of 9
There are other interesting patterns involving multiples of nine:
• 9 × 12345679 = 111111111
• 18 × 12345679 = 222222222
• 81 × 12345679 = 999999999
The difference between a base-10 positive integer and the sum of its digits is a whole multiple of nine. Examples:
• The sum of the digits of 41 is 5, and 41 − 5 = 36. The digital root of 36 is 3 + 6 = 9.
• The sum of the digits of 35967930 is 3 + 5 + 9 + 6 + 7 + 9 + 3 + 0 = 42, and 35967930 − 42 = 35967888. The digital root of 35967888 is 3 + 5 + 9 + 6 + 7 + 8 + 8 + 8 = 54, 5 + 4 = 9.
If dividing a number by the amount of 9s corresponding to its number of digits, the number is turned into a repeating decimal. (e.g. 274/999 = 0.274274274274...)
Another consequence of 9 being 10 − 1 is that it is a Kaprekar number, preceding the ninth and tenth triangle numbers, 45 and 55 (where all 9, 99, 999, 9999, ... are Keprekar numbers).[25]
Six recurring nines appear in the decimal places 762 through 767 of π. (See six nines in pi).
## Culture and mythology
### Indian culture
Nine is a number that appears often in Indian culture and mythology.[26] Some instances are enumerated below.
### Chinese culture
• Nine (; pinyin: jiǔ) is considered a good number in Chinese culture because it sounds the same as the word "long-lasting" (; pinyin: jiǔ).[30]
• Nine is strongly associated with the Chinese dragon, a symbol of magic and power. There are nine forms of the dragon, it is described in terms of nine attributes, and it has nine children. It has 117 scales – 81 yang (masculine, heavenly) and 36 yin (feminine, earthly). All three numbers are multiples of 9 (9 × 13 = 117, 9 × 9 = 81, 9 × 4 = 36)[31] as well as having the same digital root of 9.
• The dragon often symbolizes the Emperor, and the number nine can be found in many ornaments in the Forbidden City.
• The circular altar platform (Earthly Mount) of the Temple of Heaven has one circular marble plate in the center, surrounded by a ring of nine plates, then by a ring of 18 plates, and so on, for a total of nine rings, with the outermost having 81 = 9 × 9 plates.
• The name of the area called Kowloon in Hong Kong literally means: nine dragons.
• The nine-dotted line (Chinese: 南海九段线; pinyin: nánhǎi jiǔduàn xiàn; lit. 'Nine-segment line of the South China Sea') delimits certain island claims by China in the South China Sea.
• The nine-rank system was a civil service nomination system used during certain Chinese dynasties.
• 9 Points of the Heart (Heal) / Heart Master (Immortality) Channels in Traditional Chinese Medicine.
### Ancient Egypt
• The nine bows is a term used in Ancient Egypt to represent the traditional enemies of Egypt.[32]
• The Ennead is a group of nine Egyptian deities, who, in some versions of the Osiris myth, judged whether Horus or Set should inherit Egypt.
### Mesoamerican mythology
• The Lords of the Night, is a group of nine deities who each ruled over every ninth night forming a calendrical cycle.
### Aztec mythology
• Mictlan the underworld in Aztec mythology, consists of nine levels.
### Mayan mythology
• The Mayan underworld Xibalba consists of nine levels.
• El Castillo, the Mayan step-pyramid in Chichén Itzá, consists of nine steps. It is said that this was done to represent the nine levels of Xibalba.
### Australian culture
The Pintupi Nine, a group of 9 Aboriginal Australian women who remained unaware of European colonisation of Australia and lived a traditional desert-dwelling life in Australia's Gibson Desert until 1984.
## Anthropology
### Idioms
• "to go the whole nine yards-"
• "A cat-o'-nine-tails suggests perfect punishment and atonement." – Robert Ripley.
• "A cat has nine lives"
• "to be on cloud nine"
• " A stitch in time saves nine"
• "found true 9 out of 10 times"
• "possession is nine tenths of the law"
• The word "K-9" pronounces the same as canine and is used in many US police departments to denote the police dog unit. Despite not sounding like the translation of the word canine in other languages, many police and military units around the world use the same designation.
• Someone dressed "to the nines" is dressed up as much as they can be.
• In North American urban culture, "nine" is a slang word for a 9mm pistol or homicide, the latter from the Illinois Criminal Code for homicide.
## Literature
• There are nine circles of Hell in Dante's Divine Comedy.
• The Nine Bright Shiners, characters in Garth Nix's Old Kingdom trilogy. The Nine Bright Shiners was a 1930s book of poems by Anne Ridler[34] and a 1988 fiction book by Anthea Fraser;[35] the name derives from "a very curious old semi-pagan, semi-Christian" song.[36]
• The Nine Tailors is a 1934 mystery novel by British writer Dorothy L. Sayers, her ninth featuring sleuth Lord Peter Wimsey.
• Nine Unknown Men are, in occult legend, the custodians of the sciences of the world since ancient times.
• In J. R. R. Tolkien's Middle-earth, there are nine rings of power given to men, and consequently, nine ringwraiths. Additionally, The Fellowship of the Ring[broken anchor] consists of nine companions.
• In Lorien Legacies there are nine Garde sent to Earth.
• Number Nine is a character in Lorien Legacies.
• In the series A Song of Ice and Fire, there are nine regions of Westeros (the Crownlands, the North, the Riverlands, the Westerlands, the Reach, the Stormlands, the Vale of Arryn, the Iron Islands and Dorne). Additionally, there is a group of nine city-states in western Essos known collectively as the Free Cities (Braavos, Lorath, Lys, Myr, Norvos, Pentos, Qohor, Tyrosh and Volantis).
• In The Wheel of Time series, Daughter of the Nine Moons is the title given to the heir to the throne of Seanchan, and the Court of the Nine Moons serves as the throne room of the Seanchan rulers themselves. Additionally, the nation of Illian is partially governed by a body known as the Council of Nine, and the flag of Illian displays nine golden bees on it. Furthermore, in the Age of Legends, the Nine Rods of Dominion were nine regional governors who administered individual areas of the world under the ruling world government.
## Organizations
• Divine Nine – The National Pan-Hellenic Council (NPHC) is a collaborative organization of nine historically African American, international Greek-lettered fraternities and sororities.
## Religion and philosophy
### Islam
There are three verses that refer to nine in the Quran.
We surely gave Moses nine clear signs.1 ˹You, O Prophet, can˺ ask the Children of Israel. When Moses came to them, Pharaoh said to him, “I really think that you, O Moses, are bewitched.”
— Surah Al-Isra (The Night Journey/Banī Isrāʾīl):101[39]
Note 1: The nine signs of Moses are: the staff, the hand (both mentioned in Surah Ta-Ha 20:17-22), famine, shortage of crops, floods, locusts, lice, frogs, and blood (all mentioned in Surah Al-A'raf 7:130-133). These signs came as proofs for Pharaoh and the Egyptians. Otherwise, Moses had some other signs such as water gushing out of the rock after he hit it with his staff, and splitting the sea.
Now put your hand through ˹the opening of˺ your collar, it will come out ˹shining˺ white, unblemished.2 ˹These are two˺ of nine signs for Pharaoh and his people. They have truly been a rebellious people.”
— Surah Al-Naml (The Ant):12[40]
Note 2: Moses, who was dark-skinned, was asked to put his hand under his armpit. When he took it out it was shining white, but not out of a skin condition like melanoma.
And there were in the city nine ˹elite˺ men who spread corruption in the land, never doing what is right.
— Surah Al-Naml (The Ant):48[41]
### Other
• Nine, as the highest single-digit number (in base ten), symbolizes completeness in the Baháʼí Faith. In addition, the word Baháʼ in the Abjad notation has a value of 9, and a 9-pointed star is used to symbolize the religion.
• The number 9 is revered in Hinduism and considered a complete, perfected and divine number because it represents the end of a cycle in the decimal system, which originated from the Indian subcontinent as early as 3000 BC.
• In Buddhism, Gautama Buddha was believed to have nine virtues, which he was (1) Accomplished, (2) Perfectly Enlightened, (3) Endowed with knowledge and Conduct or Practice, (4) Well-gone or Well-spoken, (5) the Knower of worlds, (6) the Guide Unsurpassed of men to be tamed, (7) the Teacher of gods and men, (8) Enlightened, and (9) Blessed.
• Important Buddhist rituals usually involve nine monks.
• The first nine days of the Hebrew month of Av are collectively known as "The Nine Days" (Tisha HaYamim), and are a period of semi-mourning leading up to Tisha B'Av, the ninth day of Av on which both Temples in Jerusalem were destroyed.
• Nine is a significant number in Norse Mythology. Odin hung himself on an ash tree for nine days to learn the runes.
• The Fourth Way Enneagram is one system of knowledge which shows the correspondence between the 9 integers and the circle.
• Tian's Trigram Number, of Feng Shui, in Taoism.
## Science
### Physiology
A human pregnancy normally lasts nine months, the basis of Naegele's rule.
### Psychology
Common terminal digit in psychological pricing.
## References
1. ^ Lippman, David (12 July 2021). "6.0.2: The Hindu-Arabic Number System". Mathematics LibreTexts. Retrieved 31 March 2024.
2. ^ Sloane, N. J. A. (ed.). "Sequence A001358 (Semiprimes (or biprimes): products of two primes.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 27 February 2024.
3. ^ Cajori, Florian (1991, 5e) A History of Mathematics, AMS. ISBN 0-8218-2102-4. p.91
4. ^ Mihăilescu, Preda (2004). "Primary Cyclotomic Units and a Proof of Catalan's Conjecture". J. Reine Angew. Math. 572. Berlin: De Gruyter: 167–195. doi:10.1515/crll.2004.048. MR 2076124. S2CID 121389998.
5. ^ Metsänkylä, Tauno (2004). "Catalan's conjecture: another old Diophantine problem solved" (PDF). Bulletin of the American Mathematical Society. 41 (1). Providence, R.I.: American Mathematical Society: 43–57. doi:10.1090/S0273-0979-03-00993-5. MR 2015449. S2CID 17998831. Zbl 1081.11021.
6. ^ Sloane, N. J. A. (ed.). "Sequence A000537 (Sum of first n cubes; or n-th triangular number squared.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 19 June 2023.
7. ^ "Sloane's A049384 : a(0)=1, a(n+1) = (n+1)^a(n)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 1 June 2016.
8. ^ Sloane, N. J. A. (ed.). "Sequence A000166 (Subfactorial or rencontres numbers, or derangements: number of permutations of n elements with no fixed points.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 10 December 2022.
9. ^ Sloane, N. J. A. (ed.). "Sequence A033950 (Refactorable numbers: number of divisors of k divides k. Also known as tau numbers.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 19 June 2023.
10. ^ Davenport, H. (1939), "On Waring's problem for cubes", Acta Mathematica, 71, Somerville, MA: International Press of Boston: 123–143, doi:10.1007/BF02547752, MR 0000026, S2CID 120792546, Zbl 0021.10601
11. ^ Pace P., Nielsen (2007). "Odd perfect numbers have at least nine distinct prime factors". Mathematics of Computation. 76 (260). Providence, R.I.: American Mathematical Society: 2109–2126. arXiv:math/0602485. Bibcode:2007MaCom..76.2109N. doi:10.1090/S0025-5718-07-01990-4. MR 2336286. S2CID 2767519. Zbl 1142.11086.
12. ^ "Sloane's A001006 : Motzkin numbers". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 1 June 2016.
13. ^ William H. Richardson. "Magic Squares of Order 3". Wichita State University Dept. of Mathematics. Retrieved 6 November 2022.
14. ^ Sloane, N. J. A. (ed.). "Sequence A006003 (Also the sequence M(n) of magic constants for n X n magic squares)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 8 December 2022.
15. ^ Bryan Bunch, The Kingdom of Infinite Number. New York: W. H. Freeman & Company (2000): 93
16. ^ Robert Dixon, Mathographics. New York: Courier Dover Publications: 24
17. ^
18. ^ Grünbaum, Branko; Shepard, Geoffrey (November 1977). "Tilings by Regular Polygons" (PDF). Mathematics Magazine. 50 (5). Taylor & Francis, Ltd.: 228–234. doi:10.2307/2689529. JSTOR 2689529. S2CID 123776612. Zbl 0385.51006.
19. ^
20. ^ Webb, Robert. "Enumeration of Stellations". www.software3d.com. Archived from the original on 26 November 2022. Retrieved 15 December 2022.
21. ^ Weisstein, Eric W. "Regular Polyhedron". Mathworld -- A WolframAlpha Resource. Retrieved 27 February 2024.
22. ^ Coxeter, H. S. M. (1948). Regular Polytopes (1st ed.). London: Methuen & Co., Ltd. p. 93. ISBN 0-486-61480-8. MR 0027148. OCLC 798003.
23. ^ Coxeter, H. S. M. (1956), "Regular honeycombs in hyperbolic space", Proceedings of the International Congress of Mathematicians, vol. III, Amsterdam: North-Holland Publishing Co., pp. 167–169, MR 0087114
24. ^ Martin Gardner, A Gardner's Workout: Training the Mind and Entertaining the Spirit. New York: A. K. Peters (2001): 155
25. ^ Sloane, N. J. A. (ed.). "Sequence A006886 (Kaprekar numbers.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 27 February 2024.
26. ^ DHAMIJA, ANSHUL (16 May 2018). "The Auspiciousness Of Number 9". Forbes India. Retrieved 1 April 2024.
27. ^ "Vaisheshika | Atomism, Realism, Dualism | Britannica". www.britannica.com. Retrieved 13 April 2024.
28. ^ "Navratri | Description, Importance, Goddess, & Facts | Britannica". www.britannica.com. 11 April 2024. Retrieved 13 April 2024.
29. ^ Lochtefeld, James G. (2002). The illustrated encyclopedia of hinduism. New York: the Rosen publ. group. ISBN 978-0-8239-2287-1.
30. ^ "Lucky Number Nine, Meaning of Number 9 in Chinese Culture". www.travelchinaguide.com. Retrieved 15 January 2021.
31. ^ Donald Alexander Mackenzie (2005). Myths of China And Japan. Kessinger. ISBN 1-4179-6429-4.
32. ^ "The Global Egyptian Museum | Nine Bows". www.globalegyptianmuseum.org. Retrieved 16 November 2023.
33. ^ Mark, Joshua J. "Nine Realms of Norse Cosmology". World History Encyclopedia. Retrieved 16 November 2023.
34. ^ Jane Dowson (1996). Women's Poetry of the 1930s: A Critical Anthology. Routledge. ISBN 0-415-13095-6.
35. ^ Anthea Fraser (1988). The Nine Bright Shiners. Doubleday. ISBN 0-385-24323-5.
36. ^ Charles Herbert Malden (1905). Recollections of an Eton Colleger, 1898–1902. Spottiswoode. p. 182. nine-bright-shiners.
37. ^ Galatians 5:22–23
38. ^ "Meaning of Numbers in the Bible The Number 9". Bible Study. Archived from the original on 17 November 2007.
39. ^ "Surah Al-Isra - 101". Quran.com. Retrieved 17 August 2023.
40. ^ "Surah An-Naml - 12". Quran.com. Retrieved 17 August 2023.
41. ^ "Surah An-Naml - 48". Quran.com. Retrieved 17 August 2023.
42. ^ "Web site for NINE: A Journal of Baseball History & Culture". Archived from the original on 4 November 2009. Retrieved 20 February 2013.
43. ^ Glover, Diane (9 October 2019). "#9 Dream: John Lennon and numerology". www.beatlesstory.com. Beatles Story. Retrieved 6 November 2022. Perhaps the most significant use of the number 9 in John's music was the White Album's 'Revolution 9', an experimental sound collage influenced by the avant-garde style of Yoko Ono and composers such as Edgard Varèse and Karlheinz Stockhausen. It featured a series of tape loops including one with a recurring 'Number Nine' announcement. John said of 'Revolution 9': 'It's an unconscious picture of what I actually think will happen when it happens; just like a drawing of a revolution. One thing was an engineer's testing voice saying, 'This is EMI test series number nine.' I just cut up whatever he said and I'd number nine it. Nine turned out to be my birthday and my lucky number and everything. I didn't realise it: it was just so funny the voice saying, 'number nine'; it was like a joke, bringing number nine into it all the time, that's all it was.'
44. ^ Truax, Barry (2001). Handbook for Acoustic Ecology (Interval). Burnaby: Simon Fraser University. ISBN 1-56750-537-6..
45. ^ "The Curse of the Ninth Haunted These Composers | WQXR Editorial". WQXR. 17 October 2016. Retrieved 16 January 2022. | 7,637 | 26,242 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 22, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-30 | latest | en | 0.904504 |
https://discuss.leetcode.com/topic/103783/my-java-recursive-solution | 1,516,391,184,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888113.39/warc/CC-MAIN-20180119184632-20180119204632-00422.warc.gz | 666,708,121 | 9,689 | # My Java Recursive Solution
• The method I used is pretty like the Binary Search algorithm.
Here is my code:
``````class Solution {
private TreeNode createNode(int[] nums, int startIndex, int endIndex) {
if (startIndex > endIndex) {
return null;
}
return new TreeNode(nums[(startIndex + endIndex) / 2]);
}
private void addNode(TreeNode root, int[] nums, int startIndex, int endIndex) {
if (root != null) {
int median = (startIndex + endIndex) / 2;
root.left = createNode(nums, startIndex, median - 1);
root.right = createNode(nums, median + 1, endIndex);
addNode(root.left, nums, startIndex, median - 1);
addNode(root.right, nums, median + 1, endIndex);
}
}
public TreeNode sortedArrayToBST(int[] nums) {
int startIndex = 0;
int endIndex = nums.length - 1;
TreeNode root = createNode(nums, startIndex, endIndex); | 204 | 823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-05 | latest | en | 0.3349 |
http://www.sl4.org/archive/0804/18339.html | 1,563,561,872,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526337.45/warc/CC-MAIN-20190719182214-20190719204214-00263.warc.gz | 269,021,965 | 2,550 | # Re: Mathematical Model of GLUTs and Lookups
Date: Wed Apr 02 2008 - 05:45:36 MDT
Hum, my understanding of a GLUT was that you fed it the state of a
system, and it would give you the next one. Thus a GLUT is a function
on the states of a system; you feed it one state and it gives you
another.
So S is my set of states (taken to be polynomials, for this example),
and G is a GLUT on S - a function from S to itself.
(note: I don't actually want G to be full GLUT, but actually a simpler
rule, like differentiation; but we'll have to start with G being a
GLUT).
G is therefore a collection of ordered pairs of elements of S (the
pairs are (initial state , resulting state). The KC of such a
collection is well defined.
Now R is a rule, i.e. a restiction that the function G must obey. It
is given by an ordered collection of sets (P(n) = the "n-th degree
polynomials"), and the rule that G must map P(n) to P(n-1)).
Now the maximal KC that G can have, given this rule, is huge.
Formulating the rule R itself is of trivial KC; the process of taking
a polynomial and sending it to its degree is also of trivial KC. In
fact I have just, completely defined it in a few sentences, no matter
what the cardinality of S is.
Now we have a hash function f, mapping bijectively to the set f(S). By
the composition f G f^{-1} , we have a function (a GLUT) mapping f(S)
to itself. I called it f(G).
We also have the "rule" f(R). This is given by the requirement that
f(G) must map f(P(n)) to f(P(n-1)). That statement is of trivial KC.
But given an element p of f(S), knowing whether it is in f(P(n)) is of
very high KC: to do so, we need to calculate f^{-1}(p), and the KC of
f is maximal.
Thus the rule f(R) is of high KC, as it requires a lot of information
to describe.
With me so far?
Stuart
PS: will continue in a later message
This archive was generated by hypermail 2.1.5 : Wed Jul 17 2013 - 04:01:02 MDT | 531 | 1,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-30 | latest | en | 0.958115 |
http://betterlesson.com/lesson/reflection/18128/planning-for-differentiated-instruction | 1,477,657,058,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988722459.85/warc/CC-MAIN-20161020183842-00533-ip-10-171-6-4.ec2.internal.warc.gz | 28,624,431 | 21,544 | ## Reflection: Intervention and Extension Law of Cosines Day 2 of 2 - Section 2: Solving triangles when 3 sides are known.
When I differentiate instruction, I like to work with one or two students. I work with small groups so that I can ask students questions and learn more about when and how they are misunderstanding problems that are difficult for them. As discussed in prior lessons, in lessons like today's, I often model reading skills, to help students identify and use information in a contextual problem. I did a lot of 1-on-1 reading work with today's lesson. I also noticed that students were helping each other with reading and interpreting the problems.
Planning for Differentiated Instruction
Intervention and Extension: Planning for Differentiated Instruction
# Law of Cosines Day 2 of 2
Unit 6: Solving Problems Involving Triangles
Lesson 9 of 13
## Big Idea: Which method is the best one to use to solve the triangle?
Print Lesson
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Standards:
Subject(s):
Math, Law of Cosines, Triangles, triangles (Determining Measurements), Oblique Triangles, problem solving, pre
45 minutes
### Katharine Sparks
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https://www.scribd.com/doc/315988043/Test-on-a-Centrifugal-Pump-Complete-Report | 1,568,684,124,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572980.56/warc/CC-MAIN-20190917000820-20190917022820-00528.warc.gz | 1,018,103,266 | 62,265 | You are on page 1of 12
# TEST ON A CENTRIFUGAL PUMP
OBJECTIVES
## To measure and plot the Head-Discharge and Efficiency-Discharge
Characteristics for a centrifugal pump when operated at a constant speed
To determine the specific speed of the pump Ns
Introduction
A centrifugal pump is a rotodynamic pump that uses a rotating impeller to increase the velocity
of a fluid. Centrifugal pumps are commonly used to move liquids through a piping system. The
fluid enters the pump impeller along or near to the rotating axis and is accelerated by the
impeller, flowing radially outward into a diffuser or volute chamber, from where it exits into the
downstream piping system. Centrifugal pumps are used for large discharge through smaller
Theory
A pump is a device for supplying energy to a fluid. The effect of supplying energy can be studied via the
mechanical energy equation.
This equation neglects all shearing stresses. Note: all nomenclature is defined at the end of the section. In general,
the power originates from a pressure change (P 2 > P 1), a change in kinetic energy (v 2 2 > v 1 2 ), a change in
level (z 2 > z 1), or as friction ( F ). Since friction is always present in mechanical devices, the energy supplied goes
partly into friction and the rest to one of the other effects cited. For the pump used in the present experiment, the
- E = 0 no internal energy change
- v 2 = 0 no kinetic energy change
- Q h = 0 no heat generated or lost
- z = 0 no significant height change
The energy balance simplifies to
Thus, the actual shaft work done (W s) is the total work (W T ) minus friction losses
## Net Positive Suction Head (NPSH)
NPSH = difference between the static head at the suction inlet and the head at the inlet at the vapor pressure.
Apparatus
## The TecQuipment centrifugal pump test rig
Tachometer (for measuring and setting r.p.m.)
Discussion
Even though I personally prepare this lab report, I would like to elaborate it for better understanding
of both party. I gather this from the lab coordinator, fluid mechanics lecturer and from other suitable
source.
The maximum volume flow rate through a pump occurs when its net heat is zero H=0, this flow rate
is called the pump free delivery achieved when there is no flow restriction at the pump inlet or
outlet. In other word there is no load on the pump. At this operating point. volume flow rate is large
but H is zero, the pump efficiency is zero because there no useful work. At the other extreme, the
shutoff head is the net head that occurs when the volume flow rate is zero achieved when outlet port
of the pump is block off. Under these conditions, H is large but volume flow rate is zero. pump
efficiency is also zero. Between the two the pumps net head may increase from its shutoff value
somewhat as the flow rate increases, but H must eventually decreased to zero as volume flow rate
increases to its delivery value. The pumps efficiency reaches its maximum value somewhere
between he shutoff condition and the free delivery condition. This is called best efficiency point
(BEP).
EXPERIMENTAL EQUIPMENT
The main elements of the system are a centrifugal pump, a rotameter to measure flow, a valve to control flow, a
differential pressure (DP) cell to measure pressure, and a reservoir, as shown in the figure below. The pump is
driven by an electric motor, and the shaft connecting the motor to the pump is provided with a torque indicator.
The motor is provided with a speed controller and tachometer.
In operating the equipment, it is necessary that the intake from the reservoir should be at least 6 inches above the
intake. The liquid should also be clean and free of sediment. At start-up and at other times (as indicated by the
next paragraph), air may be in the lines. This will be observed as air bubbles flowing through the rotameter. Do
not attempt to make any measurements as long as air bubbles are observed. If the problem does not clear up
after a few minutes of operation, the system has a leak and is in need of repair. In normal operation, the speed is
to be controlled using the speed controller and tachometer, flow is to be controlled using valve V-3, pressures are
to be determined from the DP cell readings, torque is to be measured from the torque meter employing a strobe
light, and flow is to be determined from the rotameter reading.
Wattmeter
The wattmeter allows measurement of electrical power input to the motor. Determine the appropriate k factor
based on the wiring of the wattmeter. The k factors can be located in the lid of the wattmeter. Do not make any
changes in the existing connections.
Rotameter
The rotameter reads flow rate directly in gallons per minute for a fluid having a specific gravity of 1.0. If water is
used, flow can be read directly from the rotameter. The rotameter reading is obtained from the position of the
widest part of the float.
Speed Control
The speed control regulates the motor rpms. To start the motor, set the control to zero, and flip the on-off switch
to on. Flip the brake switch to forward, and adjust the control until the desired speed is indicated on the
Flow Control
The valve V-3 controls the flow. With the pump in operation, adjust the valve to produce the flow rate desired, as
indicated by the rotameter float.
Torque Meter
The torque meter is provided with a vernier scale and indicates torque in lbf -in. A stroboscopic light is necessary
for reading the meter and this stroboscope should also be used to check the indicated motor speed in rpms.
Procedure
## The electrical mains supply was switched on to the apparatus.
The suction and delivery valves were checked whether fully opened
Tachometer was checked, and torque meter to see if they are zeroed properly. If not, correction or
service are needed
Reservoir was filled to at least 6 inches above the intake to the pump. Submergence of the return
line is desirable.
Power switch was pressed
Speed control was adjusted to produce the desired pump speed (2000rpm/3000rpm) Adjust the
discharge valve to obtain the desired flow rate, obtain 8 data points covering the maximum range
attainable
Observation
Single Centrifugal pump
Pump speed 70rps
## Pump in series combination
Conclusion
The performance of a centrifugal pump can be shown graphically on a characteristic curve. A typical
characteristic curve shows the total dynamic head, brake horsepower, efficiency, and net positive
Suction head all plotted over the capacity range of the pump.
Non-dimensional curves which indicate the general shape of the characteristic curves for the various
types of pumps. They show the head, brake horsepower, and efficiency plotted as a percent of their
values at the design or best efficiency point of the pump.
flow increases. Note that the brake horsepower increases gradually over the flow range with the
maximum normally at the point of maximum flow
The performance or characteristic curve of the pump provides information on the relationship
between total head and flow rate. There are three important points on this curve. | 1,581 | 7,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-39 | latest | en | 0.881266 |
https://www.edumedia-sciences.com/en/media/850-fractions | 1,620,676,519,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991759.1/warc/CC-MAIN-20210510174005-20210510204005-00347.warc.gz | 762,810,737 | 9,307 | # FractionsHTML5
## Summary
A fraction is a number representing a set of identical parts of a whole. Example: The fraction 3/5 represents 3 parts of a whole divided into 5 equal parts.
A fraction is a division between the number of parties considered (the numerator) and the total number of parties sharing the whole (the denominator). In the fraction 3/5, the numerator is 3 and the denominator is 5.
The whole is often represented by:
• a circle divided into identical angular sections,
• a bar divided into identical rectangles,
• a collection of items (here the tomatoes) grouped by packages.
## Learning goals
• To match fractions in different picture patterns.
• To understand the notion of equivalent fractions.
• To know how to order fractions: graphic method or calculation of the LCM.
• To compare fractions using numbers and patterns.
• To develop strategies for representing and ordering fractions.
A fraction is a rational number.
If the numerator is smaller than the denominator, this number is less than 1.
If the numerator is greater than the denominator, all the parts considered… | 229 | 1,107 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-21 | latest | en | 0.877151 |
https://ch.mathworks.com/matlabcentral/cody/problems/74-balanced-number/solutions/3344393 | 1,610,970,524,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514495.52/warc/CC-MAIN-20210118092350-20210118122350-00758.warc.gz | 277,158,826 | 17,267 | Cody
Problem 74. Balanced number
Solution 3344393
Submitted on 23 Oct 2020 by Louis Repussard
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
n = 13722; assert(isequal(isBalanced(n),true))
s = -1 s = 0 tf = logical 1
2 Pass
n = 23567414; assert(isequal(isBalanced(n),true))
s = -2 s = 0 s = 1 s = 0 tf = logical 1
3 Pass
n = 20567410; assert(isequal(isBalanced(n),false))
s = 2 s = 1 s = 2 s = 1 tf = logical 0
4 Pass
n = 1; assert(isequal(isBalanced(n),true))
tf = logical 1
5 Pass
n = 11111111; assert(isequal(isBalanced(n),true))
s = 0 s = 0 s = 0 s = 0 tf = logical 1
6 Pass
n = 12345678; assert(isequal(isBalanced(n),false))
s = -7 s = -12 s = -15 s = -16 tf = logical 0
7 Pass
n = 12333; assert(isequal(isBalanced(n),false))
s = -2 s = -3 tf = logical 0
8 Pass
n = 9898; assert(isequal(isBalanced(n),true))
s = 1 s = 0 tf = logical 1
9 Pass
n = 469200; assert(isequal(isBalanced(n),false))
s = 4 s = 10 s = 17 tf = logical 0
10 Pass
n = 57666; assert(isequal(isBalanced(n),true))
s = -1 s = 0 tf = logical 1
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Start Hunting! | 467 | 1,293 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-04 | latest | en | 0.655881 |
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# Convert Watts to Volts
Power: milliwattswattskilowatts Current: microampsmilliampsampskiloamps Result in Volts: V
## Watts to Volts Electrical Conversion Calculator
Where you as a kid in the same dilemma that electronics and electrical are almost the same? I understand as I have been there. I have to agree I never really thought of digging in until now. To my surprise, they are quite different and the differences are easy to understand. Electrical will tell you about electron flow whereas electronics talks about the control of that electron flow. You must be wondering why all of a sudden we need comparisons for the two. Well, the title says it all.
Watts and Volts are related to entities that are at times common to electronics as well as electrical.
Here, you will be learning in detail about Watt and Volts in addition to their electrical conversion. The need for conversion has always been an integral part of both the fields. Make sure you leverage the information in the content below.
• What is Potential?
Are you familiar with homonyms? Well, if you are it will come handy now.
Has anyone ever told you – You are a human with the utmost potential? What he/she meant with potential was the ability or capacity to develop or do something- Human Potential. It is easy to understand in simple English language.
However, what we are dealing with here is – Electric Potential.
The word ‘electric’ itself changes it all. With that, the other two terms- Potential difference and Electromotive force is important as well study.
• Electric Potential
The term can be best explained that to move a unit charge from a reference point to a particular point against the influence of the electric field, some amount of work is needed to be done. This amount of work is referred to as the Electric Potential.
The Earth or any point beyond the electric field influence can be a reference point.
• Potential Difference & Electromotive Force
We can express difference mathematically as subtraction between two entities. Well, yes the term is as simple as it appears. Potential difference can be understood as the difference between the energy of charge carriers possessed by two different points in a circuit. We also describe the potential difference as the work being done per unit charge.
The other term, the electromotive force is similar to the potential difference. It is usually referred to as the emf. The difference between emf and the potential difference lies in one quantity- Current. The emf is the potential difference when there is no current flow between the terminals in the external circuit.
Well, the similarity is both are expressed in terms of volts.
• Volts – the unit of potential difference & emf
It becomes important to be able to measure the above quantities for which a term was introduced called Volts. Both the potential difference and emf is also known as the voltage is measured in an electrical unit – volts. A volt is indicated with an uppercase V.
1 volt is equivalent to the work done in 1 joule per coulomb of charge.
The mathematical representation is,
V = W/Q
where,
V = potential difference (volts)
W = work done (Joule)
Q = electric charge (Coulomb)
• Measurement tool- Voltmeter
To facilitate the conversion of watts to volts, one needs to know the value of voltage. The question arises how will one measure it and get the value. Well, the answer is a simple measurement tool called a Voltmeter.
The voltmeters are designed to derive the value of the potential difference between two separate points. One can surely figure it out just by the name itself.
Another key point to remember is the connection. Voltmeters are to be connected mandatorily on a different branch i.e. parallel connection to the points which are being measured.
## Relation between Watts and Volts
To get onto the conversion we need a head start which is easily provided by concluding a relationship between two quantities. In this case, we will look into the relationship between Watts and Volts.
We have,
The voltage is a result of a division of the power and the current.
I= P/V
Where
‘I’ is in Ampere.
‘P’ is in Watts,
‘V’ is in Volts
• Formula
The relation above suggests a simple formula and conversion.
Rearranging this we can get,
V= P/I ———- 1
It is used for the DC circuit.
However, we have learned that there always happens to be more than one way. So is the case here.
• For AC circuit
When the conversion is being carried on an AC circuit the above formula fails, hence we have this particular formula for you.
V = P ÷ (I x PF) ——— 2
In the above formula, PF stands for Power factor which is the ratio of reactive power to the apparent power moreover like a measurement of energy efficiency. The value will always come between 0 and 1.
• Using resistance
Another easy to go method is when resistance is known to you. With that knowledge, you can use the formula below and carry out the conversion.
V = √ (P x R) ———- 3
Remember, resistance is expressed in ohms (Ω).
## How to convert Watts to Volts?
At times just by looking at the formula, we are carried away interpreting all sorts of difficulty. Well, for that we know how to make it easy for you.
Given below are general words explanations of carrying out the conversion in the right way. In case you are not a fan of calculations, the calculator will act just like a friend who won’t betray you.
1. For the very first equation, we have to carry out simple algebraic calculations. The product of power in watts and Current in amperes will give out a result V in volts.
2. As per equation 2, you are required to substitute values in the correct symbols. With that remember the BODMAS rule, just apply that meaning solve the bracket first and then let power divided by that value.
3. For equation 3, simply select the square root symbol in the calculator and insert the symbolic values. What you are doing is, multiplying the product by the current under the square root of voltage.
The above steps are surely your way to easy and hassle-free calculations.
Benefit
The need and benefit of the conversion vary from person to person and work to work. For some, it may be an ordinary conversion while others view it as useful to their knowledge. This conversion apart from knowledge tells about electrical energy consumption. Each quantity stands for itself and contributes towards data that is useful in understanding the battery life or lowering the electricity costs.
With two quantities- current and power known, you can easily derive the third i.e. Volts. In circuits, the voltage can be directly known by a voltmeter. | 1,422 | 6,662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-40 | longest | en | 0.958324 |
https://brainkart.com/article/Exercise-10-9--Choose-the-correct-answer_41259/ | 1,669,615,582,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00011.warc.gz | 178,764,514 | 9,353 | Home | | Maths 12th Std | Exercise 10.9: Choose the correct answer
# Exercise 10.9: Choose the correct answer
Choose the correct or the most suitable answer from the given four alternatives - Maths Book back answers and solution for Exercise questions - Mathematics : Ordinary Differential Equations
EXERCISE 10.9
Choose the correct or the most suitable answer from the given four alternatives :
1. The order and degree of the differential equation are respectively
(1) 2, 3
(2) 3, 3
(3) 2, 6
(4) 2, 4
2. The differential equation representing the family of curves y = A cos(x + B), where A and B are parameters, is
3. The order and degree of the differential equation √sin x dx + dy ) = √cos x dx − dy) is
(1) 1, 2
(2) 2, 2
(3) 1, 1
(4) 2, 1
4. The order of the differential equation of all circles with centre at ( h, k ) and radius ‘a’ is
(1) 2
(2) 3
(3) 4
(4) 1
5. The differential equation of the family of curves y = Aex + Bex , where A and B are arbitrary constants is
6. The general solution of the differential equation is
(1) xy = k
(2) y = k log x
(3) y = kx
(4) log y kx
7. The solution of the differential equation 2x dy/dx − y = 3 represents
(1) straight lines
(2) circles
(3) parabola
(4) ellipse
8. The solution of dy/dx + p ( x) y = 0 is
9. The integrating factor of the differential equation is
(1) x/eλ
(2) eλ/x
(3) λex
(4) ex
10. The integrating factor of the differential equation dy/ dx + P ( x) y = Q(x) is x, then P(x)
(1) x
(2) x2/2
(3) 1/x
(4) 1/x2
11. The degree of the differential equation is
(1) 2
(2) 3
(3) 1
(3) 4
12. If p and q are the order and degree of the differential equation when
(1) p < q
(2) p = q
(3) p > q
(4) p exists and q does not exist
13. The solution of the differential equation is
(1) y + sin−1 x = c
(2) x + sin−1 y = 0
(3) y2 + 2 sin−1 x = C
(4) x2 + 2 sin−1 y = 0
14. The solution of the differential equation dy/dx = 2xy is
(1) y = Cex2
(2) y = 2x2 + C
(3) y = Ce −x2 + C
(4) y = x2 + C
15. The general solution of the differential equation = x + y is
(1) ex + ey = C
(2) ex + ey = C
(3) ex + ey = C
(4) ex + ey = C
16. The solution of is
(1) 2x + 2y = C
(2) 2x − 2y = C −
(3) 1/2x – 1/2y = C
(4) x + y = C
17. The solution of the differential equation is
18. If sin x is the integrating factor of the linear differential equation dy/dx+ Py = Q , then P is
(1) log sin x
(2) cos x
(3) tan x
(4) cot x
19. The number of arbitrary constants in the general solutions of order n and n +1are respectively
(1) n −1, n
(2) n, n +1
(3) + 1, n + 2
(4) n +1, n
20. The number of arbitrary constants in the particular solution of a differential equation of third order is
(1) 3
(2) 2
(3) 1
(4) 0
21. Integrating factor of the differential equation is
(1) 1 / x +1
(2) x +1
(3) 1 / √(+1)
(4) √(+1)
22.The population P in any year t is such that the rate of increase in the population is proportional to the population. Then
(1) P = Cekt
(2) P = Ce−kt
(3) P = Ckt
(4) P = C
23. P is the amount of certain substance left in after time t. If the rate of evaporation of the substance is proportional to the amount remaining, then
(1) P = Cekt
(2) P = Ce−kt
(3) P = Ckt
(4) Pt = C
24. If the solution of the differential equation represents a circle, then the value of is
(1) 2
(2) −2
(3) 1
(4) −1
25. The slope at any point of a curve y f (x) is given by dy/dx = 3x2 and it passes through (-1,1). Then the equation of the curve is
(1) y = x3 + 2
(2) y = 3x2 + 4
(3) y = 3x3 + 4
(4) y x3 + 5
Tags : Ordinary Differential Equations | Mathematics , 12th Maths : UNIT 10 : Ordinary Differential Equations
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
12th Maths : UNIT 10 : Ordinary Differential Equations : Exercise 10.9: Choose the correct answer | Ordinary Differential Equations | Mathematics | 1,510 | 3,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2022-49 | latest | en | 0.6359 |
https://dsp.stackexchange.com/questions/47198/fourier-transform-of-smart-phone-accelerometer-in-matlab | 1,652,775,378,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00654.warc.gz | 281,386,659 | 65,777 | fourier transform of smart-phone accelerometer in matlab
I'm new to matlab. I want to process my smart-phone accelerometer data in matlab. I know Matlab let's you connect your phone via USB cable to see accelerometer data in realtime. But according to some reasons I don't want to use this method. Instead I've developed an application that saves accelerometer data. I want this application to produce a ".m" file and use this file to represent x-y-z signals. The problem is that the accelerometer data rate is not constant. for example: "data-1" corresponds to time "0.1", "data-2" corresponds to time "0.2", and "data-3" corresponds to time "0.8". You can see the time interval can change. I need to get fourier transform of these signals, so I can not use the simple matlab vector to represent them. How can I represent these signals? How can I get the correct fourier transform of these signals?
• Don't use Arduino it is simpler way USB ACCELEROMETER ;) 🢂 tindie.com/products/10373 🢀 Feb 5, 2019 at 23:48
Two methods occur to me.
1. Perform a non-uniform DFT. You can use Google to find MATLAB implementations of it, or try to write one on your own. This file seems to be legit, although I haven't tried it myself.
2. If you have the Signal Processing Toolbox, then you can use the resample() function to interpolate your non-uniformly sampled vector linearly onto a vector of uniformly spaced instants, and then perform the usual DFT on the latter.
• y = resample(x,tx) resamples the values, x, of a signal sampled at the instants specified in vector tx. The function interpolates x linearly onto a vector of uniformly spaced instants with the same endpoints and number of samples as tx. NaNs are treated as missing data and are ignored. Feb 16, 2018 at 7:11
Creating a .m file with all your data explicitly assigned is not really efficient if you have more than a few tens of data points. You really just want to get the data into your memory space. Once you get it in your memory the way you want, you can save it as a .mat file using save and use load to bring it back in.
It sounds like your data from your phone will be in a text file and matlab has several methods including a data import tool, which isn’t hard to use.
A particularly easy way to represent data with irregular time samples is to use a timeseries object and use timeseries methods to manipulate the data. As suggested in @Tenderos answer, there is a resample method. timeseries objects have a lot of overloaded methods that are straightforward to apply.
Another advantage is that a timeseries object can be vector valued. | 601 | 2,607 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-21 | longest | en | 0.919314 |
http://maskray.me/leetcode/two-sum.cc | 1,726,494,634,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-38/segments/1725700651697.32/warc/CC-MAIN-20240916112213-20240916142213-00776.warc.gz | 17,538,853 | 890 | // Two Sum class Solution { public: vector twoSum(vector &a, int s) { vector r(a.size()); iota(r.begin(), r.end(), 0); sort(r.begin(), r.end(), [&](int x, int y) { return a[x] < a[y]; }); for (size_t i = 0, j = a.size()-1; i < j; i++) { while (j > i+1 && a[r[i]]+a[r[j]] > s) j--; if (a[r[i]]+a[r[j]] == s) { int x = r[i], y = r[j]; r.clear(); if (x > y) swap(x, y); r.push_back(x); r.push_back(y); break; } } return r; } }; | 163 | 424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.169097 |
https://www.coursehero.com/file/75717/Test-1-Take-Home/ | 1,524,695,443,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947968.96/warc/CC-MAIN-20180425213156-20180425233156-00638.warc.gz | 764,738,520 | 51,132 | {[ promptMessage ]}
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Test 1 Take Home
Test 1 Take Home - values 4(10 points Find the exact value...
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Test 1 (Review) Pledged Take-home Part Math 121-03, 04 Calculus II Fall 2007 Bring answers to class to attach to your in-class test. If you are really stumped, send me an email. 2. (10 points) What is the range of ( ) 1/(1 ) x y f x e = = + ? Hint: Consider what happens as x y - and x y + . ANSWER: As x y - the range approaches 1, and as x y + the range approaches 0. So the range is from 0 to 1, NOT including the values, the range simply approaches these
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Unformatted text preview: values. 4. (10 points) Find the exact value of 5 log (1/ 25) . ANSWER: log(1/25) log(5) = (-2) 7. (10 points) Solve 5 4 x = for x in terms of a logarithm. X=log 5 4 X= log4 log5 (0.8614) 8. (10 points) Simplify ln x x e + as far as possible. 9. (10 points) Solve ln(1 ) 5 x + = for x ....
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https://plainmath.net/7698/algebraic-multiplicity-eigenvalue-dimension-eigenspace-correspondi | 1,656,581,904,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103671290.43/warc/CC-MAIN-20220630092604-20220630122604-00166.warc.gz | 502,958,105 | 14,355 | # It can be shown that the algebraic multiplicity of an eigenvalue lambda is always greater than or equal to the dimension of the eigenspace correspondi
It can be shown that the algebraic multiplicity of an eigenvalue lambda is always greater than or equal to the dimension of the eigenspace corresponding to lambda. Find h in the matrix A below such that the eigenspace for lambda = 5 is two-dimensional: $A=\left[\begin{array}{cccc}5& -2& 6& -1\\ 0& 3& h& 0\\ 0& 0& 5& 4\\ 0& 0& 0& 1\end{array}\right]$
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1) Row reduce the augmented matrix for the equation $\left(A-5I\right)x=0$
2) $\left[\begin{array}{ccccc}0& -2& 6& -1& 0\\ 0& -2& h& 0& 0\\ 0& 0& 0& 4& 0\\ 0& 0& 0& -4& 0\end{array}\right]=\left[\begin{array}{ccccc}0& -2& 6& -1& 0\\ 0& -2& h-6& 1& 0\\ 0& 0& 0& 4& 0\\ 0& 0& 0& 4& 0\end{array}\right]=\left[\begin{array}{ccccc}0& 1& -3& 0& 0\\ 0& 0& h-6& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0\end{array}\right]$
3) For a two-dimensional eigenspace, the system above needs two free variables. his happens if and inly if $h=6$
$\left[\begin{array}{ccccc}0& 1& -3& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0\end{array}\right]$ | 573 | 1,410 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 22, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2022-27 | latest | en | 0.70227 |
http://www.algorithmic-solutions.info/leda_manual/mw_matching.html | 1,534,526,166,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212639.36/warc/CC-MAIN-20180817163057-20180817183057-00152.warc.gz | 482,085,435 | 5,076 | Next: Stable Matching ( stable_matching Up: Graph Algorithms Previous: Maximum Cardinality Matchings in Contents Index
Subsections
# General Weighted Matchings ( mw_matching )
We give functions
• to compute maximum-weight matchings,
• to compute maximum-weight or minimum-weight perfect matchings, and
• to check the optimality of weighted matchings
in general graph.
You may skip the following subsections and restrict on reading the function signatures and the corresponding comments in order to use these functions. If you are interested in technical details, or if you would like to ensure that the input data is well chosen, or if you would like to know the exact meaning of all output parameters, you should continue reading.
The functions in this section are template functions. It is intended that in the near future the template parameter NT can be instantiated with any number type. Please note that for the time being the template functions are only guaranteed to perform correctly for the number type int. In order to use the template version of the function the appropriate .h-file must be included.
#include <LEDA/graph/templates/mw_matching.h>
There are pre-instantiations for the number types int. In order to use them either
#include <LEDA/graph/mw_matching.h>
or
#include <LEDA/graph/graph_alg.h>
has to be included (the latter file includes the former). The connection between template functions and pre-instantiated functions is discussed in detail in the section Templates for Network Algorithms'' of the LEDA book. The function names of the pre-instantiated versions and the template versions only differ by an additional suffix _T in the names of the latter ones.
#### Proof of Optimality.
Most of the functions for computing maximum or minimum weighted matchings provide a proof of optimality in the form of a dual solution represented by pot, BT and b. We briefly discuss their semantics: Each node is associated with a potential which is stored in the node array pot. The array BT (type array<two_tuple<NT, int> >) is used to represent the nested family of odd cardinality sets which is constructed during the course of the algorithm. For each (non-trivial) blossom B , a two tuple (zB, pB) is stored in BT, where zB is the potential and pB is the parent index of B . The parent index pB is set to -1 if B is a surface blossom. Otherwise, pB stores the index of the entry in BT corresponding to the immediate super-blossom of B . The index range of BT is [0,..., k - 1] , where k denotes the number of (non-trivial) blossoms. Let B' be a sub-blossom of B and let the corresponding index of B' and B in BT be denoted by i' and i , respectively. Then, i' < i . In b (type node_array<int>) the parent index for each node u is stored (-1 if u is not contained in any blossom).
#### Heuristics for Initial Matching Constructions.
Each function can be asked to start with either an empty matching ( heur = 0 ), a greedy matching ( heur = 1 ) or an (adapted) fractional matching ( heur = 2 ); by default, the fractional matching heuristic is used.
#### Graph Structure.
All functions assume the underlying graph (type graph) to be connected, simple, and loopfree. They work on the underlying undirected graph of the directed graph parameter.
#### Edge Weight Restrictions.
The algorithms use divisions. In order to avoid rounding errors for the number type int, please make sure that all edge weights are multiples of 4 ; the algorithm will automatically multiply all edge weights by 4 if this condition is not met. (Then, however, the returned dual solution is valid only with respect to the modified weight function.) Moreover, in the maximum-weight (non-perfect) matching case all edge weights are assumed to be non-negative.
#### Arithmetic Demand.
The arithmetic demand for integer edge weights is as follows. Let C denote the maximal absolute value of any edge weight and let n be the number of nodes of the graph.
In the perfect weighted matching case we have for a potential pot[u] of a node u and for a potential zB of a blossom B :
In the non-perfect matching case we have for a potential pot[u] of a node u and for a potential zB of a blossom B :
The function CHECK_WEIGHTS may be used to test whether the edge weights are feasible or not. It is automatically called at the beginning of each of the algorithms provided in this chapter.
#### Single Tree vs. Multiple Tree Approach:
All functions can either run a single tree approach or a multiple tree approach. In the single tree approach, one alternating tree is grown from a free node at a time. In the multiple tree approach, multiple alternating trees are grown simultaneously from all free nodes. On large instances, the multiple tree approach is significantly faster and therefore is used by default. If #define _SST_APPROACH is defined before the template file is included all functions will run the single tree approach.
#### Worst-Case Running Time:
All functions for computing maximum or minimum weighted (perfect or non-perfect) matchings guarantee a running time of O(nm log n) , where n and m denote the number of nodes and edges, respectively.
template < class NT> list< edge> MAX_WEIGHT_MATCHING_T(const graph& G, const edge_array< NT> & w, bool check = true, int heur = 2) computes a maximum-weight matching M of the underlying undirected graph of graph G with weight function w. If check is set to true, the optimality of M is checked internally. The heuristic used for the construction of an initial matching is determined by heur. Precondition All edge weights must be non-negative. template < class NT> list< edge> MAX_WEIGHT_MATCHING_T(const graph& G, const edge_array< NT> & w, node_array< NT> & pot, array< two_tuple< NT, int> > & BT, node_array< int> & b, bool check = true, int heur = 2) computes a maximum-weight matching M of the underlying undirected graph of graph G with weight function w. The function provides a proof of optimality in the form of a dual solution given by pot, BT and b. If check is set to true, the optimality of M is checked internally. The heuristic used for the construction of an initial matching is determined by heur. Precondition All edge weights must be non-negative. template < class NT> bool CHECK_MAX_WEIGHT_MATCHING_T(const graph& G, const edge_array< NT> & w, const list< edge> & M, const node_array< NT> & pot, const array< two_tuple< NT, int> > & BT, const node_array< int> & b) checks if M together with the dual solution represented by pot, BT and b are optimal. The function returns true if M is a maximum-weight matching of G with weight function w. template < class NT> list< edge> MAX_WEIGHT_PERFECT_MATCHING_T(const graph& G, const edge_array< NT> & w, bool check = true, int heur = 2) computes a maximum-weight perfect matching M of the underlying undirected graph of graph G and weight function w. If G contains no perfect matching the empty set of edges is returned. If check is set to true, the optimality of M is checked internally. The heuristic used for the construction of an initial matching is determined by heur. template < class NT> list< edge> MAX_WEIGHT_PERFECT_MATCHING_T(const graph& G, const edge_array< NT> & w, node_array< NT> & pot, array< two_tuple< NT, int> > & BT, node_array< int> & b, bool check = true, int heur = 2) computes a maximum-weight perfect matching M of the underlying undirected graph of graph G with weight function w. If G contains no perfect matching the empty set of edges is returned. The function provides a proof of optimality in the form of a dual solution given by pot, BT and b. If check is set to true, the optimality of M is checked internally. The heuristic used for the construction of an initial matching is determined by heur. template < class NT> bool CHECK_MAX_WEIGHT_PERFECT_MATCHING_T(const graph& G, const edge_array< NT> & w, const list< edge> & M, const node_array< NT> & pot, const array< two_tuple< NT, int> > & BT, const node_array< int> & b) checks if M together with the dual solution represented by pot, BT and b are optimal. The function returns true iff M is a maximum-weight perfect matching of G with weight function w. template < class NT> list< edge> MIN_WEIGHT_PERFECT_MATCHING_T(const graph& G, const edge_array< NT> & w, bool check = true, int heur = 2) computes a minimum-weight perfect matching M of the underlying undirected graph of graph G with weight function w. If G contains no perfect matching the empty set of edges is returned. If check is set to true, the optimality of M is checked internally. The heuristic used for the construction of an initial matching is determined by heur. template < class NT> list< edge> MIN_WEIGHT_PERFECT_MATCHING_T(const graph& G, const edge_array< NT> & w, node_array< NT> & pot, array< two_tuple< NT, int> > & BT, node_array< int> & b, bool check = true, int heur = 2) computes a minimum-weight perfect matching M of the underlying undirected graph of graph G with weight function w. If G contains no perfect matching the empty set of edges is returned. The function provides a proof of optimality in the form of a dual solution given by pot, BT and b. If check is set to true, the optimality of M is checked internally. The heuristic used for the construction of an initial matching is determined by heur. template < class NT> bool CHECK_MIN_WEIGHT_PERFECT_MATCHING_T(const graph& G, const edge_array< NT> & w, const list< edge> & M, const node_array< NT> & pot, const array< two_tuple< NT, int> > & BT, const node_array< int> & b) checks if M together with the dual solution represented by pot, BT and b are optimal. The function returns true iff M is a minimum-weight matching of G with weight function w. template < class NT> bool CHECK_WEIGHTS_T(const graph& G, edge_array< NT> & w, bool perfect) returns true, if w is a feasible weight function for G; false otherwise. perfect must be set to true in the perfect matching case; otherwise it must be set to false. If the edge weights are not multiplicatives of 4 all edge weights will be scaled by a factor of 4 . The modified weight function is returned in w then. This function is automatically called by each of the maximum weighted machting algorithms provided in this chapter, the user does not have to take care of it.
Next: Stable Matching ( stable_matching Up: Graph Algorithms Previous: Maximum Cardinality Matchings in Contents Index
Christian Uhrig 2017-04-07 | 2,375 | 10,434 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-34 | longest | en | 0.881356 |
https://byjus.com/question-answer/under-a-given-correspondence-two-right-angled-triangles-are-congruent-if-the-hypotenuse-and-the/ | 1,713,252,158,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817073.16/warc/CC-MAIN-20240416062523-20240416092523-00785.warc.gz | 138,664,930 | 21,346 | 1
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Question
# Under a given correspondence, two right-angled triangles are congruent if the hypotenuse and the leg of one of the triangles are equal to the hypotenuse and the corresponding leg of the other triangle.The given statement is known as:
A
SSS congruence of two triangles
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B
ASA congruence of two triangles
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C
SAS congruence of two triangles
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D
RHS congruence of two right-angled triangles
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Solution
## The correct option is D RHS congruence of two right-angled trianglesUnder a given correspondence, two right-angled triangles are congruent if the hypotenuse and the leg of one of the triangles are equal to the hypotenuse and the corresponding leg of the other triangle. It is known as RHS congruence of two right-angled triangles.
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Join BYJU'S Learning Program | 317 | 1,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-18 | latest | en | 0.874284 |
https://googology.fandom.com/wiki/Quinquacentillion | 1,653,138,291,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662539101.40/warc/CC-MAIN-20220521112022-20220521142022-00463.warc.gz | 336,978,384 | 129,687 | 11,580
pages
"Quincentillion" redirects here. It is not to be confused with Quingentillion.
The quinquacentillion is equal to \(10^{318}\) in the short scale and \(10^{630}\) in the long scale.[1]
It is 319 digits long in short scale, and 631 digits long in long scale.
## Decimal expansion
For short scale:
1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
For long scale:
1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 | 412 | 1,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-21 | latest | en | 0.573666 |
https://mail.haskell.org/pipermail/haskell-cafe/2007-December/037047.html | 1,513,575,728,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948608836.84/warc/CC-MAIN-20171218044514-20171218070514-00479.warc.gz | 594,218,501 | 3,406 | # [Haskell-cafe] what does @ mean?.....
Nicholls, Mark Nicholls.Mark at mtvne.com
Fri Dec 28 06:21:54 EST 2007
```So in the example given...
mulNat a b
| a <= b = mulNat' a b b
| otherwise = mulNat' b a a
where
mulNat' x@(S a) y orig
| x == one = y
| otherwise = mulNat' a (addNat orig y) orig
Is equivalent to
mulNat a b
| a <= b = mulNat' a b b
| otherwise = mulNat' b a a
where
mulNat' (S a) y orig
| (S a) == one = y
| otherwise = mulNat' a (addNat orig y) orig
?
-----Original Message-----
From: Alfonso Acosta [mailto:alfonso.acosta at gmail.com]
Sent: 28 December 2007 11:20
To: Nicholls, Mark
Subject: Re: [Haskell-cafe] what does @ mean?.....
@ works as an aliasing primitive for the arguments of a function
f x@(Just y) = ...
using "x" in the body of f is equivalent to use "Just y". Perhaps in
this case is not really useful, but in some other cases it saves the
effort and space of retyping really long expressions. And what is even
more important, in case an error is made when choosing the pattern,
you only have to correct it in one place.
On Dec 28, 2007 12:05 PM, Nicholls, Mark <Nicholls.Mark at mtvne.com>
wrote:
>
>
>
>
> Hello, I wonder if someone could answer the following...
>
> The short question is what does @ mean in
>
>
>
> mulNat a b
>
> | a <= b = mulNat' a b b
>
> | otherwise = mulNat' b a a
>
> where
>
> mulNat' x@(S a) y orig
>
> | x == one = y
>
> | otherwise = mulNat' a (addNat orig y) orig
>
>
>
> The long version, explaining what everything means is....
>
>
>
> here's a definition of multiplication on natural numbers I'm reading
>
> on a blog....
>
>
>
> data Nat = Z | S Nat
>
> deriving Show
>
>
>
> one :: Nat
>
> one = (S Z)
>
>
>
> mulNat :: Nat -> Nat -> Nat
>
> mulNat _ Z = Z
>
> mulNat Z _ = Z
>
> mulNat a b
>
> | a <= b = mulNat' a b b
>
> | otherwise = mulNat' b a a
>
> where
>
> mulNat' x@(S a) y orig
>
> | x == one = y
>
> | otherwise = mulNat' a (addNat orig y) orig
>
>
>
> Haskell programmers seem to have a very irritating habit of trying to
>
> be overly concise...which makes learnign the language extremely
>
> hard...this example is actually relatively verbose....but anyway...
>
>
>
> Z looks like Zero...S is the successor function...Nat are the
>
> "Natural" numbers.....
>
>
>
> mulNat _ Z = Z
>
> mulNat Z _ = Z
>
>
>
> translates to...
>
>
>
> x * 0 = 0....fine...
>
> 0 * x = 0....fine..
>
>
>
> mulNat a b
>
> | a <= b = mulNat' a b b
>
> | otherwise = mulNat' b a a
>
> where
>
> mulNat' x@(S a) y orig
>
> | x == one = y
>
> | otherwise = mulNat' a (addNat orig y) orig
>
>
>
> is a bit more problematic...
>
> lets take a as 3 and b as 5...
>
>
>
> so now we have
>
>
>
> mulNat' 3 5 5
>
>
>
> but what does the "x@(S a)" mean? in
>
>
>
> mulNat' x@(S a) y orig
>
>
>
> ________________________________
>
>
> Sent: 21 December 2007 17:47
> To: David Menendez
> Subject: RE: [Haskell-cafe] nice simple problem for someone
struggling....
>
>
>
> Let me resend the code...as it stands....
>
>
>
> module Main where
>
>
>
> data SquareType numberType = Num numberType => SquareConstructor
numberType
>
>
>
> class ShapeInterface shape where
>
> area :: Num numberType => shape->numberType
>
>
>
> data ShapeType = forall a. ShapeInterface a => ShapeType a
>
>
>
> instance (Num a) => ShapeInterface (SquareType a) where
>
> area (SquareConstructor side) = side * side
>
>
>
>
>
> and the errors are for the instance declaration.......
>
>
>
> [1 of 1] Compiling Main ( Main.hs, C:\Documents and
>
>
>
> Main.hs:71:36:
>
> Couldn't match expected type `numberType' against inferred type
`a'
>
> `numberType' is a rigid type variable bound by
>
> the type signature for `area' at Main.hs:38:15
>
> `a' is a rigid type variable bound by
>
> the instance declaration at Main.hs:70:14
>
> In the expression: side * side
>
> In the definition of `area':
>
> area (SquareConstructor side) = side * side
>
>
>
> I'm becoming lost in errors I don't comprehend....
>
>
>
> What bamboozles me is it seemed such a minor enhancement.
>
>
> ________________________________
>
>
> From: d4ve.menendez at gmail.com [mailto:d4ve.menendez at gmail.com] On
Behalf Of
> David Menendez
> Sent: 21 December 2007 17:05
> To: Nicholls, Mark
> Subject: Re: [Haskell-cafe] nice simple problem for someone
struggling....
>
>
>
> On Dec 21, 2007 11:50 AM, Nicholls, Mark <Nicholls.Mark at mtvne.com>
wrote:
>
>
>
> Now I have....
>
> module Main where
>
> data SquareType numberType = Num numberType => SquareConstructor
> numberType
>
>
>
> This is a valid declaration, but I don't think it does what you want
it to.
> The constraint on numberType applies only to the data constructor.
>
> That is, given an unknown value of type SquareType a for some a, we
do not
> have enough information to infer Num a.
>
> For your code, you want something like:
>
> instance (Num a) => ShapeInterface (SquareType a) where
> area (SquareConstructor side) = side * side
>
>
> --
> Dave Menendez <dave at zednenem.com>
> <http://www.eyrie.org/~zednenem/ >
> _______________________________________________ | 1,663 | 5,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-51 | latest | en | 0.718038 |
https://philosophy.stackexchange.com/questions/65327/an-erotetic-argument-about-the-liar-paradox | 1,596,574,176,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735882.86/warc/CC-MAIN-20200804191142-20200804221142-00130.warc.gz | 399,790,156 | 33,927 | I noticed something about the liar paradox, when the liar sentences are taken as questions. Let’s start with the liar index, “This sentence is false.” Allow this to be questioned: “Is this sentence false?” Now, if this is a yes-or-no question, then saying, “No,” amounts to reinterpreting the sentence as, “No, this sentence is not false; it’s true.” In other words, the liar index is logically interchangeable with the honest index, as such (more on this in a little).
Next, take the recursive liar, “L: L is false.” As a question this would be, “L: L is false?” However, L in that case would be a question, and questions are not truth-apt as such. The recursive liar lacks an erotetic value.
Another example would be the liar loop, “The next sentence is true; the previous sentence is false.” But if either of the sentences is taken as a question, the loop halts: “The next sentence is true; is the previous sentence false?” doesn’t work since “is the previous sentence false?” is a question, so “The next sentence” does not refer to something that is truth-apt, now.
Let’s also consider the liar imperative. This comes from the epistemic-imperative theory of erotetic logic, where a question is converted into an epistemic imperative. Here, “L: Is L false?” = “L: Let me know whether L is false.” But L in this event is an imperative: again, a kind of sentence that is neither true nor the opposite of true (false), but just “not true” in the abstract. (I think there's a related analysis of, "Don't comply with this imperative," which is complied with if and only if it is not complied with, vs., "Comply with the imperative, 'Do x'" being reducible just to, "Do x" in the same way that, "It is truth that X," reduces to the straight assertion that X. But I can't remember how to "do it" (solve the problem) right now. It shows up in Hofstadter's GEB as, "I wish that this wish would not be granted," for whatever that's worth.)
I wrote an essay on the topic once where I covered a few more examples IIRC and they all go pretty much the same way. For example, trying to work out the liar disquotation ("This sentence is false," is true if and only if this sentence is false) can be used to exactly describe how saying "no" to the liar index converts it into the honest one (i.e. how (false:not true):no "goes to" true:true; it's a kind of subtle double-negation elimination that ends up with: "This sentence is false," is false if and only if, "This sentence is true," is true).
Tarski’s model of levels of truth-predication therefore holds as a two-place relation, between questions and answers in general. That is, since the liar sentences are “unquestionable,” if we accepted them (believed them), we would be using them as unquestionable axioms. The problem, then, is that the liar sentences are not answers to questions, so they have no erotetic form of truth, which would be the alternative truth-predicate of a t1/t2/...-predicate model.
NOTE: I am not arguing that the liar sentences are “meaningless.” The liar index is certainly not meaningless: it has an entirely regular use in natural language, as in saying the sentence while pointing at some other false sentence to which “This” actually refers, e.g. pointing at a written token of, “2 + 2 = 22,” and saying, “This sentence is false.” (Moreover, the existence of this natural usage is what allows us to commit to an equivalent "This" in the interior and exterior of the liar disquotation, instead of saying, "'This sentence is false,' is true if and only if that sentence is false." This haecceitic quasi-copying is allowable on the ground that the liar indexicals have to be adjudged only as sentence-tokens: the sentence-type is truth-inapt, since it does not actually self-refer enough to be evaluated as such.)
My last question, then, is: is this the solution to the liar paradox?
• The liar paradox already has a solution; the words "this sentence" are nebulous. If you said "He said something false," then that could be a meaningful statement, but telling us that "this sentence" is false is really telling us nothing. – David Blomstrom Aug 17 '19 at 12:48
• @KristianBerry "is this the solution to the liar paradox" Which part of your question are you referring to by "this"?! – Speakpigeon Aug 17 '19 at 14:34
• Argument: liar sentences are not answers to well-formed questions. So if we believed/used them, e.g. as evidence against the axiom of noncontradiction, we would use claims that couldn't be doubted, which seems out of the spirit of philosophy. Or: Tarski's multi-level truth-predicate idea IS correct, yet it can be simplified (I think his has possibly infinite levels?) to the difference between truth by itself and as true answers to questions. Liar sentences are disquotationally interchangeable with honest sentences, too, it seems. – Kristian Berry Aug 17 '19 at 22:37
• The liar paradox was an illustration of a problem in set theory, as such any of the solutions to that problem is also a solution the liar paradox e.g. Russell's "type theory". Also for interest: philosophy.stackexchange.com/q/54384/33787 – christo183 Aug 18 '19 at 4:55
• Declaring that Liar-like sentences are not well-formed is not a solution, a solution would be to give precise and algorithmic formation rules that allow common uses of self-reference but rule out undesirable consequences. Since this is likely impossible to everyone's satisfaction, there is no solution, see IEP Liar Paradox. – Conifold Aug 18 '19 at 5:01 | 1,309 | 5,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-34 | latest | en | 0.968425 |
http://psychology.wikia.com/wiki/Kendall_tau_distance?oldid=142375 | 1,419,048,658,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802769373.55/warc/CC-MAIN-20141217075249-00114-ip-10-231-17-201.ec2.internal.warc.gz | 235,435,098 | 24,280 | Kendall tau distance
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It has been suggested that this article or section be merged into Kendall tau rank correlation coefficient. (Discuss)
The Kendall tau distance is a metric that counts the number of pairwise disagreements between two lists. The larger the distance, the more dissimilar the two lists are. Kendall tau distance is also called bubble-sort distance since it is equivalent to the number of swaps that the bubble sort algorithm would make to place one list in the same order as the other list. The Kendall tau distance was created by Maurice Kendall.
DefinitionEdit
The Kendall tau distance between two lists $\tau_1$ and $\tau_2$ is
$K(\tau_1,\tau_2) = |(i,j): i < j, ( \tau_1(i) < \tau_1(j) \wedge \tau_2(i) > \tau_2(j) ) \vee ( \tau_1(i) > \tau_1(j) \wedge \tau_2(i) < \tau_2(j) )|.$
$K(\tau_1,\tau_2)$ will be equal to 0 if the two lists are identical and $n(n-1)/2$ (where $n$ is the list size) if one list is the reverse of the other. Often Kendall tau distance is normalized by dividing by $n(n-1)/2$ so a value of 1 indicates maximum disagreement. The normalized Kendall tau distance therefore lies in the interval [0,1].
Kendall tau distance may also be defined as
$K(\tau_1,\tau_2) = \begin{matrix} \sum_{\{i,j\}\in P} \bar{K}_{i,j}(\tau_1,\tau_2) \end{matrix}$
where
• P is the set of unordered pairs of distinct elements in $\tau_1$ and $\tau_2$
• $\bar{K}_{i,j}(\tau_1,\tau_2)$ = 0 if i and j are in the same order in $\tau_1$ and $\tau_2$
• $\bar{K}_{i,j}(\tau_1,\tau_2)$ = 1 if i and j are in the opposite order in $\tau_1$ and $\tau_2.$
Kendall tau distance can also be defined as the total number of discordant pairs.
Kendall tau distance in Rankings: A permutation (or ranking) is an array of N integers where each of the integers between 0 and N-1 appears exactly once. The Kendall tau distance between two rankings is the number of pairs that are in different order in the two rankings. For example the Kendall tau distance between 0 3 1 6 2 5 4 and 1 0 3 6 4 2 5 is four because the pairs 0-1, 3-1, 2-4, 5-4 are in different order in the two rankings, but all other pairs are in the same order. [1]
ExampleEdit
Suppose we rank a group of five people by height and by weight:
Person A B C D E
Rank by Height 1 2 3 4 5
Rank by Weight 3 4 1 2 5
Here person A is tallest and third-heaviest, and so on.
In order to calculate the Kendall tau distance, pair each person with every other person and count the number of times the values in list 1 are in the opposite order of the values in list 2.
Pair Height Weight Count
(A,B) 1 < 2 3 < 4
(A,C) 1 < 3 3 > 1 X
(A,D) 1 < 4 3 > 2 X
(A,E) 1 < 5 3 < 5
(B,C) 2 < 3 4 > 1 X
(B,D) 2 < 4 4 > 2 X
(B,E) 2 < 5 4 < 5
(C,D) 3 < 4 1 < 2
(C,E) 3 < 5 1 < 5
(D,E) 4 < 5 2 < 5
Since there are 4 pairs whose values are in opposite order, the Kendall tau distance is 4. The normalized Kendall tau distance is
$\frac{4}{5(5 - 1)/2} = 0.4.$
A value of 0.4 indicates a somewhat low agreement in the rankings. | 995 | 3,086 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 17, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2014-52 | longest | en | 0.895725 |
https://www.emathhelp.net/register-tons-to-acre-inches/ | 1,631,940,116,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056297.61/warc/CC-MAIN-20210918032926-20210918062926-00037.warc.gz | 763,327,877 | 4,935 | # Register tons to acre-inches
This free conversion calculator will convert register tons to acre-inches, i.e. register ton to ac-in. Correct conversion between different measurement scales.
Convert
The formula is $V_{\text{ac-in}} = \frac{10}{363} V_{\text{register ton}}$, where $V_{\text{register ton}} = 15$.
Therefore, $V_{\text{ac-in}} = \frac{50}{121}$.
Answer: $15 \text{register ton} = \frac{50}{121} \text{ac-in}\approx 0.413223140495868 \text{ac-in}.$ | 144 | 467 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-39 | latest | en | 0.632307 |
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# Swine Flu - May or May not be Natural
page: 1
0
share:
posted on Apr, 27 2009 @ 09:19 AM
*Yea I ranted bout the validy of users here, and I still feel the same way. But if Im gonna throw negetivity out I might as well throw in my 2 cents on sound research done.
!First off, This Swine Flu will not be deadly as its going to be in this first wave.
-The 1918 Flu Pandemic started out simultaneously in three different parts of the world.
*Europe
*Asia
*North America
*1st Wave = Spring of 1918
*2nd Wave = Fall of 1918
*3rd Wave = Winter of 1918-1919 (! This wave was the most fatal)
============================
Right now we are in the first wave, with only a 100+ deaths reported so far, with almost simultaneous outbreaks in North America(Canada, United States and Mexico).
Its 1st wave kicked off in this month of April 2009 (Spring of 2009)
If history is any type of blueprint or outline, then we should be expecting another outbreak and more deadly one in Fall time frame of this year, and round winter we should see the deadliest outbreak
The group of people that will be effected the most is as follows....
*This is per 100,000 person's in each age group (this is the 1918 chart)
-2,200 (give or take) = Age less than 1 years old
-1,000 (give or take) = Ages 25-34
-2,200 (give or take) = Ages 85+
the one's less effected by the 1918 flu epidemic was Ages 5-14 with Less than 500 people dying from it.
Total Over All death rate
The worlds population in 1918 was around 1.8 Billion, with a 50 Million death toll world wide after the 1918 flu hit.
Todays world population is 6.7 Billion. I think we will see a 300 Million or more death toll, and I come to this conclusion through simple math and the coincidences that are present in today's flu outbreak and 1918 which I will further expand upon.
======================
Time Coincidences
Since the sun has be brought into this as a suspect, lets continue that frame of thought for a moment...
Total Solar Eclipse of 1918 was on June 8, 1918
Total Solar Eclipse of 2009 will be on July 22, 2009
1st Case of 1918 flu was on/or around March 4, 1918 = Monday
1st Case of 2009 Swine flue was on/or around April 2009
!Between the 1st case of 1918 and the Total Solar Eclipse of 1918 was a 3 month difference
!Between the 1st case of 2009 swine flue and the Total Solar Eclipse of 2009 will be about a 3 months difference
Sun Coincidences
The 1918 flu epidemic fell under the 15th Solar Cycle which between the years of 1913-1923 seen just over 100 sun spots. it had a 92 consecutive stretch of spotless days spreading from April-June 1913
2009 Swine flu, we are in the 9th consecutive day of spotlessness. alot of scientist have already said they think this is our solar minimum. which if true, means that these 2 pandemics are taking place during solar minimum's.
=========================
Right now im still going over news, and archives of cdc and nasa information to get a better understanding of this. when I get more i'll post
posted on Apr, 27 2009 @ 10:12 AM
hate to break the news to ya, but,this is a "mutation" of the swine flu and almost every time a virus has mutated it has been at solar maximums with the increased radiation output from the sun was the causing factor of the mutation.so if were in solar minimum what would have caused the mutation?I mean it has a swine,avion and human genetic make-up.that is some serious mutating.can anyone say bio-engineered?
posted on Apr, 27 2009 @ 10:24 AM
Very interesting observation. You may be on to something here as there may be a connection between the Sun or changes in weather and tempreture that triggers viruses.
I come from Malaysia in South East Asia where there was an swine flu outbreak that killed over 100 villagers from local pig farming community in the late 1990s. It was somewhere between 1996-1997.
Initial speculations was that the virus strain was that of Japanese Enchepalitis (JE). The virus manage to jump across state borders resulting in several other fatalities, However, it was confined to those who had come into contact with blood of infected pigs.
Thankfully the virus did not evolved to one that could spread from human to human. A massive screening, quarantine and culling operation brought the outbeak under control within a couple of months.
However, it was months after the incident that tests (US's CDC experts were called in to assist) revealed that the outbreak was caused by a new virus strain that was aptly given a local name of Nipah (palm) Virus.
What is intriguing is that the outbreak happened about the same time that the region was undergoing the haze phenomenon caused massive forest clearing via fires in Sumatra and Borneo. The smoke from the fires blanketed several majot cities in Malaysia including Singapore. It was also the tailend of the 22 Solar Cycle.
Your mention of the minimum sunspots is quite interesting. I wonder whether it has anything to do with the recent rise in tempreture. Several parts of the world, in particular Australia suffered record heatwaves and massive devastating fires several months ago.
Ironically, the tempretures over here in Malaysia has risen between 3 celsius to 4 celcius over the past two weeks. The Sun has been blazingly hot with sheltering weather.
In that respect, I am quite certain that you may have stumbled across the possibilty of what may trigger ( the Sun or change in weather conditions) a dormant super virus or cause a non-fatal one to mutate to a global killer.
Cheers.
To None I Wish Ill But To All I Seek Peace and Love.
posted on Apr, 27 2009 @ 11:08 AM
Interesting coincidences at least. What inspired you to look into solar cycle correlations and such?
posted on Apr, 27 2009 @ 01:05 PM
Originally posted by notsoobvious
hate to break the news to ya, but,this is a "mutation" of the swine flu and almost every time a virus has mutated it has been at solar maximums with the increased radiation output from the sun was the causing factor of the mutation.so if were in solar minimum what would have caused the mutation?I mean it has a swine,avion and human genetic make-up.that is some serious mutating.can anyone say bio-engineered?
I understand it is a mutation, it is believed to have 4 different strains
*2 Human Influenza
*1 Avian
*The other escapes my mind, But Im going safely say Swine
And everytime a virus has mutated it has not been a solar maximums, this current virus is a current and great example to this.
As you have asked, if there is a solar minimum what has caused this mutation? I have a suggestion, and its a fairly easy one to grasp. Follow me along with this example....
The Human cell
*If exposed to extra-than-normal amounts of sun light the cell begins to replicate wrong and can cause skin cancer
*if not giving the proper amount of light the skin will start to suffer deficiencies.
The common factor in this is radiation.
Too much = dangerious
Too Little = dangerious
now, with this same thought in mind lets apply virus mutations to the equation....
If a virus is said to mutate under solar maximum's then that means there is too much radiation for the virus to handle and thus it mutates
Whats to say that during solar minimums that the virus doesn't receive enough radiation to "keep it in check" and it begins to mutate?
Who's to say that the earth herself isnt giving off some form of radiation?
Or the Moon, or some other celestial source(s).
And as for you claim's of Bio-engineering, was the 1918 flu bio-engineered? because it was during a solar minimum, not a maximum.
posted on Apr, 27 2009 @ 01:17 PM
I have just a slight feeling the sun is factoring into this due to its conditions almost mimicking those of 1918. But another thought that ran across my mind is that 1918 epidemic follows right after World War 1, this epidemic follows after economic crisis, global war's on terror, the changes in the sun, and some other celestial changes going on with the planet and our milky way.
Maybe it is earths immune system kicking in to weed out some of the negativity on its surface (just thinking out the box).
The CDC stated that virus have a funny way of shifting and changing even when they arent giving the proper conditions to do so.
With the lowering of solar activity the temperatures on our planet should be getting cooler, not warmer due to the inactivity of the sun's surface, now I have read somewhere (i cant recall, maybe here) that our earth is getting the vice grip treatment from gravitational influences which is causing more earth quakes, which would also (in my view) release more heat into our atmosphere and is causing temp raises.
I dont know if its dormant or not, but the cdc makes it look like virus have a universal connection the same why it is hinted humans have universal consciousness and sub consciousness.
=======================
I use to be a conspiracy nut, but now Im tryna find more plausible explanations of happenings other than "OMG BIO ATTACK FROM LAB. etc. etc." Alot of times nature will do more damage than man, as I looked into swine flu I came across 1918 swine flu and one of the articles mentioned mutations and the sun.
But basically im looking for common grounds in different areas to tie all this mess up naturally and not terror wise
posted on Apr, 27 2009 @ 02:06 PM
What Mr.Lump is saying is totally correct. The coincidence is there. The facts are pretty solid. But the fact that we can currently genetically engineer a virus is still there too. That the same people who mess with these viruses in laboratories are funded by powerful individuals. Pharmaceutical companies have the most to gain from something like this. Or the virus can just be going through a usual mutation that it naturally does from time to time. Who really knows... The drug cartels in Mexico even... Missing things from lab in Ft. Deitrichs... What about a distraction? Is someone trying to take away more rights while everyone is panicking over Swine Flu... Or (D)All of The Above... Though it sounds ridiculous, it does not mean it cant be true. We have to be vigilant and keep digging.
Good work though Mr. Lump
[edit on 27-4-2009 by Mindflux]
posted on Apr, 27 2009 @ 03:24 PM
mr lump where are you getting your figures from? 1918 was a solar maximum.you can go to the soho web page,look at the article dated april 1 2009. you can obvisly see that 1918 was a solar maximum.but you could be onto something else. with solar minimums comes less solar winds which keep cosmic rays out of our inner solar system.so in that itself leaves something to think about.im not out on a misson to prove that this thing is manmade. i just think that its kind of strange that 3 seperate strains could mutate into one deadly strain.like everybody else im trying to find a cause for this and wanted to throw my ideas out there so that maybe it will catch the attention of someone who could provide the next step.oh by the way the solar minimum in that period of time was 1913.
[edit on 27-4-2009 by notsoobvious]
new topics
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0 | 2,635 | 11,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-34 | longest | en | 0.943937 |
http://www.britannica.com/topic/inner-product-space | 1,462,527,361,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861743914.17/warc/CC-MAIN-20160428164223-00016-ip-10-239-7-51.ec2.internal.warc.gz | 408,829,687 | 12,405 | # Inner product space
Mathematics
Inner product space, In mathematics, a vector space or function space in which an operation for combining two vectors or functions (whose result is called an inner product) is defined and has certain properties. Such spaces, an essential tool of functional analysis and vector theory, allow analysis of classes of functions rather than individual functions. In mathematical analysis, an inner product space of particular importance is a Hilbert space, a generalization of ordinary space to an infinite number of dimensions. A point in a Hilbert space can be represented as an infinite sequence of coordinates or as a vector with infinitely many components. The inner product of two such vectors is the sum of the products of corresponding coordinates. When such an inner product is zero, the vectors are said to be orthogonal (see orthogonality). Hilbert spaces are an essential tool of mathematical physics. See also David Hilbert.
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https://www.dreamincode.net/forums/topic/409496-for-loop-timer/ | 1,529,529,895,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863886.72/warc/CC-MAIN-20180620202232-20180620222232-00417.warc.gz | 806,851,945 | 34,423 | # For loop timer?
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### #1 FlamingBurrito15
Reputation: 0
• Posts: 71
• Joined: 13-December 17
# For loop timer?
Posted 25 February 2018 - 12:56 PM
Hello just out of curiosity what number in a for loop would equal 10 seconds?
(Like
```for(I = 0; I < 28; I++)
```
would equal to 3 seconds)
Is This A Good Question/Topic? 0
## Replies To: For loop timer?
### #2 FlamingBurrito15
Reputation: 0
• Posts: 71
• Joined: 13-December 17
## Re: For loop timer?
Posted 25 February 2018 - 12:58 PM
Hello just out of curiosity what number in a for loop would equal 3 seconds?
(Like
```for(I = 0; I < 28; I++)
```
would equal to 4 seconds)
### #3 ndc85430
• I think you'll find it's "Dr"
Reputation: 825
• Posts: 3,337
• Joined: 13-June 14
## Re: For loop timer?
Posted 25 February 2018 - 01:03 PM
What? Your question doesn't really make any sense. What do you mean when you say that for loop "would equal to 3 seconds"? More to the point, though, what problem are you really trying to solve?
### #4 Martyr2
• Programming Theoretician
Reputation: 5226
• Posts: 14,005
• Joined: 18-April 07
## Re: For loop timer?
Posted 25 February 2018 - 01:05 PM
You can't gauge time like that. The loops execution is independent of system time. The speed of the loop would be effected by current CPU cycles, what else is running, available RAM etc. If you are looking to implement some kind of delay, then look at the sleep() function which will allow you to pause for a specific period of time.
### #5 Skydiver
• Code herder
Reputation: 6216
• Posts: 21,455
• Joined: 05-May 12
## Re: For loop timer?
Posted 25 February 2018 - 01:05 PM
There is no well defined value. Each computer runs at a different rate. Your best bet is to loop and look at the elapsed time instead of some fixed loop value.
### #6 FlamingBurrito15
Reputation: 0
• Posts: 71
• Joined: 13-December 17
## Re: For loop timer?
Posted 25 February 2018 - 01:14 PM
Ok thanks for the help
### #7 Skydiver
• Code herder
Reputation: 6216
• Posts: 21,455
• Joined: 05-May 12
## Re: For loop timer?
Posted 25 February 2018 - 01:18 PM
+1 for using sleep(). Looping waiting for the elapsed time consumes for CPU power while sleep() tells the OS to reactivate your program after at least that amount of time has passed. | 848 | 2,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-26 | latest | en | 0.883008 |
https://archive.sap.com/discussions/thread/1972701 | 1,540,305,870,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516194.98/warc/CC-MAIN-20181023132213-20181023153713-00362.warc.gz | 603,625,818 | 5,902 | Search
Search
# Key Figure Duplication Problem in Webi Report
Hi Experts,
I have a problem in creating a formula. Please find below the information.
Every Opportunity can have more than one Product Id
Opp Id Prod Id Revenue
Oppid1 Prodid1 10
Oppid1 Prodid2 10
Oppid2 Prodid1 15
Oppid3 Prodid1 20
If you see in the above table Revenue for Opp1 is shown as 10+10 =20 where as it should be 10 only(Duplication Error).
I have to calculate Total Revenue :
Revenue
13.75 is shown instead of 45
If I calculate it as ,
Total Revenue = if(Count(product id) in(opportunity id)) >1; Revenue/ Count(product id) in(opportunity id);Revenue) , its giving me 13.75 (10101520)/4 but I want 45(1010)/21520. | 200 | 703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-43 | longest | en | 0.707322 |
https://2012books.lardbucket.org/books/advanced-algebra/s10-exponential-and-logarithmic-fu.html | 1,534,763,956,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221216333.66/warc/CC-MAIN-20180820101554-20180820121554-00411.warc.gz | 594,538,254 | 84,353 | This is “Exponential and Logarithmic Functions”, chapter 7 from the book Advanced Algebra (v. 1.0). For details on it (including licensing), click here.
Has this book helped you? Consider passing it on:
Creative Commons supports free culture from music to education. Their licenses helped make this book available to you.
DonorsChoose.org helps people like you help teachers fund their classroom projects, from art supplies to books to calculators.
# Chapter 7 Exponential and Logarithmic Functions
## 7.1 Composition and Inverse Functions
### Learning Objectives
1. Perform function composition.
2. Determine whether or not given functions are inverses.
3. Use the horizontal line test.
4. Find the inverse of a one-to-one function algebraically.
## Composition of Functions
In mathematics, it is often the case that the result of one function is evaluated by applying a second function. For example, consider the functions defined by $f(x)=x2$ and $g(x)=2x+5.$ First, g is evaluated where $x=−1$ and then the result is squared using the second function, f.
This sequential calculation results in 9. We can streamline this process by creating a new function defined by $f(g(x))$, which is explicitly obtained by substituting $g(x)$ into $f(x).$
$f(g(x))=f(2x+5)=(2x+5)2=4x2+20x+25$
Therefore, $f(g(x))=4x2+20x+25$ and we can verify that when $x=−1$ the result is 9.
$f(g(−1))=4(−1)2+20(−1)+25=4−20+25=9$
The calculation above describes composition of functionsApplying a function to the results of another function., which is indicated using the composition operatorThe open dot used to indicate the function composition $(f○g)(x)=f(g(x)).$ ($○$). If given functions f and g,
$(f○g)(x)=f(g(x)) Composition of Functions$
The notation $f○g$ is read, “f composed with g.” This operation is only defined for values, x, in the domain of g such that $g(x)$ is in the domain of f.
### Example 1
Given $f(x)=x2−x+3$ and $g(x)=2x−1$ calculate:
1. $(f○g)(x).$
2. $(g○f)(x).$
Solution:
1. Substitute g into f.
$(f○g)(x)=f(g(x))=f(2x−1)=(2x−1)2−(2x−1)+3=4x2−4x+1−2x+1+3=4x2−6x+5$
2. Substitute f into g.
$(g○f)(x)=g(f(x))=g(x2−x+3)=2(x2−x+3)−1=2x2−2x+6−1=2x2−2x+5$
1. $(f○g)(x)=4x2−6x+5$
2. $(g○f)(x)=2x2−2x+5$
The previous example shows that composition of functions is not necessarily commutative.
### Example 2
Given $f(x)=x3+1$ and $g(x)=3x−13$ find $(f○g)(4).$
Solution:
Begin by finding $(f○g)(x).$
$(f○g)(x)=f(g(x))=f(3x−13)=(3x−13)3+1=3x−1+1=3x$
Next, substitute 4 in for x.
$(f○g)(x)=3x(f○g)(4)=3(4)=12$
Answer: $(f○g)(4)=12$
Functions can be composed with themselves.
### Example 3
Given $f(x)=x2−2$ find $(f○f)(x).$
Solution:
$(f○f)(x)=f(f(x))=f(x2−2)=(x2−2)2−2=x4−4x2+4−2=x4−4x2+2$
Answer: $(f○f)(x)=x4−4x2+2$
Try this! Given $f(x)=2x+3$ and $g(x)=x−1$ find $(f○g)(5).$
## Inverse Functions
Consider the function that converts degrees Fahrenheit to degrees Celsius: $C(x)=59(x−32).$ We can use this function to convert 77°F to degrees Celsius as follows.
$C(77)=59(77−32)=59(45)=25$
Therefore, 77°F is equivalent to 25°C. If we wish to convert 25°C back to degrees Fahrenheit we would use the formula: $F(x)=95x+32.$
$F(25)=95(25)+32=45+32=77$
Notice that the two functions $C$ and $F$ each reverse the effect of the other.
This describes an inverse relationship. In general, f and g are inverse functions if,
$(f○g)(x)=f(g(x))=x for all x in the domain of g and(g○f)(x)=g(f(x))=x for all x in the domain of f.$
In this example,
$C(F(25))=C(77)=25F(C(77))=F(25)=77$
### Example 4
Verify algebraically that the functions defined by $f(x)=12x−5$ and $g(x)=2x+10$ are inverses.
Solution:
Compose the functions both ways and verify that the result is x.
$(f○g)(x)=f(g(x))=f(2x+10)=12(2x+10)−5=x+5−5=x ✓$ $(g○f)(x)=g(f(x))=g(12x−5)=2(12x−5)+10=x−10+10=x ✓$
Answer: Both $(f○g)(x)=(g○f)(x)=x$; therefore, they are inverses.
Next we explore the geometry associated with inverse functions. The graphs of both functions in the previous example are provided on the same set of axes below.
Note that there is symmetry about the line $y=x$; the graphs of f and g are mirror images about this line. Also notice that the point (20, 5) is on the graph of f and that (5, 20) is on the graph of g. Both of these observations are true in general and we have the following properties of inverse functions:
1. The graphs of inverse functions are symmetric about the line $y=x.$
2. If $(a,b)$ is on the graph of a function, then $(b,a)$ is on the graph of its inverse.
Furthermore, if g is the inverse of f we use the notation $g=f−1.$ Here $f−1$ is read, “f inverse,” and should not be confused with negative exponents. In other words, $f−1(x)≠1f(x)$ and we have,
$(f○f−1)(x)=f(f−1(x))=x and(f−1○f)(x)=f−1(f(x))=x$
### Example 5
Verify algebraically that the functions defined by $f(x)=1x−2$ and $f−1(x)=1x+2$ are inverses.
Solution:
Compose the functions both ways to verify that the result is x.
$(f○f−1)(x)=f(f−1(x))=f(1x+2)=1(1x+2)−2=x+21−2=x+2−2=x ✓$ $(f−1○f)(x)=f−1(f(x))=f−1(1x−2)=1(1x−2)+2=1 1x =x ✓$
Answer: Since $(f○f−1)(x)=(f−1○f)(x)=x$ they are inverses.
Recall that a function is a relation where each element in the domain corresponds to exactly one element in the range. We use the vertical line test to determine if a graph represents a function or not. Functions can be further classified using an inverse relationship. One-to-one functionsFunctions where each value in the range corresponds to exactly one value in the domain. are functions where each value in the range corresponds to exactly one element in the domain. The horizontal line testIf a horizontal line intersects the graph of a function more than once, then it is not one-to-one. is used to determine whether or not a graph represents a one-to-one function. If a horizontal line intersects a graph more than once, then it does not represent a one-to-one function.
The horizontal line represents a value in the range and the number of intersections with the graph represents the number of values it corresponds to in the domain. The function defined by $f(x)=x3$ is one-to-one and the function defined by $f(x)=|x|$ is not. Determining whether or not a function is one-to-one is important because a function has an inverse if and only if it is one-to-one. In other words, a function has an inverse if it passes the horizontal line test.
Note: In this text, when we say “a function has an inverse,” we mean that there is another function, $f−1$, such that $(f○f−1)(x)=(f−1○f)(x)=x.$
### Example 6
Determine whether or not the given function is one-to-one.
Solution:
Answer: The given function passes the horizontal line test and thus is one-to-one.
In fact, any linear function of the form $f(x)=mx+b$ where $m≠0$, is one-to-one and thus has an inverse. The steps for finding the inverse of a one-to-one function are outlined in the following example.
### Example 7
Find the inverse of the function defined by $f(x)=32x−5.$
Solution:
Before beginning this process, you should verify that the function is one-to-one. In this case, we have a linear function where $m≠0$ and thus it is one-to-one.
• Step 1: Replace the function notation $f(x)$ with y.
$f(x)=32x−5y=32x−5$
• Step 2: Interchange x and y. We use the fact that if $(x,y)$ is a point on the graph of a function, then $(y,x)$ is a point on the graph of its inverse.
$x=32y−5$
• Step 3: Solve for y.
$x=32y−5x+5=32y23⋅(x+5)=23⋅32y23x+103=y$
• Step 4: The resulting function is the inverse of f. Replace y with $f−1(x).$
$f−1(x)=23x+103$
• Step 5: Check.
$(f○f−1)(x)=f(f−1(x))=f(23x+103)=32(23x+103)−5=x+5−5=x ✓$ $(f−1○f)(x)=f−1(f(x))=f−1(32x−5)=23(32x−5)+103=x−103+103=x ✓$
Answer: $f−1(x)=23x+103$
If a function is not one-to-one, it is often the case that we can restrict the domain in such a way that the resulting graph is one-to-one. For example, consider the squaring function shifted up one unit, $g(x)=x2+1.$ Note that it does not pass the horizontal line test and thus is not one-to-one. However, if we restrict the domain to nonnegative values, $x≥0$, then the graph does pass the horizontal line test.
On the restricted domain, g is one-to-one and we can find its inverse.
### Example 8
Find the inverse of the function defined by $g(x)=x2+1$ where $x≥0.$
Solution:
Begin by replacing the function notation $g(x)$ with y.
$g(x)=x2+1y=x2+1 where x≥0$
Interchange x and y.
$x=y2+1 where y≥0$
Solve for y.
$x=y2+1x−1=y2±x−1=y$
Since $y≥0$ we only consider the positive result.
$y=x−1g−1(x)=x−1$
Answer: $g−1(x)=x−1.$ The check is left to the reader.
The graphs in the previous example are shown on the same set of axes below. Take note of the symmetry about the line $y=x.$
### Example 9
Find the inverse of the function defined by $f(x)=2x+1x−3.$
Solution:
Use a graphing utility to verify that this function is one-to-one. Begin by replacing the function notation $f(x)$ with y.
$f(x)=2x+1x−3y=2x+1x−3$
Interchange x and y.
$x=2y+1y−3$
Solve for y.
$x=2y+1y−3x(y−3)=2y+1xy−3x=2y+1$
Obtain all terms with the variable y on one side of the equation and everything else on the other. This will enable us to treat y as a GCF.
$xy−3x=2y+1xy−2y=3x+1y(x−2)=3x+1y=3x+1x−2$
Answer: $f−1(x)=3x+1x−2.$ The check is left to the reader.
Try this! Find the inverse of $f(x)=x+13−3.$
Answer: $f−1(x)=(x+3)3−1$
### Key Takeaways
• The composition operator ($○$) indicates that we should substitute one function into another. In other words, $(f○g)(x)=f(g(x))$ indicates that we substitute $g(x)$ into $f(x).$
• If two functions are inverses, then each will reverse the effect of the other. Using notation, $(f○g)(x)=f(g(x))=x$ and $(g○f)(x)=g(f(x))=x.$
• Inverse functions have special notation. If $g$ is the inverse of $f$, then we can write $g(x)=f−1(x).$ This notation is often confused with negative exponents and does not equal one divided by $f(x).$
• The graphs of inverses are symmetric about the line $y=x.$ If $(a,b)$ is a point on the graph of a function, then $(b,a)$ is a point on the graph of its inverse.
• If each point in the range of a function corresponds to exactly one value in the domain then the function is one-to-one. Use the horizontal line test to determine whether or not a function is one-to-one.
• A one-to-one function has an inverse, which can often be found by interchanging x and y, and solving for y. This new function is the inverse of the original function.
### Part A: Composition of Functions
Given the functions defined by f and g find $(f○g)(x)$ and $(g○f)(x).$
1. $f(x)=4x−1$, $g(x)=3x$
2. $f(x)=−2x+5$, $g(x)=2x$
3. $f(x)=3x−5$, $g(x)=x−4$
4. $f(x)=5x+1$, $g(x)=2x−3$
5. $f(x)=x2−x+1$, $g(x)=2x−1$
6. $f(x)=x2−3x−2$, $g(x)=x−2$
7. $f(x)=x2+3$, $g(x)=x2−5$
8. $f(x)=2x2$, $g(x)=x2−x$
9. $f(x)=8x3+5$, $g(x)=x−53$
10. $f(x)=27x3−1$, $g(x)=x+13$
11. $f(x)=1x+5$, $g(x)=1x$
12. $f(x)=1x−3$, $g(x)=3x+3$
13. $f(x)=5x$, $g(x)=3x−2$
14. $f(x)=2x$, $g(x)=4x+1$
15. $f(x)=12x$, $g(x)=x2+8$
16. $f(x)=2x−1$, $g(x)=1x+1$
17. $f(x)=1−x2x$, $g(x)=12x+1$
18. $f(x)=2xx+1$, $g(x)=x+1x$
Given the functions defined by $f(x)=3x2−2$, $g(x)=5x+1$, and $h(x)=x$, calculate the following.
1. $(f○g)(2)$
2. $(g○f)(−1)$
3. $(g○f)(0)$
4. $(f○g)(0)$
5. $(f○h)(3)$
6. $(g○h)(16)$
7. $(h○g)(35)$
8. $(h○f)(−3)$
Given the functions defined by $f(x)=x+33$, $g(x)=8x3−3$, and $h(x)=2x−1$, calculate the following.
1. $(f○g)(1)$
2. $(g○f)(−2)$
3. $(g○f)(0)$
4. $(f○g)(−2)$
5. $(f○h)(−1)$
6. $(h○g)(−12)$
7. $(h○f)(24)$
8. $(g○h)(0)$
Given the function, determine $(f○f)(x).$
1. $f(x)=3x−1$
2. $f(x)=25x+1$
3. $f(x)=x2+5$
4. $f(x)=x2−x+6$
5. $f(x)=x3+2$
6. $f(x)=x3−x$
7. $f(x)=1x+1$
8. $f(x)=x+12x$
### Part B: Inverse Functions
Are the given functions one-to-one? Explain.
1. $f(x)=x+1$
2. $g(x)=x2+1$
3. $h(x)=|x|+1$
4. $r(x)=x3+1$
5. $f(x)=x+1$
6. $g(x)=3$
Given the graph of a one-to-one function, graph its inverse.
Verify algebraically that the two given functions are inverses. In other words, show that $(f○f−1)(x)=x$ and $(f−1○f)(x)=x.$
1. $f(x)=3x−4$, $f−1(x)=x+43$
2. $f(x)=−5x+1$, $f−1(x)=1−x5$
3. $f(x)=−23x+1$, $f−1(x)=−32x+32$
4. $f(x)=4x−13$, $f−1(x)=14x+112$
5. $f(x)=x−8$, $f−1(x)=x2+8$, $x≥0$
6. $f(x)=6x3−3$, $f−1(x)=(x+3)36$
7. $f(x)=xx+1$, $f−1(x)=x1−x$
8. $f(x)=x−33x$, $f−1(x)=31−3x$
9. $f(x)=2(x−1)3+3$, $f−1(x)=1+x−323$
10. $f(x)=5x−13+4$, $f−1(x)=(x−4)3+15$
Find the inverses of the following functions.
1. $f(x)=5x$
2. $f(x)=12x$
3. $f(x)=2x+5$
4. $f(x)=−4x+3$
5. $f(x)=−23x+13$
6. $f(x)=−12x+34$
7. $g(x)=x2+5$, $x≥0$
8. $g(x)=x2−7$, $x≥0$
9. $f(x)=(x−5)2$, $x≥5$
10. $f(x)=(x+1)2$, $x≥−1$
11. $h(x)=3x3+5$
12. $h(x)=2x3−1$
13. $f(x)=(2x−3)3$
14. $f(x)=(x+4)3−1$
15. $g(x)=2x3+1$
16. $g(x)=1x3−2$
17. $f(x)=5x+1$
18. $f(x)=12x−9$
19. $f(x)=x+5x−5$
20. $f(x)=3x−42x−1$
21. $h(x)=x−510x$
22. $h(x)=9x+13x$
23. $g(x)=5x+23$
24. $g(x)=4x−33$
25. $f(x)=x−63−4$
26. $f(x)=2x+23+5$
27. $h(x)=x+15−3$
28. $h(x)=x−85+1$
29. $f(x)=mx+b$, $m≠0$
30. $f(x)=ax2+c$, $x≥0$
31. $f(x)=ax3+d$
32. $f(x)=a(x−h)2+k$, $x≥h$
Graph the function and its inverse on the same set of axes.
1. $f(x)=x+2$
2. $f(x)=23x−4$
3. $f(x)=−2x+2$
4. $f(x)=−13x+4$
5. $g(x)=x2−2$, $x≥0$
6. $g(x)=(x−2)2$, $x≥2$
7. $h(x)=x3+1$
8. $h(x)=(x+2)3−2$
9. $f(x)=2−x$
10. $f(x)=−x+1$
### Part C: Discussion Board
1. Is composition of functions associative? Explain.
2. Explain why $C(x)=59(x−32)$ and $F(x)=95x+32$ define inverse functions. Prove it algebraically.
3. Do the graphs of all straight lines represent one-to-one functions? Explain.
4. If the graphs of inverse functions intersect, then how can we find the point of intersection? Explain.
1. $(f○g)(x)=12x−1$; $(g○f)(x)=12x−3$
2. $(f○g)(x)=3x−17$; $(g○f)(x)=3x−9$
3. $(f○g)(x)=4x2−6x+3$; $(g○f)(x)=2x2−2x+1$
4. $(f○g)(x)=x4−10x2+28$; $(g○f)(x)=x4+6x2+4$
5. $(f○g)(x)=8x−35$; $(g○f)(x)=2x$
6. $(f○g)(x)=x5x+1$; $(g○f)(x)=x+5$
7. $(f○g)(x)=53x−2$; $(g○f)(x)=15x−2$
8. $(f○g)(x)=12x2+16;$ $(g○f)(x)=1+32x24x2$
9. $(f○g)(x)=x$; $(g○f)(x)=x$
10. 361
11. −9
12. 7
13. 2
14. 2
15. 21
16. 0
17. 5
18. $(f○f)(x)=9x−4$
19. $(f○f)(x)=x4+10x2+30$
20. $(f○f)(x)=x9+6x6+12x3+10$
21. $(f○f)(x)=x+1x+2$
1. No, fails the HLT
2. Yes, passes the HLT
3. Yes, its graph passes the HLT.
4. No, its graph fails the HLT.
5. Yes, its graph passes the HLT.
6. Proof
7. Proof
8. Proof
9. Proof
10. Proof
11. $f−1(x)=x5$
12. $f−1(x)=12x−52$
13. $f−1(x)=−32x+12$
14. $g−1(x)=x−5$
15. $f−1(x)=x+5$
16. $h−1(x)=x−533$
17. $f−1(x)=x3+32$
18. $g−1(x)=2−xx3$
19. $f−1(x)=5−xx$
20. $f−1(x)=5(x+1)x−1$
21. $h−1(x)=−510x−1$
22. $g−1(x)=x3−25$
23. $f−1(x)=(x+4)3+6$
24. $h−1(x)=(x+3)5−1$
25. $f−1(x)=x−bm$
26. $f−1(x)=x−da3$
## 7.2 Exponential Functions and Their Graphs
### Learning Objectives
1. Identify and evaluate exponential functions.
2. Sketch the graph of exponential functions and determine the domain and range.
3. Identify and graph the natural exponential function.
4. Apply the formulas for compound interest.
## Exponential Functions
At this point in our study of algebra we begin to look at transcendental functions or functions that seem to “transcend” algebra. We have studied functions with variable bases and constant exponents such as $x2$ or $y−3.$ In this section we explore functions with a constant base and variable exponents. Given a real number $b>0$ where $b≠1$ an exponential functionAny function with a definition of the form $f(x)=bx$ where $b>0$ and $b≠1.$ has the form,
$f(x)=bx Exponential Function$
For example, if the base b is equal to 2, then we have the exponential function defined by $f(x)=2x.$ Here we can see the exponent is the variable. Up to this point, rational exponents have been defined but irrational exponents have not. Consider $27$, where the exponent is an irrational number in the range,
$2.64<7<2.65$
We can use these bounds to estimate $27$,
$22.64<27<22.656.23<27<6.28$
Using rational exponents in this manner, an approximation of $27$ can be obtained to any level of accuracy. On a calculator,
$2^7≈6.26$
Therefore the domain of any exponential function consists of all real numbers $(−∞,∞).$ Choose some values for x and then determine the corresponding y-values.
$xyf(x)=2xSolutions−214y=2−2=122=14(−2, 14)−112y=2−1=121=12(−1, 12)01y=20=1(0, 1)12y=21=2(1, 2)24y=22=4(2, 4)76.26y=27≈6.26(2.65, 6.26)$
Because exponents are defined for any real number we can sketch the graph using a continuous curve through these given points:
It is important to point out that as x approaches negative infinity, the results become very small but never actually attain zero. For example,
$f(−5)=2−5=125≈0.03125f(−10)=2−10=1210≈0.0009766f(−15)=2−15=12−15≈.00003052$
This describes a horizontal asymptote at $y=0$, the x-axis, and defines a lower bound for the range of the function: $(0,∞).$
The base b of an exponential function affects the rate at which it grows. Below we have graphed $y=2x$, $y=3x$, and $y=10x$ on the same set of axes.
Note that all of these exponential functions have the same y-intercept, namely $(0,1).$ This is because $f(0)=b0=1$ for any function defined using the form $f(x)=bx.$ As the functions are read from left to right, they are interpreted as increasing or growing exponentially. Furthermore, any exponential function of this form will have a domain that consists of all real numbers $(−∞,∞)$ and a range that consists of positive values $(0,∞)$ bounded by a horizontal asymptote at $y=0.$
### Example 1
Sketch the graph and determine the domain and range: $f(x)=10x+5.$
Solution:
The base 10 is used often, most notably with scientific notation. Hence, 10 is called the common base. In fact, the exponential function $y=10x$ is so important that you will find a button $10x$ dedicated to it on most modern scientific calculators. In this example, we will sketch the basic graph $y=10x$ and then shift it up 5 units.
Note that the horizontal asymptote of the basic graph $y=10x$ was shifted up 5 units to $y=5$ (shown dashed). Take a minute to evaluate a few values of x with your calculator and convince yourself that the result will never be less than 5.
Domain: $(−∞,∞)$; Range: $(5,∞)$
Next consider exponential functions with fractional bases $0 For example, $f(x)=(12)x$ is an exponential function with base $b=12.$
$xyf(x)=(12)xSolutions−24f(12)=(12)−2=1−22−2=2212=4(−2, 4)−12f(12)=(12)−1=1−12−1=2111=2(−1, 2)01f(12)=(12)0=1(0, 1)112f(12)=(12)1=12(1, 12)214f(12)=(12)2=14(2, 14)$
Plotting points we have,
Reading the graph from left to right, it is interpreted as decreasing exponentially. The base affects the rate at which the exponential function decreases or decays. Below we have graphed $y=(12)x$, $y=(13)x$, and $y=(110)x$ on the same set of axes.
Recall that $x−1=1x$ and so we can express exponential functions with fractional bases using negative exponents. For example,
$g(x)=(12)x=1x2x=12x=2−x.$
Furthermore, given that $f(x)=2x$ we can see $g(x)=f(−x)=2−x$ and can consider g to be a reflection of f about the y-axis.
In summary, given $b>0$
And for both cases,
$Domain:(−∞,∞)Range:(0,∞)y-intercept:(0,1)Asymptote:y=0$
Furthermore, note that the graphs pass the horizontal line test and thus exponential functions are one-to-one. We use these basic graphs, along with the transformations, to sketch the graphs of exponential functions.
### Example 2
Sketch the graph and determine the domain and range: $f(x)=5−x−10.$
Solution:
Begin with the basic graph $y=5−x$ and shift it down 10 units.
The y-intercept is $(0,−9)$ and the horizontal asymptote is $y=−10.$
Domain: $(−∞,∞)$; Range: $(−10,∞)$
Note: Finding the x-intercept of the graph in the previous example is left for a later section in this chapter. For now, we are more concerned with the general shape of exponential functions.
### Example 3
Sketch the graph and determine the domain and range: $g(x)=−2x−3.$
Solution:
Begin with the basic graph $y=2x$ and identify the transformations.
$y=2x Basic graphy=−2x Reflection about the x-axisy=−2x−3 Shift right 3 units$
Note that the horizontal asymptote remains the same for all of the transformations. To finish we usually want to include the y-intercept. Remember that to find the y-intercept we set $x=0.$
$g(0)=−20−3=−2−3=−123=−18$
Therefore the y-intercept is $(0, −18)$.
Domain: $(−∞,∞)$; Range: $(−∞,0)$
Try this! Sketch the graph and determine the domain and range: $f(x)=2x−1+3.$
Domain: $(−∞,∞)$; Range: $(3,∞)$
## Natural Base e
Some numbers occur often in common applications. One such familiar number is pi (π), which we know occurs when working with circles. This irrational number has a dedicated button on most calculators $π$ and approximated to five decimal places, $π≈3.14159.$ Another important number e occurs when working with exponential growth and decay models. It is an irrational number and approximated to five decimal places, $e≈2.71828.$ This constant occurs naturally in many real-world applications and thus is called the natural base. Sometimes e is called Euler’s constant in honor of Leonhard Euler (pronounced “Oiler”).
Figure 7.1
Leonhard Euler (1707–1783)
In fact, the natural exponential function, $f(x)=ex$ is so important that you will find a button $ex$ dedicated to it on any modern scientific calculator. In this section, we are interested in evaluating the natural exponential function for given real numbers and sketching its graph. To evaluate the natural exponential function, defined by $f(x)=ex$ where $x=−2$ using a calculator, you may need to apply the shift button. On many scientific calculators the caret will display as follows,
$f(−2)=e^(−2)≈0.13534$
After learning how to use your particular calculator, you can now sketch the graph by plotting points. (Round off to the nearest hundredth.)
$xyf(x)=exSolutions−20.14f(−2)=e−2=0.14(−2, 0.14)−10.37f(−1)=e−1=0.37(−1, 0.37)01f(0)=e0=1(0, 1)12.72f(1)=e1=2.72(1, 2.72)27.39f(2)=e2=7.39(2, 7.39)$
Plot the points and sketch the graph.
Note that the function is similar to the graph of $y=3x.$ The domain consists of all real numbers and the range consists of all positive real numbers. There is an asymptote at $y=0$ and a y-intercept at $(0,1).$ We can use the transformations to sketch the graph of more complicated exponential functions.
### Example 4
Sketch the graph and determine the domain and range: $g(x)=ex+2−3.$
Solution:
Identify the basic transformations.
$y=ex Basic graphy=ex+2 Shift left 2 unitsy=ex+2−3 Shift down 3 units$
To determine the y-intercept set $x=0.$
$g(0)=e0+2−3=e2−3≈4.39$
Therefore the y-intercept is $(0,e2−3).$
Domain: $(−∞,∞)$; Range: $(−3,∞)$
Try this! Sketch the graph and determine the domain and range: $f(x)=e−x+2.$
Domain: $(−∞,∞)$; Range: $(2,∞)$
## Compound Interest Formulas
Exponential functions appear in formulas used to calculate interest earned in most regular savings accounts. Compound interest occurs when interest accumulated for one period is added to the principal investment before calculating interest for the next period. The amount accrued in this manner over time is modeled by the compound interest formulaA formula that gives the amount accumulated by earning interest on principal and interest over time: $A(t)=P(1+rn)nt.$:
$A(t)=P(1+rn)nt$
Here the amount A depends on the time t in years the principal P is accumulating compound interest at an annual interest rate r. The value n represents the number of times the interest is compounded in a year.
### Example 5
An investment of \$500 is made in a 6-year CD that earns $412$% annual interest that is compounded monthly. How much will the CD be worth at the end of the 6-year term?
Solution:
Here the principal $P=500$, the interest rate $r=412%=0.045$, and because the interest is compounded monthly, $n=12.$ The investment is modeled by the following,
$A(t)=500(1+0.04512)12t$
To determine the amount in the account after 6 years evaluate $A(6)$ and round off to the nearest cent.
$A(6)=500(1+0.04512)12(6)=500(1.00375)72=654.65$
Answer: The CD will be worth \$654.65 at the end of the 6-year term.
Next we explore the effects of increasing n in the formula. For the sake of clarity we let P and r equal 1 and calculate accordingly.
Annual compounding
$(1+1n)n$
Yearly ($n=1$)
$(1+11)1=2$
Semi-annually ($n=2$)
$(1+12)2=2.25$
Quarterly ($n=4$)
$(1+14)4≈2.44140$
Monthly ($n=12$)
$(1+112)12≈2.61304$
Weekly ($n=52$)
$(1+152)52≈2.69260$
Daily ($n=365$)
$(1+1365)365≈2.71457$
Hourly ($n=8760$)
$(1+18760)8760≈2.71813$
Continuing this pattern, as n increases to say compounding every minute or even every second, we can see that the result tends toward the natural base $e≈2.71828.$ Compounding interest every instant leads to the continuously compounding interest formulaA formula that gives the amount accumulated by earning continuously compounded interest: $A(t)=Pert.$,
$A(t)=Pert$
Here P represents the initial principal amount invested, r represents the annual interest rate, and t represents the time in years the investment is allowed to accrue continuously compounded interest.
### Example 6
An investment of \$500 is made in a 6-year CD that earns $412$% annual interest that is compounded continuously. How much will the CD be worth at the end of the 6-year term?
Solution:
Here the principal $P=500$, and the interest rate $r=412%=0.045.$ Since the interest is compounded continuously we will use the formula $A(t)=Pert.$ The investment is modeled by the following,
$A(t)=500e0.045t$
To determine the amount in the account after 6 years, evaluate $A(6)$ and round off to the nearest cent.
$A(6)=500e0.045(6)=500e0.27=654.98$
Answer: The CD will be worth \$654.98 at the end of the 6-year term.
Compare the previous two examples and note that compounding continuously may not be as beneficial as it sounds. While it is better to compound interest more often, the difference is not that profound. Certainly, the interest rate is a much greater factor in the end result.
Try this! How much will a \$1,200 CD, earning 5.2% annual interest compounded continuously, be worth at the end of a 10-year term?
### Key Takeaways
• Exponential functions have definitions of the form $f(x)=bx$ where $b>0$ and $b≠1.$ The domain consists of all real numbers $(−∞,∞)$ and the range consists of positive numbers $(0,∞).$ Also, all exponential functions of this form have a y-intercept of $(0,1)$ and are asymptotic to the x-axis.
• If the base of an exponential function is greater than 1 ($b>1$), then its graph increases or grows as it is read from left to right.
• If the base of an exponential function is a proper fraction ($0), then its graph decreases or decays as it is read from left to right.
• The number 10 is called the common base and the number e is called the natural base.
• The natural exponential function defined by $f(x)=ex$ has a graph that is very similar to the graph of $g(x)=3x.$
• Exponential functions are one-to-one.
### Part A: Exponential Functions
Evaluate.
1. $f(x)=3x$ where $f(−2)$, $f(0)$, and $f(2).$
2. $f(x)=10x$ where $f(−1)$, $f(0)$, and $f(1).$
3. $g(x)=(13)x$ where $g(−1)$, $g(0)$, and $g(3).$
4. $g(x)=(34)x$ where $g(−2)$, $g(−1)$, and $g(0).$
5. $h(x)=9−x$ where $h(−1)$, $h(0)$, and $h(12).$
6. $h(x)=4−x$ where $h(−1)$, $h(−12)$, and $h(0).$
7. $f(x)=−2x+1$ where $f(−1)$, $f(0)$, and $f(3).$
8. $f(x)=2−3x$ where $f(−1)$, $f(0)$, and $f(2).$
9. $g(x)=10−x+20$ where $g(−2)$, $g(−1)$, and $g(0).$
10. $g(x)=1−2−x$ where $g(−1)$, $g(0)$, and $g(1).$
Use a calculator to approximate the following to the nearest hundredth.
1. $f(x)=2x+5$ where $f(2.5).$
2. $f(x)=3x−10$ where $f(3.2).$
3. $g(x)=4x$ where $g(2).$
4. $g(x)=5x−1$ where $g(3).$
5. $h(x)=10x$ where $h(π).$
6. $h(x)=10x+1$ where $h(π3)$
7. $f(x)=10−x−2$ where $f(1.5).$
8. $f(x)=5−x+3$ where $f(1.3).$
9. $f(x)=(23)x+1$ where $f(−2.7).$
10. $f(x)=(35)−x−1$ where $f(1.4).$
Sketch the function and determine the domain and range. Draw the horizontal asymptote with a dashed line.
1. $f(x)=4x$
2. $g(x)=3x$
3. $f(x)=4x+2$
4. $f(x)=3x−6$
5. $f(x)=2x−2$
6. $f(x)=4x+2$
7. $f(x)=3x+1−4$
8. $f(x)=10x−4+2$
9. $h(x)=2x−3−2$
10. $h(x)=3x+2+4$
11. $f(x)=(14)x$
12. $h(x)=(13)x$
13. $f(x)=(14)x−2$
14. $h(x)=(13)x+2$
15. $g(x)=2−x−3$
16. $g(x)=3−x+1$
17. $f(x)=6−10−x$
18. $g(x)=5−4−x$
19. $f(x)=5−2x$
20. $f(x)=3−3x$
### Part B: Natural Base e
Find $f(−1)$, $f(0)$, and $f(32)$ for the given function. Use a calculator where appropriate to approximate to the nearest hundredth.
1. $f(x)=ex+2$
2. $f(x)=ex−4$
3. $f(x)=5−3ex$
4. $f(x)=e−x+3$
5. $f(x)=1+e−x$
6. $f(x)=3−2e−x$
7. $f(x)=e−2x+2$
8. $f(x)=e−x2−1$
Sketch the function and determine the domain and range. Draw the horizontal asymptote with a dashed line.
1. $f(x)=ex−3$
2. $f(x)=ex+2$
3. $f(x)=ex+1$
4. $f(x)=ex−3$
5. $f(x)=ex−2+1$
6. $f(x)=ex+2−1$
7. $g(x)=−ex$
8. $g(x)=e−x$
9. $h(x)=−ex+1$
10. $h(x)=−ex+3$
### Part C: Compound Interest Formulas
1. Jim invested \$750 in a 3-year CD that earns 4.2% annual interest that is compounded monthly. How much will the CD be worth at the end of the 3-year term?
2. Jose invested \$2,450 in a 4-year CD that earns 3.6% annual interest that is compounded semi-annually. How much will the CD be worth at the end of the 4-year term?
3. Jane has her \$5,350 savings in an account earning $358$% annual interest that is compounded quarterly. How much will be in the account at the end of 5 years?
4. Bill has \$12,400 in a regular savings account earning $423$% annual interest that is compounded monthly. How much will be in the account at the end of 3 years?
5. If \$85,200 is invested in an account earning 5.8% annual interest compounded quarterly, then how much interest is accrued in the first 3 years?
6. If \$124,000 is invested in an account earning 4.6% annual interest compounded monthly, then how much interest is accrued in the first 2 years?
7. Bill invested \$1,400 in a 3-year CD that earns 4.2% annual interest that is compounded continuously. How much will the CD be worth at the end of the 3-year term?
8. Brooklyn invested \$2,850 in a 5-year CD that earns 5.3% annual interest that is compounded continuously. How much will the CD be worth at the end of the 5-year term?
9. Omar has his \$4,200 savings in an account earning $438$% annual interest that is compounded continuously. How much will be in the account at the end of $212$ years?
10. Nancy has her \$8,325 savings in an account earning $578$% annual interest that is compounded continuously. How much will be in the account at the end of $512$ years?
11. If \$12,500 is invested in an account earning 3.8% annual interest compounded continuously, then how much interest is accrued in the first 10 years?
12. If \$220,000 is invested in an account earning 4.5% annual interest compounded continuously, then how much interest is accrued in the first 2 years?
13. The population of a certain small town is growing according to the function $P(t)=12,500(1.02)t$ where t represents time in years since the last census. Use the function to determine the population on the day of the census (when t = 0) and estimate the population in 6 years from that time.
14. The population of a certain small town is decreasing according to the function $P(t)=22,300(0.982)t$ where t represents time in years since the last census. Use the function to determine the population on the day of the census (when t = 0) and estimate the population in 6 years from that time.
15. The decreasing value, in dollars, of a new car is modeled by the formula $V(t)=28,000(0.84)t$ where t represents the number of years after the car was purchased. Use the formula to determine the value of the car when it was new (t = 0) and the value after 4 years.
16. The number of unique visitors to the college website can be approximated by the formula $N(t)=410(1.32)t$ where t represents the number of years after 1997 when the website was created. Approximate the number of unique visitors to the college website in the year 2020.
17. If left unchecked, a new strain of flu virus can spread from a single person to others very quickly. The number of people affected can be modeled using the formula $P(t)=e0.22t$ where t represents the number of days the virus is allowed to spread unchecked. Estimate the number of people infected with the virus after 30 days and after 60 days.
18. If left unchecked, a population of 24 wild English rabbits can grow according to the formula $P(t)=24e0.19t$ where the time t is measured in months. How many rabbits would be present $312$ years later?
19. The population of a certain city in 1975 was 65,000 people and was growing exponentially at an annual rate of 1.7%. At the time, the population growth was modeled by the formula $P(t)=65,000e0.017t$ where t represented the number of years since 1975. In the year 2000, the census determined that the actual population was 104,250 people. What population did the model predict for the year 2000 and what was the actual error?
20. Because of radioactive decay, the amount of a 10 milligram sample of Iodine-131 decreases according to the formula $A(t)=10e−0.087t$ where t represents time measured in days. How much of the sample remains after 10 days?
21. The number of cells in a bacteria sample is approximated by the logistic growth model $N(t)=1.2×1051+9e−0.32t$ where t represents time in hours. Determine the initial number of cells and then determine the number of cells 6 hours later.
22. The market share of a product, as a percentage, is approximated by the formula $P(t)=1002+e−0.44t$ where t represents the number of months after an aggressive advertising campaign is launched. By how much can we expect the market share to increase after the first three months of advertising?
### Part D: Discussion Board
1. Why is b = 1 excluded as a base in the definition of exponential functions? Explain.
2. Explain why an exponential function of the form $y=bx$ can never be negative.
3. Research and discuss the derivation of the compound interest formula.
4. Research and discuss the logistic growth model. Provide a link to more information on this topic.
5. Research and discuss the life and contributions of Leonhard Euler.
1. $f(−2)=19$, $f(0)=1$, $f(2)=9$
2. $g(−1)=3$, $g(0)=1$, $g(3)=127$
3. $h(−1)=9$, $h(0)=1$, $h(12)=13$
4. $f(−1)=12$, $f(0)=0$, $f(3)=−7$
5. $g(−2)=120$, $g(−1)=30$, $g(0)=21$
6. 10.66
7. 7.10
8. 1385.46
9. −1.97
10. 3.99
11. Domain: $(−∞,∞)$; Range: $(0,∞)$
12. Domain: $(−∞,∞)$; Range: $(2,∞)$
13. Domain: $(−∞,∞)$; Range: $(0,∞)$
14. Domain: $(−∞,∞)$; Range: $(−4,∞)$
15. Domain: $(−∞,∞)$; Range: $(−2,∞)$
16. Domain: $(−∞,∞)$; Range: $(0,∞)$
17. Domain: $(−∞,∞)$; Range: $(−2,∞)$
18. Domain: $(−∞,∞)$; Range: $(−3,∞)$
19. Domain: $(−∞,∞)$; Range: $(−∞,6)$
20. Domain: $(−∞,∞)$; Range: $(−∞,5)$
1. $f(−1)≈2.37$, $f(0)=3$, $f(32)≈6.48$
2. $f(−1)≈3.90$, $f(0)=2$, $f(32)≈−8.45$
3. $f(−1)≈3.72$, $f(0)=2$, $f(32)≈1.22$
4. $f(−1)≈9.39$, $f(0)=3$, $f(32)≈2.05$
5. Domain: $(−∞,∞)$; Range: $(−3,∞)$
6. Domain: $(−∞,∞)$; Range: $(0,∞)$
7. Domain: $(−∞,∞)$; Range: $(1,∞)$
8. Domain: $(−∞,∞)$; Range: $(−∞,0)$
9. Domain: $(−∞,∞)$; Range: $(−∞,0)$
1. \$850.52
2. \$6,407.89
3. \$16,066.13
4. \$1,588.00
5. \$4,685.44
6. \$5,778.56
7. Initial population: 12,500; Population 6 years later: 14,077
8. New: \$28,000; In 4 years: \$13,940.40
9. After 30 days: 735 people; After 60 days: 540,365 people
10. Model: 99,423 people; error: 4,827 people
11. Initially there are 12,000 cells and 6 hours later there are 51,736 cells.
## 7.3 Logarithmic Functions and Their Graphs
### Learning Objectives
1. Define and evaluate logarithms.
2. Identify the common and natural logarithm.
3. Sketch the graph of logarithmic functions.
## Definition of the Logarithm
We begin with the exponential function defined by $f(x)=2x$ and note that it passes the horizontal line test.
Therefore it is one-to-one and has an inverse. Reflecting $y=2x$ about the line $y=x$ we can sketch the graph of its inverse. Recall that if $(x,y)$ is a point on the graph of a function, then $(y,x)$ will be a point on the graph of its inverse.
To find the inverse algebraically, begin by interchanging x and y and then try to solve for y.
$f(x)=2xy=2x ⇒ x=2y$
We quickly realize that there is no method for solving for y. This function seems to “transcend” algebra. Therefore, we define the inverse to be the base-2 logarithm, denoted $log2 x.$ The following are equivalent:
$y=log2 x ⇔ x=2y$
This gives us another transcendental function defined by $f−1(x)=log2 x$, which is the inverse of the exponential function defined by $f(x)=2x.$
The domain consists of all positive real numbers $(0,∞)$ and the range consists of all real numbers $(−∞,∞).$ John Napier is widely credited for inventing the term logarithm.
Figure 7.2
John Napier (1550–1617)
In general, given base $b>0$ where $b≠1$, the logarithm base bThe exponent to which the base b is raised in order to obtain a specific value. In other words, $y=logb x$ is equivalent to $by=x.$ is defined as follows:
$y=logb x if and only if x=by$
Use this definition to convert logarithms to exponential form and back.
Logarithmic Form
Exponential Form
$log2 16=4$
$24=16$
$log5 25=2$
$52=25$
$log6 1=0$
$60=1$
$log3 3=12$
$31/2=3$
$log7 (149)=−2$
$7−2=149$
It is useful to note that the logarithm is actually the exponent y to which the base b is raised to obtain the argument x.
### Example 1
Evaluate:
1. $log5 125$
2. $log2 (18)$
3. $log4 2$
4. $log11 1$
Solution:
1. $log5 125=3$ because $53=125.$
2. $log2 (18)=−3$ because $2−3=123=18.$
3. $log4 2=12$ because $41/2=4=2.$
4. $log11 1=0$ because $110=1.$
Note that the result of a logarithm can be negative or even zero. However, the argument of a logarithm is not defined for negative numbers or zero:
$log2 (−4)=? ⇒ 2?=−4log2 (0)=? ⇒ 2?=0$
There is no power of two that results in −4 or 0. Negative numbers and zero are not in the domain of the logarithm. At this point it may be useful to go back and review all of the rules of exponents.
### Example 2
Find x:
1. $log7 x=2$
2. $log16 x=12$
3. $log1/2 x=−5$
Solution:
Convert each to exponential form and then simplify using the rules of exponents.
1. $log7 x=2$ is equivalent to $72=x$ and thus $x=49.$
2. $log16 x=12$ is equivalent to $161/2=x$ or $16=x$ and thus $x=4.$
3. $log1/2 x=−5$ is equivalent to $(12)−5=x$ or $25=x$ and thus $x=32$
Try this! Evaluate: $log5 (153)$.
Answer: $−13$
## The Common and Natural Logarithm
A logarithm can have any positive real number, other than 1, as its base. If the base is 10, the logarithm is called the common logarithmThe logarithm base 10, denoted $log x.$.
$f(x)=log10 x=log x Common logarithm$
When a logarithm is written without a base it is assumed to be the common logarithm. (Note: This convention varies with respect to the subject in which it appears. For example, computer scientists often let $log x$ represent the logarithm base 2.)
### Example 3
Evaluate:
1. $log 105$
2. $log 10$
3. $log 0.01$
Solution:
1. $log 105=5$ because $105=105.$
2. $log 10=12$ because $101/2=10.$
3. $log 0.01=log (1100)=−2$ because $10−2=1102=1100=0.01.$
The result of a logarithm is not always apparent. For example, consider $log 75.$
$log 10=1log 75=?log 100=2$
We can see that the result of $log 75$ is somewhere between 1 and 2. On most scientific calculators there is a common logarithm button $LOG.$ Use it to find the $log 75$ as follows:
$LOG75=1.87506$
Therefore, rounded off to the nearest thousandth, $log 75≈1.875.$ As a check, we can use a calculator to verify that $10^1.875≈75.$
If the base of a logarithm is $e$, the logarithm is called the natural logarithmThe logarithm base e, denoted $ln x.$.
$f(x)=loge x=ln x Natural logarithm$
The natural logarithm is widely used and is often abbreviated $ln x.$
### Example 4
Evaluate:
1. $ln e$
2. $ln e23$
3. $ln (1e4)$
Solution:
1. $ln e=1$ because $ln e=loge e=1$ and $e1=e.$
2. $ln (e23)=23$ because $e2/3=e23.$
3. $ln (1e4)=−4$ because $e−4=1e4$
On a calculator you will find a button for the natural logarithm $LN.$
$LN 75 = 4.317488$
Therefore, rounded off to the nearest thousandth, $ln (75)≈4.317.$ As a check, we can use a calculator to verify that $e^4.317≈75.$
### Example 5
Find x. Round answers to the nearest thousandth.
1. $log x=3.2$
2. $ln x=−4$
3. $log x=−23$
Solution:
Convert each to exponential form and then use a calculator to approximate the answer.
1. $log x=3.2$ is equivalent to $103.2=x$ and thus $x≈1584.893.$
2. $ln x=−4$ is equivalent to $e−4=x$ and thus $x≈0.018.$
3. $log x=−23$ is equivalent to $10−2/3=x$ and thus $x≈0.215$
Try this! Evaluate: $ln (1e)$.
Answer: $−12$
## Graphing Logarithmic Functions
We can use the translations to graph logarithmic functions. When the base $b>1$, the graph of $f(x)=logb x$ has the following general shape:
The domain consists of positive real numbers, $(0,∞)$ and the range consists of all real numbers, $(−∞,∞).$ The y-axis, or $x=0$, is a vertical asymptote and the x-intercept is $(1,0).$ In addition, $f(b)=logb b=1$ and so $(b,1)$ is a point on the graph no matter what the base is.
### Example 6
Sketch the graph and determine the domain and range: $f(x)=log3 (x+4)−1.$
Solution:
Begin by identifying the basic graph and the transformations.
$y=log3 x Basic graphy=log3 (x+4) Shift left 4 units.y=log3 (x+4)−1 Shift down 1 unit.$
Notice that the asymptote was shifted 4 units to the left as well. This defines the lower bound of the domain. The final graph is presented without the intermediate steps.
Domain: $(−4,∞)$; Range: $(−∞,∞)$
Note: Finding the intercepts of the graph in the previous example is left for a later section in this chapter. For now, we are more concerned with the general shape of logarithmic functions.
### Example 7
Sketch the graph and determine the domain and range: $f(x)=−log (x−2).$
Solution:
Begin by identifying the basic graph and the transformations.
$y=log x Basic graphy=−log x Reflection about the x-axisy=−log (x−2) Shift right 2 units.$
Here the vertical asymptote was shifted two units to the right. This defines the lower bound of the domain.
Domain: $(2,∞)$; Range: $(−∞,∞)$
Try this! Sketch the graph and determine the domain and range: $g(x)=ln (−x)+2.$
Domain: (−∞, 0); Range: (−∞, ∞)
Next, consider exponential functions with fractional bases, such as the function defined by $f(x)=(12)x.$ The domain consists of all real numbers. Choose some values for x and then find the corresponding y-values.
$xySolutions−24f(−2)=(12)−2=22=4(−2, 4)−12f(−1)=(12)−1=21=2(−1, 2)01f(0)=(12)0=1(0, 1)112f(1)=(12)1=12(1, 12)214f(2)=(12)2=14(2, 14)$
Use these points to sketch the graph and note that it passes the horizontal line test.
Therefore this function is one-to-one and has an inverse. Reflecting the graph about the line $y=x$ we have:
which gives us a picture of the graph of $f−1(x)=log1/2 x.$ In general, when the base $b>1$, the graph of the function defined by $g(x)=log1/b x$ has the following shape.
The domain consists of positive real numbers, $(0,∞)$ and the range consists of all real numbers, $(−∞,∞).$ The y-axis, or $x=0$, is a vertical asymptote and the x-intercept is $(1,0).$ In addition, $f(b)=log1/b b=−1$ and so $(b,−1)$ is a point on the graph.
### Example 8
Sketch the graph and determine the domain and range: $f(x)=log1/3 (x+3)+2.$
Solution:
Begin by identifying the basic graph and the transformations.
$y=log1/3 x Basic graphy=log1/3 (x+3) Shift left 3 units.y=log1/3 (x+3)+2 Shift up 2 units.$
In this case the shift left 3 units moved the vertical asymptote to $x=−3$ which defines the lower bound of the domain.
Domain: $(−3,∞)$; Range: $(−∞,∞)$
In summary, if $b>1$
And for both cases,
$Domain:(0,∞)Range:(−∞,∞)x-intercept:(1,0)Asymptote:x=0$
Try this! Sketch the graph and determine the domain and range: $f(x)=log1/3 (x−1).$
### Key Takeaways
• The base-b logarithmic function is defined to be the inverse of the base-b exponential function. In other words, $y=logb x$ if and only if $by=x$ where $b>0$ and $b≠1.$
• The logarithm is actually the exponent to which the base is raised to obtain its argument.
• The logarithm base 10 is called the common logarithm and is denoted $log x.$
• The logarithm base e is called the natural logarithm and is denoted $ln x.$
• Logarithmic functions with definitions of the form $f(x)=logb x$ have a domain consisting of positive real numbers $(0,∞)$ and a range consisting of all real numbers $(−∞,∞).$ The y-axis, or $x=0$, is a vertical asymptote and the x-intercept is $(1,0).$
• To graph logarithmic functions we can plot points or identify the basic function and use the transformations. Be sure to indicate that there is a vertical asymptote by using a dashed line. This asymptote defines the boundary of the domain.
### Part A: Definition of the Logarithm
Evaluate.
1. $log3 9$
2. $log7 49$
3. $log4 4$
4. $log5 1$
5. $log5 625$
6. $log3 243$
7. $log2 (116)$
8. $log3 (19)$
9. $log5 (1125)$
10. $log2 (164)$
11. $log4 410$
12. $log9 95$
13. $log5 53$
14. $log2 2$
15. $log7 (17)$
16. $log9 (193)$
17. $log1/2 4$
18. $log1/3 27$
19. $log2/3 (23)$
20. $log3/4 (916)$
21. $log25 5$
22. $log8 2$
23. $log4 (12)$
24. $log27 (13)$
25. $log1/9 1$
26. $log3/5 (53)$
Find x.
1. $log3 x=4$
2. $log2 x=5$
3. $log5 x=−3$
4. $log6 x=−2$
5. $log12 x=0$
6. $log7 x=−1$
7. $log1/4 x=−2$
8. $log2/5 x=2$
9. $log1/9 x=12$
10. $log1/4 x=32$
11. $log1/3 x=−1$
12. $log1/5 x=0$
### Part B: The Common and Natural Logarithm
Evaluate. Round off to the nearest hundredth where appropriate.
1. $log 1000$
2. $log 100$
3. $log 0.1$
4. $log 0.0001$
5. $log 162$
6. $log 23$
7. $log 0.025$
8. $log 0.235$
9. $ln e4$
10. $ln 1$
11. $ln (1e)$
12. $ln (1e5)$
13. $ln (25)$
14. $ln (100)$
15. $ln (0.125)$
16. $ln (0.001)$
Find x. Round off to the nearest hundredth.
1. $log x=2.5$
2. $log x=1.8$
3. $log x=−1.22$
4. $log x=−0.8$
5. $ln x=3.1$
6. $ln x=1.01$
7. $ln x=−0.69$
8. $ln x=−1$
Find a without using a calculator.
1. $log3 (127)=a$
2. $ln e=a$
3. $log2 a=8$
4. $log2 25=a$
5. $log1012 =a$
6. $ln a=9$
7. $log1/8 (164)=a$
8. $log6 a=−3$
9. $ln a=15$
10. $log4/9 (23)=a$
In 1935 Charles Richter developed a scale used to measure earthquakes on a seismograph. The magnitude M of an earthquake is given by the formula,
$M=log (II0)$
Here I represents the intensity of the earthquake as measured on the seismograph 100 km from the epicenter and $I0$ is the minimum intensity used for comparison. For example, if an earthquake intensity is measured to be 100 times that of the minimum, then $I=100I0$ and
$M=log (100I0I0)=log (100)=2$
The earthquake would be said to have a magnitude 2 on the Richter scale. Determine the magnitudes of the following intensities on the Richter scale. Round off to the nearest tenth.
1. I is 3 million times that of the minimum intensity.
2. I is 6 million times that of the minimum intensity.
3. I is the same as the minimum intensity.
4. I is 30 million times that of the minimum intensity.
In chemistry, pH is a measure of acidity and is given by the formula,
$pH=−log (H+)$
Here $H+$ represents the hydrogen ion concentration (measured in moles of hydrogen per liter of solution.) Determine the pH given the following hydrogen ion concentrations.
1. Pure water: $H+=0.0000001$
2. Blueberry: $H+=0.0003162$
3. Lemon Juice: $H+=0.01$
4. Battery Acid: $H+=0.1$
### Part C: Graphing Logarithmic Functions
Sketch the function and determine the domain and range. Draw the vertical asymptote with a dashed line.
1. $f(x)=log2 (x+1)$
2. $f(x)=log3 (x−2)$
3. $f(x)=log2 x−2$
4. $f(x)=log3 x+3$
5. $f(x)=log2 (x−2)+4$
6. $f(x)=log3 (x+1)−2$
7. $f(x)=−log2 x+1$
8. $f(x)=−log3 (x+3)$
9. $f(x)=log2 (−x)+1$
10. $f(x)=2−log3 (−x)$
11. $f(x)=log x+5$
12. $f(x)=log x−1$
13. $f(x)=log (x+4)−8$
14. $f(x)=log (x−5)+10$
15. $f(x)=−log (x+2)$
16. $f(x)=−log (x−1)+2$
17. $f(x)=ln (x−3)$
18. $f(x)=ln x+3$
19. $f(x)=ln (x−2)+4$
20. $f(x)=ln (x+5)$
21. $f(x)=2−ln x$
22. $f(x)=−ln (x−1)$
23. $f(x)=log1/2 x$
24. $f(x)=log1/3 x+2$
25. $f(x)=log1/2 (x−2)$
26. $f(x)=log1/3 (x+1)−1$
27. $f(x)=2−log1/4 x$
28. $f(x)=1+log1/4 (−x)$
29. $f(x)=1−log1/3 (x−2)$
30. $f(x)=1+log1/2 (−x)$
### Part D: Discussion Board
1. Research and discuss the origins and history of the logarithm. How did students work with them before the common availability of calculators?
2. Research and discuss the history and use of the Richter scale. What does each unit on the Richter scale represent?
3. Research and discuss the life and contributions of John Napier.
1. 2
2. 1
3. 4
4. −4
5. −3
6. 10
7. $13$
8. $−12$
9. −2
10. 1
11. $12$
12. $−12$
13. 0
14. 81
15. $1125$
16. 1
17. 16
18. $13$
19. 3
1. 3
2. −1
3. 2.21
4. −1.60
5. 4
6. −1
7. 3.22
8. −2.08
9. 316.23
10. 0.06
11. 22.20
12. 0.50
13. −3
14. 256
15. 12
16. 2
17. $e5$
18. 6.5
19. 0
20. 7
21. 2
1. Domain: $(−1,∞)$; Range: $(−∞,∞)$
2. Domain: $(0,∞)$; Range: $(−∞,∞)$
3. Domain: $(2,∞)$; Range: $(−∞,∞)$
4. Domain: $(0,∞)$; Range: $(−∞,∞)$
5. Domain: $(−∞,0)$; Range: $(−∞,∞)$
6. Domain: $(0,∞)$; Range: $(−∞,∞)$
7. Domain: $(−4,∞)$; Range: $(−∞,∞)$
8. Domain: $(−2,∞)$; Range: $(−∞,∞)$
9. Domain: $(3,∞)$; Range: $(−∞,∞)$
10. Domain: $(2,∞)$; Range: $(−∞,∞)$
11. Domain: $(0,∞)$; Range: $(−∞,∞)$
12. Domain: $(0,∞)$; Range: $(−∞,∞)$
13. Domain: $(2,∞)$; Range: $(−∞,∞)$
14. Domain: $(0,∞)$; Range: $(−∞,∞)$
15. Domain: $(2, ∞)$; Range: $(−∞, ∞)$
## 7.4 Properties of the Logarithm
### Learning Objectives
1. Apply the inverse properties of the logarithm.
2. Expand logarithms using the product, quotient, and power rule for logarithms.
3. Combine logarithms into a single logarithm with coefficient 1.
## Logarithms and Their Inverse Properties
Recall the definition of the base-b logarithm: given $b>0$ where $b≠1$,
$y=logb x if and only if x=by$
Use this definition to convert logarithms to exponential form. Doing this, we can derive a few properties:
$logb 1=0 because b0=1logb b=1 because b1=blogb (1b)=−1 because b−1=1b$
### Example 1
Evaluate:
1. $log 1$
2. $ln e$
3. $log5 (15)$
Solution:
1. When the base is not written, it is assumed to be 10. This is the common logarithm,
$log 1=log10 1=0$
2. The natural logarithm, by definition, has base e,
$ln e=loge e=1$
3. Because $5−1=15$ we have,
$log5 (15)=−1$
Furthermore, consider fractional bases of the form $1/b$ where $b>1.$
$log1/b b=−1 because (1b)−1=1−1b−1=b1=b$
### Example 2
Evaluate:
1. $log1/4 4$
2. $log2/3 (32)$
Solution:
1. $log1/4 4=−1$ because $(14)−1=4$
2. $log2/3 (32)=−1$ because $(23)−1=32$
Given an exponential function defined by $f(x)=bx$, where $b>0$ and $b≠1$, its inverse is the base-b logarithm, $f−1(x)=logb x.$ And because $f(f−1(x))=x$ and $f−1(f(x))=x$, we have the following inverse properties of the logarithmGiven $b>0$ we have $logb bx=x$ and $blogb x=x$ when $x>0.$:
$f−1(f(x))=logb bx=xandf(f−1(x))=blogb x=x ,x>0$
Since $f−1(x)=logb x$ has a domain consisting of positive values $(0,∞)$, the property $blogb x=x$ is restricted to values where $x>0.$
### Example 3
Evaluate:
1. $log5 625$
2. $5log5 3$
3. $eln 5$
Solution:
Apply the inverse properties of the logarithm.
1. $log5 625=log5 54=4$
2. $5log5 3=3$
3. $eln 5=5$
In summary, when $b>0$ and $b≠1$, we have the following properties:
$logb 1=0$ $logb b=1$ $log1/b b=−1$ $logb (1b)=−1$ $logb bx=x$ $blogb x=x$, $x>0$
Try this! Evaluate: $log 0.00001$
## Product, Quotient, and Power Properties of Logarithms
In this section, three very important properties of the logarithm are developed. These properties will allow us to expand our ability to solve many more equations. We begin by assigning u and v to the following logarithms and then write them in exponential form:
$logb x=u ⇒ bu=xlogb y=v ⇒ bv=y$
Substitute $x=bu$ and $y=bv$ into the logarithm of a product $logb (xy)$ and the logarithm of a quotient $logb (xy).$ Then simplify using the rules of exponents and the inverse properties of the logarithm.
Logarithm of a Product
Logarithm of a Quotient
$logb (xy)=logb (bubv)=logb bu+v=u+v=logb x+logb y$
$logb (xy)=logb (bubv)=logb bu−v=u−v=logb x−logb y$
This gives us two essential properties: the product property of logarithms$logb (xy)=logb x+logb y;$ the logarithm of a product is equal to the sum of the logarithm of the factors.,
$logb (xy)=logb x+logb y$
and the quotient property of logarithms$logb (xy)=logb x−logb y;$ the logarithm of a quotient is equal to the difference of the logarithm of the numerator and the logarithm of the denominator.,
$logb (xy)=logb x−logb y$
In words, the logarithm of a product is equal to the sum of the logarithm of the factors. Similarly, the logarithm of a quotient is equal to the difference of the logarithm of the numerator and the logarithm of the denominator.
### Example 4
Write as a sum: $log2 (8x).$
Solution:
Apply the product property of logarithms and then simplify.
$log2 (8x)=log2 8+log2 x=log2 23+log2 x=3+log2 x$
Answer: $3+log2 x$
### Example 5
Write as a difference: $log (x10)$.
Solution:
Apply the quotient property of logarithms and then simplify.
$log (x10)=log x−log 10=log x−1$
Answer: $log x−1$
Next we begin with $logb x=u$ and rewrite it in exponential form. After raising both sides to the nth power, convert back to logarithmic form, and then back substitute.
$logb x=u ⇒ bu=x(bu)n=(x )nlogb xn=nu ⇐ bnu=xnlogb xn=nlogb x$
This leads us to the power property of logarithms$logb xn=nlogb x$; the logarithm of a quantity raised to a power is equal to that power times the logarithm of the quantity.,
$logb xn=nlogb x$
In words, the logarithm of a quantity raised to a power is equal to that power times the logarithm of the quantity.
### Example 6
Write as a product:
1. $log2 x4$
2. $log5 (x).$
Solution:
1. Apply the power property of logarithms.
$log2 x4=4log2 x$
2. Recall that a square root can be expressed using rational exponents, $x=x1/2.$ Make this replacement and then apply the power property of logarithms.
$log5 (x)=log5 x1/2=12log5 x$
In summary,
Product property of logarithms $logb (xy)=logb x+logb y$ Quotient property of logarithms $logb (xy)=logb x−logb y$ Power property of logarithms $logb xn=nlogb x$
We can use these properties to expand logarithms involving products, quotients, and powers using sums, differences and coefficients. A logarithmic expression is completely expanded when the properties of the logarithm can no further be applied.
$Caution:$ It is important to point out the following:
$log (xy)≠log x⋅log y and log (xy)≠log xlog y$
### Example 7
Expand completely: $ln (2x3).$
Solution:
Recall that the natural logarithm is a logarithm base e, $ln x=loge x.$ Therefore, all of the properties of the logarithm apply.
$ln (2x3)=ln 2+ln x3 Product rule for logarithms=ln 2+3ln x Power rule for logarithms$
Answer: $ln 2+3ln x$
### Example 8
Expand completely: $log 10xy23.$
Solution:
Begin by rewriting the cube root using the rational exponent $13$ and then apply the properties of the logarithm.
$log 10xy23=log (10xy2)1/3=13log (10xy2)=13(log 10+log x+log y2)=13(1+log x+2log y)=13+13log x+23log y$
Answer: $13+13log x+23log y$
### Example 9
Expand completely: $log2 ( (x+1)25y)$.
Solution:
When applying the product property to the denominator, take care to distribute the negative obtained from applying the quotient property.
$log2 ( (x+1)25y)=log2 (x+1)2−log2 (5y)=log2 (x+1)2−(log2 5+log2 y) Distribute.=log2 (x+1)2−log2 5−log2 y=2log2 (x+1)−log2 5−log2 y$
Answer: $2log2 (x+1)−log2 5−log2 y$
Caution: There is no rule that allows us to expand the logarithm of a sum or difference. In other words,
$log (x±y)≠log x±log y$
Try this! Expand completely: $ln (5y4x)$.
Answer: $ln 5+4ln y−12ln x$
### Example 10
Given that $log2 x=a$, $log2 y=b$, and that $log2 z=c$, write the following in terms of a, b and c:
a. $log2 (8x2y)$
b. $log2 (2x4z)$
Solution:
1. Begin by expanding using sums and coefficients and then replace a and b with the appropriate logarithm.
$log2 (8x2y)=log2 8+log2 x2+log2 y=log2 8+2log2 x+log2 y=3+2a+b$
2. Expand and then replace a, b, and c where appropriate.
$log2 (2x4z)=log2 (2x4)−log2 z1/2=log2 2+log2 x4−log2 z1/2=log2 2+4log2 x−12log2 z=1+4a−12b$
Next we will condense logarithmic expressions. As we will see, it is important to be able to combine an expression involving logarithms into a single logarithm with coefficient 1. This will be one of the first steps when solving logarithmic equations.
### Example 11
Write as a single logarithm with coefficient $1: 3log3 x−log3 y+2log3 5.$
Solution:
Begin by rewriting all of the logarithmic terms with coefficient 1. Use the power rule to do this. Then use the product and quotient rules to simplify further.
$3log3x−log3y+2log35={log3x3−log3y}+log352 quotient property={log3(x3y)+log325} product property=log3(x3y⋅25)=log3(25x3y)$
Answer: $log3 (25x3y)$
### Example 12
Write as a single logarithm with coefficient $1: 12ln x−3ln y−ln z.$
Solution:
Begin by writing the coefficients of the logarithms as powers of their argument, after which we will apply the quotient rule twice working from left to right.
$12ln x−3ln y−ln z=ln x1/2−ln y3−ln z=ln (x1/2y3)−ln z=ln (x1/2y3÷z)=ln (x1/2y3⋅1z)=ln (x1/2y3z) or =ln (xy3z)$
Answer: $ln (xy3z)$
Try this! Write as a single logarithm with coefficient $1: 3log (x+y)−6log z+2log 5.$
Answer: $log (25(x+y)3z6)$
### Key Takeaways
• Given any base $b>0$ and $b≠1$, we can say that $logb 1=0$, $logb b=1$, $log1/b b=−1$ and that $logb (1b)=−1.$
• The inverse properties of the logarithm are $logb bx=x$ and $blogb x=x$ where $x>0.$
• The product property of the logarithm allows us to write a product as a sum: $logb (xy)=logb x+logb y.$
• The quotient property of the logarithm allows us to write a quotient as a difference: $logb (xy)=logb x−logb y.$
• The power property of the logarithm allows us to write exponents as coefficients: $logb xn=nlogb x.$
• Since the natural logarithm is a base-e logarithm, $ln x=loge x$, all of the properties of the logarithm apply to it.
• We can use the properties of the logarithm to expand logarithmic expressions using sums, differences, and coefficients. A logarithmic expression is completely expanded when the properties of the logarithm can no further be applied.
• We can use the properties of the logarithm to combine expressions involving logarithms into a single logarithm with coefficient 1. This is an essential skill to be learned in this chapter.
### Part A: Logarithms and Their Inverse Properties
Evaluate:
1. $log7 1$
2. $log1/2 2$
3. $log 1014$
4. $log 10−23$
5. $log3 310$
6. $log6 6$
7. $ln e7$
8. $ln (1e)$
9. $log1/2 (12)$
10. $log1/5 5$
11. $log3/4 (43)$
12. $log2/3 1$
13. $2log2 100$
14. $3log3 1$
15. $10log 18$
16. $eln 23$
17. $eln x2$
18. $eln ex$
Find a:
1. $ln a=1$
2. $log a=−1$
3. $log9 a=−1$
4. $log12 a=1$
5. $log2 a=5$
6. $log a=13$
7. $2a=7$
8. $ea=23$
9. $loga 45=5$
10. $loga 10=1$
### Part B: Product, Quotient, and Power Properties of Logarithms
Expand completely.
1. $log4 (xy)$
2. $log (6x)$
3. $log3 (9x2)$
4. $log2 (32x7)$
5. $ln (3y2)$
6. $log (100x2)$
7. $log2 (xy2)$
8. $log5 (25x)$
9. $log (10x2y3)$
10. $log2 (2x4y5)$
11. $log3 (x3yz2)$
12. $log (xy3z2)$
13. $log5 (1x2yz)$
14. $log4 (116x2z3)$
15. $log6 [36(x+y)4]$
16. $ln [e4(x−y)3]$
17. $log7 (2xy)$
18. $ln (2xy)$
19. $log3 (x2y3z)$
20. $log (2(x+y)3z2)$
21. $log (100x3(y+10)3)$
22. $log7 (x(y+z)35)$
23. $log5 (x3yz23)$
24. $log (x2y3z25)$
Given $log3 x=a$, $log3 y=b$, and $log3 z=c$, write the following logarithms in terms of a, b, and c.
1. $log3 (27x2y3z)$
2. $log3 (xy3z)$
3. $log3 (9x2yz3)$
4. $log3 (x3yz2)$
Given $logb 2=0.43$, $logb 3=0.68$, and $logb 7=1.21$, calculate the following. (Hint: Expand using sums, differences, and quotients of the factors 2, 3, and 7.)
1. $logb 42$
2. $logb (36)$
3. $logb (289)$
4. $logb 21$
Expand using the properties of the logarithm and then approximate using a calculator to the nearest tenth.
1. $log (3.10×1025)$
2. $log (1.40×10−33)$
3. $ln (6.2e−15)$
4. $ln (1.4e22)$
Write as a single logarithm with coefficient 1.
1. $log x+log y$
2. $log3 x−log3 y$
3. $log2 5+2log2 x+log2 y$
4. $log3 4+3log3 x+12log3 y$
5. $3log2 x−2log2 y+12log2 z$
6. $4log x−log y−log 2$
7. $log 5+3log (x+y)$
8. $4log5 (x+5)+log5 y$
9. $ln x−6ln y+ln z$
10. $log3 x−2log3 y+5log3 z$
11. $7log x−log y−2log z$
12. $2ln x−3ln y−ln z$
13. $23log3 x−12(log3 y+log3 z)$
14. $15(log7 x+2log7 y)−2log7 (z+1)$
15. $1+log2 x−12log2 y$
16. $2−3log3 x+13log3 y$
17. $13log2 x+23log2 y$
18. $−2log5 x+35log5 y$
19. $−ln 2+2ln (x+y)−ln z$
20. $−3ln (x−y)−ln z+ln 5$
21. $13(ln x+2ln y)−(3ln 2+ln z)$
22. $4log 2+23log x−4log (y+z)$
23. $log2 3−2log2 x+12log2 y−4log2 z$
24. $2log5 4−log5 x−3log5 y+23log5 z$
Express as a single logarithm and simplify.
1. $log (x+1)+log (x−1)$
2. $log2 (x+2)+log2 (x+1)$
3. $ln (x2+2x+1)−ln (x+1)$
4. $ln (x2−9)−ln (x+3)$
5. $log5 (x3−8)−log5 (x−2)$
6. $log3 (x3+1)−log3 (x+1)$
7. $log x+log (x+5)−log (x2−25)$
8. $log (2x+1)+log (x−3)−log (2x2−5x−3)$
1. 0
2. 14
3. 10
4. 7
5. 1
6. −1
7. 100
8. 18
9. $x2$
10. $e$
11. $19$
12. $25=32$
13. $log2 7$
14. 4
1. $log4 x+log4 y$
2. $2+2log3 x$
3. $ln 3+2ln y$
4. $log2 x−2log2 y$
5. $1+2log x+3log y$
6. $3log3 x−log3 y−2log3 z$
7. $−2log5 x−log5 y−log5 z$
8. $2+4log6 (x+y)$
9. $log7 2+12log7 x+12log7 y$
10. $2log3 x+13log3 y−log3 z$
11. $2+3log x−3log (y+10)$
12. $3log5 x−13log5 y−23log5 z$
13. $3+2a+3b+c$
14. $2+2a+b−3c$
15. 2.32
16. 0.71
17. $log (3.1)+25≈25.5$
18. $ln (6.2)−15≈−13.2$
19. $log (xy)$
20. $log2 (5x2y)$
21. $log2 (x3zy2)$
22. $log [5(x+y)3]$
23. $ln (xzy6)$
24. $log (x7yz2)$
25. $log3 (x23yz)$
26. $log2 (2xy)$
27. $log2 (xy23)$
28. $ln ((x+y)22z)$
29. $ln (xy238z)$
30. $log2 (3yx2z4)$
31. $log (x2−1)$
32. $ln (x+1)$
33. $log5 (x2+2x+4)$
34. $log (xx−5)$
## 7.5 Solving Exponential and Logarithmic Equations
### Learning Objectives
1. Solve exponential equations.
2. Use the change of base formula to approximate logarithms.
3. Solve logarithmic equations.
## Solving Exponential Equations
An exponential equationAn equation which includes a variable as an exponent. is an equation that includes a variable as one of its exponents. In this section we describe two methods for solving exponential equations. First, recall that exponential functions defined by $f(x)=bx$ where $b>0$ and $b≠1$, are one-to-one; each value in the range corresponds to exactly one element in the domain. Therefore, $f(x)=f(y)$ implies $x=y.$ The converse is true because f is a function. This leads to the very important one-to-one property of exponential functionsGiven $b>0$ and $b≠1$ we have $bx=by$ if and only if $x=y.$:
$bx=by if and only if x=y$
Use this property to solve special exponential equations where each side can be written in terms of the same base.
### Example 1
Solve: $32x−1=27.$
Solution:
Begin by writing 27 as a power of 3.
$32x−1=2732x−1=33$
Next apply the one-to-one property of exponential functions. In other words, set the exponents equal to each other and then simplify.
$2x−1=32x=4x=2$
### Example 2
Solve: $161−3x=2.$
Solution:
Begin by writing 16 as a power of 2 and then apply the power rule for exponents.
$161−3x=2(24)1−3x=224(1−3x)=21$
Now that the bases are the same we can set the exponents equal to each other and simplify.
$4(1−3x)=14−12x=1−12x=−3x=−3−12=14$
Answer: $14$
Try this! Solve: $252x+3=125.$
Answer: $−34$
In many cases we will not be able to equate the bases. For this reason we develop a second method for solving exponential equations. Consider the following equations:
$32=93?=1233=27$
We can see that the solution to $3x=12$ should be somewhere between 2 and 3. A graphical interpretation follows.
To solve this we make use of fact that logarithms are one-to-one functions. Given $x,y>0$ the one-to-one property of logarithmsGiven $b>0$ and $b≠1$ where $x,y>0$ we have $logb x=logb y$ if and only if $x=y.$ follows:
$logb x=logb y if and only if x=y$
This property, as well as the properties of the logarithm, allows us to solve exponential equations. For example, to solve $3x=12$ apply the common logarithm to both sides and then use the properties of the logarithm to isolate the variable.
$3x=12log 3x=log 12 One-to-one property of logarithmsxlog 3=log 12 Power rule for logarithmsx=log 12log 3$
Approximating to four decimal places on a calculator.
$x=log (12)/log (3)≈2.2619$
An answer between 2 and 3 is what we expected. Certainly we can check by raising 3 to this power to verify that we obtain a good approximation of 12.
$3^2.2618≈12✓$
Note that we are not multiplying both sides by “log”; we are applying the one-to-one property of logarithmic functions — which is often expressed as “taking the log of both sides.” The general steps for solving exponential equations are outlined in the following example.
### Example 3
Solve: $52x−1+2=9.$
Solution:
• Step 1: Isolate the exponential expression.
$52x−1+2=952x−1=7$
• Step 2: Take the logarithm of both sides. In this case, we will take the common logarithm of both sides so that we can approximate our result on a calculator.
$log 52x−1=log 7$
• Step 3: Apply the power rule for logarithms and then solve.
$log 52x−1=log 7(2x−1)log 5=log 7 Distribute.2xlog 5−log 5=log 72xlog 5=log 5+log 7x=log 5+log 72log 5$
This is an irrational number which can be approximated using a calculator. Take care to group the numerator and the product in the denominator when entering this into your calculator. To do this, make use of the parenthesis buttons $($ and $)$ :
$x=(log 5+log (7))/(2*log (5))≈1.1045$
Answer: $log 5+log 72log 5≈1.1045$
### Example 4
Solve: $e5x+3=1.$
Solution:
The exponential function is already isolated and the base is e. Therefore, we choose to apply the natural logarithm to both sides.
$e5x+3=1ln e5x+3=ln 1$
Apply the power rule for logarithms and then simplify.
$ln e5x+3=ln 1(5x+3)ln e=ln 1 Recall ln e=1 and ln1=0.(5x+3)⋅1=05x+3=0x=−35$
Answer: $−35$
On most calculators there are only two logarithm buttons, the common logarithm $LOG$ and the natural logarithm $LN.$ If we want to approximate $log3 10$ we have to somehow change this base to 10 or e. The idea begins by rewriting the logarithmic function $y=loga x$, in exponential form.
$loga x=y ⇒ x=ay$
Here $x>0$ and so we can apply the one-to-one property of logarithms. Apply the logarithm base b to both sides of the function in exponential form.
$x=aylogb x=logb ay$
And then solve for y.
$logb x=ylogb alogb xlogb a=y$
Replace y into the original function and we have the very important change of base formula$loga x=logb xlogb a$; we can write any base-a logarithm in terms of base-b logarithms using this formula.:
$loga x=logb xlogb a$
We can use this to approximate $log3 10$ as follows.
$log3 10=log 10log 3≈2.0959 or log3 10=ln 10ln 3≈2.0959$
Notice that the result is independent of the choice of base. In words, we can approximate the logarithm of any given base on a calculator by dividing the logarithm of the argument by the logarithm of that given base.
### Example 5
Approximate $log7 120$ the nearest hundredth.
Solution:
Apply the change of base formula and use a calculator.
$log7 120=log 120log 7$
On a calculator,
$log (120)/log (7)≈2.46$
Try this! Solve: $23x+1−4=1.$ Give the exact and approximate answer rounded to four decimal places.
Answer: $log 5−log 23log 2≈0.4406$
## Solving Logarithmic Equations
A logarithmic equationAn equation that involves a logarithm with a variable argument. is an equation that involves a logarithm with a variable argument. Some logarithmic equations can be solved using the one-to-one property of logarithms. This is true when a single logarithm with the same base can be obtained on both sides of the equal sign.
### Example 6
Solve: $log2 (2x−5)−log2 (x−2)=0.$
Solution:
We can obtain two equal logarithms base 2 by adding $log2 (x−2)$ to both sides of the equation.
$log2 (2x−5)−log2 (x−2)=0log2 (2x−5)=log2 (x−2)$
Here the bases are the same and so we can apply the one-to-one property and set the arguments equal to each other.
$log2 (2x−5)=log2 (x−2)2x−5=x−2x=3$
Checking $x=3$ in the original equation:
$log2 (2(3)−5)=log2 ((3)−2)log2 1=log2 10=0 ✓$
When solving logarithmic equations the check is very important because extraneous solutions can be obtained. The properties of the logarithm only apply for values in the domain of the given logarithm. And when working with variable arguments, such as $log (x−2)$, the value of x is not known until the end of this process. The logarithmic expression $log (x−2)$ is only defined for values $x>2.$
### Example 7
Solve: $log (3x−4)=log (x−2).$
Solution:
Apply the one-to-one property of logarithms (set the arguments equal to each other) and then solve for x.
$log (3x−4)=log (x−2)3x−4=x−22x=2x=1$
When performing the check we encounter a logarithm of a negative number:
$log (x−2)=log (1−2)=log (−1) Undefined$
Try this on a calculator, what does it say? Here $x=1$ is not in the domain of $log (x−2).$ Therefore our only possible solution is extraneous and we conclude that there are no solutions to this equation.
Caution: Solving logarithmic equations sometimes leads to extraneous solutions — we must check our answers.
Try this! Solve: $ln (x2−15)−ln (2x)=0.$
In many cases we will not be able to obtain two equal logarithms. To solve such equations we make use of the definition of the logarithm. If $b>0$, where $b≠1$, then $logb x=y$ implies that $by=x.$ Consider the following common logarithmic equations (base 10),
$log x=0 ⇒ x=1 Because 100=1.log x=0.5 ⇒ x=?log x=1 ⇒ x=10 Because 101=10.$
We can see that the solution to $log x=0.5$ will be somewhere between 1 and 10. A graphical interpretation follows.
To find x we can apply the definition as follows.
$log10 x=0.5 ⇒ 100.5=x$
This can be approximated using a calculator,
$x=100.5=10^0.5≈3.1623$
An answer between 1 and 10 is what we expected. Check this on a calculator.
$log 3.1623≈5✓$
### Example 8
Solve: $log3 (2x−5)=2.$
Solution:
Apply the definition of the logarithm.
$log3 (2x−5)=2 ⇒ 2x−5=32$
Solve the resulting equation.
$2x−5=92x=14x=7$
Check.
$log3 (2(7)−5)=?2 log3 (9)=2 ✓$
In order to apply the definition, we will need to rewrite logarithmic expressions as a single logarithm with coefficient 1.The general steps for solving logarithmic equations are outlined in the following example.
### Example 9
Solve: $log2 (x−2)+log2 (x−3)=1.$
Solution:
• Step 1: Write all logarithmic expressions as a single logarithm with coefficient 1. In this case, apply the product rule for logarithms.
$log2 (x−2)+log2 (x−3)=1log2 [(x−2)(x−3)]=1$
• Step 2: Use the definition and rewrite the logarithm in exponential form.
$log2 [(x−2)(x−3)]=1 ⇒ (x−2)(x−3)=21$
• Step 3: Solve the resulting equation. Here we can solve by factoring.
$(x−2)(x−3)=2x2−5x+6=2x2−5x+4=0(x−4)(x−1)=0x−4=0 or x−1=0x=4x=1$
• Step 4: Check. This step is required.
Check $x=4$
Check $x=1$
$log2 (x−2)+log2 (x−3)=1log2 (4−2)+log2 (4−3)=1log2 (2)+log2 (1)=11+0=1 ✓$
$log2 (x−2)+log2 (x−3)=1log2 (1−2)+log2 (1−3)=1log2 (−1)+log2 (−2)=1 ✗$
In this example, $x=1$ is not in the domain of the given logarithmic expression and is extraneous. The only solution is $x=4.$
### Example 10
Solve: $log (x+15)−1=log (x+6).$
Solution:
Begin by writing all logarithmic expressions on one side and constants on the other.
$log (x+15)−1=log (x+6)log (x+15)−log (x+6)=1$
Apply the quotient rule for logarithms as a means to obtain a single logarithm with coefficient 1.
$log (x+15)−log (x+6)=1log (x+15x+6)=1$
This is a common logarithm; therefore use 10 as the base when applying the definition.
$x+15x+6=101x+15=10(x+6)x+15=10x+60−9x=45x=−5$
Check.
$log (x+15)−1=log (x+6)log (−5+15)−1=log (−5+6)log 10−1=log 11−1=00=0 ✓$
Try this! Solve: $log2 (x)+log2 (x−1)=1.$
### Example 11
Find the inverse: $f(x)=log2 (3x−4).$
Solution:
Begin by replacing the function notation $f(x)$ with y.
$f(x)=log2 (3x−4)y=log2 (3x−4)$
Interchange x and y and then solve for y.
$x=log2 (3y−4) ⇒ 3y−4=2x3y=2x+4y=2x+43$
The resulting function is the inverse of f. Present the answer using function notation.
Answer: $f−1(x)=2x+43$
### Key Takeaways
• If each side of an exponential equation can be expressed using the same base, then equate the exponents and solve.
• To solve a general exponential equation, first isolate the exponential expression and then apply the appropriate logarithm to both sides. This allows us to use the properties of logarithms to solve for the variable.
• The change of base formula allows us to use a calculator to calculate logarithms. The logarithm of a number is equal to the common logarithm of the number divided by the common logarithm of the given base.
• If a single logarithm with the same base can be isolated on each side of an equation, then equate the arguments and solve.
• To solve a general logarithmic equation, first isolate the logarithm with coefficient 1 and then apply the definition. Solve the resulting equation.
• The steps for solving logarithmic equations sometimes produce extraneous solutions. Therefore, the check is required.
### Part A: Solving Exponential Equations
Solve using the one-to-one property of exponential functions.
1. $3x=81$
2. $2−x=16$
3. $5x−1=25$
4. $3x+4=27$
5. $25x−2=16$
6. $23x+7=8$
7. $812x+1=3$
8. $643x−2=2$
9. $92−3x−27=0$
10. $81−5x−32=0$
11. $16x2−2=0$
12. $4x2−1−64=0$
13. $9x(x+1)=81$
14. $4x(2x+5)=64$
15. $100x2−107x−3=0$
16. $e3(3x2−1)−e=0$
Solve. Give the exact answer and the approximate answer rounded to the nearest thousandth.
1. $3x=5$
2. $7x=2$
3. $4x=9$
4. $2x=10$
5. $5x−3=13$
6. $3x+5=17$
7. $72x+5=2$
8. $35x−9=11$
9. $54x+3+6=4$
10. $107x−1−2=1$
11. $e2x−3−5=0$
12. $e5x+1−10=0$
13. $63x+1−3=7$
14. $8−109x+2=9$
15. $15−e3x=2$
16. $7+e4x+1=10$
17. $7−9e−x=4$
18. $3−6e−x=0$
19. $5x2=2$
20. $32x2−x=1$
21. $100e27x=50$
22. $6e12x=2$
23. $31+e−x=1$
24. $21+3e−x=1$
Find the x- and y-intercepts of the given function.
1. $f(x)=3x+1−4$
2. $f(x)=23x−1−1$
3. $f(x)=10x+1+2$
4. $f(x)=104x−5$
5. $f(x)=ex−2+1$
6. $f(x)=ex+4−4$
Use a u-substitution to solve the following.
1. $32x−3x−6=0$ (Hint: Let $u=3x$)
2. $22x+2x−20=0$
3. $102x+10x−12=0$
4. $102x−10x−30=0$
5. $e2x−3ex+2=0$
6. $e2x−8ex+15=0$
Use the change of base formula to approximate the following to the nearest hundredth.
1. $log2 5$
2. $log3 7$
3. $log5 (23)$
4. $log7 (15)$
5. $log1/2 10$
6. $log2/3 30$
7. $log2 5$
8. $log2 63$
9. If left unchecked, a new strain of flu virus can spread from a single person to others very quickly. The number of people affected can be modeled using the formula $P(t)=e0.22t$, where t represents the number of days the virus is allowed to spread unchecked. Estimate the number of days it will take 1,000 people to become infected.
10. The population of a certain small town is growing according to the function $P(t)=12,500(1.02)t$, where t represents time in years since the last census. Use the function to determine number of years it will take the population to grow to 25,000 people.
### Part B: Solving Logarithmic Equations
Solve using the one-to-one property of logarithms.
1. $log5 (2x+4)=log5 (3x−6)$
2. $log4 (7x)=log4 (5x+14)$
3. $log2 (x−2)−log2 (6x−5)=0$
4. $ln (2x−1)=ln (3x)$
5. $log (x+5)−log (2x+7)=0$
6. $ln (x2+4x)=2ln (x+1)$
7. $log3 2+2log3 x=log3 (7x−3)$
8. $2log x−log 36=0$
9. $ln (x+3)+ln (x+1)=ln 8$
10. $log5 (x−2)+log5 (x−5)=log5 10$
Solve.
1. $log2 (3x−7)=5$
2. $log3 (2x+1)=2$
3. $log (2x+20)=1$
4. $log4 (3x+5)=12$
5. $log3 x2=2$
6. $log (x2+3x+10)=1$
7. $ln (x2−1)=0$
8. $log5 (x2+20)−2=0$
9. $log2 (x−5)+log2 (x−9)=5$
10. $log2 (x+5)+log2 (x+1)=5$
11. $log4 x+log4 (x−6)=2$
12. $log6 x+log6 (2x−1)=2$
13. $log3 (2x+5)−log3 (x−1)=2$
14. $log2 (x+1)−log2 (x−2)=4$
15. $ln x−ln (x−1)=1$
16. $ln (2x+1)−ln x=2$
17. $2log3 x=2+log3 (2x−9)$
18. $2log2 x=3+log2 (x−2)$
19. $log2 (x−2)=2−log2 x$
20. $log2 (x+3)+log2 (x+1)−1=0$
21. $log x−log (x+1)=1$
22. $log2 (x+2)+log2 (1−x)=1+log2 (x+1)$
Find the x- and y-intercepts of the given function.
1. $f(x)=log (x+3)−1$
2. $f(x)=log (x−2)+1$
3. $f(x)=log2 (3x)−4$
4. $f(x)=log3 (x+4)−3$
5. $f(x)=ln (2x+5)−6$
6. $f(x)=ln (x+1)+2$
Find the inverse of the following functions.
1. $f(x)=log2 (x+5)$
2. $f(x)=4+log3 x$
3. $f(x)=log (x+2)−3$
4. $f(x)=ln (x−4)+1$
5. $f(x)=ln (9x−2)+5$
6. $f(x)=log6 (2x+7)−1$
7. $g(x)=e3x$
8. $g(x)=10−2x$
9. $g(x)=2x+3$
10. $g(x)=32x+5$
11. $g(x)=10x+4−3$
12. $g(x)=e2x−1+1$
Solve.
1. $log (9x+5)=1+log (x−5)$
2. $2+log2 (x2+1)=log2 13$
3. $e5x−2−e3x=0$
4. $3x2−11=70$
5. $23x−5=0$
6. $log7 (x+1)+log7 (x−1)=1$
7. $ln (4x−1)−1=ln x$
8. $log (20x+1)=log x+2$
9. $31+e2x=2$
10. $2e−3x=4$
11. $2e3x=e4x+1$
12. $2log x+log x−1=0$
13. $3log x=log (x−2)+2log x$
14. $2ln 3+ln x2=ln (x2+1)$
15. In chemistry, pH is a measure of acidity and is given by the formula $pH=−log (H+)$, where $H+$ is the hydrogen ion concentration (measured in moles of hydrogen per liter of solution.) Determine the hydrogen ion concentration if the pH of a solution is 4.
16. The volume of sound, L in decibels (dB), is given by the formula $L=10log (I/10−12)$ where I represents the intensity of the sound in watts per square meter. Determine the intensity of an alarm that emits 120 dB of sound.
### Part C: Discussion Board
1. Research and discuss the history and use of the slide rule.
2. Research and discuss real-world applications involving logarithms.
1. 4
2. 3
3. $65$
4. $−38$
5. $16$
6. $±12$
7. −2, 1
8. $12$, 3
9. $log 5log 3≈1.465$
10. $log 3log 2≈1.585$
11. $3log 5+log 13log 5≈4.594$
12. $log 2−5log 72log 7≈−2.322$
13. Ø
14. $3+ln 52≈2.305$
15. $1−log 63log 6≈0.095$
16. $ln 133≈0.855$
17. $ln 3≈1.099$
18. $±log 2log 5≈±0.656$
19. $−ln 227≈−0.026$
20. $−ln 2≈−0.693$
21. x-intercept: $(2log 2−log 3log 3,0)$; y-intercept: $(0,−1)$
22. x-intercept: None; y-intercept: $(0,12)$
23. x-intercept: None; y-intercept: $(0,1+e2e2)$
24. 1
25. $log 3$
26. 0, $ln 2$
27. 2.32
28. −0.25
29. −3.32
30. 1.16
31. Approximately 31 days
1. 10
2. $35$
3. −2
4. $12$, 3
5. 1
6. 13
7. −5
8. ±3
9. $±2$
10. 13
11. 8
12. 2
13. $ee−1$
14. 9
15. $1+5$
16. Ø
17. x-intercept: $(7,0)$; y-intercept: $(0,log 3−1)$
18. x-intercept: $(163, 0)$; y-intercept: None
19. x-intercept: $(e6−52,0)$; y-intercept: $(0,ln 5−6)$
20. $f−1(x)=2x−5$
21. $f−1(x)=10x+3−2$
22. $f−1(x)=ex−5+29$
23. $g−1(x)=ln x3$
24. $g−1(x)=log2 x−3$
25. $g−1(x)=log (x+3)−4$
26. 55
27. 1
28. $log2 53$
29. $14−e$
30. $ln (1/2)2$
31. $ln 2−1$
32. Ø
33. $10−4$ moles per liter
## 7.6 Applications
### Learning Objectives
1. Use the compound and continuous interest formulas.
2. Calculate doubling time.
3. Use the exponential growth/decay model.
4. Calculate the rate of decay given half-life.
## Compound and Continuous Interest Formulas
Recall that compound interest occurs when interest accumulated for one period is added to the principal investment before calculating interest for the next period. The amount A accrued in this manner over time t is modeled by the compound interest formula:
$A(t)=P(1+rn)nt$
Here the initial principal P is accumulating compound interest at an annual rate r where the value n represents the number of times the interest is compounded in a year.
### Example 1
Susan invested \$500 in an account earning $412$% annual interest that is compounded monthly.
1. How much will be in the account after 3 years?
2. How long will it take for the amount to grow to \$750?
Solution:
In this example, the principal $P=500$, the interest rate $r=412%=0.045$, and because the interest is compounded monthly, $n=12.$ The investment can be modeled by the following function:
$A(t)=500(1+0.04512)12tA(t)=500(1.00375)12t$
1. Use this model to calculate the amount in the account after $t=3$ years.
$A(3)=500(1.00375)12(3)=500(1.00375)36≈572.12$
Rounded off to the nearest cent, after 3 years, the amount accumulated will be \$572.12.
2. To calculate the time it takes to accumulate \$750, set $A(t)=750$ and solve for t.
$A(t)=500(1.00375)12t750=500(1.00375)12t$
This results in an exponential equation that can be solved by first isolating the exponential expression.
$750=500(1.00375)12t750500=(1.00375)12t1.5=(1.00375)12t$
At this point take the common logarithm of both sides, apply the power rule for logarithms, and then solve for t.
$log (1.5)=log (1.00375)12tlog (1.5)=12tlog (1.00375)log (1.5)12log (1.00375)=12tlog (1.00375)12log (1.00375)log (1.5)12log (1.00375)=t$
Using a calculator we can approximate the time it takes.
$t=log (1.5)/(12*log (1.00375))≈9 years$
1. \$572.12
2. Approximately 9 years.
The period of time it takes a quantity to double is called the doubling timeThe period of time it takes a quantity to double.. We next outline a technique for calculating the time it takes to double an initial investment earning compound interest.
### Example 2
Mario invested \$1,000 in an account earning 6.3% annual interest that is compounded semi-annually. How long will it take the investment to double?
Solution:
Here the principal $P=1,000$, the interest rate $r=6.3%=0.063$, and because the interest is compounded semi-annually $n=2.$ This investment can be modeled as follows:
$A(t)=1,000(1+0.0632)2tA(t)=1,000(1.0315)2t$
Since we are looking for the time it takes to double \$1,000, substitute \$2,000 for the resulting amount $A(t)$ and then solve for t.
$2,000=1,000(1.0315)2t2,0001,000=(1.0315)2t2=(1.0315)2t$
At this point we take the common logarithm of both sides.
$2=(1.0315)2tlog 2=log (1.0315)2tlog 2=2tlog (1.0315)log 22log (1.0315)=t$
Using a calculator we can approximate the time it takes:
$t=log (2)/(2*log (1.0315))≈11.17 years$
Answer: Approximately 11.17 years to double at 6.3%.
If the investment in the previous example was one million dollars, how long would it take to double? To answer this we would use $P=1,000,000$ and $A(t)=2,000,000$:
$A(t)=1,000(1.0315)2t2,000,000=1,000,000(1.0315)2t$
Dividing both sides by 1,000,000 we obtain the same exponential function as before.
$2=(1.0315)2t$
Hence, the result will be the same, about 11.17 years. In fact, doubling time is independent of the initial investment P.
Interest is typically compounded semi-annually (n = 2), quarterly (n = 4), monthly (n = 12), or daily (n = 365). However if interest is compounded every instant we obtain a formula for continuously compounding interest:
$A(t)=Pert$
Here P represents the initial principal amount invested, r represents the annual interest rate, and t represents the time in years the investment is allowed to accrue continuously compounded interest.
### Example 3
Mary invested \$200 in an account earning $534$% annual interest that is compounded continuously. How long will it take the investment to grow to \$350?
Solution:
Here the principal $P=200$ and the interest rate $r=534%=5.75%=0.0575.$ Since the interest is compounded continuously, use the formula $A(t)=Pert.$ Hence, the investment can be modeled by the following,
$A(t)=200e0.0575t$
To calculate the time it takes to accumulate to \$350, set $A(t)=350$ and solve for t.
$A(t)=200e0.0575t350=200e0.0575t$
Begin by isolating the exponential expression.
$350200=e0.0575t74=e0.0575t1.75=e0.0575t$
Because this exponential has base e, we choose to take the natural logarithm of both sides and then solve for t.
$ln (1.75)=ln e0.0575t Apply the power rule for logarithms.ln (1.75)=0.0575tln e Recall that ln e=1.ln (1.75)=0.0575t⋅1ln (1.75)0.0575=t$
Using a calculator we can approximate the time it takes:
$t=ln (1.75)/0.0575≈9.73 years$
Answer: It will take approximately 9.73 years.
When solving applications involving compound interest, look for the keyword “continuous,” or the keywords that indicate the number of annual compoundings. It is these keywords that determine which formula to choose.
Try this! Mario invested \$1,000 in an account earning 6.3% annual interest that is compounded continuously. How long will it take the investment to double?
## Modeling Exponential Growth and Decay
In the sciences, when a quantity is said to grow or decay exponentially, it is specifically meant to be modeled using the exponential growth/decay formulaA formula that models exponential growth or decay: $P(t)=P0ekt.$:
$P(t)=P0ekt$
Here $P0$, read “P naught,” or “P zero,” represents the initial amount, k represents the growth rate, and t represents the time the initial amount grows or decays exponentially. If k is negative, then the function models exponential decay. Notice that the function looks very similar to that of continuously compounding interest formula. We can use this formula to model population growth when conditions are optimal.
### Example 4
It is estimated that the population of a certain small town is 93,000 people with an annual growth rate of 2.6%. If the population continues to increase exponentially at this rate:
1. Estimate the population in 7 years’ time.
2. Estimate the time it will take for the population to reach 120,000 people.
Solution:
We begin by constructing a mathematical model based on the given information. Here the initial population $P0=93,000$ people and the growth rate $r=2.6%=0.026.$ The following model gives population in terms of time measured in years:
$P(t)=93,000e0.026t$
1. Use this function to estimate the population in $t=7$ years.
$P(t)=93,000e0.026(7)=93,000e0.182≈111,564 people$
2. Use the model to determine the time it takes to reach $P(t)=120,000$ people.
$P(t)=93,000e0.026t120,000=93,000e0.026t120,00093,000=e0.026t4031=e0.026t$
Take the natural logarithm of both sides and then solve for t.
$ln (4031)=ln e0.026tln (4031)=0.026tln eln (4031)=0.026t⋅1ln (4031)0.026=t$
Using a calculator,
$t=ln (40/31)/0.026≈9.8 years$
1. 111,564 people
2. 9.8 years
Often the growth rate k is not given. In this case, we look for some other information so that we can determine it and then construct a mathematical model. The general steps are outlined in the following example.
### Example 5
Under optimal conditions Escherichia coli (E. coli) bacteria will grow exponentially with a doubling time of 20 minutes. If 1,000 E. coli cells are placed in a Petri dish and maintained under optimal conditions, how many E. coli cells will be present in 2 hours?
Figure 7.3
Escherichia coli (E. coli) (Wikipedia)
Solution:
The goal is to use the given information to construct a mathematical model based on the formula $P(t)=P0ekt.$
• Step 1: Find the growth rate k. Use the fact that the initial amount, $P0=1,000$ cells, doubles in 20 minutes. That is, $P(t)=2,000$ cells when $t=20$ minutes.
$P(t)=P0ekt2,000=1,000ek20$
Solve for the only variable k.
$2,000=1,000ek202,0001,000=ek202=ek20ln (2)=ln ek20ln (2)=k20ln eln (2)=k20⋅1ln (2)20=k$
• Step 2: Write a mathematical model based on the given information. Here $k≈0.0347$, which is about 3.5% growth rate per minute. However, we will use the exact value for k in our model. This will allow us to avoid round-off error in the final result. Use $P0=1,000$ and $k=ln (2)/20$:
$P(t)=1,000e(ln (2)/20)t$
This equation models the number of E. coli cells in terms of time in minutes.
• Step 3: Use the function to answer the questions. In this case, we are asked to find the number of cells present in 2 hours. Because time is measured in minutes, use $t=120$ minutes to calculate the number of E. coli cells.
$P(120)=1,000e(ln (2)/20)(120)=1,000eln (2)⋅6=1,000eln 26=1,000⋅26=64,000 cells$
Answer: In two hours 64,000 cells will be present.
When the growth rate is negative the function models exponential decay. We can describe decreasing quantities using a half-lifeThe period of time it takes a quantity to decay to one-half of the initial amount., or the time it takes to decay to one-half of a given quantity.
### Example 6
Due to radioactive decay, caesium-137 has a half-life of 30 years. How long will it take a 50-milligram sample to decay to 10 milligrams?
Solution:
Use the half-life information to determine the rate of decay k. In $t=30$ years the initial amount $P0=50$ milligrams will decay to half $P(30)=25$ milligrams.
$P(t)=P0ekt25=50ek30$
Solve for the only variable, k.
$25=50ek302550=e30kln (12)=ln e30kln (12)=30kln eln 1−ln 230=k Recall that ln 1=0.−ln 230=k$
Note that $k=−ln 230≈−0.0231$ is negative. However, we will use the exact value to construct a model that gives the amount of cesium-137 with respect to time in years.
$P(t)=50e(−ln 2/30)t$
Use this model to find t when $P(t)=10$ milligrams.
$10=50e(−ln 2/30)t1050=e(−ln 2/30)tln (15)=ln e(−ln 2/30)tln 1− ln 5=(−ln 230)t ln e Recall that ln e=1.−30(ln 1−ln 5)ln 2=t−30(0−ln 5)ln 2=t30ln 5ln 2=t$
Answer: Using a calculator, it will take $t≈69.66$ years to decay to 10 milligrams.
Radiocarbon dating is a method used to estimate the age of artifacts based on the relative amount of carbon-14 present in it. When an organism dies, it stops absorbing this naturally occurring radioactive isotope, and the carbon-14 begins to decay at a known rate. Therefore, the amount of carbon-14 present in an artifact can be used to estimate the age of the artifact.
### Example 7
An ancient bone tool is found to contain 25% of the carbon-14 normally found in bone. Given that carbon-14 has a half-life of 5,730 years, estimate the age of the tool.
Solution:
Begin by using the half-life information to find k. Here the initial amount $P0$ of carbon-14 is not given, however, we know that in $t=5,730$ years, this amount decays to half, $12P0.$
$P(t)=P0ekt12P0=P0ek5,730$
Dividing both sides by $P0$ leaves us with an exponential equation in terms of k. This shows that half-life is independent of the initial amount.
$12=ek5,730$
Solve for k.
$ln (12)=ln ek5,730ln 1−ln 2=5,730k ln e0−ln 25,730=k−ln 25,730=k$
Therefore we have the model,
$P(t)=P0e(−ln 2/5,730)t$
Next we wish to the find time it takes the carbon-14 to decay to 25% of the initial amount, or $P(t)=0.25P0.$
$0.25P0=P0e(−ln 2/5,730)t$
Divide both sides by $P0$ and solve for t.
$0.25=e(−ln 2/5,730)tln (0.25)=ln e(−ln 2/5,730)tln (0.25)=(−ln 25,730)tln e−5,730 ln (0.25)ln 2=t11,460≈t$
Answer: The tool is approximately 11,460 years old.
Try this! The half-life of strontium-90 is about 28 years. How long will it take a 36 milligram sample of strontium-90 to decay to 30 milligrams?
### Key Takeaways
• When interest is compounded a given number of times per year use the formula $A(t)=P(1+rn)nt.$
• When interest is to be compounded continuously use the formula $A(t)=Pert.$
• Doubling time is the period of time it takes a given amount to double. Doubling time is independent of the principal.
• When amounts are said to be increasing or decaying exponentially, use the formula $P(t)=P0ekt.$
• Half-life is the period of time it takes a given amount to decrease to one-half. Half-life is independent of the initial amount.
• To model data using the exponential growth/decay formula, use the given information to determine the growth/decay rate k. Once k is determined, a formula can be written to model the problem. Use the formula to answer the questions.
### Part A: Compound and Continuous Interest
1. Jill invested \$1,450 in an account earning $458$% annual interest that is compounded monthly.
1. How much will be in the account after 6 years?
2. How long will it take the account to grow to \$2,200?
2. James invested \$825 in an account earning $525$% annual interest that is compounded monthly.
1. How much will be in the account after 4 years?
2. How long will it take the account to grow to \$1,500?
3. Raul invested \$8,500 in an online money market fund earning 4.8% annual interest that is compounded continuously.
1. How much will be in the account after 2 years?
2. How long will it take the account to grow to \$10,000?
4. Ian deposited \$500 in an account earning 3.9% annual interest that is compounded continuously.
1. How much will be in the account after 3 years?
2. How long will it take the account to grow to \$1,500?
5. Bill wants to grow his \$75,000 inheritance to \$100,000 before spending any of it. How long will this take if the bank is offering 5.2% annual interest compounded quarterly?
6. Mary needs \$25,000 for a down payment on a new home. If she invests her savings of \$21,350 in an account earning 4.6% annual interest that is compounded semi-annually, how long will it take to grow to the amount that she needs?
7. Joe invested his \$8,700 savings in an account earning $634$% annual interest that is compounded continuously. How long will it take to earn \$300 in interest?
8. Miriam invested \$12,800 in an account earning $514$% annual interest that is compounded monthly. How long will it take to earn \$1,200 in interest?
9. Given that the bank is offering 4.2% annual interest compounded monthly, what principal is needed to earn \$25,000 in interest for one year?
10. Given that the bank is offering 3.5% annual interest compounded continuously, what principal is needed to earn \$12,000 in interest for one year?
11. Jose invested his \$3,500 bonus in an account earning $512$% annual interest that is compounded quarterly. How long will it take to double his investment?
12. Maria invested her \$4,200 savings in an account earning $634$% annual interest that is compounded semi-annually. How long will it take to double her savings?
13. If money is invested in an account earning 3.85% annual interest that is compounded continuously, how long will it take the amount to double?
14. If money is invested in an account earning 6.82% annual interest that is compounded continuously, how long will it take the amount to double?
15. Find the annual interest rate at which an account earning continuously compounding interest has a doubling time of 9 years.
16. Find the annual interest rate at which an account earning interest that is compounded monthly has a doubling time of 10 years.
17. Alice invested her savings of \$7,000 in an account earning 4.5% annual interest that is compounded monthly. How long will it take the account to triple in value?
18. Mary invested her \$42,000 bonus in an account earning 7.2% annual interest that is compounded continuously. How long will it take the account to triple in value?
19. Calculate the doubling time of an investment made at 7% annual interest that is compounded:
1. monthly
2. continuously
20. Calculate the doubling time of an investment that is earning continuously compounding interest at an annual interest rate of:
1. 4%
2. 6%
21. Billy’s grandfather invested in a savings bond that earned 5.5% annual interest that was compounded annually. Currently, 30 years later, the savings bond is valued at \$10,000. Determine what the initial investment was.
22. In 1935 Frank opened an account earning 3.8% annual interest that was compounded quarterly. He rediscovered this account while cleaning out his garage in 2005. If the account is now worth \$11,294.30, how much was his initial deposit in 1935?
### Part B: Modeling Exponential Growth and Decay
1. The population of a small town of 24,000 people is expected grow exponentially at a rate of 1.6% per year. Construct an exponential growth model and use it to:
1. Estimate the population in 3 years’ time.
2. Estimate the time it will take for the population to reach 30,000 people.
2. During the exponential growth phase, certain bacteria can grow at a rate of 4.1% per hour. If 10,000 cells are initially present in a sample, construct an exponential growth model and use it to:
1. Estimate the population in 5 hours.
2. Estimate the time it will take for the population to reach 25,000 cells.
3. In 2000, the world population was estimated to be 6.115 billion people and in 2010 the estimate was 6.909 billion people. If the world population continues to grow exponentially, estimate the total world population in 2020.
4. In 2000, the population of the United States was estimated to be 282 million people and in 2010 the estimate was 309 million people. If the population of the United States grows exponentially, estimate the population in 2020.
5. An automobile was purchased new for \$42,500 and 2 years later it was valued at \$33,400. Estimate the value of the automobile in 5 years if it continues to decrease exponentially.
6. A new PC was purchased for \$1,200 and in 1.5 years it was worth \$520. Assume the value is decreasing exponentially and estimate the value of the PC four years after it is purchased.
7. The population of the downtown area of a certain city decreased from 12,500 people to 10,200 people in two years. If the population continues to decrease exponentially at this rate, what would we expect the population to be in two more years?
8. A new MP3 player was purchased for \$320 and in 1 year it was selling used online for \$210. If the value continues to decrease exponentially at this rate, determine the value of the MP3 player 3 years after it was purchased.
9. The half-life of radium-226 is about 1,600 years. How long will a 5-milligram sample of radium-226 take to decay to 1 milligram?
10. The half-life of plutonium-239 is about 24,000 years. How long will a 5-milligram sample of plutonium-239 take to decay to 1 milligram?
11. The half-life of radioactive iodine-131 is about 8 days. How long will it take a 28-gram initial sample of iodine-131 to decay to 12 grams?
12. The half-life of caesium-137 is about 30 years. How long will it take a 15-milligram sample of caesium-137 to decay to 5 milligrams?
13. The Rhind Mathematical Papyrus is considered to be the best example of Egyptian mathematics found to date. This ancient papyrus was found to contain 64% of the carbon-14 normally found in papyrus. Given that carbon-14 has a half-life of 5,730 years, estimate the age of the papyrus.
14. A wooden bowl artifact carved from oak was found to contain 55% of the carbon-14 normally found in oak. Given that carbon-14 has a half-life of 5,730 years, estimate the age of the bowl.
15. The half-life of radioactive iodine-131 is about 8 days. How long will it take a sample of iodine-131 to decay to 10% of the original amount?
16. The half-life of caesium-137 is about 30 years. How long will it take a sample of caesium-137 to decay to 25% of the original amount?
17. The half-life of caesium-137 is about 30 years. What percent of an initial sample will remain in 100 years?
18. The half-life of radioactive iodine-131 is about 8 days. What percent of an initial sample will remain in 30 days?
19. If a bone is 100 years old, what percent of its original amount of carbon-14 do we expect to find in it?
20. The half-life of plutonium-239 is about 24,000 years. What percent of an initial sample will remain in 1,000 years?
21. Find the amount of time it will take for 10% of an initial sample of plutonium-239 to decay. (Hint: If 10% decays, then 90% will remain.)
22. Find the amount of time it will take for 10% of an initial sample of carbon-14 to decay.
Solve for the given variable:
1. Solve for t: $A=Pert$
2. Solve for t: $A=P(1+r)t$
3. Solve for I: $M=log (II0)$
4. Solve for $H+$: $pH=−log (H+)$
5. Solve for t: $P=11+e−t$
6. Solve for I: $L=10log (I/10−12)$
7. The number of cells in a certain bacteria sample is approximated by the logistic growth model $N(t)=1.2×1051+9e−0.32t$, where t represents time in hours. Determine the time it takes the sample to grow to 24,000 cells.
8. The market share of a product, as a percentage, is approximated by the formula $P(t)=1003+e−0.44t$ where t represents the number of months after an aggressive advertising campaign is launched.
1. What was the initial market share?
2. How long would we expect to see a 3.5% increase in market share?
9. In chemistry, pH is a measure of acidity and is given by the formula $pH=−log (H+)$, where $H+$ is the hydrogen ion concentration (measured in moles of hydrogen per liter of solution.) What is the hydrogen ion concentration of seawater with a pH of 8?
10. Determine the hydrogen ion concentration of milk with a pH of 6.6.
11. The volume of sound, L in decibels (dB), is given by the formula $L=10log (I/10−12)$ where I represents the intensity of the sound in watts per square meter. Determine the sound intensity of a hair dryer that emits 70 dB of sound.
12. The volume of a chainsaw measures 110 dB. Determine the intensity of this sound.
### Part C: Discussion Board
1. Which factor affects the doubling time the most, the annual compounding n or the interest rate r? Explain.
3. Is exponential growth sustainable over an indefinite amount of time? Explain.
4. Research and discuss the half-life of radioactive materials.
1. \$1,912.73
2. 9 years
1. \$9,356.45
2. 3.4 years
1. 5.6 years
2. $12$ year
3. \$583,867
4. 12.7 years
5. 18 years
6. 7.7%
7. 24.5 years
1. 9.93 years
2. 9.90 years
8. \$2,006.44
3. 8,323 people
4. 3,715 years
5. 9.8 days
7. 26.6 days
8. 9.9%
9. 98.8%
10. 3,648 years
11. $t=ln (A)−ln (P)r$
12. $I=I0⋅10M$
13. $t=ln (P1−P)$
14. Approximately 2.5 hours
15. $10−8$ moles per liter
16. $10−5$ watts per square meter
Given f and g find $(f○g)(x)$ and $(g○f)(x).$
1. $f(x)$ | 34,738 | 101,695 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1577, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2018-34 | longest | en | 0.826926 |
https://community.powerbi.com/t5/Developer/Running-Sum-on-If-statement-in-Power-BI-Table/m-p/2054247 | 1,642,361,108,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300010.26/warc/CC-MAIN-20220116180715-20220116210715-00372.warc.gz | 251,651,605 | 99,197 | cancel
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## Running Sum on If statement in Power BI Table
Hello. I have a measure that is using an if statement for the result by month. I need a running sum of that monthly column
MonthlyCommEarnedCalc =
IF( [% Month Target] >1.0,
[Monthly Comm],
[MonthlyCommCalc]
)
I have tried several different ways to create the running sum and cant get it to work. Any help would be appreciated.
The calculation below is giving me the same value as the original measure.
RunSum Month Comm Earned =
CALCULATE(
[MonthlyCommEarnedCalc],
FILTER(
ALLSELECTED(Transactions[Fiscal Month]),
MIN(Transactions[Fiscal Month start Date]) <= MAX(Transactions[Fiscal Month End Date])
))
1 ACCEPTED SOLUTION
Resolver II
First, create the month column in numeric form.
To calculate the running total for another measure then just add that measure code ( if block ) into your running total code.
example code :
Monthly Comm running sum =
VAR month = MAX(Transactions[Fiscal Month])
VAR result = CALCULATE(
[MonthlyCommEarnedCalc],// OR Put if condtion measure code here
Transactions[Fiscal Month] <= month,
ALLSELECTED(Transactions)
)
Return result
check this tutorial :
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
4 REPLIES 4
Resolver II
First, create the month column in numeric form.
To calculate the running total for another measure then just add that measure code ( if block ) into your running total code.
example code :
Monthly Comm running sum =
VAR month = MAX(Transactions[Fiscal Month])
VAR result = CALCULATE(
[MonthlyCommEarnedCalc],// OR Put if condtion measure code here
Transactions[Fiscal Month] <= month,
ALLSELECTED(Transactions)
)
Return result
check this tutorial :
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Frequent Visitor
Thanks. Im new to Dax and im not sure exactly how to proceed.
Super User
@MKatsanevas What happens if you change your ALLSELECTED to ALL?
Check out my External Tool for Power BI Desktop! Microsoft Hates Greg's Quick Measures
Check out my latest book!
Super User
@MKatsanevas This looks like a measure aggregation problem. See my blog article about that here: https://community.powerbi.com/t5/Community-Blog/Design-Pattern-Groups-and-Super-Groups/ba-p/138149
The pattern is:
MinScoreMeasure = MINX ( SUMMARIZE ( Table, Table[Group] , "Measure",[YourMeasure] ), [Measure])
MaxScoreMeasure = MAXX ( SUMMARIZE ( Table, Table[Group] , "Measure",[YourMeasure] ), [Measure])
SumScoreMeasure = SUMX ( SUMMARIZE ( Table, Table[Group] , "Measure",[YourMeasure] ), [Measure])
etc.
Check out my External Tool for Power BI Desktop! Microsoft Hates Greg's Quick Measures
Check out my latest book!
Announcements | 674 | 2,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-05 | latest | en | 0.790965 |
https://oeis.org/A062965 | 1,632,379,068,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.10/warc/CC-MAIN-20210923044248-20210923074248-00100.warc.gz | 496,208,636 | 4,435 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A062965 Positive numbers which are one less than a perfect square that is also another power. 2
15, 63, 80, 255, 624, 728, 1023, 1295, 2400, 4095, 6560, 9999, 14640, 15624, 16383, 20735, 28560, 38415, 46655, 50624, 59048, 65535, 83520, 104975, 117648, 130320, 159999, 194480, 234255, 262143, 279840, 331775, 390624, 456975, 531440, 614655 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 REFERENCES W. Dunham, Euler: The Master of Us All, The Mathematical Association of America, Washington D.C., 1999, p. 65. L. Euler, "Variae observationes circa series infinitas," Opera Omnia, Ser. 1, Vol. 14, pp. 216-244. Fvss, Nicolao, "Demonstratio Theorematvm Qvorvndam Analyticorvm," Nova Acta Academiae Scientiarum Imperialis Petropolitanae, 8 (1794) 223-226. LINKS Leonard Euler, Variae observationes circa series infinitas Leonard Euler, Variae observationes circa series infinitas Eric Weisstein's World of Mathematics, Perfect Power FORMULA From Terry D. Grant, Oct 25 2020: (Start) a(n) = A001597(n+1)^2 - 1. Sum_{k>=1} 1/a(k) = 7/4 - Pi^2/6 = 7/4 - zeta(2). Sum_{k>=1} 1/(a(k)+1) = Sum_{k>=2} mu(k)*(1-zeta(2*k)). (End) EXAMPLE a(2)=63 because the perfect square 64 = 8^2 = 4^3. MATHEMATICA Take[ Select[ Range[ 2, 150 ], GCD@@(Last/@FactorInteger[ # ])>1& ]^2-1] (* corrected by Jon Maiga, Sep 28 2019 *) CROSSREFS Cf. A037450, A062834, A062757, A001597. Cf. A131605. Sequence in context: A067572 A066584 A065915 * A157968 A141759 A335574 Adjacent sequences: A062962 A062963 A062964 * A062966 A062967 A062968 KEYWORD nonn AUTHOR Jason Earls, Jul 16 2001 EXTENSIONS More terms from Dean Hickerson, Jul 24 2001 STATUS approved
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Last modified September 23 02:26 EDT 2021. Contains 347609 sequences. (Running on oeis4.) | 745 | 2,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-39 | latest | en | 0.579743 |
https://teacherblog.co/?p=5130 | 1,611,796,917,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835583.91/warc/CC-MAIN-20210128005448-20210128035448-00412.warc.gz | 595,443,093 | 9,887 | Great Math Products!
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# More Free and Editable AMI—Electricity
So, before the rest of the year became virtual, my students wanted to learn about electricity. I was rather disappointed that we wouldn’t get to learn about electricity, so I decided to give them my best effort at a week with AMI electricity learning. Below are my plans. You are welcome to copy, paste and adapt them for your students.
FUN AND OPTIONAL ACTIVITIES for students
1.Unsafe electric conditions.
Now, look around your house. Do you see any unsafe electricalconditions in your house? Tell your parents
2. How does electricity get to your house? Watch this.
3. Again with your parent’s permission turn on all the lights andmany items that use electricity. Go back and read the meter. Writedown the number. How did it change, and what did you notice?
4. Ask your parents if they can let you see all of the electric bills. Make a bar graph or line graph to show the cost of electricity foreach month for the last year. Which month was most expensive forelectricity? Why do you think so? Discuss this with your parents.
5. Build a lemon battery. This is a rather involved project, but youcan get all the materials at Wal-Mart. You can find alligator clipsand small light bulbs in the automotive section.
https://www.instructables.com/id/Lemon-Batteries-Lighting-an-LED-with-Lemons/
6. Read your electric meter outside. Write down the number, Beforeyou do this, turn off as many lights and appliances as possible. Askyour parents’ permission before you go outside, if you’re not surewhat your electric meter looks like or how to read it watch this.https://m.youtube.com/watch?v=I6UPkncvFhw.
7. Take a picture and post it to Dojo. Did you get the lemon battery to work? Do you think other fruits or vegetables would work? Which ones?
_____
I know number 7 is a little steep to ask of parents, but some students might accomplish this and what a much better idea than playing video games all day! | 506 | 2,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2021-04 | latest | en | 0.930137 |
https://www.proprofs.com/discuss/q/72020/susan-and-lucy-have-invested-in-venture-an-amount-the | 1,597,154,371,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738777.54/warc/CC-MAIN-20200811115957-20200811145957-00534.warc.gz | 792,943,673 | 82,670 | Susan and Lucy have invested in a venture. Susan invested an amount - ProProfs Discuss
# Susan and Lucy have invested in a venture. Susan invested an amount in the initial months and then reduced that amount by 1/4th after 4 months and then again reduced the amount by 1/2 of the currently being invested after 8 months. Lucy joined after 2 months from Susan’s initial month and invested 90 % of Susan’s last month contribution every month. Quantity A Quantity B Susan’s share in profit after Lucy’s share in profit after oneone year year
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship can not be determined from the information given.
Asked by HAYES, Last updated: Mar 30, 2020
John Smith | 174 | 750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-34 | latest | en | 0.934711 |
https://www.jiskha.com/questions/1604206/a-red-ball-is-thrown-directly-upwards-from-the-ground-with-an-initial-velocity-of-35-0 | 1,600,496,290,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400190270.10/warc/CC-MAIN-20200919044311-20200919074311-00283.warc.gz | 931,977,130 | 4,733 | # Physics
A red ball is thrown directly upwards from the ground with an initial velocity of 35.0 m/s. At the same time, a blue ball is thrown directly upwards from he ground with an initial velocity of 10.0 m/s.
1. How high is the red ball when the blue ball reaches its highest point?
2. How fast is the red ball travelling when the blue ball hits the ground?
1. 👍 0
2. 👎 0
3. 👁 242
1. 1. time to blue max ... v / g
red vel @ blu max ... v - g t
ave red vel ... (2 v - g t) / 2
red hgt ... (ave vel) * t
2. blu ... time up = time down
red vel @ blu impact
... v - g(blu flight time)
1. 👍 0
2. 👎 0
## Similar Questions
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2. ### Calculus 1
A ball is thrown vertically upward from ground level with initial velocity of 96 feet per second. Assume the acceleration of the ball is a(t) = -32 ft^2 per second. (Neglect air Resistance.) (a) How long will it take the ball to
3. ### math
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4. ### Physics
A ball is thrown vertically upward with a speed of 25.0 m/s from a height of 2.0 m. How high does the ball rise? How long does it take to reach its highest point? How long does the ball take to hit the ground after it reaches its | 980 | 3,539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-40 | latest | en | 0.904252 |
https://math.stackexchange.com/questions/1530792/equivalent-definitions-of-boolean-algebras | 1,680,267,845,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949642.35/warc/CC-MAIN-20230331113819-20230331143819-00351.warc.gz | 450,496,718 | 31,589 | # Equivalent definitions of Boolean Algebras
Assume that my definition of a Boolean algebras is the following one: I have a set $B$ with two binary operations $\vee$ and $\wedge$ which both satisfy the commutative, associative and distributive laws. Both the operations have an identity element ($0$ and $1$) and for each element $b\in B$ there exists a complement (i.e. an element $b{'}\in B$ such that $b\vee b{'}=1$ and $b\wedge b{'}=0$).
Now I found the following definition of Boolean Algebra. Let $B$ be a set on which we define a binary operation $\cap$ and an unary operation ${'}$. We also require that those operations satisfy the following rules:
1) $x\cap y=y\cap x$ for each $x,\, y\in B$
2) $(x\cap y)\cap z=x\cap (y\cap z)$
3) $x\cap y{'}=z\cap z{'}$ iff $x\cap y=x$
Are those equivalent definitions? I was trying to work the thing out in the following way.
a) First we may put $0:=z\cap z'$ by property 3).
b) We may define $x\cup y:=(x{'}\cap y{'}){'}$ and put $0{'}=:1$ and we may prove the commutative law for $\cup$.
Anyway I cannot prove the associativity of $\cup$ or either the distributivity or even the fact the $0$ and $1$ are identity elements.
In rule 3, put $x=y=z=a$ to get $$a \cap a' = a \cap a' \iff a \cap a = a$$ so $a\cap a = a$ in general. This implies $$p \cap 0 = p \cap p \cap p' = p \cap p' = 0$$ for all $p$.
Now put $x=b''$, $y=b$, $z=b'$ to get $$b''\cap b' = b'\cap b'' \iff b'' \cap b = b''$$ so $b'' \cap b = b''$ in general.
Finally put $x=c$, $y=c''$ to get $$c \cap c''' = 0 \iff c \cap c'' = c$$ However, $$c\cap c''' = c\cap c' \cap c''' = 0 \cap c''' = 0$$ so $c\cap c'' = c$ in general.
Thus $p = p\cap p'' = p''$ in general, so $'$ is an involution, which is what you need to transfer properties such as associativity from $\cap$ to $\cup$.
Now put $x=d$, $y=0'$ to get $$d\cap 0'' = 0 \iff d\cap 0' = d$$ and since $0''=0$, the RHS of this is always true, and $1:=0'$ is an identity for $\cap$. By duality, $0$ is an identity for $\cup$. | 689 | 2,003 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-14 | latest | en | 0.862822 |
https://www.jiskha.com/questions/1486826/the-non-decreasing-sequence-of-odd-integers-a1-a2-a3-1-3-3-3-5-5-5-5-5 | 1,653,590,177,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662619221.81/warc/CC-MAIN-20220526162749-20220526192749-00644.warc.gz | 949,943,694 | 5,138 | # maths
the non-
decreasing sequence of odd integers {a1, a2, a3, .
. .} = {1,3,3,3,5,5,5,5,5,...} each positive odd
integer k appears k times. it is a fact that there
are integers b, c, and d such that, for all positive
integers n, añ = b[√(n+c)] +d. Where [x] denotes the largest integer not exceeding x. Find the sum of b+c+d.
Show step too hard
1. 👍
2. 👎
3. 👁
4. ℹ️
5. 🚩
1. since the terms of the sequence are all odd numbers (2k+1), and there are k^2 terms <= the kth odd number, look for something like
2[√n]+1
That sequence is
3,3,3, 5,5,5,5,5,5,5, ...
Looks like we have to adjust it by 1 to get
2[√(n-1)]+1
So, b+c+d = 2
No, I don't have an actual proof, but maybe you can work from here. I'll think on it some.
1. 👍
2. 👎
3. ℹ️
4. 🚩
## Similar Questions
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1. ### Algerbra
f m and p are positive integers and (m + p) x m is even, which of the following must be true? A. If m is odd, then p is odd. B. If m is odd, then p is even. C. If m is even, then p is even. D. If m is even, then p is odd. I choose
2. ### algebra
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4. ### discrete math
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1. ### Math
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2. ### Discrete Mathematics
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4. ### help me plz
if a and b are both odd integers, which expression must always equal an odd integer? 1 a+b 2 a*b 3 a-b 4 a/b | 971 | 3,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-21 | latest | en | 0.871433 |
https://byjus.com/questions/integrate-log-1-plus-tan-x-from-0-to-pi-by-4/ | 1,611,253,907,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527224.75/warc/CC-MAIN-20210121163356-20210121193356-00709.warc.gz | 268,615,540 | 25,263 | # Integrate log(1+tan x) from 0 to pi/4.
$\int_0^af(x)dx=\int_0^af(a-x)dx \\ \text{Let I}=\int_0^{\pi/4} log(1+tanx)dx=\int_0^{\pi/4}log(1+tan[{\pi/4}-x]dx \\= \int_0^{\pi/4}log(1+(1-tanx)/(1+tanx))dx = \int_0^{\pi/4}log(2/(1+tanx))dx \\= \int_0^{\pi/4}log2dx-\int_0^{\pi/4}log(1+tanx)dx = \int_0^{\pi/4}log2dx-I \\2I=\int_0^{\pi/4}log2dx \ and \ I=log2*{\pi/4}*1/2\\={\pi/8}\ log2$ | 224 | 383 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-04 | latest | en | 0.45622 |
https://www.aqua-calc.com/calculate/food-calories/substance/instant-blank-white-blank-chocolate-blank-pudding-coma-and-blank-upc-column--blank-072392011223 | 1,576,241,073,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540555616.2/warc/CC-MAIN-20191213122716-20191213150716-00521.warc.gz | 601,669,638 | 7,594 | # Calories of INSTANT WHITE CHOCOLATE PUDDING, UPC: 072392011223
## food calories and nutrients calculator
### compute calories and nutrients of generic and branded foods
#### See how many calories in0.1 kg (0.22 lbs) ofINSTANT WHITE CHOCOLATEPUDDING, UPC:072392011223
From kilocalories(kcal) kilojoule(kJ) Carbohydrate 0 0 Fat 0 0 Protein 0 0 Other 360 1 506.24 Total 360 1 506.24
Nutrient (find foodsrich in nutrients) Unit Value /100 g BasicAdvancedAll Proximates Energy kcal 360 Protein g 0 Total lipid (fat) g 0 Carbohydrate,bydifference g 92 Fiber,totaldietary g 0 Sugars, total g 72 Minerals Calcium, Ca mg 80 Iron, Fe mg 0 Sodium, Na mg 1640 Vitamins Vitamin C,totalascorbic acid mg 0 Vitamin A, IU IU 0 Lipids Fatty acids,totalsaturated g 0 Cholesterol mg 0
#### Weight
gram 100 ounce 3.53 kilogram 0.1 pound 0.22 milligram 100 000
• About INSTANT WHITE CHOCOLATE PUDDING, UPC: 072392011223
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Estimate blood volume based on height, weight and gender | 1,058 | 3,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-51 | latest | en | 0.636055 |
http://stacks.math.columbia.edu/tag/07ZD | 1,500,828,522,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424575.75/warc/CC-MAIN-20170723162614-20170723182614-00279.warc.gz | 315,837,038 | 4,342 | # The Stacks Project
## Tag 07ZD
Lemma 15.8.7. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $r \geq 0$. The following are equivalent
1. $M$ is finite locally free of rank $r$ (Algebra, Definition 10.77.1),
2. $\text{Fit}_{r - 1}(M) = 0$ and $\text{Fit}_r(M) = R$, and
3. $\text{Fit}_k(M) = 0$ for $k < r$ and $\text{Fit}_k(M) = R$ for $k \geq r$.
Proof. It is immediate that (2) is equivalent to (3) because the Fitting ideals form an increasing sequence of ideals. Since the formation of $\text{Fit}_k(M)$ commutes with base change (Lemma 15.8.4) we see that (1) implies (2) by Example 15.8.5 and glueing results (Algebra, Section 10.23). Conversely, assume (2). By Lemma 15.8.6 we may assume that $M$ is generated by $r$ elements. Thus a presentation $\bigoplus_{j \in J} R \to R^{\oplus r} \to M \to 0$. But now the assumption that $\text{Fit}_{r - 1}(M) = 0$ implies that all entries of the matrix of the map $\bigoplus_{j \in J} R \to R^{\oplus r}$ are zero. Thus $M$ is free. $\square$
The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 1510–1521 (see updates for more information).
\begin{lemma}
\label{lemma-fitting-ideal-finite-locally-free}
Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $r \geq 0$.
The following are equivalent
\begin{enumerate}
\item $M$ is finite locally free of rank $r$
(Algebra, Definition \ref{algebra-definition-locally-free}),
\item $\text{Fit}_{r - 1}(M) = 0$ and $\text{Fit}_r(M) = R$, and
\item $\text{Fit}_k(M) = 0$ for $k < r$ and $\text{Fit}_k(M) = R$
for $k \geq r$.
\end{enumerate}
\end{lemma}
\begin{proof}
It is immediate that (2) is equivalent to (3) because the Fitting ideals
form an increasing sequence of ideals.
Since the formation of $\text{Fit}_k(M)$ commutes with base change
(Lemma \ref{lemma-fitting-ideal-basics}) we see that (1) implies (2) by
Example \ref{example-fitting-free}
and glueing results (Algebra, Section \ref{algebra-section-more-glueing}).
Conversely, assume (2). By
Lemma \ref{lemma-fitting-ideal-generate-locally} we may assume that $M$
is generated by $r$ elements. Thus a presentation
$\bigoplus_{j \in J} R \to R^{\oplus r} \to M \to 0$.
But now the assumption that $\text{Fit}_{r - 1}(M) = 0$ implies
that all entries of the matrix of the map
$\bigoplus_{j \in J} R \to R^{\oplus r}$ are zero.
Thus $M$ is free.
\end{proof}
Comment #1423 by Kestutis Cesnavicius on April 17, 2015 a 6:26 pm UTC
In (1), "locally free of rank $k$" ---> "locally free of rank $r$".
Comment #1436 by Johan (site) on April 22, 2015 a 1:18 pm UTC
Oops! Thanks, see here.
## Add a comment on tag 07ZD
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner). | 959 | 2,848 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2017-30 | latest | en | 0.737112 |
https://byjus.com/questions/who-introduced-the-concept-of-free-fall/ | 1,632,720,768,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00668.warc.gz | 200,190,384 | 37,446 | # Who introduced the concept of free fall?
Do you think about a free fall apple from a tree or falling of the ball when we throw it upside. The basis of a free falling object is given by Sir Isaac Newton. A free falling object falls under the influence of gravity. So a body which is under the effect of gravity is its state of free fall. It does not need any kind of force and it does not affect by air resistance. The effect of gravitational force is under the gravitational acceleration that acts on the free falling object. Its value is 9.8 m/s/s in the downward direction.
This concept of attracting gravitational force of the earth is established for all objects in the air. It is not required to apply this concept of falling downward object, it can be applied for rising upward under the gravity influence of 9.8 m/s/s. Letƒ¢¢€š¬¢€ž¢s discuss on this concept of free fall objects, its speed and velocity, and motion equations for free fall objects. | 209 | 958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-39 | latest | en | 0.925097 |
http://gmatclub.com/forum/2-2-1-2-2-1-2-4-1-2-8-1-a-2-16-1-b-39144.html?kudos=1 | 1,485,303,134,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285315.77/warc/CC-MAIN-20170116095125-00271-ip-10-171-10-70.ec2.internal.warc.gz | 119,806,062 | 42,566 | 2^2)-1))((2^2)+1))((2^4)+1))((2^8)+1))= A) (2^16)-1 B) : PS Archive
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2^2)-1))((2^2)+1))((2^4)+1))((2^8)+1))= A) (2^16)-1 B)
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((2^2)-1))((2^2)+1))((2^4)+1))((2^8)+1))=
A) (2^16)-1
B) (2^16)+1
C) (2^32)-1
D) (2^128)-1
E) 2^16((2^16)-1)
Director
Joined: 05 Feb 2006
Posts: 898
Followers: 3
Kudos [?]: 107 [0], given: 0
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26 Nov 2006, 05:34
I did apply this formula but somehow managed to make a mistake... maybe rushed a bit too hard...
Yes now it works fine.
OA is A
26 Nov 2006, 05:34
Display posts from previous: Sort by | 522 | 1,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-04 | latest | en | 0.822855 |
https://www.eclipse.org/lists/sumo-dev/msg00479.html | 1,723,186,352,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640759711.49/warc/CC-MAIN-20240809044241-20240809074241-00192.warc.gz | 599,793,269 | 8,583 | Re: [sumo-dev] Miscalculation of Rotational Kinetic Energy, Curvature Energy Loss, and Auxiliary Energy Demand
Hi,
I just fixed the main issues and I can confirm that the changes in our
tests are below 1%. The only thing with major deviations is the
aggregated emission output (meandata) when used with a different step
length than 1s (a case which had no tests at all until now).
Best regards,
Michael
Am 2020-10-16 09:40, schrieb Jakub Ševčík:
Dear all,
thank you for your comments on the energy model in SUMO. I am just
internally answering my colleague Jan to his questions but think that you
could be interested in some of my observations.
Despite bugs mentioned by Burford, their correction (that is certainly
important in SUMO) will only slightly impacts our energy results for these
reasons:
1. Change of Rotational Energy << Change of Kinetic Energy
since the internal moment of inertial << mass of the vehicle (for example
for our simulated trolleybus J_int = 0.03*mass)
2. The calculation of the Curvature Energy Loss is applicated only 3 times per trolleybus No. 13 line circulation in simulation without surrounding traffic. The reason for such behavior is that the diff of route angle is
not zero only at three time-steps of the simulation from total ca 5k.
3. Fortunately we use the default SUMO length 1 s of the time-step and
therefore, the confusion of power [W] vs energy [Ws] vanishes. (It counts both for the bug in the constat power of aux system calculation and for the order logic of energy/power calculations in SUMO's emission/energy model).
Finally, the paper [ I. Sagaama, A. Kchiche, W. Trojet and F. Kamoun,
"Proposal of More Accurate Energy Model of Electric Vehicle For SUMO] you
mentioned proposes three changes for SUMO model
b. calculation of efficiency as \eta_bat*\eta_inv*\eta_mot
I don't understand the difference between such calculation in comparison with using only one constant eta, that can be precalculated of course using
the above-mentioned formula.
Maybe in our case would be interesting to consider different values for the
efficiency of propulsion using energy from battery
(battery->inverter->motor) and from overhead wire (owhd wire -> inverter -> motor) but again fortunately these values are almost the same in practice.
a. using the proposed modified equation for calculation of the energy of auxiliary drives corresponds to the common approach in SUMO if the ambient
temperature is 20°C
c. the modification of efficiency of regenerative breaking is already
integrated into the SUMO
https://github.com/eclipse/sumo/blob/4358260686c4a331c75bce1b5d7702e3eda5cdde/src/utils/emissions/HelpersEnergy.cpp#L132
.
Best,
Jakub
On Thu, Oct 15, 2020 at 11:55 PM Jan Přikryl <prikryl@xxxxxxxxxx> wrote:
Dear Joerg,
I am just curious: Do you have (or will you have) any comparison with
measurements of real vehicles in real traffic? We are namely using the
original HelpersEnergy subsystem to compute energy flows in the ElecHybrid
device and, to be sure, we have tried to validate the outputs of SUMO
against telemetry that we have from trolleybuses -- and, given the natural
variance of the data that you get from such a vehicle in real traffic
(different saloon occupation and hence different mass, different traffic conditions, different daily temperatures resulting in heavy variations in
auxiliary power consumption and so on) and given that for the modelled
vehicle you get at least three different sets of drag and resistance
parameters (one set provided by the manufacturer of the vehicle, a second set from the provider of the chassis, and the third one from the vehicle
operator), we were quite happy with the results even with the current
implementation (the only change is that we try to limit the acceleration of
a vehicle based on its power limits).
On the other hand, there is a paper from 2018 (
https://ieeexplore.ieee.org/document/8450371) that reports a 23-29%
difference
in energy consumption compared to a test with a real electric vehicle. I
did not read it yet, I just came across it recently.
That leaves me wondering whether our comparison is correct and whether we
should not crosscheck it one more time...
Jan
On Thu, 15 Oct 2020, at 10:23 PM, joerg.schweizer@xxxxxxxx wrote:
> Thanks Michael for putting your hands on this, and sorry it took time to
> find the file.
>
> It has been quite crucial for the project as the determination of the
> energy consumption has been the main goal.
>
> Hope it gets fixed soon.
>
> Thanks again,
>
> Joerg
>
>
> On 15/10/20 21:53, Michael Behrisch wrote:
> > Hi,
> > thanks for the report and sorry for the delay. Unfortunately we have
> > several places in the code which calculate energies. I suppose you are
> > referring to HelpersEnergy::compute
> >
https://github.com/eclipse/sumo/blob/master/src/utils/emissions/HelpersEnergy.cpp#L57
> > ?
> > This function seems to have some problems indeed not only the ones you
> > mention but in total it should return power and not energy because the
> > time scaling is done outside the function when used as an emission
> > model. For the battery device however energy is expected. Sorry for the
> > hassle, it is now tracked here:
https://github.com/eclipse/sumo/issues/7695
> >
> > Thanks again for finding out!
> >
> > Best regards,
> > Michael
> >
> > Am 11.10.20 um 06:40 schrieb Burford Furman:
> >> Hello,
> >>
> >> We are running SUMO ver 1.6.0 under Windows 10, and we just discovered
> >> that the SUMO source code is calculating the Rotational Kinetic
Energy,
> >> the Curvature Energy Loss, and the Constant Auxiliary Energy Demand
> >> incorrectly. The SUMO documentation lists the correct equations, but
for
> >> some reason the source code is not correct in ver 1.6.0. Our findings
> >> are summarized below:
> >>
> >> image.png
> >>
> >> image.png
> >>
> >> Thank you.
> >>
> >> Burford Furman
> >>
> >> --
> >>
************************************************************************************
> >> Burford J. Furman
> >> Professor
> >> Department of Mechanical Engineering
> >> San Jose State University
> >> One Washington Square
> >> San Jose, CA 95192-0087
> >> Phone: (408) 924-3817 Email: Burford.Furman@xxxxxxxx
> >> <mailto:Burford.Furman@xxxxxxxx>
> >> Web page: http://www.sjsu.edu/people/burford.furman/
> >> <http://www.engr.sjsu.edu/bjfurman/>
> >>
************************************************************************************
> >> "We must remember that intelligence is not enough. Intelligence plus
> >> character--that is the goal of true education." - Dr. Martin Luther
King
> >> Jr., 1947
> >>
> >> "/Instrumental or mechanical science
> >> is the noblest and above all others, the most useful.../" - Leonardo
da
> >> Vinci
> >>
> >> _______________________________________________
> >> sumo-dev mailing list
> >> sumo-dev@xxxxxxxxxxx
> >> To unsubscribe from this list, visit
https://www.eclipse.org/mailman/listinfo/sumo-dev
> >>
> >
> _______________________________________________
> sumo-dev mailing list
> sumo-dev@xxxxxxxxxxx
> To unsubscribe from this list, visit
> https://www.eclipse.org/mailman/listinfo/sumo-dev
>
--
Jan Přikryl
prikryl@xxxxxxxxxx
• References: | 1,783 | 7,216 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-33 | latest | en | 0.925827 |
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7. (S-CP.6) Conditional probability | 695 | 1,858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-50 | latest | en | 0.512663 |
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